为记简便，本文所有的「原式」将简写为「O.F. \text{O.F.} O.F.
极限
常见的涉考极限有0 0 \frac{0}{0} 0 0 1 ∞ 1^\infty 1 ∞
定义·性质
数列极限
设有数列{ u n } \{u_n\} { u n } A A A ε > 0 \varepsilon>0 ε > 0 N ∈ N + N\in\mathbb{N}_+ N ∈ N + n > N n>N n > N ∣ u n − A ∣ < ε |u_n-A|<\varepsilon ∣ u n − A ∣ < ε A A A { u n } \{u_n\} { u n }
上述定义中ε \varepsilon ε u n u_n u n A A A N N N N N N ε \varepsilon ε n n n n n n u n u_n u n A A A ε \varepsilon ε ε \varepsilon ε { u n } \{u_n\} { u n } n n n A A A
考虑数列{ u n } \{u_n\} { u n } { u n k } \{u_{n_k}\} { u n k }
lim n → ∞ u n = A ⟺ ∀ { u n k } , lim k → ∞ u n k = A . \lim_{n\rightarrow\infty}u_n=A\iff\forall\{u_{n_k}\}, \lim_{k\rightarrow\infty}u_{n_k}=A.
n → ∞ lim u n = A ⟺ ∀ { u n k } , k → ∞ lim u n k = A .
依此结论可知“一个数列的奇数子列和偶数子列极限与母列相等”。即
lim n → ∞ u n = A ⟺ lim k → ∞ { u 2 k } = lim k → ∞ { u 2 k + 1 } = A . \lim_{n\rightarrow\infty}u_n=A\iff\lim_{k\rightarrow\infty}\{u_{2k}\}=\lim_{k\rightarrow\infty}\{u_{2k+1}\}=A.
n → ∞ lim u n = A ⟺ k → ∞ lim { u 2 k } = k → ∞ lim { u 2 k + 1 } = A .
另一方面，若{ u n } \{u_n\} { u n }
根据上述结论，考虑数列{ x n } \{x_n\} { x n } lim n → ∞ x 3 n = lim n → ∞ x 3 n + 1 = a . \lim_{n\rightarrow\infty}x_{3n}=\lim_{n\rightarrow\infty}x_{3n+1}=a. lim n → ∞ x 3 n = lim n → ∞ x 3 n + 1 = a . lim n → ∞ x 3 n + 2 \lim_{n\rightarrow\infty}x_{3n+2} lim n → ∞ x 3 n + 2 lim n → ∞ x n = a \lim_{n\rightarrow\infty}x_{n}=a lim n → ∞ x n = a
e.g.1.0 \textbf{e.g.1.0 } e.g.1.0 lim n → ∞ a n = a ≠ 0 , \lim_{n\rightarrow\infty}a_n=a≠0, lim n → ∞ a n = a = 0 , n n n ∣ a n ∣ |a_n| ∣ a n ∣ ∣ a ∣ 2 \frac{|a|}{2} 2 ∣ a ∣
Solution \textbf{Solution } Solution
∵ lim n → ∞ a n = a , ∴ ∀ ε , ∣ a n − a ∣ < ε . Additionally ∵ ∣ ∣ a n ∣ − ∣ a ∣ ∣ ≤ ∣ a n − a ∣ , ∴ ∣ ∣ a n ∣ − ∣ a ∣ ∣ < ε . ∴ lim n → ∞ ∣ a n ∣ = ∣ a ∣ . \begin{align*}
\because&\lim_{n\rightarrow\infty}a_n=a,\\
\therefore&\forall\varepsilon,|a_n-a|<\varepsilon.\\
\text{Additionally}\\
\because&\big||a_n|-|a|\big|\leq|a_n-a|,\\
\therefore&\big||a_n|-|a|\big|<\varepsilon.\\
\therefore&\lim_{n\rightarrow\infty}|a_n|=|a|.
\end{align*}
∵ ∴ Additionally ∵ ∴ ∴ n → ∞ lim a n = a , ∀ ε , ∣ a n − a ∣ < ε . ∣ a n ∣ − ∣ a ∣ ≤ ∣ a n − a ∣ , ∣ a n ∣ − ∣ a ∣ < ε . n → ∞ lim ∣ a n ∣ = ∣ a ∣.
而由极限定义知∀ ε , ∣ ∣ a n ∣ − ∣ a ∣ ∣ < ε , \forall\varepsilon,\big||a_n|-|a|\big|<\varepsilon, ∀ ε , ∣ a n ∣ − ∣ a ∣ < ε , ε = ∣ a ∣ 2 . \varepsilon=\frac{|a|}{2}. ε = 2 ∣ a ∣ .
∣ ∣ a n ∣ − ∣ a ∣ ∣ < ∣ a ∣ 2 ⇒ − ∣ a ∣ 2 < ∣ a n ∣ − ∣ a ∣ < ∣ a ∣ 2 ⇒ ∣ a ∣ 2 < ∣ a n ∣ < 3 ∣ a ∣ 2 . \big||a_n|-|a|\big|<\frac{|a|}{2}\Rightarrow-\frac{|a|}{2}<|a_n|-|a|<\frac{|a|}{2}\Rightarrow\frac{|a|}{2}<|a_n|<\frac{3|a|}{2}.
∣ a n ∣ − ∣ a ∣ < 2 ∣ a ∣ ⇒ − 2 ∣ a ∣ < ∣ a n ∣ − ∣ a ∣ < 2 ∣ a ∣ ⇒ 2 ∣ a ∣ < ∣ a n ∣ < 2 3∣ a ∣ .
函数极限
对函数y = f ( x ) y=f(x) y = f ( x ) A A A ε > 0 \varepsilon>0 ε > 0 M > 0 , M>0, M > 0 , ∣ x ∣ > M |x|>M ∣ x ∣ > M ∣ f ( x ) − A ∣ < ε |f(x)-A|<\varepsilon ∣ f ( x ) − A ∣ < ε y = f ( x ) y=f(x) y = f ( x ) x → ∞ x\rightarrow\infty x → ∞ A . A. A . y = f ( x ) y=f(x) y = f ( x ) A A A ε > 0 \varepsilon>0 ε > 0 δ > 0 , \delta>0, δ > 0 , 0 < ∣ x − x 0 ∣ < δ 0<|x-x_0|<\delta 0 < ∣ x − x 0 ∣ < δ ∣ f ( x ) − A ∣ < ε |f(x)-A|<\varepsilon ∣ f ( x ) − A ∣ < ε x → x 0 x\rightarrow x_0 x → x 0 f ( x ) f(x) f ( x ) A . A. A .
考虑下面几个极限的性质。
（唯一性）若一函数/数列极限存在，则其极限值唯一。lim x → x 0 f ( x ) = A \lim_{x\rightarrow x_0}f(x)=A lim x → x 0 f ( x ) = A A > 0 A>0 A > 0 A < 0 A<0 A < 0 f ( x ) f(x) f ( x ) x 0 x_0 x 0 f ( x ) > 0 f(x)>0 f ( x ) > 0 f ( x ) < 0 f(x)<0 f ( x ) < 0
现在考虑下面4个函数的有界性。
观察函数x sin 1 x , x ∈ ( 0 , + ∞ ) . x\sin\frac{1}{x},x\in(0,+\infty). x sin x 1 , x ∈ ( 0 , + ∞ ) . ( 0 , + ∞ ) (0,+\infty) ( 0 , + ∞ ) ( 0 , 0 + δ ) (0,0+\delta) ( 0 , 0 + δ ) [ 0 + δ , + x ] [0+\delta, +x] [ 0 + δ , + x ] ( + x , + ∞ ) (+x,+\infty) ( + x , + ∞ ) lim x → 0 + x sin 1 x \lim_{x\rightarrow0^+}x\sin\frac{1}{x} lim x → 0 + x sin x 1 lim x → + ∞ x sin 1 x \lim_{x\rightarrow+\infty}x\sin\frac{1}{x} lim x → + ∞ x sin x 1 × \times × sin 1 x ∼ 1 x \sin\frac{1}{x}\sim\frac{1}{x} sin x 1 ∼ x 1
观察函数1 x sin 1 x , x ∈ ( 0 , + ∞ ) . \frac{1}{x}\sin\frac{1}{x},x\in(0,+\infty). x 1 sin x 1 , x ∈ ( 0 , + ∞ ) . x → 0 + x\rightarrow0^+ x → 0 + x = 1 n π + π 2 , x=\frac{1}{n\pi+\frac{\pi}{2}}, x = nπ + 2 π 1 , 1 x sin 1 x = n π + π 2 , \frac{1}{x}\sin\frac{1}{x}=n\pi+\frac{\pi}{2}, x 1 sin x 1 = nπ + 2 π , n n n x x x x → 0 + x\rightarrow0^+ x → 0 +
观察函数1 x sin x , x ∈ ( 0 , + ∞ ) . \frac{1}{x}\sin x,x\in(0,+\infty). x 1 sin x , x ∈ ( 0 , + ∞ ) . x sin 1 x , x ∈ ( 0 , + ∞ ) x\sin\frac{1}{x},x\in(0,+\infty) x sin x 1 , x ∈ ( 0 , + ∞ ) lim x → 0 + 1 x sin x = 1 , lim x → + ∞ 1 x sin x = 0 , \lim_{x\rightarrow0^+}\frac{1}{x}\sin x=1,\lim_{x\rightarrow+\infty}\frac{1}{x}\sin x=0, lim x → 0 + x 1 sin x = 1 , lim x → + ∞ x 1 sin x = 0 ,
观察函数∫ 0 x sin t t d t , x ∈ ( 0 , 2049 ) . \int_0^x\frac{\sin t}{t}\text{d}t,x\in(0,2049). ∫ 0 x t s i n t d t , x ∈ ( 0 , 2049 ) . sin x x \frac{\sin x}{x} x s i n x x ∈ ( 0 , 2049 ) x\in(0,2049) x ∈ ( 0 , 2049 ) sin x x \frac{\sin x}{x} x s i n x x ∈ ( 0 , + ∞ ) x\in(0,+\infty) x ∈ ( 0 , + ∞ ) sin x x \frac{\sin x}{x} x s i n x x ∈ ( 0 , 2049 ) x\in(0,2049) x ∈ ( 0 , 2049 ) f ′ ( x ) f'(x) f ′ ( x ) f ( x ) f(x) f ( x ) x ∈ ( 0 , 2049 ) x\in(0,2049) x ∈ ( 0 , 2049 )
求极限必备技能
可以使用L’Hospital法则求出的极限
0 0 \frac{0}{0} 0 0
e.g.1.1 \textbf{e.g.1.1 } e.g.1.1 lim x → 0 sin x x . \lim_{x\rightarrow0}\frac{\sin x}{x}. lim x → 0 x s i n x .
Solution \textbf{Solution } Solution
sin x ∼ x ⇒ O.F. = 1. \begin{align*}
\sin x\sim x\Rightarrow\text{O.F.}=1.
\end{align*}
sin x ∼ x ⇒ O.F. = 1.
∞ ∞ \frac{\infty}{\infty} ∞ ∞
e.g.1.2 \textbf{e.g.1.2 } e.g.1.2 lim x → ∞ ∑ i = 0 n a i x i ∑ j = 0 m b j x j . \lim_{x\rightarrow\infty}\frac{\sum_{i=0}^{n}a_ix^i}{\sum_{j=0}^{m}b_jx^j}. lim x → ∞ ∑ j = 0 m b j x j ∑ i = 0 n a i x i .
Solution \textbf{Solution } Solution
lim x → ∞ ∑ i = 0 n a i x i ∑ j = 0 m b j x j = lim x → ∞ a n x n b m x m . \lim_{x\rightarrow\infty}\frac{\sum_{i=0}^{n}a_ix^i}{\sum_{j=0}^{m}b_jx^j}=\lim_{x\rightarrow\infty}\frac{a_nx^n}{b_mx^m}.
x → ∞ lim ∑ j = 0 m b j x j ∑ i = 0 n a i x i = x → ∞ lim b m x m a n x n .
因此各情况对应的极限如下：
1 ∘ n = m , lim x → ∞ ∑ i = 0 n a i x i ∑ j = 0 m b j x j = a n b m . 2 ∘ n < m , lim x → ∞ ∑ i = 0 n a i x i ∑ j = 0 m b j x j = 0. 3 ∘ n > m , lim x → ∞ ∑ i = 0 n a i x i ∑ j = 0 m b j x j does not exist. \begin{align*}
1^\circ n=m,& \lim_{x\rightarrow\infty}\frac{\sum_{i=0}^{n}a_ix^i}{\sum_{j=0}^{m}b_jx^j}=\frac{a_n}{b_m}.\\
2^\circ n<m,& \lim_{x\rightarrow\infty}\frac{\sum_{i=0}^{n}a_ix^i}{\sum_{j=0}^{m}b_jx^j}=0.\\
3^\circ n>m,& \lim_{x\rightarrow\infty}\frac{\sum_{i=0}^{n}a_ix^i}{\sum_{j=0}^{m}b_jx^j}\text{ does not exist.}
\end{align*}
1 ∘ n = m , 2 ∘ n < m , 3 ∘ n > m , x → ∞ lim ∑ j = 0 m b j x j ∑ i = 0 n a i x i = b m a n . x → ∞ lim ∑ j = 0 m b j x j ∑ i = 0 n a i x i = 0. x → ∞ lim ∑ j = 0 m b j x j ∑ i = 0 n a i x i does not exist.
e.g.1.3 \textbf{e.g.1.3 } e.g.1.3 lim x → − ∞ 4 x 2 − x + 1 − x + 1 x 2 + sin x . \lim_{x\rightarrow-\infty}\frac{\sqrt{4x^2-x+1}-x+1}{\sqrt{x^2+\sin x}}. lim x → − ∞ x 2 + s i n x 4 x 2 − x + 1 − x + 1 .
Solution \textbf{Solution } Solution
O.F. = lim x → − ∞ ∣ 2 x ∣ − x ∣ x ∣ = lim x → − ∞ − 2 x − x − x = 3. \text{O.F.}=\lim_{x\rightarrow-\infty}\frac{|2x|-x}{|x|}=\lim_{x\rightarrow-\infty}\frac{-2x-x}{-x}=3.
O.F. = x → − ∞ lim ∣ x ∣ ∣2 x ∣ − x = x → − ∞ lim − x − 2 x − x = 3.
∞ ∞ \frac{\infty}{\infty} ∞ ∞
x → + ∞ : ln α x ( α > 0 ) ≪ x β ( β > 0 ) ≪ a x ( a > 1 ) ≪ x x n → + ∞ : ln α n ( α > 0 ) ≪ n β ( β > 0 ) ≪ a n ( a > 1 ) ≪ n ! ≪ n n \begin{align*}
x\rightarrow+\infty:&\ln ^\alpha x(\alpha>0)\ll x^\beta(\beta>0)\ll a^x(a>1)\ll x^x\\
n\rightarrow+\infty:&\ln ^\alpha n(\alpha>0)\ll n^\beta(\beta>0)\ll a^n(a>1)\ll n!\ll n^n
\end{align*}
x → + ∞ : n → + ∞ : ln α x ( α > 0 ) ≪ x β ( β > 0 ) ≪ a x ( a > 1 ) ≪ x x ln α n ( α > 0 ) ≪ n β ( β > 0 ) ≪ a n ( a > 1 ) ≪ n ! ≪ n n
其中n ∈ N + . n\in\mathbb{N}_+. n ∈ N + .
e.g.1.4 \textbf{e.g.1.4 } e.g.1.4 lim x → ∞ x 2 e x . \lim_{x\rightarrow\infty}\frac{x^2}{e^x}. lim x → ∞ e x x 2 .
Solution \textbf{Solution } Solution
O.F. = 0. \text{O.F.}=0.
O.F. = 0.
0 ⋅ ∞ 0\cdot\infty 0 ⋅ ∞
∞ − ∞ \infty-\infty ∞ − ∞ 0 0 \frac{0}{0} 0 0 ∞ ∞ \frac{\infty}{\infty} ∞ ∞
e.x.1.1 \textbf{e.x.1.1 } e.x.1.1 lim x → + ∞ ln ( 1 + 2 x ) ln ( 1 + 3 x ) . \lim_{x\rightarrow+\infty}\frac{\ln(1+2^x)}{\ln(1+3^x)}. lim x → + ∞ l n ( 1 + 3 x ) l n ( 1 + 2 x ) .
Solution \textbf{Solution } Solution
O.F. = lim x → + ∞ x ln 2 x ln 3 = ln 2 ln 3 . \text{O.F.}=\lim_{x\rightarrow+\infty}\frac{x\ln2}{x\ln3}=\frac{\ln2}{\ln3}.
O.F. = x → + ∞ lim x ln 3 x ln 2 = ln 3 ln 2 .
e.x.1.2 \textbf{e.x.1.2 } e.x.1.2 lim x → + ∞ ln ( x + 1 + x 2 ) ln ( x + x e x ) . \lim_{x\rightarrow+\infty}\frac{\ln(x+\sqrt{1+x^2})}{\ln(x+xe^x)}. lim x → + ∞ l n ( x + x e x ) l n ( x + 1 + x 2 ) .
Solution \textbf{Solution } Solution
O.F. = lim x → + ∞ ln ( 2 x ) ln ( x e x ) = ln 2 + ln x ln x + x = ln x x = 0. \text{O.F.}=\lim_{x\rightarrow+\infty}\frac{\ln(2x)}{\ln(xe^x)}=\frac{\ln2+\ln x}{\ln x+x}=\frac{\ln x}{x}=0.
O.F. = x → + ∞ lim ln ( x e x ) ln ( 2 x ) = ln x + x ln 2 + ln x = x ln x = 0.
已知极限求参数值
从已知极限式成立的必要条件去寻找参数的关系。
一般有以下几个结论：
若lim x → a f ( x ) g ( x ) \lim_{x\rightarrow a}\frac{f(x)}{g(x)} lim x → a g ( x ) f ( x ) lim x → a g ( x ) = 0 ⇒ lim x → a f ( x ) = 0. \lim_{x\rightarrow a}g(x)=0\Rightarrow \lim_{x\rightarrow a}f(x)=0. lim x → a g ( x ) = 0 ⇒ lim x → a f ( x ) = 0.
若lim x → a f ( x ) g ( x ) \lim_{x\rightarrow a}\frac{f(x)}{g(x)} lim x → a g ( x ) f ( x ) 且不为0 ，则lim x → a f ( x ) = 0 ⇒ lim x → a g ( x ) = 0. \lim_{x\rightarrow a}f(x)=0\Rightarrow \lim_{x\rightarrow a}g(x)=0. lim x → a f ( x ) = 0 ⇒ lim x → a g ( x ) = 0.
若lim x → a ( f ( x ) − g ( x ) ) \lim_{x\rightarrow a}(f(x)-g(x)) lim x → a ( f ( x ) − g ( x )) 同阶无穷大 且lim x → a f ( x ) = ∞ ⟺ lim x → a g ( x ) = ∞ . \lim_{x\rightarrow a}f(x)=\infty\iff\lim_{x\rightarrow a}g(x)=\infty. lim x → a f ( x ) = ∞ ⟺ lim x → a g ( x ) = ∞.
e.g.1.5 \textbf{e.g.1.5 } e.g.1.5 lim x → − ∞ ( x 2 − x + 1 − a x − b ) = 0 \lim_{x\rightarrow-\infty}(\sqrt{x^2-x+1}-ax-b)=0 lim x → − ∞ ( x 2 − x + 1 − a x − b ) = 0 a , b a,b a , b
Solution \textbf{Solution } Solution x < 0 x<0 x < 0 x 2 = − x , \sqrt{x^2}=-x, x 2 = − x , a = − 1. a=-1. a = − 1.
b = lim x → − ∞ ( x 2 − x + 1 + x ) = lim x → − ∞ x 2 − x + 1 − x 2 x 2 − x + 1 − x = lim x → − ∞ − x ∣ x ∣ − x = 1 2 . b=\lim_{x\rightarrow-\infty}(\sqrt{x^2-x+1}+x)=\lim_{x\rightarrow-\infty}\frac{x^2-x+1-x^2}{\sqrt{x^2-x+1}-x}=\lim_{x\rightarrow-\infty}\frac{-x}{|x|-x}=\frac{1}{2}.
b = x → − ∞ lim ( x 2 − x + 1 + x ) = x → − ∞ lim x 2 − x + 1 − x x 2 − x + 1 − x 2 = x → − ∞ lim ∣ x ∣ − x − x = 2 1 .
e.g.1.6 \textbf{e.g.1.6 } e.g.1.6 lim x → + ∞ ( ( x 5 + 7 x 4 + 2 ) α − x ) = β ≠ 0 \lim_{x\rightarrow+\infty}((x^5+7x^4+2)^\alpha-x)=\beta≠0 lim x → + ∞ (( x 5 + 7 x 4 + 2 ) α − x ) = β = 0 α , β \alpha,\beta α , β
Solution \textbf{Solution } Solution α = 1 5 . \alpha=\frac{1}{5}. α = 5 1 .
β = lim x → − ∞ ( x ( ( 1 + 7 x + 2 x 5 ) 1 5 − 1 ) ) = lim x → − ∞ ( x ( 1 5 ( 7 x + 2 x 5 ) ) ) = 7 5 . \beta=\lim_{x\rightarrow-\infty}(x((1+\frac{7}{x}+\frac{2}{x^5})^\frac{1}{5}-1))=\lim_{x\rightarrow-\infty}(x(\frac{1}{5}(\frac{7}{x}+\frac{2}{x^5})))=\frac{7}{5}.
β = x → − ∞ lim ( x (( 1 + x 7 + x 5 2 ) 5 1 − 1 )) = x → − ∞ lim ( x ( 5 1 ( x 7 + x 5 2 ))) = 5 7 .
在例1.6中使用了结论“( 1 + x ) α − 1 ∼ α x ( x → 0 ) (1+x)^\alpha-1\sim\alpha x(x\rightarrow0) ( 1 + x ) α − 1 ∼ αx ( x → 0 )
lim x → ∞ ( ( ∑ i = 0 n a i x i ) 1 n − x ) = a n − 1 n . \lim_{x\rightarrow\infty}((\sum_{i=0}^na_ix^i)^\frac{1}{n}-x)=\frac{a_{n-1}}{n}.
x → ∞ lim (( i = 0 ∑ n a i x i ) n 1 − x ) = n a n − 1 .
e.x.1.3 \textbf{e.x.1.3 } e.x.1.3 lim x → + ∞ ( x 6 + x 5 6 − x 6 − x 5 6 ) . \lim_{x\rightarrow+\infty}(\sqrt[6]{x^6+x^5}-\sqrt[6]{x^6-x^5}). lim x → + ∞ ( 6 x 6 + x 5 − 6 x 6 − x 5 ) .
Solution \textbf{Solution } Solution lim x → + ∞ ( x 6 + x 5 ) 1 6 − x = 1 6 , lim x → + ∞ ( x 6 − x 5 ) 1 6 − x = − 1 6 , \lim_{x\rightarrow+\infty}(x^6+x^5)^\frac{1}{6}-x=\frac{1}{6}, \lim_{x\rightarrow+\infty}(x^6-x^5)^\frac{1}{6}-x=-\frac{1}{6}, lim x → + ∞ ( x 6 + x 5 ) 6 1 − x = 6 1 , lim x → + ∞ ( x 6 − x 5 ) 6 1 − x = − 6 1 ,
O.F. = lim x → + ∞ ( ( x 6 + x 5 6 − x ) − ( x 6 − x 5 6 − x ) ) = 1 6 − ( − 1 6 ) = 1 3 . \text{O.F.}=\lim_{x\rightarrow+\infty}((\sqrt[6]{x^6+x^5}-x)-(\sqrt[6]{x^6-x^5}-x))=\frac{1}{6}-(-\frac{1}{6})=\frac{1}{3}.
O.F. = x → + ∞ lim (( 6 x 6 + x 5 − x ) − ( 6 x 6 − x 5 − x )) = 6 1 − ( − 6 1 ) = 3 1 .
现在考虑一个结论
lim x → ∞ x α − ( x − 1 ) α = α x α − 1 . \lim_{x\rightarrow\infty}x^\alpha-(x-1)^\alpha=\alpha x^{\alpha-1}.
x → ∞ lim x α − ( x − 1 ) α = α x α − 1 .
这可以理解为幂函数f ( x ) = x α f(x)=x^\alpha f ( x ) = x α x x x 1 1 1 f ( x ) f(x) f ( x ) f ( x − 1 ) f(x-1) f ( x − 1 ) f ( x ) f(x) f ( x )
e.g.1.7 \textbf{e.g.1.7 } e.g.1.7 lim x → ∞ x 2013 x α − ( x − 1 ) α = β ≠ 0 \lim_{x\rightarrow\infty}\frac{x^{2013}}{x^\alpha-(x-1)^\alpha}=\beta≠0 lim x → ∞ x α − ( x − 1 ) α x 2013 = β = 0 α , β \alpha,\beta α , β
Solution \textbf{Solution } Solution α = 2014. \alpha=2014. α = 2014.
β = lim x → ∞ x 2013 2014 x 2013 = 1 2014 . \beta=\lim_{x\rightarrow\infty}\frac{x^{2013}}{2014x^{2013}}=\frac{1}{2014}.
β = x → ∞ lim 2014 x 2013 x 2013 = 2014 1 .
e.x.1.4 \textbf{e.x.1.4 } e.x.1.4 lim x → + ∞ ( ( a x + b ) e 1 x − x ) = 2 \lim_{x\rightarrow+\infty}((ax+b)e^\frac{1}{x}-x)=2 lim x → + ∞ (( a x + b ) e x 1 − x ) = 2 a , b a,b a , b
Solution \textbf{Solution } Solution a = 1. a=1. a = 1.
2 = lim x → + ∞ ( ( x + b ) e 1 x − x ) = lim t → 0 + ( ( 1 t + b ) e t − 1 t ) = lim t → 0 + e t − 1 t + lim t → 0 + b e t = 1 + b 2=\lim_{x\rightarrow+\infty}((x+b)e^\frac{1}{x}-x)=\lim_{t\rightarrow0^+}((\frac{1}{t}+b)e^t-\frac{1}{t})=\lim_{t\rightarrow0^+}\frac{e^t-1}{t}+\lim_{t\rightarrow0^+}be^t=1+b
2 = x → + ∞ lim (( x + b ) e x 1 − x ) = t → 0 + lim (( t 1 + b ) e t − t 1 ) = t → 0 + lim t e t − 1 + t → 0 + lim b e t = 1 + b
因此b = 1 b=1 b = 1
需要注意的是，习题1.4中的b b b ( x + b ) ⋅ 1 − x = 2 (x+b)\cdot1-x=2 ( x + b ) ⋅ 1 − x = 2 e 1 x . e^\frac{1}{x}. e x 1 .
Taylor公式求极限
0 0 \frac{0}{0} 0 0
常用的8个Taylor公式推得的极限包括：①sin x = x − 1 6 x 3 + o ( x 3 ) ; \sin x=x-\frac{1}{6}x^3+o(x^3); sin x = x − 6 1 x 3 + o ( x 3 ) ; arcsin x = x + 1 6 x 3 + o ( x 3 ) ; \arcsin x=x+\frac{1}{6}x^3+o(x^3); arcsin x = x + 6 1 x 3 + o ( x 3 ) ; tan x = x + 1 3 x 3 + o ( x 3 ) ; \tan x=x+\frac{1}{3}x^3+o(x^3); tan x = x + 3 1 x 3 + o ( x 3 ) ; arctan x = x − 1 3 x 3 + o ( x 3 ) ; \arctan x=x-\frac{1}{3}x^3+o(x^3); arctan x = x − 3 1 x 3 + o ( x 3 ) ; ln ( 1 + x ) = x − 1 2 x 2 + o ( x 2 ) ; \ln(1+x)=x-\frac{1}{2}x^2+o(x^2); ln ( 1 + x ) = x − 2 1 x 2 + o ( x 2 ) ; cos x = 1 − 1 2 x 2 + 1 4 x 4 + o ( x 4 ) ; \cos x=1-\frac{1}{2}x^2+\frac{1}{4}x^4+o(x^4); cos x = 1 − 2 1 x 2 + 4 1 x 4 + o ( x 4 ) ; e x = 1 + x + 1 2 x 2 + o ( x 2 ) ; e^x=1+x+\frac{1}{2}x^2+o(x^2); e x = 1 + x + 2 1 x 2 + o ( x 2 ) ; ( 1 + x ) α = 1 + α x + α ( α − 1 ) 2 x 2 + o ( x 2 ) . (1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2}x^2+o(x^2). ( 1 + x ) α = 1 + αx + 2 α ( α − 1 ) x 2 + o ( x 2 ) . m < n m<n m < n o ( x m ) + o ( x n ) = o ( x m ) o(x^m)+o(x^n)=o(x^m) o ( x m ) + o ( x n ) = o ( x m )
e.g.1.8 \textbf{e.g.1.8 } e.g.1.8 lim x → 0 1 − cos x cos 2 x cos 3 x a x n = 1 , \lim_{x\rightarrow0}\frac{1-\cos x\cos 2x\cos 3x}{ax^n}=1, lim x → 0 a x n 1 − c o s x c o s 2 x c o s 3 x = 1 , a , b . a,b. a , b .
Solution \textbf{Solution } Solution cos x \cos x cos x cos 2 x \cos 2x cos 2 x cos 3 x \cos 3x cos 3 x
cos x cos 2 x cos 3 x = ( 1 − 1 2 ! x 2 + 1 4 ! x 4 − … ) ( 1 − 1 2 ! ( 2 x ) 2 + … ) ( 1 − 1 2 ! ( 3 x ) 2 + … ) = 1 − ( 1 2 + 2 + 9 2 ) x 2 + o ( x 2 ) = 1 − 7 x 2 + o ( x 2 ) \begin{align*}
\cos x\cos 2x\cos 3x&=(1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\dots)(1-\frac{1}{2!}(2x)^2+\dots)(1-\frac{1}{2!}(3x)^2+\dots)\\
&=1-(\frac{1}{2}+2+\frac{9}{2})x^2+o(x^2)=1-7x^2+o(x^2)
\end{align*}
cos x cos 2 x cos 3 x = ( 1 − 2 ! 1 x 2 + 4 ! 1 x 4 − … ) ( 1 − 2 ! 1 ( 2 x ) 2 + … ) ( 1 − 2 ! 1 ( 3 x ) 2 + … ) = 1 − ( 2 1 + 2 + 2 9 ) x 2 + o ( x 2 ) = 1 − 7 x 2 + o ( x 2 )
因此1 − cos x cos 2 x cos 3 x = 7 x 2 + o ( x 2 ) ∼ 7 x 2 ⇒ lim x → 0 1 − cos x cos 2 x cos 3 x a x n = lim x → 0 7 x 2 a x n = 1 ⇒ a = 7 , n = 2. 1-\cos x\cos 2x\cos 3x=7x^2+o(x^2)\sim 7x^2\Rightarrow\lim_{x\rightarrow0}\frac{1-\cos x\cos 2x\cos 3x}{ax^n}=\lim_{x\rightarrow0}\frac{7x^2}{ax^n}=1\Rightarrow a=7,n=2. 1 − cos x cos 2 x cos 3 x = 7 x 2 + o ( x 2 ) ∼ 7 x 2 ⇒ lim x → 0 a x n 1 − c o s x c o s 2 x c o s 3 x = lim x → 0 a x n 7 x 2 = 1 ⇒ a = 7 , n = 2.
另外需要掌握Taylor公式的展开原则。
当所展开的函数是待求极限式中的分子（或分母）且分母（子）是x x x k k k 上下同阶 。
k k k
e.g.1.9 \textbf{e.g.1.9 } e.g.1.9 lim x → 0 1 + tan x − 1 + sin x x ln ( 1 + x ) − x 2 . \lim_{x\rightarrow0}\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x\ln(1+x)-x^2}. lim x → 0 x l n ( 1 + x ) − x 2 1 + t a n x − 1 + s i n x .
Solution \textbf{Solution } Solution
O.F. = lim x → 0 tan x − sin x x ( ln ( 1 + x ) − x ) ( 1 + tan x + 1 + sin x ) = lim x → 0 1 2 x 3 x ⋅ ( − 1 2 x 2 ) ⋅ 2 = − 1 2 . \begin{align*}\text{O.F.}&=\lim_{x\rightarrow0}\frac{\tan x-\sin x}{x(\ln(1+x)-x)(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\
&=\lim_{x\rightarrow 0}\frac{\frac{1}{2}x^3}{x\cdot (-\frac{1}{2}x^2)\cdot 2}&=-\frac{1}{2}.
\end{align*} O.F. = x → 0 lim x ( ln ( 1 + x ) − x ) ( 1 + tan x + 1 + sin x ) tan x − sin x = x → 0 lim x ⋅ ( − 2 1 x 2 ) ⋅ 2 2 1 x 3 = − 2 1 .
考虑下方的例子，可以体会什么叫做“全部展开”。
e.g.1.10 \textbf{e.g.1.10 } e.g.1.10 lim x → 0 ln ( 1 + sin 2 x ) − x 2 x 4 . \lim_{x\rightarrow0}\frac{\ln(1+\sin^2 x)-x^2}{x^4}. lim x → 0 x 4 l n ( 1 + s i n 2 x ) − x 2 .
Solution \textbf{Solution } Solution ln ( 1 + sin 2 x ) \ln(1+\sin^2 x) ln ( 1 + sin 2 x )
ln ( 1 + sin 2 x ) = sin 2 x − 1 2 sin 4 x + o ( sin 4 x ) = ( x − 1 6 x 3 + o ( x 3 ) ) 2 − 1 2 ( x − 1 6 x 3 + o ( x 3 ) ) 4 + o ( x 4 ) = ( x 2 − 1 3 x 4 + o ( x 4 ) ) − 1 2 ( x 4 + o ( x 4 ) ) + o ( x 4 ) = x 2 − 5 6 x 4 + o ( x 4 ) ∼ x 2 − 5 6 x 4 , \begin{align*}
\ln(1+\sin^2x)&=\sin^2x-\frac{1}{2}\sin^4x+o(\sin^4x)\\
&=(x-\frac{1}{6}x^3+o(x^3))^2-\frac{1}{2}(x-\frac{1}{6}x^3+o(x^3))^4+o(x^4)\\
&=(x^2-\frac{1}{3}x^4+o(x^4))-\frac{1}{2}(x^4+o(x^4))+o(x^4)\\
&=x^2-\frac{5}{6}x^4+o(x^4)&\sim x^2-\frac{5}{6}x^4,
\end{align*}
ln ( 1 + sin 2 x ) = sin 2 x − 2 1 sin 4 x + o ( sin 4 x ) = ( x − 6 1 x 3 + o ( x 3 ) ) 2 − 2 1 ( x − 6 1 x 3 + o ( x 3 ) ) 4 + o ( x 4 ) = ( x 2 − 3 1 x 4 + o ( x 4 )) − 2 1 ( x 4 + o ( x 4 )) + o ( x 4 ) = x 2 − 6 5 x 4 + o ( x 4 ) ∼ x 2 − 6 5 x 4 ,
因此
O.F. = lim x → 0 ( x 2 − 7 6 x 4 ) − x 2 x 4 = − 7 6 . \begin{align*}\text{O.F.}&=\lim_{x\rightarrow0}\frac{(x^2-\frac{7}{6}x^4)-x^2}{x^4}=-\frac{7}{6}.
\end{align*} O.F. = x → 0 lim x 4 ( x 2 − 6 7 x 4 ) − x 2 = − 6 7 .
在上述例子中毫无疑问我们需要寻找到ln ( 1 + sin 2 x ) \ln(1+\sin^2x) ln ( 1 + sin 2 x ) x x x sin x \sin x sin x sin x \sin x sin x x x x x 4 x^4 x 4
e.x.1.5 \textbf{e.x.1.5 } e.x.1.5 lim x → 0 1 + x 2 3 − e x 2 3 ln ( 1 + 3 x 2 ) − 3 x 2 cos x . \lim_{x\rightarrow0}\frac{\sqrt[3]{1+x^2}-e^{\frac{x^2}{3}}}{\ln(1+3x^2)-3x^2\cos x}. lim x → 0 l n ( 1 + 3 x 2 ) − 3 x 2 c o s x 3 1 + x 2 − e 3 x 2 .
Solution \textbf{Solution } Solution
1 + x 2 3 − e x 2 3 = ( 1 + 1 3 x 2 + 1 3 ( 1 3 − 1 ) 2 x 4 + o ( x 4 ) ) − ( ( 1 + x 2 3 + 1 2 ( x 2 3 ) 2 ) + o ( x 4 ) ) = − 1 6 x 4 + o ( x 4 ) ∼ − 1 6 x 4 . \begin{align*}
\sqrt[3]{1+x^2}-e^{\frac{x^2}{3}}&=(1+\frac{1}{3}x^2+\frac{\frac{1}{3}(\frac{1}{3}-1)}{2}x^4+o(x^4))-((1+\frac{x^2}{3}+\frac{1}{2}(\frac{x^2}{3})^2)+o(x^4))\\
&=-\frac{1}{6}x^4+o(x^4)&\sim-\frac{1}{6}x^4.
\end{align*}
3 1 + x 2 − e 3 x 2 = ( 1 + 3 1 x 2 + 2 3 1 ( 3 1 − 1 ) x 4 + o ( x 4 )) − (( 1 + 3 x 2 + 2 1 ( 3 x 2 ) 2 ) + o ( x 4 )) = − 6 1 x 4 + o ( x 4 ) ∼ − 6 1 x 4 .
另一方面，对分母展开得
ln ( 1 + 3 x 2 ) − 3 x 2 cos x = ( 3 x 2 − 1 2 ( 3 x 2 ) 2 + o ( x 4 ) ) − 3 x 2 ( 1 − 1 2 x 2 + o ( x 2 ) ) = ( 3 x 2 − 9 2 x 4 + o ( x 4 ) ) − ( 3 x 2 − 3 2 x 4 + o ( x 4 ) ) = − 3 x 4 + o ( x 4 ) ∼ − 3 x 4 . \begin{align*}
\ln(1+3x^2)-3x^2\cos x&=(3x^2-\frac{1}{2}(3x^2)^2+o(x^4))-3x^2(1-\frac{1}{2}x^2+o(x^2))\\
&=(3x^2-\frac{9}{2}x^4+o(x^4))-(3x^2-\frac{3}{2}x^4+o(x^4))\\
&=-3x^4+o(x^4)&\sim-3x^4.
\end{align*}
ln ( 1 + 3 x 2 ) − 3 x 2 cos x = ( 3 x 2 − 2 1 ( 3 x 2 ) 2 + o ( x 4 )) − 3 x 2 ( 1 − 2 1 x 2 + o ( x 2 )) = ( 3 x 2 − 2 9 x 4 + o ( x 4 )) − ( 3 x 2 − 2 3 x 4 + o ( x 4 )) = − 3 x 4 + o ( x 4 ) ∼ − 3 x 4 .
因此
O.F. = lim x → 0 − 1 6 x 4 − 3 x 4 = 1 18 . \begin{align*}\text{O.F.}&=\lim_{x\rightarrow0}\frac{-\frac{1}{6}x^4}{-3x^4}=\frac{1}{18}.
\end{align*} O.F. = x → 0 lim − 3 x 4 − 6 1 x 4 = 18 1 .
仔细考量练习1.5和上方警示。如果对分母的cos x \cos x cos x 1 1 1
e.x.1.6 \textbf{e.x.1.6 } e.x.1.6 lim x → 0 ( 1 ln ( x + 1 + x 2 ) − 1 ln ( 1 + x ) ) . \lim_{x\rightarrow0}(\frac{1}{\ln(x+\sqrt{1+x^2})}-\frac{1}{\ln(1+x)}). lim x → 0 ( l n ( x + 1 + x 2 ) 1 − l n ( 1 + x ) 1 ) .
Solution \textbf{Solution } Solution ∞ − ∞ \infty-\infty ∞ − ∞
O.F. = lim x → 0 ln 1 + x x + 1 + x 2 ln ( x + 1 + x 2 ) ln ( 1 + x ) \text{O.F.}=\lim_{x\rightarrow0}\frac{\ln\frac{1+x}{x+\sqrt{1+x^2}}}{\ln(x+\sqrt{1+x^2})\ln(1+x)}
O.F. = x → 0 lim ln ( x + 1 + x 2 ) ln ( 1 + x ) ln x + 1 + x 2 1 + x
而( ln ( x + 1 + x 2 ) ) ′ = 1 1 + x 2 = ( 1 + x 2 ) − 1 2 = 1 + ( − 1 2 ) x 2 + o ( x 2 ) ⇒ ln ( x + 1 + x 2 ) = ∫ ( 1 − 1 2 x 2 + o ( x 2 ) ) d x = x − 1 6 x 3 + o ( x 3 ) , (\ln(x+\sqrt{1+x^2}))'=\frac{1}{\sqrt{1+x^2}}=(1+x^2)^{-\frac{1}{2}}=1+(-\frac{1}{2})x^2+o(x^2)\Rightarrow\ln(x+\sqrt{1+x^2})=\int(1-\frac{1}{2}x^2+o(x^2))\text{d}x=x-\frac{1}{6}x^3+o(x^3), ( ln ( x + 1 + x 2 ) ) ′ = 1 + x 2 1 = ( 1 + x 2 ) − 2 1 = 1 + ( − 2 1 ) x 2 + o ( x 2 ) ⇒ ln ( x + 1 + x 2 ) = ∫ ( 1 − 2 1 x 2 + o ( x 2 )) d x = x − 6 1 x 3 + o ( x 3 ) , ln ( x + 1 + x 2 ) ln ( 1 + x ) ∼ x 2 , \ln(x+\sqrt{1+x^2})\ln(1+x)\sim x^2, ln ( x + 1 + x 2 ) ln ( 1 + x ) ∼ x 2 ,
O.F. = ( x − 1 2 x 2 + o ( x 2 ) ) − ( x + o ( x 2 ) ) x 2 = − 1 2 x 2 x 2 = − 1 2 . \begin{align*}
\text{O.F.}&=\frac{(x-\frac{1}{2}x^2+o(x^2))-(x+o(x^2))}{x^2}\\
&=\frac{-\frac{1}{2}x^2}{x^2}&=-\frac{1}{2}.
\end{align*}
O.F. = x 2 ( x − 2 1 x 2 + o ( x 2 )) − ( x + o ( x 2 )) = x 2 − 2 1 x 2 = − 2 1 .
注意展开的目的是便于运算，因此需要适当调节选取的高阶无穷小；同一个式子哪怕是分子和分母都有可能展开时的阶数不一样。
等价无穷小代换求极限
在x → 0 x\rightarrow0 x → 0
e x − 1 ∼ x e^x-1\sim x e x − 1 ∼ x e f ( x ) − e g ( x ) e^{f(x)}-e^{g(x)} e f ( x ) − e g ( x ) e g ( x ) ( e f ( x ) − g ( x ) − 1 ) ∼ e g ( x ) ( f ( x ) − g ( x ) ) . e^g(x)(e^{f(x)-g(x)}-1)\sim e^g(x)(f(x)-g(x)). e g ( x ) ( e f ( x ) − g ( x ) − 1 ) ∼ e g ( x ) ( f ( x ) − g ( x )) . ln ( x + 1 ) ∼ x \ln(x+1)\sim x ln ( x + 1 ) ∼ x ln f ( x ) = ln ( ( f ( x ) − 1 ) + 1 ) ∼ f ( x ) − 1. \ln f(x)=\ln((f(x)-1)+1)\sim f(x)-1. ln f ( x ) = ln (( f ( x ) − 1 ) + 1 ) ∼ f ( x ) − 1. ( 1 + x ) α − 1 ∼ x (1+x)^\alpha-1\sim x ( 1 + x ) α − 1 ∼ x α \alpha α 1 + x sin x 3 − 1 ∼ 1 3 x sin x ∼ 1 3 x 2 . \sqrt[3]{1+x\sin x}-1\sim\frac{1}{3}x\sin x\sim\frac{1}{3}x^2. 3 1 + x sin x − 1 ∼ 3 1 x sin x ∼ 3 1 x 2 .
e.g.1.11 \textbf{e.g.1.11 } e.g.1.11 lim x → 0 ln cos x x 2 \lim_{x\rightarrow0}\frac{\ln\cos x}{x^2} lim x → 0 x 2 l n c o s x
Solution \textbf{Solution } Solution
O.F. = lim x → 0 ln ( 1 + ( cos x − 1 ) ) x 2 = lim x → 0 ln ( 1 + ( − 1 2 x 2 ) ) x 2 = lim x → 0 − 1 2 x 2 x 2 = − 1 2 . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow0}\frac{\ln(1+(\cos x-1))}{x^2}&=\lim_{x\rightarrow0}\frac{\ln(1+(-\frac{1}{2}x^2))}{x^2}\\
&=\lim_{x\rightarrow0}\frac{-\frac{1}{2}x^2}{x^2}&&=-\frac{1}{2}.
\end{align*}
O.F. = x → 0 lim x 2 ln ( 1 + ( cos x − 1 )) = x → 0 lim x 2 − 2 1 x 2 = x → 0 lim x 2 ln ( 1 + ( − 2 1 x 2 )) = − 2 1 .
e.g.1.12 \textbf{e.g.1.12 } e.g.1.12 lim x → 0 ln sin x x x 2 \lim_{x\rightarrow0}\frac{\ln\frac{\sin x}{x}}{x^2} lim x → 0 x 2 l n x s i n x
Solution \textbf{Solution } Solution
O.F. = lim x → 0 ln ( 1 + ( sin x x − 1 ) ) x 2 = lim x → 0 sin x x − 1 x 2 = lim x → 0 sin x − x x 3 = ( x − 1 6 x 3 + o ( x 3 ) ) − x x 3 = − 1 6 . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow0}\frac{\ln(1+(\frac{\sin x}{x}-1))}{x^2}&=\lim_{x\rightarrow0}\frac{\frac{\sin x}{x}-1}{x^2}\\
&=\lim_{x\rightarrow0}\frac{\sin x-x}{x^3}&=\frac{(x-\frac{1}{6}x^3+o(x^3))-x}{x^3}&=-\frac{1}{6}.
\end{align*}
O.F. = x → 0 lim x 2 ln ( 1 + ( x s i n x − 1 )) = x → 0 lim x 3 sin x − x = x → 0 lim x 2 x s i n x − 1 = x 3 ( x − 6 1 x 3 + o ( x 3 )) − x = − 6 1 .
例1.12过程中不可以采用逻辑ln sin x x = ln sin x − ln x ∼ ( sin x − 1 ) − ( x − 1 ) = sin x − x . \ln\frac{\sin x}{x}=\ln\sin x-\ln x\sim(\sin x-1)-(x-1)=\sin x-x. ln x s i n x = ln sin x − ln x ∼ ( sin x − 1 ) − ( x − 1 ) = sin x − x .
e.x.1.7 \textbf{e.x.1.7 } e.x.1.7 lim x → 0 1 + x arcsin x 3 − cos x x 2 \lim_{x\rightarrow0}\frac{\sqrt[3]{1+x\arcsin x}-\cos x}{x^2} lim x → 0 x 2 3 1 + x a r c s i n x − c o s x
Solution \textbf{Solution } Solution
O.F. = lim x → 0 ( 1 + x arcsin x 3 − 1 ) + ( 1 − cos x ) x 2 = 1 3 x arcsin x + 1 2 x 2 x 2 = 1 3 x ⋅ x + 1 2 x 2 x 2 = 5 6 x 2 x 2 = 5 6 . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow0}\frac{(\sqrt[3]{1+x\arcsin x}-1)+(1-\cos x)}{x^2}&=\frac{\frac{1}{3}x\arcsin x+\frac{1}{2}x^2}{x^2}\\
&=\frac{\frac{1}{3}x\cdot x+\frac{1}{2}x^2}{x^2}&=\frac{\frac{5}{6}x^2}{x^2}&=\frac{5}{6}.
\end{align*}
O.F. = x → 0 lim x 2 ( 3 1 + x arcsin x − 1 ) + ( 1 − cos x ) = x 2 3 1 x ⋅ x + 2 1 x 2 = x 2 3 1 x arcsin x + 2 1 x 2 = x 2 6 5 x 2 = 6 5 .
e.x.1.8 \textbf{e.x.1.8 } e.x.1.8 lim x → 0 1 + x + 1 − x − 2 x 2 \lim_{x\rightarrow0}\frac{\sqrt{1+x}+\sqrt{1-x}-2}{x^2} lim x → 0 x 2 1 + x + 1 − x − 2
Solution \textbf{Solution } Solution
O.F. = lim x → 0 ( 1 + x − 1 ) + ( 1 − x − 1 ) x 2 = 1 2 x + 1 2 ( − x ) x 2 = 0. \text{O.F.}=\lim_{x\rightarrow0}\frac{(\sqrt{1+x}-1)+(\sqrt{1-x}-1)}{x^2}=\frac{\frac{1}{2}x+\frac{1}{2}(-x)}{x^2}=0.
O.F. = x → 0 lim x 2 ( 1 + x − 1 ) + ( 1 − x − 1 ) = x 2 2 1 x + 2 1 ( − x ) = 0.
等价无穷小的本质是对Taylor展开后的结果保留较低阶（如x 3 + x 5 ∼ x 3 x^3+x^5\sim x^3 x 3 + x 5 ∼ x 3
e.x.1.9 \textbf{e.x.1.9 } e.x.1.9 lim x → 0 a tan x + b ( 1 − cos x ) c ln ( 1 − 2 x ) + d ( 1 − e − x 2 ) = 2 \lim_{x\rightarrow0}\frac{a\tan x+b(1-\cos x)}{c\ln(1-2x)+d(1-e^{-x^2})}=2 lim x → 0 c l n ( 1 − 2 x ) + d ( 1 − e − x 2 ) a t a n x + b ( 1 − c o s x ) = 2 a 2 + c 2 ≠ 0 , a^2+c^2≠0, a 2 + c 2 = 0 , a a a c c c b b b d d d
Solution \textbf{Solution } Solution a tan x ∼ a x , b ( 1 − cos x ) ∼ b ( 1 2 x 2 ) , a\tan x\sim ax, b(1-\cos x)\sim b(\frac{1}{2}x^2), a tan x ∼ a x , b ( 1 − cos x ) ∼ b ( 2 1 x 2 ) , a x . ax. a x .
同理得分母保留一阶项c ln ( 1 − 2 x ) ∼ c ⋅ ( − 2 x ) . c\ln(1-2x)\sim c\cdot(-2x). c ln ( 1 − 2 x ) ∼ c ⋅ ( − 2 x ) .
左式 = a x − 2 c x = − a 2 c . \text{左式}=\frac{ax}{-2cx}=-\frac{a}{2c}.
左式 = − 2 c x a x = − 2 c a .
而左式=2，因此− a 2 c = 2 ⇒ a = − 4 c . -\frac{a}{2c}=2\Rightarrow a=-4c. − 2 c a = 2 ⇒ a = − 4 c .
b b b d d d
练习1.9中的式子本应该使用Taylor展开，同样可以得到分子为a x + o ( x ) ax+o(x) a x + o ( x ) c ( − 2 x ) + o ( x ) . c(-2x)+o(x). c ( − 2 x ) + o ( x ) .
e.x.1.10 \textbf{e.x.1.10 } e.x.1.10 lim x → 0 e − e cos x 1 + x 2 3 − 1 . \lim_{x\rightarrow0}\frac{e-e^{\cos x}}{\sqrt[3]{1+x^2}-1}. lim x → 0 3 1 + x 2 − 1 e − e c o s x .
Solution \textbf{Solution } Solution
O.F. = lim x → 0 e ( 1 − e cos x − 1 ) 1 3 x 2 = lim x → 0 e ( 1 − cos x ) 1 3 x 2 = lim x → 0 e ( 1 2 x 2 ) 1 3 x 2 = 1 2 e 1 3 = 3 2 e . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow0}\frac{e(1-e^{\cos x-1})}{\frac{1}{3}x^2}&=\lim_{x\rightarrow0}\frac{e(1-\cos x)}{\frac{1}{3}x^2}\\
&=\lim_{x\rightarrow0}\frac{e(\frac{1}{2}x^2)}{\frac{1}{3}x^2}&=\frac{\frac{1}{2}e}{\frac{1}{3}}&=\frac{3}{2}e.
\end{align*}
O.F. = x → 0 lim 3 1 x 2 e ( 1 − e c o s x − 1 ) = x → 0 lim 3 1 x 2 e ( 2 1 x 2 ) = x → 0 lim 3 1 x 2 e ( 1 − cos x ) = 3 1 2 1 e = 2 3 e .
对于待求极限式中的幂指函数（即一个含有x x x x x x f ( x ) g ( x ) {f(x)}^{g(x)} f ( x ) g ( x ) x x x^x x x e x − 1 ∼ x e^x-1\sim x e x − 1 ∼ x ln ( x + 1 ) ∼ x \ln(x+1)\sim x ln ( x + 1 ) ∼ x x x x
在上述提示中所谓的“对数的性质”主要停留在ln ( f ( x ) g ( x ) ) = g ( x ) ln f ( x ) . \ln({f(x)}^{g(x)})=g(x)\ln f(x). ln ( f ( x ) g ( x ) ) = g ( x ) ln f ( x ) . f ( x ) g ( x ) = e ln ( f ( x ) g ( x ) ) ∼ ln ( f ( x ) g ( x ) ) + 1 , {f(x)}^{g(x)}=e^{\ln({f(x)}^{g(x)})}\sim \ln({f(x)}^{g(x)})+1, f ( x ) g ( x ) = e l n ( f ( x ) g ( x ) ) ∼ ln ( f ( x ) g ( x ) ) + 1 , ln ( f ( x ) g ( x ) ) + 1 = g ( x ) ln f ( x ) + 1 = g ( x ) ln ( ( f ( x ) − 1 ) + 1 ) + 1 ∼ g ( x ) ( f ( x ) − 1 ) + 1 , \ln({f(x)}^{g(x)})+1=g(x)\ln f(x)+1=g(x)\ln((f(x)-1)+1)+1\sim g(x)(f(x)-1)+1, ln ( f ( x ) g ( x ) ) + 1 = g ( x ) ln f ( x ) + 1 = g ( x ) ln (( f ( x ) − 1 ) + 1 ) + 1 ∼ g ( x ) ( f ( x ) − 1 ) + 1 , f ( x ) g ( x ) ∼ g ( x ) ( f ( x ) − 1 ) + 1. {f(x)}^{g(x)}\sim g(x)(f(x)-1)+1. f ( x ) g ( x ) ∼ g ( x ) ( f ( x ) − 1 ) + 1.
e.g.1.13 \textbf{e.g.1.13 } e.g.1.13 lim x → 0 1 x 3 ( ( 2 + cos x 3 ) x − 1 ) . \lim_{x\rightarrow0}\frac{1}{x^3}((\frac{2+\cos x}{3})^x-1). lim x → 0 x 3 1 (( 3 2 + c o s x ) x − 1 ) .
Solution \textbf{Solution } Solution x x x ( 2 + cos x 3 ) x (\frac{2+\cos x}{3})^x ( 3 2 + c o s x ) x e ln ( ( 2 + cos x 3 ) x ) = e x ln ( 2 + cos x 3 ) e^{\ln((\frac{2+\cos x}{3})^x)}=e^{x\ln(\frac{2+\cos x}{3})} e l n (( 3 2 + c o s x ) x ) = e x l n ( 3 2 + c o s x )
O.F. = lim x → 0 e x ln ( 2 + cos x 3 ) − 1 x 3 = lim x → 0 x ln ( 2 + cos x 3 ) x 3 = lim x → 0 ln ( 2 + cos x 3 ) x 2 = lim x → 0 ln ( 1 + cos x − 1 3 ) x 2 = lim x → 0 cos x − 1 3 x 2 = lim x → 0 − 1 2 x 2 3 x 2 = − 1 6 . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow0}\frac{e^{x\ln(\frac{2+\cos x}{3})}-1}{x^3}&=\lim_{x\rightarrow0}\frac{x\ln(\frac{2+\cos x}{3})}{x^3}&=\lim_{x\rightarrow0}\frac{\ln(\frac{2+\cos x}{3})}{x^2}\\
&=\lim_{x\rightarrow0}\frac{\ln(1+\frac{\cos x-1}{3})}{x^2}&=\lim_{x\rightarrow0}\frac{\frac{\cos x-1}{3}}{x^2}&=\lim_{x\rightarrow0}\frac{-\frac{1}{2}x^2}{3x^2}&=-\frac{1}{6}.
\end{align*}
O.F. = x → 0 lim x 3 e x l n ( 3 2 + c o s x ) − 1 = x → 0 lim x 2 ln ( 1 + 3 c o s x − 1 ) = x → 0 lim x 3 x ln ( 3 2 + c o s x ) = x → 0 lim x 2 3 c o s x − 1 = x → 0 lim x 2 ln ( 3 2 + c o s x ) = x → 0 lim 3 x 2 − 2 1 x 2 = − 6 1 .
e.x.1.11 \textbf{e.x.1.11 } e.x.1.11 lim x → 0 ( 1 + x ) 1 x − e x . \lim_{x\rightarrow0}\frac{(1+x)^\frac{1}{x}-e}{x}. lim x → 0 x ( 1 + x ) x 1 − e .
Solution \textbf{Solution } Solution
O.F. = lim x → 0 e ln ( ( 1 + x ) 1 x ) − e x = lim x → 0 e 1 x ln ( 1 + x ) − e x = lim x → 0 e ( e 1 x ln ( 1 + x ) − 1 − 1 ) x = e lim x → 0 1 x ln ( 1 + x ) − 1 x = e lim x → 0 ln ( 1 + x ) − x x 2 = e − 1 2 x 2 x 2 = − 1 2 e . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow0}\frac{e^{\ln((1+x)^{\frac{1}{x}})}-e}{x}&=\lim_{x\rightarrow0}\frac{e^{\frac{1}{x}\ln(1+x)}-e}{x}&=\lim_{x\rightarrow0}\frac{e(e^{\frac{1}{x}\ln(1+x)-1}-1)}{x}\\
&=e\lim_{x\rightarrow0}\frac{\frac{1}{x}\ln(1+x)-1}{x}&=e\lim_{x\rightarrow0}\frac{\ln(1+x)-x}{x^2}&=e\frac{-\frac{1}{2}x^2}{x^2}=-\frac{1}{2}e.
\end{align*}
O.F. = x → 0 lim x e l n (( 1 + x ) x 1 ) − e = e x → 0 lim x x 1 ln ( 1 + x ) − 1 = x → 0 lim x e x 1 l n ( 1 + x ) − e = e x → 0 lim x 2 ln ( 1 + x ) − x = x → 0 lim x e ( e x 1 l n ( 1 + x ) − 1 − 1 ) = e x 2 − 2 1 x 2 = − 2 1 e .
L’Hospital法则求极限
为记简便，本章节所有的「L’Hospital法则」将简写为「L’H」。
对待求极限分式的分子和分母同时求导，得到的式子极限与原式相等。这一法则是对0 0 \frac{0}{0} 0 0 ∞ ∞ \frac{\infty}{\infty} ∞ ∞
不是所有的0 0 \frac{0}{0} 0 0 ∞ ∞ \frac{\infty}{\infty} ∞ ∞
上方的警示并不是因为使用逻辑存在问题而不能使用。考虑lim x → + ∞ e x + e − x e x − e − x \lim_{x\rightarrow +\infty}\frac{e^x+e^{-x}}{e^x-e^{-x}} lim x → + ∞ e x − e − x e x + e − x lim x → + ∞ x + sin x x − cos x \lim_{x\rightarrow +\infty}\frac{x+\sin x}{x-\cos x} lim x → + ∞ x − c o s x x + s i n x ∞ ∞ \frac{\infty}{\infty} ∞ ∞ ( e x + e − x ) ′ ( e x − e − x ) ′ = e x − e − x e x + e − x , \frac{(e^x+e^{-x})'}{(e^x-e^{-x})'}=\frac{e^x-e^{-x}}{e^x+e^{-x}}, ( e x − e − x ) ′ ( e x + e − x ) ′ = e x + e − x e x − e − x , 1 + cos x 1 + sin x → L ′ H − sin x cos x → L ′ H − cos x − sin x → L ′ H sin x − cos x → L ′ H cos x sin x → L ′ H − sin x cos x → L ′ H … \frac{1+\cos x}{1+\sin x}\xrightarrow{L'H}\frac{-\sin x}{\cos x}\xrightarrow{L'H}\frac{-\cos x}{-\sin x}\xrightarrow{L'H}\frac{\sin x}{-\cos x}\xrightarrow{L'H}\frac{\cos x}{\sin x}\xrightarrow{L'H}\frac{-\sin x}{\cos x}\xrightarrow{L'H}\dots 1 + s i n x 1 + c o s x L ′ H c o s x − s i n x L ′ H − s i n x − c o s x L ′ H − c o s x s i n x L ′ H s i n x c o s x L ′ H c o s x − s i n x L ′ H … 使用L’H之后并不能简化运算 ，因此才不能使用L’H。
事实上，使用“抓大 ”的方法轻而易举就能得到lim x → ∞ x + sin x x − cos x = 1. \lim_{x\rightarrow\infty}\frac{x+\sin x}{x-\cos x}=1. lim x → ∞ x − c o s x x + s i n x = 1. x → 0 x\rightarrow 0 x → 0 sin 1 x \sin\frac{1}{x} sin x 1 cos 1 x \cos\frac{1}{x} cos x 1
f ′ ( x ) g ′ ( x ) = a \frac{f'(x)}{g'(x)}=a g ′ ( x ) f ′ ( x ) = a f ( x ) g ( x ) = a \frac{f(x)}{g(x)}=a g ( x ) f ( x ) = a
依然考虑lim x → + ∞ x + sin x x − cos x . \lim_{x\rightarrow +\infty}\frac{x+\sin x}{x-\cos x}. lim x → + ∞ x − c o s x x + s i n x . 1 1 1 1 + cos x 1 + sin x \frac{1+\cos x}{1+\sin x} 1 + s i n x 1 + c o s x f ′ ( x ) g ′ ( x ) = a \frac{f'(x)}{g'(x)}=a g ′ ( x ) f ′ ( x ) = a f ( x ) g ( x ) = a \frac{f(x)}{g(x)}=a g ( x ) f ( x ) = a f ( x ) g ( x ) = a \frac{f(x)}{g(x)}=a g ( x ) f ( x ) = a f ′ ( x ) g ′ ( x ) = a ; f ′ ( x ) g ′ ( x ) \frac{f'(x)}{g'(x)}=a;\frac{f'(x)}{g'(x)} g ′ ( x ) f ′ ( x ) = a ; g ′ ( x ) f ′ ( x ) f ( x ) g ( x ) \frac{f(x)}{g(x)} g ( x ) f ( x )
当一次L’H后的极限不存在时需要使用其他方法求极限。
若仅已知f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) f ( x ) g ( x ) \frac{f(x)}{g(x)} g ( x ) f ( x )
使用L’H的前提是f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) 使用导数的定义求极限 。
一般求解极限时需要L’H与等价无穷小代换等其他方法结合使用，但仍需要注意这些非L’H方法的使用条件。
e.g.1.14 \textbf{e.g.1.14 } e.g.1.14 lim x → + ∞ ∫ 1 x ( t 2 ( e 1 t − 1 ) − t ) d t x 2 ln ( 1 + 1 x ) . \lim_{x\rightarrow+\infty}\frac{\int_1^x(t^2(e^{\frac{1}{t}}-1)-t)\text{d}t}{x^2\ln(1+\frac{1}{x})}. lim x → + ∞ x 2 l n ( 1 + x 1 ) ∫ 1 x ( t 2 ( e t 1 − 1 ) − t ) d t .
Solution \textbf{Solution } Solution ln ( 1 + 1 x ) ∼ 1 x , \ln(1+\frac{1}{x})\sim \frac{1}{x}, ln ( 1 + x 1 ) ∼ x 1 , x 2 ln ( 1 + 1 x ) ∼ x . x^2\ln(1+\frac{1}{x})\sim x. x 2 ln ( 1 + x 1 ) ∼ x .
O.F. = lim x → + ∞ ∫ 1 x ( t 2 ( e 1 t − 1 ) − t ) x = L ′ H lim x → + ∞ x 2 ( e 1 x − 1 ) − x 1 = t = 1 x e t − 1 − t t 2 = 1 2 t 2 t 2 = 1 2 . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow+\infty}\frac{\int_1^x(t^2(e^{\frac{1}{t}}-1)-t)}{x}&\stackrel{L'H}{=}\lim_{x\rightarrow+\infty}\frac{x^2(e^{\frac{1}{x}}-1)-x}{1}&\stackrel{t=\frac{1}{x}}{=}\frac{e^t-1-t}{t^2}=\frac{\frac{1}{2}t^2}{t^2}=\frac{1}{2}.
\end{align*}
O.F. = x → + ∞ lim x ∫ 1 x ( t 2 ( e t 1 − 1 ) − t ) = L ′ H x → + ∞ lim 1 x 2 ( e x 1 − 1 ) − x = t = x 1 t 2 e t − 1 − t = t 2 2 1 t 2 = 2 1 .
e.x.1.12 \textbf{e.x.1.12 } e.x.1.12 lim x → + ∞ ( π 2 − arctan x ) 1 ln x . \lim_{x\rightarrow+\infty}(\frac{\pi}{2}-\arctan x)^\frac{1}{\ln x}. lim x → + ∞ ( 2 π − arctan x ) l n x 1 .
Solution \textbf{Solution } Solution arcctg x + arctan x = π 2 \arcctg x+\arctan x=\frac{\pi}{2} arcctg x + arctan x = 2 π
O.F. = lim x → + ∞ e ln ( ( π 2 − arctan x ) 1 ln x ) = lim x → + ∞ e 1 ln x ln ( π 2 − arctan x ) = lim x → + ∞ e ln ( π 2 − arctan x ) ln x = lim x → + ∞ e ln arcctg x ln x = L ′ H lim x → + ∞ e 1 arcctg x ( − 1 1 + x 2 ) 1 x = lim x → + ∞ e − x ( 1 + x 2 ) arcctg x = lim x → + ∞ e − 1 x arcctg x = L ′ H lim x → + ∞ e − − 1 x 2 − 1 1 + x 2 = lim x → + ∞ e − 1 + x 2 x 2 = e − 1 . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow+\infty}e^{\ln((\frac{\pi}{2}-\arctan x)^\frac{1}{\ln x})}&=\lim_{x\rightarrow+\infty}e^{\frac{1}{\ln x}\ln(\frac{\pi}{2}-\arctan x)}&=\lim_{x\rightarrow+\infty}e^{\frac{\ln(\frac{\pi}{2}-\arctan x)}{\ln x}}\\
&=\lim_{x\rightarrow+\infty}e^\frac{\ln\arcctg x}{\ln x}&\stackrel{L'H}{=}\lim_{x\rightarrow+\infty}e^\frac{\frac{1}{\arcctg x}(-\frac{1}{1+x^2})}{\frac{1}{x}}&=\lim_{x\rightarrow+\infty}e^{-\frac{x}{(1+x^2)\arcctg x}}\\
&=\lim_{x\rightarrow+\infty}e^{-\frac{\frac{1}{x}}{\arcctg x}}&\stackrel{L'H}{=}\lim_{x\rightarrow+\infty}e^{-\frac{-\frac{1}{x^2}}{-\frac{1}{1+x^2}}}&=\lim_{x\rightarrow+\infty}e^{-\frac{1+x^2}{x^2}}&=e^{-1}.
\end{align*}
O.F. = x → + ∞ lim e l n (( 2 π − a r c t a n x ) l n x 1 ) = x → + ∞ lim e l n x l n a r c c t g x = x → + ∞ lim e − a r c c t g x x 1 = x → + ∞ lim e l n x 1 l n ( 2 π − a r c t a n x ) = L ′ H x → + ∞ lim e x 1 a r c c t g x 1 ( − 1 + x 2 1 ) = L ′ H x → + ∞ lim e − − 1 + x 2 1 − x 2 1 = x → + ∞ lim e l n x l n ( 2 π − a r c t a n x ) = x → + ∞ lim e − ( 1 + x 2 ) a r c c t g x x = x → + ∞ lim e − x 2 1 + x 2 = e − 1 .
从练习1.12可以看出，在使用L’H时未必要一次把式子处理到底；中途在L’H若干步后可以进行一些简化，之后再继续运用L’H。为了方便运算，此练习中使用的幂指函数处理反而没有将e ln ( π 2 − arctan x ) ln x e^\frac{\ln(\frac{\pi}{2}-\arctan x)}{\ln x} e l n x l n ( 2 π − a r c t a n x ) ln ( π 2 − arctan x ) ln x + 1. \frac{\ln(\frac{\pi}{2}-\arctan x)}{\ln x}+1. l n x l n ( 2 π − a r c t a n x ) + 1.
导数定义求极限
对0 0 \frac{0}{0} 0 0 f ( x ) f(x) f ( x ) x = x 0 x=x_0 x = x 0 n n n x → x 0 x\rightarrow x_0 x → x 0 n − 1 n-1 n − 1
显然对形如lim x → 0 f ( a + x ) − f ( a ) x \lim_{x\rightarrow 0}\frac{f(a+x)-f(a)}{x} lim x → 0 x f ( a + x ) − f ( a ) f ′ ( x 0 ) = lim h → 0 f ( x 0 + h ) − f ( x 0 ) h f'(x_0)=\lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h} f ′ ( x 0 ) = lim h → 0 h f ( x 0 + h ) − f ( x 0 )
e.g.1.15 \textbf{e.g.1.15 } e.g.1.15 f ( x ) f(x) f ( x ) x 0 x_0 x 0 f ′ ( x 0 ) , f'(x_0), f ′ ( x 0 ) , lim x → x 0 x f ( x 0 ) − x 0 f ( x ) x − x 0 . \lim_{x\rightarrow x_0}\frac{xf(x_0)-x_0f(x)}{x-x_0}. lim x → x 0 x − x 0 x f ( x 0 ) − x 0 f ( x ) .
Solution \textbf{Solution } Solution
O.F. = lim x → x 0 f ( x 0 ) x 0 − f ( x ) x 1 x 0 − 1 x = lim x → x 0 f ( x ) x − f ( x 0 ) x 0 1 x − 1 x 0 , \text{O.F.}=\lim_{x\rightarrow x_0}\frac{\frac{f(x_0)}{x_0}-\frac{f(x)}{x}}{\frac{1}{x_0}-\frac{1}{x}}=\lim_{x\rightarrow x_0}\frac{\frac{f(x)}{x}-\frac{f(x_0)}{x_0}}{\frac{1}{x}-\frac{1}{x_0}},
O.F. = x → x 0 lim x 0 1 − x 1 x 0 f ( x 0 ) − x f ( x ) = x → x 0 lim x 1 − x 0 1 x f ( x ) − x 0 f ( x 0 ) ,
故记g ( x ) = x f ( 1 x ) , g(x)=xf(\frac{1}{x}), g ( x ) = x f ( x 1 ) , g ( 1 x ) = f ( x ) x , g ( 1 x 0 ) = f ( x 0 ) x 0 . g(\frac{1}{x})=\frac{f(x)}{x},g(\frac{1}{x_0})=\frac{f(x_0)}{x_0}. g ( x 1 ) = x f ( x ) , g ( x 0 1 ) = x 0 f ( x 0 ) .
O.F. = lim x → x 0 g ( 1 x ) − g ( 1 x 0 ) 1 x − 1 x 0 = g ′ ( 1 x 0 ) . \text{O.F.}=\lim_{x\rightarrow x_0}\frac{g(\frac{1}{x})-g(\frac{1}{x_0})}{\frac{1}{x}-\frac{1}{x_0}}=g'(\frac{1}{x_0}).
O.F. = x → x 0 lim x 1 − x 0 1 g ( x 1 ) − g ( x 0 1 ) = g ′ ( x 0 1 ) .
换言之只要求g ( x ) = x f ( 1 x ) g(x)=xf(\frac{1}{x}) g ( x ) = x f ( x 1 ) x = 1 x 0 x=\frac{1}{x_0} x = x 0 1
g ′ ( x ) = f ( 1 x ) + x f ′ ( 1 x ) ⋅ ( 1 x ) ′ = f ( 1 x ) + x f ′ ( 1 x ) ⋅ ( − 1 x 2 ) = f ( 1 x ) − f ′ ( 1 x ) x , g'(x)=f(\frac{1}{x})+xf'(\frac{1}{x})\cdot(\frac{1}{x})'=f(\frac{1}{x})+xf'(\frac{1}{x})\cdot(-\frac{1}{x^2})=f(\frac{1}{x})-\frac{f'(\frac{1}{x})}{x},
g ′ ( x ) = f ( x 1 ) + x f ′ ( x 1 ) ⋅ ( x 1 ) ′ = f ( x 1 ) + x f ′ ( x 1 ) ⋅ ( − x 2 1 ) = f ( x 1 ) − x f ′ ( x 1 ) ,
故
O.F. = f ( x 0 ) − f ′ ( x 0 ) x 0 . \text{O.F.}=f(x_0)-\frac{f'(x_0)}{x_0}.
O.F. = f ( x 0 ) − x 0 f ′ ( x 0 ) .
e.g.1.16 \textbf{e.g.1.16 } e.g.1.16 f ( x ) f(x) f ( x ) x 0 x_0 x 0 f ′ ′ ( x 0 ) , f''(x_0), f ′′ ( x 0 ) , lim h → 0 f ( x 0 + h ) + f ( x 0 − h ) − 2 f ( x 0 ) h 2 . \lim_{h\rightarrow 0}\frac{f(x_0+h)+f(x_0-h)-2f(x_0)}{h^2}. lim h → 0 h 2 f ( x 0 + h ) + f ( x 0 − h ) − 2 f ( x 0 ) .
Solution \textbf{Solution } Solution h h h
O.F. = L ′ H lim h → 0 f ′ ( x 0 + h ) − f ′ ( x 0 − h ) 2 h = lim h → 0 f ′ ( x 0 + h ) − f ′ ( x 0 − h ) ( x 0 + h ) − ( x 0 − h ) = lim h → 0 ( f ′ ( x 0 + h ) − f ′ ( x 0 ) ) + ( f ′ ( x 0 ) − f ′ ( x 0 − h ) ) 2 h = lim h → 0 f ′ ( x 0 + h ) − f ′ ( x 0 ) 2 h + lim h → 0 f ′ ( x 0 ) − f ′ ( x 0 − h ) 2 h . = 1 2 f ′ ′ ( x 0 ) + 1 2 f ′ ′ ( x 0 ) = f ′ ′ ( x 0 ) . \begin{align*}
\text{O.F.}&\stackrel{L'H}{=}\lim_{h\rightarrow 0}\frac{f'(x_0+h)-f'(x_0-h)}{2h}=\lim_{h\rightarrow 0}\frac{f'(x_0+h)-f'(x_0-h)}{(x_0+h)-(x_0-h)}\\
&=\lim_{h\rightarrow 0}\frac{(f'(x_0+h)-f'(x_0))+(f'(x_0)-f'(x_0-h))}{2h}\\
&=\lim_{h\rightarrow 0}\frac{f'(x_0+h)-f'(x_0)}{2h}+\lim_{h\rightarrow 0}\frac{f'(x_0)-f'(x_0-h)}{2h}.\\
&=\frac{1}{2}f''(x_0)+\frac{1}{2}f''(x_0)&=f''(x_0).
\end{align*}
O.F. = L ′ H h → 0 lim 2 h f ′ ( x 0 + h ) − f ′ ( x 0 − h ) = h → 0 lim ( x 0 + h ) − ( x 0 − h ) f ′ ( x 0 + h ) − f ′ ( x 0 − h ) = h → 0 lim 2 h ( f ′ ( x 0 + h ) − f ′ ( x 0 )) + ( f ′ ( x 0 ) − f ′ ( x 0 − h )) = h → 0 lim 2 h f ′ ( x 0 + h ) − f ′ ( x 0 ) + h → 0 lim 2 h f ′ ( x 0 ) − f ′ ( x 0 − h ) . = 2 1 f ′′ ( x 0 ) + 2 1 f ′′ ( x 0 ) = f ′′ ( x 0 ) .
注意在例1.16中只能使用一次L’Hospital法则。这是因为，“f ( x ) f(x) f ( x ) f ′ ′ ( x ) f''(x) f ′′ ( x ) f ′ ( x ) f'(x) f ′ ( x ) f ′ ′ ( x ) f''(x) f ′′ ( x )
特别地，若导数定义中的h h h 1 n ( n ∈ N + ) , \frac{1}{n}(n\in\mathbb{N}_+), n 1 ( n ∈ N + ) ,
f ′ ( x 0 ) = lim n → ∞ f ( x 0 + 1 n ) − f ( x 0 ) 1 n = lim n → ∞ ( n ( f ( x 0 + 1 n ) − f ( x 0 ) ) ) , f'(x_0)=\lim_{n\rightarrow\infty}\frac{f(x_0+\frac{1}{n})-f(x_0)}{\frac{1}{n}}=\lim_{n\rightarrow\infty}(n(f(x_0+\frac{1}{n})-f(x_0))),
f ′ ( x 0 ) = n → ∞ lim n 1 f ( x 0 + n 1 ) − f ( x 0 ) = n → ∞ lim ( n ( f ( x 0 + n 1 ) − f ( x 0 ))) ,
而更特殊地，若x 0 = 0 , f ( x 0 ) = a , x_0=0, f(x_0)=a, x 0 = 0 , f ( x 0 ) = a , f ′ ( 0 ) = lim n → ∞ ( n ( f ( 1 n ) − a ) ) . f'(0)=\lim_{n\rightarrow\infty}(n(f(\frac{1}{n})-a)). f ′ ( 0 ) = lim n → ∞ ( n ( f ( n 1 ) − a )) .
e.x.1.13 \textbf{e.x.1.13 } e.x.1.13 y = f ( x ) y=f(x) y = f ( x ) y = ∫ 0 arctan x e − t 2 d t y=\int_0^{\arctan x}e^{-t^2}\text{d}t y = ∫ 0 a r c t a n x e − t 2 d t ( 0 , 0 ) (0,0) ( 0 , 0 ) lim n → ∞ n f ( 2 n ) . \lim_{n\rightarrow\infty}nf(\frac{2}{n}). lim n → ∞ n f ( n 2 ) .
Solution \textbf{Solution } Solution
f ( 0 ) = 0 , f(0)=0,
f ( 0 ) = 0 ,
f ′ ( 0 ) = ( ∫ 0 arctan x e − t 2 d t ) ′ ∣ x = 0 = e − arctan 2 x ⋅ 1 1 + x 2 ∣ x = 0 = 1 , f'(0)=(\int_0^{\arctan x}e^{-t^2}\text{d}t)'\bigg|_{x=0}=e^{-\arctan^2x}\cdot\frac{1}{1+x^2}\bigg|_{x=0}=1,
f ′ ( 0 ) = ( ∫ 0 a r c t a n x e − t 2 d t ) ′ x = 0 = e − a r c t a n 2 x ⋅ 1 + x 2 1 x = 0 = 1 ,
因此
O.F. = lim n → ∞ f ( 2 n ) 2 n × 2 = 2 lim n → ∞ f ( 0 + 2 n ) − f ( 0 ) 2 n = 2 f ′ ( 0 ) = 2. \begin{align*}
\text{O.F.}=\lim_{n\rightarrow\infty}\frac{f(\frac{2}{n})}{\frac{2}{n}}\times2=2\lim_{n\rightarrow\infty}\frac{f(0+\frac{2}{n})-f(0)}{\frac{2}{n}}=2f'(0)=2.
\end{align*}
O.F. = n → ∞ lim n 2 f ( n 2 ) × 2 = 2 n → ∞ lim n 2 f ( 0 + n 2 ) − f ( 0 ) = 2 f ′ ( 0 ) = 2.
Lagrange中值定理求极限
Lagrange中值定理告诉我们，若f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) ε ∈ ( a , b ) \varepsilon\in(a,b) ε ∈ ( a , b ) f ′ ( ε ) = f ( b ) − f ( a ) b − a f'(\varepsilon)=\frac{f(b)-f(a)}{b-a} f ′ ( ε ) = b − a f ( b ) − f ( a ) f ( b ) − f ( a ) = f ′ ( ε ) ( b − a ) f(b)-f(a)=f'(\varepsilon)(b-a) f ( b ) − f ( a ) = f ′ ( ε ) ( b − a ) ε \varepsilon ε
e.g.1.17 \textbf{e.g.1.17 } e.g.1.17 lim n → ∞ ( n 2 ( arctan a n − arctan a n + 1 ) ) . \lim_{n\rightarrow\infty}(n^2(\arctan\frac{a}{n}-\arctan\frac{a}{n+1})). lim n → ∞ ( n 2 ( arctan n a − arctan n + 1 a )) .
Solution \textbf{Solution } Solution arctan x \arctan x arctan x f ( x ) = arctan x f(x)=\arctan x f ( x ) = arctan x f ( x ) f(x) f ( x ) [ a n + 1 , a n ] [\frac{a}{n+1},\frac{a}{n}] [ n + 1 a , n a ] ( a n + 1 , a n ) (\frac{a}{n+1},\frac{a}{n}) ( n + 1 a , n a ) ε ∈ ( a n + 1 , a n ) , \varepsilon\in(\frac{a}{n+1},\frac{a}{n}), ε ∈ ( n + 1 a , n a ) ,
arctan a n − arctan a n + 1 = Lagrange 1 1 + ε 2 ( a n − a n + 1 ) \arctan\frac{a}{n}-\arctan\frac{a}{n+1}\xlongequal{\text{Lagrange}}\frac{1}{1+\varepsilon^2}(\frac{a}{n}-\frac{a}{n+1})
arctan n a − arctan n + 1 a Lagrange 1 + ε 2 1 ( n a − n + 1 a )
成立。故
O.F. = lim n → ∞ ( n 2 1 1 + ε 2 ( a n − a n + 1 ) ) = lim n → ∞ ( 1 1 + ε 2 ⋅ n 2 a n ( n + 1 ) ) = lim n → ∞ a 1 + ε 2 . \text{O.F.}=\lim_{n\rightarrow\infty}(n^2\frac{1}{1+\varepsilon^2}(\frac{a}{n}-\frac{a}{n+1}))=\lim_{n\rightarrow\infty}(\frac{1}{1+\varepsilon^2}\cdot\frac{n^2a}{n(n+1)})=\lim_{n\rightarrow\infty}\frac{a}{1+\varepsilon^2}.
O.F. = n → ∞ lim ( n 2 1 + ε 2 1 ( n a − n + 1 a )) = n → ∞ lim ( 1 + ε 2 1 ⋅ n ( n + 1 ) n 2 a ) = n → ∞ lim 1 + ε 2 a .
现在考虑ε \varepsilon ε n → ∞ n\rightarrow\infty n → ∞ a n + 1 → 0 , a n → 0 , \frac{a}{n+1}\rightarrow0, \frac{a}{n}\rightarrow0, n + 1 a → 0 , n a → 0 , ε \varepsilon ε ε → 0. \varepsilon\rightarrow0. ε → 0.
O.F. = lim n → ∞ a 1 + ε 2 = a 1 + 0 2 = a . \text{O.F.}=\lim_{n\rightarrow\infty}\frac{a}{1+\varepsilon^2}=\frac{a}{1+0^2}=a.
O.F. = n → ∞ lim 1 + ε 2 a = 1 + 0 2 a = a .
在例1.17中选择的f ( x ) f(x) f ( x ) f ( x ) = arctan a x f(x)=\arctan\frac{a}{x} f ( x ) = arctan x a
e.x.1.14 \textbf{e.x.1.14 } e.x.1.14 lim x → + ∞ ( x 2 ( ln arctan ( x + 1 ) − ln arctan x ) ) . \lim_{x\rightarrow+\infty}(x^2(\ln\arctan(x+1)-\ln\arctan x)). lim x → + ∞ ( x 2 ( ln arctan ( x + 1 ) − ln arctan x )) .
Solution \textbf{Solution } Solution f ( x ) = ln arctan x f(x)=\ln\arctan x f ( x ) = ln arctan x f ( x ) f(x) f ( x ) [ x , x + 1 ] [x,x+1] [ x , x + 1 ] ( x , x + 1 ) (x,x+1) ( x , x + 1 ) ε ∈ ( x , x + 1 ) , \varepsilon\in(x,x+1), ε ∈ ( x , x + 1 ) ,
ln arctan ( x + 1 ) − ln arctan x = Lagrange ( 1 arctan ε ⋅ 1 1 + ε 2 ) ( ( x + 1 ) − x ) = 1 arctan ε ( 1 + ε 2 ) \ln\arctan(x+1)-\ln\arctan x\xlongequal{\text{Lagrange}}\bigg(\frac{1}{\arctan\varepsilon}\cdot\frac{1}{1+\varepsilon^2}\bigg)((x+1)-x)=\frac{1}{\arctan\varepsilon(1+\varepsilon^2)}
ln arctan ( x + 1 ) − ln arctan x Lagrange ( arctan ε 1 ⋅ 1 + ε 2 1 ) (( x + 1 ) − x ) = arctan ε ( 1 + ε 2 ) 1
成立。故
O.F. = lim x → + ∞ x 2 arctan ε ( 1 + ε 2 ) = lim x → + ∞ ( x 2 1 + ε 2 ⋅ 1 arctan ε ) . \text{O.F.}=\lim_{x\rightarrow+\infty}\frac{x^2}{\arctan\varepsilon(1+\varepsilon^2)}=\lim_{x\rightarrow+\infty}\bigg(\frac{x^2}{1+\varepsilon^2}\cdot\frac{1}{\arctan\varepsilon}\bigg).
O.F. = x → + ∞ lim arctan ε ( 1 + ε 2 ) x 2 = x → + ∞ lim ( 1 + ε 2 x 2 ⋅ arctan ε 1 ) .
现在考虑ε \varepsilon ε x → + ∞ x\rightarrow+\infty x → + ∞ ε → x . \varepsilon\rightarrow x. ε → x .
O.F. = lim x → + ∞ ( x 2 1 + x 2 1 arctan x ) = lim x → + ∞ 1 arctan x = 1 π 2 = 2 π . \text{O.F.}=\lim_{x\rightarrow+\infty}\bigg(\frac{x^2}{1+x^2}\frac{1}{\arctan x}\bigg)=\lim_{x\rightarrow+\infty}\frac{1}{\arctan x}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}.
O.F. = x → + ∞ lim ( 1 + x 2 x 2 arctan x 1 ) = x → + ∞ lim arctan x 1 = 2 π 1 = π 2 .
利用Lagrange中值定理求极限时关键在于函数的构造。一般选取的函数要么方便求导，要么方便ε \varepsilon ε
考虑下方的练习1.15。
e.x.1.15 \textbf{e.x.1.15 } e.x.1.15 lim x → + ∞ ( sin x + 1 − sin x ) . \lim_{x\rightarrow+\infty}(\sin\sqrt{x+1}-\sin\sqrt{x}). lim x → + ∞ ( sin x + 1 − sin x ) .
Solution \textbf{Solution } Solution f ( x ) = sin x f(x)=\sin\sqrt{x} f ( x ) = sin x f ( x ) f(x) f ( x ) [ x , x + 1 ] [x,x+1] [ x , x + 1 ] ( x , x + 1 ) (x,x+1) ( x , x + 1 ) ε ∈ ( x , x + 1 ) , \varepsilon\in(x,x+1), ε ∈ ( x , x + 1 ) ,
sin x + 1 − sin x = Lagrange ( cos ε ⋅ 1 2 ε ) ( ( x + 1 ) − x ) = cos ε ε \sin\sqrt{x+1}-\sin\sqrt{x}\xlongequal{\text{Lagrange}}\bigg(\cos\sqrt{\varepsilon}\cdot\frac{1}{2\sqrt{\varepsilon}}\bigg)((x+1)-x)=\frac{\cos\sqrt{\varepsilon}}{\sqrt{\varepsilon}}
sin x + 1 − sin x Lagrange ( cos ε ⋅ 2 ε 1 ) (( x + 1 ) − x ) = ε cos ε
成立。故
O.F. = lim x → + ∞ cos ε ε . \text{O.F.}=\lim_{x\rightarrow+\infty}\frac{\cos\sqrt{\varepsilon}}{\sqrt{\varepsilon}}.
O.F. = x → + ∞ lim ε cos ε .
现在考虑ε \varepsilon ε x → + ∞ x\rightarrow+\infty x → + ∞ ε → x , \varepsilon\rightarrow x, ε → x , cos ε \cos\sqrt{\varepsilon} cos ε cos ε ∈ [ − 1 , 1 ] \cos\sqrt{\varepsilon}\in[-1,1] cos ε ∈ [ − 1 , 1 ] 1 ε \frac{1}{\sqrt{\varepsilon}} ε 1
O.F. = lim x → + ∞ cos ε ε = lim x → + ∞ ( cos ε ⋅ 1 ε ) = 0. \text{O.F.}=\lim_{x\rightarrow+\infty}\frac{\cos\sqrt{\varepsilon}}{\sqrt{\varepsilon}}=\lim_{x\rightarrow+\infty}\bigg(\cos\sqrt{\varepsilon}\cdot\frac{1}{\sqrt{\varepsilon}}\bigg)=0.
O.F. = x → + ∞ lim ε cos ε = x → + ∞ lim ( cos ε ⋅ ε 1 ) = 0.
上方给出的是方便最后代换的例子；也可以取f ( x ) = sin x , f(x)=\sin x, f ( x ) = sin x , f ( x ) f(x) f ( x ) ε \varepsilon ε sin x + 1 − sin x = Lagrange cos ε ( x + 1 − x ) \sin\sqrt{x+1}-\sin\sqrt{x}\xlongequal{\text{Lagrange}}\cos\varepsilon(\sqrt{x+1}-\sqrt{x}) sin x + 1 − sin x Lagrange cos ε ( x + 1 − x ) O.F. = lim x → + ∞ ( cos ε ⋅ 1 x + 1 + x ) = 0 \text{O.F.}=\lim_{x\rightarrow+\infty}\big(\cos\varepsilon\cdot\frac{1}{\sqrt{x+1}+\sqrt{x}}\big)=0 O.F. = lim x → + ∞ ( cos ε ⋅ x + 1 + x 1 ) = 0
e.x.1.16 \textbf{e.x.1.16 } e.x.1.16 lim n → ∞ ( n ( arctan ln ( n + 1 ) − arctan ln n ) ) . \lim_{n\rightarrow\infty}(n(\arctan\ln(n+1)-\arctan\ln n)). lim n → ∞ ( n ( arctan ln ( n + 1 ) − arctan ln n )) .
Solution \textbf{Solution } Solution f ( x ) = arctan ln x f(x)=\arctan\ln x f ( x ) = arctan ln x f ( x ) f(x) f ( x ) [ n , n + 1 ] [n,n+1] [ n , n + 1 ] ( n , n + 1 ) (n,n+1) ( n , n + 1 ) ε ∈ ( n , n + 1 ) , \varepsilon\in(n,n+1), ε ∈ ( n , n + 1 ) ,
arctan ln ( x + 1 ) − arctan ln x = Lagrange ( 1 1 + ln 2 ε ⋅ 1 ε ) ( ( x + 1 ) − x ) = 1 ε ( 1 + ln 2 ε ) \arctan\ln(x+1)-\arctan\ln x\xlongequal{\text{Lagrange}}(\frac{1}{1+\ln^2\varepsilon}\cdot\frac{1}{\varepsilon})((x+1)-x)=\frac{1}{\varepsilon(1+\ln^2\varepsilon)}
arctan ln ( x + 1 ) − arctan ln x Lagrange ( 1 + ln 2 ε 1 ⋅ ε 1 ) (( x + 1 ) − x ) = ε ( 1 + ln 2 ε ) 1
成立。故
O.F. = lim n → ∞ 1 ε ( 1 + ln 2 ε ) . \text{O.F.}=\lim_{n\rightarrow\infty}\frac{1}{\varepsilon(1+\ln^2\varepsilon)}.
O.F. = n → ∞ lim ε ( 1 + ln 2 ε ) 1 .
现在考虑ε \varepsilon ε n → ∞ n\rightarrow\infty n → ∞ ε → n \varepsilon\rightarrow n ε → n
O.F. = lim n → ∞ 1 n ( 1 + ln 2 n ) = 0. \text{O.F.}=\lim_{n\rightarrow\infty}\frac{1}{n(1+\ln^2n)}=0.
O.F. = n → ∞ lim n ( 1 + ln 2 n ) 1 = 0.
不是所有含有f ( b ) − f ( a ) f(b)-f(a) f ( b ) − f ( a )
考虑lim x → 0 cos 5 x − cos x x 2 . \lim_{x\rightarrow0}\frac{\cos 5x-\cos x}{x^2}. lim x → 0 x 2 c o s 5 x − c o s x . x < ε < 5 x x<\varepsilon< 5x x < ε < 5 x 1 < ε x < 5 1<\frac{\varepsilon}{x}< 5 1 < x ε < 5 ε \varepsilon ε Taylor展开 才能求解。
对拥有f ( b ) − f ( a ) f(b)-f(a) f ( b ) − f ( a ) a ∼ b a\sim b a ∼ b a a a b b b
不是所有含有f ( b ) − f ( a ) f(b)-f(a) f ( b ) − f ( a )
考虑下方练习1.17。
e.x.1.17 \textbf{e.x.1.17 } e.x.1.17 lim x → 0 e x 2 − e 2 − 2 cos x x 4 . \lim_{x\rightarrow0}\frac{e^{x^2}-e^{2-2\cos x}}{x^4}. lim x → 0 x 4 e x 2 − e 2 − 2 c o s x .
Solution \textbf{Solution } Solution
O.F. = lim x → 0 e 2 − 2 cos x ( e x 2 − 2 + 2 cos x − 1 ) x 4 = e t − 1 ∼ t lim x → 0 e 2 − 2 cos x ( x 2 − 2 + 2 cos x ) x 4 = cos x = 1 − 1 2 x 2 + 1 4 x 4 + o ( x 4 ) Taylor lim x → 0 e 2 − 2 × 1 ( x 2 − 2 + 2 ( 1 − 1 2 x 2 + 1 4 ! x 4 ) ) x 4 = lim x → 0 2 ( 1 4 ! x 4 ) x 4 = 2 4 ! = 1 12 . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow0}\frac{e^{2-2\cos x}(e^{x^2-2+2\cos x}-1)}{x^4}\xlongequal{e^t-1\sim t}\lim_{x\rightarrow0}\frac{e^{2-2\cos x}(x^2-2+2\cos x)}{x^4}\\
&\xlongequal[\cos x=1-\frac{1}{2}x^2+\frac{1}{4}x^4+o(x^4)]{\text{Taylor}}\lim_{x\rightarrow0}\frac{e^{2-2\times1}(x^2-2+2(1-\frac{1}{2}x^2+\frac{1}{4!}x^4))}{x^4}\\
&=\lim_{x\rightarrow0}\frac{2(\frac{1}{4!}x^4)}{x^4}=\frac{2}{4!}&=\frac{1}{12}.
\end{align*}
O.F. = x → 0 lim x 4 e 2 − 2 c o s x ( e x 2 − 2 + 2 c o s x − 1 ) e t − 1 ∼ t x → 0 lim x 4 e 2 − 2 c o s x ( x 2 − 2 + 2 cos x ) Taylor c o s x = 1 − 2 1 x 2 + 4 1 x 4 + o ( x 4 ) x → 0 lim x 4 e 2 − 2 × 1 ( x 2 − 2 + 2 ( 1 − 2 1 x 2 + 4 ! 1 x 4 )) = x → 0 lim x 4 2 ( 4 ! 1 x 4 ) = 4 ! 2 = 12 1 .
显然这一案例中可以定义f ( x ) = e x f(x)=e^x f ( x ) = e x
求幂指函数极限
幂指函数主要以1 ∞ 1^\infty 1 ∞ ∞ 0 \infty^0 ∞ 0 0 0 0^0 0 0
现在考虑1 ∞ 1^\infty 1 ∞
本章节围绕1 ∞ 1^\infty 1 ∞
lim x → x 0 f ( x ) g ( x ) = e ( f ( x ) − 1 ) g ( x ) . \lim_{x\rightarrow x_0}{f(x)}^{g(x)}=e^{(f(x)-1)g(x)}.
x → x 0 lim f ( x ) g ( x ) = e ( f ( x ) − 1 ) g ( x ) .
依此结论，先后可以得到
lim x → ∞ ( a x + b a x + c ) h x + k = e ( b − c ) h a lim x → 0 ( ∑ i = 1 m a i x n ) 1 x = ∏ i = 1 m a i n \begin{align}
\lim_{x\rightarrow\infty}(\frac{ax+b}{ax+c})^{hx+k}&=&e^{\frac{(b-c)h}{a}}\tag{1}\\
\lim_{x\rightarrow0}(\frac{\sum_{i=1}^m{a_i}^x}{n})^\frac{1}{x}&=&\sqrt[n]{\prod_{i=1}^ma_i}\tag{2}
\end{align}
x → ∞ lim ( a x + c a x + b ) h x + k x → 0 lim ( n ∑ i = 1 m a i x ) x 1 = = e a ( b − c ) h n i = 1 ∏ m a i ( 1 ) ( 2 )
这两个结论。这三个结论在实际解题中可以非常常用，能够实现求极限的大提速。
e.g.1.18 \textbf{e.g.1.18 } e.g.1.18 lim x → x 0 f ( x ) = 1 , lim x → x 0 g ( x ) = ∞ , \lim_{x\rightarrow x_0}f(x)=1,\lim_{x\rightarrow x_0}g(x)=\infty, lim x → x 0 f ( x ) = 1 , lim x → x 0 g ( x ) = ∞ , lim x → x 0 f ( x ) g ( x ) = e ( f ( x ) − 1 ) g ( x ) . \lim_{x\rightarrow x_0}{f(x)}^{g(x)}=e^{(f(x)-1)g(x)}. lim x → x 0 f ( x ) g ( x ) = e ( f ( x ) − 1 ) g ( x ) .
Proof \textbf{Proof } Proof
lim x → x 0 f ( x ) g ( x ) = lim x → x 0 ( 1 + ( f ( x ) − 1 ) ) 1 f ( x ) − 1 ⋅ ( f ( x ) − 1 ) g ( x ) = lim x → x 0 ( ( 1 + ( f ( x ) − 1 ) ) 1 f ( x ) − 1 ) ( f ( x ) − 1 ) g ( x ) \lim_{x\rightarrow x_0}{f(x)}^{g(x)}=\lim_{x\rightarrow x_0}(1+(f(x)-1))^{\frac{1}{f(x)-1}\cdot(f(x)-1)g(x)}=\lim_{x\rightarrow x_0}((1+(f(x)-1))^{\frac{1}{f(x)-1}})^{(f(x)-1)g(x)}
x → x 0 lim f ( x ) g ( x ) = x → x 0 lim ( 1 + ( f ( x ) − 1 ) ) f ( x ) − 1 1 ⋅ ( f ( x ) − 1 ) g ( x ) = x → x 0 lim (( 1 + ( f ( x ) − 1 ) ) f ( x ) − 1 1 ) ( f ( x ) − 1 ) g ( x )
而我们知道lim t → 0 ( 1 + t ) 1 t = e , \lim_{t\rightarrow 0}(1+t)^\frac{1}{t}=e, lim t → 0 ( 1 + t ) t 1 = e ,
lim x → x 0 f ( x ) g ( x ) = lim x → x 0 ( ( 1 + ( f ( x ) − 1 ) ) 1 f ( x ) − 1 ) ( f ( x ) − 1 ) g ( x ) = lim x → x 0 e ( f ( x ) − 1 ) g ( x ) . \lim_{x\rightarrow x_0}{f(x)}^{g(x)}=\lim_{x\rightarrow x_0}((1+(f(x)-1))^{\frac{1}{f(x)-1}})^{(f(x)-1)g(x)}=\lim_{x\rightarrow x_0}e^{(f(x)-1)g(x)}.
x → x 0 lim f ( x ) g ( x ) = x → x 0 lim (( 1 + ( f ( x ) − 1 ) ) f ( x ) − 1 1 ) ( f ( x ) − 1 ) g ( x ) = x → x 0 lim e ( f ( x ) − 1 ) g ( x ) .
Q.E.D. \textbf{Q.E.D.} Q.E.D.
例1.18中给出了重要结论。利用这个结论可以直接求大多数1 ∞ 1^\infty 1 ∞
e.g.1.19 \textbf{e.g.1.19 } e.g.1.19 lim x → ∞ ( a x + b a x + c ) h x + k . \lim_{x\rightarrow\infty}(\frac{ax+b}{ax+c})^{hx+k}. lim x → ∞ ( a x + c a x + b ) h x + k .
Solution \textbf{Solution} Solution
O.F. = lim x → x 0 e ( a x + b a x + c − 1 ) ( h x + k ) = lim x → x 0 e ( b − c ) ( h x + k ) a x + d = e ( b − c ) h a . \text{O.F.}=\lim_{x\rightarrow x_0}e^{(\frac{ax+b}{ax+c}-1)(hx+k)}=\lim_{x\rightarrow x_0}e^{\frac{(b-c)(hx+k)}{ax+d}}=e^{\frac{(b-c)h}{a}}.
O.F. = x → x 0 lim e ( a x + c a x + b − 1 ) ( h x + k ) = x → x 0 lim e a x + d ( b − c ) ( h x + k ) = e a ( b − c ) h .
例1.19是利用1.18所给结论推出的另一个重要结论。利用此也可快速求出一部分极限。
e.x.1.18 \textbf{e.x.1.18 } e.x.1.18 lim x → ∞ ( x + 3 x + 6 ) x − 1 2 . \lim_{x\rightarrow\infty}(\frac{x+3}{x+6})^{\frac{x-1}{2}}. lim x → ∞ ( x + 6 x + 3 ) 2 x − 1 .
Solution \textbf{Solution} Solution
O.F. = e ( 3 − 6 ) 1 2 1 = e − 3 2 . \text{O.F.}=e^{\frac{(3-6)\frac{1}{2}}{1}}=e^{-\frac{3}{2}}.
O.F. = e 1 ( 3 − 6 ) 2 1 = e − 2 3 .
e.x.1.19 \textbf{e.x.1.19 } e.x.1.19 lim x → ∞ ( x 2 ( x − a ) ( x + b ) ) x . \lim_{x\rightarrow\infty}(\frac{x^2}{(x-a)(x+b)})^x. lim x → ∞ ( ( x − a ) ( x + b ) x 2 ) x .
Solution \textbf{Solution} Solution
O.F. = lim x → ∞ ( x x − a ) x ( x x + b ) x = ( e ( 0 − ( − a ) ) × 1 1 ) ( e ( 0 − b ) × 1 1 ) = e a ⋅ e − b = e a − b . \text{O.F.}=\lim_{x\rightarrow\infty}(\frac{x}{x-a})^x(\frac{x}{x+b})^x=(e^{\frac{(0-(-a))\times1}{1}})(e^{\frac{(0-b)\times1}{1}})=e^a\cdot e^{-b}=e^{a-b}.
O.F. = x → ∞ lim ( x − a x ) x ( x + b x ) x = ( e 1 ( 0 − ( − a )) × 1 ) ( e 1 ( 0 − b ) × 1 ) = e a ⋅ e − b = e a − b .
现在继续利用例1.18中的结论考虑下例。
e.g.1.20 \textbf{e.g.1.20 } e.g.1.20 lim x → 0 ( a x + b x + c x 3 ) 1 x . \lim_{x\rightarrow0}(\frac{a^x+b^x+c^x}{3})^\frac{1}{x}. lim x → 0 ( 3 a x + b x + c x ) x 1 .
Solution \textbf{Solution} Solution
O.F. = lim x → 0 e ( a x + b x + c x 3 − 1 ) 1 x = lim x → 0 e a x + b x + c x − 3 3 x = L’Hospital lim x → 0 e a x ln a + b x ln b + c x ln c 3 = e ln a + ln b + ln c 3 = e ln ( ( a b c ) 1 3 ) = a b c 3 . \text{O.F.}=\lim_{x\rightarrow0}e^{(\frac{a^x+b^x+c^x}{3}-1)\frac{1}{x}}=\lim_{x\rightarrow0}e^{\frac{a^x+b^x+c^x-3}{3x}}\xlongequal{\text{L'Hospital}}\lim_{x\rightarrow0}e^{\frac{a^x\ln a+b^x\ln b+c^x\ln c}{3}}=e^{\frac{\ln a+\ln b+\ln c}{3}}=e^{\ln((abc)^\frac{1}{3})}=\sqrt[3]{abc}.
O.F. = x → 0 lim e ( 3 a x + b x + c x − 1 ) x 1 = x → 0 lim e 3 x a x + b x + c x − 3 L’Hospital x → 0 lim e 3 a x l n a + b x l n b + c x l n c = e 3 l n a + l n b + l n c = e l n (( ab c ) 3 1 ) = 3 ab c .
与例1.20中的解法同理，可以得到lim x → 0 ( a x + b x + c x n ) 1 x = a b c n . \lim_{x\rightarrow0}(\frac{a^x+b^x+c^x}{n})^\frac{1}{x}=\sqrt[n]{abc}. lim x → 0 ( n a x + b x + c x ) x 1 = n ab c .
lim x → 0 ( ∑ i = 1 m a i x n ) 1 x = ∏ i = 1 m a i n . \lim_{x\rightarrow0}(\frac{\sum_{i=1}^m{a_i}^x}{n})^\frac{1}{x}=\sqrt[n]{\prod_{i=1}^ma_i}.
x → 0 lim ( n ∑ i = 1 m a i x ) x 1 = n i = 1 ∏ m a i .
e.x.1.20 \textbf{e.x.1.20 } e.x.1.20 lim x → 0 ( 1 + 2 x 2 ) 1 x . \lim_{x\rightarrow0}(\frac{1+2^x}{2})^\frac{1}{x}. lim x → 0 ( 2 1 + 2 x ) x 1 .
Solution \textbf{Solution} Solution
O.F. = 2 . \text{O.F.}=\sqrt{2}.
O.F. = 2 .
例1.20的结论有变形，对应的结论本质上与原结论相同，如lim x → 0 ( a 1 x + b 1 x + c 1 x 3 ) x = a b c 3 \lim_{x\rightarrow0}(\frac{a^\frac{1}{x}+b^\frac{1}{x}+c^\frac{1}{x}}{3})^x=\sqrt[3]{abc} lim x → 0 ( 3 a x 1 + b x 1 + c x 1 ) x = 3 ab c
左右极限
求极限时待求极限式中有以下函数时需要考虑左右极限。
e x e^x e x x → ∞ x\rightarrow\infty x → ∞ lim x → + ∞ e x = + ∞ \lim_{x\rightarrow+\infty}e^x=+\infty lim x → + ∞ e x = + ∞ lim x → − ∞ e x = 0 \lim_{x\rightarrow-\infty}e^x=0 lim x → − ∞ e x = 0 arctan x \arctan x arctan x arcctg x \arcctg x arcctg x x → ∞ x\rightarrow\infty x → ∞ lim x → − ∞ arctan x = − π 2 , lim x → + ∞ arctan x = π 2 ; \lim_{x\rightarrow-\infty}\arctan x=-\frac{\pi}{2},\lim_{x\rightarrow+\infty}\arctan x=\frac{\pi}{2}; lim x → − ∞ arctan x = − 2 π , lim x → + ∞ arctan x = 2 π ; lim x → − ∞ arcctg x = 0 , lim x → + ∞ arcctg x = π . \lim_{x\rightarrow-\infty}\arcctg x=0, \lim_{x\rightarrow+\infty}\arcctg x=\pi. lim x → − ∞ arcctg x = 0 , lim x → + ∞ arcctg x = π . [ x ] [x] [ x ] x → n x\rightarrow n x → n n ∈ N + n\in\mathbb{N}_+ n ∈ N + lim x → n + [ x ] = n , lim x → n − [ x ] = n − 1. \lim_{x\rightarrow n^+}[x]=n,\lim_{x\rightarrow n^-}[x]=n-1. lim x → n + [ x ] = n , lim x → n − [ x ] = n − 1. x n \sqrt[n]{x} n x n ≡ 0 ( m o d 2 ) , x → ∞ n\equiv 0\pmod{2},x\rightarrow\infty n ≡ 0 ( mod 2 ) , x → ∞ x → − ∞ x\rightarrow-\infty x → − ∞ ∣ x − a ∣ |x-a| ∣ x − a ∣ x → a x\rightarrow a x → a lim x → a − = a − x , lim x → a + = x − a . \lim_{x\rightarrow a^-}=a-x,\lim_{x\rightarrow a^+}=x-a. lim x → a − = a − x , lim x → a + = x − a . 分段函数（x x x lim x → x i − = f i ( x i ) , lim x → x i + = f i + 1 ( x i ) . \lim_{x\rightarrow{x_i}^-}=f_i(x_i),\lim_{x\rightarrow{x_i}^+}=f_{i+1}(x_i). lim x → x i − = f i ( x i ) , lim x → x i + = f i + 1 ( x i ) .
f ( x ) = { f 1 ( x ) , x < x 1 f 2 ( x ) , x 1 ≤ x < x 2 f 3 ( x ) , x 2 ≤ x < x 3 … f i ( x ) , x i − 1 ≤ x < x i … f n ( x ) , x n − 1 ≤ x < x n f n + 1 ( x ) , x ≥ x n . f(x)=
\begin{cases}
f_1(x), &x<x_1\\
f_2(x), &x_1\leq x<x_2\\
f_3(x), &x_2\leq x<x_3\\
&\dots\\
f_i(x), &x_{i-1}\leq x<x_i\\
&\dots\\
f_n(x), &x_{n-1}\leq x<x_n\\
f_{n+1}(x), &x\geq x_n
\end{cases}
.
f ( x ) = ⎩ ⎨ ⎧ f 1 ( x ) , f 2 ( x ) , f 3 ( x ) , f i ( x ) , f n ( x ) , f n + 1 ( x ) , x < x 1 x 1 ≤ x < x 2 x 2 ≤ x < x 3 … x i − 1 ≤ x < x i … x n − 1 ≤ x < x n x ≥ x n .
显然出题人不会简单地罗列上方各种左右极限不一致的情况进行考核。对于一个组合了上述函数中各种的待求极限式，因各组件的左右极限不同而可能造成整体极限类型不同，进而导致整体左右极限的求法可能相异。
e.g.1.21 \textbf{e.g.1.21 } e.g.1.21 lim x → 0 ( ln ( 1 + e 2 x ) ln ( 1 + e 1 x ) + a [ x ] ) \lim_{x\rightarrow0}\big(\frac{\ln(1+e^\frac{2}{x})}{\ln(1+e^\frac{1}{x})}+a[x]\big) lim x → 0 ( l n ( 1 + e x 1 ) l n ( 1 + e x 2 ) + a [ x ] ) a , b a,b a , b
Solution \textbf{Solution } Solution ln ( 1 + e 2 x ) \ln(1+e^\frac{2}{x}) ln ( 1 + e x 2 ) ln ( 1 + e 1 x ) \ln(1+e^\frac{1}{x}) ln ( 1 + e x 1 ) [ x ] [x] [ x ] x → 0 x\rightarrow0 x → 0
lim x → 0 a [ x ] = { − a , x = 0 − 0 , x = 0 + , \lim_{x\rightarrow0}a[x]=
\begin{cases}
-a, &x=0^-\\
0, &x=0^+
\end{cases},
x → 0 lim a [ x ] = { − a , 0 , x = 0 − x = 0 + ,
lim x → 0 ln ( 1 + e 1 x ) = { ln ( 1 + e − ∞ ) = ln 1 = 0 , x = 0 − ln ( 1 + e + ∞ ) = ln ( + ∞ ) = + ∞ , x = 0 + , \lim_{x\rightarrow0}\ln(1+e^\frac{1}{x})=
\begin{cases}
\ln(1+e^{-\infty})&=\ln 1&=0, &x=0^-\\
\ln(1+e^{+\infty})&=\ln(+\infty)&=+\infty, &x=0^+
\end{cases},
x → 0 lim ln ( 1 + e x 1 ) = { ln ( 1 + e − ∞ ) ln ( 1 + e + ∞ ) = ln 1 = ln ( + ∞ ) = 0 , = + ∞ , x = 0 − x = 0 + ,
lim x → 0 ln ( 1 + e 2 x ) = { ln ( 1 + e − ∞ ) = ln 1 = 0 , x = 0 − ln ( 1 + e + ∞ ) = ln ( + ∞ ) = + ∞ , x = 0 + , \lim_{x\rightarrow0}\ln(1+e^\frac{2}{x})=
\begin{cases}
\ln(1+e^{-\infty})&=\ln 1&=0, &x=0^-\\
\ln(1+e^{+\infty})&=\ln(+\infty)&=+\infty, &x=0^+
\end{cases},
x → 0 lim ln ( 1 + e x 2 ) = { ln ( 1 + e − ∞ ) ln ( 1 + e + ∞ ) = ln 1 = ln ( + ∞ ) = 0 , = + ∞ , x = 0 − x = 0 + ,
故lim x → 0 ln ( 1 + e 2 x ) ln ( 1 + e 1 x ) \lim_{x\rightarrow 0}\frac{\ln(1+e^\frac{2}{x})}{\ln(1+e^\frac{1}{x})} lim x → 0 l n ( 1 + e x 1 ) l n ( 1 + e x 2 ) x → 0 − x\rightarrow0^- x → 0 − 0 0 \frac{0}{0} 0 0 x → 0 + x\rightarrow0^+ x → 0 + ∞ ∞ \frac{\infty}{\infty} ∞ ∞
lim x → 0 ( ln ( 1 + e 2 x ) ln ( 1 + e 1 x ) + a [ x ] ) = { lim ( ln ( 1 + e 2 x ) ln ( 1 + e 1 x ) − a ) = ln ( 1 + t ) ∼ t lim e 2 x e 1 x − a = − a , x → 0 − lim ( ln ( 1 + e 2 x ) ln ( 1 + e 1 x ) + 0 ) = lim 2 x 1 x = 2 , x → 0 + , \lim_{x\rightarrow0}\bigg(\frac{\ln(1+e^\frac{2}{x})}{\ln(1+e^\frac{1}{x})}+a[x]\bigg)=
\begin{cases}
\lim(\frac{\ln(1+e^\frac{2}{x})}{\ln(1+e^\frac{1}{x})}-a)&\xlongequal{\ln(1+t)\sim t}\lim\frac{e^\frac{2}{x}}{e^\frac{1}{x}}-a&=-a, &x\rightarrow0^-\\
\lim(\frac{\ln(1+e^\frac{2}{x})}{\ln(1+e^\frac{1}{x})}+0)&=\lim\frac{\frac{2}{x}}{\frac{1}{x}}&=2, &x\rightarrow0^+
\end{cases},
x → 0 lim ( ln ( 1 + e x 1 ) ln ( 1 + e x 2 ) + a [ x ] ) = ⎩ ⎨ ⎧ lim ( l n ( 1 + e x 1 ) l n ( 1 + e x 2 ) − a ) lim ( l n ( 1 + e x 1 ) l n ( 1 + e x 2 ) + 0 ) l n ( 1 + t ) ∼ t lim e x 1 e x 2 − a = lim x 1 x 2 = − a , = 2 , x → 0 − x → 0 + ,
因此当且仅当− a = 2 -a=2 − a = 2 a = − 2 ; a=-2; a = − 2 ; b = 2. b=2. b = 2.
上述例子中x → 0 − x\rightarrow0^- x → 0 − x → 0 + x\rightarrow0^+ x → 0 + e 1 x e^\frac{1}{x} e x 1 e 2 x e^\frac{2}{x} e x 2 ln ( 1 + e 2 x ) ln ( 1 + e 1 x ) \frac{\ln(1+e^\frac{2}{x})}{\ln(1+e^\frac{1}{x})} l n ( 1 + e x 1 ) l n ( 1 + e x 2 )
运用
讨论函数连续性
极限也可用于讨论函数连续性。
lim x → x 0 − f ( x ) = f ( x 0 ) = lim x → x 0 + f ( x ) ⟺ lim x → x 0 f ( x ) = f ( x 0 ) ⟺ f ( x ) 在 x = x 0 处连续 . \lim_{x\rightarrow {x_0}^-}f(x)=f(x_0)=\lim_{x\rightarrow {x_0}^+}f(x)\iff\lim_{x\rightarrow x_0}f(x)=f(x_0)\iff f(x)\text{在}x=x_0\text{处连续}.
x → x 0 − lim f ( x ) = f ( x 0 ) = x → x 0 + lim f ( x ) ⟺ x → x 0 lim f ( x ) = f ( x 0 ) ⟺ f ( x ) 在 x = x 0 处连续 .
上述若lim x → x 0 − f ( x ) = f ( x 0 ) \lim_{x\rightarrow {x_0}^-}f(x)=f(x_0) lim x → x 0 − f ( x ) = f ( x 0 ) f ( x ) f(x) f ( x ) x = x 0 x=x_0 x = x 0 lim x → x 0 + f ( x ) = f ( x 0 ) \lim_{x\rightarrow {x_0}^+}f(x)=f(x_0) lim x → x 0 + f ( x ) = f ( x 0 ) f ( x ) f(x) f ( x ) x = x 0 x=x_0 x = x 0
结合函数图像考虑可去间断点和跳跃间断点这两种第一类间断点。显然，当在x = x 0 x=x_0 x = x 0 x 0 x_0 x 0 x = x 0 x=x_0 x = x 0 f ( x 0 − ) ≠ f ( x 0 + ) f({x_0}^-)≠f({x_0}^+) f ( x 0 − ) = f ( x 0 + ) x 0 x_0 x 0 f ( x 0 − ) f({x_0}^-) f ( x 0 − ) f ( x 0 + ) f({x_0}^+) f ( x 0 + ) x 0 x_0 x 0 ∞ , \infty, ∞ ,
对连续函数，间断点的来源只有使分母为0的点和使函数无意义的点。
e.g.1.22 \textbf{e.g.1.22 } e.g.1.22 f ( x ) = ln ∣ x ∣ ∣ x − 1 ∣ sin x f(x)=\frac{\ln|x|}{|x-1|}\sin x f ( x ) = ∣ x − 1∣ l n ∣ x ∣ sin x
Solution \textbf{Solution } Solution f ( x ) f(x) f ( x ) x = 1 x=1 x = 1 x = 0 x=0 x = 0 ln ∣ x ∣ \ln|x| ln ∣ x ∣
注意到
lim x → 0 x ln x = lim x → 0 ln x 1 x = L ′ H lim x → 0 1 x − 1 x 2 = lim x → 0 ( − x ) = 0 ⇒ lim x → 0 x ln ∣ x ∣ = 0 \lim_{x\rightarrow0}x\ln x=\lim_{x\rightarrow0}\frac{\ln x}{\frac{1}{x}}\xlongequal{L'H}\lim_{x\rightarrow0}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\rightarrow0}(-x)=0\Rightarrow\lim_{x\rightarrow0}x\ln|x|=0
x → 0 lim x ln x = x → 0 lim x 1 ln x L ′ H x → 0 lim − x 2 1 x 1 = x → 0 lim ( − x ) = 0 ⇒ x → 0 lim x ln ∣ x ∣ = 0
进而
lim x → 0 f ( x ) = ∣ x − 1 ∣ ∣ x = 0 = 1 lim x → 0 ln ∣ x ∣ sin x = sin x ∼ x lim x → 0 x ln ∣ x ∣ = 0 , \lim_{x\rightarrow0}f(x)\xlongequal{|x-1|\big|_{x=0}=1}\lim_{x\rightarrow0}\ln|x|\sin x\xlongequal{\sin x\sim x}\lim_{x\rightarrow0}x\ln|x|=0,
x → 0 lim f ( x ) ∣ x − 1∣ x = 0 = 1 x → 0 lim ln ∣ x ∣ sin x s i n x ∼ x x → 0 lim x ln ∣ x ∣ = 0 ,
所以x = 0 x=0 x = 0
lim x → 1 − f ( x ) = lim x → 1 − ln x 1 − x sin x = ln ( 1 + t ) ∼ t x − 1 1 − x sin 1 = − sin 1 , lim x → 1 + f ( x ) = lim x → 1 + ln x x − 1 sin x = ln ( 1 + t ) ∼ t x − 1 x − 1 sin 1 = sin 1 , \begin{align*}
\lim_{x\rightarrow1^-}f(x)&=\lim_{x\rightarrow1^-}\frac{\ln x}{1-x}\sin x\xlongequal{\ln(1+t)\sim t}\frac{x-1}{1-x}\sin 1=-\sin 1,\\
\lim_{x\rightarrow1^+}f(x)&=\lim_{x\rightarrow1^+}\frac{\ln x}{x-1}\sin x\xlongequal{\ln(1+t)\sim t}\frac{x-1}{x-1}\sin 1=\sin 1,
\end{align*}
x → 1 − lim f ( x ) x → 1 + lim f ( x ) = x → 1 − lim 1 − x ln x sin x l n ( 1 + t ) ∼ t 1 − x x − 1 sin 1 = − sin 1 , = x → 1 + lim x − 1 ln x sin x l n ( 1 + t ) ∼ t x − 1 x − 1 sin 1 = sin 1 ,
因此x = 1 x=1 x = 1
综上所述，f ( x ) f(x) f ( x )
渐近线
常见的是铅直渐近线、水平渐近线和斜渐近线三种。
若lim x → x 0 f ( x ) = ∞ , \lim_{x\rightarrow{x_0}}f(x)=\infty, lim x → x 0 f ( x ) = ∞ , x = x 0 x=x_0 x = x 0 y = f ( x ) y=f(x) y = f ( x ) lim x → − ∞ f ( x ) = b , \lim_{x\rightarrow-\infty}f(x)=b, lim x → − ∞ f ( x ) = b , y = b y=b y = b y = f ( x ) y=f(x) y = f ( x ) lim x → + ∞ f ( x ) = b , \lim_{x\rightarrow+\infty}f(x)=b, lim x → + ∞ f ( x ) = b , y = b y=b y = b y = f ( x ) y=f(x) y = f ( x ) y = b y=b y = b y = f ( x ) y=f(x) y = f ( x ) y = f ( x ) y=f(x) y = f ( x ) lim x → ∞ f ( x ) = b \lim_{x\rightarrow\infty}f(x)=b lim x → ∞ f ( x ) = b y = b y=b y = b y = f ( x ) y=f(x) y = f ( x )
现在考虑函数f ( x ) = 1 x . f(x)=\frac{1}{x}. f ( x ) = x 1 . lim x → 0 + = + ∞ , lim x → 0 − = − ∞ , lim x → + ∞ = 0 , lim x → − ∞ = 0. \lim_{x\rightarrow0^+}=+\infty,\lim_{x\rightarrow0^-}=-\infty,\lim_{x\rightarrow+\infty}=0,\lim_{x\rightarrow-\infty}=0. lim x → 0 + = + ∞ , lim x → 0 − = − ∞ , lim x → + ∞ = 0 , lim x → − ∞ = 0. lim x → 0 = ∞ , lim x → ∞ = 0 , \lim_{x\rightarrow0}=\infty,\lim_{x\rightarrow\infty}=0, lim x → 0 = ∞ , lim x → ∞ = 0 , x = 0 x=0 x = 0 y = 0. y=0. y = 0.
若f ( x ) f(x) f ( x ) m m m n n n P m ( x ) P_m(x) P m ( x ) P n ( x ) P_n(x) P n ( x ) f ( x ) = P n ( x ) P m ( x ) f(x)=\frac{P_n(x)}{P_m(x)} f ( x ) = P m ( x ) P n ( x ) n ≤ m n\leq m n ≤ m f ( x ) f(x) f ( x ) n = m + 1 n=m+1 n = m + 1 f ( x ) f(x) f ( x )
对于上方警示中的前半部分，显然根据“n ≤ m n\leq m n ≤ m n = m + 1 n=m+1 n = m + 1
求一个函数的斜渐近线可以通过下面两种方法实现。
定义法k = lim x → ∞ f ( x ) x k=\lim_{x\rightarrow\infty}\frac{f(x)}{x} k = lim x → ∞ x f ( x ) b = f ( x ) − k x b=f(x)-kx b = f ( x ) − k x y = k x + b . y=kx+b. y = k x + b .
定理法f ( x ) = k x + b + α ( x ) f(x)=kx+b+\alpha(x) f ( x ) = k x + b + α ( x ) y = k x + b ; y=kx+b; y = k x + b ; α ( x ) \alpha(x) α ( x ) x x x
上述两种办法都有使用场景。显然第二种方法是基于第一种方法的原理得到的；其一般通过Taylor展开得到α ( x ) . \alpha(x). α ( x ) . x → 0 x\rightarrow0 x → 0 x → ∞ x\rightarrow\infty x → ∞ x x x 1 x . \frac{1}{x}. x 1 .
考虑下方四个函数。
y = x + sin x y=x+\sin x y = x + sin x k = lim x → ∞ x + sin x x = lim x → ∞ 1 + sin x x . k=\lim_{x\rightarrow\infty}\frac{x+\sin x}{x}=\lim_{x\rightarrow\infty}1+\frac{\sin x}{x}. k = lim x → ∞ x x + s i n x = lim x → ∞ 1 + x s i n x . x → ∞ x\rightarrow\infty x → ∞ sin x x \frac{\sin x}{x} x s i n x y = x 2 + sin x y=x^2+\sin x y = x 2 + sin x k = lim x → ∞ x 2 + sin x x = lim x → ∞ x + sin x x , k=\lim_{x\rightarrow\infty}\frac{x^2+\sin x}{x}=\lim_{x\rightarrow\infty}x+\frac{\sin x}{x}, k = lim x → ∞ x x 2 + s i n x = lim x → ∞ x + x s i n x , x → ∞ x\rightarrow\infty x → ∞ x + sin x x x+\frac{\sin x}{x} x + x s i n x y = x + sin 1 x y=x+\sin\frac{1}{x} y = x + sin x 1 k = lim x → ∞ x + sin 1 x x = lim x → ∞ ( 1 + sin 1 x x ) = 1 + 0 = 1 , k=\lim_{x\rightarrow\infty}\frac{x+\sin\frac{1}{x}}{x}=\lim_{x\rightarrow\infty}(1+\frac{\sin\frac{1}{x}}{x})=1+0=1, k = lim x → ∞ x x + s i n x 1 = lim x → ∞ ( 1 + x s i n x 1 ) = 1 + 0 = 1 , b = y − k x = lim x → ∞ sin 1 x = 0. b=y-kx=\lim_{x\rightarrow\infty}\sin\frac{1}{x}=0. b = y − k x = lim x → ∞ sin x 1 = 0. y = x . y=x. y = x . y = x 2 + sin 1 x y=x^2+\sin\frac{1}{x} y = x 2 + sin x 1 k = lim x → ∞ x 2 + sin 1 x x = lim x → ∞ ( x + sin 1 x x ) , k=\lim_{x\rightarrow\infty}\frac{x^2+\sin\frac{1}{x}}{x}=\lim_{x\rightarrow\infty}(x+\frac{\sin\frac{1}{x}}{x}), k = lim x → ∞ x x 2 + s i n x 1 = lim x → ∞ ( x + x s i n x 1 ) , x → ∞ x\rightarrow\infty x → ∞ x + sin 1 x x x+\frac{\sin\frac{1}{x}}{x} x + x s i n x 1
e.g.1.23 \textbf{e.g.1.23 } e.g.1.23 C C C
{ x = t 2 t − 1 y = t t 2 − 1 \begin{cases}
x&=\frac{t^2}{t-1}\\
y&=\frac{t}{t^2-1}
\end{cases}
{ x y = t − 1 t 2 = t 2 − 1 t
Solution \textbf{Solution } Solution t t t x → ∞ . x\rightarrow\infty. x → ∞. x = t 2 t − 1 = ( t + 1 ) + 1 t − 1 , x=\frac{t^2}{t-1}=(t+1)+\frac{1}{t-1}, x = t − 1 t 2 = ( t + 1 ) + t − 1 1 , t → 1 t\rightarrow1 t → 1 x → ∞ . x\rightarrow\infty. x → ∞.
求渐近线斜率，得
k = lim x → ∞ y x = lim t → 1 t t 2 − 1 t 2 t − 1 = lim t → 1 1 t ( t + 1 ) = 1 2 . k=\lim_{x\rightarrow\infty}\frac{y}{x}=\lim_{t\rightarrow1}\frac{\frac{t}{t^2-1}}{\frac{t^2}{t-1}}=\lim_{t\rightarrow1}\frac{1}{t(t+1)}=\frac{1}{2}.
k = x → ∞ lim x y = t → 1 lim t − 1 t 2 t 2 − 1 t = t → 1 lim t ( t + 1 ) 1 = 2 1 .
因此渐近线纵截距
b = lim x → ∞ ( y − k x ) = lim t → 1 ( t t 2 − 1 − t 2 2 ( t − 1 ) ) = lim t → 1 2 t − t 2 ( t + 1 ) 2 ( t 2 − 1 ) = lim t → 1 2 t − t 3 − t 2 2 t 2 − 2 = L ′ H lim t → 1 2 − 3 t 2 − 2 t 4 t = 2 − 3 × 1 2 − 2 × 1 4 × 1 = − 3 4 . \begin{align*}
b&=\lim_{x\rightarrow\infty}(y-kx)=\lim_{t\rightarrow1}\big(\frac{t}{t^2-1}-\frac{t^2}{2(t-1)}\big)=\lim_{t\rightarrow1}\frac{2t-t^2(t+1)}{2(t^2-1)}\\
&=\lim_{t\rightarrow1}\frac{2t-t^3-t^2}{2t^2-2}\xlongequal{L'H}\lim_{t\rightarrow1}\frac{2-3t^2-2t}{4t}=\frac{2-3\times 1^2-2\times 1}{4\times 1}&=-\frac{3}{4}.
\end{align*}
b = x → ∞ lim ( y − k x ) = t → 1 lim ( t 2 − 1 t − 2 ( t − 1 ) t 2 ) = t → 1 lim 2 ( t 2 − 1 ) 2 t − t 2 ( t + 1 ) = t → 1 lim 2 t 2 − 2 2 t − t 3 − t 2 L ′ H t → 1 lim 4 t 2 − 3 t 2 − 2 t = 4 × 1 2 − 3 × 1 2 − 2 × 1 = − 4 3 .
故所求渐近线为y = 1 2 x − 3 4 . y=\frac{1}{2}x-\frac{3}{4}. y = 2 1 x − 4 3 .
求一个函数的斜渐近线除了上述根据定义「若lim x → ∞ ( f ( x ) − ( k x + b ) ) = 0 , \lim_{x\rightarrow\infty}\big(f(x)-(kx+b)\big)=0, lim x → ∞ ( f ( x ) − ( k x + b ) ) = 0 , y = k x + b y=kx+b y = k x + b f ( x ) f(x) f ( x )
不妨考虑一个满足lim x → x 0 f ( x ) = A \lim_{x\rightarrow x_0}f(x)=A lim x → x 0 f ( x ) = A f ( x ) , f(x), f ( x ) , f ( x ) = A + α ( x ) , f(x)=A+\alpha(x), f ( x ) = A + α ( x ) , α ( x ) \alpha(x) α ( x ) x x x f ( x ) − ( k x + b ) = 0 + α ( x ) , f(x)-(kx+b)=0+\alpha(x), f ( x ) − ( k x + b ) = 0 + α ( x ) ,
f ( x ) = k x + b + α ( x ) . f(x)=kx+b+\alpha(x).
f ( x ) = k x + b + α ( x ) .
换言之，只要能够将待求渐近线函数展开为形如“k x + b + 无穷小 kx+b+\text{无穷小} k x + b + 无穷小 y = k x + b y=kx+b y = k x + b
e.x.1.21 \textbf{e.x.1.21 } e.x.1.21 y = x 2 + x ( x > 0 ) y=\sqrt{x^2+x}(x>0) y = x 2 + x ( x > 0 )
Solution \textbf{Solution } Solution
y = x 2 + x = x ( 1 + 1 x ) 1 2 = 1 x → 0 Taylor x ( 1 + 1 2 ⋅ 1 x + 1 2 ( 1 2 − 1 ) 2 ( 1 x ) 2 + o ( ( 1 x ) 2 ) ) = x + 1 2 + α ( x ) , y=\sqrt{x^2+x}=x\bigg(1+\frac{1}{x}\bigg)^\frac{1}{2}\xlongequal[\frac{1}{x}\rightarrow0]{\text{Taylor}}x\bigg(1+\frac{1}{2}\cdot\frac{1}{x}+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2}(\frac{1}{x})^2+o\big((\frac{1}{x})^2\big)\bigg)=x+\frac{1}{2}+\alpha(x),
y = x 2 + x = x ( 1 + x 1 ) 2 1 Taylor x 1 → 0 x ( 1 + 2 1 ⋅ x 1 + 2 2 1 ( 2 1 − 1 ) ( x 1 ) 2 + o ( ( x 1 ) 2 ) ) = x + 2 1 + α ( x ) ,
因此斜渐近线为y = x + 1 2 . y=x+\frac{1}{2}. y = x + 2 1 .
e.x.1.22 \textbf{e.x.1.22 } e.x.1.22 y = ( 1 + x ) 3 2 x y=\frac{(1+x)^\frac{3}{2}}{\sqrt{x}} y = x ( 1 + x ) 2 3
Solution \textbf{Solution } Solution
y = ( 1 + x ) 3 2 x = x 3 2 x ( 1 x + 1 ) 3 2 = 1 x → 0 Taylor x ( 1 + 3 2 ⋅ 1 x + 3 2 ( 3 2 − 1 ) x ( 1 x ) 2 + o ( ( 1 x ) 2 ) ) = x + 3 2 + α ( x ) y=\frac{(1+x)^\frac{3}{2}}{\sqrt{x}}=\frac{x^{\frac{3}{2}}}{\sqrt{x}}\bigg(\frac{1}{x}+1\bigg)^\frac{3}{2}\xlongequal[\frac{1}{x}\rightarrow0]{\text{Taylor}}x\bigg(1+\frac{3}{2}\cdot\frac{1}{x}+\frac{\frac{3}{2}(\frac{3}{2}-1)}{x}\big(\frac{1}{x}\big)^2+o\big((\frac{1}{x})^2\big)\bigg)=x+\frac{3}{2}+\alpha(x)
y = x ( 1 + x ) 2 3 = x x 2 3 ( x 1 + 1 ) 2 3 Taylor x 1 → 0 x ( 1 + 2 3 ⋅ x 1 + x 2 3 ( 2 3 − 1 ) ( x 1 ) 2 + o ( ( x 1 ) 2 ) ) = x + 2 3 + α ( x )
因此斜渐近线为y = x + 3 2 . y=x+\frac{3}{2}. y = x + 2 3 .
用这种基于「去极限号定理」的渐近线求法回看本章节最初考虑的四个函数。
再次考虑下方四个函数。
y = x + sin x y=x+\sin x y = x + sin x k = 1 , b = 0 k=1,b=0 k = 1 , b = 0 k x + b kx+b k x + b sin x \sin x sin x x → ∞ x\rightarrow\infty x → ∞ x x x y = x 2 + sin x y=x^2+\sin x y = x 2 + sin x x 2 x^2 x 2 k x + b kx+b k x + b y = x + sin 1 x y=x+\sin\frac{1}{x} y = x + sin x 1 1 x → 0 \frac{1}{x}\rightarrow0 x 1 → 0 y = x + ( 1 x − 1 6 ( 1 x ) 3 + o ( ( 1 x ) 3 ) ) = x + α ( x ) , y=x+\bigg(\frac{1}{x}-\frac{1}{6}\big(\frac{1}{x}\big)^3+o\big((\frac{1}{x})^3\big)\bigg)=x+\alpha(x), y = x + ( x 1 − 6 1 ( x 1 ) 3 + o ( ( x 1 ) 3 ) ) = x + α ( x ) , y = x . y=x. y = x . y = x 2 + sin 1 x y=x^2+\sin\frac{1}{x} y = x 2 + sin x 1 y = x 2 + sin x y=x^2+\sin x y = x 2 + sin x
夹逼定理与定积分定义结合求数列极限
对三个函数f ( x ) , g ( x ) , h ( x ) , f(x),g(x),h(x), f ( x ) , g ( x ) , h ( x ) , g ( x ) ≤ f ( x ) ≤ h ( x ) g(x)\leq f(x)\leq h(x) g ( x ) ≤ f ( x ) ≤ h ( x ) lim x → x 0 g ( x ) = lim x → x 0 h ( x ) = A , \lim_{x\rightarrow x_0}g(x)=\lim_{x\rightarrow x_0}h(x)=A, lim x → x 0 g ( x ) = lim x → x 0 h ( x ) = A , lim x → x 0 f ( x ) = A . \lim_{x\rightarrow x_0}f(x)=A. lim x → x 0 f ( x ) = A .
夹逼准则多数情况下和定积分同时出现。现在考虑定积分
∫ a b f ( x ) d x = lim n → ∞ ∑ i = 1 n ( f ( a + ( b − a ) i n ) ⋅ b − a n ) . \int_a^bf(x)\text{d}x=\lim_{n\rightarrow\infty}\sum_{i=1}^n\Bigg(f\bigg(a+\frac{(b-a)i}{n}\bigg)\cdot\frac{b-a}{n}\Bigg).
∫ a b f ( x ) d x = n → ∞ lim i = 1 ∑ n ( f ( a + n ( b − a ) i ) ⋅ n b − a ) .
对上述定积分，若a = 0 , b = 1 , a=0,b=1, a = 0 , b = 1 , f ( x ) f(x) f ( x ) 1 n \frac{1}{n} n 1 f ( i − 1 n ) f\big(\frac{i-1}{n}\big) f ( n i − 1 ) f ( i n ) f\big(\frac{i}{n}\big) f ( n i ) f ( 2 i − 1 2 n ) f\big(\frac{2i-1}{2n}\big) f ( 2 n 2 i − 1 )
1 ∘ 取左端点： ∫ 0 1 f ( x ) d x = lim n → ∞ 1 n ∑ i = 0 n − 1 f ( i n ) 2 ∘ 取右端点： ∫ 0 1 f ( x ) d x = lim n → ∞ 1 n ∑ i = 1 n f ( i n ) 3 ∘ 取中点： ∫ 0 1 f ( x ) d x = lim n → ∞ 1 n ∑ i = 1 n f ( 2 i − 1 2 n ) \begin{align*}
1^\circ&\text{取左端点：}&\int^1_0f(x)\text{d}x&=&\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=0}^{n-1}f\bigg(\frac{i}{n}\bigg)\\
2^\circ&\text{取右端点：}&\int^1_0f(x)\text{d}x&=&\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}f\bigg(\frac{i}{n}\bigg)\\
3^\circ&\text{取中点：}&\int^1_0f(x)\text{d}x&=&\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}f\bigg(\frac{2i-1}{2n}\bigg)
\end{align*}
1 ∘ 2 ∘ 3 ∘ 取左端点： 取右端点： 取中点： ∫ 0 1 f ( x ) d x ∫ 0 1 f ( x ) d x ∫ 0 1 f ( x ) d x = = = n → ∞ lim n 1 i = 0 ∑ n − 1 f ( n i ) n → ∞ lim n 1 i = 1 ∑ n f ( n i ) n → ∞ lim n 1 i = 1 ∑ n f ( 2 n 2 i − 1 )
虽然上方三种取点方法对应的式子不一致，但是本质上它们的结果应当相同。上方的1 n \frac{1}{n} n 1
现在考虑利用这种定积分方法求数列极限。
e.g.1.24 \textbf{e.g.1.24 } e.g.1.24 lim n → ∞ ∑ k = 1 n k n 2 ln ( 1 + k n ) . \lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{k}{n^2}\ln\big(1+\frac{k}{n}). lim n → ∞ ∑ k = 1 n n 2 k ln ( 1 + n k ) .
Solution \textbf{Solution } Solution
O.F. = lim n → ∞ 1 n ∑ k = 1 n k n ln ( 1 + k n ) = ∫ 0 1 x ln ( 1 + x ) d x = ( x 2 ln ( 1 + x ) 2 − ln ( 1 + x ) 2 − x 2 4 + x 2 ) ∣ 0 1 = 1 4 . \text{O.F.}=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\frac{k}{n}\ln(1+\frac{k}{n})=\int_0^1x\ln(1+x)\text{d}x=\bigg(\frac{x^2\ln(1+x)}{2}-\frac{\ln(1+x)}{2}-\frac{x^2}{4}+\frac{x}{2}\bigg)\bigg|_0^1=\frac{1}{4}.
O.F. = n → ∞ lim n 1 k = 1 ∑ n n k ln ( 1 + n k ) = ∫ 0 1 x ln ( 1 + x ) d x = ( 2 x 2 ln ( 1 + x ) − 2 ln ( 1 + x ) − 4 x 2 + 2 x ) 0 1 = 4 1 .
e.x.1.23 \textbf{e.x.1.23 } e.x.1.23 lim n → ∞ n ( 1 1 + n 2 + 1 2 2 + n 2 + ⋯ + 1 n 2 + n 2 ) . \lim_{n\rightarrow\infty}n\big(\frac{1}{1+n^2}+\frac{1}{2^2+n^2}+\dots+\frac{1}{n^2+n^2}\big). lim n → ∞ n ( 1 + n 2 1 + 2 2 + n 2 1 + ⋯ + n 2 + n 2 1 ) .
Solution \textbf{Solution } Solution
O.F. = lim n → ∞ n ∑ k = 1 n 1 k 2 + n 2 = lim n → ∞ 1 n ∑ k = 1 n n 2 k 2 + n 2 = lim n → ∞ 1 n ∑ k = 1 n 1 ( k n ) 2 + 1 = ∫ 0 1 1 x 2 + 1 d x = arctan x ∣ 0 1 = π 4 . \text{O.F.}=\lim_{n\rightarrow\infty}n\sum_{k=1}^n\frac{1}{k^2+n^2}=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\frac{n^2}{k^2+n^2}=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{\big(\frac{k}{n})^2+1}=\int_0^1\frac{1}{x^2+1}\text{d}x=\arctan x\big|_0^1=\frac{\pi}{4}.
O.F. = n → ∞ lim n k = 1 ∑ n k 2 + n 2 1 = n → ∞ lim n 1 k = 1 ∑ n k 2 + n 2 n 2 = n → ∞ lim n 1 k = 1 ∑ n ( n k ) 2 + 1 1 = ∫ 0 1 x 2 + 1 1 d x = arctan x 0 1 = 4 π .
现在考虑夹逼定理与定积分的结合。一部分数列极限很可能不是规则的定积分定义形式，因此需要选取合适的定积分定义形式函数作为这个极限的两端进行夹逼。在这种情况下，需要先用夹逼定理再进行定积分运算。
e.g.1.25 \textbf{e.g.1.25 } e.g.1.25 lim n → ∞ ∑ k = 1 n n + k n n + 1 . \lim_{n\rightarrow\infty}\frac{\sum_{k=1}^n\sqrt{n+k}}{n\sqrt{n}+1}. lim n → ∞ n n + 1 ∑ k = 1 n n + k .
Solution \textbf{Solution } Solution
n + 1 + n + 2 + ⋯ + n + n n n + n ≤ ∑ k = 1 n n + k n n + 1 ≤ n + 1 + n + 2 + ⋯ + n + n n n . \frac{\sqrt{n+1}+\sqrt{n+2}+\dots+\sqrt{n+n}}{n\sqrt{n}+n}\leq\frac{\sum_{k=1}^n\sqrt{n+k}}{n\sqrt{n}+1}\leq\frac{\sqrt{n+1}+\sqrt{n+2}+\dots+\sqrt{n+n}}{n\sqrt{n}}.
n n + n n + 1 + n + 2 + ⋯ + n + n ≤ n n + 1 ∑ k = 1 n n + k ≤ n n n + 1 + n + 2 + ⋯ + n + n .
而对不等式右端有
lim n → ∞ n + 1 + n + 2 + ⋯ + n + n n n = lim n → ∞ 1 n ∑ k = 1 n n + k n = lim n → ∞ 1 n ∑ k = 1 n 1 + k n = ∫ 0 1 1 + x d x , \lim_{n\rightarrow\infty}\frac{\sqrt{n+1}+\sqrt{n+2}+\dots+\sqrt{n+n}}{n\sqrt{n}}=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{n+k}{n}}=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n\sqrt{1+\frac{k}{n}}=\int_0^1\sqrt{1+x}\text{d}x,
n → ∞ lim n n n + 1 + n + 2 + ⋯ + n + n = n → ∞ lim n 1 k = 1 ∑ n n n + k = n → ∞ lim n 1 k = 1 ∑ n 1 + n k = ∫ 0 1 1 + x d x ,
对不等式左端有
lim n → ∞ n + 1 + n + 2 + ⋯ + n + n n n + n = lim n → ∞ ( ( n + 1 + n + 2 + ⋯ + n + n n n ) ⋅ ( n n n n + n ) ) = ∫ 0 1 1 + x d x ⋅ 1 = ∫ 0 1 1 + x d x , \begin{align*}
\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}+\sqrt{n+2}+\dots+\sqrt{n+n}}{n\sqrt{n}+n}&=\lim_{n\rightarrow\infty}\Bigg(\bigg(\frac{\sqrt{n+1}+\sqrt{n+2}+\dots+\sqrt{n+n}}{n\sqrt{n}}\bigg)\cdot\bigg(\frac{n\sqrt{n}}{n\sqrt{n}+n}\bigg)\Bigg)\\
&=\int_0^1\sqrt{1+x}\text{d}x\cdot 1=\int_0^1\sqrt{1+x}\text{d}x,
\end{align*}
n → ∞ lim n n + n n + 1 + n + 2 + ⋯ + n + n = n → ∞ lim ( ( n n n + 1 + n + 2 + ⋯ + n + n ) ⋅ ( n n + n n n ) ) = ∫ 0 1 1 + x d x ⋅ 1 = ∫ 0 1 1 + x d x ,
故由夹逼定理，得
O.F = ∫ 0 1 1 + x d x = 2 ( x + 1 ) 3 2 3 ∣ 0 1 = 4 2 − 2 3 . \text{O.F}=\int_0^1\sqrt{1+x}\text{d}x=\frac{2(x+1)^\frac{3}{2}}{3}\bigg|^1_0=\frac{4\sqrt{2}-2}{3}.
O.F = ∫ 0 1 1 + x d x = 3 2 ( x + 1 ) 2 3 0 1 = 3 4 2 − 2 .
e.x.1.24 \textbf{e.x.1.24 } e.x.1.24 lim n → ∞ ∑ i = 1 k a i n n ( ∀ k ∈ N + , a k > 0 ) . \lim_{n\rightarrow\infty}\sqrt[n]{\sum_{i=1}^k{a_i}^n}(\forall k\in\mathbb{N}_+,a_k>0). lim n → ∞ n ∑ i = 1 k a i n ( ∀ k ∈ N + , a k > 0 ) .
Solution \textbf{Solution } Solution
( max ( a i ) ) n n ≤ ∑ i = 1 k a i n n ≤ k ( max ( a i ) ) n n \sqrt[n]{(\max(a_i))^n}\leq\sqrt[n]{\sum_{i=1}^k{a_i}^n}\leq\sqrt[n]{k(\max(a_i))^n}
n ( max ( a i ) ) n ≤ n i = 1 ∑ k a i n ≤ n k ( max ( a i ) ) n
而对不等式右端有
lim n → ∞ k ( max ( a i ) ) n n = lim n → ∞ max ( a i ) k n = max ( a i ) × 1 = max ( a i ) \lim_{n\rightarrow\infty}\sqrt[n]{k(\max(a_i))^n}=\lim_{n\rightarrow\infty}\max(a_i)\sqrt[n]{k}=\max(a_i)\times 1=\max(a_i)
n → ∞ lim n k ( max ( a i ) ) n = n → ∞ lim max ( a i ) n k = max ( a i ) × 1 = max ( a i )
对不等式左端有
( max ( a i ) ) n n = max ( a i ) ⇒ lim n → ∞ ( max ( a i ) ) n n = max ( a i ) \sqrt[n]{(\max(a_i))^n}=\max(a_i)\Rightarrow\lim_{n\rightarrow\infty}\sqrt[n]{(\max(a_i))^n}=\max(a_i)
n ( max ( a i ) ) n = max ( a i ) ⇒ n → ∞ lim n ( max ( a i ) ) n = max ( a i )
故由夹逼定理，得
O.F = max ( a i ) = max { a 1 , a 2 , … , a k } . \text{O.F}=\max(a_i)=\max\{a_1,a_2,\dots,a_k\}.
O.F = max ( a i ) = max { a 1 , a 2 , … , a k } .
上方练习1.24中得到一个重要结论，即lim n → ∞ ∑ i = 1 k a i n n = max { a 1 , a 2 , … , a k } . \lim_{n\rightarrow\infty}\sqrt[n]{\sum_{i=1}^k{a_i}^n}=\max\{a_1,a_2,\dots,a_k\}. lim n → ∞ n ∑ i = 1 k a i n = max { a 1 , a 2 , … , a k } .
e.x.1.25(1) \textbf{e.x.1.25(1) } e.x.1.25(1) ∫ 0 1 ∣ ln t ∣ ln t ( 1 + t ) d t \int_0^1|\ln t|\ln^t(1+t)\text{d}t ∫ 0 1 ∣ ln t ∣ ln t ( 1 + t ) d t ∫ 0 1 t n ⋅ ∣ ln t ∣ d t \int_0^1t^n\cdot|\ln t|\text{d}t ∫ 0 1 t n ⋅ ∣ ln t ∣ d t
Solution(1) \textbf{Solution(1) } Solution(1) ∣ ln t ∣ , |\ln t|, ∣ ln t ∣ , ln ( x + 1 ) \ln(x+1) ln ( x + 1 ) x x x x > 0 x>0 x > 0
令g ( x ) = x − ln ( x + 1 ) , g(x)=x-\ln(x+1), g ( x ) = x − ln ( x + 1 ) , g ′ ( x ) = 1 − 1 x + 1 > 0 , g'(x)=1-\frac{1}{x+1}>0, g ′ ( x ) = 1 − x + 1 1 > 0 , g ( x ) g(x) g ( x ) x > 0 x>0 x > 0 g ( x ) > g ( 0 ) = 0 , g(x)>g(0)=0, g ( x ) > g ( 0 ) = 0 , x − ln ( x + 1 ) > 0 , x-\ln(x+1)>0, x − ln ( x + 1 ) > 0 , x > ln ( x + 1 ) . x>\ln(x+1). x > ln ( x + 1 ) .
因此
∫ 0 1 ∣ ln t ∣ [ ln ( 1 + t ) ] n d t < ∫ 0 1 t n ⋅ ∣ ln t ∣ d t . \int_0^1|\ln t|[\ln(1+t)]^n\text{d}t<\int_0^1t^n\cdot|\ln t|\text{d}t.
∫ 0 1 ∣ ln t ∣ [ ln ( 1 + t ) ] n d t < ∫ 0 1 t n ⋅ ∣ ln t ∣ d t .
e.x.1.25(2) \textbf{e.x.1.25(2) } e.x.1.25(2) lim n → ∞ ∫ 0 1 ∣ ln t ∣ ln t ( 1 + t ) d t . \lim_{n\rightarrow\infty}\int_0^1|\ln t|\ln^t(1+t)\text{d}t. lim n → ∞ ∫ 0 1 ∣ ln t ∣ ln t ( 1 + t ) d t .
Solution(2) \textbf{Solution(2) } Solution(2) ( 1 ) (1) ( 1 ) ∫ 0 1 ∣ ln t ∣ ln t ( 1 + t ) d t ≤ ∫ 0 1 t n ∣ ln t ∣ d t . \int_0^1|\ln t|\ln^t(1+t)\text{d}t\leq\int_0^1t^n|\ln t|\text{d}t. ∫ 0 1 ∣ ln t ∣ ln t ( 1 + t ) d t ≤ ∫ 0 1 t n ∣ ln t ∣ d t .
∫ 0 1 t n ∣ ln t ∣ d t = − ∫ 0 1 t n ln t d t = − 1 n + 1 ∫ 0 1 ln t d t n + 1 = − 1 n + 1 ( t n + 1 ln t ∣ 0 1 − ∫ 0 1 t n + 1 1 t d t ) = − 1 n + 1 ( 0 − ∫ 0 1 t n d t ) = 1 ( n + 1 ) 2 ⇒ lim n → ∞ ∫ 0 1 t n ∣ ln t ∣ d t = 0. \begin{align*}
&\int_0^1t^n|\ln t|\text{d}t&=&-\int_0^1t^n\ln t\text{d}t=-\frac{1}{n+1}\int_0^1\ln t\text{d}t^{n+1}\\
&&=&-\frac{1}{n+1}\Bigg(t^{n+1}\ln t\bigg|^1_0-\int_0^1t^{n+1}\frac{1}{t}\text{d}t\Bigg)\\
&&=&-\frac{1}{n+1}(0-\int_0^1t^{n}\text{d}t)&=\frac{1}{(n+1)^2}\\
\Rightarrow&\lim_{n\rightarrow\infty}\int_0^1t^n|\ln t|\text{d}t&=&0.
\end{align*}
⇒ ∫ 0 1 t n ∣ ln t ∣ d t n → ∞ lim ∫ 0 1 t n ∣ ln t ∣ d t = = = = − ∫ 0 1 t n ln t d t = − n + 1 1 ∫ 0 1 ln t d t n + 1 − n + 1 1 ( t n + 1 ln t 0 1 − ∫ 0 1 t n + 1 t 1 d t ) − n + 1 1 ( 0 − ∫ 0 1 t n d t ) 0. = ( n + 1 ) 2 1
而显然0 ≤ ∫ 0 1 ∣ ln t ∣ ln t ( 1 + t ) d t 0\leq\int_0^1|\ln t|\ln^t(1+t)\text{d}t 0 ≤ ∫ 0 1 ∣ ln t ∣ ln t ( 1 + t ) d t O.F. = 0. \text{O.F.}=0. O.F. = 0.
上方练习1.25中有结论l n ( x + 1 ) < x , ln(x+1)<x, l n ( x + 1 ) < x , x x + 1 < ln ( x + 1 ) . \frac{x}{x+1}<\ln(x+1). x + 1 x < ln ( x + 1 ) .
数列极限存在性证明
核心在于证明数列的单调性和有界性 。
证明单调性
毫无疑问根据定义知道a n + 1 − a n a_{n+1}-a{n} a n + 1 − a n a n + 1 a n \frac{a_{n+1}}{a_n} a n a n + 1 { a n } \{a_n\} { a n }
（基本不等式）∏ i = 1 n x i n ≤ 1 n ∑ i = 1 n x i . \sqrt[n]{\prod_{i=1}^nx_i}\leq\frac{1}{n}\sum_{i=1}^nx_i. n ∏ i = 1 n x i ≤ n 1 ∑ i = 1 n x i .
sin x ≤ x ≤ tan x , x ∈ ( 0 , π 2 ) . \sin x\leq x\leq\tan x, x\in\bigg(0,\frac{\pi}{2}\bigg). sin x ≤ x ≤ tan x , x ∈ ( 0 , 2 π ) . x x + 1 < ln ( x + 1 ) < x . \frac{x}{x+1}<\ln(x+1)<x. x + 1 x < ln ( x + 1 ) < x . x = 1 n x=\frac{1}{n} x = n 1 1 n + 1 < ln ( 1 + 1 n ) < 1 n \frac{1}{n+1}<\ln(1+\frac{1}{n})<\frac{1}{n} n + 1 1 < ln ( 1 + n 1 ) < n 1
也可以用数学归纳法或求导等方法确定这个数列的单调性。
lim n → ∞ a n 存在 ⟺ ∑ n = 1 ∞ ( a n + 1 − a n ) 收敛 . \lim_{n\rightarrow\infty}a_n\text{存在}\iff\sum_{n=1}^\infty(a_{n+1}-a_n)\text{收敛}.
n → ∞ lim a n 存在 ⟺ n = 1 ∑ ∞ ( a n + 1 − a n ) 收敛 .
考虑这个等价关系。右侧条件的∑ n = 1 ∞ ( a n + 1 − a n ) \sum_{n=1}^\infty(a_{n+1}-a_n) ∑ n = 1 ∞ ( a n + 1 − a n ) lim n → ∞ ( a n + 1 − a 1 ) , \lim_{n\rightarrow\infty}(a_{n+1}-a_1), lim n → ∞ ( a n + 1 − a 1 ) , a 1 a_1 a 1 a n a_n a n
此外，若a n + 1 − a n < k ( a n − a n − 1 ) , a_{n+1}-a_n<k(a_n-a_{n-1}), a n + 1 − a n < k ( a n − a n − 1 ) , { a n } \{a_n\} { a n }
证明有界性
参考“函数极限 ”章节中对有界性的讨论。
对已知递推关系要求极限的数列，需要先证明这个极限存在再求出它。单调递减的数列只要求出这个数列的下界（单调递增的数列只要求出这个数列的上界）即可证明极限存在。
e.g.1.26 \textbf{e.g.1.26 } e.g.1.26 { x n } \{x_n\} { x n } x n + 1 = 1 4 ( 3 x n + a x n 3 ) x_{n+1}=\frac{1}{4}\big(3x_n+\frac{a}{x_n^3}\big) x n + 1 = 4 1 ( 3 x n + x n 3 a ) a > 0 , x 1 > 0 , a>0,x_1>0, a > 0 , x 1 > 0 , lim n → ∞ x n . \lim_{n\rightarrow\infty}x_n. lim n → ∞ x n .
Solution \textbf{Solution } Solution
x n + 1 = 1 4 ( 3 x n + a x n 3 ) = 1 4 ( x n + x n + x n + a x n 3 ) ≥ 基本不等式 1 4 × ( 4 × x n x n x n a x n 3 4 ) = a 4 , x_{n+1}=\frac{1}{4}\big(3x_n+\frac{a}{x_n^3}\big)=\frac{1}{4}\big(x_n+x_n+x_n+\frac{a}{x_n^3}\big)\stackrel{\text{基本不等式}}{\geq}\frac{1}{4}\times\bigg(4\times\sqrt[4]{x_nx_nx_n\frac{a}{x_n^3}}\bigg)=\sqrt[4]{a},
x n + 1 = 4 1 ( 3 x n + x n 3 a ) = 4 1 ( x n + x n + x n + x n 3 a ) ≥ 基本不等式 4 1 × ( 4 × 4 x n x n x n x n 3 a ) = 4 a ,
因此{ x n } \{x_n\} { x n } a 4 . \sqrt[4]{a}. 4 a .
x n + 1 x n = 1 4 ( 3 + a x n 4 ) ≤ 1 4 ( 3 + a ( a 4 ) 4 ) = 3 4 + 1 4 = 1 , \frac{x_{n+1}}{x_n}=\frac{1}{4}\bigg(3+\frac{a}{x_n^4}\bigg)\leq\frac{1}{4}\bigg(3+\frac{a}{(\sqrt[4]{a})^4}\bigg)=\frac{3}{4}+\frac{1}{4}=1,
x n x n + 1 = 4 1 ( 3 + x n 4 a ) ≤ 4 1 ( 3 + ( 4 a ) 4 a ) = 4 3 + 4 1 = 1 ,
即x n + 1 x n ≤ 1 ⇒ { x n } \frac{x_{n+1}}{x_n}\leq1\Rightarrow\{x_n\} x n x n + 1 ≤ 1 ⇒ { x n }
综上，{ x n } \{x_n\} { x n } lim n → ∞ x n = A , \lim_{n\rightarrow\infty}x_n=A, lim n → ∞ x n = A ,
A = 1 4 ( 3 A + a A 3 ) ⇒ A 4 = a ⇒ A = a 4 . A=\frac{1}{4}\bigg(3A+\frac{a}{A^3}\bigg)\Rightarrow A^4=a\Rightarrow A=\sqrt[4]{a}.
A = 4 1 ( 3 A + A 3 a ) ⇒ A 4 = a ⇒ A = 4 a .
e.x.1.26 \textbf{e.x.1.26 } e.x.1.26 { a n } \{a_n\} { a n } a n = 1 + 1 2 + ⋯ + 1 n − ln n , a_n=1+\frac{1}{2}+\dots+\frac{1}{n}-\ln n, a n = 1 + 2 1 + ⋯ + n 1 − ln n , { a n } \{a_n\} { a n } Proof \textbf{Proof } Proof { a n } \{a_n\} { a n }
据题知
a n + 1 − a n = 1 n + 1 − ln ( n + 1 ) + ln n = 1 n + 1 − ln ( n + 1 n ) = 1 n + 1 − ln ( 1 + 1 n ) , a_{n+1}-a_n=\frac{1}{n+1}-\ln(n+1)+\ln n=\frac{1}{n+1}-\ln(\frac{n+1}{n})=\frac{1}{n+1}-\ln(1+\frac{1}{n}),
a n + 1 − a n = n + 1 1 − ln ( n + 1 ) + ln n = n + 1 1 − ln ( n n + 1 ) = n + 1 1 − ln ( 1 + n 1 ) ,
而（下面是对常用不等式x x + 1 < ln ( 1 + x ) < x \frac{x}{x+1}<\ln(1+x)<x x + 1 x < ln ( 1 + x ) < x 对f ( t ) = ln t f(t)=\ln t f ( t ) = ln t
∃ ε ∈ [ 1 , 1 + x ] , ln ( 1 + x ) = f ( 1 + x ) − f ( 1 ) = Lagrange f ′ ( ε ) ( ( 1 + x ) − 1 ) = x ⋅ 1 ε , \exists\varepsilon\in[1,1+x],\ln(1+x)=f(1+x)-f(1)\xlongequal{\text{Lagrange}}f'(\varepsilon)((1+x)-1)=x\cdot\frac{1}{\varepsilon},
∃ ε ∈ [ 1 , 1 + x ] , ln ( 1 + x ) = f ( 1 + x ) − f ( 1 ) Lagrange f ′ ( ε ) (( 1 + x ) − 1 ) = x ⋅ ε 1 ,
此时1 < ε < 1 + x , 1<\varepsilon<1+x, 1 < ε < 1 + x , 1 1 + x < 1 ε < 1 , \frac{1}{1+x}<\frac{1}{\varepsilon}<1, 1 + x 1 < ε 1 < 1 , 1 1 + x < ln ( 1 + x ) x < 1 \frac{1}{1+x}<\frac{\ln(1+x)}{x}<1 1 + x 1 < x l n ( 1 + x ) < 1
x 1 + x < ln ( 1 + x ) < x , \frac{x}{1+x}<\ln(1+x)<x,
1 + x x < ln ( 1 + x ) < x ,
进而知
a n + 1 − a n = 1 n + 1 − ln ( 1 + 1 n ) = 1 n 1 + 1 n − ln ( 1 + 1 n ) < 0 , a_{n+1}-a_n=\frac{1}{n+1}-\ln(1+\frac{1}{n})=\frac{\frac{1}{n}}{1+\frac{1}{n}}-\ln(1+\frac{1}{n})<0,
a n + 1 − a n = n + 1 1 − ln ( 1 + n 1 ) = 1 + n 1 n 1 − ln ( 1 + n 1 ) < 0 ,
因此{ a n } \{a_n\} { a n }
a n = ∑ i = 1 n 1 i − ln n > ∑ i = 1 n ln ( 1 + 1 i ) − ln n = ∑ i = 1 n ln ( i + 1 i ) − ln n = ln ( ∏ i = 1 n i + 1 i ) − ln n = ln ( n + 1 ) − ln n = ln n + 1 n = ln ( 1 + 1 n ) > ln 1 = 0 , \begin{align*}
a_n&=\sum_{i=1}^n\frac{1}{i}-\ln n&>\sum_{i=1}^n\ln\bigg(1+\frac{1}{i}\bigg)-\ln n&=\sum_{i=1}^n\ln\bigg(\frac{i+1}{i}\bigg)-\ln n&=\ln\bigg(\prod_{i=1}^n{i+1}{i}\bigg)-\ln n\\
&&=\ln (n+1)-\ln n&=\ln \frac{n+1}{n}=\ln\bigg(1+\frac{1}{n}\bigg)&>\ln 1&=0,
\end{align*}
a n = i = 1 ∑ n i 1 − ln n > i = 1 ∑ n ln ( 1 + i 1 ) − ln n = ln ( n + 1 ) − ln n = i = 1 ∑ n ln ( i i + 1 ) − ln n = ln n n + 1 = ln ( 1 + n 1 ) = ln ( i = 1 ∏ n i + 1 i ) − ln n > ln 1 = 0 ,
因此{ a n } \{a_n\} { a n } 0. 0. 0.
综上，{ a n } \{a_n\} { a n }
一元微分
导数·定义
在“极限”章节的一个部分 已经对此进行过导数定义的陈述，所以不再赘述。有时导数定义会变为f ′ ( x 0 ) = lim Δ x → 0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x f'(x_0)=\lim_{\Delta x\rightarrow0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x} f ′ ( x 0 ) = lim Δ x → 0 Δ x f ( x 0 + Δ x ) − f ( x 0 )
f ′ ( x 0 − ) = f ′ ( x 0 + ) ⇔ f ′ ( x 0 ) exists. f'({x_0}^-)=f'({x_0}^+)\Leftrightarrow f'(x_0)\text{ exists.} f ′ ( x 0 − ) = f ′ ( x 0 + ) ⇔ f ′ ( x 0 ) exists.
考虑上方互为充分必要的两个条件。如果一个函数在一点处的左右导数存在且相等那么这个函数在该点可导；反过来，如果一个函数在某个点可导那么这个点的左右导数存在也相等。显然解体时不会把如此浅显的结论的直接提供，因此下面考虑可导的充分条件和必要条件。
考虑导数的下面两种定义式。
f ′ ( x 0 ) = lim Δ → 0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x f ′ ( x 0 ) = lim x → x 0 f ( x ) − f ( x 0 ) x − x 0 \begin{align}
f'(x_0)&=\lim_{\Delta\rightarrow0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\
f'(x_0)&=\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}
\end{align}
f ′ ( x 0 ) f ′ ( x 0 ) = Δ → 0 lim Δ x f ( x 0 + Δ x ) − f ( x 0 ) = x → x 0 lim x − x 0 f ( x ) − f ( x 0 )
不论是上面的定义式中的哪一种，显然式中的分子都呈现出“一个动点减一个固定点”的姿势；另一方面，毫无疑问f ( x ) f(x) f ( x ) f ( x ) f(x) f ( x )
给定一个可能会是式( 1 ) (1) ( 1 ) lim Δ x → 0 f ( g 1 ( Δ x ) ) − f ( g 2 ( Δ x ) ) g 3 ( Δ x ) . \lim_{\Delta x\rightarrow0}\frac{f\big(g_1(\Delta x)\big)-f\big(g_2(\Delta x)\big)}{g_3(\Delta x)}.
Δ x → 0 lim g 3 ( Δ x ) f ( g 1 ( Δ x ) ) − f ( g 2 ( Δ x ) ) .
那么此时的分子两个被减式中应该要有一项为f ( x 0 ) . f(x_0). f ( x 0 ) . f ( g 2 ( Δ x ) ) = f ( x 0 ) ; f\big(g_2(\Delta x)\big)=f(x_0); f ( g 2 ( Δ x ) ) = f ( x 0 ) ; Δ x \Delta x Δ x 0 − 0^- 0 − 0 + 0^+ 0 + g 1 ( Δ x ) g_1(\Delta x) g 1 ( Δ x ) x 0 − ⇆ x 0 + . {x_0}^-\leftrightarrows{x_0}^+. x 0 − ⇆ x 0 + . g 1 ( Δ x ) − g 2 ( Δ x ) g_1(\Delta x)-g_2(\Delta x) g 1 ( Δ x ) − g 2 ( Δ x ) g 3 ( Δ x ) g_3(\Delta x) g 3 ( Δ x )
给定一个可能会是式( 2 ) (2) ( 2 ) lim x → x 0 f ( g 1 ( x ) ) − f ( g 2 ( x ) ) g 3 ( x ) . \lim_{x\rightarrow x_0}\frac{f\big(g_1(x)\big)-f\big(g_2(x)\big)}{g_3 (x)}.
x → x 0 lim g 3 ( x ) f ( g 1 ( x ) ) − f ( g 2 ( x ) ) .
那么此时的分子两个被减式中应该要有一项为f ( x 0 ) . f(x_0). f ( x 0 ) . f ( g 2 ( x ) ) = f ( x 0 ) ; f\big(g_2(x)\big)=f(x_0); f ( g 2 ( x ) ) = f ( x 0 ) ; x x x 0 − 0^- 0 − 0 + 0^+ 0 + g 1 ( x ) g_1(x) g 1 ( x ) x 0 − ⇆ x 0 + . {x_0}^-\leftrightarrows{x_0}^+. x 0 − ⇆ x 0 + . g 1 ( x ) − g 2 ( x ) g_1(x)-g_2(x) g 1 ( x ) − g 2 ( x ) g 3 ( x ) g_3(x) g 3 ( x )
上述是两种定义式各自的可导充分条件。
e.g.2.1 \textbf{e.g.2.1 } e.g.2.1 lim h → 0 f ( 2 h ) − f ( h ) h \lim_{h\rightarrow0}\frac{f(2h)-f(h)}{h} lim h → 0 h f ( 2 h ) − f ( h ) f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 Solution \textbf{Solution } Solution h h h 不一定 。
现在考虑充分条件中的“g 1 ( x ) − g 2 ( x ) g_1(x)-g_2(x) g 1 ( x ) − g 2 ( x ) g 3 ( x ) g_3(x) g 3 ( x ) g 1 ( x ) − g 2 ( x ) , g_1(x)-g_2(x), g 1 ( x ) − g 2 ( x ) , g 3 ( x ) g_3(x) g 3 ( x ) lim x → x 0 f ( g 1 ( x ) ) − f ( g 2 ( x ) ) g 3 ( x ) = lim x → x 0 ( f ( g 1 ( x ) ) − f ( g 2 ( x ) ) g 1 ( x ) − g 2 ( x ) ⋅ g 1 ( x ) − g 2 ( x ) g 3 ( x ) ) . \lim_{x\rightarrow x_0}\frac{f\big(g_1(x)\big)-f\big(g_2(x)\big)}{g_3(x)}=\lim_{x\rightarrow x_0}\bigg(\frac{f\big(g_1(x)\big)-f\big(g_2(x)\big)}{g_1(x)-g_2(x)}\cdot\frac{g_1(x)-g_2(x)}{g_3(x)}\bigg). lim x → x 0 g 3 ( x ) f ( g 1 ( x ) ) − f ( g 2 ( x ) ) = lim x → x 0 ( g 1 ( x ) − g 2 ( x ) f ( g 1 ( x ) ) − f ( g 2 ( x ) ) ⋅ g 3 ( x ) g 1 ( x ) − g 2 ( x ) ) . g 1 ( x ) − g 2 ( x ) g 3 ( x ) \frac{g_1(x)-g_2(x)}{g_3(x)} g 3 ( x ) g 1 ( x ) − g 2 ( x ) g 1 ( x ) − g 2 ( x ) g_1(x)-g_2(x) g 1 ( x ) − g 2 ( x ) g 3 ( x ) g_3(x) g 3 ( x ) g 3 ( x ) g_3(x) g 3 ( x )
e.g.2.2 \textbf{e.g.2.2 } e.g.2.2 lim h → 0 f ( 1 − cos h ) − f ( 0 ) h 2 \lim_{h\rightarrow0}\frac{f(1-\cos h)-f(0)}{h^2} lim h → 0 h 2 f ( 1 − c o s h ) − f ( 0 ) f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 Solution \textbf{Solution } Solution
lim h → 0 f ( 1 − cos h ) − f ( 0 ) h 2 = lim h → 0 ( f ( 0 + ( 1 − cos h ) ) − f ( 0 ) 1 − cos h ⋅ 1 − cos h h 2 ) . \lim_{h\rightarrow0}\frac{f(1-\cos h)-f(0)}{h^2}=\lim_{h\rightarrow0}\bigg(\frac{f\big(0+(1-\cos h)\big)-f(0)}{1-\cos h}\cdot\frac{1-\cos h}{h^2}\bigg).
h → 0 lim h 2 f ( 1 − cos h ) − f ( 0 ) = h → 0 lim ( 1 − cos h f ( 0 + ( 1 − cos h ) ) − f ( 0 ) ⋅ h 2 1 − cos h ) .
而显然1 − cos h h 2 \frac{1-\cos h}{h^2} h 2 1 − c o s h h → 0 h\rightarrow 0 h → 0 f ( 0 + ( 1 − cos h ) ) − f ( 0 ) 1 − cos h \frac{f\big(0+(1-\cos h)\big)-f(0)}{1-\cos h} 1 − c o s h f ( 0 + ( 1 − c o s h ) ) − f ( 0 ) h → 0 h\rightarrow 0 h → 0 又注意到lim x → 0 ( 1 − cos h ) > 0 , \lim_{x\rightarrow 0}(1-\cos h)>0, lim x → 0 ( 1 − cos h ) > 0 , 1 − cos h 1-\cos h 1 − cos h Δ x \Delta x Δ x lim Δ x → 0 + f ( 0 + Δ x ) − f ( 0 ) Δ x \lim_{\Delta x\rightarrow 0^+}\frac{f(0+\Delta x)-f(0)}{\Delta x} lim Δ x → 0 + Δ x f ( 0 + Δ x ) − f ( 0 ) f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 但无法知道f ( x ) f(x) f ( x ) 不一定 。
考虑上面例2.2中的高亮内容和充分条件的“在g 2 ( x ) = x 0 g_2(x)=x_0 g 2 ( x ) = x 0 x x x 0 − 0^- 0 − 0 + 0^+ 0 + g 1 ( x ) g_1(x) g 1 ( x ) x 0 − ⇆ x 0 + {x_0}^-\leftrightarrows{x_0}^+ x 0 − ⇆ x 0 + g 1 ( x ) − g 2 ( x ) g_1(x)-g_2(x) g 1 ( x ) − g 2 ( x )
e.x.2.1 \textbf{e.x.2.1 } e.x.2.1 lim h → 0 f ( 1 − e h ) − f ( 0 ) h \lim_{h\rightarrow0}\frac{f(1-e^h)-f(0)}{h} lim h → 0 h f ( 1 − e h ) − f ( 0 ) f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 Solution \textbf{Solution } Solution
lim h → 0 f ( 1 − e h ) − f ( 0 ) h = lim h → 0 ( f ( 0 + ( 1 − e h ) ) − f ( 0 ) 1 − e h ⋅ 1 − e h h ) . \lim_{h\rightarrow0}\frac{f(1-e^h)-f(0)}{h}=\lim_{h\rightarrow0}\bigg(\frac{f\big(0+(1-e^h)\big)-f(0)}{1-e^h}\cdot\frac{1-e^h}{h}\bigg).
h → 0 lim h f ( 1 − e h ) − f ( 0 ) = h → 0 lim ( 1 − e h f ( 0 + ( 1 − e h ) ) − f ( 0 ) ⋅ h 1 − e h ) .
而显然1 − e h h \frac{1-e^h}{h} h 1 − e h h → 0 h\rightarrow 0 h → 0 f ( 0 + ( 1 − e h ) ) − f ( 0 ) 1 − e h \frac{f\big(0+(1-e^h)\big)-f(0)}{1-e^h} 1 − e h f ( 0 + ( 1 − e h ) ) − f ( 0 ) h → 0 h\rightarrow 0 h → 0 1 − e h 1-e^h 1 − e h Δ x \Delta x Δ x lim Δ x → 0 f ( 0 + Δ x ) − f ( 0 ) Δ x \lim_{\Delta x\rightarrow 0}\frac{f(0+\Delta x)-f(0)}{\Delta x} lim Δ x → 0 Δ x f ( 0 + Δ x ) − f ( 0 ) f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 可导 。
现在考虑必要条件。如果f ( x ) f(x) f ( x ) x = x 0 x=x_0 x = x 0
对式子lim Δ x → 0 f ( g 1 ( Δ x ) ) − f ( g 2 ( Δ x ) ) g 3 ( Δ x ) \lim_{\Delta x\rightarrow0}\frac{f\big(g_1(\Delta x)\big)-f\big(g_2(\Delta x)\big)}{g_3(\Delta x)} lim Δ x → 0 g 3 ( Δ x ) f ( g 1 ( Δ x ) ) − f ( g 2 ( Δ x ) ) g 1 ( Δ x ) − g 2 ( Δ x ) = g 3 ( Δ x ) g_1(\Delta x)-g_2(\Delta x)=g_3(\Delta x) g 1 ( Δ x ) − g 2 ( Δ x ) = g 3 ( Δ x ) f ′ ( x 0 ) . f'(x_0). f ′ ( x 0 ) . g 1 ( Δ x ) − g 2 ( Δ x ) g_1(\Delta x)-g_2(\Delta x) g 1 ( Δ x ) − g 2 ( Δ x )
对式子lim x → x 0 f ( g 1 ( x ) ) − f ( g 2 ( x ) ) g 3 ( x ) \lim_{x\rightarrow x_0}\frac{f\big(g_1(x)\big)-f\big(g_2(x)\big)}{g_3(x)} lim x → x 0 g 3 ( x ) f ( g 1 ( x ) ) − f ( g 2 ( x ) ) g 1 ( x ) − g 2 ( x ) = g 3 ( x ) g_1(x)-g_2(x)=g_3(x) g 1 ( x ) − g 2 ( x ) = g 3 ( x ) f ′ ( x 0 ) . f'(x_0). f ′ ( x 0 ) . g 1 ( x ) − g 2 ( x ) g_1(x)-g_2(x) g 1 ( x ) − g 2 ( x )
利用导数定义求导往往发生在分段函数的分段点或复杂函数（如多个式子连续乘/除）中。
e.g.2.3 \textbf{e.g.2.3 } e.g.2.3 f ( x ) = ∏ i = 1 100 ( tan ( π 4 x i ) − i ) f(x)=\prod_{i=1}^{100}\big(\tan(\frac{\pi}{4}x^i)-i\big) f ( x ) = ∏ i = 1 100 ( tan ( 4 π x i ) − i ) f ′ ( 1 ) . f'(1). f ′ ( 1 ) . Solution \textbf{Solution } Solution
O.F. = lim x → 1 f ( x ) − f ( 1 ) x − 1 . \text{O.F.}=\lim_{x\rightarrow 1}\frac{f(x)-f(1)}{x-1}.
O.F. = x → 1 lim x − 1 f ( x ) − f ( 1 ) .
而显然f ( 1 ) = 0 f(1)=0 f ( 1 ) = 0 x → 1 x\rightarrow 1 x → 1
( tan ( π 4 x 2 ) − 2 ) ≠ 0 , ( tan ( π 4 x 3 ) − 3 ) ≠ 0 , ⋯ ( tan ( π 4 x 100 ) − 100 ) ≠ 0 , ⇒ ∏ i = 2 100 ( tan ( π 4 x i ) − i ) ≠ 0 , \begin{align*}
&\bigg(\tan\Big(\frac{\pi}{4}x^2\Big)-2\bigg)&≠&0,\\
&\bigg(\tan\Big(\frac{\pi}{4}x^3\Big)-3\bigg)&≠&0,\\
&\cdots\\
&\bigg(\tan\Big(\frac{\pi}{4}x^{100}\Big)-100\bigg)&≠&0,\\
\Rightarrow&\prod_{i=2}^{100}\bigg(\tan\Big(\frac{\pi}{4}x^i\Big)-i\bigg)&≠&0,
\end{align*}
⇒ ( tan ( 4 π x 2 ) − 2 ) ( tan ( 4 π x 3 ) − 3 ) ⋯ ( tan ( 4 π x 100 ) − 100 ) i = 2 ∏ 100 ( tan ( 4 π x i ) − i ) = = = = 0 , 0 , 0 , 0 ,
故
lim x → 1 f ( x ) − f ( 1 ) x − 1 = lim x → 1 f ( x ) x − 1 = lim x → 1 ( tan ( π 4 x 2 ) − 2 ) ⋅ ⋯ ⋅ ( tan ( π 4 x 100 ) − 100 ) ⋅ tan ( π 4 x ) − 1 x − 1 = ( − 1 ) × ( − 2 ) × ⋯ × ( − 99 ) lim x → 1 tan ( π 4 x ) − 1 x − 1 = L ′ H − 99 ! × lim x → 1 π 4 sec 2 ( π 4 x ) 1 = − 99 ! × π 4 × 1 cos 2 π 4 = − 99 ! π 2 \begin{align*}
\lim_{x\rightarrow 1}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow 1}\frac{f(x)}{x-1}&=\lim_{x\rightarrow 1}\bigg(\tan\Big(\frac{\pi}{4}x^2\Big)-2\bigg)\cdot\cdots\cdot\bigg(\tan\Big(\frac{\pi}{4}x^{100}\Big)-100\bigg)\cdot\frac{\tan(\frac{\pi}{4}x)-1}{x-1}\\
&=(-1)\times(-2)\times\cdots\times(-99)\lim_{x\rightarrow 1}\frac{\tan(\frac{\pi}{4}x)-1}{x-1}\xlongequal{L'H}-99!\times\lim_{x\rightarrow 1}\frac{\frac{\pi}{4}\sec^2(\frac{\pi}{4}x)}{1}\\
&=-99!\times\frac{\pi}{4}\times\frac{1}{\cos^2\frac{\pi}{4}}=-\frac{99!\pi}{2}
\end{align*}
x → 1 lim x − 1 f ( x ) − f ( 1 ) = x → 1 lim x − 1 f ( x ) = x → 1 lim ( tan ( 4 π x 2 ) − 2 ) ⋅ ⋯ ⋅ ( tan ( 4 π x 100 ) − 100 ) ⋅ x − 1 tan ( 4 π x ) − 1 = ( − 1 ) × ( − 2 ) × ⋯ × ( − 99 ) x → 1 lim x − 1 tan ( 4 π x ) − 1 L ′ H − 99 ! × x → 1 lim 1 4 π sec 2 ( 4 π x ) = − 99 ! × 4 π × cos 2 4 π 1 = − 2 99 ! π
因此f ′ ( 1 ) = − 99 ! 2 π . f'(1)=-\frac{99!}{2}\pi. f ′ ( 1 ) = − 2 99 ! π .
e.x.2.2 \textbf{e.x.2.2 } e.x.2.2 f ( x ) = x ∣ x ∣ f(x)=x|x| f ( x ) = x ∣ x ∣ f ′ ( x ) . f'(x). f ′ ( x ) . Solution \textbf{Solution } Solution
f ( x ) = { − x 2 , x ≤ 0 x 2 , x > 0 , f(x)=\begin{cases}
-x^2&,x\leq0\\
x^2&,x>0
\end{cases},
f ( x ) = { − x 2 x 2 , x ≤ 0 , x > 0 ,
因此不难知道
f ′ ( x ) = { − 2 x , x < 0 2 x , x > 0 , f'(x)=\begin{cases}
-2x&,x<0\\
2x&,x>0
\end{cases},
f ′ ( x ) = { − 2 x 2 x , x < 0 , x > 0 ,
故只要考虑f ( x ) f(x) f ( x ) x = 0 x=0 x = 0
对于f ( x ) f(x) f ( x ) x = 0 x=0 x = 0
f ′ ( 0 − ) = lim x → 0 − f ( x ) − f ( 0 ) x = lim x → 0 − − x 2 − 0 x = lim x → 0 − ( − x ) = 0 , f'(0^-)=\lim_{x\rightarrow 0^-}\frac{f(x)-f(0)}{x}=\lim_{x\rightarrow 0^-}\frac{-x^2-0}{x}=\lim_{x\rightarrow 0^-}(-x)=0,
f ′ ( 0 − ) = x → 0 − lim x f ( x ) − f ( 0 ) = x → 0 − lim x − x 2 − 0 = x → 0 − lim ( − x ) = 0 ,
同理知道f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 f ′ ( 0 + ) = 0 , f'(0^+)=0, f ′ ( 0 + ) = 0 , f ′ ( x ) = 0 , f'(x)=0, f ′ ( x ) = 0 ,
f ′ ( x ) = { − 2 x , x < 0 0 , x = 0 2 x , x > 0 = { − 2 x , x ≤ 0 2 x , x > 0 . f'(x)=\begin{cases}
-2x&,x<0\\
0&,x=0\\
2x&,x>0
\end{cases}
=\begin{cases}
-2x&,x\leq0\\
2x&,x>0
\end{cases}.
f ′ ( x ) = ⎩ ⎨ ⎧ − 2 x 0 2 x , x < 0 , x = 0 , x > 0 = { − 2 x 2 x , x ≤ 0 , x > 0 .
对于可导性有下面几个重要结论。
f ( x 0 ) = 0 ⟺ F ( x ) = f ( x ) g ( x ) f(x_0)=0\iff F(x)=f(x)g(x) f ( x 0 ) = 0 ⟺ F ( x ) = f ( x ) g ( x ) x = x 0 x=x_0 x = x 0 g ( x ) g(x) g ( x ) x = x 0 x=x_0 x = x 0 f ( x ) = ( x − x 0 ) k ∣ x − x 0 ∣ f(x)=(x-x_0)^k|x-x_0| f ( x ) = ( x − x 0 ) k ∣ x − x 0 ∣ x = x 0 x=x_0 x = x 0 k k k k = 0 k=0 k = 0 若在x = x 0 x=x_0 x = x 0 φ ( x ) \varphi(x) φ ( x ) f ( x ) = ∣ x − x 0 ∣ φ ( x ) , f(x)=|x-x_0|\varphi(x), f ( x ) = ∣ x − x 0 ∣ φ ( x ) ,
f ( x ) f(x) f ( x ) x 0 x_0 x 0 ⟺ φ ( x 0 ) = 0 ; \iff\varphi(x_0)=0; ⟺ φ ( x 0 ) = 0 ; f ( x ) f(x) f ( x ) x 0 x_0 x 0 ⟺ φ ( x 0 ) ≠ 0. \iff\varphi(x_0)≠0. ⟺ φ ( x 0 ) = 0.
奇函数的导数是偶函数，偶函数的导数是奇函数。
（导数极限定理）若f ( x ) f(x) f ( x ) x 0 x_0 x 0 x 0 x_0 x 0 x 0 x_0 x 0 a a a f ′ ( x 0 ) = a . f'(x_0)=a. f ′ ( x 0 ) = a .
（Darboux定理）在[ a , b ] [a,b] [ a , b ] f ( x ) f(x) f ( x ) f ′ ( a + ) f ′ ( b − ) < 0 , f'(a^+)f'(b^-)<0, f ′ ( a + ) f ′ ( b − ) < 0 , ∃ ξ ∈ ( a , b ) s.t. f ′ ( ξ ) = 0. \exists\xi\in(a,b) \text{ s.t. }f'(\xi)=0. ∃ ξ ∈ ( a , b ) s.t. f ′ ( ξ ) = 0.
e.g.2.4 \textbf{e.g.2.4 } e.g.2.4 f ( x ) = ∣ x − x 2 ∣ ( e x − 1 ) f(x)=|x-x^2|(e^x-1) f ( x ) = ∣ x − x 2 ∣ ( e x − 1 ) Solution \textbf{Solution } Solution g ( x ) = ∣ x ∣ , h ( x ) = ∣ 1 − x ∣ ( e x − 1 ) g(x)=|x|,h(x)=|1-x|(e^x-1) g ( x ) = ∣ x ∣ , h ( x ) = ∣1 − x ∣ ( e x − 1 ) f ( x ) = g ( x ) h ( x ) f(x)=g(x)h(x) f ( x ) = g ( x ) h ( x ) g ( x ) g(x) g ( x ) x = 0 x=0 x = 0 h ( x ) h(x) h ( x ) h ( 0 ) = 0 h(0)=0 h ( 0 ) = 0 f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 g ( x ) = ∣ 1 − x ∣ g(x)=|1-x| g ( x ) = ∣1 − x ∣ h ( x ) = ∣ x ∣ ( e x − 1 ) h(x)=|x|(e^x-1) h ( x ) = ∣ x ∣ ( e x − 1 ) f ( x ) f(x) f ( x ) x = 1 x=1 x = 1 f ( x ) f(x) f ( x ) 1个 。
e.x.2.3 \textbf{e.x.2.3 } e.x.2.3 f ( x ) = ( x − 3 ) tan ( x + 3 ) ∣ x 2 − 4 ∣ f(x)=(x-3)\tan(x+3)|x^2-4| f ( x ) = ( x − 3 ) tan ( x + 3 ) ∣ x 2 − 4∣ Solution \textbf{Solution } Solution f ( x ) f(x) f ( x ) x = 2 x=2 x = 2 x = − 2. x=-2. x = − 2. { g ( x ) = ∣ x − 2 ∣ h ( x ) = ∣ x + 2 ∣ ( x − 3 ) tan ( x + 3 ) \begin{cases}g(x)&=|x-2|\\h(x)&=|x+2|(x-3)\tan(x+3)\end{cases} { g ( x ) h ( x ) = ∣ x − 2∣ = ∣ x + 2∣ ( x − 3 ) tan ( x + 3 ) { g ( x ) = ∣ x + 2 ∣ h ( x ) = ∣ x − 2 ∣ ( x − 3 ) tan ( x + 3 ) \begin{cases}g(x)&=|x+2|\\h(x)&=|x-2|(x-3)\tan(x+3)\end{cases} { g ( x ) h ( x ) = ∣ x + 2∣ = ∣ x − 2∣ ( x − 3 ) tan ( x + 3 ) f ( x ) f(x) f ( x ) x = ± 2 x=\pm 2 x = ± 2 2个 。
e.x.2.4 \textbf{e.x.2.4 } e.x.2.4 f ( x ) = ( 3 x 3 ) + x 2 ∣ x ∣ f(x)=(3x^3)+x^2|x| f ( x ) = ( 3 x 3 ) + x 2 ∣ x ∣ x = 0 x=0 x = 0 n n n Solution \textbf{Solution } Solution x 2 x^2 x 2 x x x x 3 x^3 x 3 n = 2. n=2. n = 2.
e.x.2.5 \textbf{e.x.2.5 } e.x.2.5 f ( x ) = 1 1 + x 2 f(x)=\frac{1}{1+x^2} f ( x ) = 1 + x 2 1 x = 0 x=0 x = 0 f ( 3 ) ( 0 ) . f^{(3)}(0). f ( 3 ) ( 0 ) . Solution \textbf{Solution } Solution f ( − x ) = 1 1 + x 2 = f ( x ) f(-x)=\frac{1}{1+x^2}=f(x) f ( − x ) = 1 + x 2 1 = f ( x ) f ( x ) f(x) f ( x ) R \mathbb{R} R f ( 3 ) ( x ) f^{(3)}(x) f ( 3 ) ( x ) R \mathbb{R} R f ( 3 ) ( − 0 ) = − f ( 3 ) ( 0 ) ⇒ f ( 3 ) ( 0 ) = 0. f^{(3)}(-0)=-f^{(3)}(0)\Rightarrow f^{(3)}(0)=0. f ( 3 ) ( − 0 ) = − f ( 3 ) ( 0 ) ⇒ f ( 3 ) ( 0 ) = 0.
e.g.2.5 \textbf{e.g.2.5 } e.g.2.5 Proof \textbf{Proof } Proof f ( x ) f(x) f ( x ) x 0 x_0 x 0 ∃ F ( x ) , s.t. F ′ ( x ) = f ( x ) , \exists F(x),\text{ s.t. }F'(x)=f(x), ∃ F ( x ) , s.t. F ′ ( x ) = f ( x ) , F ( x ) F(x) F ( x ) f ( x ) f(x) f ( x )
lim x → x 0 − g ( x ) = lim x → x 0 − G ′ ( x ) = G ′ ( x 0 − ) = Def. 1 st Discontinuity G ′ ( x 0 ) = g ( x 0 ) , \lim_{x\rightarrow {x_0}^-}g(x)=\lim_{x\rightarrow {x_0}^-}G'(x)=G'({x_0}^-)\xlongequal{\text{Def. 1}^\text{st}\text{ Discontinuity}}G'(x_0)=g(x_0),
x → x 0 − lim g ( x ) = x → x 0 − lim G ′ ( x ) = G ′ ( x 0 − ) Def. 1 st Discontinuity G ′ ( x 0 ) = g ( x 0 ) ,
类似根据第一类间断点定义也知
lim x → x 0 + g ( x ) = g ( x 0 ) , \lim_{x\rightarrow {x_0}^+}g(x)=g(x_0),
x → x 0 + lim g ( x ) = g ( x 0 ) ,
对应有lim x → x 0 − g ( x ) = g ( x 0 ) = lim x → x 0 + g ( x ) , \lim_{x\rightarrow {x_0}^-}g(x)=g(x_0)=\lim_{x\rightarrow {x_0}^+}g(x), lim x → x 0 − g ( x ) = g ( x 0 ) = lim x → x 0 + g ( x ) , g ( x ) g(x) g ( x ) x 0 x_0 x 0
Q.E.D. \textbf{Q.E.D.} Q.E.D.
例2.5中的待证明题也可以理解为结论“一个函数最多只有第二类间断点”。
一元隐／反函数求导
隐函数／参数方程求导
对形如F ( x , y ) = 0 F(x,y)=0 F ( x , y ) = 0 y y y d y d x \frac{\text{d}y}{\text{d}x} d x d y
把y y y x x x y ( x ) y(x) y ( x ) x x x y ′ y' y ′ y ′ , y', y ′ , y ′ y' y ′ x , y x,y x , y
对F ( x , y ) F(x,y) F ( x , y ) d y d x = − F x ′ ( x , y ) F y ′ ( x , y ) . \frac{\text{d}y}{\text{d}x}=-\frac{F'_x(x,y)}{F'_y(x,y)}. d x d y = − F y ′ ( x , y ) F x ′ ( x , y ) .
对方程两端求微分后从求得的等式中解出d y d x . \frac{\text{d}y}{\text{d}x}. d x d y .
对这个隐函数求二阶导数时首先求其一阶导函数φ ( x , y ) \varphi(x,y) φ ( x , y ) φ ( x , y ) = φ ( x , y ( x ) ) \varphi(x,y)=\varphi(x,y(x)) φ ( x , y ) = φ ( x , y ( x )) x x x y ′ y' y ′ φ ( x , y ) \varphi(x,y) φ ( x , y ) d 2 y d x 2 . \frac{\text{d}^2y}{\text{d}x^2}. d x 2 d 2 y .
参数方程{ x = x ( t ) y = y ( t ) \begin{cases}x&=x(t)\\y&=y(t)\end{cases} { x y = x ( t ) = y ( t ) y = y ( x ) . y=y(x). y = y ( x ) .
d y d x = y ′ ( t ) x ′ ( t ) . d 2 y d x 2 = d ( d y d x ) d x = d ( d y d x ) d t ⋅ d t d x = y ′ ′ ( t ) x ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) ( x ′ ( t ) ) 2 ⋅ 1 x ′ ( t ) = y ′ ′ ( t ) x ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) ( x ′ ( t ) ) 3 . \begin{align*}
\frac{\text{d}y}{\text{d}x}&=\frac{y'(t)}{x'(t)}.\\
\frac{\text{d}^2y}{\text{d}x^2}&=\frac{\text{d}(\frac{\text{d}y}{\text{d}x})}{\text{d}x}=\frac{\text{d}(\frac{\text{d}y}{\text{d}x})}{\text{d}t}\cdot\frac{\text{d}t}{\text{d}x}=\frac{y''(t)x'(t)-x''(t)y'(t)}{(x'(t))^2}\cdot\frac{1}{x'(t)}=\frac{y''(t)x'(t)-x''(t)y'(t)}{(x'(t))^3}.
\end{align*}
d x d y d x 2 d 2 y = x ′ ( t ) y ′ ( t ) . = d x d ( d x d y ) = d t d ( d x d y ) ⋅ d x d t = ( x ′ ( t ) ) 2 y ′′ ( t ) x ′ ( t ) − x ′′ ( t ) y ′ ( t ) ⋅ x ′ ( t ) 1 = ( x ′ ( t ) ) 3 y ′′ ( t ) x ′ ( t ) − x ′′ ( t ) y ′ ( t ) .
利用此可以求参数方程确定的函数的导数。
反函数求导
若y = f ( x ) y=f(x) y = f ( x ) x = φ ( y ) , x=\varphi(y), x = φ ( y ) , d x d y = 1 d x d y ( = 1 y ′ ) , \frac{\text{d}x}{\text{d}y}=\frac{1}{\frac{\text{d}x}{\text{d}y}}(=\frac{1}{y'}), d y d x = d y d x 1 ( = y ′ 1 ) ,
d 2 x d y 2 = d ( d x d y ) d y = d ( d x d y ) d x ⋅ d x d y = d ( 1 y ′ ) d x ⋅ d x d y = ( 1 y ′ ) ′ ⋅ d x d y = 0 − y ′ ′ ( y ′ ) 2 ⋅ 1 y ′ = − y ′ ′ ( y ′ ) 3 = − f ′ ′ ( x ) ( f ′ ( x ) ) 3 . \frac{\text{d}^2x}{\text{d}y^2}=\frac{\text{d}(\frac{\text{d}x}{\text{d}y})}{\text{d}y}=\frac{\text{d}(\frac{\text{d}x}{\text{d}y})}{\text{d}x}\cdot\frac{\text{d}x}{\text{d}y}=\frac{\text{d}(\frac{1}{y'})}{\text{d}x}\cdot\frac{\text{d}x}{\text{d}y}=\bigg(\frac{1}{y'}\bigg)'\cdot\frac{\text{d}x}{\text{d}y}=\frac{0-y''}{(y')^2}\cdot\frac{1}{y'}=-\frac{y''}{(y')^3}=-\frac{f''(x)}{\big(f'(x)\big)^3}.
d y 2 d 2 x = d y d ( d y d x ) = d x d ( d y d x ) ⋅ d y d x = d x d ( y ′ 1 ) ⋅ d y d x = ( y ′ 1 ) ′ ⋅ d y d x = ( y ′ ) 2 0 − y ′′ ⋅ y ′ 1 = − ( y ′ ) 3 y ′′ = − ( f ′ ( x ) ) 3 f ′′ ( x ) .
求高阶导
一般考虑使用数学归纳法、Leibniz公式或Taylor公式。
数学归纳法求高阶导数主要是针对各阶导数存在一定规律的情况使用的。一般先求出函数的前几阶导数后分析所得的结果即可找到规律，然后写出n n n
Leibniz公式是针对乘积型函数求高阶导的泛用方法。利用( u v ) ( n ) = ∑ k = 0 n C n k u ( n − k ) v ( k ) , (uv)^{(n)}=\sum_{k=0}^nC_n^ku^{(n-k)}v^{(k)}, ( uv ) ( n ) = ∑ k = 0 n C n k u ( n − k ) v ( k ) , ( u v ) ′ = u v ′ + u ′ v (uv)'=uv'+u'v ( uv ) ′ = u v ′ + u ′ v
e.g.2.6 \textbf{e.g.2.6 } e.g.2.6 y = x 3 sin x y=x^3\sin x y = x 3 sin x x = 0 x=0 x = 0 y ( 6 ) ( 0 ) . y^{(6)}(0). y ( 6 ) ( 0 ) .
Solution \textbf{Solution } Solution sin x \sin x sin x ( sin x ) ( n ) = { sin x , n ≡ 0 ( m o d 4 ) cos x , n ≡ 1 ( m o d 4 ) − sin x , n ≡ 2 ( m o d 4 ) − cos x , n ≡ 3 ( m o d 4 ) , (\sin x)^{(n)}=\begin{cases}\sin x&,n\equiv 0\pmod{4}\\\cos x&, n\equiv 1\pmod{4}\\-\sin x&, n\equiv 2\pmod{4}\\-\cos x&, n\equiv 3\pmod{4}\end{cases}, ( sin x ) ( n ) = ⎩ ⎨ ⎧ sin x cos x − sin x − cos x , n ≡ 0 ( mod 4 ) , n ≡ 1 ( mod 4 ) , n ≡ 2 ( mod 4 ) , n ≡ 3 ( mod 4 ) ,
y ( 6 ) ( x ) = ∑ i = 0 6 C 6 i ( x 3 ) ( i ) ( sin x ) ( 6 − i ) = C 6 0 ( x 3 ) ⋅ ( sin x ) ( 6 ) + C 6 1 ( x 3 ) ′ ⋅ ( sin x ) ( 5 ) + C 6 2 ( x 3 ) ′ ′ ⋅ ( sin x ) ( 4 ) + C 6 3 ( x 3 ) ′ ′ ′ ⋅ ( sin x ) ′ ′ ′ + C 6 4 ( x 3 ) ( 4 ) ⋅ ( sin x ) ′ ′ + C 6 5 ( x 3 ) ( 5 ) ⋅ ( sin x ) ′ + C 6 6 ( x 3 ) ( 6 ) ⋅ ( sin x ) = x 3 ( − sin x ) + 6 × ( 3 x 2 ) cos x + 15 × ( 6 x ) sin x + 20 × 6 ( − cos x ) + 0 + 0 + 0 = − x 3 sin x + 18 x 2 cos x + 90 x sin x − 120 cos x . \begin{align*}
y^{(6)}(x)&=&\sum_{i=0}^6C_6^i(x^3)^{(i)}(\sin x)^{(6-i)}&\\
&=&C_6^0(x^3)\cdot(\sin x)^{(6)}+C_6^1(x^3)'\cdot(\sin x)^{(5)}+C_6^2(x^3)''\cdot(\sin x)^{(4)}+C_6^3(x^3)'''\cdot(\sin x)'''\\
&&+C_6^4(x^3)^{(4)}\cdot(\sin x)''+C_6^5(x^3)^{(5)}\cdot(\sin x)'+C_6^6(x^3)^{(6)}\cdot(\sin x)\\
&=&x^3(-\sin x)+6\times(3x^2)\cos x+15\times(6x)\sin x+20\times 6(-\cos x)+0+0+0\\
&=&-x^3\sin x+18x^2\cos x+90x\sin x-120\cos x.
\end{align*}
y ( 6 ) ( x ) = = = = i = 0 ∑ 6 C 6 i ( x 3 ) ( i ) ( sin x ) ( 6 − i ) C 6 0 ( x 3 ) ⋅ ( sin x ) ( 6 ) + C 6 1 ( x 3 ) ′ ⋅ ( sin x ) ( 5 ) + C 6 2 ( x 3 ) ′′ ⋅ ( sin x ) ( 4 ) + C 6 3 ( x 3 ) ′′′ ⋅ ( sin x ) ′′′ + C 6 4 ( x 3 ) ( 4 ) ⋅ ( sin x ) ′′ + C 6 5 ( x 3 ) ( 5 ) ⋅ ( sin x ) ′ + C 6 6 ( x 3 ) ( 6 ) ⋅ ( sin x ) x 3 ( − sin x ) + 6 × ( 3 x 2 ) cos x + 15 × ( 6 x ) sin x + 20 × 6 ( − cos x ) + 0 + 0 + 0 − x 3 sin x + 18 x 2 cos x + 90 x sin x − 120 cos x .
进而所求为
y ( 6 ) ( 0 ) = − x 3 sin x + 18 x 2 cos x + 90 x sin x − 120 cos x ∣ x = 0 = − 120 cos 0 = − 120. y^{(6)}(0)=-x^3\sin x+18x^2\cos x+90x\sin x-120\cos x\big|_{x=0}=-120\cos 0=-120.
y ( 6 ) ( 0 ) = − x 3 sin x + 18 x 2 cos x + 90 x sin x − 120 cos x x = 0 = − 120 cos 0 = − 120.
Taylor公式求函数在某点的各阶导数时一般按下面3步。
首先，写出f ( x ) f(x) f ( x ) x = x 0 x=x_0 x = x 0
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n + ⋯ , f(x)=f(x_0)+f'(x_0)(x-x_0)+\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+\cdots,
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + ⋯ + n ! f ( n ) ( x 0 ) ( x − x 0 ) n + ⋯ ,
然后通过化简、变量代换或其他已知Taylor公式将f ( x ) f(x) f ( x )
f ( x ) = a 0 + a 1 ( x − x 0 ) + a 2 ( x − x 0 ) 2 + ⋯ + a n ( x − x 0 ) n + ⋯ . f(x)=a_0+a_1(x-x_0)+a_2(x-x_0)^2+\cdots+a_n(x-x_0)^n+\cdots.
f ( x ) = a 0 + a 1 ( x − x 0 ) + a 2 ( x − x 0 ) 2 + ⋯ + a n ( x − x 0 ) n + ⋯ .
这时根据函数Taylor公式的唯一性可以知道相同的次幂项的系数应当相同，所以f ( n ) ( x 0 ) n ! = a n , \frac{f^{(n)}(x_0)}{n!}=a_n, n ! f ( n ) ( x 0 ) = a n , f ( n ) ( x 0 ) = n ! a n . f^{(n)}(x_0)=n!a_n. f ( n ) ( x 0 ) = n ! a n .
考虑Taylor方法求高阶导，重新尝试解决例2.6中的问题。
e.g.2.6 \textbf{e.g.2.6 } e.g.2.6 y = x 3 sin x y=x^3\sin x y = x 3 sin x x = 0 x=0 x = 0 y ( 6 ) ( 0 ) . y^{(6)}(0). y ( 6 ) ( 0 ) .
Solution (Another Story) \textbf{Solution (Another Story) } Solution (Another Story) y y y x = 0 x=0 x = 0
y = ∑ i = 0 ∞ f ( i ) ( 0 ) i ! ( x − 0 ) i = f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) 2 ! x 2 + f ( 3 ) ( 0 ) 3 ! x 3 + f ( 4 ) ( 0 ) 4 ! x 4 + f ( 5 ) ( 0 ) 5 ! x 5 + f ( 6 ) ( 0 ) 6 ! x 6 + ⋯ y=\sum_{i=0}^\infty\frac{f^{(i)}(0)}{i!}(x-0)^i=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4+\frac{f^{(5)}(0)}{5!}x^5+\frac{f^{(6)}(0)}{6!}x^6+\cdots
y = i = 0 ∑ ∞ i ! f ( i ) ( 0 ) ( x − 0 ) i = f ( 0 ) + f ′ ( 0 ) x + 2 ! f ′′ ( 0 ) x 2 + 3 ! f ( 3 ) ( 0 ) x 3 + 4 ! f ( 4 ) ( 0 ) x 4 + 5 ! f ( 5 ) ( 0 ) x 5 + 6 ! f ( 6 ) ( 0 ) x 6 + ⋯
而我们本就知道sin x = x − 1 6 x 3 + ⋯ , \sin x=x-\frac{1}{6}x^3+\cdots, sin x = x − 6 1 x 3 + ⋯ , y y y y = x 3 ( x − 1 6 x 3 + ⋯ ) , y=x^3(x-\frac{1}{6}x^3+\cdots), y = x 3 ( x − 6 1 x 3 + ⋯ ) , x 6 x^6 x 6
f ( 6 ) ( 0 ) 6 ! = − 1 6 , \frac{f^{(6)}(0)}{6!}=-\frac{1}{6},
6 ! f ( 6 ) ( 0 ) = − 6 1 ,
解得f ( 6 ) ( 0 ) = − 1 6 × 6 ! = − 5 ! = − 120. f^{(6)}(0)=-\frac{1}{6}\times{6!}=-5!=-120. f ( 6 ) ( 0 ) = − 6 1 × 6 ! = − 5 ! = − 120. y ( 6 ) ( 0 ) = − 120. y^{(6)}(0)=-120. y ( 6 ) ( 0 ) = − 120.
e.x.2.6 \textbf{e.x.2.6 } e.x.2.6 f ( x ) = x 2 ln ( x + 1 ) f(x)=x^2\ln(x+1) f ( x ) = x 2 ln ( x + 1 ) x = 0 x=0 x = 0 f ( n ) ( 0 ) . f^{(n)}(0). f ( n ) ( 0 ) .
Solution \textbf{Solution } Solution f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 f ( x ) = f ( 0 ) + f ′ ( 0 ) x + ⋯ + f ( n ) ( 0 ) n ! x n . f(x)=f(0)+f'(0)x+\cdots+\frac{f^{(n)}(0)}{n!}x^n. f ( x ) = f ( 0 ) + f ′ ( 0 ) x + ⋯ + n ! f ( n ) ( 0 ) x n . ln ( x + 1 ) = x − 1 2 x 2 + ⋯ + ( − 1 ) n − 1 x n n , \ln(x+1)=x-\frac{1}{2}x^2+\cdots+(-1)^{n-1}\frac{x^n}{n}, ln ( x + 1 ) = x − 2 1 x 2 + ⋯ + ( − 1 ) n − 1 n x n , x 2 ln ( x + 1 ) = x 3 − 1 2 x 4 + ⋯ + ( − 1 ) n − 3 x n n − 2 + ( − 1 ) n − 2 x n + 1 n − 1 + ( − 1 ) n − 1 x n + 2 n , x^2\ln(x+1)=x^3-\frac{1}{2}x^4+\cdots+(-1)^{n-3}\frac{x^n}{n-2}+(-1)^{n-2}\frac{x^{n+1}}{n-1}+(-1)^{n-1}\frac{x^{n+2}}{n}, x 2 ln ( x + 1 ) = x 3 − 2 1 x 4 + ⋯ + ( − 1 ) n − 3 n − 2 x n + ( − 1 ) n − 2 n − 1 x n + 1 + ( − 1 ) n − 1 n x n + 2 ,
f ( n ) ( 0 ) n ! = ( − 1 ) n − 3 n − 2 . \frac{f^{(n)}(0)}{n!}=\frac{(-1)^{n-3}}{n-2}.
n ! f ( n ) ( 0 ) = n − 2 ( − 1 ) n − 3 .
因此所求为
f ( n ) ( 0 ) = ( − 1 ) n − 3 n ! n − 2 . f^{(n)}(0)=\frac{(-1)^{n-3}n!}{n-2}.
f ( n ) ( 0 ) = n − 2 ( − 1 ) n − 3 n ! .
e.g.2.7 \textbf{e.g.2.7 } e.g.2.7 f ( x ) = arctan x − x 1 + a x 2 f(x)=\arctan x-\frac{x}{1+ax^2} f ( x ) = arctan x − 1 + a x 2 x f ′ ′ ′ ( 0 ) = 1 , f'''(0)=1, f ′′′ ( 0 ) = 1 , a a a
Solution \textbf{Solution } Solution f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 f ( x ) = ⋯ + f ′ ′ ′ ( 0 ) 3 ! x 3 + ⋯ . f(x)=\cdots+\frac{f'''(0)}{3!}x^3+\cdots. f ( x ) = ⋯ + 3 ! f ′′′ ( 0 ) x 3 + ⋯ . arctan x = x − 1 3 x 2 + ⋯ , \arctan x=x-\frac{1}{3}x^2+\cdots, arctan x = x − 3 1 x 2 + ⋯ , 1 1 + a x 2 = ∑ i = 0 ∞ ( ( − a x 2 ) i ) , \frac{1}{1+ax^2}=\sum_{i=0}^\infty((-ax^2)^i), 1 + a x 2 1 = ∑ i = 0 ∞ (( − a x 2 ) i ) ,
f ( x ) = − x 1 + a x 2 + ( x − 1 3 + ⋯ ) = ( x − 1 3 x 3 + ⋯ ) − x ( 1 − a x 2 + ⋯ ) = ( a − 1 3 ) x 3 + ⋯ f(x)=-\frac{x}{1+ax^2}+\bigg(x-\frac{1}{3}+\cdots)=(x-\frac{1}{3}x^3+\cdots)-x(1-ax^2+\cdots)=(a-\frac{1}{3})x^3+\cdots
f ( x ) = − 1 + a x 2 x + ( x − 3 1 + ⋯ ) = ( x − 3 1 x 3 + ⋯ ) − x ( 1 − a x 2 + ⋯ ) = ( a − 3 1 ) x 3 + ⋯
因此f ′ ′ ′ ( 0 ) 3 ! = a − 1 3 . \frac{f'''(0)}{3!}=a-\frac{1}{3}. 3 ! f ′′′ ( 0 ) = a − 3 1 . f ′ ′ ′ ( 0 ) = 1 f'''(0)=1 f ′′′ ( 0 ) = 1 a = 1 6 + 1 3 = 1 2 . a=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}. a = 6 1 + 3 1 = 2 1 .
中值定理
在讨论中值定理之前，先回想一下连续函数应该具备的性质。
如果f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] M M M m m m ∀ x ∈ [ a , b ] , m ≤ f ( x ) ≤ M . \forall x\in [a,b], m\leq f(x)\leq M. ∀ x ∈ [ a , b ] , m ≤ f ( x ) ≤ M . f ( a ) = A , f ( b ) = b f(a)=A,f(b)=b f ( a ) = A , f ( b ) = b ∀ C ∈ ( A , B ) , ∃ ξ ∈ ( a , b ) , s.t. f ( ξ ) = C . \forall C\in (A,B), \exists\xi\in(a,b), \text{ s.t. } f(\xi)=C. ∀ C ∈ ( A , B ) , ∃ ξ ∈ ( a , b ) , s.t. f ( ξ ) = C . m ≤ μ ≤ M , m\leq\mu\leq M, m ≤ μ ≤ M , ∃ ζ ∈ [ a , b ] , s.t. f ( ζ ) = μ . \exists\zeta\in[a,b],\text{ s.t. }f(\zeta)=\mu. ∃ ζ ∈ [ a , b ] , s.t. f ( ζ ) = μ . f ( a ) f ( b ) < 0 ⇒ ∃ ε ∈ ( a , b ) , f ( ε ) = 0. f(a)f(b)<0\Rightarrow\exists\varepsilon\in(a,b),f(\varepsilon)=0. f ( a ) f ( b ) < 0 ⇒ ∃ ε ∈ ( a , b ) , f ( ε ) = 0. ∃ δ ∈ [ a , b ] , ∫ a b f ( x ) d x = f ( δ ) ( b − a ) . \exists\delta\in[a,b],\int_a^bf(x)\text{d}x=f(\delta)(b-a). ∃ δ ∈ [ a , b ] , ∫ a b f ( x ) d x = f ( δ ) ( b − a ) .
在上方性质中，积分中值定理是可以通过介值定理证明的。（证明过程可参考例2.9第一部分 ）
e.g.2.8 \textbf{e.g.2.8 } e.g.2.8 x = a sin x + b ( a > 0 , b > 0 ) x=a\sin x+b(a>0,b>0) x = a sin x + b ( a > 0 , b > 0 ) a + b a+b a + b
Proof \textbf{Proof } Proof f ( x ) = a sin x − x + b . f(x)=a\sin x-x+b. f ( x ) = a sin x − x + b . f ( 0 ) = b > 0 , f ( a + b ) = a sin x − a ≤ 0 ⇒ f ( 0 ) f ( a + b ) ≤ 0. f(0)=b>0,f(a+b)=a\sin x-a\leq0\Rightarrow f(0)f(a+b)\leq 0. f ( 0 ) = b > 0 , f ( a + b ) = a sin x − a ≤ 0 ⇒ f ( 0 ) f ( a + b ) ≤ 0. f ( x ) = 0 f(x)=0 f ( x ) = 0 0 < x ≤ a + b 0<x\leq a+b 0 < x ≤ a + b f ( x ) = 0 f(x)=0 f ( x ) = 0 a + b a+b a + b
Q.E.D. \textbf{Q.E.D.} Q.E.D.
e.x.2.7 \textbf{e.x.2.7 } e.x.2.7 [ a , b ] [a,b] [ a , b ] f ( x ) f(x) f ( x ) a < x 1 < x 2 < … < x n < b , a< x_1< x_2< \ldots< x_n< b, a < x 1 < x 2 < … < x n < b , ∃ ξ ∈ [ x 1 , x n ] , s.t. f ( ξ ) = 1 n ∑ i = 1 n f ( x i ) . \exists\xi\in[x_1, x_n], \text{ s.t. }f(\xi)=\frac{1}{n}\sum_{i=1}^nf(x_i). ∃ ξ ∈ [ x 1 , x n ] , s.t. f ( ξ ) = n 1 ∑ i = 1 n f ( x i ) .
Proof \textbf{Proof } Proof f ( x ) f(x) f ( x ) [ x 1 , x n ] [x_1,x_n] [ x 1 , x n ] m m m M M M
m ≤ f ( x 1 ) ≤ M m ≤ f ( x 2 ) ≤ M ⋮ ≤ ⋮ ≤ ⋮ m ≤ f ( x n ) ≤ M ⇒ n m ≤ f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x n ) ≤ n M ⇒ m ≤ 1 n ∑ i = 1 n f ( x i ) ≤ M \begin{align*}
&m &\leq &f(x_1) &\leq &M\\
&m &\leq &f(x_2) &\leq &M\\
&\vdots &\leq &\vdots &\leq &\vdots\\
&m &\leq &f(x_n) &\leq &M\\
\Rightarrow&nm &\leq &f(x_1)+f(x_2)+\cdots+f(x_n) &\leq &nM\\
\Rightarrow&m &\leq &\frac{1}{n}\sum_{i=1}^nf(x_i) &\leq &M
\end{align*}
⇒ ⇒ m m ⋮ m nm m ≤ ≤ ≤ ≤ ≤ ≤ f ( x 1 ) f ( x 2 ) ⋮ f ( x n ) f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x n ) n 1 i = 1 ∑ n f ( x i ) ≤ ≤ ≤ ≤ ≤ ≤ M M ⋮ M n M M
所以根据介值定理，∃ ξ ∈ [ x 1 , x n ] , f ( ξ ) = 1 n ∑ i = 1 n f ( x i ) . \exists\xi\in[x_1,x_n],f(\xi)=\frac{1}{n}\sum_{i=1}^nf(x_i). ∃ ξ ∈ [ x 1 , x n ] , f ( ξ ) = n 1 ∑ i = 1 n f ( x i ) .
Q.E.D. \textbf{Q.E.D.} Q.E.D.
有时函数在区间上连续不会明示。在一部分情况下，其会通过“函数的n n n
中值定理都是在函数连续的情况下存在的。
Rolle定理 ⇋ 推广 f ( a ) = f ( b ) Lagrange中值定理 ⇋ 推广 g ( x ) = x Cauchy中值定理 n = 0 ↿ ⇂ 推广 Taylor中值定理 \begin{align*}
\fbox{\text{Rolle定理}}&\xleftrightharpoons[\text{推广}]{f(a)=f(b)}&\fbox{\text{Lagrange中值定理}}&\xleftrightharpoons[\text{推广}]{g(x)=x}&\fbox{\text{Cauchy中值定理}}\\
&&n=0\upharpoonleft\downharpoonright{\text{推广}}&&\\
&&\fbox{\text{Taylor中值定理}}&&
\end{align*}
Rolle 定理 f ( a ) = f ( b ) 推广 Lagrange 中值定理 n = 0 ↿⇂ 推广 Taylor 中值定理 g ( x ) = x 推广 Cauchy 中值定理
（Rolle）在[ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) f ( x ) f(x) f ( x ) f ( a ) = f ( b ) , f(a)=f(b), f ( a ) = f ( b ) , ∃ ξ ∈ ( a , b ) , s.t. f ′ ( ξ ) = 0. \exists\xi\in(a,b),\text{ s.t. }f'(\xi)=0. ∃ ξ ∈ ( a , b ) , s.t. f ′ ( ξ ) = 0.
（Lagrange）在[ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) f ( x ) f(x) f ( x ) ∃ ξ ∈ ( a , b ) , s.t. f ′ ( ξ ) = f ( b ) − f ( a ) b − a . \exists\xi\in(a,b),\text{ s.t. }f'(\xi)=\frac{f(b)-f(a)}{b-a}. ∃ ξ ∈ ( a , b ) , s.t. f ′ ( ξ ) = b − a f ( b ) − f ( a ) .
（Cauchy）在[ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) ∃ ξ ∈ ( a , b ) , s.t. f ′ ( ξ ) g ′ ( ξ ) = f ( b ) − f ( a ) g ( b ) − g ( a ) . \exists\xi\in(a,b),\text{ s.t. }\frac{f'(\xi)}{g'(\xi)}=\frac{f(b)-f(a)}{g(b)-g(a)}. ∃ ξ ∈ ( a , b ) , s.t. g ′ ( ξ ) f ′ ( ξ ) = g ( b ) − g ( a ) f ( b ) − f ( a ) .
现在考虑这三个中值定理。
Rolle定理
Rolle定理可以用来证明形如“一个函数的导数至少有两个不同零点”的命题。在寻找到f ( x ) f(x) f ( x )
e.g.2.9 \textbf{e.g.2.9 } e.g.2.9 f ( x ) f(x) f ( x ) [ 0 , 3 ] [0,3] [ 0 , 3 ] ( 0 , 3 ) (0,3) ( 0 , 3 )
2 f ( 0 ) = ∫ 0 2 f ( x ) d x = f ( 2 ) + f ( 3 ) , 2f(0)=\int_0^2f(x)\text{d}x=f(2)+f(3),
2 f ( 0 ) = ∫ 0 2 f ( x ) d x = f ( 2 ) + f ( 3 ) ,
求证：
(1)∃ η ∈ ( 0 , 2 ) , s.t. f ( η ) = f ( 0 ) ; \exists\eta\in(0,2),\text{ s.t. }f(\eta)=f(0); ∃ η ∈ ( 0 , 2 ) , s.t. f ( η ) = f ( 0 ) ; ∃ ξ ∈ ( 0 , 3 ) , s.t. f ′ ′ ( ξ ) = 0. \exists\xi\in(0,3),\text{ s.t. }f''(\xi)=0. ∃ ξ ∈ ( 0 , 3 ) , s.t. f ′′ ( ξ ) = 0.
Proof \textbf{Proof } Proof
(1)显然「f ( x ) f(x) f ( x ) [ 0 , 3 ] [0,3] [ 0 , 3 ] f ( x ) f(x) f ( x ) [ 0 , 2 ] [0,2] [ 0 , 2 ] 由积分中值定理得
（这一部分是在证明积分中值定理。若要将这一部分含有值的证明过程一般化，只要将0 0 0 2 2 2 a a a b . b. b . f ( x ) f(x) f ( x ) [ 0 , 2 ] [0,2] [ 0 , 2 ] M M M m m m
m ≤ f ( x ) ≤ M ⇒ ∫ 0 2 m d x ≤ ∫ 0 2 f ( x ) d x ≤ ∫ 0 2 M d x ⇒ m ( 2 − 0 ) ≤ ∫ 0 2 f ( x ) d x ≤ M ( 2 − 0 ) ⇒ m ≤ ∫ 0 2 f ( x ) d x 2 − 0 ≤ M . \begin{align*}
&m &\leq &f(x) &\leq &M\\
\Rightarrow&\int_0^2m\text{d}x&\leq &\int_0^2f(x)\text{d}x &\leq &\int_0^2M\text{d}x\\
\Rightarrow&m(2-0)&\leq&\int_0^2f(x)\text{d}x&\leq&M(2-0)\\
\Rightarrow&m&\leq&\frac{\int_0^2f(x)\text{d}x}{2-0}&\leq&M.
\end{align*}
⇒ ⇒ ⇒ m ∫ 0 2 m d x m ( 2 − 0 ) m ≤ ≤ ≤ ≤ f ( x ) ∫ 0 2 f ( x ) d x ∫ 0 2 f ( x ) d x 2 − 0 ∫ 0 2 f ( x ) d x ≤ ≤ ≤ ≤ M ∫ 0 2 M d x M ( 2 − 0 ) M .
因此由介值定理得
∃ η ∈ ( 0 , 2 ) , s.t. f ( η ) = ∫ 0 2 f ( x ) d x 2 − 0 . \exists\eta\in(0,2),\text{ s.t. }f(\eta)=\frac{\int_0^2f(x)\text{d}x}{2-0}.
∃ η ∈ ( 0 , 2 ) , s.t. f ( η ) = 2 − 0 ∫ 0 2 f ( x ) d x .
此时对应有2 f ( 0 ) = 已知条件 ∫ 0 2 f ( x ) d x = 2 f ( η ) . 2f(0)\xlongequal{\text{已知条件}}\int_0^2f(x)\text{d}x=2f(\eta). 2 f ( 0 ) 已知条件 ∫ 0 2 f ( x ) d x = 2 f ( η ) . f ( η ) = f ( 0 ) . f(\eta)=f(0). f ( η ) = f ( 0 ) .
Q.E.D. \textbf{Q.E.D.} Q.E.D.
(2)由(1)知f ( 0 ) = f ( η ) . f(0)=f(\eta). f ( 0 ) = f ( η ) .
而f ( x ) f(x) f ( x ) [ 0 , 3 ] [0,3] [ 0 , 3 ] ⇒ 函数平均值定理 ∃ ζ ∈ ( 2 , 3 ) , s.t. f ( ζ ) = f ( 2 ) + f ( 3 ) 2 = 已知条件 2 f ( 0 ) 2 = f ( 0 ) , \xRightarrow{\text{函数平均值定理}}\exists\zeta\in(2,3),\text{ s.t. }f(\zeta)=\frac{f(2)+f(3)}{2}\xlongequal{\text{已知条件}}\frac{2f(0)}{2}=f(0), 函数平均值定理 ∃ ζ ∈ ( 2 , 3 ) , s.t. f ( ζ ) = 2 f ( 2 ) + f ( 3 ) 已知条件 2 2 f ( 0 ) = f ( 0 ) ,
f ( 0 ) = f ( ζ ) = f ( η ) . ⇒ Rolle ∃ ε 1 ∈ ( 0 , η ) , ε 2 ∈ ( η , ζ ) , s.t. f ′ ( ε 1 ) = 0 = f ′ ( ε 2 ) . ⇒ Rolle ∃ ξ ∈ ( ε 1 , ε 2 ) , s.t. f ′ ′ ( ξ ) = 0. \begin{align*}
&f(0)=f(\zeta)=f(\eta).\\
\xRightarrow{\text{Rolle}}&\exists\varepsilon_1\in(0,\eta),\varepsilon_2\in(\eta,\zeta), \text{ s.t. }f'(\varepsilon_1)=0=f'(\varepsilon_2).\\
\xRightarrow{\text{Rolle}}&\exists\xi\in(\varepsilon_1,\varepsilon_2),\text{ s.t. }f''(\xi)=0.
\end{align*}
Rolle Rolle f ( 0 ) = f ( ζ ) = f ( η ) . ∃ ε 1 ∈ ( 0 , η ) , ε 2 ∈ ( η , ζ ) , s.t. f ′ ( ε 1 ) = 0 = f ′ ( ε 2 ) . ∃ ξ ∈ ( ε 1 , ε 2 ) , s.t. f ′′ ( ξ ) = 0.
显然区间( ε 1 , ε 2 ) (\varepsilon_1, \varepsilon_2) ( ε 1 , ε 2 ) ( 0 , 3 ) (0,3) ( 0 , 3 ) ∃ ξ ∈ ( 0 , 3 ) , s.t. f ′ ′ ( ξ ) = 0 \exists\xi\in(0,3),\text{ s.t. }f''(\xi)=0 ∃ ξ ∈ ( 0 , 3 ) , s.t. f ′′ ( ξ ) = 0
Q.E.D. \textbf{Q.E.D.} Q.E.D.
Rolle定理一般用来证明形如“F ( ξ , f ( ξ ) , f ′ ( ξ ) ) = 0 F\big(\xi,f(\xi),f'(\xi)\big)=0 F ( ξ , f ( ξ ) , f ′ ( ξ ) ) = 0
构造辅助函数的本质是对多个函数相乘/除的复合函数求导结果进行还原。下面列举了一些常见的含f ( x ) f(x) f ( x )
待证结论
辅助函数
f ′ ( ξ ) = k f'(\xi)=k f ′ ( ξ ) = k
F ( x ) = f ( x ) − k x F(x)=f(x)-kx F ( x ) = f ( x ) − k x
f ′ ( ξ ) g ( ξ ) + f ( ξ ) g ′ ( ξ ) = 0 f'(\xi)g(\xi)+f(\xi)g'(\xi)=0 f ′ ( ξ ) g ( ξ ) + f ( ξ ) g ′ ( ξ ) = 0
F ( x ) = f ( x ) g ( x ) F(x)=f(x)g(x) F ( x ) = f ( x ) g ( x )
f ′ ( ξ ) g ( ξ ) − g ′ ( ξ ) f ( ξ ) = 0 f'(\xi)g(\xi)-g'(\xi)f(\xi)=0 f ′ ( ξ ) g ( ξ ) − g ′ ( ξ ) f ( ξ ) = 0
F ( x ) = f ( x ) g ( x ) F(x)=\frac{f(x)}{g(x)} F ( x ) = g ( x ) f ( x )
f ′ ( ξ ) + λ f ( ξ ) = 0 f'(\xi)+\lambda f(\xi)=0 f ′ ( ξ ) + λ f ( ξ ) = 0
F ( x ) = f ( x ) e λ x F(x)=f(x)e^{\lambda x} F ( x ) = f ( x ) e λ x
ξ f ′ ( ξ ) + k f ( ξ ) = 0 \xi f'(\xi)+kf(\xi)=0 ξ f ′ ( ξ ) + k f ( ξ ) = 0
F ( x ) = f ( x ) x k F(x)=f(x)x^k F ( x ) = f ( x ) x k
f ′ ( ξ ) + g ( ξ ) f ( ξ ) = 0 f'(\xi)+g(\xi)f(\xi)=0 f ′ ( ξ ) + g ( ξ ) f ( ξ ) = 0
F ( x ) = e ∫ g ( x ) d x f ( x ) F(x)=e^{\int g(x)\text{d}x}f(x) F ( x ) = e ∫ g ( x ) d x f ( x )
ξ f ′ ( ξ ) + k f ( ξ ) = 0 \xi f'(\xi)+kf(\xi)=0 ξ f ′ ( ξ ) + k f ( ξ ) = 0
F ( x ) = x k f ( x ) F(x)=x^kf(x) F ( x ) = x k f ( x )
只要让上方表格中的辅助函数F ′ ( x ) = 0 , F'(x)=0, F ′ ( x ) = 0 ,
e.x.2.8 \textbf{e.x.2.8 } e.x.2.8 f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) [ a , b ] [a,b] [ a , b ] g ′ ′ ( x ) ≠ 0 , f ( a ) = f ( b ) = g ( a ) = g ( b ) = 0 , g''(x)≠0,f(a)=f(b)=g(a)=g(b)=0, g ′′ ( x ) = 0 , f ( a ) = f ( b ) = g ( a ) = g ( b ) = 0 ,
(1)∀ x ∈ ( a , b ) , g ( x ) ≠ 0 ; \forall x\in(a,b), g(x)≠0; ∀ x ∈ ( a , b ) , g ( x ) = 0 ;
(2)∃ ξ ∈ ( a , b ) , s.t. f ( ξ ) g ( ξ ) = f ′ ′ ( ξ ) g ′ ′ ( ξ ) . \exists\xi\in(a,b),\text{ s.t. }\frac{f(\xi)}{g(\xi)}=\frac{f''(\xi)}{g''(\xi)}. ∃ ξ ∈ ( a , b ) , s.t. g ( ξ ) f ( ξ ) = g ′′ ( ξ ) f ′′ ( ξ ) .
Analyses \textbf{Analyses } Analyses
(1)思考在g ( x ) = 0 g(x)=0 g ( x ) = 0 g ( a ) = g ( b ) = 0 g(a)=g(b)=0 g ( a ) = g ( b ) = 0
(2)待证命题的等式可以变形为f ( x ) g ′ ′ ( x ) − g ( x ) f ′ ′ ( x ) = 0. f(x)g''(x)-g(x)f''(x)=0. f ( x ) g ′′ ( x ) − g ( x ) f ′′ ( x ) = 0. f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) ( u v ) ′ = u ′ v + u v ′ , (uv)'=u'v+uv', ( uv ) ′ = u ′ v + u v ′ , f ( x ) g ′ ′ ( x ) f(x)g''(x) f ( x ) g ′′ ( x ) ( f ( x ) g ′ ( x ) ) ′ \big(f(x)g'(x)\big)' ( f ( x ) g ′ ( x ) ) ′ u v ′ uv' u v ′ g ( x ) f ′ ′ ( x ) g(x)f''(x) g ( x ) f ′′ ( x ) ( g ( x ) f ′ ( x ) ) ′ \big(g(x)f'(x)\big)' ( g ( x ) f ′ ( x ) ) ′ u v ′ uv' u v ′ u ′ v u'v u ′ v F ( x ) = f ( x ) g ′ ( x ) − f ′ ( x ) g ( x ) F(x)=f(x)g'(x)-f'(x)g(x) F ( x ) = f ( x ) g ′ ( x ) − f ′ ( x ) g ( x ) F ′ ( ξ ) = 0 F'(\xi)=0 F ′ ( ξ ) = 0
Proof \textbf{Proof } Proof
(1)假设∃ m ∈ ( a , b ) , g ( m ) = 0. \exists m\in(a,b), g(m)=0. ∃ m ∈ ( a , b ) , g ( m ) = 0. g ( a ) = g ( m ) = g ( b ) = 0 ; g(a)=g(m)=g(b)=0; g ( a ) = g ( m ) = g ( b ) = 0 ; ∃ ε 1 ∈ ( a , m ) , ε 2 ∈ ( m , b ) , s.t. g ′ ( ε 1 ) = g ′ ( ε 2 ) = 0. \exists\varepsilon_1\in(a,m),\varepsilon_2\in(m,b),\text{ s.t. }g'(\varepsilon_1)=g'(\varepsilon_2)=0. ∃ ε 1 ∈ ( a , m ) , ε 2 ∈ ( m , b ) , s.t. g ′ ( ε 1 ) = g ′ ( ε 2 ) = 0. ∃ ζ ∈ ( ε 1 , ε 2 ) , s.t. g ′ ′ ( ζ ) = 0 , \exists\zeta\in(\varepsilon_1,\varepsilon_2),\text{ s.t. }g''(\zeta)=0, ∃ ζ ∈ ( ε 1 , ε 2 ) , s.t. g ′′ ( ζ ) = 0 , g ′ ′ ( x ) ≠ 0 g''(x)≠0 g ′′ ( x ) = 0
Q.E.D. \textbf{Q.E.D.} Q.E.D.
(2)构造函数F ( x ) = f ( x ) g ′ ( x ) − g ( x ) f ′ ( x ) . F(x)=f(x)g'(x)-g(x)f'(x). F ( x ) = f ( x ) g ′ ( x ) − g ( x ) f ′ ( x ) . F ( a ) = F ( b ) = 0 , F(a)=F(b)=0, F ( a ) = F ( b ) = 0 ,
∃ ξ ∈ ( a , b ) , s.t. F ′ ( ξ ) = 0. \exists\xi\in(a,b),\text{ s.t. }F'(\xi)=0.
∃ ξ ∈ ( a , b ) , s.t. F ′ ( ξ ) = 0.
此时可知
F ′ ( ξ ) = 0 ⇒ f ′ ( x ) g ′ ( x ) + f ( x ) g ′ ′ ( x ) − ( g ′ ( x ) f ′ ( x ) + g ( x ) f ′ ′ ( x ) ) = 0 ⇒ f ( x ) g ′ ′ ( x ) − g ( x ) f ′ ′ ( x ) = 0 ⇒ f ( x ) g ′ ′ ( x ) = g ( x ) f ′ ′ ( x ) ⇒ f ( x ) g ( x ) = f ′ ′ ( x ) g ′ ′ ( x ) . \begin{align*}
F'(\xi)=0&\Rightarrow f'(x)g'(x)+f(x)g''(x)-(g'(x)f'(x)+g(x)f''(x))=0\\
&\Rightarrow f(x)g''(x)-g(x)f''(x)=0\Rightarrow f(x)g''(x)=g(x)f''(x)\\
&\Rightarrow \frac{f(x)}{g(x)}=\frac{f''(x)}{g''(x)}.
\end{align*}
F ′ ( ξ ) = 0 ⇒ f ′ ( x ) g ′ ( x ) + f ( x ) g ′′ ( x ) − ( g ′ ( x ) f ′ ( x ) + g ( x ) f ′′ ( x )) = 0 ⇒ f ( x ) g ′′ ( x ) − g ( x ) f ′′ ( x ) = 0 ⇒ f ( x ) g ′′ ( x ) = g ( x ) f ′′ ( x ) ⇒ g ( x ) f ( x ) = g ′′ ( x ) f ′′ ( x ) .
故原命题得证。
Q.E.D. \textbf{Q.E.D.} Q.E.D.
e.x.2.9 \textbf{e.x.2.9 } e.x.2.9 f ( x ) f(x) f ( x ) [ 0 , 1 ] [0,1] [ 0 , 1 ] ( 0 , 1 ) (0,1) ( 0 , 1 ) f ( 1 ) = 3 ∫ 0 1 3 e 1 − x 2 f ( x ) d x , f(1)=3\int_0^\frac{1}{3}e^{1-x^2}f(x)\text{d}x, f ( 1 ) = 3 ∫ 0 3 1 e 1 − x 2 f ( x ) d x , ∃ ξ ∈ ( 0 , 1 ) , f ′ ( ξ ) = 2 ξ f ( ξ ) . \exists\xi\in(0,1),f'(\xi)=2\xi f(\xi). ∃ ξ ∈ ( 0 , 1 ) , f ′ ( ξ ) = 2 ξ f ( ξ ) .
Analyses \textbf{Analyses } Analyses f ′ ( ξ ) − 2 ξ f ( ξ ) = 0. f'(\xi)-2\xi f(\xi)=0. f ′ ( ξ ) − 2 ξ f ( ξ ) = 0. ( f ′ ( ξ ) − 2 ξ f ( ξ ) ) ( e − ξ 2 ) = ( f ( ξ ) e − ξ 2 ) ′ , \big(f'(\xi)-2\xi f(\xi)\big)(e^{-\xi^2})=(f(\xi)e^{-\xi^2})', ( f ′ ( ξ ) − 2 ξ f ( ξ ) ) ( e − ξ 2 ) = ( f ( ξ ) e − ξ 2 ) ′ , F ( x ) = e − x 2 f ( x ) F(x)=e^{-x^2}f(x) F ( x ) = e − x 2 f ( x ) ∃ ξ ∈ ( 0 , 1 ) , s.t. F ′ ( ξ ) = 0 \exists\xi\in(0,1),\text{ s.t. }F'(\xi)=0 ∃ ξ ∈ ( 0 , 1 ) , s.t. F ′ ( ξ ) = 0 Proof \textbf{Proof } Proof g ( x ) = e 1 − x 2 f ( x ) g(x)=e^{1-x^2}f(x) g ( x ) = e 1 − x 2 f ( x ) η ∈ ( 0 , 1 3 ) \eta\in(0,\frac{1}{3}) η ∈ ( 0 , 3 1 ) g ( η ) ( b − a ) = ∫ a b g ( x ) d x . g(\eta)(b-a)=\int_a^bg(x)\text{d}x. g ( η ) ( b − a ) = ∫ a b g ( x ) d x .
f ( 1 ) = 3 ∫ 0 1 3 e 1 − x 2 f ( x ) d x ⇒ ∃ η ∈ ( 0 , 1 ) , s.t. f ( 1 ) = 3 ( 1 3 − 0 ) e 1 − η 2 f ( η ) = e 1 − η 2 f ( η ) . f(1)=3\int_0^\frac{1}{3}e^{1-x^2}f(x)\text{d}x\Rightarrow\exists\eta\in(0,1),\text{ s.t. }f(1)=3(\frac{1}{3}-0)e^{1-\eta^2}f(\eta)=e^{1-\eta^2}f(\eta).
f ( 1 ) = 3 ∫ 0 3 1 e 1 − x 2 f ( x ) d x ⇒ ∃ η ∈ ( 0 , 1 ) , s.t. f ( 1 ) = 3 ( 3 1 − 0 ) e 1 − η 2 f ( η ) = e 1 − η 2 f ( η ) .
令F ( x ) = e − x 2 f ( x ) . F(x)=e^{-x^2}f(x). F ( x ) = e − x 2 f ( x ) .
F ( 1 ) = e − 1 f ( 1 ) = e − 1 ⋅ ( e 1 − η 2 f ( η ) ) = e − η 2 f ( η ) = F ( η ) . F(1)=e^{-1}f(1)=e^{-1}\cdot(e^{1-\eta^2}f(\eta))=e^{-\eta^2}f(\eta)=F(\eta).
F ( 1 ) = e − 1 f ( 1 ) = e − 1 ⋅ ( e 1 − η 2 f ( η )) = e − η 2 f ( η ) = F ( η ) .
也即∃ η ∈ ( 0 , 1 ) , s.t. F ( 1 ) = F ( η ) . \exists\eta\in(0,1),\text{ s.t. }F(1)=F(\eta). ∃ η ∈ ( 0 , 1 ) , s.t. F ( 1 ) = F ( η ) .
∃ ξ ∈ ( η , 1 ) , s.t. F ′ ( ξ ) = 0 , \exists\xi\in(\eta,1),\text{ s.t. }F'(\xi)=0,
∃ ξ ∈ ( η , 1 ) , s.t. F ′ ( ξ ) = 0 ,
此时也就有( e − ξ 2 f ( ξ ) ) ′ = 0 , (e^{-\xi^2}f(\xi))'=0, ( e − ξ 2 f ( ξ ) ) ′ = 0 , f ′ ( ξ ) e − ξ 2 − 2 ξ f ( ξ ) e − ξ 2 = 0. f'(\xi)e^{-\xi^2}-2\xi f(\xi)e^{-\xi^2}=0. f ′ ( ξ ) e − ξ 2 − 2 ξ f ( ξ ) e − ξ 2 = 0. f ′ ( ξ ) − 2 ξ f ( ξ ) = 0 , f'(\xi)-2\xi f(\xi)=0, f ′ ( ξ ) − 2 ξ f ( ξ ) = 0 ,
Q.E.D. \textbf{Q.E.D.} Q.E.D.
e.g.2.10 \textbf{e.g.2.10 } e.g.2.10 f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) ∃ ε ∈ ( a , b ) , s.t. f ′ ( ε ) = f ( ε ) − f ( a ) b − ε . \exists\varepsilon\in(a,b),\text{ s.t. }f'(\varepsilon)=\frac{f(\varepsilon)-f(a)}{b-\varepsilon}. ∃ ε ∈ ( a , b ) , s.t. f ′ ( ε ) = b − ε f ( ε ) − f ( a ) .
Analyses \textbf{Analyses } Analyses ( b − ε ) f ′ ( ε ) − ( f ( ε ) − f ( a ) ) = 0. (b-\varepsilon)f'(\varepsilon)-(f(\varepsilon)-f(a))=0. ( b − ε ) f ′ ( ε ) − ( f ( ε ) − f ( a )) = 0. ( f ( ε ) − f ( a ) ) ′ ( b − ε ) + ( b − ε ) ′ ( f ( ε ) − f ( a ) ) = ( ( f ( ε ) − f ( a ) ) ( b − ε ) ) ′ , (f(\varepsilon)-f(a))'(b-\varepsilon)+(b-\varepsilon)'(f(\varepsilon)-f(a))=\bigg(\big(f(\varepsilon)-f(a)\big)(b-\varepsilon)\bigg)', ( f ( ε ) − f ( a ) ) ′ ( b − ε ) + ( b − ε ) ′ ( f ( ε ) − f ( a )) = ( ( f ( ε ) − f ( a ) ) ( b − ε ) ) ′ , F ( x ) = ( f ( x ) − f ( a ) ) ( b − x ) ; F(x)=\big(f(x)-f(a)\big)\big(b-x\big); F ( x ) = ( f ( x ) − f ( a ) ) ( b − x ) ; ∃ ε , s.t. F ′ ( ε ) = 0 , \exists\varepsilon,\text{ s.t. }F'(\varepsilon)=0, ∃ ε , s.t. F ′ ( ε ) = 0 ,
Proof \textbf{Proof } Proof F ( x ) = ( f ( x ) − f ( a ) ) ( b − x ) . F(x)=\big(f(x)-f(a)\big)\big(b-x\big). F ( x ) = ( f ( x ) − f ( a ) ) ( b − x ) . F ( x ) F(x) F ( x ) ( a , b ) (a,b) ( a , b ) [ a , b ] [a,b] [ a , b ] F ( a ) = 0 = F ( b ) , F(a)=0=F(b), F ( a ) = 0 = F ( b ) ,
∃ ε ∈ ( a , b ) , s.t. F ′ ( ε ) = 0. \exists\varepsilon\in(a,b),\text{ s.t. }F'(\varepsilon)=0.
∃ ε ∈ ( a , b ) , s.t. F ′ ( ε ) = 0.
对应有f ′ ( ε ) ( b − ε ) − ( f ( ε ) − f ( a ) ) = 0 , f'(\varepsilon)(b-\varepsilon)-\big(f(\varepsilon)-f(a)\big)=0, f ′ ( ε ) ( b − ε ) − ( f ( ε ) − f ( a ) ) = 0 ,
Q.E.D. \textbf{Q.E.D.} Q.E.D.
一部分待证明等式中并没有很明显的“含函数项+含导数项”特征，但本质仍然相同。例如，等式中可能会是“含函数项+含积分项”或“含一阶导数项+含二阶导数项”；这些情况利用各种定义知道它们仍然属于“含函数项+含导数项”的变形。
e.g.2.11 \textbf{e.g.2.11 } e.g.2.11 f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] ∃ ε ∈ ( a , b ) , s.t. ( ε − b ) f ( ε ) + ∫ a ε f ( x ) d x = 0. \exists\varepsilon\in(a,b),\text{ s.t. }(\varepsilon-b)f(\varepsilon)+\int_a^\varepsilon f(x)\text{d}x=0. ∃ ε ∈ ( a , b ) , s.t. ( ε − b ) f ( ε ) + ∫ a ε f ( x ) d x = 0.
Analyses \textbf{Analyses } Analyses ( ∫ a ε f ( x ) d x ) ′ = f ( x ) ∣ x = ε \big(\int_a^\varepsilon f(x)\text{d}x\big)'=f(x)\big|_{x=\varepsilon} ( ∫ a ε f ( x ) d x ) ′ = f ( x ) x = ε a a a x = a x=a x = a f ( x ) f(x) f ( x ) F ( x ) = ( x − b ) ∫ a x f ( x ) d x . F(x)=(x-b)\int^x_af(x)\text{d}x. F ( x ) = ( x − b ) ∫ a x f ( x ) d x . ∃ ε ∈ ( a , b ) s.t. F ′ ( ε ) = 0 \exists\varepsilon\in(a,b)\text{ s.t. }F'(\varepsilon)=0 ∃ ε ∈ ( a , b ) s.t. F ′ ( ε ) = 0
Proof \textbf{Proof } Proof F ( x ) = ( x − b ) ∫ a x f ( x ) d x , F(x)=(x-b)\int^x_af(x)\text{d}x, F ( x ) = ( x − b ) ∫ a x f ( x ) d x , F ( a ) = 0 = F ( b ) . F(a)=0=F(b). F ( a ) = 0 = F ( b ) .
∃ ε ∈ ( a , b ) , s.t. F ′ ( ε ) = 0. \exists\varepsilon\in(a,b),\text{ s.t. }F'(\varepsilon)=0.
∃ ε ∈ ( a , b ) , s.t. F ′ ( ε ) = 0.
对等式F ′ ( ε ) = 0 F'(\varepsilon)=0 F ′ ( ε ) = 0
( ( ε − b ) ∫ a ε f ( x ) d x ) ′ = 0 ⇒ ( ε − b ) ( ∫ a ε f ( x ) d x ) ′ + ( ε − b ) ′ ∫ a ε f ( x ) d x = 0 ⇒ ( ε − b ) ( f ( ε ) − 0 ) + ∫ a ε f ( x ) d x = 0 ⇒ ( ε − b ) f ( ε ) + ∫ a ε f ( x ) d x = 0. \begin{align*}
\bigg((\varepsilon-b)\int^\varepsilon_af(x)\text{d}x\bigg)'=0&\Rightarrow(\varepsilon-b)\bigg(\int^\varepsilon_af(x)\text{d}x\bigg)'+(\varepsilon-b)'\int^\varepsilon_af(x)\text{d}x=0\\
&\Rightarrow(\varepsilon-b)(f(\varepsilon)-0)+\int^\varepsilon_af(x)\text{d}x=0\\
&\Rightarrow(\varepsilon-b)f(\varepsilon)+\int^\varepsilon_af(x)\text{d}x=0.
\end{align*}
( ( ε − b ) ∫ a ε f ( x ) d x ) ′ = 0 ⇒ ( ε − b ) ( ∫ a ε f ( x ) d x ) ′ + ( ε − b ) ′ ∫ a ε f ( x ) d x = 0 ⇒ ( ε − b ) ( f ( ε ) − 0 ) + ∫ a ε f ( x ) d x = 0 ⇒ ( ε − b ) f ( ε ) + ∫ a ε f ( x ) d x = 0.
从而原命题得证。
Q.E.D. \textbf{Q.E.D.} Q.E.D.
e.x.2.10 \textbf{e.x.2.10 } e.x.2.10 f ( x ) f(x) f ( x ) [ 0 , 1 ] [0,1] [ 0 , 1 ] f ( 0 ) = f ( 1 ) , f(0)=f(1), f ( 0 ) = f ( 1 ) , ∃ ξ ∈ ( 0 , 1 ) , s.t. 2 f ′ ( ξ ) + ( ξ − 1 ) f ′ ′ ( ξ ) = 0. \exists\xi\in(0,1),\text{ s.t. }2f'(\xi)+(\xi-1)f''(\xi)=0. ∃ ξ ∈ ( 0 , 1 ) , s.t. 2 f ′ ( ξ ) + ( ξ − 1 ) f ′′ ( ξ ) = 0.
Analyses \textbf{Analyses } Analyses g ( x ) = f ′ ( x ) , g(x)=f'(x), g ( x ) = f ′ ( x ) , 2 ξ − 1 g ( ξ ) + g ′ ( ξ ) = 0. \frac{2}{\xi-1}g(\xi)+g'(\xi)=0. ξ − 1 2 g ( ξ ) + g ′ ( ξ ) = 0. F ( x ) = e ∫ 2 x − 1 d x g ( x ) , F(x)=e^{\int\frac{2}{x-1}\text{d}x}g(x), F ( x ) = e ∫ x − 1 2 d x g ( x ) , F ′ ( ξ ) = 0. F'(\xi)=0. F ′ ( ξ ) = 0.
Proof \textbf{Proof } Proof f ( 0 ) = f ( 1 ) f(0)=f(1) f ( 0 ) = f ( 1 ) f ( x ) f(x) f ( x )
∃ ζ ∈ ( 0 , 1 ) , s.t. f ′ ( ζ ) = 0. \exists\zeta\in(0,1),\text{ s.t. }f'(\zeta)=0.
∃ ζ ∈ ( 0 , 1 ) , s.t. f ′ ( ζ ) = 0.
随后构造函数F ( x ) = e ∫ 2 x − 1 d x f ′ ( x ) , F(x)=e^{\int\frac{2}{x-1}\text{d}x}f'(x), F ( x ) = e ∫ x − 1 2 d x f ′ ( x ) , F ( x ) = e 2 ln ( x − 1 ) f ′ ( x ) , F(x)=e^{2\ln(x-1)}f'(x), F ( x ) = e 2 l n ( x − 1 ) f ′ ( x ) , F ( x ) = ( x − 1 ) 2 f ′ ( x ) . F(x)=(x-1)^2f'(x). F ( x ) = ( x − 1 ) 2 f ′ ( x ) . F ( ζ ) = 0 = F ( 1 ) , F(\zeta)=0=F(1), F ( ζ ) = 0 = F ( 1 ) ,
∃ ξ ∈ ( ζ , 1 ) , s.t. F ′ ( ξ ) = 0. ⇒ ∃ ξ ∈ ( 0 , 1 ) , s.t. F ′ ( ξ ) = 0. \begin{align*}
&\exists\xi\in(\zeta,1),\text{ s.t. }F'(\xi)=0.\\
\Rightarrow&\exists\xi\in(0,1),\text{ s.t. }F'(\xi)=0.
\end{align*}
⇒ ∃ ξ ∈ ( ζ , 1 ) , s.t. F ′ ( ξ ) = 0. ∃ ξ ∈ ( 0 , 1 ) , s.t. F ′ ( ξ ) = 0.
这时对应有2 ( ξ − 1 ) f ′ ( ξ ) + ( ξ − 1 ) 2 f ′ ′ ( ξ ) = 0 , 2(\xi-1)f'(\xi)+(\xi-1)^2f''(\xi)=0, 2 ( ξ − 1 ) f ′ ( ξ ) + ( ξ − 1 ) 2 f ′′ ( ξ ) = 0 , ξ − 1 \xi-1 ξ − 1 2 f ′ ( ξ ) + ( ξ − 1 ) f ′ ′ ( ξ ) = 0. 2f'(\xi)+(\xi-1)f''(\xi)=0. 2 f ′ ( ξ ) + ( ξ − 1 ) f ′′ ( ξ ) = 0.
Q.E.D. \textbf{Q.E.D.} Q.E.D.
e.x.2.11 \textbf{e.x.2.11 } e.x.2.11 f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) f ( a ) = f ( b ) = 0 , f ′ ( a ) f ′ ( b ) > 0 , f(a)=f(b)=0,f'(a)f'(b)>0, f ( a ) = f ( b ) = 0 , f ′ ( a ) f ′ ( b ) > 0 , ∃ ξ ∈ ( a , b ) , s.t. f ′ ′ ( ξ ) = f ( ξ ) . \exists\xi\in(a,b),\text{ s.t. }f''(\xi)=f(\xi). ∃ ξ ∈ ( a , b ) , s.t. f ′′ ( ξ ) = f ( ξ ) .
Analyses \textbf{Analyses } Analyses f ( ξ ) − f ′ ′ ( ξ ) = 0. f(\xi)-f''(\xi)=0. f ( ξ ) − f ′′ ( ξ ) = 0. f ( ξ ) − f ′ ′ ( ξ ) = ( f ( ξ ) − f ′ ( ξ ) ) + ( f ′ ( ξ ) − f ′ ′ ( ξ ) ) ; f(\xi)-f''(\xi)=\big(f(\xi)-f'(\xi)\big)+\big(f'(\xi)-f''(\xi)\big); f ( ξ ) − f ′′ ( ξ ) = ( f ( ξ ) − f ′ ( ξ ) ) + ( f ′ ( ξ ) − f ′′ ( ξ ) ) ; g ( x ) = f ( x ) − f ′ ( x ) , g(x)=f(x)-f'(x), g ( x ) = f ( x ) − f ′ ( x ) , g ( ξ ) + g ′ ( ξ ) = 0 , g(\xi)+g'(\xi)=0, g ( ξ ) + g ′ ( ξ ) = 0 , G ( x ) = e x g ( x ) G(x)=e^xg(x) G ( x ) = e x g ( x ) a , b a,b a , b G ( x ) G(x) G ( x ) 0 0 0 g ( x ) = 0 , g(x)=0, g ( x ) = 0 , ∃ ε ∈ ( a , b ) , s.t. f ( x ) = f ′ ( x ) \exists\varepsilon\in(a,b),\text{ s.t. }f(x)=f'(x) ∃ ε ∈ ( a , b ) , s.t. f ( x ) = f ′ ( x ) F ( x ) = e − x f ( x ) , F(x)=e^{-x}f(x), F ( x ) = e − x f ( x ) , G ( x ) G(x) G ( x ) a , b a,b a , b
Proof \textbf{Proof } Proof f ′ ( a ) f ′ ( b ) > 0 , f'(a)f'(b)>0, f ′ ( a ) f ′ ( b ) > 0 , f ′ ( a + ) f ′ ( b − ) > 0 f'(a^+)f'(b^-)>0 f ′ ( a + ) f ′ ( b − ) > 0 f ′ ( a + ) > 0 , f'(a^+)>0, f ′ ( a + ) > 0 , f ′ ( b − ) > 0. f'(b^-)>0. f ′ ( b − ) > 0. lim x → a + f ( x ) − f ( a ) x − a > 0 , \lim_{x\rightarrow a^+}\frac{f(x)-f(a)}{x-a}>0, lim x → a + x − a f ( x ) − f ( a ) > 0 ,
∃ μ ∈ ( a , a + δ 1 ) , s.t. f ( μ ) − f ( a ) μ − a > 0. ⇒ f ( μ ) − f ( a ) > 0 ⇒ f ( μ ) > f ( a ) = 0. \exists\mu\in(a,a+\delta_1),\text{ s.t. }\frac{f(\mu)-f(a)}{\mu-a}>0.\Rightarrow f(\mu)-f(a)>0\Rightarrow f(\mu)>f(a)=0.
∃ μ ∈ ( a , a + δ 1 ) , s.t. μ − a f ( μ ) − f ( a ) > 0. ⇒ f ( μ ) − f ( a ) > 0 ⇒ f ( μ ) > f ( a ) = 0.
同理，∃ η ∈ ( b − δ 2 , b ) , s.t. f ( η ) < 0. \exists\eta\in(b-\delta_2,b),\text{ s.t. }f(\eta)<0. ∃ η ∈ ( b − δ 2 , b ) , s.t. f ( η ) < 0.
∃ ζ ∈ ( μ , η ) , s.t. f ( ζ ) = 0. \exists\zeta\in(\mu,\eta),\text{ s.t. }f(\zeta)=0.
∃ ζ ∈ ( μ , η ) , s.t. f ( ζ ) = 0.
现在构造函数F ( x ) = e − x f ( x ) , g ( x ) = f ( x ) + f ′ ( x ) , G ( x ) = e x g ( x ) . F(x)=e^{-x}f(x),g(x)=f(x)+f'(x),G(x)=e^xg(x). F ( x ) = e − x f ( x ) , g ( x ) = f ( x ) + f ′ ( x ) , G ( x ) = e x g ( x ) . F ( a ) = F ( ζ ) = F ( b ) = 0 F(a)=F(\zeta)=F(b)=0 F ( a ) = F ( ζ ) = F ( b ) = 0
∃ ε 1 ∈ ( a , ζ ) , ε 2 ∈ ( ζ , b ) , s.t. F ′ ( ε 1 ) = 0 = F ′ ( ε 2 ) . \exists\varepsilon_1\in(a,\zeta),\varepsilon_2\in(\zeta,b),\text{ s.t. }F'(\varepsilon_1)=0=F'(\varepsilon_2).
∃ ε 1 ∈ ( a , ζ ) , ε 2 ∈ ( ζ , b ) , s.t. F ′ ( ε 1 ) = 0 = F ′ ( ε 2 ) .
这分别对应有e − x f ′ ( ε 1 ) − e − x f ( ε 1 ) = 0 ⇒ f ( ε 1 ) = f ′ ( ε 1 ) , e − x f ′ ( ε 2 ) − e − x f ( ε 2 ) = 0 ⇒ f ( ε 2 ) = f ′ ( ε 2 ) . e^{-x}f'(\varepsilon_1)-e^{-x}f(\varepsilon_1)=0\Rightarrow f(\varepsilon_1)=f'(\varepsilon_1),e^{-x}f'(\varepsilon_2)-e^{-x}f(\varepsilon_2)=0\Rightarrow f(\varepsilon_2)=f'(\varepsilon_2). e − x f ′ ( ε 1 ) − e − x f ( ε 1 ) = 0 ⇒ f ( ε 1 ) = f ′ ( ε 1 ) , e − x f ′ ( ε 2 ) − e − x f ( ε 2 ) = 0 ⇒ f ( ε 2 ) = f ′ ( ε 2 ) . G ( ε 1 ) = G ( ε 2 ) . G(\varepsilon_1)=G(\varepsilon_2). G ( ε 1 ) = G ( ε 2 ) .
∃ ξ ∈ ( ε 1 , ε 2 ) , G ′ ( ξ ) = 0. \exists\xi\in(\varepsilon_1,\varepsilon_2),G'(\xi)=0.
∃ ξ ∈ ( ε 1 , ε 2 ) , G ′ ( ξ ) = 0.
此时e − ξ ( f ′ ′ ( ξ ) − f ( ξ ) ) = 0 e^{-\xi}\big(f''(\xi)-f(\xi)\big)=0 e − ξ ( f ′′ ( ξ ) − f ( ξ ) ) = 0
Q.E.D. \textbf{Q.E.D.} Q.E.D.
Lagrange中值定理
毫无疑问Lagrange中值定理可以证明定理式本身或其变形的式子。
e.g.2.12 \textbf{e.g.2.12 } e.g.2.12 f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) ∃ ξ ∈ ( a , b ) , s.t. b f ( b ) − a f ( a ) b − a = f ( ξ ) + ξ f ′ ( ξ ) . \exists\xi\in(a,b),\text{ s.t. }\frac{bf(b)-af(a)}{b-a}=f(\xi)+\xi f'(\xi). ∃ ξ ∈ ( a , b ) , s.t. b − a b f ( b ) − a f ( a ) = f ( ξ ) + ξ f ′ ( ξ ) .
Analyses \textbf{Analyses } Analyses f ( ξ ) + ξ f ′ ( ξ ) = ( x f ( x ) ) ′ ∣ x = ξ , f(\xi)+\xi f'(\xi)=\big(xf(x)\big)'\Big|_{x=\xi}, f ( ξ ) + ξ f ′ ( ξ ) = ( x f ( x ) ) ′ x = ξ ,
Proof \textbf{Proof } Proof g ( x ) = x f ( x ) . g(x)=xf(x). g ( x ) = x f ( x ) . b f ( b ) − a f ( a ) b − a = g ( b ) − g ( a ) b − a . \frac{bf(b)-af(a)}{b-a}=\frac{g(b)-g(a)}{b-a}. b − a b f ( b ) − a f ( a ) = b − a g ( b ) − g ( a ) .
∃ ξ ∈ ( a , b ) , s.t. g ( b ) − g ( a ) b − a = g ′ ( ξ ) , \exists\xi\in(a,b),\text{ s.t. }\frac{g(b)-g(a)}{b-a}=g'(\xi),
∃ ξ ∈ ( a , b ) , s.t. b − a g ( b ) − g ( a ) = g ′ ( ξ ) ,
因此b f ( b ) − a f ( a ) b − a = g ′ ( ξ ) = f ( ξ ) + ξ f ′ ( ξ ) . \frac{bf(b)-af(a)}{b-a}=g'(\xi)=f(\xi)+\xi f'(\xi). b − a b f ( b ) − a f ( a ) = g ′ ( ξ ) = f ( ξ ) + ξ f ′ ( ξ ) .
Q.E.D. \textbf{Q.E.D.} Q.E.D.
e.x.2.12 \textbf{e.x.2.12 } e.x.2.12 f ( x ) f(x) f ( x ) [ 0 , 1 ] [0,1] [ 0 , 1 ] ∃ ξ ∈ ( 0 , 1 ) , s.t. π 4 ( 1 + ξ 2 ) f ( 1 ) = f ( ξ ) + ( 1 + ξ 2 ) arctan ξ ⋅ f ′ ( ξ ) . \exists\xi\in(0,1),\text{ s.t. }\frac{\pi}{4}(1+\xi^2)f(1)=f(\xi)+(1+\xi^2)\arctan\xi\cdot f'(\xi). ∃ ξ ∈ ( 0 , 1 ) , s.t. 4 π ( 1 + ξ 2 ) f ( 1 ) = f ( ξ ) + ( 1 + ξ 2 ) arctan ξ ⋅ f ′ ( ξ ) .
Analyses \textbf{Analyses } Analyses π 4 f ( 1 ) = 1 1 + ξ 2 f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . \frac{\pi}{4}f(1)=\frac{1}{1+\xi^2}f(\xi)+\arctan\xi\cdot f'(\xi). 4 π f ( 1 ) = 1 + ξ 2 1 f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . ( f ( x ) arctan x ) ′ ∣ x = ξ , \big(f(x)\arctan x\big)'\Big|_{x=\xi}, ( f ( x ) arctan x ) ′ x = ξ ,
Proof \textbf{Proof } Proof F ( x ) = f ( x ) arctan x . F(x)=f(x)\arctan x. F ( x ) = f ( x ) arctan x .
∃ ξ ∈ ( 0 , 1 ) , s.t. F ( 1 ) − F ( 0 ) 1 − 0 = F ′ ( ξ ) . ⇒ ∃ ξ ∈ ( 0 , 1 ) , s.t. f ( 1 ) arctan 1 − f ( 0 ) arctan 0 = ( arctan ξ ) ′ f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . ⇒ ∃ ξ ∈ ( 0 , 1 ) , s.t. π 4 f ( 1 ) − 0 = 1 1 + ξ 2 f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . \begin{align*}
&\exists\xi\in(0,1),\text{ s.t. }\frac{F(1)-F(0)}{1-0}=F'(\xi).\\
\Rightarrow&\exists\xi\in(0,1),\text{ s.t. }f(1)\arctan 1-f(0)\arctan 0=(\arctan\xi)'f(\xi)+\arctan\xi\cdot f'(\xi).\\
\Rightarrow&\exists\xi\in(0,1),\text{ s.t. }\frac{\pi}{4}f(1)-0=\frac{1}{1+\xi^2}f(\xi)+\arctan\xi\cdot f'(\xi).
\end{align*}
⇒ ⇒ ∃ ξ ∈ ( 0 , 1 ) , s.t. 1 − 0 F ( 1 ) − F ( 0 ) = F ′ ( ξ ) . ∃ ξ ∈ ( 0 , 1 ) , s.t. f ( 1 ) arctan 1 − f ( 0 ) arctan 0 = ( arctan ξ ) ′ f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . ∃ ξ ∈ ( 0 , 1 ) , s.t. 4 π f ( 1 ) − 0 = 1 + ξ 2 1 f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) .
在上方最后得到的等式的两端同时乘以1 + ξ 2 , 1+\xi^2, 1 + ξ 2 , π 4 ( 1 + ξ 2 ) f ( 1 ) = f ( ξ ) + ( 1 + ξ 2 ) arctan ξ ⋅ f ′ ( ξ ) . \frac{\pi}{4}(1+\xi^2)f(1)=f(\xi)+(1+\xi^2)\arctan\xi\cdot f'(\xi). 4 π ( 1 + ξ 2 ) f ( 1 ) = f ( ξ ) + ( 1 + ξ 2 ) arctan ξ ⋅ f ′ ( ξ ) .
Q.E.D. \textbf{Q.E.D.} Q.E.D.
除此之外，Lagrange中值定理还可以证明某函数的导数为某个非零常数。
e.g.2.13 \textbf{e.g.2.13 } e.g.2.13 f ( x ) f(x) f ( x ) [ 0 , 1 ] [0,1] [ 0 , 1 ] ( 0 , 1 ) (0,1) ( 0 , 1 ) f ( 0 ) = 0 , f ( 1 ) = 1 2 , f(0)=0,f(1)=\frac{1}{2}, f ( 0 ) = 0 , f ( 1 ) = 2 1 , ∃ ξ ∈ ( 0 , 1 ) , η ∈ ( 0 , 1 ) , ξ ≠ η , s.t. f ′ ( ξ ) + f ′ ( η ) = ξ + η . \exists\xi\in(0,1),\eta\in(0,1),\xi≠\eta,\text{ s.t. }f'(\xi)+f'(\eta)=\xi+\eta. ∃ ξ ∈ ( 0 , 1 ) , η ∈ ( 0 , 1 ) , ξ = η , s.t. f ′ ( ξ ) + f ′ ( η ) = ξ + η .
Analyses \textbf{Analyses } Analyses ( f ′ ( ξ ) − ξ ) + ( f ′ ( η ) − η ) = 0. (f'(\xi)-\xi)+(f'(\eta)-\eta)=0. ( f ′ ( ξ ) − ξ ) + ( f ′ ( η ) − η ) = 0. f ′ ( x ) − x = ( f ( x ) − 1 2 x 2 ) ′ , f'(x)-x=\big(f(x)-\frac{1}{2}x^2\big)', f ′ ( x ) − x = ( f ( x ) − 2 1 x 2 ) ′ ,
Proof \textbf{Proof } Proof F ( x ) = f ( x ) − 1 2 x 2 , F(x)=f(x)-\frac{1}{2}x^2, F ( x ) = f ( x ) − 2 1 x 2 , F ( 0 ) = 0 = F ( 1 ) . F(0)=0=F(1). F ( 0 ) = 0 = F ( 1 ) .
∃ ξ ∈ ( 0 , 1 2 ) , F ( 1 2 ) − F ( 0 ) 1 2 − 0 = F ′ ( ξ ) ⇒ ∃ ξ ∈ ( 0 , 1 2 ) , F ( 1 2 ) = 1 2 F ′ ( ξ ) , ∃ η ∈ ( 1 2 , 1 ) , F ( 1 ) − F ( 1 2 ) 1 − 1 2 = F ′ ( η ) ⇒ ∃ ξ ∈ ( 1 2 , 1 ) , − F ( 1 2 ) = 1 2 F ′ ( η ) , \begin{align*}
&\exists\xi\in(0,\frac{1}{2}),\frac{F(\frac{1}{2})-F(0)}{\frac{1}{2}-0}=F'(\xi)\\
\Rightarrow&\exists\xi\in(0,\frac{1}{2}),F(\frac{1}{2})=\frac{1}{2}F'(\xi),\\
&\exists\eta\in(\frac{1}{2},1),\frac{F(1)-F(\frac{1}{2})}{1-\frac{1}{2}}=F'(\eta)\\
\Rightarrow&\exists\xi\in(\frac{1}{2},1),-F(\frac{1}{2})=\frac{1}{2}F'(\eta),\\
\end{align*}
⇒ ⇒ ∃ ξ ∈ ( 0 , 2 1 ) , 2 1 − 0 F ( 2 1 ) − F ( 0 ) = F ′ ( ξ ) ∃ ξ ∈ ( 0 , 2 1 ) , F ( 2 1 ) = 2 1 F ′ ( ξ ) , ∃ η ∈ ( 2 1 , 1 ) , 1 − 2 1 F ( 1 ) − F ( 2 1 ) = F ′ ( η ) ∃ ξ ∈ ( 2 1 , 1 ) , − F ( 2 1 ) = 2 1 F ′ ( η ) ,
这两个式子相加，得到0 = 1 2 ( F ′ ( ξ ) + F ′ ( η ) ) , 0=\frac{1}{2}\big(F'(\xi)+F'(\eta)\big), 0 = 2 1 ( F ′ ( ξ ) + F ′ ( η ) ) ,
F ′ ( ξ ) + F ′ ( η ) = 0 ⇒ f ′ ( ξ ) − ξ + f ′ ( η ) − η = 0 ⇒ f ′ ( ξ ) + f ′ ( η ) = ξ + η . \begin{align*}
F'(\xi)+F'(\eta)=0&\Rightarrow f'(\xi)-\xi+f'(\eta)-\eta=0\\
&\Rightarrow f'(\xi)+f'(\eta)=\xi+\eta.
\end{align*}
F ′ ( ξ ) + F ′ ( η ) = 0 ⇒ f ′ ( ξ ) − ξ + f ′ ( η ) − η = 0 ⇒ f ′ ( ξ ) + f ′ ( η ) = ξ + η .
故原命题得证。
Q.E.D. \textbf{Q.E.D.} Q.E.D.
对于像上方例2.13的这种双中值问题，其核心在于将原有的区间一分为二，在两个子空间分别取一个符合题意的存在性结论后通过某种变换来得到原命题的式子。由于Lagrange中值定理没有太多前提条件，这里取的“区间划分点”本就可以根据需要任意选择；在上方例子中，为了相加后消去所以选择了1 2 . \frac{1}{2}. 2 1 .
e.x.2.13 \textbf{e.x.2.13 } e.x.2.13 f ( x ) f(x) f ( x ) [ 0 , 1 ] [0,1] [ 0 , 1 ] ( 0 , 1 ) (0,1) ( 0 , 1 ) f ( 0 ) = 0 , f ( 1 ) = 1 , f(0)=0,f(1)=1, f ( 0 ) = 0 , f ( 1 ) = 1 ,
(1)∃ a ∈ ( 0 , 1 ) , s.t. f ( a ) = 1 − a ; \exists a\in(0,1), \text{ s.t. }f(a)=1-a; ∃ a ∈ ( 0 , 1 ) , s.t. f ( a ) = 1 − a ; ∃ ξ ∈ ( 0 , 1 ) , η ∈ ( 0 , 1 ) , s.t. f ′ ( ξ ) f ′ ( η ) = 1. \exists\xi\in(0,1),\eta\in(0,1),\text{ s.t. }f'(\xi)f'(\eta)=1. ∃ ξ ∈ ( 0 , 1 ) , η ∈ ( 0 , 1 ) , s.t. f ′ ( ξ ) f ′ ( η ) = 1.
Analyses \textbf{Analyses } Analyses f ′ ( ξ ) ≠ 0 , f ′ ( η ) ≠ 0 , f'(\xi)≠0,f'(\eta)≠0, f ′ ( ξ ) = 0 , f ′ ( η ) = 0 , f ( a ) = 1 − a , f(a)=1-a, f ( a ) = 1 − a , a a a
Proof \textbf{Proof } Proof F ( x ) = f ( x ) − ( 1 − x ) , F(x)=f(x)-(1-x), F ( x ) = f ( x ) − ( 1 − x ) , F ( 0 ) = − 1 < 0 , F ( 1 ) = 1 > 0. F(0)=-1<0, F(1)=1>0. F ( 0 ) = − 1 < 0 , F ( 1 ) = 1 > 0.
∃ a ∈ ( 0 , 1 ) , s.t. F ( a ) = 0. \exists a\in(0,1), \text{ s.t. }F(a)=0.
∃ a ∈ ( 0 , 1 ) , s.t. F ( a ) = 0.
此时对应也就有f ( a ) − ( 1 − a ) = 0 , f(a)-(1-a)=0, f ( a ) − ( 1 − a ) = 0 , f ( a ) = 1 − a . f(a)=1-a. f ( a ) = 1 − a .
Q.E.D. \textbf{Q.E.D.} Q.E.D.
(2)由Lagrange中值定理，得
∃ ξ ∈ ( 0 , a ) , s.t. f ( a ) − f ( 0 ) a − 0 = f ′ ( ξ ) , \exists\xi\in(0,a),\text{ s.t. }\frac{f(a)-f(0)}{a-0}=f'(\xi),
∃ ξ ∈ ( 0 , a ) , s.t. a − 0 f ( a ) − f ( 0 ) = f ′ ( ξ ) ,
∃ η ∈ ( a , 1 ) , s.t. f ( 1 ) − f ( a ) 1 − a = f ′ ( η ) , \exists\eta\in(a,1),\text{ s.t. }\frac{f(1)-f(a)}{1-a}=f'(\eta),
∃ η ∈ ( a , 1 ) , s.t. 1 − a f ( 1 ) − f ( a ) = f ′ ( η ) ,
两式相乘可得
f ′ ( ξ ) f ′ ( η ) = f ( a ) − f ( 0 ) a − 0 ⋅ f ( 1 ) − f ( a ) 1 − a = ( 1 − a ) − 0 a ⋅ 1 − ( 1 − a ) 1 − a = 1 − a a ⋅ a 1 − a = 1. f'(\xi)f'(\eta)=\frac{f(a)-f(0)}{a-0}\cdot\frac{f(1)-f(a)}{1-a}=\frac{(1-a)-0}{a}\cdot\frac{1-(1-a)}{1-a}=\frac{1-a}{a}\cdot\frac{a}{1-a}=1.
f ′ ( ξ ) f ′ ( η ) = a − 0 f ( a ) − f ( 0 ) ⋅ 1 − a f ( 1 ) − f ( a ) = a ( 1 − a ) − 0 ⋅ 1 − a 1 − ( 1 − a ) = a 1 − a ⋅ 1 − a a = 1.
Q.E.D. \textbf{Q.E.D.} Q.E.D.
一些题目不会直接把Lagrange中值定理应有的“函数两值之差=导数的若干倍”形式呈现出来。其可能是含有积分或多阶导的形式；在这种情况下把这些积分或者低阶导看作函数，其函数或高阶导看作导数即可。
思考本质仍然是Lagrange中值定理的结论
∃ ξ ∈ ( a , b ) , s.t. ∫ a b f ( x ) d x = f ( ξ ) ( b − a ) . \exists\xi\in(a,b),\text{ s.t. }\int_a^bf(x)\text{d}x=f(\xi)(b-a).
∃ ξ ∈ ( a , b ) , s.t. ∫ a b f ( x ) d x = f ( ξ ) ( b − a ) .
和先前提及的积分中值定理相比，其自变量取值范围由闭区间变为了开区间；在寻找一个函数取相同值的两点时，积分中值定理所对应的闭区间无法保证区间中央的等值点不等于端点之一，但用这个结论就可以避免这种争议，因此这个结论的泛用性就比积分中值定理高了很多。
此外Lagrange中值定理还可以用以证明形如「f ′ ′ ( ξ ) f''(\xi) f ′′ ( ξ ) f ′ ′ ( ξ ) = f ′ ( b ) − f ′ ( a ) b − a , f''(\xi)=\frac{f'(b)-f'(a)}{b-a}, f ′′ ( ξ ) = b − a f ′ ( b ) − f ′ ( a ) , b > a , b>a, b > a , f ′ ′ ( ξ ) > 0 f''(\xi)>0 f ′′ ( ξ ) > 0 < 0 <0 < 0 f ′ ( b ) > 0 , f ′ ( a ) < 0 f'(b)>0,f'(a)<0 f ′ ( b ) > 0 , f ′ ( a ) < 0 f ′ ( b ) < 0 , f ′ ( a ) > 0 f'(b)<0,f'(a)>0 f ′ ( b ) < 0 , f ′ ( a ) > 0
Lagrange中值定理也可以证明函数恒等式。在Lagrange中值定理的公式f ( b ) − f ( a ) b − a = f ′ ( ξ ) \frac{f(b)-f(a)}{b-a}=f'(\xi) b − a f ( b ) − f ( a ) = f ′ ( ξ ) f ′ ( ξ ) = 0 f'(\xi)=0 f ′ ( ξ ) = 0 f ( b ) − f ( a ) = 0 , f(b)-f(a)=0, f ( b ) − f ( a ) = 0 , f ( x ) = C f(x)=C f ( x ) = C
除此之外Lagrange中值定理也可以解决一部分围绕中值点ξ \xi ξ f ( b ) − f ( a ) = f ′ ( ξ ) ( b − a ) , f(b)-f(a)=f'(\xi)(b-a), f ( b ) − f ( a ) = f ′ ( ξ ) ( b − a ) , ξ − a b − a = θ ∈ ( 0 , 1 ) , \frac{\xi-a}{b-a}=\theta\in(0,1), b − a ξ − a = θ ∈ ( 0 , 1 ) , ξ = a + θ ( b − a ) ; \xi=a+\theta(b-a); ξ = a + θ ( b − a ) ; f ( b ) − f ( a ) = f ′ ( a + θ ( b − a ) ) ( b − a ) . f(b)-f(a)=f'\big(a+\theta(b-a)\big)(b-a). f ( b ) − f ( a ) = f ′ ( a + θ ( b − a ) ) ( b − a ) . a = 0 , b = x a=0,b=x a = 0 , b = x ξ = θ x , \xi=\theta x, ξ = θ x , θ \theta θ ξ \xi ξ x x x θ = θ ( x ) , \theta=\theta(x), θ = θ ( x ) , f ( x ) − f ( 0 ) x = f ′ ′ ( x ⋅ θ ( x ) ) \frac{f(x)-f(0)}{x}=f''\big(x\cdot\theta(x)\big) x f ( x ) − f ( 0 ) = f ′′ ( x ⋅ θ ( x ) )
e.g.2.14 \textbf{e.g.2.14 } e.g.2.14 y = f ( x ) y=f(x) y = f ( x ) ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) f ′ ′ ( x ) ≠ 0 , f''(x)≠0, f ′′ ( x ) = 0 ,
(1)∀ x ∈ ( − 1 , 0 ) ∪ ( 0 , 1 ) , ∃ θ ( x ) ∈ ( 0 , 1 ) , s.t. f ( x ) = f ( 0 ) + x f ′ ( θ ( x ) x ) ; \forall x\in(-1,0)\cup(0,1),\exists \theta(x)\in(0,1), \text{ s.t. }f(x)=f(0)+xf'\big(\theta(x)x\big); ∀ x ∈ ( − 1 , 0 ) ∪ ( 0 , 1 ) , ∃ θ ( x ) ∈ ( 0 , 1 ) , s.t. f ( x ) = f ( 0 ) + x f ′ ( θ ( x ) x ) ; lim x → 0 θ ( x ) = 1 2 . \lim_{x\rightarrow 0}\theta(x)=\frac{1}{2}. lim x → 0 θ ( x ) = 2 1 .
Analyses \textbf{Analyses } Analyses f ( x ) − f ( 0 ) x − 0 = f ′ ( θ ( x ) x ) . \frac{f(x)-f(0)}{x-0}=f'\big(\theta(x)x\big). x − 0 f ( x ) − f ( 0 ) = f ′ ( θ ( x ) x ) . ξ \xi ξ θ ( x ) x , \theta(x) x, θ ( x ) x , θ ( x ) = ξ x . \theta(x)=\frac{\xi}{x}. θ ( x ) = x ξ .
Proof \textbf{Proof } Proof
∃ ξ ∈ ( − 1 , 1 ) , s.t. f ( x ) − f ( 0 ) x − 0 = f ′ ( ξ ) . \exists\xi\in(-1,1),\text{ s.t. }\frac{f(x)-f(0)}{x-0}=f'(\xi).
∃ ξ ∈ ( − 1 , 1 ) , s.t. x − 0 f ( x ) − f ( 0 ) = f ′ ( ξ ) .
也即∃ ξ ∈ ( − 1 , 1 ) , s.t. f ( x ) = f ( 0 ) + x f ′ ( ξ ) . \exists\xi\in(-1,1),\text{ s.t. }f(x)=f(0)+xf'(\xi). ∃ ξ ∈ ( − 1 , 1 ) , s.t. f ( x ) = f ( 0 ) + x f ′ ( ξ ) . θ ( x ) = ξ x , \theta(x)=\frac{\xi}{x}, θ ( x ) = x ξ , f ( x ) = f ( 0 ) + x f ′ ( θ ( x ) x ) . f(x)=f(0)+xf'\big(\theta(x)x\big). f ( x ) = f ( 0 ) + x f ′ ( θ ( x ) x ) . x x x θ ( x ) \theta(x) θ ( x ) ξ \xi ξ x ∈ ( − 1 , 0 ) ∪ ( 0 , 1 ) x\in(-1,0)\cup(0,1) x ∈ ( − 1 , 0 ) ∪ ( 0 , 1 )
Q.E.D. \textbf{Q.E.D.} Q.E.D.
(2)根据(1)可得
f ( x ) − f ( 0 ) x = f ′ ( θ ( x ) x ) ⇒ f ( x ) − f ( 0 ) x − f ′ ( 0 ) = f ′ ( θ ( x ) x ) − f ′ ( 0 ) ⇒ f ( x ) − f ( 0 ) x − f ′ ( 0 ) x = f ′ ( θ ( x ) x ) − f ′ ( 0 ) x ⇒ f ( x ) − f ( 0 ) x − f ′ ( 0 ) x = f ′ ( θ ( x ) x ) − f ′ ( 0 ) θ ( x ) x ⋅ θ ( x ) . \begin{align*}
\frac{f(x)-f(0)}{x}=f'\big(\theta(x)x\big)&\Rightarrow\frac{f(x)-f(0)}{x}-f'(0)=f'\big(\theta(x)x\big)-f'(0)\\
&\Rightarrow\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}=\frac{f'\big(\theta(x)x\big)-f'(0)}{x}\\
&\Rightarrow\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}=\frac{f'\big(\theta(x)x\big)-f'(0)}{\theta(x)x}\cdot\theta(x).
\end{align*}
x f ( x ) − f ( 0 ) = f ′ ( θ ( x ) x ) ⇒ x f ( x ) − f ( 0 ) − f ′ ( 0 ) = f ′ ( θ ( x ) x ) − f ′ ( 0 ) ⇒ x x f ( x ) − f ( 0 ) − f ′ ( 0 ) = x f ′ ( θ ( x ) x ) − f ′ ( 0 ) ⇒ x x f ( x ) − f ( 0 ) − f ′ ( 0 ) = θ ( x ) x f ′ ( θ ( x ) x ) − f ′ ( 0 ) ⋅ θ ( x ) .
再根据导数定义，
f ( x ) − f ( 0 ) x − f ′ ( 0 ) x = f ′ ( θ ( x ) x ) − f ′ ( 0 ) θ ( x ) x ⋅ θ ( x ) ⇒ lim x → 0 f ( x ) − f ( 0 ) x − f ′ ( 0 ) x = lim x → 0 f ′ ( θ ( x ) x ) − f ′ ( 0 ) θ ( x ) x ⋅ θ ( x ) ⇒ lim x → 0 ( f ( x ) − f ( 0 ) ) − x f ′ ( 0 ) x 2 = f ′ ′ ( 0 ) lim x → 0 θ ( x ) . \begin{align*}
\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}=\frac{f'\big(\theta(x)x\big)-f'(0)}{\theta(x)x}\cdot\theta(x)&\Rightarrow\lim_{x\rightarrow 0}\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}=\lim_{x\rightarrow 0}\frac{f'\big(\theta(x)x\big)-f'(0)}{\theta(x)x}\cdot\theta(x)\\
&\Rightarrow\lim_{x\rightarrow 0}\frac{\big(f(x)-f(0)\big)-xf'(0)}{x^2}=f''(0)\lim_{x\rightarrow 0}\theta(x).
\end{align*}
x x f ( x ) − f ( 0 ) − f ′ ( 0 ) = θ ( x ) x f ′ ( θ ( x ) x ) − f ′ ( 0 ) ⋅ θ ( x ) ⇒ x → 0 lim x x f ( x ) − f ( 0 ) − f ′ ( 0 ) = x → 0 lim θ ( x ) x f ′ ( θ ( x ) x ) − f ′ ( 0 ) ⋅ θ ( x ) ⇒ x → 0 lim x 2 ( f ( x ) − f ( 0 ) ) − x f ′ ( 0 ) = f ′′ ( 0 ) x → 0 lim θ ( x ) .
而
lim x → 0 ( f ( x ) − f ( 0 ) ) − x f ′ ( 0 ) x 2 = L ′ H lim x → 0 f ′ ( x ) − f ′ ( 0 ) 2 x = Def. Derivative 1 2 f ′ ′ ( 0 ) , \begin{align*}
\lim_{x\rightarrow 0}\frac{\big(f(x)-f(0)\big)-xf'(0)}{x^2}&\xlongequal{L'H}\lim_{x\rightarrow 0}\frac{f'(x)-f'(0)}{2x}\\
&\xlongequal{\text{Def. Derivative}}\frac{1}{2}f''(0),
\end{align*}
x → 0 lim x 2 ( f ( x ) − f ( 0 ) ) − x f ′ ( 0 ) L ′ H x → 0 lim 2 x f ′ ( x ) − f ′ ( 0 ) Def. Derivative 2 1 f ′′ ( 0 ) ,
故
1 2 f ′ ′ ( 0 ) = f ′ ′ ( 0 ) lim x → 0 θ ( x ) ⇒ f ′ ′ ( x ) ≠ 0 lim x → 0 θ ( x ) = 1 2 . \frac{1}{2}f''(0)=f''(0)\lim_{x\rightarrow 0}\theta(x)\xRightarrow{f''(x)≠0}\lim_{x\rightarrow 0}\theta(x)=\frac{1}{2}.
2 1 f ′′ ( 0 ) = f ′′ ( 0 ) x → 0 lim θ ( x ) f ′′ ( x ) = 0 x → 0 lim θ ( x ) = 2 1 .
Q.E.D. \textbf{Q.E.D.} Q.E.D.
有些情况下Lagrange中值定理还会用于分析渐近性或有界性。
e.g.2.15 \textbf{e.g.2.15 } e.g.2.15 f ( x ) f(x) f ( x ) x ≥ a x\geq a x ≥ a lim x → + ∞ f ( x ) \lim_{x\rightarrow+\infty}f(x) lim x → + ∞ f ( x ) lim x → + ∞ f ′ ( x ) \lim_{x\rightarrow+\infty}f'(x) lim x → + ∞ f ′ ( x ) lim x → + ∞ f ′ ( x ) . \lim_{x\rightarrow+\infty}f'(x). lim x → + ∞ f ′ ( x ) .
Solution \textbf{Solution } Solution lim x → + ∞ f ( x ) = A , lim x → + ∞ f ′ ( x ) = B . \lim_{x\rightarrow+\infty}f(x)=A,\lim_{x\rightarrow+\infty}f'(x)=B. lim x → + ∞ f ( x ) = A , lim x → + ∞ f ′ ( x ) = B .
A = lim x → + ∞ f ( x ) = lim x → + ∞ e x f ( x ) e x = L ′ H lim x → + ∞ e x f ( x ) + e x f ′ ( x ) e x = lim x → + ∞ ( f ( x ) + f ′ ( x ) ) = A + B . \begin{align*}
A&=\lim_{x\rightarrow+\infty}f(x)=\lim_{x\rightarrow +\infty}\frac{e^xf(x)}{e^x}\\
&\xlongequal{L'H}\lim_{x\rightarrow+\infty}\frac{e^xf(x)+e^xf'(x)}{e^x}=\lim_{x\rightarrow+\infty}\big(f(x)+f'(x)\big)=A+B.
\end{align*}
A = x → + ∞ lim f ( x ) = x → + ∞ lim e x e x f ( x ) L ′ H x → + ∞ lim e x e x f ( x ) + e x f ′ ( x ) = x → + ∞ lim ( f ( x ) + f ′ ( x ) ) = A + B .
因此B = 0. B=0. B = 0. lim x → + ∞ f ′ ( x ) = 0. \lim_{x\rightarrow+\infty}f'(x)=0. lim x → + ∞ f ′ ( x ) = 0.
Cauchy中值定理
只要在待证结论看到形如“f ′ ( x ) g ′ ( x ) \frac{f'(x)}{g'(x)} g ′ ( x ) f ′ ( x )
用下面两个例子尝试理解「形如“f ′ ( x ) g ′ ( x ) \frac{f'(x)}{g'(x)} g ′ ( x ) f ′ ( x )
考虑形如f ( b ) − f ( a ) b 2 − a 2 = f ′ ( ξ ) 2 ξ \frac{f(b)-f(a)}{b^2-a^2}=\frac{f'(\xi)}{2\xi} b 2 − a 2 f ( b ) − f ( a ) = 2 ξ f ′ ( ξ ) g ( x ) g(x) g ( x ) x 2 x^2 x 2
考虑待证结论a e b − b e a a − b = e ξ − ξ e ξ \frac{ae^b-be^a}{a-b}=e^\xi-\xi e^\xi a − b a e b − b e a = e ξ − ξ e ξ e b b − e a a 1 b − 1 a = ξ e ξ − e ξ ξ 2 − 1 ξ 2 , \frac{\frac{e^b}{b}-\frac{e^a}{a}}{\frac{1}{b}-\frac{1}{a}}=\frac{\frac{\xi e^\xi-e^\xi}{\xi^2}}{-\frac{1}{\xi^2}}, b 1 − a 1 b e b − a e a = − ξ 2 1 ξ 2 ξ e ξ − e ξ , f ( x ) = e x x , g ( x ) = 1 x f(x)=\frac{e^x}{x},g(x)=\frac{1}{x} f ( x ) = x e x , g ( x ) = x 1
e.x.2.12 \textbf{e.x.2.12 } e.x.2.12 f ( x ) f(x) f ( x ) [ 0 , 1 ] [0,1] [ 0 , 1 ] ∃ ξ ∈ ( 0 , 1 ) , s.t. π 4 ( 1 + ξ 2 ) f ( 1 ) = f ( ξ ) + ( 1 + ξ 2 ) arctan ξ ⋅ f ′ ( ξ ) . \exists\xi\in(0,1),\text{ s.t. }\frac{\pi}{4}(1+\xi^2)f(1)=f(\xi)+(1+\xi^2)\arctan\xi\cdot f'(\xi). ∃ ξ ∈ ( 0 , 1 ) , s.t. 4 π ( 1 + ξ 2 ) f ( 1 ) = f ( ξ ) + ( 1 + ξ 2 ) arctan ξ ⋅ f ′ ( ξ ) .
Analyses \textbf{Analyses } Analyses π 4 f ( 1 ) = 1 1 + ξ 2 f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . \frac{\pi}{4}f(1)=\frac{1}{1+\xi^2}f(\xi)+\arctan\xi\cdot f'(\xi). 4 π f ( 1 ) = 1 + ξ 2 1 f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . ( f ( x ) arctan x ) ′ ∣ x = ξ , \big(f(x)\arctan x\big)'\Big|_{x=\xi}, ( f ( x ) arctan x ) ′ x = ξ ,
Proof \textbf{Proof } Proof F ( x ) = f ( x ) arctan x . F(x)=f(x)\arctan x. F ( x ) = f ( x ) arctan x .
∃ ξ ∈ ( 0 , 1 ) , s.t. F ( 1 ) − F ( 0 ) 1 − 0 = F ′ ( ξ ) . ⇒ ∃ ξ ∈ ( 0 , 1 ) , s.t. f ( 1 ) arctan 1 − f ( 0 ) arctan 0 = ( arctan ξ ) ′ f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . ⇒ ∃ ξ ∈ ( 0 , 1 ) , s.t. π 4 f ( 1 ) − 0 = 1 1 + ξ 2 f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . \begin{align*}
&\exists\xi\in(0,1),\text{ s.t. }\frac{F(1)-F(0)}{1-0}=F'(\xi).\\
\Rightarrow&\exists\xi\in(0,1),\text{ s.t. }f(1)\arctan 1-f(0)\arctan 0=(\arctan\xi)'f(\xi)+\arctan\xi\cdot f'(\xi).\\
\Rightarrow&\exists\xi\in(0,1),\text{ s.t. }\frac{\pi}{4}f(1)-0=\frac{1}{1+\xi^2}f(\xi)+\arctan\xi\cdot f'(\xi).
\end{align*}
⇒ ⇒ ∃ ξ ∈ ( 0 , 1 ) , s.t. 1 − 0 F ( 1 ) − F ( 0 ) = F ′ ( ξ ) . ∃ ξ ∈ ( 0 , 1 ) , s.t. f ( 1 ) arctan 1 − f ( 0 ) arctan 0 = ( arctan ξ ) ′ f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) . ∃ ξ ∈ ( 0 , 1 ) , s.t. 4 π f ( 1 ) − 0 = 1 + ξ 2 1 f ( ξ ) + arctan ξ ⋅ f ′ ( ξ ) .
在上方最后得到的等式的两端同时乘以1 + ξ 2 , 1+\xi^2, 1 + ξ 2 , π 4 ( 1 + ξ 2 ) f ( 1 ) = f ( ξ ) + ( 1 + ξ 2 ) arctan ξ ⋅ f ′ ( ξ ) . \frac{\pi}{4}(1+\xi^2)f(1)=f(\xi)+(1+\xi^2)\arctan\xi\cdot f'(\xi). 4 π ( 1 + ξ 2 ) f ( 1 ) = f ( ξ ) + ( 1 + ξ 2 ) arctan ξ ⋅ f ′ ( ξ ) .
Q.E.D. \textbf{Q.E.D.} Q.E.D.
除此之外，Lagrange中值定理还可以证明某函数的导数为某个非零常数。
e.x.2.15 \textbf{e.x.2.15 } e.x.2.15 f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) 0 < a < b , 0<a<b, 0 < a < b , ∃ ξ ∈ ( a , b ) , s.t. f ( b ) − f ( a ) b − a = ( a 2 + a b + b 2 ) f ′ ( ξ ) 3 ξ 2 . \exists\xi\in(a,b),\text{ s.t. }\frac{f(b)-f(a)}{b-a}=(a^2+ab+b^2)\frac{f'(\xi)}{3\xi^2}. ∃ ξ ∈ ( a , b ) , s.t. b − a f ( b ) − f ( a ) = ( a 2 + ab + b 2 ) 3 ξ 2 f ′ ( ξ ) .
Analyses \textbf{Analyses } Analyses f ′ ( ξ ) f'(\xi) f ′ ( ξ ) ξ 3 \xi^3 ξ 3 a 2 + a b + b 2 a^2+ab+b^2 a 2 + ab + b 2 b 3 − a 3 b^3-a^3 b 3 − a 3
Proof \textbf{Proof } Proof f ( x ) f(x) f ( x ) g ( x ) = x 3 , g(x)=x^3, g ( x ) = x 3 ,
∃ ξ ∈ ( a , b ) , s.t. f ( b ) − f ( a ) g ( b ) − g ( a ) = f ′ ( ξ ) g ′ ( ξ ) . \exists\xi\in(a,b),\text{ s.t. }\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)}.
∃ ξ ∈ ( a , b ) , s.t. g ( b ) − g ( a ) f ( b ) − f ( a ) = g ′ ( ξ ) f ′ ( ξ ) .
而此时
f ( b ) − f ( a ) g ( b ) − g ( a ) = f ′ ( ξ ) g ′ ( ξ ) ⇒ f ( b ) − f ( a ) b 3 − a 3 = f ′ ( ξ ) 3 ξ 2 ⇒ f ( b ) − f ( a ) ( b − a ) ( a 2 + a b + b 2 ) = f ′ ( ξ ) 3 ξ 2 ⇒ f ( b ) − f ( a ) b − a = ( a 2 + a b + b 2 ) f ′ ( ξ ) 3 ξ 2 . \begin{align*}
\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\xi)}{g'(\xi)}&\Rightarrow\frac{f(b)-f(a)}{b^3-a^3}=\frac{f'(\xi)}{3\xi^2}\\
&\Rightarrow\frac{f(b)-f(a)}{(b-a)(a^2+ab+b^2)}=\frac{f'(\xi)}{3\xi^2}\\
&\Rightarrow\frac{f(b)-f(a)}{b-a}=(a^2+ab+b^2)\frac{f'(\xi)}{3\xi^2}.
\end{align*}
g ( b ) − g ( a ) f ( b ) − f ( a ) = g ′ ( ξ ) f ′ ( ξ ) ⇒ b 3 − a 3 f ( b ) − f ( a ) = 3 ξ 2 f ′ ( ξ ) ⇒ ( b − a ) ( a 2 + ab + b 2 ) f ( b ) − f ( a ) = 3 ξ 2 f ′ ( ξ ) ⇒ b − a f ( b ) − f ( a ) = ( a 2 + ab + b 2 ) 3 ξ 2 f ′ ( ξ ) .
故原命题得证。
Q.E.D. \textbf{Q.E.D.} Q.E.D.
e.x.2.16 \textbf{e.x.2.16 } e.x.2.16 f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) 0 < a < b , 0<a<b, 0 < a < b , ∃ x 1 , x 2 , x 3 ∈ ( a , b ) , s.t. f ′ ( x 1 ) 2 x 1 = ( b 2 + a 2 ) f ′ ( x 2 ) 4 x 2 3 = ln b a b 2 − a 2 x 3 f ′ ( x 3 ) . \exists x_1,x_2,x_3\in(a,b),\text{ s.t. }\frac{f'(x_1)}{2x_1}=(b^2+a^2)\frac{f'(x_2)}{4x_2^3}=\frac{\ln\frac{b}{a}}{b^2-a^2}x_3f'(x_3). ∃ x 1 , x 2 , x 3 ∈ ( a , b ) , s.t. 2 x 1 f ′ ( x 1 ) = ( b 2 + a 2 ) 4 x 2 3 f ′ ( x 2 ) = b 2 − a 2 l n a b x 3 f ′ ( x 3 ) .
Analyses \textbf{Analyses } Analyses f ′ ( x i ) f'(x_i) f ′ ( x i ) 2 x 1 = ( x 1 2 ) ′ , 4 x 2 3 = ( x 2 4 ) ′ , 1 x 3 = ( ln x 3 ) ′ . 2x_1=({x_1}^2)',4{x_2}^3=({x_2}^4)',\frac{1}{x_3}=(\ln x_3)'. 2 x 1 = ( x 1 2 ) ′ , 4 x 2 3 = ( x 2 4 ) ′ , x 3 1 = ( ln x 3 ) ′ . ( b 2 + a 2 ) ( b 2 − a 2 ) = b 4 − a 4 , ln b a = ln b − ln a , (b^2+a^2)(b^2-a^2)=b^4-a^4,\ln\frac{b}{a}=\ln b-\ln a, ( b 2 + a 2 ) ( b 2 − a 2 ) = b 4 − a 4 , ln a b = ln b − ln a ,
Proof \textbf{Proof } Proof f ( x ) f(x) f ( x ) g 1 ( x ) = x 2 , g 2 ( x ) = x 4 , g 3 ( x ) = ln x , g_1(x)=x^2,g_2(x)=x^4,g_3(x)=\ln x, g 1 ( x ) = x 2 , g 2 ( x ) = x 4 , g 3 ( x ) = ln x ,
∃ x 1 ∈ ( a , b ) , s.t. f ( b ) − f ( a ) g 1 ( b ) − g 1 ( a ) = f ′ ( x 1 ) g 1 ′ ( x 1 ) ; ∃ x 2 ∈ ( a , b ) , s.t. f ( b ) − f ( a ) g 2 ( b ) − g 2 ( a ) = f ′ ( x 2 ) g 1 ′ ( x 2 ) ; ∃ x 3 ∈ ( a , b ) , s.t. f ( b ) − f ( a ) g 3 ( b ) − g 3 ( a ) = f ′ ( x 3 ) g 1 ′ ( x 3 ) ; \begin{align*}
\exists x_1\in(a,b),\text{ s.t. }\frac{f(b)-f(a)}{g_1(b)-g_1(a)}=\frac{f'(x_1)}{g_1'(x_1)};\\
\exists x_2\in(a,b),\text{ s.t. }\frac{f(b)-f(a)}{g_2(b)-g_2(a)}=\frac{f'(x_2)}{g_1'(x_2)};\\
\exists x_3\in(a,b),\text{ s.t. }\frac{f(b)-f(a)}{g_3(b)-g_3(a)}=\frac{f'(x_3)}{g_1'(x_3)};\\
\end{align*}
∃ x 1 ∈ ( a , b ) , s.t. g 1 ( b ) − g 1 ( a ) f ( b ) − f ( a ) = g 1 ′ ( x 1 ) f ′ ( x 1 ) ; ∃ x 2 ∈ ( a , b ) , s.t. g 2 ( b ) − g 2 ( a ) f ( b ) − f ( a ) = g 1 ′ ( x 2 ) f ′ ( x 2 ) ; ∃ x 3 ∈ ( a , b ) , s.t. g 3 ( b ) − g 3 ( a ) f ( b ) − f ( a ) = g 1 ′ ( x 3 ) f ′ ( x 3 ) ;
也即
f ( b ) − f ( a ) = f ′ ( x 1 ) 2 x 1 ( b 2 − a 2 ) , f ( b ) − f ( a ) = f ′ ( x 2 ) 4 x 2 3 ( b 4 − a 4 ) , f ( b ) − f ( a ) = f ′ ( x 3 ) 1 x 3 ( ln b − ln a ) . \begin{align*}
f(b)-f(a)=\frac{f'(x_1)}{2x_1}(b^2-a^2),\\
f(b)-f(a)=\frac{f'(x_2)}{4{x_2}^3}(b^4-a^4),\\
f(b)-f(a)=\frac{f'(x_3)}{\frac{1}{x_3}}(\ln b-\ln a).
\end{align*}
f ( b ) − f ( a ) = 2 x 1 f ′ ( x 1 ) ( b 2 − a 2 ) , f ( b ) − f ( a ) = 4 x 2 3 f ′ ( x 2 ) ( b 4 − a 4 ) , f ( b ) − f ( a ) = x 3 1 f ′ ( x 3 ) ( ln b − ln a ) .
因此f ′ ( x 1 ) 2 x 1 ( b 2 − a 2 ) = f ′ ( x 2 ) 4 x 2 3 ( b 4 − a 4 ) = f ′ ( x 3 ) 1 x 3 ( ln b − ln a ) . \frac{f'(x_1)}{2x_1}(b^2-a^2)=\frac{f'(x_2)}{4{x_2}^3}(b^4-a^4)=\frac{f'(x_3)}{\frac{1}{x_3}}(\ln b-\ln a). 2 x 1 f ′ ( x 1 ) ( b 2 − a 2 ) = 4 x 2 3 f ′ ( x 2 ) ( b 4 − a 4 ) = x 3 1 f ′ ( x 3 ) ( ln b − ln a ) . b 2 − a 2 , b^2-a^2, b 2 − a 2 ,
f ′ ( x 1 ) 2 x 1 = f ′ ( x 2 ) 4 x 2 3 ( b 2 + a 2 ) = f ′ ( x 3 ) 1 x 3 ln b − ln a b 2 − a 2 . \frac{f'(x_1)}{2x_1}=\frac{f'(x_2)}{4{x_2}^3}(b^2+a^2)=\frac{f'(x_3)}{\frac{1}{x_3}}\frac{\ln b-\ln a}{b^2-a^2}.
2 x 1 f ′ ( x 1 ) = 4 x 2 3 f ′ ( x 2 ) ( b 2 + a 2 ) = x 3 1 f ′ ( x 3 ) b 2 − a 2 ln b − ln a .
也即
f ′ ( x 1 ) 2 x 1 = ( b 2 + a 2 ) f ′ ( x 2 ) 4 x 2 3 = ln b a b 2 − a 2 x 3 f ′ ( x 3 ) . \frac{f'(x_1)}{2x_1}=(b^2+a^2)\frac{f'(x_2)}{4x_2^3}=\frac{\ln\frac{b}{a}}{b^2-a^2}x_3f'(x_3).
2 x 1 f ′ ( x 1 ) = ( b 2 + a 2 ) 4 x 2 3 f ′ ( x 2 ) = b 2 − a 2 ln a b x 3 f ′ ( x 3 ) .
Q.E.D. \textbf{Q.E.D.} Q.E.D.
Taylor中值定理
设f ( x ) f(x) f ( x ) x = a x=a x = a ( a − δ 1 , a + δ 2 ) (a-\delta_1,a+\delta_2) ( a − δ 1 , a + δ 2 ) n + 1 n+1 n + 1
∀ x ∈ ( a − δ 1 , a + δ 2 ) , ∃ ξ ∈ ( x , a ) , s.t. f ( x ) = ∑ i = 0 n ( f ( n ) ( a ) n ! ( x − a ) n ) + f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − a ) n + 1 . \forall x\in(a-\delta_1,a+\delta_2),\exists\xi\in(x,a),\text{ s.t. }f(x)=\sum_{i=0}^n\Big(\frac{f^{(n)}(a)}{n!}(x-a)^n\Big)+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}.
∀ x ∈ ( a − δ 1 , a + δ 2 ) , ∃ ξ ∈ ( x , a ) , s.t. f ( x ) = i = 0 ∑ n ( n ! f ( n ) ( a ) ( x − a ) n ) + ( n + 1 )! f ( n + 1 ) ( ξ ) ( x − a ) n + 1 .
这个R n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − a ) n + 1 R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1} R n ( x ) = ( n + 1 )! f ( n + 1 ) ( ξ ) ( x − a ) n + 1 f ( x ) f(x) f ( x ) a a a a = 0 a=0 a = 0
f ( x ) = ∑ i = 0 n ( f ( n ) ( 0 ) n ! x n ) + f ( n + 1 ) ( ξ ) ( n + 1 ) ! x n + 1 ( ξ ∈ ( 0 , x ) ) . f(x)=\sum_{i=0}^n\Big(\frac{f^{(n)}(0)}{n!}x^n\Big)+\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}(\xi\in(0,x)).
f ( x ) = i = 0 ∑ n ( n ! f ( n ) ( 0 ) x n ) + ( n + 1 )! f ( n + 1 ) ( ξ ) x n + 1 ( ξ ∈ ( 0 , x )) .
这个式子也就成为带有Lagrange余项的McLaurin公式。
在使用Taylor中值定理进行解题或证明时一般需要列出对应的公式，随后选择x , a x,a x , a x x x
e.g.2.17 \textbf{e.g.2.17 } e.g.2.17 f ( x ) f(x) f ( x ) [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] f ( − 1 ) = 0 , f ( 1 ) = 1 , f ′ ( 0 ) = 0 f(-1)=0,f(1)=1,f'(0)=0 f ( − 1 ) = 0 , f ( 1 ) = 1 , f ′ ( 0 ) = 0 ∃ ξ ∈ ( − 1 , 1 ) , s.t. f ′ ′ ′ ( ξ ) = 3. \exists\xi\in(-1,1),\text{ s.t. }f'''(\xi)=3. ∃ ξ ∈ ( − 1 , 1 ) , s.t. f ′′′ ( ξ ) = 3.
Analyses \textbf{Analyses } Analyses x , a x,a x , a
Proof \textbf{Proof } Proof f ( x ) f(x) f ( x )
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + f ′ ′ ′ ( ξ ) 3 ! ( x − x 0 ) 3 . f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\frac{f'''(\xi)}{3!}(x-x_0)^3.
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + 2 ! f ′′ ( x 0 ) ( x − x 0 ) 2 + 3 ! f ′′′ ( ξ ) ( x − x 0 ) 3 .
不妨取x 0 = 0 , x = 1 , x_0=0,x=1, x 0 = 0 , x = 1 ,
f ( 1 ) = f ( 0 ) + f ′ ( 0 ) + f ′ ′ ( 0 ) 2 + f ′ ′ ′ ( ξ ) 6 ⇒ 1 = f ( 0 ) + f ′ ′ ( 0 ) 2 + f ′ ′ ′ ( ξ ) 6 \begin{align*}
&f(1)=f(0)+f'(0)+\frac{f''(0)}{2}+\frac{f'''(\xi)}{6}
\Rightarrow&1=f(0)+\frac{f''(0)}{2}+\frac{f'''(\xi)}{6}\tag{1}
\end{align*}
f ( 1 ) = f ( 0 ) + f ′ ( 0 ) + 2 f ′′ ( 0 ) + 6 f ′′′ ( ξ ) ⇒ 1 = f ( 0 ) + 2 f ′′ ( 0 ) + 6 f ′′′ ( ξ ) ( 1 )
同理取x 0 = 0 , x = − 1 , x_0=0,x=-1, x 0 = 0 , x = − 1 ,
0 = f ( 0 ) + f ′ ′ ( 0 ) 2 − f ′ ′ ′ ( ξ ) 6 \begin{align*}
0=f(0)+\frac{f''(0)}{2}-\frac{f'''(\xi)}{6}\tag{2}
\end{align*}
0 = f ( 0 ) + 2 f ′′ ( 0 ) − 6 f ′′′ ( ξ ) ( 2 )
联立( 1 ) , ( 2 ) (1),(2) ( 1 ) , ( 2 ) { f ( 0 ) + f ′ ′ ( 0 ) 2 = 1 2 , f ′ ′ ′ ( ξ ) 6 = 1 2 . \begin{cases}f(0)+\frac{f''(0)}{2}&=&\frac{1}{2},\\\frac{f'''(\xi)}{6}&=&\frac{1}{2}\end{cases}. { f ( 0 ) + 2 f ′′ ( 0 ) 6 f ′′′ ( ξ ) = = 2 1 , 2 1 .
f ′ ′ ′ ( ξ ) = f ′ ′ ′ ( ξ ) 6 × 6 = 1 2 × 6 = 3. f'''(\xi)=\frac{f'''(\xi)}{6}\times6=\frac{1}{2}\times6=3.
f ′′′ ( ξ ) = 6 f ′′′ ( ξ ) × 6 = 2 1 × 6 = 3.
Q.E.D. \textbf{Q.E.D.} Q.E.D.
e.x.2.17 \textbf{e.x.2.17 } e.x.2.17 f ( x ) f(x) f ( x ) [ − a , a ] ( a > 0 ) [-a,a](a>0) [ − a , a ] ( a > 0 ) f ( 0 ) = 0. f(0)=0. f ( 0 ) = 0.
(1)求f ( x ) f(x) f ( x ) ∃ η ∈ [ − a , a ] , s.t. a 3 f ′ ′ ( η ) = 3 ∫ − a a f ( x ) d x . \exists\eta\in[-a,a],\text{ s.t. }a^3f''(\eta)=3\int_{-a}^af(x)\text{d}x. ∃ η ∈ [ − a , a ] , s.t. a 3 f ′′ ( η ) = 3 ∫ − a a f ( x ) d x .
Analyses \textbf{Analyses } Analyses
Proof \textbf{Proof } Proof
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + R 1 ( x ) , R 1 ( x ) = f ′ ′ ( ξ ) 2 ! ( x − x 0 ) 2 . f(x)=f(x_0)+f'(x_0)(x-x_0)+R_1(x),R_1(x)=\frac{f''(\xi)}{2!}(x-x_0)^2.
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + R 1 ( x ) , R 1 ( x ) = 2 ! f ′′ ( ξ ) ( x − x 0 ) 2 .
(2)记F ( x ) F(x) F ( x ) f ( x ) f(x) f ( x ) , F ( a ) − F ( − a ) = ∫ − a a f ( x ) d x , f ′ ′ ( x ) = F ′ ′ ′ ( x ) . ,F(a)-F(-a)=\int_{-a}^af(x)\text{d}x,f''(x)=F'''(x). , F ( a ) − F ( − a ) = ∫ − a a f ( x ) d x , f ′′ ( x ) = F ′′′ ( x ) . F ( x ) F(x) F ( x )
F ( x ) = F ( x 0 ) + F ′ ( x 0 ) ( x − x 0 ) + F ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + F ′ ′ ′ ( η ) 3 ! ( x − x 0 ) 3 . F(x)=F(x_0)+F'(x_0)(x-x_0)+\frac{F''(x_0)}{2!}(x-x_0)^2+\frac{F'''(\eta)}{3!}(x-x_0)^3.
F ( x ) = F ( x 0 ) + F ′ ( x 0 ) ( x − x 0 ) + 2 ! F ′′ ( x 0 ) ( x − x 0 ) 2 + 3 ! F ′′′ ( η ) ( x − x 0 ) 3 .
即
F ( x ) = F ( x 0 ) + f ( x 0 ) ( x − x 0 ) + f ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + f ′ ′ ( η ) 3 ! ( x − x 0 ) 3 . F(x)=F(x_0)+f(x_0)(x-x_0)+\frac{f'(x_0)}{2!}(x-x_0)^2+\frac{f''(\eta)}{3!}(x-x_0)^3.
F ( x ) = F ( x 0 ) + f ( x 0 ) ( x − x 0 ) + 2 ! f ′ ( x 0 ) ( x − x 0 ) 2 + 3 ! f ′′ ( η ) ( x − x 0 ) 3 .
不妨取x 0 = 0 , x = a x_0=0,x=a x 0 = 0 , x = a F ( a ) = F ( 0 ) + a f ( 0 ) + a 2 f ′ ( 0 ) 2 ! + a 3 f ′ ′ ( η ) 6 F(a)=F(0)+af(0)+a^2\frac{f'(0)}{2!}+a^3\frac{f''(\eta)}{6} F ( a ) = F ( 0 ) + a f ( 0 ) + a 2 2 ! f ′ ( 0 ) + a 3 6 f ′′ ( η )
F ( a ) = F ( 0 ) + a 2 f ′ ( 0 ) 2 ! + a 3 f ′ ′ ( η ) 6 . (1) F(a)=F(0)+a^2\frac{f'(0)}{2!}+a^3\frac{f''(\eta)}{6}.\tag{1}
F ( a ) = F ( 0 ) + a 2 2 ! f ′ ( 0 ) + a 3 6 f ′′ ( η ) . ( 1 )
同理取x 0 = 0 , x = − a x_0=0,x=-a x 0 = 0 , x = − a
F ( − a ) = F ( 0 ) + a 2 f ′ ( 0 ) 2 ! − a 3 f ′ ′ ( η ) 6 . (2) F(-a)=F(0)+a^2\frac{f'(0)}{2!}-a^3\frac{f''(\eta)}{6}.\tag{2}
F ( − a ) = F ( 0 ) + a 2 2 ! f ′ ( 0 ) − a 3 6 f ′′ ( η ) . ( 2 )
将( 1 ) , ( 2 ) (1),(2) ( 1 ) , ( 2 )
F ( a ) − F ( − a ) = 2 a 3 f ′ ′ ( η ) 6 , F(a)-F(-a)=2a^3\frac{f''(\eta)}{6},
F ( a ) − F ( − a ) = 2 a 3 6 f ′′ ( η ) ,
即
∫ − a a f ( x ) d x = a 3 f ′ ′ ( η ) 3 . \int_{-a}^af(x)\text{d}x=a^3\frac{f''(\eta)}{3}.
∫ − a a f ( x ) d x = a 3 3 f ′′ ( η ) .
也就是a 3 f ′ ′ ( η ) = 3 ∫ − a a f ( x ) d x , a^3f''(\eta)=3\int_{-a}^af(x)\text{d}x, a 3 f ′′ ( η ) = 3 ∫ − a a f ( x ) d x ,
Q.E.D. \textbf{Q.E.D.} Q.E.D.
函数性态（三点二性）
主要是拐点、极值点、这「三点」和单调性、凹凸性这「二性」。
根据高中知识就知道一个函数的导数正负直接表现出其单调增减。在实际考题中，对单调性的讨论很有可能会按照一些导数的变形呈现出来。
用下面两个例子尝试理解「一些导数的变形」为何物。
如果我们知道f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) < 0 , f'(x)g(x)-f(x)g'(x)<0, f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) < 0 , ( f ( x ) g ( x ) ) ′ = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) ( g ( x ) ) 2 < 0 , \big(\frac{f(x)}{g(x)}\big)'=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}<0, ( g ( x ) f ( x ) ) ′ = ( g ( x ) ) 2 f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) < 0 , f ( x ) g ( x ) \frac{f(x)}{g(x)} g ( x ) f ( x )
如果我们知道f ′ ( x ) f ( x ) > 0 , f'(x)f(x)>0, f ′ ( x ) f ( x ) > 0 , 2 f ′ ( x ) f ( x ) > 0 , 2f'(x)f(x)>0, 2 f ′ ( x ) f ( x ) > 0 , ( ( f ( x ) ) 2 ) ′ > 0. \Big(\big(f(x)\big)^2\Big)'>0. ( ( f ( x ) ) 2 ) ′ > 0. ( f ( x ) ) 2 \big(f(x)\big)^2 ( f ( x ) ) 2
一个函数的导数零点叫做这个函数的驻点。一个函数的极值点只可能出现在这个函数的不可导点或驻点处，至于这些点是否是极值点还需要进一步判断。
考虑上方警示。对f ( x ) = x 3 , f(x)=x^3, f ( x ) = x 3 , x = 0 x=0 x = 0 f ( x ) = ∣ x ∣ , f(x)=|x|, f ( x ) = ∣ x ∣ , x = 0 x=0 x = 0
极值点的求法高中已经学过，这里不再赘述。总的来说，如果一个导数是0的点左右的导数同号，那么这个点不是极值点；否则应该就是。当然极值点也可以通过一二阶导数联合判定——极值点处的一阶导为零而二阶导不为零；此时二阶导为正对应极小值点，为负对应极大值点。
e.g.2.18 \textbf{e.g.2.18 } e.g.2.18 x 2 f ′ ( x ) + x f ( x ) = ln x , f ( e ) = 1 e , x^2f'(x)+xf(x)=\ln x,f(e)=\frac{1}{e}, x 2 f ′ ( x ) + x f ( x ) = ln x , f ( e ) = e 1 , f ( x ) f(x) f ( x ) ( 0 , + ∞ ) (0,+\infty) ( 0 , + ∞ ) f ( x ) f(x) f ( x ) ( 0 , + ∞ ) (0,+\infty) ( 0 , + ∞ ) f ( x ) f(x) f ( x )
Solution \textbf{Solution } Solution x x x x f ′ ( x ) + f ( x ) = ln x x , xf'(x)+f(x)=\frac{\ln x}{x}, x f ′ ( x ) + f ( x ) = x l n x , ( x f ( x ) ) ′ = ln 2 x 2 . \big(xf(x)\big)'=\frac{\ln^2x}{2}. ( x f ( x ) ) ′ = 2 l n 2 x . x f ( x ) = ln 2 x 2 + C , xf(x)=\frac{\ln^2x}{2}+C, x f ( x ) = 2 l n 2 x + C , f ( x ) = ln 2 x 2 x + C x . f(x)=\frac{\ln^2x}{2x}+\frac{C}{x}. f ( x ) = 2 x l n 2 x + x C . f ( e ) = 1 e , f(e)=\frac{1}{e}, f ( e ) = e 1 ,
ln 2 e 2 e + C e = 1 e ⇒ C = 1 2 . \frac{\ln^2e}{2e}+\frac{C}{e}=\frac{1}{e}\Rightarrow C=\frac{1}{2}.
2 e ln 2 e + e C = e 1 ⇒ C = 2 1 .
因此f ( x ) = ln 2 x + 1 2 x . f(x)=\frac{\ln^2x+1}{2x}. f ( x ) = 2 x l n 2 x + 1 . f ′ ( x ) = 1 x ⋅ 2 ln x ⋅ 2 x − 2 ( ln 2 x + 1 ) ( 2 x ) 2 = 2 ln x − ln 2 x − 1 2 x 2 = − 1 2 ( ln x − 1 x ) 2 ≤ 0. f'(x)=\frac{\frac{1}{x}\cdot 2\ln x\cdot 2x-2(\ln^2x+1)}{(2x)^2}=\frac{2\ln x-\ln^2x-1}{2x^2}=-\frac{1}{2}\big(\frac{\ln x-1}{x}\big)^2\leq0. f ′ ( x ) = ( 2 x ) 2 x 1 ⋅ 2 l n x ⋅ 2 x − 2 ( l n 2 x + 1 ) = 2 x 2 2 l n x − l n 2 x − 1 = − 2 1 ( x l n x − 1 ) 2 ≤ 0. f ( x ) f(x) f ( x ) ( 0 , + ∞ ) (0,+\infty) ( 0 , + ∞ )
函数图像的凹凸是其对应函数的凹凸性的最直观体现。简单地量化就是f ( x 1 + x 2 2 ) > f ( x 1 ) + f ( x 2 ) 2 f(\frac{x_1+x_2}{2})>\frac{f(x_1)+f(x_2)}{2} f ( 2 x 1 + x 2 ) > 2 f ( x 1 ) + f ( x 2 )
对于一个函数，如果其在一个点左右的凹凸性相反，那么这个点就是这个函数的拐点。
根据上方对拐点的描述不难看出，点x = x 0 x=x_0 x = x 0
f ( x ) f(x) f ( x ) x = x 0 x=x_0 x = x 0 x 0 x_0 x 0 x 0 x_0 x 0 f ′ ′ ( x ) f''(x) f ′′ ( x ) f ′ ′ ( x 0 ) = 0 , f ′ ′ ′ ( x 0 ) ≠ 0. f''(x_0)=0,f'''(x_0)≠0. f ′′ ( x 0 ) = 0 , f ′′′ ( x 0 ) = 0. f ( x ) f(x) f ( x ) x = x 0 x=x_0 x = x 0 n n n n > 2 , n ≡ 1 ( m o d 2 ) n>2, n\equiv 1\pmod{2} n > 2 , n ≡ 1 ( mod 2 ) f ′ ′ ( x 0 ) = ⋯ = f ( n − 1 ) ( x 0 ) , f ( n ) ( x 0 ) ≠ 0. f''(x_0)=\cdots=f^{(n-1)}(x_0),f^{(n)}(x_0)≠0. f ′′ ( x 0 ) = ⋯ = f ( n − 1 ) ( x 0 ) , f ( n ) ( x 0 ) = 0.
e.g.2.19 \textbf{e.g.2.19 } e.g.2.19 y = ( x − 1 ) ( x − 2 ) 2 ( x − 3 ) 3 ( x − 4 ) 4 y=(x-1)(x-2)^2(x-3)^3(x-4)^4 y = ( x − 1 ) ( x − 2 ) 2 ( x − 3 ) 3 ( x − 4 ) 4
Solution \textbf{Solution } Solution y = 0 y=0 y = 0 x = 1 x=1 x = 1 x = 2 x=2 x = 2 x = 3 x=3 x = 3 x = 4. x=4. x = 4. y ′ = 0 y'=0 y ′ = 0 x = 2 x=2 x = 2 x = 3 x=3 x = 3 x = 4 x=4 x = 4 ( 1 , 2 ) , ( 2 , 3 ) , ( 3 , 4 ) (1,2),(2,3),(3,4) ( 1 , 2 ) , ( 2 , 3 ) , ( 3 , 4 ) x = ξ 1 , x = ξ 2 , x = ξ 3 . x=\xi_1,x=\xi_2,x=\xi_3. x = ξ 1 , x = ξ 2 , x = ξ 3 . y ′ ′ = 0 y''=0 y ′′ = 0 x = 3 x=3 x = 3 x = 4 x=4 x = 4 ( ξ 1 , 2 ) , ( 2 , ξ 2 ) , ( ξ 2 , 3 ) , ( 3 , ξ 3 ) , ( ξ 3 , 4 ) (\xi_1,2),(2,\xi_2),(\xi_2,3),(3,\xi_3),(\xi_3,4) ( ξ 1 , 2 ) , ( 2 , ξ 2 ) , ( ξ 2 , 3 ) , ( 3 , ξ 3 ) , ( ξ 3 , 4 ) ζ 1 , ζ 2 , ζ 3 , ζ 4 , ζ 5 . \zeta_1,\zeta_2,\zeta_3,\zeta_4,\zeta_5. ζ 1 , ζ 2 , ζ 3 , ζ 4 , ζ 5 . y y y 6. 6. 6.
在上方例子中，二重根之所以不取是因为y ′ ′ = 0 y''=0 y ′′ = 0
e.x.2.18 \textbf{e.x.2.18 } e.x.2.18 y = ( x − 1 ) 2 ( x − 3 ) 2 y=(x-1)^2(x-3)^2 y = ( x − 1 ) 2 ( x − 3 ) 2
Solution \textbf{Solution } Solution y = 0 y=0 y = 0 x 1 = x 2 = 1 x_1=x_2=1 x 1 = x 2 = 1 x 3 = x 4 = 3. x_3=x_4=3. x 3 = x 4 = 3. y ′ = 2 ( x − 1 ) ( x − 3 ) 2 + 2 ( x − 3 ) ( x − 1 ) 2 = 2 ( x − 1 ) ( x − 3 ) ( 2 x − 4 ) , y'=2(x-1)(x-3)^2+2(x-3)(x-1)^2=2(x-1)(x-3)(2x-4), y ′ = 2 ( x − 1 ) ( x − 3 ) 2 + 2 ( x − 3 ) ( x − 1 ) 2 = 2 ( x − 1 ) ( x − 3 ) ( 2 x − 4 ) , y ′ = 0 y'=0 y ′ = 0 x 1 = 1 , x 2 = 2 , x 3 = 3. x_1=1,x_2=2,x_3=3. x 1 = 1 , x 2 = 2 , x 3 = 3. f ′ ( 1 ) = f ′ ( 2 ) = f ′ ( 3 ) f'(1)=f'(2)=f'(3) f ′ ( 1 ) = f ′ ( 2 ) = f ′ ( 3 )
∃ ξ 1 ∈ ( 1 , 2 ) , ξ 2 ∈ ( 2 , 3 ) , s.t. y ′ ′ ∣ x = ξ 1 = 0 = y ′ ′ ∣ x = ξ 2 . \exists\xi_1\in(1,2),\xi_2\in(2,3),\text{ s.t. }y''\big|_{x=\xi_1}=0=y''\big|_{x=\xi_2}.
∃ ξ 1 ∈ ( 1 , 2 ) , ξ 2 ∈ ( 2 , 3 ) , s.t. y ′′ x = ξ 1 = 0 = y ′′ x = ξ 2 .
因此y ′ ′ = ( x − ξ 1 ) ( x − ξ 2 ) , y''=(x-\xi_1)(x-\xi_2), y ′′ = ( x − ξ 1 ) ( x − ξ 2 ) , y ′ ′ y'' y ′′ ξ 1 \xi_1 ξ 1 ξ 2 \xi_2 ξ 2 2. 2. 2.
函数曲率R = ∣ y ′ ′ ∣ ( 1 + y ′ 2 ) 3 2 R=\frac{|y''|}{(1+y'^2)^\frac{3}{2}} R = ( 1 + y ′2 ) 2 3 ∣ y ′′ ∣
相关变化率
这侧重于实际应用场景或者在函数图线中的一些分析。只要构造出两个变量之间的关系后将这两个变量分别看作关于另一个参数的函数，随后对其关于这个参数求导，就可以得到这两个导数之间的关系，之后就可以解出变化率。
e.x.2.19 \textbf{e.x.2.19 } e.x.2.19 L : y = 4 9 x 2 ( x ≥ 0 ) L:y=\frac{4}{9}x^2(x\geq0) L : y = 9 4 x 2 ( x ≥ 0 ) O ( 0 , 0 ) , A ( 0 , 1 ) . O(0,0),A(0,1). O ( 0 , 0 ) , A ( 0 , 1 ) . L L L P , S P,S P , S O A OA O A A P AP A P L L L P P P ( 3 , 4 ) (3,4) ( 3 , 4 ) S S S t t t
Solution \textbf{Solution } Solution t t t P P P ( x ( t ) , 4 9 ( x ( t ) ) 2 ) \Big(x(t),\frac{4}{9}\big(x(t)\big)^2\Big) ( x ( t ) , 9 4 ( x ( t ) ) 2 )
S = 1 2 ( 4 9 ( x ( t ) ) 2 + 1 ) x ( t ) − ∫ 0 x ( t ) 4 9 u 2 d u = 1 2 x ( t ) + 2 27 ( x ( t ) ) 3 . \begin{align*}
S&=\frac{1}{2}\Bigg(\frac{4}{9}\bigg(x(t)\bigg)^2+1\Bigg)x(t)-\int_0^{x(t)}\frac{4}{9}u^2\text{d}u\\
&=\frac{1}{2}x(t)+\frac{2}{27}\big(x(t)\big)^3.
\end{align*}
S = 2 1 ( 9 4 ( x ( t ) ) 2 + 1 ) x ( t ) − ∫ 0 x ( t ) 9 4 u 2 d u = 2 1 x ( t ) + 27 2 ( x ( t ) ) 3 .
因此d S d t = 1 2 d x d t + 2 9 ( x ( t ) ) 2 d x d t . \frac{\text{d}S}{\text{d}t}=\frac{1}{2}\frac{\text{d}x}{\text{d}t}+\frac{2}{9}\big(x(t)\big)^2\frac{\text{d}x}{\text{d}t}. d t d S = 2 1 d t d x + 9 2 ( x ( t ) ) 2 d t d x . d x d t ∣ x = 3 = 4 , \frac{\text{d}x}{\text{d}t}\big|_{x=3}=4, d t d x x = 3 = 4 ,
d S d t ∣ x = 3 = 10. \frac{\text{d}S}{\text{d}t}\bigg|_{x=3}=10.
d t d S x = 3 = 10.
不等式
围绕一元微分学的不等式以证明题为主。这种证明一般可以利用函数本身或辅助函数的单调性、最值、Lagrange中值定理、Taylor公式、凹凸性这六种办法来实现；其中又以前三种为核心。
我们根据单调性和最值的定义就知道它可以证明不等式；关键在于如何构造出可以进行比较且能得到待证不等式的式子。现在考虑下面几个例子。
e.g.2.20 \textbf{e.g.2.20 } e.g.2.20 x ln 1 + x 1 − x + cos x ≥ 1 + x 2 2 ( x ∈ ( − 1 , 1 ) ) . x\ln\frac{1+x}{1-x}+\cos x\geq1+\frac{x^2}{2}\big(x\in(-1,1)\big). x ln 1 − x 1 + x + cos x ≥ 1 + 2 x 2 ( x ∈ ( − 1 , 1 ) ) .
Analyses \textbf{Analyses } Analyses x ln 1 + x 1 − x + cos x − 1 − x 2 2 ≥ 0. x\ln\frac{1+x}{1-x}+\cos x-1-\frac{x^2}{2}\geq 0. x ln 1 − x 1 + x + cos x − 1 − 2 x 2 ≥ 0. x ln 1 + x 1 − x + cos x − 1 − x 2 2 ∣ x = 0 = 0 , x\ln\frac{1+x}{1-x}+\cos x-1-\frac{x^2}{2}\big|_{x=0}=0, x ln 1 − x 1 + x + cos x − 1 − 2 x 2 x = 0 = 0 , x = 0 x=0 x = 0 f ( x ) = x ln 1 + x 1 − x + cos x − 1 − x 2 2 f(x)=x\ln\frac{1+x}{1-x}+\cos x-1-\frac{x^2}{2} f ( x ) = x ln 1 − x 1 + x + cos x − 1 − 2 x 2 ( − 1 , 1 ) (-1,1) ( − 1 , 1 )
Proof \textbf{Proof } Proof f ( x ) = x ln 1 + x 1 − x + cos x − 1 − x 2 2 . f(x)=x\ln\frac{1+x}{1-x}+\cos x-1-\frac{x^2}{2}. f ( x ) = x ln 1 − x 1 + x + cos x − 1 − 2 x 2 .
f ( x ) = x ln ( 1 + x ) − x ln ( 1 − x ) + cos x − 1 − x 2 2 . ⇒ f ′ ( x ) = ( ln ( 1 + x ) + x 1 + x ) − ( ln ( 1 − x ) − x 1 − x ) − sin x − x = ln 1 + x 1 − x + 2 x 1 − x 2 − sin x − x ⇒ f ′ ′ ( x ) = 2 1 − x 2 + 2 ( 1 − x 2 ) + 2 x ⋅ 2 x ( 1 − 2 x ) 2 − cos x − 1 = 4 ( 1 − x 2 ) 2 − cos x − 1. \begin{align*}
&f(x)&=&x\ln(1+x)-x\ln(1-x)+\cos x-1-\frac{x^2}{2}.\\
\Rightarrow&f'(x)&=&(\ln(1+x)+\frac{x}{1+x})-(\ln(1-x)-\frac{x}{1-x})-\sin x-x\\
&&=&\ln\frac{1+x}{1-x}+\frac{2x}{1-x^2}-\sin x-x\\
\Rightarrow&f''(x)&=&\frac{2}{1-x^2}+\frac{2(1-x^2)+2x\cdot 2x}{(1-2x)^2}-\cos x-1\\
&&=&\frac{4}{(1-x^2)^2}-\cos x-1.
\end{align*}
⇒ ⇒ f ( x ) f ′ ( x ) f ′′ ( x ) = = = = = x ln ( 1 + x ) − x ln ( 1 − x ) + cos x − 1 − 2 x 2 . ( ln ( 1 + x ) + 1 + x x ) − ( ln ( 1 − x ) − 1 − x x ) − sin x − x ln 1 − x 1 + x + 1 − x 2 2 x − sin x − x 1 − x 2 2 + ( 1 − 2 x ) 2 2 ( 1 − x 2 ) + 2 x ⋅ 2 x − cos x − 1 ( 1 − x 2 ) 2 4 − cos x − 1.
显然x ∈ ( − 1 , 1 ) x\in(-1,1) x ∈ ( − 1 , 1 ) 1 ( 1 − x 2 ) 2 ≥ 1 ⇒ f ′ ′ ( x ) = 4 ( 1 − x 2 ) 2 − cos x − 1 ≥ 4 − 2 = 2 > 0. \frac{1}{(1-x^2)^2}\geq 1\Rightarrow f''(x)=\frac{4}{(1-x^2)^2}-\cos x-1\geq 4-2=2>0. ( 1 − x 2 ) 2 1 ≥ 1 ⇒ f ′′ ( x ) = ( 1 − x 2 ) 2 4 − cos x − 1 ≥ 4 − 2 = 2 > 0. f ′ ( x ) f'(x) f ′ ( x ) ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) f ′ ( x ) f'(x) f ′ ( x ) f ( x ) f(x) f ( x ) x ∈ ( 0 , 1 ) x\in(0,1) x ∈ ( 0 , 1 ) f ′ ( x ) f'(x) f ′ ( x ) f ′ ( x ) > f ′ ( 0 ) = 0 , f'(x)>f'(0)=0, f ′ ( x ) > f ′ ( 0 ) = 0 , f ( x ) f(x) f ( x ) ( 0 , 1 ) (0,1) ( 0 , 1 ) f ( 0 ) f(0) f ( 0 ) ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) f ( x ) f(x) f ( x )
f ( x ) ≥ f ( 0 ) = 0 ⇒ x ln 1 + x 1 − x + cos x − 1 − x 2 2 ≥ 0 ⇒ x ln 1 + x 1 − x + cos x ≥ 1 + x 2 2 . \begin{align*}
f(x)\geq f(0)=0&\Rightarrow x\ln\frac{1+x}{1-x}+\cos x-1-\frac{x^2}{2}\geq 0\\
&\Rightarrow x\ln\frac{1+x}{1-x}+\cos x\geq 1+\frac{x^2}{2}.
\end{align*}
f ( x ) ≥ f ( 0 ) = 0 ⇒ x ln 1 − x 1 + x + cos x − 1 − 2 x 2 ≥ 0 ⇒ x ln 1 − x 1 + x + cos x ≥ 1 + 2 x 2 .
故原命题得证。
Q.E.D. \textbf{Q.E.D.} Q.E.D.
从上面的例子就已经能感觉到了，并非所有函数都直接依赖其定义域考虑取值范围。我们也需要考虑它可能存在的奇偶性；这可以大幅度减轻运算负担。
e.x.2.20 \textbf{e.x.2.20 } e.x.2.20 a b < a − b ln a − ln b < a + b 2 ( b > a > 0 ) . \sqrt{ab}<\frac{a-b}{\ln a-\ln b}<\frac{a+b}{2}(b>a>0). ab < l n a − l n b a − b < 2 a + b ( b > a > 0 ) .
Analyses \textbf{Analyses } Analyses
(1)证明a − b ln a − ln b < a + b 2 . \frac{a-b}{\ln a-\ln b}<\frac{a+b}{2}. l n a − l n b a − b < 2 a + b . ln b − ln a b − a > 2 a + b , \frac{\ln b-\ln a}{b-a}>\frac{2}{a+b}, b − a l n b − l n a > a + b 2 , ( ln b − ln a ) ( a + b ) − 2 ( b − a ) > 0. (\ln b-\ln a)(a+b)-2(b-a)>0. ( ln b − ln a ) ( a + b ) − 2 ( b − a ) > 0. f ( x ) = ( ln x − ln a ) ( a + x ) − 2 ( x − a ) f(x)=(\ln x-\ln a)(a+x)-2(x-a) f ( x ) = ( ln x − ln a ) ( a + x ) − 2 ( x − a ) f ( b ) > f ( a ) f(b)>f(a) f ( b ) > f ( a )
(2)证明a b < a − b ln a − ln b . \sqrt{ab}<\frac{a-b}{\ln a-\ln b}. ab < l n a − l n b a − b . ln b a ≤ b − a a b , \ln\frac{b}{a}\leq\frac{b-a}{\sqrt{ab}}, ln a b ≤ ab b − a , ln b a ≤ b a − a b . \ln\frac{b}{a}\leq\frac{\sqrt{b}}{\sqrt{a}}-\frac{\sqrt{a}}{\sqrt{b}}. ln a b ≤ a b − b a . ln ( t 2 ) − t + 1 t ≥ 0 \ln(t^2)-t+\frac{1}{t}\geq 0 ln ( t 2 ) − t + t 1 ≥ 0 t > 1 t>1 t > 1
Proof \textbf{Proof } Proof f ( x ) = ( ln x − ln a ) ( a + x ) − 2 ( x − a ) ( x > a > 0 ) , f(x)=(\ln x-\ln a)(a+x)-2(x-a)(x>a>0), f ( x ) = ( ln x − ln a ) ( a + x ) − 2 ( x − a ) ( x > a > 0 ) ,
f ′ ( x ) = ( x + a ) ⋅ 1 x + ( ln x − ln a ) − 2 = a x + ln x − ln a − 1. ⇒ f ′ ′ ( x ) = − a x 2 + 1 x = 1 x ( 1 − a x ) . \begin{align*}
&f'(x)&=&(x+a)\cdot\frac{1}{x}+(\ln x-\ln a)-2\\
&&=&\frac{a}{x}+\ln x-\ln a-1.\\
\Rightarrow&f''(x)&=&-\frac{a}{x^2}+\frac{1}{x}\\
&&=&\frac{1}{x}(1-\frac{a}{x}).
\end{align*}
⇒ f ′ ( x ) f ′′ ( x ) = = = = ( x + a ) ⋅ x 1 + ( ln x − ln a ) − 2 x a + ln x − ln a − 1. − x 2 a + x 1 x 1 ( 1 − x a ) .
因此x ∈ ( a , + ∞ ) x\in(a,+\infty) x ∈ ( a , + ∞ ) f ′ ′ ( x ) > 0 , f''(x)>0, f ′′ ( x ) > 0 , f ′ ( x ) f'(x) f ′ ( x ) f ′ ( x ) > f ′ ( a ) = 0 , f'(x)>f'(a)=0, f ′ ( x ) > f ′ ( a ) = 0 , f ( x ) f(x) f ( x ) f ( x ) > f ( a ) = 0 , f(x)>f(a)=0, f ( x ) > f ( a ) = 0 ,
∀ b ∈ ( a , + ∞ ) , s.t. ( ln b − ln a ) ( a + b ) − 2 ( b − a ) > 0. \forall b\in(a,+\infty),\text{ s.t. }(\ln b-\ln a)(a+b)-2(b-a)>0.
∀ b ∈ ( a , + ∞ ) , s.t. ( ln b − ln a ) ( a + b ) − 2 ( b − a ) > 0.
这个不等式可以化为ln b − ln a b − a > 2 a + b , \frac{\ln b-\ln a}{b-a}>\frac{2}{a+b}, b − a l n b − l n a > a + b 2 , a − b ln a − ln b < a + b 2 . \frac{a-b}{\ln a-\ln b}<\frac{a+b}{2}. l n a − l n b a − b < 2 a + b .
另一方面，记g ( x ) = 2 ln x − x + 1 x ( x ∈ ( 1 , + ∞ ) ) . g(x)=2\ln x-x+\frac{1}{x}(x\in(1,+\infty)). g ( x ) = 2 ln x − x + x 1 ( x ∈ ( 1 , + ∞ )) . g ′ ( x ) = 2 x − 1 − 1 x 2 = − ( 1 x − 1 ) 2 ≤ 0 , g'(x)=\frac{2}{x}-1-\frac{1}{x^2}=-(\frac{1}{x}-1)^2\leq 0, g ′ ( x ) = x 2 − 1 − x 2 1 = − ( x 1 − 1 ) 2 ≤ 0 , g ( x ) g(x) g ( x ) ( 1 , + ∞ ) (1,+\infty) ( 1 , + ∞ ) g ( x ) < g ( 1 ) = 0 , g(x)<g(1)=0, g ( x ) < g ( 1 ) = 0 , 2 ln x < x − 1 x . 2\ln x<x-\frac{1}{x}. 2 ln x < x − x 1 .
不妨取x = b a . x=\sqrt{\frac{b}{a}}. x = a b .
2 ln b a < b a − a b ⇒ ln b a < b a − a b ⇒ ln b − ln a < b − a a b ⇒ a b < b − a ln b − ln a . \begin{align*}
2\ln\sqrt{\frac{b}{a}}<\sqrt{\frac{b}{a}}-\sqrt{\frac{a}{b}}&\Rightarrow\ln\frac{b}{a}<\frac{\sqrt{b}}{\sqrt{a}}-\frac{\sqrt{a}}{\sqrt{b}}\\
&\Rightarrow\ln b-\ln a<\frac{b-a}{\sqrt{ab}}\\
&\Rightarrow\sqrt{ab}<\frac{b-a}{\ln b-\ln a}.
\end{align*}
2 ln a b < a b − b a ⇒ ln a b < a b − b a ⇒ ln b − ln a < ab b − a ⇒ ab < ln b − ln a b − a .
即a b < a − b ln a − ln b . \sqrt{ab}<\frac{a-b}{\ln a-\ln b}. ab < l n a − l n b a − b .
综合上述两方面，原命题得证。
Q.E.D. \textbf{Q.E.D.} Q.E.D.
思考利用Lagrange中值定理证明不等式的基本原理。
我们本就知道一个形如“一个函数的两值之差”的式子可以用一个带有这个函数的导数在某处的取值的式子来表示，也就是f ( b ) − f ( a ) = f ′ ( ξ ) ( b − a ) . f(b)-f(a)=f'(\xi)(b-a). f ( b ) − f ( a ) = f ′ ( ξ ) ( b − a ) . ξ \xi ξ ξ ∈ ( a , b ) , \xi\in(a,b), ξ ∈ ( a , b ) ,
这里举一个x 1 + x < ln ( 1 + x ) < x \frac{x}{1+x}<\ln(1+x)<x 1 + x x < ln ( 1 + x ) < x ln ( 1 + x ) − ln 1 , \ln(1+x)-\ln 1, ln ( 1 + x ) − ln 1 , f ( x ) = ln x , f(x)=\ln x, f ( x ) = ln x , ln ( 1 + x ) − ln 1 = 1 ξ ( 1 + x − 1 ) , \ln(1+x)-\ln 1=\frac{1}{\xi}(1+x-1), ln ( 1 + x ) − ln 1 = ξ 1 ( 1 + x − 1 ) , ξ ∈ ( 1 , 1 + x ) , \xi\in(1,1+x), ξ ∈ ( 1 , 1 + x ) , 1 ξ ∈ ( 1 1 + x , 1 ) . \frac{1}{\xi}\in(\frac{1}{1+x},1). ξ 1 ∈ ( 1 + x 1 , 1 ) .
ln ( 1 + x ) − ln ( 1 ) = x ξ ∈ ( x 1 + x , x ) . \ln(1+x)-\ln(1)=\frac{x}{\xi}\in\Big(\frac{x}{1+x},x\Big).
ln ( 1 + x ) − ln ( 1 ) = ξ x ∈ ( 1 + x x , x ) .
这也就是原不等式。
e.g.2.21 \textbf{e.g.2.21 } e.g.2.21 f ( x ) f(x) f ( x ) f ′ ′ ( x ) < 0 , f ( 0 ) = 0 , f''(x)<0,f(0)=0, f ′′ ( x ) < 0 , f ( 0 ) = 0 , ∀ x 1 > 0 , x 2 > 0 , s.t. f ( x 1 ) + f ( x 2 ) > f ( x 1 + x 2 ) . \forall x_1>0, x_2>0, \text{ s.t. }f(x_1)+f(x_2)>f(x_1+x_2). ∀ x 1 > 0 , x 2 > 0 , s.t. f ( x 1 ) + f ( x 2 ) > f ( x 1 + x 2 ) .
Analyses \textbf{Analyses } Analyses f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 f ( x 1 + x 2 ) − f ( x 2 ) < f ( x 1 ) − f ( 0 ) . f(x_1+x_2)-f(x_2)<f(x_1)-f(0). f ( x 1 + x 2 ) − f ( x 2 ) < f ( x 1 ) − f ( 0 ) .
Proof \textbf{Proof } Proof
∃ ξ 1 ∈ ( 0 , x 1 ) , s.t. f ( x 1 ) − f ( 0 ) = f ′ ( ξ 1 ) ( x 1 − 0 ) , ∃ ξ 2 ∈ ( x 1 , x 2 ) , s.t. f ( x 1 + x 2 ) − f ( x 2 ) = f ′ ( ξ 2 ) ( x 1 + x 2 − x 2 ) . ⇒ f ( x 1 ) − f ( 0 ) = x 1 f ′ ( ξ 1 ) , f ( x 1 + x 2 ) − f ( x 2 ) = x 1 f ′ ( ξ 2 ) . \begin{align*}
&\exists\xi_1\in(0,x_1),\text{ s.t. }f(x_1)-f(0)=f'(\xi_1)(x_1-0),\\
&\exists\xi_2\in(x_1,x_2),\text{ s.t. }f(x_1+x_2)-f(x_2)=f'(\xi_2)(x_1+x_2-x_2).\\
\Rightarrow&f(x_1)-f(0)=x_1f'(\xi_1),\\
&f(x_1+x_2)-f(x_2)=x_1f'(\xi_2).
\end{align*}
⇒ ∃ ξ 1 ∈ ( 0 , x 1 ) , s.t. f ( x 1 ) − f ( 0 ) = f ′ ( ξ 1 ) ( x 1 − 0 ) , ∃ ξ 2 ∈ ( x 1 , x 2 ) , s.t. f ( x 1 + x 2 ) − f ( x 2 ) = f ′ ( ξ 2 ) ( x 1 + x 2 − x 2 ) . f ( x 1 ) − f ( 0 ) = x 1 f ′ ( ξ 1 ) , f ( x 1 + x 2 ) − f ( x 2 ) = x 1 f ′ ( ξ 2 ) .
显然上述有ξ 1 < ξ 2 . \xi_1<\xi_2. ξ 1 < ξ 2 . f ′ ′ ( x ) < 0 ⇒ f ′ ( x ) f''(x)<0\Rightarrow f'(x) f ′′ ( x ) < 0 ⇒ f ′ ( x ) f ′ ( ξ 1 ) > f ′ ( ξ 2 ) , f'(\xi_1)>f'(\xi_2), f ′ ( ξ 1 ) > f ′ ( ξ 2 ) ,
f ( x 1 + x 2 ) − f ( x 2 ) = x 1 f ′ ( ξ 2 ) < x 1 f ′ ( ξ 1 ) = f ( x 1 ) − f ( 0 ) . ⇒ f ( x 1 + x 2 ) − f ( x 2 ) < f ( x 1 ) − f ( 0 ) = f ( x 1 ) . ⇒ f ( x 1 + x 2 ) < f ( x 1 ) + f ( x 2 ) . \begin{align*}
&f(x_1+x_2)-f(x_2)=x_1f'(\xi_2)<x_1f'(\xi_1)=f(x_1)-f(0).\\
\Rightarrow&f(x_1+x_2)-f(x_2)<f(x_1)-f(0)=f(x_1).\\
\Rightarrow&f(x_1+x_2)<f(x_1)+f(x_2).
\end{align*}
⇒ ⇒ f ( x 1 + x 2 ) − f ( x 2 ) = x 1 f ′ ( ξ 2 ) < x 1 f ′ ( ξ 1 ) = f ( x 1 ) − f ( 0 ) . f ( x 1 + x 2 ) − f ( x 2 ) < f ( x 1 ) − f ( 0 ) = f ( x 1 ) . f ( x 1 + x 2 ) < f ( x 1 ) + f ( x 2 ) .
原命题因此得证。
Q.E.D. \textbf{Q.E.D.} Q.E.D.
思考利用Taylor公式证明不等式的基本原理。
我们知道Taylor公式展开后含有若干项。那么根据数学公理就知道，当一个式子是这个展开式的其中一部分时其一定无法大于这个展开式所对应的原函数。这就是一个不等式。
思考利用凹凸性证明不等式的基本原理。
现在以凹函数为例。我们已经知道f ( a ) + f ( b ) 2 ≥ f ( a + b 2 ) \frac{f(a)+f(b)}{2}\geq f(\frac{a+b}{2}) 2 f ( a ) + f ( b ) ≥ f ( 2 a + b ) f ( x ) = e x f(x)=e^x f ( x ) = e x e a + e b 2 ≥ e a + b 2 . \frac{e^a+e^b}{2}\geq e^{\frac{a+b}{2}}. 2 e a + e b ≥ e 2 a + b .
方程根
想要讨论方程根，自然是讨论它的存在性和唯一性。
根据初中知识就知道对于存在性我们可以根据零点定理。但对于找不到f ( a ) f ( b ) < 0 f(a)f(b)<0 f ( a ) f ( b ) < 0 f ( x ) f(x) f ( x ) F ( x ) , F(x), F ( x ) , F ( c ) = F ( d ) F(c)=F(d) F ( c ) = F ( d ) f ( x ) = 0 f(x)=0 f ( x ) = 0 ( c , d ) (c,d) ( c , d )
一元n n n
a 0 + a 1 x + a 2 x 2 + ⋯ + a n x n = 0 ( a 0 ≠ 0 ) a_0+a_1x+a_2x^2+\cdots+a_nx^n=0(a_0≠0)
a 0 + a 1 x + a 2 x 2 + ⋯ + a n x n = 0 ( a 0 = 0 )
在n n n
反证法（假设某个区间方程有第二个实根）或利用函数f ( x ) f(x) f ( x ) Rolle定理推论的逆否命题（若f ( x ) f(x) f ( x ) n n n f ( x ) f(x) f ( x ) n n n 求函数的根数上限后证明。
这里以n = 2 n=2 n = 2 f ′ ′ ( 0 ) = 0 f''(0)=0 f ′′ ( 0 ) = 0 f ( x ) = 0 f(x)=0 f ( x ) = 0
Proof \textbf{Proof } Proof f ( x ) f(x) f ( x ) x 1 < x 2 < x 3 . x_1<x_2<x_3. x 1 < x 2 < x 3 .
∃ ζ 1 ∈ ( x 1 , x 2 ) , s.t. f ′ ( ζ 1 ) = 0 , \exists\zeta_1\in(x_1,x_2),\text{ s.t. }f'(\zeta_1)=0,
∃ ζ 1 ∈ ( x 1 , x 2 ) , s.t. f ′ ( ζ 1 ) = 0 ,
∃ ζ 2 ∈ ( x 2 , x 3 ) , s.t. f ′ ( ζ 2 ) = 0. \exists\zeta_2\in(x_2,x_3),\text{ s.t. }f'(\zeta_2)=0.
∃ ζ 2 ∈ ( x 2 , x 3 ) , s.t. f ′ ( ζ 2 ) = 0.
这时根据Rolle定理又能得到
∃ ξ ∈ ( ζ 1 , ζ 2 ) , s.t. f ′ ′ ( ξ ) = 0. \exists\xi\in(\zeta_1,\zeta_2),\text{ s.t. }f''(\xi)=0.
∃ ξ ∈ ( ζ 1 , ζ 2 ) , s.t. f ′′ ( ξ ) = 0.
这就和已知的“f ′ ′ ( 0 ) f''(0) f ′′ ( 0 )
Q.E.D. \textbf{Q.E.D.} Q.E.D.
e.g.2.22 \textbf{e.g.2.22 } e.g.2.22 3 a 2 − 5 b < 0 , 3a^2-5b<0, 3 a 2 − 5 b < 0 , x 5 + 2 a x 3 + 3 b x + 4 c = 0 x^5+2ax^3+3bx+4c=0 x 5 + 2 a x 3 + 3 b x + 4 c = 0
Solution \textbf{Solution } Solution b > 3 5 a 2 . b>\frac{3}{5}a^2. b > 5 3 a 2 . f ( x ) = x 5 + 2 a x 3 + 3 b x + 4 c , f(x)=x^5+2ax^3+3bx+4c, f ( x ) = x 5 + 2 a x 3 + 3 b x + 4 c ,
f ′ ( x ) = 5 x 4 + 6 a x 2 + 3 b > 5 x 4 + 6 a x 2 + 9 5 a 2 = ( 5 x 2 + 3 a 5 ) 2 ≥ 0. \begin{align*}
f'(x)&=5x^4+6ax^2+3b
&>5x^4+6ax^2+\frac{9}{5}a^2=(\sqrt{5}x^2+\frac{3a}{\sqrt{5}})^2\geq 0.
\end{align*}
f ′ ( x ) = 5 x 4 + 6 a x 2 + 3 b > 5 x 4 + 6 a x 2 + 5 9 a 2 = ( 5 x 2 + 5 3 a ) 2 ≥ 0.
所以f ( x ) f(x) f ( x ) f ( x ) = 0 f(x)=0 f ( x ) = 0 1 .
e.x.2.21 \textbf{e.x.2.21 } e.x.2.21 f ( x ) = ∫ x 1 1 + t 2 d t + ∫ 1 x 2 1 + t d t f(x)=\int_x^1\sqrt{1+t^2}\text{d}t+\int_1^{x^2}\sqrt{1+t}\text{d}t f ( x ) = ∫ x 1 1 + t 2 d t + ∫ 1 x 2 1 + t d t
Solution \textbf{Solution } Solution
f ′ ( x ) = 2 x 1 + x 2 − 1 + x 2 = 1 + x 2 ( 2 x − 1 ) . \begin{align*}
f'(x)&=2x\sqrt{1+x^2}-\sqrt{1+x^2}
&=\sqrt{1+x^2}(2x-1).
\end{align*}
f ′ ( x ) = 2 x 1 + x 2 − 1 + x 2 = 1 + x 2 ( 2 x − 1 ) .
因此f ′ ( x ) = 0 f'(x)=0 f ′ ( x ) = 0 x = 1 2 . x=\frac{1}{2}. x = 2 1 . f ( x ) f(x) f ( x ) ( − ∞ , 1 2 ) (-\infty,\frac{1}{2}) ( − ∞ , 2 1 ) ( 1 2 , + ∞ ) (\frac{1}{2},+\infty) ( 2 1 , + ∞ ) f ( 1 ) = 0 , f ( − 1 ) = ∫ − 1 1 1 + t 2 d t > 0 , f(1)=0,f(-1)=\int_{-1}^1\sqrt{1+t^2}\text{d}t>0, f ( 1 ) = 0 , f ( − 1 ) = ∫ − 1 1 1 + t 2 d t > 0 , f ( 1 2 ) < 0. f\big(\frac{1}{2}\big)<0. f ( 2 1 ) < 0. f ( x ) f(x) f ( x ) ( − 1 , 1 2 ) (-1,\frac{1}{2}) ( − 1 , 2 1 ) ( 1 2 , + ∞ ) (\frac{1}{2},+\infty) ( 2 1 , + ∞ ) 2 .
$\textbf{e.x.2.22 }$求$\int_0^xe^{-t^2}\text{d}t=x^3-x$的实根个数。
Solution \textbf{Solution } Solution f ( x ) = ∫ 0 x e − t 2 d t − x 3 + x . f(x)=\int_0^xe^{-t^2}\text{d}t-x^3+x. f ( x ) = ∫ 0 x e − t 2 d t − x 3 + x . f ′ ( x ) = e − x 2 − 3 x 2 + 1. f'(x)=e^{-x^2}-3x^2+1. f ′ ( x ) = e − x 2 − 3 x 2 + 1. x = 0 ; x=0; x = 0 ; f ′ ′ ( x ) = − 2 x e − x 2 − 6 x = − 2 x ( e − x 2 + 3 ) , f''(x)=-2xe^{-x^2}-6x=-2x(e^{-x^2}+3), f ′′ ( x ) = − 2 x e − x 2 − 6 x = − 2 x ( e − x 2 + 3 ) , e − x 2 + 3 > 0 + 3 = 3 , e^{-x^2}+3>0+3=3, e − x 2 + 3 > 0 + 3 = 3 , x = 0. x=0. x = 0. f ′ ( x ) f'(x) f ′ ( x ) ( − ∞ , 0 ) (-\infty,0) ( − ∞ , 0 ) ( 0 , + ∞ ) (0,+\infty) ( 0 , + ∞ ) f ′ ( x ) ≤ f ′ ( 0 ) = 0. f'(x)\leq f'(0)=0. f ′ ( x ) ≤ f ′ ( 0 ) = 0. f ( x ) f(x) f ( x ) f ( x ) = 0 f(x)=0 f ( x ) = 0 f ( 0 ) = 0 , f(0)=0, f ( 0 ) = 0 , x = 0. x=0. x = 0. 0 .
一元积分
原函数
F ′ ( x ) = f ( x ) ⟺ F ( x ) F'(x)=f(x)\iff F(x) F ′ ( x ) = f ( x ) ⟺ F ( x ) f ( x ) f(x) f ( x )
这是人尽皆知的原函数定义，因此在这方面很少考简单的概念而是原函数存在定理。
原函数存在定理
函数f ( x ) f(x) f ( x ) F ( x ) F(x) F ( x ) f ( x ) f(x) f ( x ) f ( x ) f(x) f ( x ) ⇒ F ( x ) \Rightarrow F(x) ⇒ F ( x ) f ( x ) f(x) f ( x ) ⇒ F ( x ) \Rightarrow F(x) ⇒ F ( x ) f ( x ) f(x) f ( x ) ⇒ F ( x ) \Rightarrow F(x) ⇒ F ( x )
考虑原函数存在定理的第三个情况。我们举函数f ( x ) = { 2 x cos 1 x + sin 1 x , x ≠ 0 0 , x = 0 f(x)=\begin{cases}2x\cos\frac{1}{x}+\sin\frac{1}{x}&,x≠0\\0&,x=0\end{cases} f ( x ) = { 2 x cos x 1 + sin x 1 0 , x = 0 , x = 0 lim x → 0 ( 2 x cos 1 x + sin 1 x ) \lim_{x\rightarrow 0}(2x\cos\frac{1}{x}+\sin\frac{1}{x}) lim x → 0 ( 2 x cos x 1 + sin x 1 ) x = 0 x=0 x = 0 f ( x ) f(x) f ( x ) F ( x ) = { x 2 cos 1 x + C 1 , x ≠ 0 C 2 , x = 0 . F(x)=\begin{cases}x^2\cos\frac{1}{x}+C_1&,x≠0\\C_2&,x=0\end{cases}. F ( x ) = { x 2 cos x 1 + C 1 C 2 , x = 0 , x = 0 .
对于形如∫ a x f ( t ) d t \int_a^xf(t)\text{d}t ∫ a x f ( t ) d t
（微积分基本原理）若f ( t ) f(t) f ( t ) f ( t ) f(t) f ( t ) F ′ ( x ) = f ( x ) . F'(x)=f(x). F ′ ( x ) = f ( x ) .
若f ( t ) f(t) f ( t ) F ( x ) F(x) F ( x )
对第二个性质，考虑在[ − 1 , 1 ] [-1,1] [ − 1 , 1 ] f ( t ) = sgn t . f(t)=\text{sgn }t. f ( t ) = sgn t . F ( x ) = ∫ − 1 x f ( t ) d t = ∣ x ∣ − 1 , F(x)=\int_{-1}^xf(t)\text{d}t=|x|-1, F ( x ) = ∫ − 1 x f ( t ) d t = ∣ x ∣ − 1 , x = 0 x=0 x = 0
根据上述性质也就能知道
e.g.3.1 \textbf{e.g.3.1 } e.g.3.1 g ( x ) = ∫ 0 x f ( u ) d u , g(x)=\int_0^xf(u)\text{d}u, g ( x ) = ∫ 0 x f ( u ) d u , f ( x ) = { 1 2 ( x 2 + 1 ) , 0 ≤ x < 1 1 3 ( x − 1 ) , 1 ≤ x < 2 , f(x)=\begin{cases}\frac{1}{2}(x^2+1)&,0\leq x<1\\\frac{1}{3}(x-1)&,1\leq x<2\end{cases}, f ( x ) = { 2 1 ( x 2 + 1 ) 3 1 ( x − 1 ) , 0 ≤ x < 1 , 1 ≤ x < 2 , g ( x ) g(x) g ( x ) ( 0 , 2 ) (0,2) ( 0 , 2 )
A. 无界
B. 递减
C. 不连续
D. 连续
Draft \textbf{Draft } Draft g ( x ) g(x) g ( x ) Answer \textbf{Answer } Answer
e.x.3.1 \textbf{e.x.3.1 } e.x.3.1 f ( x ) = { sin x , 0 ≤ x < π 2 , π ≤ x ≤ 2 π , f(x)=\begin{cases}\sin x&,0\leq x<\pi\\2&,\pi\leq x\leq2\pi\end{cases}, f ( x ) = { sin x 2 , 0 ≤ x < π , π ≤ x ≤ 2 π , F ( x ) = ∫ 0 x f ( t ) d t F(x)=\int_0^xf(t)\text{d}t F ( x ) = ∫ 0 x f ( t ) d t x = π x=\pi x = π
A. 此处是跳跃间断点
B. 此处是可去间断点
C. 函数在此处连续但不可导
D. 函数在此处可导
Draft \textbf{Draft } Draft F ( x ) F(x) F ( x ) F ( x ) F(x) F ( x ) f ( x ) f(x) f ( x ) x = π x=\pi x = π f ( π − ) ≠ f ( π + ) f(\pi^-)≠f(\pi^+) f ( π − ) = f ( π + ) F ( x ) F(x) F ( x ) x = π x=\pi x = π
Answer \textbf{Answer } Answer
思考定积分和不定积分的联系和区别。我们知道一个函数f ( x ) f(x) f ( x )
考虑函数F ( x ) = { x 2 sin 1 x 2 , x ≠ 0 0 , x = 0 . F(x)=\begin{cases}x^2\sin\frac{1}{x^2}&,x≠0\\0&,x=0\end{cases}. F ( x ) = { x 2 sin x 2 1 0 , x = 0 , x = 0 . f ( x ) = F ′ ( x ) = { 2 x sin 1 x 2 − 2 x cos 1 x 2 , x ≠ 0 0 , x = 0 , f(x)=F'(x)=\begin{cases}2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2}&,x≠0\\0&,x=0\end{cases}, f ( x ) = F ′ ( x ) = { 2 x sin x 2 1 − x 2 cos x 2 1 0 , x = 0 , x = 0 , [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] f ( x ) f(x) f ( x )
再考虑符号函数f ( x ) = sgn x = { 1 , x > 0 0 , x = 0 − 1 , x < 0 . f(x)=\text{sgn }x=\begin{cases}1&,x>0\\0&,x=0\\-1&,x<0\end{cases}. f ( x ) = sgn x = ⎩ ⎨ ⎧ 1 0 − 1 , x > 0 , x = 0 , x < 0 . x = 0 x=0 x = 0 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
但不定积分本质上也就是两个原函数之差，这也是其和定积分之间的潜在联系。
不定积分
这里围绕讨论的是求不定积分的方法。
公式法
利用常用的积分公式求不定积分。
我们本就掌握了基本函数的导数。根据这些导数自然也就能反推回原函数，这也就是常用的积分公式：
∫ x α d x = 1 1 + α x 1 + α + C ; ( α ≠ − 1 ) \int x^\alpha\text{d}x=\frac{1}{1+\alpha}x^{1+\alpha}+C;(\alpha≠-1) ∫ x α d x = 1 + α 1 x 1 + α + C ; ( α = − 1 ) ∫ 1 x d x = ln ∣ x ∣ + C ; \int\frac{1}{x}\text{d}x=\ln |x|+C; ∫ x 1 d x = ln ∣ x ∣ + C ; ∫ a x d x = x a ln a + C ; \int a^x\text{d}x=\frac{x^a}{\ln a}+C; ∫ a x d x = l n a x a + C ; ∫ cos x d x = sin x + C ; \int\cos x\text{d}x=\sin x+C; ∫ cos x d x = sin x + C ; ∫ sin x d x = − cos x + C ; \int\sin x\text{d}x=-\cos x+C; ∫ sin x d x = − cos x + C ; ∫ 1 a 2 + x 2 d x = 1 a arctan x a + C ; \int\frac{1}{a^2+x^2}\text{d}x=\frac{1}{a}\arctan\frac{x}{a}+C; ∫ a 2 + x 2 1 d x = a 1 arctan a x + C ; ∫ sec 2 x d x = ∫ 1 cos 2 x d x = tan x + C ; \int\sec^2x\text{d}x=\int\frac{1}{\cos^2x}\text{d}x=\tan x+C; ∫ sec 2 x d x = ∫ c o s 2 x 1 d x = tan x + C ; ∫ csc 2 x d x = ∫ 1 sin 2 x d x = − cot x + C ; \int\csc^2x\text{d}x=\int\frac{1}{\sin^2x}\text{d}x=-\cot x+C; ∫ csc 2 x d x = ∫ s i n 2 x 1 d x = − cot x + C ; ∫ 1 1 − x 2 d x = arcsin x + C ; \int\frac{1}{\sqrt{1-x^2}}\text{d}x=\arcsin x+C; ∫ 1 − x 2 1 d x = arcsin x + C ; ∫ 1 a 2 − x 2 d x = 1 2 a ln ∣ a + x a − x ∣ + C ; \int\frac{1}{a^2-x^2}\text{d}x=\frac{1}{2a}\ln|\frac{a+x}{a-x}|+C; ∫ a 2 − x 2 1 d x = 2 a 1 ln ∣ a − x a + x ∣ + C ; ∫ 1 x 2 ± a 2 d x = ln ∣ x + x 2 ± a 2 ∣ + C . \int\frac{1}{\sqrt{x^2\pm a^2}}\text{d}x=\ln|x+\sqrt{x^2\pm a^2}|+C. ∫ x 2 ± a 2 1 d x = ln ∣ x + x 2 ± a 2 ∣ + C .
除了我们根据常用导数表反推就能得到的简单函数的原函数公式之外，还可以使用下面三个常用的公式：
∫ sec x d x = ln ∣ sec x + tan x ∣ + C ; \int\sec x\text{d}x=\ln|\sec x+\tan x|+C; ∫ sec x d x = ln ∣ sec x + tan x ∣ + C ; ∫ sec 3 x d x = 1 2 sec x tan x + 1 2 ln ∣ sec x + tan x ∣ + C ; \int\sec^3x\text{d}x=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\sec x+\tan x|+C; ∫ sec 3 x d x = 2 1 sec x tan x + 2 1 ln ∣ sec x + tan x ∣ + C ; ∫ csc x d x = ln ∣ csc x − cot x ∣ + C . \int\csc x\text{d}x=\ln|\csc x-\cot x|+C. ∫ csc x d x = ln ∣ csc x − cot x ∣ + C .
e.g.3.2 \textbf{e.g.3.2 } e.g.3.2 ∫ e 2 x 3 + e 4 x d x . \int\frac{e^{2x}}{3+e^{4x}}\text{d}x. ∫ 3 + e 4 x e 2 x d x . Solution \textbf{Solution } Solution
O.F. = t = e 2 x ∫ 1 2 d ( e 2 x ) ( 3 ) 2 + ( e 2 x ) 2 = ∫ 1 a 2 + t 2 d t = 1 a arctan t a + C 1 2 ( 1 3 arctan e 2 x 3 + C ) = 3 6 arctan 3 e 2 x 3 + C . \begin{align*}
\text{O.F.}&\xlongequal{t=e^{2x}}\int\frac{\frac{1}{2}\text{d}(e^{2x})}{(\sqrt{3})^2+(e^{2x})^2}\\
&\xlongequal{\int\frac{1}{a^2+t^2}\text{d}t=\frac{1}{a}\arctan\frac{t}{a}+C}\frac{1}{2}\big(\frac{1}{\sqrt{3}}\arctan\frac{e^{2x}}{\sqrt{3}}+C\big)\\
&=\frac{\sqrt{3}}{6}\arctan\frac{\sqrt{3}e^{2x}}{3}+C.
\end{align*}
O.F. t = e 2 x ∫ ( 3 ) 2 + ( e 2 x ) 2 2 1 d ( e 2 x ) ∫ a 2 + t 2 1 d t = a 1 a r c t a n a t + C 2 1 ( 3 1 arctan 3 e 2 x + C ) = 6 3 arctan 3 3 e 2 x + C .
e.g.3.3 \textbf{e.g.3.3 } e.g.3.3 ∫ 1 − x 1 + x d x . \int\sqrt{\frac{1-x}{1+x}}\text{d}x. ∫ 1 + x 1 − x d x . Solution \textbf{Solution } Solution
O.F. = ∫ 1 − x 1 − x 2 d x = ∫ 1 1 − x 2 d x − ∫ x 1 − x 2 d x = arcsin x − ∫ ( − 1 2 ) d ( 1 − x 2 ) 1 − x 2 = arcsin x + ∫ 1 2 1 − x 2 d ( 1 − x 2 ) = arcsin x + 1 − x 2 + C . \begin{align*}
\text{O.F.}&=\int\frac{1-x}{\sqrt{1-x^2}}\text{d}x=\int\frac{1}{\sqrt{1-x^2}}\text{d}x-\int\frac{x}{\sqrt{1-x^2}}\text{d}x\\
&=\arcsin x-\int\frac{(-\frac{1}{2})\text{d}(1-x^2)}{\sqrt{1-x^2}}=\arcsin x+\int\frac{1}{2\sqrt{1-x^2}}\text{d}(1-x^2)\\
&=\arcsin x+\sqrt{1-x^2}+C.
\end{align*}
O.F. = ∫ 1 − x 2 1 − x d x = ∫ 1 − x 2 1 d x − ∫ 1 − x 2 x d x = arcsin x − ∫ 1 − x 2 ( − 2 1 ) d ( 1 − x 2 ) = arcsin x + ∫ 2 1 − x 2 1 d ( 1 − x 2 ) = arcsin x + 1 − x 2 + C .
e.x.3.2 \textbf{e.x.3.2 } e.x.3.2 ∫ e 2 x 3 − e 4 x d x . \int\frac{e^{2x}}{3-e^{4x}}\text{d}x. ∫ 3 − e 4 x e 2 x d x . Solution \textbf{Solution } Solution
O.F. = t = e 2 x ∫ 1 2 d ( e 2 x ) ( 3 ) 2 − ( e 2 x ) 2 = ∫ 1 a 2 − x 2 d x = 1 2 a ln ∣ a + x a − x ∣ + C 1 2 ⋅ 1 2 3 ln ∣ 3 + e 2 x 3 − e 2 x ∣ + C = 3 12 ln ∣ 3 + e 2 x 3 − e 2 x ∣ + C . \begin{align*}
\text{O.F.}&\xlongequal{t=e^{2x}}\int\frac{\frac{1}{2}\text{d}(e^{2x})}{(\sqrt{3})^2-(e^{2x})^2}\\
&\xlongequal{\int\frac{1}{a^2-x^2}\text{d}x=\frac{1}{2a}\ln|\frac{a+x}{a-x}|+C}\frac{1}{2}\cdot\frac{1}{2\sqrt{3}}\ln\bigg|\frac{\sqrt{3}+e^{2x}}{\sqrt{3}-e^{2x}}\bigg|+C\\
&=\frac{\sqrt{3}}{12}\ln\bigg|\frac{\sqrt{3}+e^{2x}}{\sqrt{3}-e^{2x}}\bigg|+C.
\end{align*}
O.F. t = e 2 x ∫ ( 3 ) 2 − ( e 2 x ) 2 2 1 d ( e 2 x ) ∫ a 2 − x 2 1 d x = 2 a 1 l n ∣ a − x a + x ∣ + C 2 1 ⋅ 2 3 1 ln 3 − e 2 x 3 + e 2 x + C = 12 3 ln 3 − e 2 x 3 + e 2 x + C .
凑微分法
我们知道对于一个有导函数f ( x ) f(x) f ( x ) F ( x ) F(x) F ( x ) g ( x ) g(x) g ( x ) ( F ( g ( x ) ) ) ′ = f ( g ( x ) ) g ′ ( x ) . \Big(F\big(g(x)\big)\Big)'=f\big(g(x)\big)g'(x). ( F ( g ( x ) ) ) ′ = f ( g ( x ) ) g ′ ( x ) .
∫ f ( g ( x ) ) g ′ ( x ) d x = F ( g ( x ) ) + C . \int f\big(g(x)\big)g'(x)\text{d}x=F\big(g(x)\big)+C.
∫ f ( g ( x ) ) g ′ ( x ) d x = F ( g ( x ) ) + C .
凑微分法的核心是找到一个中间变量u = g ( x ) , u=g(x), u = g ( x ) , u u u x x x
e.g.3.4 \textbf{e.g.3.4 } e.g.3.4 ∫ e x 1 + e 2 x d x . \int\frac{e^x}{1+e^{2x}}\text{d}x. ∫ 1 + e 2 x e x d x . Solution \textbf{Solution } Solution
O.F. = ∫ 1 1 + ( e x ) 2 d e x = arctan e x + C . \begin{align*}
\text{O.F.}&=\int\frac{1}{1+(e^x)^2}\text{d}e^x\\
&=\arctan e^x+C.
\end{align*}
O.F. = ∫ 1 + ( e x ) 2 1 d e x = arctan e x + C .
e.x.3.3 \textbf{e.x.3.3 } e.x.3.3 ∫ 1 1 + x + ( 1 + x ) 3 d x . \int\frac{1}{\sqrt{1+x}+(\sqrt{1+x})^3}\text{d}x. ∫ 1 + x + ( 1 + x ) 3 1 d x . Solution \textbf{Solution } Solution
O.F. = ∫ d ( 1 + x ) 1 + x ( 1 + ( 1 + x ) 2 ) = ∫ 2 d 1 + x 1 + ( 1 + x ) 2 = 2 arctan 1 + x + C . \begin{align*}
\text{O.F.}&=\int\frac{\text{d}(1+x)}{\sqrt{1+x}(1+(\sqrt{1+x})^2)}=\int\frac{2\text{d}\sqrt{1+x}}{1+(\sqrt{1+x})^2}\\
&=2\arctan\sqrt{1+x}+C.
\end{align*}
O.F. = ∫ 1 + x ( 1 + ( 1 + x ) 2 ) d ( 1 + x ) = ∫ 1 + ( 1 + x ) 2 2 d 1 + x = 2 arctan 1 + x + C .
对于可以使用凑微分法的不定积分式并不一定能一眼望出来变换方法；在求解之前还需要进行一定的变换。
e.g.3.5 \textbf{e.g.3.5 } e.g.3.5 ∫ 2 x ⋅ 3 x 9 x + 4 x d x . \int\frac{2^x\cdot3^x}{9^x+4^x}\text{d}x. ∫ 9 x + 4 x 2 x ⋅ 3 x d x . Solution \textbf{Solution } Solution
O.F. = ∫ ( 2 3 ) x 1 + ( 4 9 ) x d x = ∫ d ( 2 3 ) x ln 2 3 1 + ( ( 2 3 ) x ) 2 = 1 ln 2 − ln 3 arctan ( ( 2 3 ) x ) \begin{align*}
\text{O.F.}&=\int\frac{\big(\frac{2}{3}\big)^x}{1+\big(\frac{4}{9}\big)^x}\text{d}x=\int\frac{\frac{\text{d}(\frac{2}{3})^x}{\ln\frac{2}{3}}}{1+\Big(\big(\frac{2}{3}\big)^x\Big)^2}\\
&=\frac{1}{\ln 2-\ln 3}\arctan\bigg(\Big(\frac{2}{3}\Big)^x\bigg)
\end{align*}
O.F. = ∫ 1 + ( 9 4 ) x ( 3 2 ) x d x = ∫ 1 + ( ( 3 2 ) x ) 2 l n 3 2 d ( 3 2 ) x = ln 2 − ln 3 1 arctan ( ( 3 2 ) x )
e.x.3.4 \textbf{e.x.3.4 } e.x.3.4 ∫ sin x cos x sin 4 x + cos 4 x d x . \int\frac{\sin x\cos x}{\sin^4x+\cos^4x}\text{d}x. ∫ s i n 4 x + c o s 4 x s i n x c o s x d x . Solution \textbf{Solution } Solution
O.F. = ∫ 1 2 sin 2 x 1 2 ( ( cos 2 x − sin 2 x ) 2 + ( cos 2 x + sin 2 x ) 2 ) d x = ∫ − 1 2 cos 2 2 x + 1 d ( cos 2 x ) = − 1 2 arctan ( cos 2 x ) + C . \begin{align*}
\text{O.F.}&=\int\frac{\frac{1}{2}\sin2x}{\frac{1}{2}\big((\cos^2x-\sin^2x)^2+(\cos^2x+\sin^2x)^2\big)}\text{d}x\\
&=\int\frac{-\frac{1}{2}}{\cos^22x+1}\text{d}(\cos2x)\\
&=-\frac{1}{2}\arctan(\cos2x)+C.
\end{align*}
O.F. = ∫ 2 1 ( ( cos 2 x − sin 2 x ) 2 + ( cos 2 x + sin 2 x ) 2 ) 2 1 sin 2 x d x = ∫ cos 2 2 x + 1 − 2 1 d ( cos 2 x ) = − 2 1 arctan ( cos 2 x ) + C .
e.x.3.5 \textbf{e.x.3.5 } e.x.3.5 ∫ 1 + ln x x − x + x x d x . \int\frac{1+\ln x}{x^{-x}+x^x}\text{d}x. ∫ x − x + x x 1 + l n x d x . Solution \textbf{Solution } Solution ( x x ) ′ = ( e x ln x ) ′ = e x ln x ( ln x + 1 ) = x x ( ln x + 1 ) . (x^x)'=(e^{x\ln x})'=e^{x\ln x}(\ln x+1)=x^x(\ln x+1). ( x x ) ′ = ( e x l n x ) ′ = e x l n x ( ln x + 1 ) = x x ( ln x + 1 ) .
O.F. = ∫ x x ( 1 + ln x ) 1 + x 2 x d x = ∫ 1 1 + ( x x ) 2 d ( x x ) = arctan ( x x ) + C . \begin{align*}
\text{O.F.}&=\int\frac{x^x(1+\ln x)}{1+x^{2x}}\text{d}x=\int\frac{1}{1+(x^x)^2}\text{d}(x^x)\\
&=\arctan(x^x)+C.
\end{align*}
O.F. = ∫ 1 + x 2 x x x ( 1 + ln x ) d x = ∫ 1 + ( x x ) 2 1 d ( x x ) = arctan ( x x ) + C .
另外有一些可以先经过换元然后再进行凑微分求不定积分的情况。
e.g.3.6 \textbf{e.g.3.6 } e.g.3.6 ∫ 1 ( 2 − x ) 1 − x d x . \int\frac{1}{(2-x)\sqrt{1-x}}\text{d}x. ∫ ( 2 − x ) 1 − x 1 d x . Solution \textbf{Solution } Solution 1 − x = t , \sqrt{1-x}=t, 1 − x = t , x = 1 − t 2 . x=1-t^2. x = 1 − t 2 .
O.F. = ∫ − 2 t d t t ( 1 + t 2 ) = − 2 ∫ 1 1 + t 2 d t = − 2 arctan t + C = − 2 arctan 1 − x + C . \begin{align*}
\text{O.F.}&=\int\frac{-2t\text{d}t}{t(1+t^2)}=-2\int\frac{1}{1+t^2}\text{d}t\\
&=-2\arctan t+C=-2\arctan\sqrt{1-x}+C.
\end{align*}
O.F. = ∫ t ( 1 + t 2 ) − 2 t d t = − 2 ∫ 1 + t 2 1 d t = − 2 arctan t + C = − 2 arctan 1 − x + C .
显然换元的构造并不会经常这么简单，因此有时还是使用比较原始的凑微分方法。在上方的例3.6中，也可以通过两步将待求不定积分化为∫ 1 ( 2 − x ) 1 − x d x = − ∫ d ( 1 − x ) ( 2 − x ) 1 − x = − 2 ∫ d 1 − x 1 + ( 1 − x ) 2 , \int\frac{1}{(2-x)\sqrt{1-x}}\text{d}x=-\int\frac{\text{d}(1-x)}{(2-x)\sqrt{1-x}}=-2\int\frac{\text{d}\sqrt{1-x}}{1+(\sqrt{1-x})^2}, ∫ ( 2 − x ) 1 − x 1 d x = − ∫ ( 2 − x ) 1 − x d ( 1 − x ) = − 2 ∫ 1 + ( 1 − x ) 2 d 1 − x ,
第二类换元法
除了以简化凑微分法求不定积分的流程为目的之外，换元法本身也可以作为求不定积分的重要思路之一。
考虑通过三角函数的定义和定理进行换元。既然使用三角函数定义，就可以有如下的换元：
x = { a sin t , 式中有 a 2 − x 2 a tan t , 式中有 a 2 + x 2 a sec t , 式中有 x 2 − a 2 x=\begin{cases}
a\sin t &,\text{式中有}\sqrt{a^2-x^2}\\
a\tan t &,\text{式中有}\sqrt{a^2+x^2}\\
a\sec t &,\text{式中有}\sqrt{x^2-a^2}
\end{cases}
x = ⎩ ⎨ ⎧ a sin t a tan t a sec t , 式中有 a 2 − x 2 , 式中有 a 2 + x 2 , 式中有 x 2 − a 2
再考虑通过代换根式进行换元。如果待求积分式中有a x + b n , \sqrt[n]{ax+b}, n a x + b , t = a x + b n ; t=\sqrt[n]{ax+b}; t = n a x + b ; a x + b m , a x + b n \sqrt[m]{ax+b},\sqrt[n]{ax+b} m a x + b , n a x + b t = a x + b k t=\sqrt[k]{ax+b} t = k a x + b k k k m , n m,n m , n a x + b c x + d n \sqrt[n]{\frac{ax+b}{cx+d}} n c x + d a x + b
现在考虑分式的换元。如果分母和分子关于x x x x = 1 t x=\frac{1}{t} x = t 1 ∫ d x x a 2 ± x 2 , ∫ d x x x 2 − a 2 , ∫ d x x 2 a 2 ± x 2 , ∫ d x x 2 x 2 − a 2 , ∫ a 2 ± x 2 x 4 d x \int\frac{\text{d}x}{x\sqrt{a^2\pm x^2}},\int\frac{\text{d}x}{x\sqrt{x^2-a^2}},\int\frac{\text{d}x}{x^2\sqrt{a^2\pm x^2}},\int\frac{\text{d}x}{x^2\sqrt{x^2-a^2}},\int\frac{\sqrt{a^2\pm x^2}}{x^4}\text{d}x ∫ x a 2 ± x 2 d x , ∫ x x 2 − a 2 d x , ∫ x 2 a 2 ± x 2 d x , ∫ x 2 x 2 − a 2 d x , ∫ x 4 a 2 ± x 2 d x ∫ x 2 − a 2 x 4 d x \int\frac{\sqrt{x^2-a^2}}{x^4}\text{d}x ∫ x 4 x 2 − a 2 d x
e.g.3.7 \textbf{e.g.3.7 } e.g.3.7
(1)∫ 4 − x x d x ; \int\sqrt{\frac{4-x}{x}}\text{d}x; ∫ x 4 − x d x ;
(2)∫ x 1 + x 1 − x d x ; \int x\sqrt{\frac{1+x}{1-x}}\text{d}x; ∫ x 1 − x 1 + x d x ;
(3)∫ 1 x 1 − x 1 + x d x ; \int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\text{d}x; ∫ x 1 1 + x 1 − x d x ;
Solution \textbf{Solution } Solution
∫ 4 − x x d x = ∫ 4 − x 4 x − x 2 d x = ∫ ( 2 − x ) + 2 4 − ( 2 − x ) 2 d x = t = 2 − x ⇒ x = 2 − t ∫ t + 2 4 − t 2 ⋅ ( − 1 ) d t = − ∫ t 4 − t 2 d t − ∫ 2 4 − t 2 d t = − ∫ − 1 2 d ( 4 − t 2 ) 4 − t 2 − 2 × ∫ 1 2 1 − ( t 2 ) 2 d t = − ∫ − 1 2 d ( 4 − t 2 ) 4 − t 2 − 2 × ∫ 1 1 − ( t 2 ) 2 d ( t 2 ) = 4 − t 2 − 2 arcsin t 2 + C = 4 − ( 2 − x ) 2 − 2 arcsin 2 − x 2 + C = 4 x − x 2 − 2 arcsin ( 1 − x 2 ) + C . ∫ x 1 + x 1 − x d x = ∫ x ( 1 + x ) 1 − x 2 d x = ∫ x 1 − x 2 d x + ∫ x 2 1 − x 2 d x = ∫ − 1 2 1 − x 2 d ( 1 − x 2 ) + ∫ x 2 1 − x 2 d x = − 1 − x 2 + ∫ x 2 d ( arcsin x ) = x = sin t − 1 − x 2 + ∫ sin 2 t d t = ∫ 1 − cos 2 t 2 d t − 1 − x 2 = 1 2 t − 1 2 sin 2 t 2 − 1 − x 2 + C = t 2 − sin t cos t 2 − 1 − x 2 + C = t = arcsin x arcsin x 2 − x cos ( arcsin x ) 2 − 1 − x 2 + C = Def. of Trigonometric Func. arcsin x 2 − x 1 − x 2 2 − 1 − x 2 + C = arcsin x − ( x + 2 ) 1 − x 2 2 + C . ∫ 1 x 1 − x 1 + x d x = ∫ 1 − x x 1 − x 2 d x = x = sin t ∫ 1 − sin t sin t d t = ∫ csc t − 1 d t = ln ∣ csc t − cot t ∣ − t + C = t = arcsin x ln ∣ csc ( arcsin x ) − cot ( arcsin x ) ∣ − arcsin x + C = Def. of Trigonometric Func. ln ∣ 1 x − 1 − x 2 x ∣ − arcsin x + C = ln ∣ 1 − 1 − x 2 x ∣ − arcsin x + C . \begin{align*}
\int\sqrt{\frac{4-x}{x}}\text{d}x&=\int\frac{4-x}{\sqrt{4x-x^2}}\text{d}x=\int\frac{(2-x)+2}{\sqrt{4-(2-x)^2}}\text{d}x\\
&\xlongequal{t=2-x\Rightarrow x=2-t}\int\frac{t+2}{\sqrt{4-t^2}}\cdot(-1)\text{d}t=-\int\frac{t}{\sqrt{4-t^2}}\text{d}t-\int\frac{2}{\sqrt{4-t^2}}\text{d}t\\
&=-\int\frac{-\frac{1}{2}\text{d}(4-t^2)}{\sqrt{4-t^2}}-2\times\int\frac{1}{2\sqrt{1-(\frac{t}{2})^2}}\text{d}t=-\int\frac{-\frac{1}{2}\text{d}(4-t^2)}{\sqrt{4-t^2}}-2\times\int\frac{1}{\sqrt{1-(\frac{t}{2})^2}}\text{d}\bigg(\frac{t}{2}\bigg)\\
&=\sqrt{4-t^2}-2\arcsin\frac{t}{2}+C=\sqrt{4-(2-x)^2}-2\arcsin\frac{2-x}{2}+C\\
&=\sqrt{4x-x^2}-2\arcsin(1-\frac{x}{2})+C.\tag{1}\\
\int x\sqrt{\frac{1+x}{1-x}}\text{d}x&=\int\frac{x(1+x)}{\sqrt{1-x^2}}\text{d}x=\int\frac{x}{\sqrt{1-x^2}}\text{d}x+\int\frac{x^2}{\sqrt{1-x^2}}\text{d}x\\
&=\int\frac{-\frac{1}{2}}{\sqrt{1-x^2}}\text{d}(1-x^2)+\int\frac{x^2}{\sqrt{1-x^2}}\text{d}x=-\sqrt{1-x^2}+\int x^2\text{d}(\arcsin x)\\
&\xlongequal{x=\sin t}-\sqrt{1-x^2}+\int\sin^2t\text{d}t=\int\frac{1-\cos2t}{2}\text{d}t-\sqrt{1-x^2}\\
&=\frac{1}{2}t-\frac{\frac{1}{2}\sin2t}{2}-\sqrt{1-x^2}+C=\frac{t}{2}-\frac{\sin t\cos t}{2}-\sqrt{1-x^2}+C\\
&\xlongequal{t=\arcsin x}\frac{\arcsin x}{2}-\frac{x\cos(\arcsin x)}{2}-\sqrt{1-x^2}+C\\
&\xlongequal{\text{Def. of Trigonometric Func.}}\frac{\arcsin x}{2}-\frac{x\sqrt{1-x^2}}{2}-\sqrt{1-x^2}+C\\
&=\frac{\arcsin x-(x+2)\sqrt{1-x^2}}{2}+C.\tag{2}\\
\int\frac{1}{x}\sqrt{\frac{1-x}{1+x}}\text{d}x&=\int\frac{1-x}{x\sqrt{1-x^2}}\text{d}x\xlongequal{x=\sin t}\int\frac{1-\sin t}{\sin t}\text{d}t=\int\csc t-1\text{d}t\\
&=\ln|\csc t-\cot t|-t+C\xlongequal{t=\arcsin x}\ln|\csc(\arcsin x)-\cot(\arcsin x)|-\arcsin x+C\\
&\xlongequal{\text{Def. of Trigonometric Func.}}\ln\bigg|\frac{1}{x}-\frac{\sqrt{1-x^2}}{x}\bigg|-\arcsin x+C \\
&=\ln\bigg|\frac{1-\sqrt{1-x^2}}{x}\bigg|-\arcsin x+C.\tag{3}
\end{align*}
∫ x 4 − x d x ∫ x 1 − x 1 + x d x ∫ x 1 1 + x 1 − x d x = ∫ 4 x − x 2 4 − x d x = ∫ 4 − ( 2 − x ) 2 ( 2 − x ) + 2 d x t = 2 − x ⇒ x = 2 − t ∫ 4 − t 2 t + 2 ⋅ ( − 1 ) d t = − ∫ 4 − t 2 t d t − ∫ 4 − t 2 2 d t = − ∫ 4 − t 2 − 2 1 d ( 4 − t 2 ) − 2 × ∫ 2 1 − ( 2 t ) 2 1 d t = − ∫ 4 − t 2 − 2 1 d ( 4 − t 2 ) − 2 × ∫ 1 − ( 2 t ) 2 1 d ( 2 t ) = 4 − t 2 − 2 arcsin 2 t + C = 4 − ( 2 − x ) 2 − 2 arcsin 2 2 − x + C = 4 x − x 2 − 2 arcsin ( 1 − 2 x ) + C . = ∫ 1 − x 2 x ( 1 + x ) d x = ∫ 1 − x 2 x d x + ∫ 1 − x 2 x 2 d x = ∫ 1 − x 2 − 2 1 d ( 1 − x 2 ) + ∫ 1 − x 2 x 2 d x = − 1 − x 2 + ∫ x 2 d ( arcsin x ) x = s i n t − 1 − x 2 + ∫ sin 2 t d t = ∫ 2 1 − cos 2 t d t − 1 − x 2 = 2 1 t − 2 2 1 sin 2 t − 1 − x 2 + C = 2 t − 2 sin t cos t − 1 − x 2 + C t = a r c s i n x 2 arcsin x − 2 x cos ( arcsin x ) − 1 − x 2 + C Def. of Trigonometric Func. 2 arcsin x − 2 x 1 − x 2 − 1 − x 2 + C = 2 arcsin x − ( x + 2 ) 1 − x 2 + C . = ∫ x 1 − x 2 1 − x d x x = s i n t ∫ sin t 1 − sin t d t = ∫ csc t − 1 d t = ln ∣ csc t − cot t ∣ − t + C t = a r c s i n x ln ∣ csc ( arcsin x ) − cot ( arcsin x ) ∣ − arcsin x + C Def. of Trigonometric Func. ln x 1 − x 1 − x 2 − arcsin x + C = ln x 1 − 1 − x 2 − arcsin x + C . ( 1 ) ( 2 ) ( 3 )
从上述例3.7中给出的3个不定积分的求法就可以看出，第二类换元法还是需要对待求不定积分式先进行化简，尤其是含有根式的待求不定积分。从实际操作经验来看含根式函数的求导远困难于其他函数，这也是为什么一般会先对这些根式有理化降低复杂度后再换元。
有时第二类换元法可能会对指数进行换元，但并不常见。
分部积分法
分部积分法的原理是∫ u d v = u v − ∫ v d u . \int u\text{d}v=uv-\int v\text{d}u. ∫ u d v = uv − ∫ v d u .
∫ u v ′ d x = u v − ∫ v u ′ d x . \int uv'\text{d}x=uv-\int vu'\text{d}x.
∫ u v ′ d x = uv − ∫ v u ′ d x .
下面考虑分部积分法的作用。
首先考虑分部积分法的「消幂功能」。我们可以通过不断的分部积分得到
∫ u v ( n + 1 ) d x = ( ∑ i = 0 n ( − 1 ) i u ( i ) v ( n − i ) ) + ( − 1 ) n + 1 ∫ u ( n + 1 ) v d x \int uv^{(n+1)}\text{d}x=\Bigg(\sum_{i=0}^n(-1)^iu^{(i)}v^{(n-i)}\Bigg)+(-1)^{n+1}\int u^{(n+1)}v\text{d}x
∫ u v ( n + 1 ) d x = ( i = 0 ∑ n ( − 1 ) i u ( i ) v ( n − i ) ) + ( − 1 ) n + 1 ∫ u ( n + 1 ) v d x
对两个有n + 1 n+1 n + 1 u = u ( x ) u=u(x) u = u ( x ) v = v ( x ) v=v(x) v = v ( x )
再考虑其「循环功能」。在不断分部积分后我们很有可能发现I = F ( x ) + λ I ( λ ≠ 1 ) , I=F(x)+\lambda I(\lambda≠1), I = F ( x ) + λ I ( λ = 1 ) , I = 1 1 − λ F ( x ) + C . I=\frac{1}{1-\lambda}F(x)+C. I = 1 − λ 1 F ( x ) + C .
此外分部积分法还可以用于「去分母」、「抵消」或「递推」。
根据下面的例3.8～例3.10考虑上面所提到的分部积分法的五个功能。
e.g.3.8 \textbf{e.g.3.8 } e.g.3.8 I = ∫ x e x e x − 1 d x . I=\int\frac{xe^x}{\sqrt{e^x-1}}\text{d}x. I = ∫ e x − 1 x e x d x .
Solution \textbf{Solution } Solution
I = ∫ x d ( e x − 1 ) e x − 1 = 2 ∫ x d ( e x − 1 ) = 分部积分 2 ( x e x − 1 − ∫ e x − 1 d x ) . \begin{align*}
I&=\int\frac{x\text{d}(e^x-1)}{\sqrt{e^x-1}}=2\int x\text{d}(\sqrt{e^x-1})\\
&\xlongequal{\text{分部积分}}2(x\sqrt{e^x-1}-\int\sqrt{e^x-1}\text{d}x).
\end{align*}
I = ∫ e x − 1 x d ( e x − 1 ) = 2 ∫ x d ( e x − 1 ) 分部积分 2 ( x e x − 1 − ∫ e x − 1 d x ) .
而
∫ e x − 1 d x = t = e x − 1 ⇒ x = ln ( 1 + t 2 ) ∫ t ⋅ 2 t 1 + t 2 d t = 2 t − 2 arctan t + C = 2 e x − 1 − 2 arctan e x − 1 + C , \begin{align*}
\int\sqrt{e^x-1}\text{d}x&\xlongequal{t=\sqrt{e^x-1}\Rightarrow x=\ln(1+t^2)}\int\frac{t\cdot2t}{1+t^2}\text{d}t\\
&=2t-2\arctan t+C=2\sqrt{e^x-1}-2\arctan\sqrt{e^x-1}+C,
\end{align*}
∫ e x − 1 d x t = e x − 1 ⇒ x = l n ( 1 + t 2 ) ∫ 1 + t 2 t ⋅ 2 t d t = 2 t − 2 arctan t + C = 2 e x − 1 − 2 arctan e x − 1 + C ,
所以所求为I = ( 2 x − 4 ) e x − 1 + 4 arctan e x − 1 + C . I=(2x-4)\sqrt{e^x-1}+4\arctan\sqrt{e^x-1}+C. I = ( 2 x − 4 ) e x − 1 + 4 arctan e x − 1 + C .
例3.8中是「去分母」功能的代表。通过对分子中的e x d x e^x\text{d}x e x d x
e.x.3.7 \textbf{e.x.3.7 } e.x.3.7 I = ∫ x cos 4 x 2 sin 3 x d x . I=\int\frac{x\cos^4\frac{x}{2}}{\sin^3x}\text{d}x. I = ∫ s i n 3 x x c o s 4 2 x d x .
Solution \textbf{Solution } Solution
I = ∫ x cos 4 x 2 ( 2 sin x 2 cos x 2 ) 3 d x = 1 8 ∫ x cos x 2 sin 3 x 2 d x = 1 8 ∫ 2 x sin − 3 x 2 d ( sin x 2 ) = 1 4 ∫ ( − 1 2 ) x d ( sin − 2 x 2 ) = − 1 8 ∫ x d ( sin − 2 x 2 ) = 分部积分 − 1 8 ( x sin − 2 x 2 − ∫ sin − 2 x 2 d x ) = 1 8 ( ∫ csc 2 x 2 d x − x csc 2 x 2 ) . \begin{align*}
I&=\int\frac{x\cos^4\frac{x}{2}}{(2\sin\frac{x}{2}\cos\frac{x}{2})^3}\text{d}x\\
&=\frac{1}{8}\int\frac{x\cos\frac{x}{2}}{\sin^3\frac{x}{2}}\text{d}x=\frac{1}{8}\int2x\sin^{-3}\frac{x}{2}\text{d}\bigg(\sin\frac{x}{2}\bigg)\\
&=\frac{1}{4}\int\bigg(-\frac{1}{2}\bigg)x\text{d}\bigg(\sin^{-2}\frac{x}{2}\bigg)\\
&=-\frac{1}{8}\int x\text{d}\bigg(\sin^{-2}\frac{x}{2}\bigg)\\
&\xlongequal{\text{分部积分}}-\frac{1}{8}\bigg(x\sin^{-2}\frac{x}{2}-\int\sin^{-2}\frac{x}{2}\text{d}x\bigg)\\
&=\frac{1}{8}(\int\csc^2\frac{x}{2}\text{d}x-x\csc^2\frac{x}{2}).
\end{align*}
I = ∫ ( 2 sin 2 x cos 2 x ) 3 x cos 4 2 x d x = 8 1 ∫ sin 3 2 x x cos 2 x d x = 8 1 ∫ 2 x sin − 3 2 x d ( sin 2 x ) = 4 1 ∫ ( − 2 1 ) x d ( sin − 2 2 x ) = − 8 1 ∫ x d ( sin − 2 2 x ) 分部积分 − 8 1 ( x sin − 2 2 x − ∫ sin − 2 2 x d x ) = 8 1 ( ∫ csc 2 2 x d x − x csc 2 2 x ) .
而
∫ csc 2 x 2 d x = − 2 cot x 2 + C , \int\csc^2\frac{x}{2}\text{d}x=-2\cot\frac{x}{2}+C,
∫ csc 2 2 x d x = − 2 cot 2 x + C ,
所以所求为I = − 1 4 cot x 2 − 1 8 x csc 2 x 2 + C . I=-\frac{1}{4}\cot\frac{x}{2}-\frac{1}{8}x\csc^2\frac{x}{2}+C. I = − 4 1 cot 2 x − 8 1 x csc 2 2 x + C .
e.g.3.9 \textbf{e.g.3.9 } e.g.3.9 I = ∫ e 2 x ( 1 + tan x ) 2 d x . I=\int e^{2x}(1+\tan x)^2\text{d}x. I = ∫ e 2 x ( 1 + tan x ) 2 d x .
Solution \textbf{Solution } Solution
I = ∫ e 2 x ( 1 + 2 tan x + tan 2 x ) d x = ∫ e 2 x ( 1 + tan 2 x ) d x + 2 ∫ e 2 x ( tan x ) d x = ∫ e 2 x cos 2 x + sin 2 x cos 2 x d x + 2 ∫ e 2 x tan x d x = ∫ e 2 x sec 2 x d x + 2 ∫ e 2 x tan x d x = ∫ e 2 x d tan x + 2 ∫ e 2 x tan x d x = 前半项 分部积分 e 2 x tan x − ∫ tan x d e x + 2 ∫ e 2 x tan x d x = e 2 x tan x − ∫ tan x 2 e 2 x d x + 2 ∫ e 2 x tan x d x = e 2 x tan x + C . \begin{align*}
I&=\int e^{2x}(1+2\tan x+\tan^2x)\text{d}x=\int e^{2x}(1+\tan^2x)\text{d}x+2\int e^{2x}(\tan x)\text{d}x\\
&=\int e^{2x}\frac{\cos^2x+\sin^2x}{\cos^2x}\text{d}x+2\int e^{2x}\tan x\text{d}x\\
&=\int e^{2x}\sec^2x\text{d}x+2\int e^{2x}\tan x\text{d}x\\
&=\int e^{2x}\text{d}\tan x+2\int e^{2x}\tan x\text{d}x\\
&\xlongequal[\text{前半项}]{\text{分部积分}}e^{2x}\tan x-\int\tan x\text{d}e^x+2\int e^{2x}\tan x\text{d}x\\
&=e^{2x}\tan x-\int\tan x2e^{2x}\text{d}x+2\int e^{2x}\tan x\text{d}x\\
&=e^{2x}\tan x+C.
\end{align*}
I = ∫ e 2 x ( 1 + 2 tan x + tan 2 x ) d x = ∫ e 2 x ( 1 + tan 2 x ) d x + 2 ∫ e 2 x ( tan x ) d x = ∫ e 2 x cos 2 x cos 2 x + sin 2 x d x + 2 ∫ e 2 x tan x d x = ∫ e 2 x sec 2 x d x + 2 ∫ e 2 x tan x d x = ∫ e 2 x d tan x + 2 ∫ e 2 x tan x d x 分部积分 前半项 e 2 x tan x − ∫ tan x d e x + 2 ∫ e 2 x tan x d x = e 2 x tan x − ∫ tan x 2 e 2 x d x + 2 ∫ e 2 x tan x d x = e 2 x tan x + C .
例3.9显然是「抵消」功能的体现。在对原积分式一分为二后，这两个分式中的一项利用分部积分进一步展开后通过变换就可以与另一项积分式消除，从而得到结果，大大减轻了运算负担。
e.x.3.8 \textbf{e.x.3.8 } e.x.3.8 I = ∫ 1 + sin x 1 + cos x e x d x . I=\int\frac{1+\sin x}{1+\cos x}e^x\text{d}x. I = ∫ 1 + c o s x 1 + s i n x e x d x .
Solution \textbf{Solution } Solution
I = ( 1 + sin x ) ( 1 − cos x ) 1 − cos 2 x e x d x = ( 1 − cos x + sin x − sin x cos x ) e x sin 2 x d x = ∫ e x sin 2 x d x − ∫ e x cos x sin 2 x d x + ∫ e x sin x d x − ∫ e x cos x sin x d x = − ∫ e x d cot x − ∫ e x sin 2 x d sin x + ∫ e x sin x d x − ∫ e x cos x sin x d x = − ∫ e x d cot x + ∫ e x d 1 sin x + ∫ e x sin x d x − ∫ e x cot x d x = 前两项 分部积分 − ( e x cot x − ∫ cot x d e x ) + ( e x sin x − ∫ 1 sin x d e x ) + ∫ e x sin x d x − ∫ e x cot x d x = − e x cot x + ∫ cot x e x d x + e x sin x − ∫ e x sin x d x + ∫ e x sin x d x − ∫ e x cot x d x = e x sin x − e x cot x + C . \begin{align*}
I&=\frac{(1+\sin x)(1-\cos x)}{1-\cos^2x}e^x\text{d}x=\frac{(1-\cos x+\sin x-\sin x\cos x)e^x}{\sin^2x}\text{d}x\\
&=\int\frac{e^x}{\sin^2x}\text{d}x-\int\frac{e^x\cos x}{\sin^2x}\text{d}x+\int\frac{e^x}{\sin x}\text{d}x-\int\frac{e^x\cos x}{\sin x}\text{d}x\\
&=-\int e^x\text{d}\cot x-\int\frac{e^x}{\sin^2x}\text{d}\sin x+\int\frac{e^x}{\sin x}\text{d}x-\int\frac{e^x\cos x}{\sin x}\text{d}x\\
&=-\int e^x\text{d}\cot x+\int e^x\text{d}\frac{1}{\sin x}+\int\frac{e^x}{\sin x}\text{d}x-\int e^x\cot x\text{d}x\\
&\xlongequal[\text{前两项}]{\text{分部积分}}-(e^x\cot x-\int\cot x\text{d}e^x)+(\frac{e^x}{\sin x}-\int\frac{1}{\sin x}\text{d}e^x)+\int\frac{e^x}{\sin x}\text{d}x-\int e^x\cot x\text{d}x\\
&=-e^x\cot x+\int\cot xe^x\text{d}x+\frac{e^x}{\sin x}-\int\frac{e^x}{\sin x}\text{d}x+\int\frac{e^x}{\sin x}\text{d}x-\int e^x\cot x\text{d}x\\
&=\frac{e^x}{\sin x}-e^x\cot x+C.
\end{align*}
I = 1 − cos 2 x ( 1 + sin x ) ( 1 − cos x ) e x d x = sin 2 x ( 1 − cos x + sin x − sin x cos x ) e x d x = ∫ sin 2 x e x d x − ∫ sin 2 x e x cos x d x + ∫ sin x e x d x − ∫ sin x e x cos x d x = − ∫ e x d cot x − ∫ sin 2 x e x d sin x + ∫ sin x e x d x − ∫ sin x e x cos x d x = − ∫ e x d cot x + ∫ e x d sin x 1 + ∫ sin x e x d x − ∫ e x cot x d x 分部积分 前两项 − ( e x cot x − ∫ cot x d e x ) + ( sin x e x − ∫ sin x 1 d e x ) + ∫ sin x e x d x − ∫ e x cot x d x = − e x cot x + ∫ cot x e x d x + sin x e x − ∫ sin x e x d x + ∫ sin x e x d x − ∫ e x cot x d x = sin x e x − e x cot x + C .
e.g.3.10 \textbf{e.g.3.10 } e.g.3.10 I n = ∫ tan n x d x I_n=\int\tan^nx\text{d}x I n = ∫ tan n x d x I n = 1 n − 1 tan n − 1 x − I n − 2 . I_n=\frac{1}{n-1}\tan^{n-1}x-I_{n-2}. I n = n − 1 1 tan n − 1 x − I n − 2 .
Proof \textbf{Proof } Proof
I n = ∫ tan n − 2 x ⋅ tan 2 x d x = ∫ tan n − 2 x ⋅ ( sec 2 x − 1 ) d x = ∫ tan n − 2 x ⋅ sec 2 x d x − ∫ tan n − 2 x d x = ∫ tan n − 2 x d ( tan x ) − ∫ tan n − 2 x d x = tan n − 1 x n − 1 − ∫ tan n − 2 x d x = 1 n − 1 tan n − 1 x − I n − 2 . \begin{align*}
I_n&=\int\tan^{n-2}x\cdot\tan^2x\text{d}x=\int\tan^{n-2}x\cdot(\sec^2x-1)\text{d}x\\
&=\int\tan^{n-2}x\cdot\sec^2x\text{d}x-\int\tan^{n-2}x\text{d}x\\
&=\int\tan^{n-2}x\text{d}(\tan x)-\int\tan^{n-2}x\text{d}x=\frac{\tan^{n-1}x}{n-1}-\int\tan^{n-2}x\text{d}x\\
&=\frac{1}{n-1}\tan^{n-1}x-I_{n-2}.
\end{align*}
I n = ∫ tan n − 2 x ⋅ tan 2 x d x = ∫ tan n − 2 x ⋅ ( sec 2 x − 1 ) d x = ∫ tan n − 2 x ⋅ sec 2 x d x − ∫ tan n − 2 x d x = ∫ tan n − 2 x d ( tan x ) − ∫ tan n − 2 x d x = n − 1 tan n − 1 x − ∫ tan n − 2 x d x = n − 1 1 tan n − 1 x − I n − 2 .
所以原命题得证。
例3.10就是「递推」的一种。虽然本题没有使用分部积分，但是思路类似——对于递推公式，我们应当将n n n
e.x.3.9 \textbf{e.x.3.9 } e.x.3.9 ∫ sec n x d x . \int\sec^nx\text{d}x. ∫ sec n x d x .
Solution \textbf{Solution } Solution O.F. = I n . \text{O.F.}=I_n. O.F. = I n .
I n = ∫ sec n − 2 x ⋅ sec 2 x d x = ∫ sec n − 2 x d ( tan x ) = 分部积分 tan x sec n − 2 x − ∫ tan x d ( sec n − 2 x ) = tan x sec n − 2 x − ∫ tan x ⋅ ( ( n − 2 ) sec n − 3 x ⋅ sin x cos 2 x ) d x = tan x sec n − 2 x − ( n − 2 ) ∫ sec n − 2 x tan 2 x d x = tan x sec n − 2 x − ( n − 2 ) ∫ sec n − 2 x ( sec 2 x − 1 ) d x = tan x sec n − 2 x − ( n − 2 ) ( ∫ sec n x d x − ∫ sec n − 2 x d x ) = tan x sec n − 2 x − ( n − 2 ) I n + ( n − 2 ) I n − 2 . \begin{align*}
I_n&=\int\sec^{n-2}x\cdot\sec^2x\text{d}x=\int\sec^{n-2}x\text{d}(\tan x)\\
&\xlongequal{\text{分部积分}}\tan x\sec^{n-2}x-\int\tan x\text{d}(\sec^{n-2}x)\\
&=\tan x\sec^{n-2}x-\int\tan x\cdot\big((n-2)\sec^{n-3}x\cdot\frac{\sin x}{\cos^2x}\big)\text{d}x\\
&=\tan x\sec^{n-2}x-(n-2)\int\sec^{n-2}x\tan^2x\text{d}x\\
&=\tan x\sec^{n-2}x-(n-2)\int\sec^{n-2}x(\sec^2x-1)\text{d}x\\
&=\tan x\sec^{n-2}x-(n-2)\bigg(\int\sec^{n}x\text{d}x-\int\sec^{n-2}x\text{d}x\bigg)\\
&=\tan x\sec^{n-2}x-(n-2)I_n+(n-2)I_{n-2}.
\end{align*}
I n = ∫ sec n − 2 x ⋅ sec 2 x d x = ∫ sec n − 2 x d ( tan x ) 分部积分 tan x sec n − 2 x − ∫ tan x d ( sec n − 2 x ) = tan x sec n − 2 x − ∫ tan x ⋅ ( ( n − 2 ) sec n − 3 x ⋅ cos 2 x sin x ) d x = tan x sec n − 2 x − ( n − 2 ) ∫ sec n − 2 x tan 2 x d x = tan x sec n − 2 x − ( n − 2 ) ∫ sec n − 2 x ( sec 2 x − 1 ) d x = tan x sec n − 2 x − ( n − 2 ) ( ∫ sec n x d x − ∫ sec n − 2 x d x ) = tan x sec n − 2 x − ( n − 2 ) I n + ( n − 2 ) I n − 2 .
所以
( n − 1 ) I n = tan x sec n − 2 x + ( n − 2 ) I n − 2 ⇒ I n = tan x sec n − 2 x + ( n − 2 ) I n − 2 n − 1 . \begin{align*}
&(n-1)I_n=\tan x\sec^{n-2}x+(n-2)I_{n-2}\\
\Rightarrow &I_n=\frac{\tan x\sec^{n-2}x+(n-2)I_{n-2}}{n-1}.
\end{align*}
⇒ ( n − 1 ) I n = tan x sec n − 2 x + ( n − 2 ) I n − 2 I n = n − 1 tan x sec n − 2 x + ( n − 2 ) I n − 2 .
有理函数积分
形如或可以化为形如∫ A x − a d x , ∫ A ( x − a ) n d x , ∫ M x + N x 2 + p x + q d x \int\frac{A}{x-a}\text{d}x,\int\frac{A}{(x-a)^n}\text{d}x,\int\frac{Mx+N}{x^2+px+q}\text{d}x ∫ x − a A d x , ∫ ( x − a ) n A d x , ∫ x 2 + p x + q M x + N d x ∫ M x + N ( x 2 + p x + q ) n d x \int\frac{Mx+N}{(x^2+px+q)^n}\text{d}x ∫ ( x 2 + p x + q ) n M x + N d x n ∈ N + ; A , M , N , a , p , q n\in\mathbb{N}_+;A,M,N,a,p,q n ∈ N + ; A , M , N , a , p , q 4 q − p 2 > 0. 4q-p^2>0. 4 q − p 2 > 0.
如果待求积分式是假分式则需先化为有理数和真分式的和；此时的真分式应当也可以化为上方这四种积分中的某一种或某几种。
e.x.3.10 \textbf{e.x.3.10 } e.x.3.10 ∫ 1 x ( 1 − x ) 2 d x . \int\frac{1}{x(1-x)^2}\text{d}x. ∫ x ( 1 − x ) 2 1 d x .
Solution \textbf{Solution } Solution
O.F. = ∫ x + ( 1 − x ) x ( 1 − x ) 2 d x = ∫ 1 ( 1 − x ) 2 d x + ∫ 1 x ( 1 − x ) d x = ∫ 1 ( 1 − x ) 2 d x + ∫ x + ( 1 − x ) x ( 1 − x ) d x = ∫ 1 ( 1 − x ) 2 d x + ∫ 1 1 − x d x + ∫ 1 x d x = 1 1 − x − ln ( 1 − x ) + ln x + C . \begin{align*}
\text{O.F.}&=\int\frac{x+(1-x)}{x(1-x)^2}\text{d}x=\int\frac{1}{(1-x)^2}\text{d}x+\int\frac{1}{x(1-x)}\text{d}x\\
&=\int\frac{1}{(1-x)^2}\text{d}x+\int\frac{x+(1-x)}{x(1-x)}\text{d}x=\int\frac{1}{(1-x)^2}\text{d}x+\int\frac{1}{1-x}\text{d}x+\int\frac{1}{x}\text{d}x\\
&=\frac{1}{1-x}-\ln(1-x)+\ln{x}+C.
\end{align*}
O.F. = ∫ x ( 1 − x ) 2 x + ( 1 − x ) d x = ∫ ( 1 − x ) 2 1 d x + ∫ x ( 1 − x ) 1 d x = ∫ ( 1 − x ) 2 1 d x + ∫ x ( 1 − x ) x + ( 1 − x ) d x = ∫ ( 1 − x ) 2 1 d x + ∫ 1 − x 1 d x + ∫ x 1 d x = 1 − x 1 − ln ( 1 − x ) + ln x + C .
再次考虑上面的练习3.10。显然这个式子我们可以一眼望出拆分方法——事实上，我们在观察时已经通过对分母进行因式分解之后得到这个待积分分式拆分后可能的若干个分式了。对于稍微复杂一些的情况，我们就需要基于这些拆分后的分式的分母利用待定系数法进行分解。此时不妨重新考虑这个练习3.10。
e.x.3.10 \textbf{e.x.3.10 } e.x.3.10 ∫ 1 x ( 1 − x ) 2 d x . \int\frac{1}{x(1-x)^2}\text{d}x. ∫ x ( 1 − x ) 2 1 d x .
Solution (The Another Story) \textbf{Solution (The Another Story) } Solution (The Another Story) 1 x ( 1 − x ) 2 = α x + β 1 − x + γ ( 1 − x ) 2 . \frac{1}{x(1-x)^2}=\frac{\alpha}{x}+\frac{\beta}{1-x}+\frac{\gamma}{(1-x)^2}. x ( 1 − x ) 2 1 = x α + 1 − x β + ( 1 − x ) 2 γ .
α = lim x → 0 1 ( 1 − x ) 2 = 1 , γ = lim x → 1 1 x = 1. \begin{align*}
\alpha&=\lim_{x\rightarrow 0}\frac{1}{(1-x)^2}=1,\\
\gamma&=\lim_{x\rightarrow 1}\frac{1}{x}=1.
\end{align*}
α γ = x → 0 lim ( 1 − x ) 2 1 = 1 , = x → 1 lim x 1 = 1.
那么1 x ( 1 − x ) 2 = 1 x + β 1 − x + 1 ( 1 − x ) 2 . \frac{1}{x(1-x)^2}=\frac{1}{x}+\frac{\beta}{1-x}+\frac{1}{(1-x)^2}. x ( 1 − x ) 2 1 = x 1 + 1 − x β + ( 1 − x ) 2 1 . ( 1 − x ) 2 , (1-x)^2, ( 1 − x ) 2 ,
1 x = ( 1 − x ) 2 x + β ( 1 − x ) + 1 ⇒ 等式两边求导 − 1 x 2 = 1 − 1 x 2 − β \begin{align*}
&\frac{1}{x}&=&\frac{(1-x)^2}{x}+\beta(1-x)+1\\
\xRightarrow{\text{等式两边求导}}&-\frac{1}{x^2}&=&1-\frac{1}{x^2}-\beta
\end{align*}
等式两边求导 x 1 − x 2 1 = = x ( 1 − x ) 2 + β ( 1 − x ) + 1 1 − x 2 1 − β
对等式两边同时取x = 1 , x=1, x = 1 , β = 1. \beta=1. β = 1. 1 x ( 1 − x ) 2 = 1 x + 1 1 − x + 1 ( 1 − x ) 2 . \frac{1}{x(1-x)^2}=\frac{1}{x}+\frac{1}{1-x}+\frac{1}{(1-x)^2}. x ( 1 − x ) 2 1 = x 1 + 1 − x 1 + ( 1 − x ) 2 1 .
O.F. = ∫ 1 x d x + ∫ 1 1 − x d x + ∫ 1 ( 1 − x ) 2 d x = ln x − ln ( 1 − x ) + 1 1 − x + C . \begin{align*}
\text{O.F.}&=\int\frac{1}{x}\text{d}x+\int\frac{1}{1-x}\text{d}x+\int\frac{1}{(1-x)^2}\text{d}x\\
&=\ln{x}-\ln(1-x)+\frac{1}{1-x}+C.
\end{align*}
O.F. = ∫ x 1 d x + ∫ 1 − x 1 d x + ∫ ( 1 − x ) 2 1 d x = ln x − ln ( 1 − x ) + 1 − x 1 + C .
在上述重新思考中α \alpha α γ \gamma γ x → 0 x\rightarrow 0 x → 0 x x x x → 1 x\rightarrow 1 x → 1 ( 1 − x ) 2 . (1-x)^2. ( 1 − x ) 2 .
三角函数有理式积分
我们高中时即接触过「万能代换公式」
tan x 2 = u ⇒ { sin x = 2 u 1 + u 2 cos x = 1 − u 2 1 + u 2 d x = 2 1 + u 2 d u , \tan\frac{x}{2}=u\Rightarrow\begin{cases}
\sin x=\frac{2u}{1+u^2}\\
\cos x=\frac{1-u^2}{1+u^2}\\
\text{d}x=\frac{2}{1+u^2}\text{d}u
\end{cases},
tan 2 x = u ⇒ ⎩ ⎨ ⎧ sin x = 1 + u 2 2 u cos x = 1 + u 2 1 − u 2 d x = 1 + u 2 2 d u ,
因此关于含x x x sin x \sin x sin x cos x \cos x cos x R ( sin x , cos x ) R(\sin x, \cos x) R ( sin x , cos x ) ∫ 1 a sin x + b cos x + c d x \int\frac{1}{a\sin x+b\cos x+c}\text{d}x ∫ a s i n x + b c o s x + c 1 d x
e.g.3.11 \textbf{e.g.3.11 } e.g.3.11 ∫ d x 2 + sin x . \int\frac{\text{d}x}{2+\sin x}. ∫ 2 + s i n x d x .
Solution \textbf{Solution } Solution tan x 2 = u . \tan\frac{x}{2}=u. tan 2 x = u . x = 2 arctan u , x=2\arctan u, x = 2 arctan u ,
O.F. = ∫ 2 1 + u 2 2 + 2 u 1 + u 2 d u = ∫ 1 u 2 + u + 1 d u = ∫ 1 ( u + 1 2 ) 2 + 3 4 = 2 3 arctan u + 1 2 3 2 + C = 2 3 arctan 2 tan x 2 + 1 3 + C . \begin{align*}
\text{O.F.}&=\int\frac{\frac{2}{1+u^2}}{2+\frac{2u}{1+u^2}}\text{d}u\\
&=\int\frac{1}{u^2+u+1}\text{d}u=\int\frac{1}{\big(u+\frac{1}{2}\big)^2+\frac{3}{4}}=\frac{2}{\sqrt{3}}\arctan\frac{u+\frac{1}{2}}{\frac{\sqrt{3}}{2}}+C\\
&=\frac{2}{\sqrt{3}}\arctan\frac{2\tan\frac{x}{2}+1}{\sqrt{3}}+C.
\end{align*}
O.F. = ∫ 2 + 1 + u 2 2 u 1 + u 2 2 d u = ∫ u 2 + u + 1 1 d u = ∫ ( u + 2 1 ) 2 + 4 3 1 = 3 2 arctan 2 3 u + 2 1 + C = 3 2 arctan 3 2 tan 2 x + 1 + C .
e.x.3.11 \textbf{e.x.3.11 } e.x.3.11 ∫ d x 1 + cos x . \int\frac{\text{d}x}{1+\cos x}. ∫ 1 + c o s x d x .
Solution \textbf{Solution } Solution tan x 2 = u . \tan\frac{x}{2}=u. tan 2 x = u . x = 2 arctan u , x=2\arctan u, x = 2 arctan u ,
O.F. = ∫ 2 1 + u 2 1 + 1 − u 2 1 + u 2 d u = ∫ 1 d u = u + C = tan x 2 + C . \begin{align*}
\text{O.F.}&=\int\frac{\frac{2}{1+u^2}}{1+\frac{1-u^2}{1+u^2}}\text{d}u=\int1\text{d}u\\
&=u+C=\tan\frac{x}{2}+C.
\end{align*}
O.F. = ∫ 1 + 1 + u 2 1 − u 2 1 + u 2 2 d u = ∫ 1 d u = u + C = tan 2 x + C .
事实上，我们还是利用更早时学的倍角公式、和差化积/积化和差公式更多一些。前者可以有效降低含三角函数项的幂数。
现在重新考虑练习3.11。我们看到1 + cos x 1+\cos x 1 + cos x
e.x.3.11 \textbf{e.x.3.11 } e.x.3.11 ∫ d x 1 + cos x . \int\frac{\text{d}x}{1+\cos x}. ∫ 1 + c o s x d x .
Solution (The Another Story) \textbf{Solution (The Another Story) } Solution (The Another Story) 1 + cos x = ( sin 2 x 2 + cos 2 x 2 ) + ( cos 2 x 2 − sin 2 x 2 ) = 2 cos 2 x 2 . 1+\cos x=\big(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\big)+\big(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\big)=2\cos^2\frac{x}{2}. 1 + cos x = ( sin 2 2 x + cos 2 2 x ) + ( cos 2 2 x − sin 2 2 x ) = 2 cos 2 2 x .
O.F. = ∫ 1 2 cos 2 x 2 d x = tan x 2 + C . \begin{align*}
\text{O.F.}&=\int\frac{1}{2\cos^2\frac{x}{2}}\text{d}x=\tan\frac{x}{2}+C.
\end{align*}
O.F. = ∫ 2 cos 2 2 x 1 d x = tan 2 x + C .
显然这个解法的复杂度也不高。事实上，多数含三角函数的待求积分式都可以用之前提到过的常规不定积分求法解决。
定积分
∫ a b f ( x ) d x = F ( b ) − F ( a ) . \int^b_af(x)\text{d}x=F(b)-F(a). ∫ a b f ( x ) d x = F ( b ) − F ( a ) .
我们都知道这是定积分的定义（其中F ( x ) F(x) F ( x ) f ( x ) f(x) f ( x ) x x x b b b a a a
对于定积分有几个比较常用的公式。
如果f ( x ) f(x) f ( x ) [ − a , a ] [-a,a] [ − a , a ] a > 0 a>0 a > 0 f ( x ) f(x) f ( x ) ∫ − a a f ( x ) d x = 0 ; \int_{-a}^af(x)\text{d}x=0; ∫ − a a f ( x ) d x = 0 ; f ( x ) f(x) f ( x ) ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x . \int_{-a}^af(x)\text{d}x=2\int_0^af(x)\text{d}x. ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x . f ( x ) = f ( x ) + f ( − x ) 2 + f ( x ) − f ( − x ) 2 f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2} f ( x ) = 2 f ( x ) + f ( − x ) + 2 f ( x ) − f ( − x ) ∫ − a a f ( x ) d x = ∫ 0 a ( f ( x ) + f ( − x ) ) d x . \int_{-a}^af(x)\text{d}x=\int_0^a\big(f(x)+f(-x)\big)\text{d}x. ∫ − a a f ( x ) d x = ∫ 0 a ( f ( x ) + f ( − x ) ) d x .
f ( x + T ) = f ( x ) ⇒ ∫ 0 T f ( x ) d x = ∫ a a + T f ( x ) d x . f(x+T)=f(x)\Rightarrow\int_0^Tf(x)\text{d}x=\int_a^{a+T}f(x)\text{d}x. f ( x + T ) = f ( x ) ⇒ ∫ 0 T f ( x ) d x = ∫ a a + T f ( x ) d x . ∫ 0 π 2 f ( sin x ) d x = ∫ 0 π 2 f ( cos x ) d x . \int_0^\frac{\pi}{2}f(\sin x)\text{d}x=\int_0^\frac{\pi}{2}f(\cos x)\text{d}x. ∫ 0 2 π f ( sin x ) d x = ∫ 0 2 π f ( cos x ) d x . ∫ 0 π 2 x f ( sin x ) d x = π 2 ∫ 0 π f ( sin x ) d x . \int_0^\frac{\pi}{2}xf(\sin x)\text{d}x=\frac{\pi}{2}\int_0^\pi f(\sin x)\text{d}x. ∫ 0 2 π x f ( sin x ) d x = 2 π ∫ 0 π f ( sin x ) d x . ∫ 0 π 2 sin n x d x = ∫ 0 π 2 cos n x d x = { π 2 ⋅ ( n − 1 ) ! ! n ! ! , n ≡ 0 ( m o d 2 ) ( n − 1 ) ! ! n ! ! , n ≡ 1 ( m o d 2 ) . \int_0^\frac{\pi}{2}\sin^nx\text{d}x=\int_0^\frac{\pi}{2}\cos^nx\text{d}x=\begin{cases}\frac{\pi}{2}\cdot\frac{(n-1)!!}{n!!}&,n\equiv 0\pmod{2}\\\frac{(n-1)!!}{n!!}&,n\equiv 1\pmod{2}\end{cases}. ∫ 0 2 π sin n x d x = ∫ 0 2 π cos n x d x = { 2 π ⋅ n !! ( n − 1 )!! n !! ( n − 1 )!! , n ≡ 0 ( mod 2 ) , n ≡ 1 ( mod 2 ) . ∫ a b f ( x ) d x = ∫ a b ( a + b − x ) d x . \int_a^bf(x)\text{d}x=\int_a^b(a+b-x)\text{d}x. ∫ a b f ( x ) d x = ∫ a b ( a + b − x ) d x .
除此之外，由于定积分本身的几何意义，还有结论
∫ 0 a 2 a x − x 2 d x = 1 4 π a 2 \int_0^a\sqrt{2ax-x^2}\text{d}x=\frac{1}{4}\pi a^2
∫ 0 a 2 a x − x 2 d x = 4 1 π a 2
成立。
e.g.3.12 \textbf{e.g.3.12 } e.g.3.12 ∫ − 1 1 2 x 2 + x cos x 1 + 1 − x 2 d x . \int_{-1}^1\frac{2x^2+x\cos x}{1+\sqrt{1-x^2}}\text{d}x. ∫ − 1 1 1 + 1 − x 2 2 x 2 + x c o s x d x .
Solution \textbf{Solution } Solution f ( x ) = 2 x 2 1 + 1 − x 2 f(x)=\frac{2x^2}{1+\sqrt{1-x^2}} f ( x ) = 1 + 1 − x 2 2 x 2 g ( x ) = x cos x 1 + 1 − x 2 g(x)=\frac{x\cos x}{1+\sqrt{1-x^2}} g ( x ) = 1 + 1 − x 2 x c o s x
O.F. = ∫ − 1 1 f ( x ) d x + ∫ − 1 1 g ( x ) d x = 4 ∫ 0 1 x 2 1 + 1 − x 2 d x + 0 = 4 ∫ 0 1 x 2 ( 1 − 1 − x 2 ) ( 1 + 1 − x 2 ) ( 1 − 1 − x 2 ) d x = 4 ∫ 0 1 1 − 1 − x 2 d x = 4 ( x ∣ 0 1 ) − 4 ∫ 0 1 1 − x 2 d x = 4 − 4 × π 4 = 4 − π . \begin{align*}
\text{O.F.}&=\int_{-1}^1f(x)\text{d}x+\int_{-1}^1g(x)\text{d}x=4\int_0^1\frac{x^2}{1+\sqrt{1-x^2}}\text{d}x+0\\
&=4\int_0^1\frac{x^2(1-\sqrt{1-x^2})}{(1+\sqrt{1-x^2})(1-\sqrt{1-x^2})}\text{d}x=4\int_0^11-\sqrt{1-x^2}\text{d}x\\
&=4\bigg(x\Big|_0^1\bigg)-4\int_0^1{\sqrt{1-x^2}}\text{d}x=4-4\times\frac{\pi}{4}\\
&=4-\pi.
\end{align*}
O.F. = ∫ − 1 1 f ( x ) d x + ∫ − 1 1 g ( x ) d x = 4 ∫ 0 1 1 + 1 − x 2 x 2 d x + 0 = 4 ∫ 0 1 ( 1 + 1 − x 2 ) ( 1 − 1 − x 2 ) x 2 ( 1 − 1 − x 2 ) d x = 4 ∫ 0 1 1 − 1 − x 2 d x = 4 ( x 0 1 ) − 4 ∫ 0 1 1 − x 2 d x = 4 − 4 × 4 π = 4 − π .
e.x.3.12 \textbf{e.x.3.12 } e.x.3.12 ∫ − 1 1 cos x ln ( x + 1 + x 2 ) d x . \int_{-1}^1\cos x\ln(x+\sqrt{1+x^2})\text{d}x. ∫ − 1 1 cos x ln ( x + 1 + x 2 ) d x .
Solution \textbf{Solution } Solution f ( x ) = ln ( x + 1 + x 2 ) f(x)=\ln(x+\sqrt{1+x^2}) f ( x ) = ln ( x + 1 + x 2 )
f ( − x ) = ln ( 1 + ( − x ) 2 − x ) = ln ( 1 + x 2 − x ) = ln ( 1 + x 2 − x ) ( 1 + x 2 + x ) 1 + x 2 + x = ln 1 x + 1 + x 2 = − ln ( x + 1 + x 2 ) = − f ( x ) , \begin{align*}
f(-x)&=\ln(\sqrt{1+(-x)^2}-x)=\ln(\sqrt{1+x^2}-x)\\
&=\ln\frac{(\sqrt{1+x^2}-x)(\sqrt{1+x^2}+x)}{\sqrt{1+x^2}+x}=\ln\frac{1}{x+\sqrt{1+x^2}}\\
&=-\ln(x+\sqrt{1+x^2})=-f(x),
\end{align*}
f ( − x ) = ln ( 1 + ( − x ) 2 − x ) = ln ( 1 + x 2 − x ) = ln 1 + x 2 + x ( 1 + x 2 − x ) ( 1 + x 2 + x ) = ln x + 1 + x 2 1 = − ln ( x + 1 + x 2 ) = − f ( x ) ,
所以f ( x ) f(x) f ( x ) g ( x ) = cos x g(x)=\cos x g ( x ) = cos x h ( x ) = f ( x ) g ( x ) h(x)=f(x)g(x) h ( x ) = f ( x ) g ( x ) O.F. = 0. \text{O.F.}=0. O.F. = 0.
e.g.3.13 \textbf{e.g.3.13 } e.g.3.13 ∫ 0 2 π sin 99 x d x . \int_0^{2\pi}\sin^{99}x\text{d}x. ∫ 0 2 π sin 99 x d x .
Solution \textbf{Solution } Solution sin 99 x \sin^{99}x sin 99 x T = 2 π T=2\pi T = 2 π
O.F. = ∫ − π π sin 99 x d x = 0. \begin{align*}
\text{O.F.}&=\int_{-\pi}^\pi\sin^{99}x\text{d}x=0.
\end{align*}
O.F. = ∫ − π π sin 99 x d x = 0.
e.x.3.13 \textbf{e.x.3.13 } e.x.3.13
(1)∫ 0 π ( e cos x − e − cos x ) d x . \int_0^\pi(e^{\cos x}-e^{-\cos x})\text{d}x. ∫ 0 π ( e c o s x − e − c o s x ) d x .
(2)∫ 0 1 x 5 1 − x 2 d x . \int_0^1x^5\sqrt{1-x^2}\text{d}x. ∫ 0 1 x 5 1 − x 2 d x .
Solution \textbf{Solution } Solution
∫ 0 π ( e cos x − e − cos x ) d x = x = arccos t t = cos x ∫ − 1 1 ( e t − e − t ) 1 1 − t 2 d t = 可证待求积分的函数是奇函数 0. ∫ 0 1 x 5 1 − x 2 d x = x = sin t cos t = 1 − x 2 ∫ 0 π 2 sin 5 t cos t ⋅ cos t d t = ∫ 0 π 2 sin 5 t cos 2 t d t = ∫ 0 π 2 sin 5 t − sin 7 t d t = 常用公式 4 5 × 2 3 × 1 − 6 7 × 4 5 × 2 3 × 1 = 8 105 . \begin{align*}
\int_0^\pi(e^{\cos x}-e^{-\cos x})\text{d}x&\xlongequal[x=\arccos t]{t=\cos x}\int_{-1}^1(e^t-e^{-t})\frac{1}{\sqrt{1-t^2}}\text{d}t\\
&\xlongequal{\text{可证待求积分的函数是奇函数}}0.\\
\int_0^1x^5\sqrt{1-x^2}\text{d}x&\xlongequal[x=\sin t]{\cos t=\sqrt{1-x^2}}\int_0^\frac{\pi}{2}\sin^5t\cos t\cdot\cos t\text{d}t=\int_0^\frac{\pi}{2}\sin^5t\cos^2t\text{d}t=\int_0^\frac{\pi}{2}\sin^5t-\sin^7t\text{d}t\\
&\xlongequal{\text{常用公式}}\frac{4}{5}\times\frac{2}{3}\times1-\frac{6}{7}\times\frac{4}{5}\times\frac{2}{3}\times1=\frac{8}{105}.
\end{align*}
∫ 0 π ( e c o s x − e − c o s x ) d x ∫ 0 1 x 5 1 − x 2 d x t = c o s x x = a r c c o s t ∫ − 1 1 ( e t − e − t ) 1 − t 2 1 d t 可证待求积分的函数是奇函数 0. c o s t = 1 − x 2 x = s i n t ∫ 0 2 π sin 5 t cos t ⋅ cos t d t = ∫ 0 2 π sin 5 t cos 2 t d t = ∫ 0 2 π sin 5 t − sin 7 t d t 常用公式 5 4 × 3 2 × 1 − 7 6 × 5 4 × 3 2 × 1 = 105 8 .
求定积分时也会遇到分段函数；一部分是隐性分段函数。对于这种，可以利用∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x \int_a^bf(x)\text{d}x=\int_a^cf(x)\text{d}x+\int_c^bf(x)\text{d}x ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x
e.g.3.14 \textbf{e.g.3.14 } e.g.3.14 ∫ 0 1 [ e x ] d x . \int_0^1[e^x]\text{d}x. ∫ 0 1 [ e x ] d x . [ x ] [x] [ x ] x x x
Solution \textbf{Solution } Solution
O.F. = ∫ 0 ln 2 1 d x + ∫ ln 2 1 2 d x = ln 2 + ( 2 − 2 ln 2 ) = 2 − ln 2. \begin{align*}
\text{O.F.}&=\int_0^{\ln2}1\text{d}x+\int_{\ln 2}^12\text{d}x\\
&=\ln 2+(2-2\ln 2)=2-\ln 2.
\end{align*}
O.F. = ∫ 0 l n 2 1 d x + ∫ l n 2 1 2 d x = ln 2 + ( 2 − 2 ln 2 ) = 2 − ln 2.
e.x.3.14 \textbf{e.x.3.14 } e.x.3.14 ∫ − 1 x f ( t ) d t . \int_{-1}^xf(t)\text{d}t. ∫ − 1 x f ( t ) d t . f ( x ) = { 2 x + 3 2 x 2 , − 1 ≤ x < 0 x e x ( e x + 1 ) 2 , 0 ≤ x ≤ 1 . f(x)=\begin{cases}2x+\frac{3}{2}x^2&,-1\leq x<0\\\frac{xe^x}{(e^x+1)^2}&,0\leq x\leq 1\end{cases}. f ( x ) = { 2 x + 2 3 x 2 ( e x + 1 ) 2 x e x , − 1 ≤ x < 0 , 0 ≤ x ≤ 1 .
Solution − 1 ≤ x < 0 \textbf{Solution }-1\leq x<0 Solution − 1 ≤ x < 0 ∫ f ( x ) d x = x 2 + 1 2 x 3 . \int f(x)\text{d}x=x^2+\frac{1}{2}x^3. ∫ f ( x ) d x = x 2 + 2 1 x 3 .
O.F. = ( t 2 + 1 2 t 3 ) − 1 x = x 2 + 1 2 x 3 − 1 2 . \begin{align*}
\text{O.F.}&=\Big(t^2+\frac{1}{2}t^3\Big)_{-1}^x=x^2+\frac{1}{2}x^3-\frac{1}{2}.
\end{align*}
O.F. = ( t 2 + 2 1 t 3 ) − 1 x = x 2 + 2 1 x 3 − 2 1 .
所以∫ − 1 0 f ( x ) d x = − 1 2 . x > 0 \int_{-1}^0f(x)\text{d}x=-\frac{1}{2}.x>0 ∫ − 1 0 f ( x ) d x = − 2 1 . x > 0
∫ f ( x ) d x = ∫ x ( e x + 1 ) 2 d e x = γ = e x ⇒ x = ln γ ∫ ln γ ( γ + 1 ) 2 d γ = − ∫ ln γ d ( 1 γ + 1 ) = 分部积分 − ( ln γ γ + 1 − ∫ 1 γ ( γ + 1 ) d γ ) = ∫ 1 γ − 1 γ + 1 d γ − ln γ γ + 1 = ln ∣ γ ∣ − ln ∣ γ + 1 ∣ − ln γ γ + 1 = x − ln ∣ e x + 1 ∣ − x e x + 1 , \begin{align*}
\int f(x)\text{d}x&=\int\frac{x}{(e^x+1)^2}\text{d}e^x\xlongequal{\gamma=e^x\Rightarrow x=\ln\gamma}\int\frac{\ln\gamma}{(\gamma+1)^2}\text{d}\gamma\\
&=-\int\ln\gamma\text{d}\Big(\frac{1}{\gamma+1}\Big)\xlongequal{\text{分部积分}}-\bigg(\frac{\ln\gamma}{\gamma+1}-\int\frac{1}{\gamma(\gamma+1)}\text{d}\gamma\bigg)\\
&=\int\frac{1}{\gamma}-\frac{1}{\gamma+1}\text{d}\gamma-\frac{\ln\gamma}{\gamma+1}=\ln|\gamma|-\ln|\gamma+1|-\frac{\ln\gamma}{\gamma+1}\\
&=x-\ln|e^x+1|-\frac{x}{e^x+1},
\end{align*}
∫ f ( x ) d x = ∫ ( e x + 1 ) 2 x d e x γ = e x ⇒ x = l n γ ∫ ( γ + 1 ) 2 ln γ d γ = − ∫ ln γ d ( γ + 1 1 ) 分部积分 − ( γ + 1 ln γ − ∫ γ ( γ + 1 ) 1 d γ ) = ∫ γ 1 − γ + 1 1 d γ − γ + 1 ln γ = ln ∣ γ ∣ − ln ∣ γ + 1∣ − γ + 1 ln γ = x − ln ∣ e x + 1∣ − e x + 1 x ,
所以此时
∫ − 1 x f ( x ) d x = ∫ − 1 0 f ( x ) d x + ∫ 0 x f ( x ) d x = − 1 2 + ( x − ln ∣ e x + 1 ∣ − x e x + 1 + ln 2 ) , \begin{align*}
\int_{-1}^xf(x)\text{d}x&=\int_{-1}^0f(x)\text{d}x+\int_0^xf(x)\text{d}x\\
&=-\frac{1}{2}+\bigg(x-\ln|e^x+1|-\frac{x}{e^x+1}+\ln 2\bigg),
\end{align*}
∫ − 1 x f ( x ) d x = ∫ − 1 0 f ( x ) d x + ∫ 0 x f ( x ) d x = − 2 1 + ( x − ln ∣ e x + 1∣ − e x + 1 x + ln 2 ) ,
故
O.F. = { 1 2 x 3 + x 2 − 1 2 , − 1 ≤ x < 0 x − ln ∣ e x + 1 ∣ − x e x + 1 + ln 2 − 1 2 , 0 ≤ x ≤ 1 . \text{O.F.}=\begin{cases}
\frac{1}{2}x^3+x^2-\frac{1}{2}&,-1\leq x<0\\
x-\ln|e^x+1|-\frac{x}{e^x+1}+\ln 2-\frac{1}{2}&,0\leq x\leq 1
\end{cases}.
O.F. = { 2 1 x 3 + x 2 − 2 1 x − ln ∣ e x + 1∣ − e x + 1 x + ln 2 − 2 1 , − 1 ≤ x < 0 , 0 ≤ x ≤ 1 .
对于待求积分式中还有积分的定积分（即二重积分），考虑使用分部积分等方法先将其化为一重积分，随后再进行运算。
e.g.3.15 \textbf{e.g.3.15 } e.g.3.15 ∫ 0 π ( ∫ 0 x sin t π − t d t ) d x . \int_0^\pi\big(\int_0^x\frac{\sin t}{\pi-t}\text{d}t\big)\text{d}x. ∫ 0 π ( ∫ 0 x π − t s i n t d t ) d x .
Solution \textbf{Solution } Solution
O.F. = ∫ 0 π ( ∫ t π sin t π − t d x ) dt = ∫ 0 π sin t d t = 2. \begin{align*}
\text{O.F.}&=\int_0^\pi\bigg(\int_t^\pi\frac{\sin t}{\pi-t}\text{d}x\bigg)\text{dt}\\
&=\int_0^\pi\sin t\text{d}t=2.
\end{align*}
O.F. = ∫ 0 π ( ∫ t π π − t sin t d x ) dt = ∫ 0 π sin t d t = 2.
e.x.3.15 \textbf{e.x.3.15 } e.x.3.15 ∫ 0 1 1 x ( ∫ 1 x ln ( 1 + t ) t d t ) d x . \int_0^1\frac{1}{\sqrt{x}}\big(\int_1^x\frac{\ln(1+t)}{t}\text{d}t\big)\text{d}x. ∫ 0 1 x 1 ( ∫ 1 x t l n ( 1 + t ) d t ) d x .
Solution \textbf{Solution } Solution
O.F. = 2 ∫ 0 1 ( ∫ 1 x ln ( 1 + t ) t ) d ( x ) = 分部积分 2 ( ( x ∫ 1 x ln ( 1 + t ) t d t ) 0 1 − ∫ 0 1 x ln ( 1 + x ) x d x ) = 0 − 2 ∫ 0 1 ln ( 1 + x ) x d x = − 4 ∫ 0 1 ln ( x + 1 ) d ( x ) = 分部积分 − 4 ( ( x ln ( 1 + x ) ) 0 1 − ∫ 0 1 x x + 1 d x ) = 4 ∫ 0 1 x x + 1 d x − 4 ln 2 = θ = x ⇒ x = θ 2 4 ∫ 0 1 θ ⋅ 2 θ 1 + θ 2 d θ − 4 ln 2 = 8 ∫ 0 1 ( 1 − 1 θ 2 + 1 ) d θ − 4 ln 2 = 8 ( 1 − arctan 1 ) − 4 ln 2 = 8 − 2 π − 4 ln 2. \begin{align*}
\text{O.F.}&=2\int_0^1\bigg(\int_1^x\frac{\ln(1+t)}{t}\bigg)\text{d}(\sqrt{x})\\
&\xlongequal{\text{分部积分}}2\bigg(\Big(\sqrt{x}\int_1^x\frac{\ln(1+t)}{t}\text{d}t\Big)_0^1-\int_0^1\frac{\sqrt{x}\ln(1+x)}{x}\text{d}x\bigg)=0-2\int_0^1\frac{\ln(1+x)}{\sqrt{x}}\text{d}x\\
&=-4\int_0^1\ln(x+1)\text{d}(\sqrt{x})\xlongequal{\text{分部积分}}-4\bigg(\big(\sqrt{x}\ln(1+x)\big)_0^1-\int_0^1\frac{\sqrt{x}}{x+1}\text{d}x\bigg)\\
&=4\int_0^1\frac{\sqrt{x}}{x+1}\text{d}x-4\ln 2\xlongequal{\theta=\sqrt{x}\Rightarrow x=\theta^2}4\int_0^1\frac{\theta\cdot2\theta}{1+\theta^2}\text{d}\theta-4\ln 2\\
&=8\int_0^1\bigg(1-\frac{1}{\theta^2+1}\bigg)\text{d}\theta-4\ln 2=8(1-\arctan 1)-4\ln 2\\
&=8-2\pi-4\ln 2.
\end{align*}
O.F. = 2 ∫ 0 1 ( ∫ 1 x t ln ( 1 + t ) ) d ( x ) 分部积分 2 ( ( x ∫ 1 x t ln ( 1 + t ) d t ) 0 1 − ∫ 0 1 x x ln ( 1 + x ) d x ) = 0 − 2 ∫ 0 1 x ln ( 1 + x ) d x = − 4 ∫ 0 1 ln ( x + 1 ) d ( x ) 分部积分 − 4 ( ( x ln ( 1 + x ) ) 0 1 − ∫ 0 1 x + 1 x d x ) = 4 ∫ 0 1 x + 1 x d x − 4 ln 2 θ = x ⇒ x = θ 2 4 ∫ 0 1 1 + θ 2 θ ⋅ 2 θ d θ − 4 ln 2 = 8 ∫ 0 1 ( 1 − θ 2 + 1 1 ) d θ − 4 ln 2 = 8 ( 1 − arctan 1 ) − 4 ln 2 = 8 − 2 π − 4 ln 2.
变限积分函数
对一个含有积分的函数，可以利用积分的定义求出（或直接写出）这个变限积分函数的导数。
考虑上面这句话。我们本就知道( ∫ a x f ( t ) d t ) ′ = f ( x ) ; \big(\int_a^xf(t)\text{d}t\big)'=f(x); ( ∫ a x f ( t ) d t ) ′ = f ( x ) ; ( ∫ a g ( x ) f ( t ) d t ) ′ = f ( g ( x ) ) g ′ ( x ) , \big(\int_a^{g(x)}f(t)\text{d}t\big)'=f\big(g(x)\big)g'(x), ( ∫ a g ( x ) f ( t ) d t ) ′ = f ( g ( x ) ) g ′ ( x ) , ( ∫ h ( x ) g ( x ) f ( t ) d t ) ′ = f ( g ( x ) ) g ′ ( x ) − f ( h ( x ) ) h ′ ( x ) . \big(\int_{h(x)}^{g(x)}f(t)\text{d}t\big)'=f\big(g(x)\big)g'(x)-f\big(h(x)\big)h'(x). ( ∫ h ( x ) g ( x ) f ( t ) d t ) ′ = f ( g ( x ) ) g ′ ( x ) − f ( h ( x ) ) h ′ ( x ) .
对于更复杂的变限积分函数之导，有
( ∫ a x f ( t ) g ( x ) d t ) ′ = ( g ( x ) ∫ a x f ( t ) d t ) ′ = g ′ ( x ) ∫ a x f ( t ) d t + g ( x ) f ( x ) . \bigg(\int_a^xf(t)g(x)\text{d}t\bigg)'=\bigg(g(x)\int_a^xf(t)\text{d}t\bigg)'=g'(x)\int_a^xf(t)\text{d}t+g(x)f(x).
( ∫ a x f ( t ) g ( x ) d t ) ′ = ( g ( x ) ∫ a x f ( t ) d t ) ′ = g ′ ( x ) ∫ a x f ( t ) d t + g ( x ) f ( x ) .
换言之，对于∫ a x f ( t , x ) d t \int_a^xf(t,x)\text{d}t ∫ a x f ( t , x ) d t x x x
对于由两层变限积分表示的函数，可以采用「由外层向内直接对变上限求导」「设内侧为新函数后求导」「通过分部积分法将其化为单层积分」三种办法求其导数。
一般地，当x → 0 x\rightarrow 0 x → 0 g ( x ) , f ( x ) g(x),f(x) g ( x ) , f ( x ) x x x n n n m m m ∫ 0 g ( x ) f ( t ) d t \int^{g(x)}_0f(t)\text{d}t ∫ 0 g ( x ) f ( t ) d t x x x ( m + 1 ) n (m+1)n ( m + 1 ) n
e.g.3.16 \textbf{e.g.3.16 } e.g.3.16 f ( x ) f(x) f ( x ) f ( 0 ) ≠ 0 , f(0)≠0, f ( 0 ) = 0 , lim x → 0 ∫ 0 x ( x − t ) f ( t ) d t x ∫ 0 x f ( x − t ) d t . \lim_{x\rightarrow 0}\frac{\int_0^x(x-t)f(t)\text{d}t}{x\int_0^xf(x-t)\text{d}t}. lim x → 0 x ∫ 0 x f ( x − t ) d t ∫ 0 x ( x − t ) f ( t ) d t .
Solution \textbf{Solution } Solution
∫ 0 x f ( x − t ) d t = u = x − t ⇒ t = − u + x ∫ x 0 − f ( u ) d u = ∫ 0 x f ( u ) d u , \begin{align*}
\int_0^xf(x-t)\text{d}t&\xlongequal{u=x-t\Rightarrow t=-u+x}\int_x^0-f(u)\text{d}u=\int_0^xf(u)\text{d}u,
\end{align*}
∫ 0 x f ( x − t ) d t u = x − t ⇒ t = − u + x ∫ x 0 − f ( u ) d u = ∫ 0 x f ( u ) d u ,
因此
O.F. = lim x → 0 x ∫ 0 x f ( t ) d t − ∫ 0 x t f ( t ) d t x ∫ 0 x f ( u ) d u = L ′ H lim x → 0 ∫ 0 x f ( t ) d t + x f ( x ) − x f ( x ) ∫ 0 x f ( u ) d u + x f ( x ) = lim x → 0 ∫ 0 x f ( t ) d t ∫ 0 x f ( u ) d u + x f ( x ) = L ′ H f ( x ) f ( x ) + f ( x ) + x f ′ ( x ) = f ( x ) 2 f ( x ) + x f ′ ( x ) = 1 2 . \begin{align*}
\text{O.F.}&=\lim_{x\rightarrow 0}\frac{x\int_0^xf(t)\text{d}t-\int_0^xtf(t)\text{d}t}{x\int_0^xf(u)\text{d}u}\\
&\xlongequal{L'H}\lim_{x\rightarrow 0}\frac{\int_0^xf(t)\text{d}t+xf(x)-xf(x)}{\int_0^xf(u)\text{d}u+xf(x)}=\lim_{x\rightarrow 0}\frac{\int_0^xf(t)\text{d}t}{\int_0^xf(u)\text{d}u+xf(x)}\\
&\xlongequal{L'H}\frac{f(x)}{f(x)+f(x)+xf'(x)}=\frac{f(x)}{2f(x)+xf'(x)}\\
&=\frac{1}{2}.
\end{align*}
O.F. = x → 0 lim x ∫ 0 x f ( u ) d u x ∫ 0 x f ( t ) d t − ∫ 0 x t f ( t ) d t L ′ H x → 0 lim ∫ 0 x f ( u ) d u + x f ( x ) ∫ 0 x f ( t ) d t + x f ( x ) − x f ( x ) = x → 0 lim ∫ 0 x f ( u ) d u + x f ( x ) ∫ 0 x f ( t ) d t L ′ H f ( x ) + f ( x ) + x f ′ ( x ) f ( x ) = 2 f ( x ) + x f ′ ( x ) f ( x ) = 2 1 .
e.x.3.16 \textbf{e.x.3.16 } e.x.3.16 f ( x ) f(x) f ( x ) ∫ 0 x f ( x − t ) d t = ∫ 0 x ( x − t ) f ( t ) d t + e − x − 1 , \int_0^xf(x-t)\text{d}t=\int_0^x(x-t)f(t)\text{d}t+e^{-x}-1, ∫ 0 x f ( x − t ) d t = ∫ 0 x ( x − t ) f ( t ) d t + e − x − 1 , f ( x ) . f(x). f ( x ) .
Solution \textbf{Solution } Solution ∫ 0 x f ( x − t ) d t = x ∫ 0 x f ( t ) d t − ∫ 0 x t f ( t ) d t + e − x − 1. \int_0^xf(x-t)\text{d}t=x\int_0^xf(t)\text{d}t-\int_0^xtf(t)\text{d}t+e^{-x}-1. ∫ 0 x f ( x − t ) d t = x ∫ 0 x f ( t ) d t − ∫ 0 x t f ( t ) d t + e − x − 1.
f ( x ) = ∫ 0 x f ( t ) d t + x f ( x ) − x f ( x ) − e − x , f(x)=\int_0^xf(t)\text{d}t+xf(x)-xf(x)-e^{-x},
f ( x ) = ∫ 0 x f ( t ) d t + x f ( x ) − x f ( x ) − e − x ,
也即
f ( x ) = ∫ 0 x f ( t ) d t − e − x . f(x)=\int_0^xf(t)\text{d}t-e^{-x}.
f ( x ) = ∫ 0 x f ( t ) d t − e − x .
对这个等式取x = 0 x=0 x = 0 f ( 0 ) = − 1. f(0)=-1. f ( 0 ) = − 1.
f ′ ( x ) = f ( x ) + e − x ⇒ f ( x ) = − e x + e − x 2 + C . f'(x)=f(x)+e^{-x}\Rightarrow f(x)=-\frac{e^x+e^{-x}}{2}+C.
f ′ ( x ) = f ( x ) + e − x ⇒ f ( x ) = − 2 e x + e − x + C .
代入f ( 0 ) = − 1 , f(0)=-1, f ( 0 ) = − 1 , f ( x ) = − e x + e − x 2 . f(x)=-\frac{e^x+e^{-x}}{2}. f ( x ) = − 2 e x + e − x .
反常积分
上下限中存在无限量或在上下限中某一处的函数值为无限量的定积分为反常积分。
对于含有无限量作为上下限的函数积分，将这个无限量视为一个趋近于无限的变量即可。
e.g.3.17 \textbf{e.g.3.17 } e.g.3.17 ∫ 0 + ∞ x e − x ( 1 + e − x ) 2 d x . \int_0^{+\infty}\frac{xe^{-x}}{(1+e^{-x})^2}\text{d}x. ∫ 0 + ∞ ( 1 + e − x ) 2 x e − x d x .
Solution \textbf{Solution } Solution
O.F. = ∫ 0 + ∞ − x ( 1 + e − x ) 2 d e − x = t = e − x ⇒ x = − ln t ∫ 1 0 ln t ( 1 + t ) 2 d t = − ∫ 1 0 ln t d ( 1 1 + t ) = 分部积分 − ( ln t 1 + t ) 1 0 + ∫ 1 0 1 t ( t + 1 ) = ∫ 1 0 1 t − 1 t + 1 d t − ( ln t 1 + t ) 1 0 = ( ln ∣ t ∣ − ln ∣ t + 1 ∣ − ln t t + 1 ) 1 0 = ( ln t 1 + t + ln ∣ t + 1 ∣ − ln ∣ t ∣ ) 0 1 = ln 2. \begin{align*}
\text{O.F.}&=\int_0^{+\infty}\frac{-x}{(1+e^{-x})^2}\text{d}e^{-x}\xlongequal{t=e^{-x}\Rightarrow x=-\ln t}\int_1^0\frac{\ln t}{(1+t)^2}\text{d}t\\
&=-\int_1^0\ln t\text{d}\Big(\frac{1}{1+t}\Big)\xlongequal{\text{分部积分}}-\Big(\frac{\ln t}{1+t}\Big)_1^0+\int_1^0\frac{1}{t(t+1)}\\
&=\int_1^0\frac{1}{t}-\frac{1}{t+1}\text{d}t-\Big(\frac{\ln t}{1+t}\Big)_1^0=\Big(\ln|t|-\ln|t+1|-\frac{\ln t}{t+1}\Big)_1^0\\
&=\Big(\frac{\ln t}{1+t}+\ln|t+1|-\ln|t|\Big)_0^1=\ln 2.
\end{align*}
O.F. = ∫ 0 + ∞ ( 1 + e − x ) 2 − x d e − x t = e − x ⇒ x = − l n t ∫ 1 0 ( 1 + t ) 2 ln t d t = − ∫ 1 0 ln t d ( 1 + t 1 ) 分部积分 − ( 1 + t ln t ) 1 0 + ∫ 1 0 t ( t + 1 ) 1 = ∫ 1 0 t 1 − t + 1 1 d t − ( 1 + t ln t ) 1 0 = ( ln ∣ t ∣ − ln ∣ t + 1∣ − t + 1 ln t ) 1 0 = ( 1 + t ln t + ln ∣ t + 1∣ − ln ∣ t ∣ ) 0 1 = ln 2.
考虑例3.17的最后一步。显然x = 0 x=0 x = 0 ln t 1 + t \frac{\ln t}{1+t} 1 + t l n t ln ∣ t ∣ \ln|t| ln ∣ t ∣ x = 0 x=0 x = 0 x → 0 x\rightarrow 0 x → 0
e.x.3.17 \textbf{e.x.3.17 } e.x.3.17 ∫ − ∞ 1 1 x 2 + 2 x + 5 d x . \int_{-\infty}^1\frac{1}{x^2+2x+5}\text{d}x. ∫ − ∞ 1 x 2 + 2 x + 5 1 d x .
Solution \textbf{Solution } Solution
O.F. = ∫ − ∞ 1 1 ( x + 1 ) 2 + 4 d x = ∫ − ∞ 1 1 ( x + 1 ) 2 + 4 d ( x + 1 ) = ( 1 2 arctan x + 1 2 ) − ∞ 1 = π 8 − ( − π 4 ) = 3 π 8 . \begin{align*}
\text{O.F.}&=\int_{-\infty}^1\frac{1}{(x+1)^2+4}\text{d}x=\int_{-\infty}^1\frac{1}{(x+1)^2+4}\text{d}(x+1)\\
&=\Big(\frac{1}{2}\arctan\frac{x+1}{2}\Big)_{-\infty}^1=\frac{\pi}{8}-\Big(-\frac{\pi}{4}\Big)\\
&=\frac{3\pi}{8}.
\end{align*}
O.F. = ∫ − ∞ 1 ( x + 1 ) 2 + 4 1 d x = ∫ − ∞ 1 ( x + 1 ) 2 + 4 1 d ( x + 1 ) = ( 2 1 arctan 2 x + 1 ) − ∞ 1 = 8 π − ( − 4 π ) = 8 3 π .
考虑下面四个常见反常积分的收敛性。
∫ a + ∞ 1 x p d x ( a > 0 ) \int_a^{+\infty}\frac{1}{x^p}\text{d}x(a>0) ∫ a + ∞ x p 1 d x ( a > 0 ) p > 1 p>1 p > 1 p ≤ 1 p\leq 1 p ≤ 1 ∫ a + ∞ 1 x ln p x d x ( a > 0 ) \int_a^{+\infty}\frac{1}{x\ln^px}\text{d}x(a>0) ∫ a + ∞ x l n p x 1 d x ( a > 0 ) p > 1 p>1 p > 1 p ≤ 1 p\leq 1 p ≤ 1 ∫ 0 + ∞ x k e − λ x d x ( k ≥ 0 ) \int_0^{+\infty}x^ke^{-\lambda x}\text{d}x(k\geq 0) ∫ 0 + ∞ x k e − λ x d x ( k ≥ 0 ) λ > 0 \lambda>0 λ > 0 λ ≤ 0 \lambda\leq 0 λ ≤ 0 ∫ a 0 1 x p d x ( a > 0 ) \int_a^0\frac{1}{x^p}\text{d}x(a>0) ∫ a 0 x p 1 d x ( a > 0 ) p < 1 p<1 p < 1 p ≥ 1 p\geq 1 p ≥ 1
e.g.3.18 \textbf{e.g.3.18 } e.g.3.18 f ( x ) f(x) f ( x ) ∫ 2 + ∞ f ( x ) d x \int_2^{+\infty}f(x)\text{d}x ∫ 2 + ∞ f ( x ) d x
A. 1 x \frac{1}{\sqrt{x}} x 1
B. ln x x \frac{\ln x}{x} x l n x
C.1 x ln x \frac{1}{x\ln x} x l n x 1
Answer \textbf{Answer } Answer
e.x.3.18 \textbf{e.x.3.18 } e.x.3.18 f ( x ) = { 1 ( x − 1 ) α − 1 , 1 < x < e 1 x ln α + 1 x , x ≥ e f(x)=\begin{cases}\frac{1}{(x-1)^{\alpha-1}}&,1<x<e\\\frac{1}{x\ln^{\alpha+1}x}&,x\geq e\end{cases} f ( x ) = { ( x − 1 ) α − 1 1 x l n α + 1 x 1 , 1 < x < e , x ≥ e ∫ 1 + ∞ f ( x ) d x \int_1^{+\infty}f(x)\text{d}x ∫ 1 + ∞ f ( x ) d x α \alpha α
Solution \textbf{Solution } Solution ∫ 1 e f ( x ) d x \int_1^ef(x)\text{d}x ∫ 1 e f ( x ) d x ∫ e + ∞ f ( x ) d x \int_e^{+\infty}f(x)\text{d}x ∫ e + ∞ f ( x ) d x α + 1 > 1 , α − 1 < 1 , \alpha+1>1,\alpha-1<1, α + 1 > 1 , α − 1 < 1 , 0 < α < 2. 0<\alpha<2. 0 < α < 2.
如果0 < f ( x ) < g ( x ) , 0<f(x)<g(x), 0 < f ( x ) < g ( x ) , f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x ) g ( x ) g(x) g ( x ) f ( x ) f(x) f ( x )
如果lim x → + ∞ f ( x ) g ( x ) = λ ≠ 0 , \lim_{x\rightarrow +\infty}\frac{f(x)}{g(x)}=\lambda≠0, lim x → + ∞ g ( x ) f ( x ) = λ = 0 , f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x )
特别地，∫ − ∞ + ∞ e − x 2 d x = π . \int_{-\infty}^{+\infty}e^{-x^2}\text{d}x=\sqrt{\pi}. ∫ − ∞ + ∞ e − x 2 d x = π .
一元积分应用
由于定积分本身的实际意义就是一个函数和横坐标轴所围面积，所以在讨论一元积分的实际应用时很难不想到这一功能。
e.g.3.19 \textbf{e.g.3.19 } e.g.3.19 y = e x y=e^x y = e x x x x
Solution \textbf{Solution } Solution y = e x y=e^x y = e x y = e x , y=ex, y = e x , ( 1 , e ) . (1,e). ( 1 , e ) .
∫ − ∞ 1 e x − 1 × e 2 = ( e x ) − ∞ 1 − e 2 = e 2 . \begin{align*}
\int_{-\infty}^1e^x-\frac{1\times e}{2}&=(e^x)_{-\infty}^1-\frac{e}{2}\\
&=\frac{e}{2}.
\end{align*}
∫ − ∞ 1 e x − 2 1 × e = ( e x ) − ∞ 1 − 2 e = 2 e .
e.x.3.19 \textbf{e.x.3.19 } e.x.3.19 y = x 3 − 4 x 2 + 3 x y=x^3-4x^2+3x y = x 3 − 4 x 2 + 3 x x x x
Solution \textbf{Solution } Solution y = x 3 − 4 x 2 + 3 x y=x^3-4x^2+3x y = x 3 − 4 x 2 + 3 x y = x ( x − 1 ) ( x − 3 ) , y=x(x-1)(x-3), y = x ( x − 1 ) ( x − 3 ) , x x x ( 0 , 0 ) , ( 1 , 0 ) (0,0),(1,0) ( 0 , 0 ) , ( 1 , 0 ) ( 3 , 0 ) . (3,0). ( 3 , 0 ) .
∣ ∫ 0 1 x 3 − 4 x 2 + 3 x d x ∣ + ∣ ∫ 1 3 x 3 − 4 x 2 + 3 x d x ∣ = ∣ ( 1 4 x 4 − 4 3 x 3 + 3 2 x 2 ) 0 1 ∣ + ∣ ( 1 4 x 4 − 4 3 x 3 + 3 2 x 2 ) 1 3 ∣ = 5 12 + 8 3 = 37 12 . \begin{align*}
\bigg|\int_0^1x^3-4x^2+3x\text{d}x\bigg|+\bigg|\int_1^3x^3-4x^2+3x\text{d}x\bigg|&=\bigg|\Big(\frac{1}{4}x^4-\frac{4}{3}x^3+\frac{3}{2}x^2\Big)_0^1\bigg|+\bigg|\Big(\frac{1}{4}x^4-\frac{4}{3}x^3+\frac{3}{2}x^2\Big)_1^3\bigg|\\
&=\frac{5}{12}+\frac{8}{3}=\frac{37}{12}.
\end{align*}
∫ 0 1 x 3 − 4 x 2 + 3 x d x + ∫ 1 3 x 3 − 4 x 2 + 3 x d x = ( 4 1 x 4 − 3 4 x 3 + 2 3 x 2 ) 0 1 + ( 4 1 x 4 − 3 4 x 3 + 2 3 x 2 ) 1 3 = 12 5 + 3 8 = 12 37 .
事实上，定积分并不仅限于对平面坐标系上的曲线有用。对极坐标系，依然可以通过求定积分来确定围成的图形面积；唯一不同的在于面积公式是$S=\frac{1}{2}\int_Ir^2\text{d}\theta$。
e.g.3.20 \textbf{e.g.3.20 } e.g.3.20 r = cos 3 θ ( − π 6 ≤ θ ≤ π 6 ) r=\cos3\theta(-\frac{\pi}{6}\leq\theta\leq\frac{\pi}{6}) r = cos 3 θ ( − 6 π ≤ θ ≤ 6 π ) L L L
Solution \textbf{Solution } Solution
1 2 ∫ − π 6 π 6 cos 2 3 θ d θ = ∫ 0 π 6 1 + cos 6 θ 2 d θ = ( 1 2 θ + 1 12 sin 6 θ ) 0 π 6 = π 12 . \begin{align*}
\frac{1}{2}\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\cos^23\theta\text{d}\theta&=\int_0^{\frac{\pi}{6}}\frac{1+\cos6\theta}{2}\text{d}\theta\\
&=\Big(\frac{1}{2}\theta+\frac{1}{12}\sin6\theta\Big)_0^{\frac{\pi}{6}}\\
&=\frac{\pi}{12}.
\end{align*}
2 1 ∫ − 6 π 6 π cos 2 3 θ d θ = ∫ 0 6 π 2 1 + cos 6 θ d θ = ( 2 1 θ + 12 1 sin 6 θ ) 0 6 π = 12 π .
e.x.3.20 \textbf{e.x.3.20 } e.x.3.20 r = a ( 1 − cos θ ) r=a(1-\cos\theta) r = a ( 1 − cos θ ) L L L
Solution \textbf{Solution } Solution
1 2 ∫ − π π a 2 ( 1 − cos θ ) 2 d θ = 3 2 π a 2 . \begin{align*}
\frac{1}{2}\int_{-\pi}^{\pi}a^2(1-\cos\theta)^2\text{d}\theta&=\frac{3}{2}\pi a^2.
\end{align*}
2 1 ∫ − π π a 2 ( 1 − cos θ ) 2 d θ = 2 3 π a 2 .
根据「点动成线、线动成面、面动成体」的思想，积分的积分自然而然就具有了体积的实际意义。另一方面，函数曲线绕某个轴旋转一圈后它的轨迹也就会成为这个立体图形的表面积。
e.g.3.21 \textbf{e.g.3.21 } e.g.3.21 { x = a cos 3 t y = a sin 3 t \begin{cases}x&=a\cos^3t\\y&=a\sin^3t\end{cases} { x y = a cos 3 t = a sin 3 t S S S x x x V . V. V .
Solution \textbf{Solution } Solution x 2 3 + y 2 3 = a 2 3 . x^{\frac{2}{3}}+y^\frac{2}{3}=a^\frac{2}{3}. x 3 2 + y 3 2 = a 3 2 .
S = 4 ∫ 0 a y d x = 4 ∫ π 2 0 ( a sin 3 t ) ( 3 a cos 2 t ( − sin t ) ) d t = 12 a 2 ∫ 0 π 2 sin 4 t ( 1 − sin 2 t ) d t = 12 a 2 ( 3 4 × 1 2 × π 2 − 5 6 × 3 4 × 1 2 × π 2 ) = 3 8 π a 2 , V = 2 ∫ 0 a y 2 d x = 2 ∫ π 2 0 a 2 sin 6 t 3 a cos 2 t ( − sin t ) d t = 32 105 π a 3 . \begin{align*}
S&=4\int_0^ay\text{d}x=4\int_{\frac{\pi}{2}}^0(a\sin^3t)\big(3a\cos^2t(-\sin t)\big)\text{d}t\\
&=12a^2\int_0^\frac{\pi}{2}\sin^4t(1-\sin^2t)\text{d}t\\
&=12a^2(\frac{3}{4}\times\frac{1}{2}\times\frac{\pi}{2}-\frac{5}{6}\times\frac{3}{4}\times\frac{1}{2}\times\frac{\pi}{2})\\
&=\frac{3}{8}\pi a^2,\\
V&=2\int_0^ay^2\text{d}x=2\int_\frac{\pi}{2}^0a^2\sin^6t3a\cos^2t(-\sin t)\text{d}t\\
&=\frac{32}{105}\pi a^3.
\end{align*}
S V = 4 ∫ 0 a y d x = 4 ∫ 2 π 0 ( a sin 3 t ) ( 3 a cos 2 t ( − sin t ) ) d t = 12 a 2 ∫ 0 2 π sin 4 t ( 1 − sin 2 t ) d t = 12 a 2 ( 4 3 × 2 1 × 2 π − 6 5 × 4 3 × 2 1 × 2 π ) = 8 3 π a 2 , = 2 ∫ 0 a y 2 d x = 2 ∫ 2 π 0 a 2 sin 6 t 3 a cos 2 t ( − sin t ) d t = 105 32 π a 3 .
e.x.3.21 \textbf{e.x.3.21 } e.x.3.21 f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) [ a , b ] [a,b] [ a , b ] g ( x ) < f ( x ) < m g(x)<f(x)<m g ( x ) < f ( x ) < m m m m y = g ( x ) , y = f ( x ) , x = a , x = b y=g(x),y=f(x),x=a,x=b y = g ( x ) , y = f ( x ) , x = a , x = b y = m y=m y = m V . V. V .
Solution \textbf{Solution } Solution
V = ∫ a b π ( m − g ( x ) ) 2 − π ( m − f ( x ) ) 2 d x = π ∫ a b ( m − g ( x ) ) 2 − ( m − f ( x ) ) 2 d x = π ∫ a b ( ( m − g ( x ) ) + ( m − f ( x ) ) ) ( ( m − g ( x ) ) − ( m − f ( x ) ) ) d x = ∫ a b π ( 2 m − f ( x ) − g ( x ) ) ( f ( x ) − g ( x ) ) d x . \begin{align*}
V&=\int_a^b\pi\big(m-g(x)\big)^2-\pi\big(m-f(x)\big)^2\text{d}x\\
&=\pi\int_a^b\big(m-g(x)\big)^2-\big(m-f(x)\big)^2\text{d}x\\
&=\pi\int_a^b\Big(\big(m-g(x)\big)+\big(m-f(x)\big)\Big)\Big(\big(m-g(x)\big)-\big(m-f(x)\big)\Big)\text{d}x\\
&=\int_a^b\pi\big(2m-f(x)-g(x)\big)\big(f(x)-g(x)\big)\text{d}x.
\end{align*}
V = ∫ a b π ( m − g ( x ) ) 2 − π ( m − f ( x ) ) 2 d x = π ∫ a b ( m − g ( x ) ) 2 − ( m − f ( x ) ) 2 d x = π ∫ a b ( ( m − g ( x ) ) + ( m − f ( x ) ) ) ( ( m − g ( x ) ) − ( m − f ( x ) ) ) d x = ∫ a b π ( 2 m − f ( x ) − g ( x ) ) ( f ( x ) − g ( x ) ) d x .
对一方程为( x − a ) 2 + y 2 = b 2 (x-a)^2+y^2=b^2 ( x − a ) 2 + y 2 = b 2 x x x
V = ( π b 2 ) ⋅ ( 2 π a ) = 2 π 2 a b 2 . V=(\pi b^2)\cdot(2\pi a)=2\pi^2ab^2.
V = ( π b 2 ) ⋅ ( 2 πa ) = 2 π 2 a b 2 .
常微分方程
一阶微分方程
可分离变量方程
形如d y d x = f ( x ) g ( y ) \frac{\text{d}y}{\text{d}x}=f(x)g(y) d x d y = f ( x ) g ( y )
∫ 1 g ( y ) d y = ∫ f ( x ) d x . \int\frac{1}{g(y)}\text{d}y=\int f(x)\text{d}x.
∫ g ( y ) 1 d y = ∫ f ( x ) d x .
如此即可得到原函数。
齐次方程
形如d y d x = f ( y x ) \frac{\text{d}y}{\text{d}x}=f(\frac{y}{x}) d x d y = f ( x y ) u = y x , u=\frac{y}{x}, u = x y ,
d u f ( u ) − u = d x x . \frac{\text{d}u}{f(u)-u}=\frac{\text{d}x}{x}.
f ( u ) − u d u = x d x .
如此对等式两端积分即可得到原函数。
e.g.4.1 \textbf{e.g.4.1 } e.g.4.1 { ( y + x 2 + y 2 ) d x − x d y = 0 y ∣ x = 1 = 0 \begin{cases}(y+\sqrt{x^2+y^2})\text{d}x-x\text{d}y=0\\y|_{x=1}=0\end{cases} { ( y + x 2 + y 2 ) d x − x d y = 0 y ∣ x = 1 = 0
Solution \textbf{Solution } Solution ( y + x 2 + y 2 ) d x − x d y = 0 (y+\sqrt{x^2+y^2})\text{d}x-x\text{d}y=0 ( y + x 2 + y 2 ) d x − x d y = 0
d y d x = y x + x 2 + y 2 x , \frac{\text{d}y}{\text{d}x}=\frac{y}{x}+\frac{\sqrt{x^2+y^2}}{x},
d x d y = x y + x x 2 + y 2 ,
即
d y d x = y x + 1 + ( y x ) 2 . \frac{\text{d}y}{\text{d}x}=\frac{y}{x}+\sqrt{1+\Big(\frac{y}{x}\Big)^2}.
d x d y = x y + 1 + ( x y ) 2 .
因此不妨令u = y x , u=\frac{y}{x}, u = x y , y = u x ⇒ d y d x = y ′ = u + u ′ x , y=ux\Rightarrow \frac{\text{d}y}{\text{d}x}=y'=u+u'x, y = ux ⇒ d x d y = y ′ = u + u ′ x ,
u + u ′ x = u + 1 + u 2 ⇒ u ′ x = 1 + u 2 ⇒ x d u d x = 1 + u 2 ⇒ d u 1 + u 2 = d x x ⇒ ∫ d u 1 + u 2 = ∫ d x x ⇒ ln ( u + 1 + u 2 ) = ln x + ln C ⇒ u + 1 + u 2 = C x . \begin{align*}
u+u'x=u+\sqrt{1+u^2}&\Rightarrow u'x=\sqrt{1+u^2}\\
&\Rightarrow\frac{x\text{d}u}{\text{d}x}=\sqrt{1+u^2}\\
&\Rightarrow\frac{\text{d}u}{\sqrt{1+u^2}}=\frac{\text{d}x}{x}\\
&\Rightarrow\int\frac{\text{d}u}{\sqrt{1+u^2}}=\int\frac{\text{d}x}{x}\\
&\Rightarrow\ln(u+\sqrt{1+u^2})=\ln x+\ln C\Rightarrow u+\sqrt{1+u^2}=Cx.
\end{align*}
u + u ′ x = u + 1 + u 2 ⇒ u ′ x = 1 + u 2 ⇒ d x x d u = 1 + u 2 ⇒ 1 + u 2 d u = x d x ⇒ ∫ 1 + u 2 d u = ∫ x d x ⇒ ln ( u + 1 + u 2 ) = ln x + ln C ⇒ u + 1 + u 2 = C x .
而我们知道y ∣ x = 1 = 0 ⇒ u ∣ x = 1 = 0 , y|_{x=1}=0\Rightarrow u|_{x=1}=0, y ∣ x = 1 = 0 ⇒ u ∣ x = 1 = 0 , C = 1. C=1. C = 1. u + 1 + u 2 = x , u+\sqrt{1+u^2}=x, u + 1 + u 2 = x ,
y x + x 2 + y 2 x 2 = x . \frac{y}{x}+\frac{\sqrt{x^2+y^2}}{x^2}=x.
x y + x 2 x 2 + y 2 = x .
实际上，有一些形如齐次方程的微分方程不一定非要用这种方法解决。我们本就知道复合函数导数公式( u v ) ′ = u v ′ + u ′ v , (uv)'=uv'+u'v, ( uv ) ′ = u v ′ + u ′ v , u , v u,v u , v y y y x x x
e.x.4.1 \textbf{e.x.4.1 } e.x.4.1 ( 1 2 , 0 ) \big(\frac{1}{2},0\big) ( 2 1 , 0 ) y ′ arcsin x + y 1 − x 2 = 1 y'\arcsin x+\frac{y}{\sqrt{1-x^2}}=1 y ′ arcsin x + 1 − x 2 y = 1
Solution \textbf{Solution } Solution y ′ arcsin x + y 1 − x 2 = 1 y'\arcsin x+\frac{y}{\sqrt{1-x^2}}=1 y ′ arcsin x + 1 − x 2 y = 1 ( y arcsin x ) ′ = 1. (y\arcsin x)'=1. ( y arcsin x ) ′ = 1. y arcsin x = x + C . y\arcsin x=x+C. y arcsin x = x + C . y ∣ x = 1 2 = 0 , y|_{x=\frac{1}{2}}=0, y ∣ x = 2 1 = 0 , C = − 1 2 . C=-\frac{1}{2}. C = − 2 1 .
y arcsin x = x − 1 2 . y\arcsin x=x-\frac{1}{2}.
y arcsin x = x − 2 1 .
一阶线性（非齐次）微分方程
形如y ′ + p ( x ) y = q ( x ) y'+p(x)y=q(x) y ′ + p ( x ) y = q ( x )
这种形式有几种解法。其一，可以先求齐次方程y ′ + p ( x ) y = 0 y'+p(x)y=0 y ′ + p ( x ) y = 0 y = c e − ∫ p ( x ) d x . y=ce^{-\int p(x)\text{d}x}. y = c e − ∫ p ( x ) d x . y = c ( x ) e − ∫ p ( x ) d x y=c(x)e^{-\int p(x)\text{d}x} y = c ( x ) e − ∫ p ( x ) d x c ( x ) = ∫ q ( x ) e ∫ p ( x ) d x d x + C . c(x)=\int q(x)e^{\int p(x)\text{d}x}\text{d}x+C. c ( x ) = ∫ q ( x ) e ∫ p ( x ) d x d x + C . y y y
y = e − ∫ p ( x ) d x ( ∫ q ( x ) e ∫ p ( x ) d x d x + C ) . y=e^{-\int p(x)\text{d}x}\Big(\int q(x)e^{\int p(x)\text{d}x}\text{d}x+C\Big).
y = e − ∫ p ( x ) d x ( ∫ q ( x ) e ∫ p ( x ) d x d x + C ) .
但这个方法显然太计算困难了，所以又有其二。只要我们把y ′ + p ( x ) y y'+p(x)y y ′ + p ( x ) y e ∫ p ( x ) d x , e^{\int p(x)\text{d}x}, e ∫ p ( x ) d x ,
e.g.4.2 \textbf{e.g.4.2 } e.g.4.2 ( y + x 3 ) d x − 2 x d y = 0 (y+x^3)\text{d}x-2x\text{d}y=0 ( y + x 3 ) d x − 2 x d y = 0 y ∣ x = 1 = 6 5 y|_{x=1}=\frac{6}{5} y ∣ x = 1 = 5 6
Solution \textbf{Solution } Solution d y d x − 1 2 x ⋅ y = 1 2 x 2 . \frac{\text{d}y}{\text{d}x}-\frac{1}{2x}\cdot y=\frac{1}{2}x^2. d x d y − 2 x 1 ⋅ y = 2 1 x 2 . y ′ − 1 2 x y = 0 y'-\frac{1}{2x}y=0 y ′ − 2 x 1 y = 0 y = c x . y=c\sqrt{x}. y = c x .
设原微分方程的解为y = c ( x ) x ; y=c(x)\sqrt{x}; y = c ( x ) x ;
c ′ ( x ) x + c ( x ) ⋅ 1 2 x − 1 2 x ⋅ c ( x ) x = 1 2 x 2 ⇒ c ′ ( x ) x = 1 2 x 2 ⇒ c ( x ) = ∫ 1 2 x 3 2 d x ⇒ c ( x ) = 1 5 x 5 2 + C . \begin{align*}
c'(x)\sqrt{x}+c(x)\cdot\frac{1}{2\sqrt{x}}-\frac{1}{2x}\cdot c(x)\sqrt{x}=\frac{1}{2}x^2&\Rightarrow c'(x)\sqrt{x}=\frac{1}{2}x^2\\
&\Rightarrow c(x)=\int\frac{1}{2}x^\frac{3}{2}\text{d}x\\
&\Rightarrow c(x)=\frac{1}{5}x^\frac{5}{2}+C.
\end{align*}
c ′ ( x ) x + c ( x ) ⋅ 2 x 1 − 2 x 1 ⋅ c ( x ) x = 2 1 x 2 ⇒ c ′ ( x ) x = 2 1 x 2 ⇒ c ( x ) = ∫ 2 1 x 2 3 d x ⇒ c ( x ) = 5 1 x 2 5 + C .
因此y = x ⋅ ( 1 5 x 5 2 + C ) , y=\sqrt{x}\cdot\Big(\frac{1}{5}x^\frac{5}{2}+C\Big), y = x ⋅ ( 5 1 x 2 5 + C ) , y = 1 5 x 3 + C x . y=\frac{1}{5}x^3+C\sqrt{x}. y = 5 1 x 3 + C x . y ∣ x = 1 = 6 5 , y|_{x=1}=\frac{6}{5}, y ∣ x = 1 = 5 6 , C = 1. C=1. C = 1.
故所求特解为y = 1 5 x 3 + x . y=\frac{1}{5}x^3+\sqrt{x}. y = 5 1 x 3 + x .
有时求这种非线性微分方程不一定非要寻找y , y ′ ; y,y'; y , y ′ ; x , x ′ , x,x', x , x ′ ,
e.x.4.2 \textbf{e.x.4.2 } e.x.4.2 d y d x = y 2 x − y 2 \frac{\text{d}y}{\text{d}x}=\frac{y}{2x-y^2} d x d y = 2 x − y 2 y
Solution \textbf{Solution } Solution d x d y = 2 x − y 2 y , \frac{\text{d}x}{\text{d}y}=\frac{2x-y^2}{y}, d y d x = y 2 x − y 2 , d x d y − 2 y x = − y . \frac{\text{d}x}{\text{d}y}-\frac{2}{y}x=-y. d y d x − y 2 x = − y . x ′ − 2 y x = 0 x'-\frac{2}{y}x=0 x ′ − y 2 x = 0 x ′ x' x ′ x x x y y y x = c y 2 . x=cy^2. x = c y 2 .
设原微分方程解为x = c ( y ) y 2 . x=c(y)y^2. x = c ( y ) y 2 . c ′ ( y ) = − 1 y , c'(y)=-\frac{1}{y}, c ′ ( y ) = − y 1 , c ( y ) = − ln y + C . c(y)=-\ln y+C. c ( y ) = − ln y + C .
x = ( C − ln y ) y 2 . x=(C-\ln y)y^2.
x = ( C − ln y ) y 2 .
Bernoulli方程
形如y ′ + p ( x ) y = q ( x ) y n y'+p(x)y=q(x)y^n y ′ + p ( x ) y = q ( x ) y n
这个方程的两端同时除以y n y^n y n z = y 1 − n , z=y^{1-n}, z = y 1 − n , z z z
d z d x + ( 1 − n ) p ( x ) z = ( 1 − n ) q ( x ) \frac{\text{d}z}{\text{d}x}+(1-n)p(x)z=(1-n)q(x)
d x d z + ( 1 − n ) p ( x ) z = ( 1 − n ) q ( x )
然后按正常一阶线性微分方程求法解决即可。
全微分方程
形如P ( x , y ) d x + Q ( x , y ) d y = 0 , ∂ Q ∂ x = ∂ P ∂ y P(x,y)\text{d}x+Q(x,y)\text{d}y=0,\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y} P ( x , y ) d x + Q ( x , y ) d y = 0 , ∂ x ∂ Q = ∂ y ∂ P
∫ x 0 x P ( x , y 0 ) d x + ∫ y 0 y Q ( x , y ) d y = C \int_{x_0}^xP(x,y_0)\text{d}x+\int_{y_0}^yQ(x,y)\text{d}y=C
∫ x 0 x P ( x , y 0 ) d x + ∫ y 0 y Q ( x , y ) d y = C
或
∫ x 0 x P ( x , y ) d x + ∫ y 0 y Q ( x 0 , y ) d y = C . \int_{x_0}^xP(x,y)\text{d}x+\int_{y_0}^yQ(x_0,y)\text{d}y=C.
∫ x 0 x P ( x , y ) d x + ∫ y 0 y Q ( x 0 , y ) d y = C .
可降阶高阶微分方程
通过变量代换实现对高阶微分方程的降阶。一般考核的式型包括y ( n ) = f ( x ) , y ′ ′ = f ( x , y ′ ) , y ′ ′ = f ( y , y ′ ) y^{(n)}=f(x),y''=f(x,y'),y''=f(y,y') y ( n ) = f ( x ) , y ′′ = f ( x , y ′ ) , y ′′ = f ( y , y ′ ) y y y x x x
对于第一种，显然直接对等式两侧进行n n n
对于不显含y y y p = y ′ , p=y', p = y ′ , y ′ ′ = d p d x , y''=\frac{\text{d}p}{\text{d}x}, y ′′ = d x d p ,
e.g.4.3 \textbf{e.g.4.3 } e.g.4.3 ( 1 + x 2 ) y ′ ′ = 2 x y ′ (1+x^2)y''=2xy' ( 1 + x 2 ) y ′′ = 2 x y ′ y ∣ x = 0 = 1 , y ′ ∣ x = 0 = 3 y|_{x=0}=1,y'|_{x=0}=3 y ∣ x = 0 = 1 , y ′ ∣ x = 0 = 3
Solution \textbf{Solution } Solution p = y ′ , p=y', p = y ′ , y ′ ′ = d p d x . y''=\frac{\text{d}p}{\text{d}x}. y ′′ = d x d p .
( 1 + x 2 ) d p d x = 2 x p ⇒ 1 p d p = 2 x 1 + x 2 d x ⇒ ∫ 1 p d p = ∫ 2 x 1 + x 2 d x ⇒ ln p = ln ( 1 + x 2 ) + ln C 1 ⇒ y ′ = C 1 ( 1 + x 2 ) ⇒ y = ∫ C 1 ( 1 + x 2 ) d x ⇒ y = C 1 x + C 1 3 x 3 + C 2 . \begin{align*}
(1+x^2)\frac{\text{d}p}{\text{d}x}=2xp&\Rightarrow\frac{1}{p}\text{d}p=\frac{2x}{1+x^2}\text{d}x\\
&\Rightarrow\int\frac{1}{p}\text{d}p=\int\frac{2x}{1+x^2}\text{d}x\\
&\Rightarrow\ln p=\ln(1+x^2)+\ln C_1\Rightarrow y'=C_1(1+x^2)\\
&\Rightarrow y=\int C_1(1+x^2)\text{d}x\Rightarrow y=C_1x+\frac{C_1}{3}x^3+C_2.
\end{align*}
( 1 + x 2 ) d x d p = 2 x p ⇒ p 1 d p = 1 + x 2 2 x d x ⇒ ∫ p 1 d p = ∫ 1 + x 2 2 x d x ⇒ ln p = ln ( 1 + x 2 ) + ln C 1 ⇒ y ′ = C 1 ( 1 + x 2 ) ⇒ y = ∫ C 1 ( 1 + x 2 ) d x ⇒ y = C 1 x + 3 C 1 x 3 + C 2 .
代入初始条件可得{ C 1 = 3 C 2 = 1 . \begin{cases}C_1&=3\\C_2&=1\end{cases}. { C 1 C 2 = 3 = 1 .
y = x 3 + 3 x + 1. y=x^3+3x+1.
y = x 3 + 3 x + 1.
对于不显含x x x p = y ′ , p=y', p = y ′ , y ′ ′ = d p d x = d p d y ⋅ d y d x = p d p d y , y''=\frac{\text{d}p}{\text{d}x}=\frac{\text{d}p}{\text{d}y}\cdot\frac{\text{d}y}{\text{d}x}=p\frac{\text{d}p}{\text{d}y}, y ′′ = d x d p = d y d p ⋅ d x d y = p d y d p ,
e.x.4.3 \textbf{e.x.4.3 } e.x.4.3 y y ′ ′ + ( y ′ ) 2 = 0 yy''+(y')^2=0 y y ′′ + ( y ′ ) 2 = 0 y ∣ x = 0 = 1 , y ′ ∣ x = 0 = 1 2 y|_{x=0}=1,y'|_{x=0}=\frac{1}{2} y ∣ x = 0 = 1 , y ′ ∣ x = 0 = 2 1
Solution \textbf{Solution } Solution p = y ′ , p=y', p = y ′ , y ′ ′ = d p d x = d p d y ⋅ d y d x = p d p d y . y''=\frac{\text{d}p}{\text{d}x}=\frac{\text{d}p}{\text{d}y}\cdot\frac{\text{d}y}{\text{d}x}=p\frac{\text{d}p}{\text{d}y}. y ′′ = d x d p = d y d p ⋅ d x d y = p d y d p .
y p d p d y + p 2 = 0 ⇒ y d p d y + p = 0 ⇒ y d p d y = − p ⇒ y d y = − p d p ⇒ d y y = − d p p ⇒ ∫ 1 y d y = − ∫ 1 p d p ⇒ ln y = − ln p + ln C 1 ⇒ y = C 1 y ′ . \begin{align*}
yp\frac{\text{d}p}{\text{d}y}+p^2=0&\Rightarrow y\frac{\text{d}p}{\text{d}y}+p=0\\
&\Rightarrow y\frac{\text{d}p}{\text{d}y}=-p\Rightarrow \frac{y}{\text{d}y}=-\frac{p}{\text{d}p}\\
&\Rightarrow \frac{\text{d}y}{y}=-\frac{\text{d}p}{p}\Rightarrow\int\frac{1}{y}\text{d}y=-\int\frac{1}{p}\text{d}p\\
&\Rightarrow \ln y=-\ln p+\ln C_1\Rightarrow y=\frac{C_1}{y'}.
\end{align*}
y p d y d p + p 2 = 0 ⇒ y d y d p + p = 0 ⇒ y d y d p = − p ⇒ d y y = − d p p ⇒ y d y = − p d p ⇒ ∫ y 1 d y = − ∫ p 1 d p ⇒ ln y = − ln p + ln C 1 ⇒ y = y ′ C 1 .
代入初始条件可得C 1 = y ∣ x = 0 ⋅ y ′ ∣ x = 0 = 1 2 ; C_1=y|_{x=0}\cdot y'|_{x=0}=\frac{1}{2}; C 1 = y ∣ x = 0 ⋅ y ′ ∣ x = 0 = 2 1 ; y = 1 2 y ′ y=\frac{1}{2y'} y = 2 y ′ 1
y = 1 2 ⋅ d x d y ⇒ 1 2 d x = y d y ⇒ ∫ 1 2 d x = ∫ y d y ⇒ 1 2 x = 1 2 y 2 + C 2 . \begin{align*}
y=\frac{1}{2}\cdot\frac{\text{d}x}{\text{d}y}&\Rightarrow\frac{1}{2}\text{d}x=y\text{d}y
&\Rightarrow\int\frac{1}{2}\text{d}x=\int y\text{d}y\Rightarrow \frac{1}{2}x=\frac{1}{2}y^2+C_2.
\end{align*}
y = 2 1 ⋅ d y d x ⇒ 2 1 d x = y d y ⇒ ∫ 2 1 d x = ∫ y d y ⇒ 2 1 x = 2 1 y 2 + C 2 .
代入初始条件可得C 2 = − 1 2 . C_2=-\frac{1}{2}. C 2 = − 2 1 . 1 2 x = 1 2 y 2 − 1 2 , \frac{1}{2}x=\frac{1}{2}y^2-\frac{1}{2}, 2 1 x = 2 1 y 2 − 2 1 ,
x = y 2 − 1. x=y^2-1.
x = y 2 − 1.
线性微分方程解的性质与结构
和线性代数中「线性方程组阶的性质与结构」类似。
对于方程
y ′ ′ + p ( x ) y ′ + q ( x ) y = 0 , y''+p(x)y'+q(x)y=0,
y ′′ + p ( x ) y ′ + q ( x ) y = 0 ,
若y 1 ( x ) , y 2 ( x ) y_1(x),y_2(x) y 1 ( x ) , y 2 ( x ) y = c 1 y 1 ( x ) + c 2 y 2 ( x ) y=c_1y_1(x)+c_2y_2(x) y = c 1 y 1 ( x ) + c 2 y 2 ( x ) y 1 ( x ) , y 2 ( x ) y_1(x),y_2(x) y 1 ( x ) , y 2 ( x ) y = c 1 y 1 ( x ) + c 2 y 2 ( x ) y=c_1y_1(x)+c_2y_2(x) y = c 1 y 1 ( x ) + c 2 y 2 ( x )
对于非齐次线性方程
y ′ ′ + p ( x ) y ′ + q ( x ) y = f ( x ) , y''+p(x)y'+q(x)y=f(x),
y ′′ + p ( x ) y ′ + q ( x ) y = f ( x ) ,
其通解结构会是y = Y + y ∗ ; y=Y+y^*; y = Y + y ∗ ; Y Y Y y ∗ y^* y ∗
若y ′ ′ + p ( x ) y ′ + q ( x ) y = f 1 ( x ) y''+p(x)y'+q(x)y=f_1(x) y ′′ + p ( x ) y ′ + q ( x ) y = f 1 ( x ) y 1 ( x ) , y ′ ′ + p ( x ) y ′ + q ( x ) y = f 2 ( x ) y_1(x),y''+p(x)y'+q(x)y=f_2(x) y 1 ( x ) , y ′′ + p ( x ) y ′ + q ( x ) y = f 2 ( x ) y 2 ( x ) , y_2(x), y 2 ( x ) , α y 1 ( x ) + β y 2 ( x ) \alpha y_1(x)+\beta y_2(x) α y 1 ( x ) + β y 2 ( x ) y ′ ′ + p ( x ) y ′ + q ( x ) y = α f 1 ( x ) + β f 2 ( x ) y''+p(x)y'+q(x)y=\alpha f_1(x)+\beta f_2(x) y ′′ + p ( x ) y ′ + q ( x ) y = α f 1 ( x ) + β f 2 ( x )
非齐次线性方程y ′ ′ + p ( x ) y ′ + q ( x ) y = f ( x ) y''+p(x)y'+q(x)y=f(x) y ′′ + p ( x ) y ′ + q ( x ) y = f ( x ) y ′ ′ + p ( x ) y ′ + q ( x ) y = 0 y''+p(x)y'+q(x)y=0 y ′′ + p ( x ) y ′ + q ( x ) y = 0
e.g.4.4 \textbf{e.g.4.4 } e.g.4.4 y 1 , y 2 y_1,y_2 y 1 , y 2 y ′ + p ( x ) y = q ( x ) y'+p(x)y=q(x) y ′ + p ( x ) y = q ( x ) λ , μ \lambda,\mu λ , μ λ y 1 + μ y 2 \lambda y_1+\mu y_2 λ y 1 + μ y 2 λ y 1 − μ y 2 \lambda y_1-\mu y_2 λ y 1 − μ y 2 λ , μ \lambda,\mu λ , μ
Solution \textbf{Solution } Solution { λ + μ = 1 λ = μ . \begin{cases}\lambda+\mu=1\\\lambda=\mu\end{cases}. { λ + μ = 1 λ = μ .
{ λ = 1 2 μ = 1 2 . \begin{cases}
\lambda&=\frac{1}{2}\\
\mu&=\frac{1}{2}
\end{cases}.
{ λ μ = 2 1 = 2 1 .
e.x.4.4 \textbf{e.x.4.4 } e.x.4.4 y = ( 1 + x 2 ) 2 − 1 + x 2 , y = ( 1 + x 2 ) 2 + 1 + x 2 y=(1+x^2)^2-\sqrt{1+x^2},y=(1+x^2)^2+\sqrt{1+x^2} y = ( 1 + x 2 ) 2 − 1 + x 2 , y = ( 1 + x 2 ) 2 + 1 + x 2 y ′ + p ( x ) y = q ( x ) y'+p(x)y=q(x) y ′ + p ( x ) y = q ( x ) q ( x ) . q(x). q ( x ) .
Solution \textbf{Solution } Solution ( 1 + x 2 ) 2 − 1 + x 2 = y 1 , ( 1 + x 2 ) 2 + 1 + x 2 = y 2 . (1+x^2)^2-\sqrt{1+x^2}=y_1,(1+x^2)^2+\sqrt{1+x^2}=y_2. ( 1 + x 2 ) 2 − 1 + x 2 = y 1 , ( 1 + x 2 ) 2 + 1 + x 2 = y 2 . y = y 1 + y 2 2 = ( 1 + x 2 ) 2 y=\frac{y_1+y_2}{2}=(1+x^2)^2 y = 2 y 1 + y 2 = ( 1 + x 2 ) 2 y ′ + p ( x ) y = 0 y'+p(x)y=0 y ′ + p ( x ) y = 0 y = y 2 − y 1 = 2 1 + x 2 , y=y_2-y_1=2\sqrt{1+x^2}, y = y 2 − y 1 = 2 1 + x 2 ,
{ ( ( 1 + x 2 ) 2 ) ′ + p ( x ) ( 1 + x 2 ) 2 = q ( x ) ( 2 1 + x 2 ) ′ + p ( x ) ( 2 1 + x 2 ) = 0 ⇒ { p ( x ) = − x 1 + x 2 q ( x ) = 3 x ( 1 + x 2 ) . \begin{cases}
\big((1+x^2)^2\big)'+p(x)(1+x^2)^2&=q(x)\\
\big(2\sqrt{1+x^2}\big)'+p(x)(2\sqrt{1+x^2})&=0
\end{cases}\Rightarrow\begin{cases}
p(x)&=-\frac{x}{1+x^2}\\
q(x)&=3x(1+x^2)
\end{cases}.
{ ( ( 1 + x 2 ) 2 ) ′ + p ( x ) ( 1 + x 2 ) 2 ( 2 1 + x 2 ) ′ + p ( x ) ( 2 1 + x 2 ) = q ( x ) = 0 ⇒ { p ( x ) q ( x ) = − 1 + x 2 x = 3 x ( 1 + x 2 ) .
所以q ( x ) = 3 x ( 1 + x 2 ) q(x)=3x(1+x^2) q ( x ) = 3 x ( 1 + x 2 )
常系数线性微分方程
齐次线性微分方程通解
形如y ′ ′ + p y ′ + q y = 0 y''+py'+qy=0 y ′′ + p y ′ + q y = 0
这个方程对应有特征方程λ 2 + p λ + q = 0 , \lambda^2+p\lambda+q=0, λ 2 + p λ + q = 0 ,
对于λ 1 , λ 2 \lambda_1,\lambda_2 λ 1 , λ 2 y = c 1 e λ 1 x + c 2 e λ 2 x . y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x}. y = c 1 e λ 1 x + c 2 e λ 2 x .
对于λ 1 = λ 2 = λ \lambda_1=\lambda_2=\lambda λ 1 = λ 2 = λ y = c 1 e λ x + c 2 x e λ x . y=c_1e^{\lambda x}+c_2xe^{\lambda x}. y = c 1 e λ x + c 2 x e λ x .
对于无实根的情况依然要求出这个虚根λ ± ω i \lambda\pm\omega i λ ± ωi y = e λ x ( c 1 cos ω x + c 2 sin ω x ) . y=e^{\lambda x}(c_1\cos\omega x+c_2\sin\omega x). y = e λ x ( c 1 cos ω x + c 2 sin ω x ) .
n n n
e.g.4.5 \textbf{e.g.4.5 } e.g.4.5 y ( 4 ) − 2 y ( 3 ) + 5 y ′ ′ = 0 y^{(4)}-2y^{(3)}+5y''=0 y ( 4 ) − 2 y ( 3 ) + 5 y ′′ = 0
Solution \textbf{Solution } Solution λ 4 − 2 λ 3 + 5 λ 2 = 0 \lambda^4-2\lambda^3+5\lambda^2=0 λ 4 − 2 λ 3 + 5 λ 2 = 0 λ 1 = λ 2 = 0 , λ 3 = 1 + 2 i , λ 4 = 1 − 2 i . \lambda_1=\lambda_2=0,\lambda_3=1+2i,\lambda_4=1-2i. λ 1 = λ 2 = 0 , λ 3 = 1 + 2 i , λ 4 = 1 − 2 i . y = c 1 e 0 x + c 2 x e 0 x + e x ( c 3 cos 2 x + c 4 sin 2 x ) , y=c_1e^{0x}+c_2xe^{0x}+e^x(c_3\cos 2x+c_4\sin 2x), y = c 1 e 0 x + c 2 x e 0 x + e x ( c 3 cos 2 x + c 4 sin 2 x ) ,
y = c 1 + c 2 x + e x ( c 3 cos 2 x + c 4 sin 2 x ) . y=c_1+c_2x+e^x(c_3\cos 2x+c_4\sin 2x).
y = c 1 + c 2 x + e x ( c 3 cos 2 x + c 4 sin 2 x ) .
非齐次线性微分方程通解
形如y ′ ′ + p y ′ + q y = f ( x ) y''+py'+qy=f(x) y ′′ + p y ′ + q y = f ( x )
先求对应齐次线性微分方程的通解Y ; Y; Y ; y ∗ . y^*. y ∗ .
这个「待定系数法」的构造方式是根据f ( x ) f(x) f ( x ) f ( x ) = Q ( x ) e λ x , f(x)=Q(x)e^{\lambda x}, f ( x ) = Q ( x ) e λ x , y ∗ = x k P ( x ) e λ x y^*=x^kP(x)e^{\lambda x} y ∗ = x k P ( x ) e λ x P ( x ) P(x) P ( x ) Q ( x ) Q(x) Q ( x ) λ \lambda λ k k k
e.x.4.5 \textbf{e.x.4.5 } e.x.4.5 y ′ ′ − 5 y ′ + 6 y = x e 2 x y''-5y'+6y=xe^{2x} y ′′ − 5 y ′ + 6 y = x e 2 x
Solution \textbf{Solution } Solution y ′ ′ − 5 y ′ + 6 y = 0 y''-5y'+6y=0 y ′′ − 5 y ′ + 6 y = 0 Y . Y. Y . λ 2 − 5 λ + 6 = 0 \lambda^2-5\lambda+6=0 λ 2 − 5 λ + 6 = 0 λ 1 = 2 , λ 2 = 3 , \lambda_1=2,\lambda_2=3, λ 1 = 2 , λ 2 = 3 ,
Y = c 1 e 2 x + c 2 e 3 x . Y=c_1e^{2x}+c_2e^{3x}.
Y = c 1 e 2 x + c 2 e 3 x .
现在考虑这个线性非齐次微分方程的特解y ∗ . y^*. y ∗ . e e e 2 x 2x 2 x 2 = λ 1 2=\lambda_1 2 = λ 1 y ∗ = x ( a x + b ) e 2 x , y^*=x(ax+b)e^{2x}, y ∗ = x ( a x + b ) e 2 x , ( y ∗ ) ′ ′ − 5 ( y ∗ ) ′ + 6 y ∗ = x e 2 x . (y^*)''-5(y^*)'+6y^*=xe^{2x}. ( y ∗ ) ′′ − 5 ( y ∗ ) ′ + 6 y ∗ = x e 2 x .
y ∗ = ( a x 2 + b x ) e 2 x ⇒ ( y ∗ ) ′ = e 2 x ( 2 a x + b + 2 a x 2 + 2 b x ) ⇒ ( y ∗ ) ′ ′ = e 2 x ( 4 a x + 2 a + 2 b + 4 a x 2 + 4 a x + 2 b + 4 b x ) , \begin{align*}
y^*=(ax^2+bx)e^{2x}&\Rightarrow (y^*)'=e^{2x}(2ax+b+2ax^2+2bx)\\
&\Rightarrow (y^*)''=e^{2x}(4ax+2a+2b+4ax^2+4ax+2b+4bx),
\end{align*}
y ∗ = ( a x 2 + b x ) e 2 x ⇒ ( y ∗ ) ′ = e 2 x ( 2 a x + b + 2 a x 2 + 2 b x ) ⇒ ( y ∗ ) ′′ = e 2 x ( 4 a x + 2 a + 2 b + 4 a x 2 + 4 a x + 2 b + 4 b x ) ,
因此
( 4 a x 2 + ( 8 a + 4 b ) x + 2 a + 4 b ) − 5 ( 2 a x 2 + ( 2 a + 2 b ) x + b ) + 6 ( a x 2 + b x ) = x ⇒ ( 4 a − 10 a + 6 a ) x 2 + ( 8 a + 4 b − 10 a − 10 b + 6 b ) x + 2 a + 4 b − 5 b = x ⇒ − 2 a x + 2 a − b = x . \begin{align*}
&\big(4ax^2+(8a+4b)x+2a+4b\big)-5\big(2ax^2+(2a+2b)x+b\big)+6(ax^2+bx)=x\\
\Rightarrow&(4a-10a+6a)x^2+(8a+4b-10a-10b+6b)x+2a+4b-5b=x\\
\Rightarrow&-2ax+2a-b=x.
\end{align*}
⇒ ⇒ ( 4 a x 2 + ( 8 a + 4 b ) x + 2 a + 4 b ) − 5 ( 2 a x 2 + ( 2 a + 2 b ) x + b ) + 6 ( a x 2 + b x ) = x ( 4 a − 10 a + 6 a ) x 2 + ( 8 a + 4 b − 10 a − 10 b + 6 b ) x + 2 a + 4 b − 5 b = x − 2 a x + 2 a − b = x .
所以
{ − 2 a = 1 2 a − b = 0 ⇒ { a = − 1 2 b = − 1 . \begin{cases}
-2a&=1\\
2a-b&=0
\end{cases}\Rightarrow\begin{cases}
a&=-\frac{1}{2}\\
b&=-1
\end{cases}.
{ − 2 a 2 a − b = 1 = 0 ⇒ { a b = − 2 1 = − 1 .
故y ∗ = ( − 1 2 x 2 − x ) e 2 x , y^*=\big(-\frac{1}{2}x^2-x\big)e^{2x}, y ∗ = ( − 2 1 x 2 − x ) e 2 x ,
y = Y + y ∗ = ( c 1 − 1 2 x 2 − x ) e 2 x + c 2 e 3 x . y=Y+y^*=\Big(c_1-\frac{1}{2}x^2-x\Big)e^{2x}+c_2e^{3x}.
y = Y + y ∗ = ( c 1 − 2 1 x 2 − x ) e 2 x + c 2 e 3 x .
如果f ( x ) = e λ x ( p l ( x ) cos ω x + p n ( x ) sin ω x ) , f(x)=e^{\lambda x}\big(p_l(x)\cos\omega x+p_n(x)\sin\omega x\big), f ( x ) = e λ x ( p l ( x ) cos ω x + p n ( x ) sin ω x ) , y ∗ = x k e λ x ( R max ( n , l ) ( x ) cos ω x + R max ( n , l ) sin ω x ) , y^*=x^ke^{\lambda x}\big(R_{\max(n,l)}(x)\cos\omega x+R_{\max(n,l)}\sin\omega x\big), y ∗ = x k e λ x ( R m a x ( n , l ) ( x ) cos ω x + R m a x ( n , l ) sin ω x ) , P h ( x ) , R j ( x ) P_h(x),R_j(x) P h ( x ) , R j ( x ) x x x h h h j j j λ ± i ω \lambda\pm i\omega λ ± iω k k k
求特解的过程本身有一种更快的方法微分算子法 ；这个方法只要记D = d d x , 1 D D=\frac{\text{d}}{\text{d}x},\frac{1}{D} D = d x d , D 1 D 2 y + p D y + q y = f ( x ) , D^2y+pDy+qy=f(x), D 2 y + p Dy + q y = f ( x ) , y ∗ = 1 D 2 + p D + q ⋅ f ( x ) = 1 L ( D ) ⋅ f ( x ) . y^*=\frac{1}{D^2+pD+q}\cdot f(x)=\frac{1}{L(D)}\cdot f(x). y ∗ = D 2 + p D + q 1 ⋅ f ( x ) = L ( D ) 1 ⋅ f ( x ) . D D D
对于f ( x ) = e a x , f(x)=e^{ax}, f ( x ) = e a x , L ( a ) ≠ 0 L(a)≠0 L ( a ) = 0 a a a D D D L L L D D D D = a D=a D = a
e.g.4.6 \textbf{e.g.4.6 } e.g.4.6 y ′ ′ − 3 y ′ + 2 y = e 3 x y''-3y'+2y=e^{3x} y ′′ − 3 y ′ + 2 y = e 3 x
Solution \textbf{Solution } Solution D = d d x . D=\frac{\text{d}}{\text{d}x}. D = d x d . y ∗ = 1 D 2 − 3 D + 2 ⋅ e 3 x . y^*=\frac{1}{D^2-3D+2}\cdot e^{3x}. y ∗ = D 2 − 3 D + 2 1 ⋅ e 3 x . D = 3 , D=3, D = 3 , y ∗ = 1 2 e 3 x y^*=\frac{1}{2}e^{3x} y ∗ = 2 1 e 3 x
e.x.4.6 \textbf{e.x.4.6 } e.x.4.6 y ′ ′ ′ − 3 y ′ ′ + 3 y ′ − y = e x y'''-3y''+3y'-y=e^x y ′′′ − 3 y ′′ + 3 y ′ − y = e x
Solution \textbf{Solution } Solution D = d d x . D=\frac{\text{d}}{\text{d}x}. D = d x d .
y ∗ = 1 D 3 − 3 D 2 + 3 D − 1 ⋅ e x = 1 ( D − 1 ) 3 ⋅ e x = L ( 0 ) = 0 1 3 ( D − 1 ) 2 ⋅ x e x = L ′ ( 0 ) = 0 1 6 ( D − 1 ) ⋅ x 2 e x = L ′ ′ ( 0 ) = 0 1 6 ⋅ x 3 e x . \begin{align*}
y^*&=\frac{1}{D^3-3D^2+3D-1}\cdot e^x=\frac{1}{(D-1)^3}\cdot e^x\\
&\xlongequal{L(0)=0}\frac{1}{3(D-1)^2}\cdot xe^x\\
&\xlongequal{L'(0)=0}\frac{1}{6(D-1)}\cdot x^2e^x\\
&\xlongequal{L''(0)=0}\frac{1}{6}\cdot x^3e^x.
\end{align*}
y ∗ = D 3 − 3 D 2 + 3 D − 1 1 ⋅ e x = ( D − 1 ) 3 1 ⋅ e x L ( 0 ) = 0 3 ( D − 1 ) 2 1 ⋅ x e x L ′ ( 0 ) = 0 6 ( D − 1 ) 1 ⋅ x 2 e x L ′′ ( 0 ) = 0 6 1 ⋅ x 3 e x .
故所求为y ∗ = x 3 e x 6 . y^*=\frac{x^3e^x}{6}. y ∗ = 6 x 3 e x .
对于f ( x ) = cos α x f(x)=\cos\alpha x f ( x ) = cos αx f ( x ) = sin α x f(x)=\sin\alpha x f ( x ) = sin αx p = 0 p=0 p = 0 y ∗ = 1 D 2 + q f ( x ) . y^*=\frac{1}{D^2+q}f(x). y ∗ = D 2 + q 1 f ( x ) . D 2 = − α 2 D^2=-\alpha^2 D 2 = − α 2
e.g.4.7 \textbf{e.g.4.7 } e.g.4.7 y ′ ′ + y = 3 cos 2 x y''+y=3\cos 2x y ′′ + y = 3 cos 2 x
Solution \textbf{Solution } Solution D = d d x . D=\frac{\text{d}}{\text{d}x}. D = d x d . y ∗ = 1 D 2 + 1 ⋅ 3 cos 2 x . y^*=\frac{1}{D^2+1}\cdot 3\cos 2x. y ∗ = D 2 + 1 1 ⋅ 3 cos 2 x . D 2 = − 2 2 = − 4 , D^2=-2^2=-4, D 2 = − 2 2 = − 4 , y ∗ = − cos 2 x y^*=-\cos 2x y ∗ = − cos 2 x
e.x.4.7 \textbf{e.x.4.7 } e.x.4.7 y ′ ′ + y = cos x y''+y=\cos x y ′′ + y = cos x
Solution \textbf{Solution } Solution D = d d x . D=\frac{\text{d}}{\text{d}x}. D = d x d . y ∗ = 1 D 2 + 1 cos x . y^*=\frac{1}{D^2+1}\cos x. y ∗ = D 2 + 1 1 cos x . D 2 = − 1 2 = − 1 , D^2=-1^2=-1, D 2 = − 1 2 = − 1 ,
y ∗ = x ⋅ cos x 2 D = 1 D = ∫ d x x 2 ∫ cos x d x = x 2 sin x . \begin{align*}
y^*&=x\cdot\frac{\cos x}{2D}\\
&\xlongequal{\frac{1}{D}=\int \text{d}x}\frac{x}{2}\int\cos x\text{d}x\\
&=\frac{x}{2}\sin x.
\end{align*}
y ∗ = x ⋅ 2 D cos x D 1 = ∫ d x 2 x ∫ cos x d x = 2 x sin x .
所以所求为y ∗ = x sin x 2 . y^*=\frac{x\sin x}{2}. y ∗ = 2 x s i n x .
对于f ( x ) = e a x p m ( x ) , f(x)=e^{ax}p_m(x), f ( x ) = e a x p m ( x ) , y ∗ = 1 L ( D ) f ( x ) = 1 L ( D + a ) f ( x ) . y^*=\frac{1}{L(D)}f(x)=\frac{1}{L(D+a)}f(x). y ∗ = L ( D ) 1 f ( x ) = L ( D + a ) 1 f ( x ) .
e.g.4.8 \textbf{e.g.4.8 } e.g.4.8 y ′ ′ + 2 y ′ + 2 y = x 2 e − x y''+2y'+2y=x^2e^{-x} y ′′ + 2 y ′ + 2 y = x 2 e − x
Solution \textbf{Solution } Solution D = d d x . D=\frac{\text{d}}{\text{d}x}. D = d x d . y ∗ = 1 D 2 + 2 D + 2 x 2 e − x . y^*=\frac{1}{D^2+2D+2}x^2e^{-x}. y ∗ = D 2 + 2 D + 2 1 x 2 e − x . D D D D − 1 D-1 D − 1
y ∗ = e − x ⋅ 1 D 2 + 1 x 2 = 大除法 e − x ⋅ ( 1 − D 2 ) x 2 = e − x ( x 2 − D 2 x 2 ) = Def. of D e − x ( x 2 − ( x 2 ) ′ ′ ) = e − x ( x 2 − 2 ) . \begin{align*}
y^*&=e^{-x}\cdot\frac{1}{D^2+1}x^2\\
&\xlongequal{\text{大除法}}e^{-x}\cdot(1-D^2)x^2=e^{-x}(x^2-D^2x^2)\\
&\xlongequal{\text{Def. of }D}e^{-x}\big(x^2-(x^2)''\big)=e^{-x}(x^2-2).
\end{align*}
y ∗ = e − x ⋅ D 2 + 1 1 x 2 大除法 e − x ⋅ ( 1 − D 2 ) x 2 = e − x ( x 2 − D 2 x 2 ) Def. of D e − x ( x 2 − ( x 2 ) ′′ ) = e − x ( x 2 − 2 ) .
故所求为y ∗ = e − x ( x 2 − 2 ) . y^*=e^{-x}(x^2-2). y ∗ = e − x ( x 2 − 2 ) .
变量代换求微分方程
利用变量代换的思路求微分方程。
我们本就知道Euler方程x n y ( n ) + p 1 x n − 1 y ( n − 1 ) + ⋯ + p n − 1 x y ′ + p n y = f ( x ) x^ny^{(n)}+p_1x^{n-1}y^{(n-1)}+\cdots+p_{n-1}xy'+p_ny=f(x) x n y ( n ) + p 1 x n − 1 y ( n − 1 ) + ⋯ + p n − 1 x y ′ + p n y = f ( x ) p i p_i p i n = 2 n=2 n = 2 x 2 y ′ ′ + p 1 x y ′ + p 2 y = f ( x ) , x^2y''+p_1xy'+p_2y=f(x), x 2 y ′′ + p 1 x y ′ + p 2 y = f ( x ) , x = e t x=e^t x = e t
e.x.4.8 \textbf{e.x.4.8 } e.x.4.8 x 2 d 2 y d x 2 + 4 x d y d x + 2 y = 0 ( x > 0 ) x^2\frac{\text{d}^2y}{\text{d}x^2}+4x\frac{\text{d}y}{\text{d}x}+2y=0(x>0) x 2 d x 2 d 2 y + 4 x d x d y + 2 y = 0 ( x > 0 )
Solution \textbf{Solution } Solution x = e t . x=e^t. x = e t . d y d x = e − t d y d t , d 2 y d x 2 = d y ′ d t ⋅ d t d x = ( − e − t d y d t + e − t d 2 y d t 2 ) e − t . \frac{\text{d}y}{\text{d}x}=e^{-t}\frac{\text{d}y}{\text{d}t},\frac{\text{d}^2y}{\text{d}x^2}=\frac{\text{d}y'}{\text{d}t}\cdot\frac{\text{d}t}{\text{d}x}=(-e^{-t}\frac{\text{d}y}{\text{d}t}+e^{-t}\frac{\text{d}^2y}{\text{d}t^2})e^{-t}. d x d y = e − t d t d y , d x 2 d 2 y = d t d y ′ ⋅ d x d t = ( − e − t d t d y + e − t d t 2 d 2 y ) e − t .
e 2 t ( − e − 2 t d y d t + e − 2 t d 2 y d t 2 ) + 4 e t e − t d y d t + 2 y = 0. e^{2t}(-e^{-2t}\frac{\text{d}y}{\text{d}t}+e^{-2t}\frac{\text{d}^2y}{\text{d}t^2})+4e^te^{-t}\frac{\text{d}y}{\text{d}t}+2y=0.
e 2 t ( − e − 2 t d t d y + e − 2 t d t 2 d 2 y ) + 4 e t e − t d t d y + 2 y = 0.
因此d 2 y d t 2 + 3 d y d t + 2 y = 0. \frac{\text{d}^2y}{\text{d}t^2}+3\frac{\text{d}y}{\text{d}t}+2y=0. d t 2 d 2 y + 3 d t d y + 2 y = 0. y = c 1 e − t + c 2 e − 2 t , y=c_1e^{-t}+c_2e^{-2t}, y = c 1 e − t + c 2 e − 2 t , y = c 1 x + c 2 x 2 . y=\frac{c_1}{x}+\frac{c_2}{x^2}. y = x c 1 + x 2 c 2 .
微分方程应用
微分方程经常和定积分与偏导联合应用。有时，其还会与无穷级数联系。
e.g.4.9 \textbf{e.g.4.9 } e.g.4.9 ∫ 0 x x f ( x − t ) d t = x + ∫ 0 x f ( t ) d t \int_0^xxf(x-t)\text{d}t=x+\int_0^xf(t)\text{d}t ∫ 0 x x f ( x − t ) d t = x + ∫ 0 x f ( t ) d t f ( x ) . f(x). f ( x ) .
Solution \textbf{Solution } Solution u = x − t . u=x-t. u = x − t . = x ∫ x 0 − f ( u ) d u = x ∫ 0 x f ( u ) d u . =x\int_x^0-f(u)\text{d}u=x\int_0^xf(u)\text{d}u. = x ∫ x 0 − f ( u ) d u = x ∫ 0 x f ( u ) d u . x ∫ 0 x f ( u ) d u = x + ∫ 0 x f ( t ) d t ; x\int_0^xf(u)\text{d}u=x+\int_0^xf(t)\text{d}t; x ∫ 0 x f ( u ) d u = x + ∫ 0 x f ( t ) d t ;
∫ 0 x f ( u ) d u + x f ( x ) = 1 + f ( x ) . \int_0^xf(u)\text{d}u+xf(x)=1+f(x).
∫ 0 x f ( u ) d u + x f ( x ) = 1 + f ( x ) .
对这个等式取x = 0 , x=0, x = 0 , f ( 0 ) = − 1. f(0)=-1. f ( 0 ) = − 1.
2 f ( x ) + x f ′ ( x ) = f ′ ( x ) , 2f(x)+xf'(x)=f'(x),
2 f ( x ) + x f ′ ( x ) = f ′ ( x ) ,
也就是( x − 1 ) f ′ ( x ) + 2 f ( x ) = 0. (x-1)f'(x)+2f(x)=0. ( x − 1 ) f ′ ( x ) + 2 f ( x ) = 0. f ( x ) = C ( 1 − x ) 2 ; f(x)=\frac{C}{(1-x)^2}; f ( x ) = ( 1 − x ) 2 C ; f ( 0 ) = − 1 , f(0)=-1, f ( 0 ) = − 1 , C = − 1 , C=-1, C = − 1 ,
f ( x ) = − 1 ( 1 − x ) 2 . f(x)=-\frac{1}{(1-x)^2}.
f ( x ) = − ( 1 − x ) 2 1 .
e.x.4.9 \textbf{e.x.4.9 } e.x.4.9 f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) f ′ ( x ) = g ( x ) , g ′ ( x ) = 2 e x − f ( x ) , f ( 0 ) = 0 , g ( 0 ) = 2 , f'(x)=g(x),g'(x)=2e^x-f(x),f(0)=0,g(0)=2, f ′ ( x ) = g ( x ) , g ′ ( x ) = 2 e x − f ( x ) , f ( 0 ) = 0 , g ( 0 ) = 2 , ∫ 0 π g ( x ) 1 + x − f ( x ) ( 1 + x ) 2 d x . \int_0^\pi\frac{g(x)}{1+x}-\frac{f(x)}{(1+x)^2}\text{d}x. ∫ 0 π 1 + x g ( x ) − ( 1 + x ) 2 f ( x ) d x .
Solution \textbf{Solution } Solution
O.F. = ∫ 0 π f ′ ( x ) 1 + x − f ( x ) ( 1 + x ) 2 d x = ∫ 0 π ( 1 + x ) f ′ ( x ) − f ( x ) ( 1 + x ) 2 d x = ∫ 0 π ( f ( x ) 1 + x ) ′ d x = f ( x ) 1 + x ∣ 0 π , \begin{align*}
\text{O.F.}&=\int_0^\pi\frac{f'(x)}{1+x}-\frac{f(x)}{(1+x)^2}\text{d}x=\int_0^\pi\frac{(1+x)f'(x)-f(x)}{(1+x)^2}\text{d}x\\
&=\int_0^\pi\Big(\frac{f(x)}{1+x}\Big)'\text{d}x=\frac{f(x)}{1+x}\Big|_0^\pi,
\end{align*}
O.F. = ∫ 0 π 1 + x f ′ ( x ) − ( 1 + x ) 2 f ( x ) d x = ∫ 0 π ( 1 + x ) 2 ( 1 + x ) f ′ ( x ) − f ( x ) d x = ∫ 0 π ( 1 + x f ( x ) ) ′ d x = 1 + x f ( x ) 0 π ,
因此只要求f ( x ) , f(x), f ( x ) , g ′ ( x ) = 2 e x − f ( x ) ⇒ f ′ ′ ( x ) = 2 e x − f ( x ) , g'(x)=2e^x-f(x)\Rightarrow f''(x)=2e^x-f(x), g ′ ( x ) = 2 e x − f ( x ) ⇒ f ′′ ( x ) = 2 e x − f ( x ) ,
令y = f ( x ) . y=f(x). y = f ( x ) . y ′ ′ + y = 2 e x ; y''+y=2e^x; y ′′ + y = 2 e x ; y ′ ′ + y = 0 y''+y=0 y ′′ + y = 0 Y = c 1 sin x + c 2 cos x ; Y=c_1\sin x+c_2\cos x; Y = c 1 sin x + c 2 cos x ; y ∗ = e x , y^*=e^x, y ∗ = e x , f ( x ) = y = c 1 sin x + c 2 cos x + e x . f(x)=y=c_1\sin x+c_2\cos x+e^x. f ( x ) = y = c 1 sin x + c 2 cos x + e x . f ( 0 ) = 0 , f ′ ( 0 ) = g ( 0 ) = 2 f(0)=0,f'(0)=g(0)=2 f ( 0 ) = 0 , f ′ ( 0 ) = g ( 0 ) = 2
{ c 2 + 1 = 0 c 1 + 1 = 2 ⇒ { c 1 = 1 c 2 = − 1 \begin{cases}
c_2+1=0\\
c_1+1=2
\end{cases}\Rightarrow\begin{cases}
c_1&=1\\
c_2&=-1
\end{cases}
{ c 2 + 1 = 0 c 1 + 1 = 2 ⇒ { c 1 c 2 = 1 = − 1
因此f ( x ) = sin x − cos x + e x f(x)=\sin x-\cos x+e^x f ( x ) = sin x − cos x + e x
O.F. = sin x − cos x + e x 1 + x ∣ 0 π = e π + 1 1 + π − 0 = e π + 1 π + 1 . \begin{align*}
\text{O.F.}&=\frac{\sin x-\cos x+e^x}{1+x}\Big|_0^\pi\\
&=\frac{e^\pi+1}{1+\pi}-0=\frac{e^\pi+1}{\pi+1}.
\end{align*}
O.F. = 1 + x sin x − cos x + e x 0 π = 1 + π e π + 1 − 0 = π + 1 e π + 1 .
微分方程的另一个应用是讨论函数的周期和有界性相关问题。
多元微分
二元函数
显而易见，z = f ( x , y ) z=f(x,y) z = f ( x , y )
如果f ( x , y ) f(x,y) f ( x , y ) x 0 , y 0 x_0,y_0 x 0 , y 0 ∀ ε > 0 , ∃ δ > 0 , 0 < ( x − x 0 ) 2 + ( y − y 0 ) 2 < δ , s.t. ∣ f ( x , y ) − A ∣ < ε , \forall\varepsilon>0,\exists\delta>0,0<\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta,\text{s.t. }|f(x,y)-A|<\varepsilon, ∀ ε > 0 , ∃ δ > 0 , 0 < ( x − x 0 ) 2 + ( y − y 0 ) 2 < δ , s.t. ∣ f ( x , y ) − A ∣ < ε , lim ( x , y ) → ( x 0 , y 0 ) f ( x , y ) = A . \lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=A. lim ( x , y ) → ( x 0 , y 0 ) f ( x , y ) = A . lim x → x 0 lim y → y 0 f ( x , y ) \lim_{x\rightarrow x_0}\lim_{y\rightarrow y_0}f(x,y) lim x → x 0 lim y → y 0 f ( x , y ) 无关 。
二元函数的连续和有界性定义和一元函数类似。对于偏导数，只要将没有被求偏导的变量视为常量即可。
d z = δ f δ x d x + δ f δ y d y \text{d}z=\frac{\delta f}{\delta x}\text{d}x+\frac{\delta f}{\delta y}\text{d}y d z = δ x δ f d x + δy δ f d y z = f ( x , y ) z=f(x,y) z = f ( x , y ) z = f ( x , y ) z=f(x,y) z = f ( x , y ) ( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 )
lim ( Δ x , Δ y ) → ( 0 , 0 ) f ( x 0 + Δ x , y 0 + Δ y ) − f ( x 0 , y 0 ) − ∂ f ∂ x Δ x − ∂ f ∂ y Δ y ( Δ x ) 2 + ( Δ y ) 2 = 0 \lim_{(\Delta x,\Delta y)\rightarrow (0,0)}\frac{f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0)-\frac{\partial f}{\partial x}\Delta x-\frac{\partial f}{\partial y}\Delta y}{\sqrt{(\Delta x)^2+(\Delta y)^2}}=0
( Δ x , Δ y ) → ( 0 , 0 ) lim ( Δ x ) 2 + ( Δ y ) 2 f ( x 0 + Δ x , y 0 + Δ y ) − f ( x 0 , y 0 ) − ∂ x ∂ f Δ x − ∂ y ∂ f Δ y = 0
或者
lim ( x , y ) → ( x 0 , y 0 ) f ( x , y ) − f ( x 0 , y 0 ) − ∂ f ∂ x ( x − x 0 ) − ∂ f ∂ y ( y − y 0 ) ( x − x 0 ) 2 + ( y − y 0 ) 2 = 0. \lim_{(x,y)\rightarrow (x_0,y_0)}\frac{f(x,y)-f(x_0,y_0)-\frac{\partial f}{\partial x}(x-x_0)-\frac{\partial f}{\partial y}(y-y_0)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0.
( x , y ) → ( x 0 , y 0 ) lim ( x − x 0 ) 2 + ( y − y 0 ) 2 f ( x , y ) − f ( x 0 , y 0 ) − ∂ x ∂ f ( x − x 0 ) − ∂ y ∂ f ( y − y 0 ) = 0.
e.g.5.1 \textbf{e.g.5.1 } e.g.5.1 f ( x , y ) f(x,y) f ( x , y ) ( 0 , 0 ) (0,0) ( 0 , 0 )
A. lim ( x , y ) → ( 0 , 0 ) ( f ( x , y ) − f ( 0 , 0 ) ) = 0 \lim_{(x,y)\rightarrow(0,0)}\big(f(x,y)-f(0,0)\big)=0 lim ( x , y ) → ( 0 , 0 ) ( f ( x , y ) − f ( 0 , 0 ) ) = 0
B. lim x → 0 f ( x , 0 ) − f ( 0 , 0 ) x = 0 \lim_{x\rightarrow 0}\frac{f(x,0)-f(0,0)}{x}=0 lim x → 0 x f ( x , 0 ) − f ( 0 , 0 ) = 0 lim y → 0 f ( 0 , y ) − f ( 0 , 0 ) y = 0 \lim_{y\rightarrow 0}\frac{f(0,y)-f(0,0)}{y}=0 lim y → 0 y f ( 0 , y ) − f ( 0 , 0 ) = 0
C. lim ( x , y ) → ( 0 , 0 ) f ( x , y ) − f ( 0 , 0 ) x 2 + y 2 = 0 \lim_{(x,y)\rightarrow(0,0)}\frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=0 lim ( x , y ) → ( 0 , 0 ) x 2 + y 2 f ( x , y ) − f ( 0 , 0 ) = 0
D. lim x → 0 ( f x ′ ( x , 0 ) − f x ′ ( 0 , 0 ) ) = 0 \lim_{x\rightarrow 0}\big(f'_x(x,0)-f'_x(0,0)\big)=0 lim x → 0 ( f x ′ ( x , 0 ) − f x ′ ( 0 , 0 ) ) = 0 lim y → 0 ( f y ′ ( 0 , y ) − f ′ y ( 0 , 0 ) ) = 0 \lim_{y\rightarrow 0}\big(f'_y(0,y)-f'y(0,0)\big)=0 lim y → 0 ( f y ′ ( 0 , y ) − f ′ y ( 0 , 0 ) ) = 0
Choice \textbf{Choice } Choice
事实上，上方例题5.1中的C选项也可以理解成“式f ( x , y ) x 2 + y 2 \frac{f(x,y)}{\sqrt{x^2+y^2}} x 2 + y 2 f ( x , y )
e.x.5.1 \textbf{e.x.5.1 } e.x.5.1 ( 0 , 1 ) (0,1) ( 0 , 1 ) z = f ( x , y ) z=f(x,y) z = f ( x , y ) lim ( x , y ) → ( 0 , 1 ) f ( x , y ) − 2 x + y − 2 x 2 + ( y − 1 ) 2 = 0 , \lim_{(x,y)\rightarrow(0,1)}\frac{f(x,y)-2x+y-2}{\sqrt{x^2+(y-1)^2}}=0, lim ( x , y ) → ( 0 , 1 ) x 2 + ( y − 1 ) 2 f ( x , y ) − 2 x + y − 2 = 0 , d z ∣ ( 0 , 1 ) . \text{d}z\big|_{(0,1)}. d z ( 0 , 1 ) .
Solution \textbf{Solution } Solution f ( 0 , 1 ) − 2 × 0 + 1 − 2 = 0 , f(0,1)-2\times 0+1-2=0, f ( 0 , 1 ) − 2 × 0 + 1 − 2 = 0 , f ( 0 , 1 ) = 1. f(0,1)=1. f ( 0 , 1 ) = 1.
lim ( x , y ) → ( 0 , 1 ) ( f ( x , y ) − f ( 0 , 1 ) ) − 2 ( x − 0 ) − ( − 1 ( y − 1 ) ) x 2 + ( y − 1 ) 2 = 0. \lim_{(x,y)\rightarrow(0,1)}\frac{\big(f(x,y)-f(0,1)\big)-2(x-0)-\big(-1(y-1)\big)}{\sqrt{x^2+(y-1)^2}}=0.
( x , y ) → ( 0 , 1 ) lim x 2 + ( y − 1 ) 2 ( f ( x , y ) − f ( 0 , 1 ) ) − 2 ( x − 0 ) − ( − 1 ( y − 1 ) ) = 0.
因此根据全微分存在的条件公式也就知道∂ z ∂ x = 2 , ∂ z ∂ y = − 1. \frac{\partial z}{\partial x}=2,\frac{\partial z}{\partial y}=-1. ∂ x ∂ z = 2 , ∂ y ∂ z = − 1.
d z ∣ ( 0 , 1 ) = 2 d x − d y . \text{d}z\big|_{(0,1)}=2\text{d}x-\text{d}y.
d z ( 0 , 1 ) = 2 d x − d y .
显然，一直利用定义在先求偏导数之后代入点的坐标值得到全微分的方法计算量太大了。可以尝试直接利用f x ′ ( x 0 , y 0 ) = d d x f ( x , y 0 ) ∣ x = x 0 , f y ′ ( x 0 , y 0 ) = d d y f ( x 0 , y ) ∣ y = y 0 f'_x(x_0,y_0)=\frac{\text{d}}{\text{d}x}f(x,y_0)\big|_{x=x_0},f'_y(x_0,y_0)=\frac{\text{d}}{\text{d}y}f(x_0,y)\big|_{y=y_0} f x ′ ( x 0 , y 0 ) = d x d f ( x , y 0 ) x = x 0 , f y ′ ( x 0 , y 0 ) = d y d f ( x 0 , y ) y = y 0
e.g.5.2 \textbf{e.g.5.2 } e.g.5.2 f ( x , y ) = x 2 ( y − 3 ) + ( x − 1 ) arctan x y , f(x,y)=x^2(y-3)+(x-1)\arctan\sqrt{\frac{x}{y}}, f ( x , y ) = x 2 ( y − 3 ) + ( x − 1 ) arctan y x , f x ′ ( 1 , 3 ) , f y ′ ( 1 , 3 ) f'_x(1,3),f'_y(1,3) f x ′ ( 1 , 3 ) , f y ′ ( 1 , 3 ) f ′ ( 1 , 3 ) . f'(1,3). f ′ ( 1 , 3 ) .
Solution \textbf{Solution } Solution
f x ′ ( 1 , 3 ) = d f ( x , 3 ) d x ∣ x = 1 = ( ( x − 1 ) arctan x 3 ) ′ ∣ x = 1 = lim x → 1 ( x − 1 ) arctan x 3 − f ( 1 , 3 ) x − 1 = arctan 1 3 = π 6 , f y ′ ( 1 , 3 ) = d f ( 1 , y ) d y ∣ y = 3 = ( y − 3 ) ′ ∣ y = 3 = 1 , \begin{align*}
f'_x(1,3)&=\frac{\text{d}f(x,3)}{\text{d}x}\Big|_{x=1}\\
&=\Bigg((x-1)\arctan\sqrt{\frac{x}{3}}\Bigg)'\Bigg|_{x=1}=\lim_{x\rightarrow 1}\frac{(x-1)\arctan\sqrt{\frac{x}{3}}-f(1,3)}{x-1}\\
&=\arctan\sqrt{\frac{1}{3}}=\frac{\pi}{6},\\
f'_y(1,3)&=\frac{\text{d}f(1,y)}{\text{d}y}\Big|_{y=3}\\
&=(y-3)'\big|_{y=3}=1,\\
\end{align*}
f x ′ ( 1 , 3 ) f y ′ ( 1 , 3 ) = d x d f ( x , 3 ) x = 1 = ( ( x − 1 ) arctan 3 x ) ′ x = 1 = x → 1 lim x − 1 ( x − 1 ) arctan 3 x − f ( 1 , 3 ) = arctan 3 1 = 6 π , = d y d f ( 1 , y ) y = 3 = ( y − 3 ) ′ y = 3 = 1 ,
因此据全微分定义得f ′ ( 1 , 3 ) = π 6 d x + d y . f'(1,3)=\frac{\pi}{6}\text{d}x+\text{d}y. f ′ ( 1 , 3 ) = 6 π d x + d y .
在上方例5.2中，并没有先求偏导的函数表达式，而是先将求偏导时不使用的变量代入数值，使求偏导的过程简化为求一元函数导数，大大提升计算速度。
e.x.5.2 \textbf{e.x.5.2 } e.x.5.2 f ( x , y , z ) = x y z f(x,y,z)=\sqrt[z]{\frac{x}{y}} f ( x , y , z ) = z y x ( 1 , 1 , 1 ) (1,1,1) ( 1 , 1 , 1 ) d f ( 1 , 1 , 1 ) . \text{d}f(1,1,1). d f ( 1 , 1 , 1 ) .
Solution \textbf{Solution } Solution
f x ′ ( 1 , 1 , 1 ) = f ′ ( x , 1 , 1 ) ∣ x = 1 = x ′ ∣ x = 1 = 1 , f y ′ ( 1 , 1 , 1 ) = f ′ ( 1 , y , 1 ) ∣ y = 1 = ( 1 y ) ′ ∣ y = 1 = − 1 , f z ′ ( 1 , 1 , 1 ) = f ′ ( 1 , 1 , z ) ∣ z = 1 = ( 1 z ) ′ ∣ z = 1 = 0 , \begin{align*}
f'_x(1,1,1)&=f'(x,1,1)\big|_{x=1}=x'\big|_{x=1}=1,\\
f'_y(1,1,1)&=f'(1,y,1)\big|_{y=1}=\Big(\frac{1}{y}\Big)'\Big|_{y=1}=-1,\\
f'_z(1,1,1)&=f'(1,1,z)\big|_{z=1}=\big(\sqrt[z]{1}\big)'\big|_{z=1}=0,
\end{align*}
f x ′ ( 1 , 1 , 1 ) f y ′ ( 1 , 1 , 1 ) f z ′ ( 1 , 1 , 1 ) = f ′ ( x , 1 , 1 ) x = 1 = x ′ x = 1 = 1 , = f ′ ( 1 , y , 1 ) y = 1 = ( y 1 ) ′ y = 1 = − 1 , = f ′ ( 1 , 1 , z ) z = 1 = ( z 1 ) ′ z = 1 = 0 ,
因此所求为d f ( 1 , 1 , 1 ) = d x − d y . \text{d}f(1,1,1)=\text{d}x-\text{d}y. d f ( 1 , 1 , 1 ) = d x − d y .
对复合有函数u ( x , y ) u(x,y) u ( x , y ) v ( x , y ) v(x,y) v ( x , y ) z = f ( u , v ) , z=f(u,v), z = f ( u , v ) , { ∂ z ∂ x = ∂ z ∂ u ⋅ ∂ u ∂ x + ∂ z ∂ v ⋅ ∂ v ∂ x ∂ z ∂ y = ∂ z ∂ u ⋅ ∂ u ∂ y + ∂ z ∂ v ⋅ ∂ v ∂ y . \begin{cases}\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial x}\\\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial y}\end{cases}. { ∂ x ∂ z = ∂ u ∂ z ⋅ ∂ x ∂ u + ∂ v ∂ z ⋅ ∂ x ∂ v ∂ y ∂ z = ∂ u ∂ z ⋅ ∂ y ∂ u + ∂ v ∂ z ⋅ ∂ y ∂ v .
对函数u = f ( x , y , z ) u=f(x,y,z) u = f ( x , y , z ) z = z ( x , y ) z=z(x,y) z = z ( x , y ) { ∂ u ∂ x = f x ′ + f z ′ ⋅ ∂ z ∂ x ∂ u ∂ y = f y ′ + f z ′ ⋅ ∂ z ∂ y . \begin{cases}\frac{\partial u}{\partial x}=f'_x+f'_z\cdot\frac{\partial z}{\partial x}\\\frac{\partial u}{\partial y}=f'_y+f'_z\cdot\frac{\partial z}{\partial y}\end{cases}. { ∂ x ∂ u = f x ′ + f z ′ ⋅ ∂ x ∂ z ∂ y ∂ u = f y ′ + f z ′ ⋅ ∂ y ∂ z .
对函数u = f ( x , y , z ) u=f(x,y,z) u = f ( x , y , z ) y = y ( x ) , z = z ( x ) y=y(x),z=z(x) y = y ( x ) , z = z ( x ) d u d x = f x ′ + f y ′ ⋅ y ′ ( x ) + f z ′ ⋅ z ′ ( x ) . \frac{\text{d}u}{\text{d}x}=f'_x+f'_y\cdot y'(x)+f'_z\cdot z'(x). d x d u = f x ′ + f y ′ ⋅ y ′ ( x ) + f z ′ ⋅ z ′ ( x ) .
对复合有函数u ( x , y , z ) u(x,y,z) u ( x , y , z ) v = v ( x , y , z ) v=v(x,y,z) v = v ( x , y , z ) w = f ( u , v ) , w=f(u,v), w = f ( u , v ) , { ∂ w ∂ x = f u ′ ∂ u ∂ x + f v ′ ∂ v ∂ x ∂ w ∂ y = f u ′ ∂ u ∂ y + f v ′ ∂ v ∂ y ∂ w ∂ z = f u ′ ∂ u ∂ z + f v ′ ∂ v ∂ z . \begin{cases}\frac{\partial w}{\partial x}=f'_u\frac{\partial u}{\partial x}+f'_v\frac{\partial v}{\partial x}\\\frac{\partial w}{\partial y}=f'_u\frac{\partial u}{\partial y}+f'_v\frac{\partial v}{\partial y}\\\frac{\partial w}{\partial z}=f'_u\frac{\partial u}{\partial z}+f'_v\frac{\partial v}{\partial z}\end{cases}. ⎩ ⎨ ⎧ ∂ x ∂ w = f u ′ ∂ x ∂ u + f v ′ ∂ x ∂ v ∂ y ∂ w = f u ′ ∂ y ∂ u + f v ′ ∂ y ∂ v ∂ z ∂ w = f u ′ ∂ z ∂ u + f v ′ ∂ z ∂ v .
高阶偏导数本质是对一个偏导再求若干次偏导。
e.g.5.3 \textbf{e.g.5.3 } e.g.5.3 z = f ( e x sin y , x 2 + y 2 ) z=f(e^x\sin y,x^2+y^2) z = f ( e x sin y , x 2 + y 2 ) ∂ 2 z ∂ x ∂ y . \frac{\partial^2 z}{\partial x\partial y}. ∂ x ∂ y ∂ 2 z .
Solution \textbf{Solution } Solution u ( x , y ) = e x sin y , v ( x , y ) = x 2 + y 2 , u(x,y)=e^x\sin y, v(x,y)=x^2+y^2, u ( x , y ) = e x sin y , v ( x , y ) = x 2 + y 2 ,
∂ z ∂ x = ∂ z ∂ u ⋅ ∂ u ∂ x + ∂ z ∂ v ⋅ ∂ v ∂ x = f 1 ′ ⋅ ( e x sin y ) + f 2 ′ ⋅ ( 2 x ) \frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial x}=f'_1\cdot(e^x\sin y)+f'_2\cdot(2x)
∂ x ∂ z = ∂ u ∂ z ⋅ ∂ x ∂ u + ∂ v ∂ z ⋅ ∂ x ∂ v = f 1 ′ ⋅ ( e x sin y ) + f 2 ′ ⋅ ( 2 x )
因此
∂ 2 z ∂ x ∂ y = ∂ ∂ y ⋅ ∂ z ∂ x = ∂ ∂ y ( sin y f 1 ′ e x + 2 x f 2 ′ ) = ( cos y e x f 1 ′ + sin y e x ( f 11 ′ ′ e x cos y + f 12 ′ ′ 2 y ) ) + ( 2 x ( f 21 ′ ′ e x cos y + f 22 ′ ′ 2 y ) ) . \begin{align*}
\frac{\partial^2z}{\partial x\partial y}&=\frac{\partial}{\partial y}\cdot\frac{\partial z}{\partial x}=\frac{\partial}{\partial y}(\sin yf'_1e^x+2xf'_2)\\
&=\big(\cos ye^xf'_1+\sin ye^x(f''_{11}e^x\cos y+f''_{12}2y)\big)+\big(2x(f''_{21}e^x\cos y+f''_{22}2y)\big).\\
\end{align*}
∂ x ∂ y ∂ 2 z = ∂ y ∂ ⋅ ∂ x ∂ z = ∂ y ∂ ( sin y f 1 ′ e x + 2 x f 2 ′ ) = ( cos y e x f 1 ′ + sin y e x ( f 11 ′′ e x cos y + f 12 ′′ 2 y ) ) + ( 2 x ( f 21 ′′ e x cos y + f 22 ′′ 2 y ) ) .
e.x.5.3 \textbf{e.x.5.3 } e.x.5.3 z = f ( x , y ) z=f(x,y) z = f ( x , y ) f ( 1 , 1 ) = 1 , ∂ f ∂ x ∣ ( 1 , 1 ) = 2 , ∂ f ∂ y ∣ ( 1 , 1 ) = 3 , φ ( x ) = f ( x , f ( x , f ( x , x ) ) ) , f(1,1)=1,\frac{\partial f}{\partial x}\big|_{(1,1)}=2,\frac{\partial f}{\partial y}\big|_{(1,1)}=3,\varphi(x)=f\Big(x,f\big(x,f(x,x)\big)\Big), f ( 1 , 1 ) = 1 , ∂ x ∂ f ( 1 , 1 ) = 2 , ∂ y ∂ f ( 1 , 1 ) = 3 , φ ( x ) = f ( x , f ( x , f ( x , x ) ) ) , d d x φ 3 ( x ) ∣ x = 1 . \frac{\text{d}}{\text{d}x}\varphi^3(x)\big|_{x=1}. d x d φ 3 ( x ) x = 1 .
Solution \textbf{Solution } Solution
φ ′ ( x ) = f 1 ′ ⋅ 1 + f 2 ′ ⋅ ( f 1 ′ ⋅ 1 + f 2 ′ ⋅ ( f 1 ′ ⋅ 1 + f 2 ′ ⋅ 1 ) ) = f 1 ′ + f 2 ′ ( f 1 ′ + f 2 ′ ( f 1 ′ + f 2 ′ ) ) . \begin{align*}
\varphi'(x)&=f'_1\cdot 1+f'_2\cdot\big(f'_1\cdot 1+f'_2\cdot(f'_1\cdot 1+f'_2\cdot 1)\big)\\
&=f'_1+f'_2\big(f'_1+f'_2(f'_1+f'_2)\big).
\end{align*}
φ ′ ( x ) = f 1 ′ ⋅ 1 + f 2 ′ ⋅ ( f 1 ′ ⋅ 1 + f 2 ′ ⋅ ( f 1 ′ ⋅ 1 + f 2 ′ ⋅ 1 ) ) = f 1 ′ + f 2 ′ ( f 1 ′ + f 2 ′ ( f 1 ′ + f 2 ′ ) ) .
而x = 1 x=1 x = 1 f 1 ′ = 2 , f 2 ′ = 3 f'_1=2,f'_2=3 f 1 ′ = 2 , f 2 ′ = 3 φ ′ ( x ) ∣ x = 1 = 2 + 3 × ( 2 + 3 × ( 2 + 3 ) ) = 53. \varphi'(x)\big|{x=1}=2+3\times\big(2+3\times(2+3)\big)=53. φ ′ ( x ) x = 1 = 2 + 3 × ( 2 + 3 × ( 2 + 3 ) ) = 53. f ( 1 , 1 ) = 1 ⇒ φ ( 1 ) = 1 , f(1,1)=1\Rightarrow \varphi(1)=1, f ( 1 , 1 ) = 1 ⇒ φ ( 1 ) = 1 ,
O.F. = 3 φ 2 ( x ) φ ′ ( x ) = 3 × 1 2 × 53 = 159. \text{O.F.}=3\varphi^2(x)\varphi'(x)=3\times 1^2\times 53=159.
O.F. = 3 φ 2 ( x ) φ ′ ( x ) = 3 × 1 2 × 53 = 159.
有时偏导会与微分方程共同考核。考虑下面的练习5.4。
e.x.5.4 \textbf{e.x.5.4 } e.x.5.4 f ( u ) f(u) f ( u ) z = f ( e x cos y ) z=f(e^x\cos y) z = f ( e x cos y ) ∂ 2 z ∂ x 2 + ∂ 2 z ∂ y 2 = ( 4 z + e x cos y ) e 2 x , \frac{\partial^2z}{\partial x^2}+\frac{\partial^2z}{\partial y^2}=(4z+e^x\cos y)e^{2x}, ∂ x 2 ∂ 2 z + ∂ y 2 ∂ 2 z = ( 4 z + e x cos y ) e 2 x , f ( 0 ) = f ′ ( 0 ) = 0 , f(0)=f'(0)=0, f ( 0 ) = f ′ ( 0 ) = 0 , f ( u ) f(u) f ( u )
Solution \textbf{Solution } Solution ∂ z ∂ x = f ′ ( e x cos y ) e x cos y ⇒ ∂ 2 z ∂ x 2 = f ′ ′ ( e x cos y ) e 2 x cos 2 y + f ′ ( e x cos y ) e x cos y , ∂ z ∂ y = − f ′ ( e x cos y ) e x sin y ⇒ ∂ 2 z ∂ y 2 = f ′ ′ ( e x cos y ) e 2 x sin 2 y − f ′ ( e x cos y ) e x cos y . \frac{\partial z}{\partial x}=f'(e^x\cos y)e^x\cos y\Rightarrow\frac{\partial^2z}{\partial x^2}=f''(e^x\cos y)e^{2x}\cos^2y+f'(e^x\cos y)e^x\cos y,\frac{\partial z}{\partial y}=-f'(e^x\cos y)e^x\sin y\Rightarrow\frac{\partial^2z}{\partial y^2}=f''(e^x\cos y)e^{2x}\sin^2y-f'(e^x\cos y)e^x\cos y. ∂ x ∂ z = f ′ ( e x cos y ) e x cos y ⇒ ∂ x 2 ∂ 2 z = f ′′ ( e x cos y ) e 2 x cos 2 y + f ′ ( e x cos y ) e x cos y , ∂ y ∂ z = − f ′ ( e x cos y ) e x sin y ⇒ ∂ y 2 ∂ 2 z = f ′′ ( e x cos y ) e 2 x sin 2 y − f ′ ( e x cos y ) e x cos y .
f ′ ′ ( e x cos y ) e 2 x cos 2 y + f ′ ( e x cos y ) e x cos y + f ′ ′ ( e x cos y ) e 2 x sin 2 y − f ′ ( e x cos y ) e x cos y = ( 4 z + e x cos y ) e 2 x , f''(e^x\cos y)e^{2x}\cos^2y+f'(e^x\cos y)e^x\cos y+f''(e^x\cos y)e^{2x}\sin^2y-f'(e^x\cos y)e^x\cos y=(4z+e^x\cos y)e^{2x},
f ′′ ( e x cos y ) e 2 x cos 2 y + f ′ ( e x cos y ) e x cos y + f ′′ ( e x cos y ) e 2 x sin 2 y − f ′ ( e x cos y ) e x cos y = ( 4 z + e x cos y ) e 2 x ,
即
f ′ ′ ( e x cos y ) e 2 x ( cos 2 y + sin 2 y ) = ( 4 z + e x cos y ) e 2 x , f''(e^x\cos y)e^{2x}(\cos^2y+\sin^2y)=(4z+e^x\cos y)e^{2x},
f ′′ ( e x cos y ) e 2 x ( cos 2 y + sin 2 y ) = ( 4 z + e x cos y ) e 2 x ,
因此f ′ ′ ( e x cos y ) = 4 z + e x cos y . f''(e^x\cos y)=4z+e^x\cos y. f ′′ ( e x cos y ) = 4 z + e x cos y . e x cos y e^x\cos y e x cos y u , u, u , f ′ ′ ( u ) − 4 f ( u ) = u . f''(u)-4f(u)=u. f ′′ ( u ) − 4 f ( u ) = u .
解这个微分方程的特征方程λ 2 − 4 = 0 \lambda^2-4=0 λ 2 − 4 = 0 λ 1 = 2 , λ 2 = − 2. \lambda_1=2,\lambda_2=-2. λ 1 = 2 , λ 2 = − 2. f ′ ′ ( u ) − 4 f ( u ) = 0 f''(u)-4f(u)=0 f ′′ ( u ) − 4 f ( u ) = 0 φ = c 1 e 2 u + c 2 e − 2 u . \varphi=c_1e^{2u}+c_2e^{-2u}. φ = c 1 e 2 u + c 2 e − 2 u . f ∗ ( u ) ; f^*(u); f ∗ ( u ) ; f ∗ ( u ) = a u + b , f^*(u)=au+b, f ∗ ( u ) = a u + b ,
0 − 4 ( a u + b ) = u ⇒ { a = − 1 4 b = 0 . 0-4(au+b)=u\Rightarrow\begin{cases}a=-\frac{1}{4}\\b=0\end{cases}.
0 − 4 ( a u + b ) = u ⇒ { a = − 4 1 b = 0 .
所以f ∗ ( u ) = − 1 4 u . f^*(u)=-\frac{1}{4}u. f ∗ ( u ) = − 4 1 u . f ( u ) = φ + f ∗ ( u ) = c 1 e 2 u + c 2 e − 2 u − 1 4 u , f(u)=\varphi+f^*(u)=c_1e^{2u}+c_2e^{-2u}-\frac{1}{4}u, f ( u ) = φ + f ∗ ( u ) = c 1 e 2 u + c 2 e − 2 u − 4 1 u , f ′ ( u ) = 2 c 1 e 2 u − 2 c 2 e − 2 u − 1 4 . f'(u)=2c_1e^{2u}-2c_2e^{-2u}-\frac{1}{4}. f ′ ( u ) = 2 c 1 e 2 u − 2 c 2 e − 2 u − 4 1 . f ( 0 ) = f ′ ( 0 ) = 0 , f(0)=f'(0)=0, f ( 0 ) = f ′ ( 0 ) = 0 ,
{ c 1 + c 2 = 0 2 c 1 − 2 c 2 − 1 4 = 0 . ⇒ { c 1 = 1 16 c 2 = − 1 16 . \begin{cases}
c_1+c_2&=&0\\
2c_1-2c_2-\frac{1}{4}&=&0
\end{cases}.\Rightarrow\begin{cases}
c_1&=\frac{1}{16}\\
c_2&=-\frac{1}{16}
\end{cases}.
{ c 1 + c 2 2 c 1 − 2 c 2 − 4 1 = = 0 0 . ⇒ { c 1 c 2 = 16 1 = − 16 1 .
故所求为
f ( u ) = 1 16 ( e 2 u − e − 2 u − 4 u ) . f(u)=\frac{1}{16}(e^{2u}-e^{-2u}-4u).
f ( u ) = 16 1 ( e 2 u − e − 2 u − 4 u ) .
多元函数隐函数
我们本就知道一元函数的隐函数存在定理。如果在点P( x 0 , y 0 ) (x_0,y_0) ( x 0 , y 0 ) F ( x , y ) F(x,y) F ( x , y ) F ( x 0 , y 0 ) = 0 , F x ′ F(x_0,y_0)=0,F'_x F ( x 0 , y 0 ) = 0 , F x ′ F y ′ F'_y F y ′ F y ′ ( x 0 , y 0 ) ≠ 0 , F'_y(x_0,y_0)≠0, F y ′ ( x 0 , y 0 ) = 0 , F ( x , y ) F(x,y) F ( x , y ) y = f ( x ) y=f(x) y = f ( x ) d y d x = − F x ′ F y ′ ; \frac{\text{d}y}{\text{d}x}=-\frac{F'_x}{F'_y}; d x d y = − F y ′ F x ′ ; F ( x , y , z ) F(x,y,z) F ( x , y , z ) ( x 0 , y 0 , z 0 ) (x_0,y_0,z_0) ( x 0 , y 0 , z 0 ) F ( x 0 , y 0 , z 0 ) = 0 , F z ′ ( x 0 , y 0 , z 0 ) ≠ 0 , F(x_0,y_0,z_0)=0,F'_z(x_0,y_0,z_0)≠0, F ( x 0 , y 0 , z 0 ) = 0 , F z ′ ( x 0 , y 0 , z 0 ) = 0 , F ( x , y , z ) = 0 F(x,y,z)=0 F ( x , y , z ) = 0 z = f ( x , y ) , z=f(x,y), z = f ( x , y ) , z 0 = f ( x 0 , y 0 ) . z_0=f(x_0,y_0). z 0 = f ( x 0 , y 0 ) .
多元函数的隐函数存在定理实际上暗含玄机。“F z ′ ≠ 0 F'_z≠0 F z ′ = 0 z = f ( x , y ) z=f(x,y) z = f ( x , y ) F x ′ ≠ 0 F'_x≠0 F x ′ = 0 x = f ( y , z ) ; F y ′ ≠ 0 x=f(y,z);F'_y≠0 x = f ( y , z ) ; F y ′ = 0
e.g.5.4 \textbf{e.g.5.4 } e.g.5.4 x y − z ln y + e x z = 1. xy-z\ln y+e^{xz}=1. x y − z ln y + e x z = 1. ( 0 , 1 , 1 ) (0,1,1) ( 0 , 1 , 1 )
A. 在这个邻域内只能确定一个具有连续偏导数的隐函数z = z ( x , y ) z=z(x,y) z = z ( x , y )
B. 可确定两个具有连续偏导数的隐函数y = y ( x , z ) y=y(x,z) y = y ( x , z ) z = z ( x , y ) z=z(x,y) z = z ( x , y )
C. 可确定两个具有连续偏导数的隐函数x = x ( y , z ) x=x(y,z) x = x ( y , z ) z = z ( x , y ) z=z(x,y) z = z ( x , y )
D. 可确定两个具有连续偏导数的隐函数x = x ( y , z ) x=x(y,z) x = x ( y , z ) y = y ( x , z ) y=y(x,z) y = y ( x , z )
Analyses \textbf{Analyses } Analyses F ( x ) = x y − z ln y + e x z − 1. F(x)=xy-z\ln y+e^{xz}-1. F ( x ) = x y − z ln y + e x z − 1. x , y , z x,y,z x , y , z F F F
∂ F ∂ x = y + e x z ⋅ z , ∂ F ∂ y = x − z y , ∂ F ∂ z = − ln y + e x z ⋅ x , \begin{align*}
\frac{\partial F}{\partial x}&=y+e^{xz}\cdot z,\\
\frac{\partial F}{\partial y}&=x-\frac{z}{y},\\
\frac{\partial F}{\partial z}&=-\ln y+e^{xz}\cdot x,
\end{align*}
∂ x ∂ F ∂ y ∂ F ∂ z ∂ F = y + e x z ⋅ z , = x − y z , = − ln y + e x z ⋅ x ,
再代入点( 0 , 1 , 1 ) (0,1,1) ( 0 , 1 , 1 ) F ′ ( x , y , z ) = 2 d x − d y . F'(x,y,z)=2\text{d}x-\text{d}y. F ′ ( x , y , z ) = 2 d x − d y . ∂ F ∂ x \frac{\partial F}{\partial x} ∂ x ∂ F ∂ F ∂ y \frac{\partial F}{\partial y} ∂ y ∂ F x x x y . y. y . Choice \textbf{Choice } Choice
多元函数的隐函数求导方法与一元函数隐函数类似。对F ( x , y , z ) F(x,y,z) F ( x , y , z ) ∂ z ∂ x = − F x ′ F z ′ , ∂ z ∂ y = − F y ′ F z ′ ; \frac{\partial z}{\partial x}=-\frac{F'_x}{F'_z},\frac{\partial z}{\partial y}=-\frac{F'_y}{F'_z}; ∂ x ∂ z = − F z ′ F x ′ , ∂ y ∂ z = − F z ′ F y ′ ; F x ′ + F z ′ ∂ z ∂ x = 0 , F y ′ + F z ′ ∂ z ∂ y = 0. F'_x+F'_z\frac{\partial z}{\partial x}=0,F'_y+F'_z\frac{\partial z}{\partial y}=0. F x ′ + F z ′ ∂ x ∂ z = 0 , F y ′ + F z ′ ∂ y ∂ z = 0. F x ′ d x + F y ′ d y + F z ′ d z = 0 F'_x\text{d}x+F'_y\text{d}y+F'_z\text{d}z=0 F x ′ d x + F y ′ d y + F z ′ d z = 0
e.g.5.5 \textbf{e.g.5.5 } e.g.5.5 e x + 2 y + 3 z + x y z = 1 e^{x+2y+3z}+xyz=1 e x + 2 y + 3 z + x yz = 1 z = z ( x , y ) , z=z(x,y), z = z ( x , y ) , d z ∣ ( 0 , 0 ) . \text{d}z\big|_{(0,0)}. d z ( 0 , 0 ) .
Solution \textbf{Solution } Solution
e x + 2 y + 3 z d ( x + 2 y + 3 z ) + y z d x + x z d y + x y d z = 0 , e^{x+2y+3z}\text{d}(x+2y+3z)+yz\text{d}x+xz\text{d}y+xy\text{d}z=0,
e x + 2 y + 3 z d ( x + 2 y + 3 z ) + yz d x + x z d y + x y d z = 0 ,
即
e x + 2 y + 3 z d x + e x + 2 y + 3 z d 2 y + e x + 2 y + 3 z d 3 z + y z d x + x z d y + x y d z = 0 e^{x+2y+3z}\text{d}x+e^{x+2y+3z}\text{d}2y+e^{x+2y+3z}\text{d}3z+yz\text{d}x+xz\text{d}y+xy\text{d}z=0
e x + 2 y + 3 z d x + e x + 2 y + 3 z d 2 y + e x + 2 y + 3 z d 3 z + yz d x + x z d y + x y d z = 0
而将{ x = 0 y = 0 \begin{cases}x&=0\\y&=0\end{cases} { x y = 0 = 0 z = 0 , z=0, z = 0 , ( 0 , 0 ) (0,0) ( 0 , 0 )
d x + 2 d y + 3 d z = 0. \text{d}x+2\text{d}y+3\text{d}z=0.
d x + 2 d y + 3 d z = 0.
因此所求为d z ∣ ( 0 , 0 ) = − 1 3 d x − 2 3 d y . \text{d}z\big|_{(0,0)}=-\frac{1}{3}\text{d}x-\frac{2}{3}\text{d}y. d z ( 0 , 0 ) = − 3 1 d x − 3 2 d y .
e.x.5.5 \textbf{e.x.5.5 } e.x.5.5 e z + x y z + x + cos x = 2 e^z+xyz+x+\cos x=2 e z + x yz + x + cos x = 2 z = z ( x , y ) , z=z(x,y), z = z ( x , y ) , d z ∣ ( 0 , 1 ) . \text{d}z\big|_{(0,1)}. d z ( 0 , 1 ) .
Solution \textbf{Solution } Solution
e z d z + y z d x + x z d y + x y d z + d x − sin x d x = 0. e^z\text{d}z+yz\text{d}x+xz\text{d}y+xy\text{d}z+\text{d}x-\sin x\text{d}x=0.
e z d z + yz d x + x z d y + x y d z + d x − sin x d x = 0.
即
( y z + 1 − sin x ) d x + ( x z ) d y + ( e z + x y ) d z = 0. (yz+1-\sin x)\text{d}x+(xz)\text{d}y+(e^z+xy)\text{d}z=0.
( yz + 1 − sin x ) d x + ( x z ) d y + ( e z + x y ) d z = 0.
而将{ x = 0 y = 1 \begin{cases}x&=0\\y&=1\end{cases} { x y = 0 = 1 z = 0 , z=0, z = 0 , ( 0 , 1 ) (0,1) ( 0 , 1 )
d x + d z = 0. \text{d}x+\text{d}z=0.
d x + d z = 0.
因此所求为d z ∣ ( 0 , 1 ) = − d x . \text{d}z\big|_{(0,1)}=-\text{d}x. d z ( 0 , 1 ) = − d x .
二重积分
∬ D ( x , y ) d x d y \iint_D(x,y)\text{d}x\text{d}y ∬ D ( x , y ) d x d y f ( x , y ) f(x,y) f ( x , y ) D D D
二重积分的几何意义是以区域D D D D D D f f f D D D
∬ D f ( x , y ) d x d y = lim n → ∞ ∑ i = 1 n ∑ j = 1 n 1 n 2 f ( i n , j n ) . \iint_Df(x,y)\text{d}x\text{d}y=\lim_{n\rightarrow\infty}\sum_{i=1}^n\sum_{j=1}^n\frac{1}{n^2}f\Big(\frac{i}{n},\frac{j}{n}\Big).
∬ D f ( x , y ) d x d y = n → ∞ lim i = 1 ∑ n j = 1 ∑ n n 2 1 f ( n i , n j ) .
e.g.6.1 lim n → ∞ ∑ i = 1 n ∑ j = 1 n n ( n + i ) ( n 2 + j 2 ) = \textbf{e.g.6.1 }\lim_{n\rightarrow\infty}\sum_{i=1}^n\sum_{j=1}^n\frac{n}{(n+i)(n^2+j^2)}= e.g.6.1 lim n → ∞ ∑ i = 1 n ∑ j = 1 n ( n + i ) ( n 2 + j 2 ) n =
A. ∫ 0 1 d x ∫ 0 x 1 ( 1 + x ) ( 1 + y 2 ) d y \int_0^1\text{d}x\int_0^x\frac{1}{(1+x)(1+y^2)}\text{d}y ∫ 0 1 d x ∫ 0 x ( 1 + x ) ( 1 + y 2 ) 1 d y
B. ∫ 0 1 d x ∫ 0 x 1 ( 1 + x ) ( 1 + y ) d y \int_0^1\text{d}x\int_0^x\frac{1}{(1+x)(1+y)}\text{d}y ∫ 0 1 d x ∫ 0 x ( 1 + x ) ( 1 + y ) 1 d y
C. ∫ 0 1 d x ∫ 0 1 1 ( 1 + x ) ( 1 + y ) d y \int_0^1\text{d}x\int_0^1\frac{1}{(1+x)(1+y)}\text{d}y ∫ 0 1 d x ∫ 0 1 ( 1 + x ) ( 1 + y ) 1 d y
D. ∫ 0 1 d x ∫ 0 1 1 ( 1 + x ) ( 1 + y 2 ) d y \int_0^1\text{d}x\int_0^1\frac{1}{(1+x)(1+y^2)}\text{d}y ∫ 0 1 d x ∫ 0 1 ( 1 + x ) ( 1 + y 2 ) 1 d y
Analyses \textbf{Analyses } Analyses
O.F. = lim n → ∞ ∑ i = 1 n ∑ j = 1 n 1 n 2 n 3 ( n + i ) ( n 2 + j 2 ) = lim n → ∞ ∑ i = 1 n ∑ j = 1 n 1 n 2 1 n + i n ⋅ n 2 + j 2 n 2 = lim n → ∞ ∑ i = 1 n ∑ j = 1 n 1 n 2 1 ( 1 + i n ) ( 1 + ( j n ) 2 ) = ∬ D 1 ( 1 + x ) ( 1 + y 2 ) d x d y . \begin{align*}
\text{O.F.}&=\lim_{n\rightarrow\infty}\sum_{i=1}^n\sum_{j=1}^n\frac{1}{n^2}\frac{n^3}{(n+i)(n^2+j^2)}=\lim_{n\rightarrow\infty}\sum_{i=1}^n\sum_{j=1}^n\frac{1}{n^2}\frac{1}{\frac{n+i}{n}\cdot\frac{n^2+j^2}{n^2}}\\
&=\lim_{n\rightarrow\infty}\sum_{i=1}^n\sum_{j=1}^n\frac{1}{n^2}\frac{1}{(1+\frac{i}{n})\Big(1+\big(\frac{j}{n}\big)^2\Big)}=\iint_D\frac{1}{(1+x)(1+y^2)}\text{d}x\text{d}y.
\end{align*}
O.F. = n → ∞ lim i = 1 ∑ n j = 1 ∑ n n 2 1 ( n + i ) ( n 2 + j 2 ) n 3 = n → ∞ lim i = 1 ∑ n j = 1 ∑ n n 2 1 n n + i ⋅ n 2 n 2 + j 2 1 = n → ∞ lim i = 1 ∑ n j = 1 ∑ n n 2 1 ( 1 + n i ) ( 1 + ( n j ) 2 ) 1 = ∬ D ( 1 + x ) ( 1 + y 2 ) 1 d x d y .
而根据极限式和二重积分式互相转写的条件，可知D D D O.F. = ∫ 0 1 d x ∫ 0 x 1 ( 1 + x ) ( 1 + y 2 ) d y \text{O.F.}=\int_0^1\text{d}x\int_0^x\frac{1}{(1+x)(1+y^2)}\text{d}y O.F. = ∫ 0 1 d x ∫ 0 x ( 1 + x ) ( 1 + y 2 ) 1 d y Choice \textbf{Choice } Choice
二重积分性质
与定积分性质类似。特别地，若在D D D f ( x , y ) f(x,y) f ( x , y ) g ( x , y ) g(x,y) g ( x , y ) D D D f ( x , y ) ≤ g ( x , y ) f(x,y)\leq g(x,y) f ( x , y ) ≤ g ( x , y ) ∬ D f ( x , y ) d x d y ≤ ∬ D g ( x , y ) d x d y . \iint_Df(x,y)\text{d}x\text{d}y\leq\iint_Dg(x,y)\text{d}x\text{d}y. ∬ D f ( x , y ) d x d y ≤ ∬ D g ( x , y ) d x d y .
（二重积分中值定理）若二元函数f ( x , y ) f(x,y) f ( x , y ) δ \delta δ D D D ∃ ( ξ , η ) ∈ D , s.t. ∬ D f ( x , y ) d x d y = f ( ξ , η ) δ . \exists(\xi,\eta)\in D,\text{ s.t. }\iint_Df(x,y)\text{d}x\text{d}y=f(\xi,\eta)\delta. ∃ ( ξ , η ) ∈ D , s.t. ∬ D f ( x , y ) d x d y = f ( ξ , η ) δ .
二重积分计算方法
直角坐标系下二重积分的计算方法
考虑先对x x x y y y f ( x , y ) f(x,y) f ( x , y ) [ a , b ] [a,b] [ a , b ] [ c , d ] [c,d] [ c , d ] x = a , x = b , x=a,x=b, x = a , x = b , f f f φ 1 ( x ) \varphi_1(x) φ 1 ( x ) φ 2 ( x ) \varphi_2(x) φ 2 ( x )
∬ D f ( x , y ) ( d y = ∫ a b d x ∫ φ 1 ( x ) φ 2 ( x ) f ( x , y ) d y ) d x . \iint_Df(x,y)\Big(\text{d}y=\int_a^b\text{d}x\int_{\varphi_1(x)}^{\varphi_2(x)}f(x,y)\text{d}y\Big)\text{d}x.
∬ D f ( x , y ) ( d y = ∫ a b d x ∫ φ 1 ( x ) φ 2 ( x ) f ( x , y ) d y ) d x .
类似地，若作直线y = c , y = d y=c,y=d y = c , y = d ψ 1 ( y ) \psi_1(y) ψ 1 ( y ) ψ 2 ( y ) \psi_2(y) ψ 2 ( y )
∬ D f ( x , y ) ( d x d y = ∫ c d d y ∫ ψ 1 ( y ) ψ 2 ( y ) f ( x , y ) d x ) d y . \iint_Df(x,y)\Big(\text{d}x\text{d}y=\int_c^d\text{d}y\int_{\psi_1(y)}^{\psi_2(y)}f(x,y)\text{d}x\Big)\text{d}y.
∬ D f ( x , y ) ( d x d y = ∫ c d d y ∫ ψ 1 ( y ) ψ 2 ( y ) f ( x , y ) d x ) d y .
在这种计算方法中，每层积分的下限应该小于上限。一般而言，内层积分的上下限都应该关于外层积分的变量（或者是常数），而外层积分的上下限一定是常数。对于先积x x x y y y
如果这个函数关于某个变量的原函数不是初等函数的被积函数，需要改变积分次序，先对另一个变量积分，从而让内层积分更容易被求出。
观察函数f ( x , y ) = e y x . f(x,y)=e^{\frac{y}{x}}. f ( x , y ) = e x y . x x x y y y
观察函数f ( x , y ) = g ( x ) e y 2 . f(x,y)=g(x)e^{y^2}. f ( x , y ) = g ( x ) e y 2 . y y y x x x
观察函数f ( x , y ) = x 1 − x 2 + y 2 . f(x,y)=x\sqrt{1-x^2+y^2}. f ( x , y ) = x 1 − x 2 + y 2 . x x x
若积分区域用平行于x x x D , D, D , D D D x x x y y y D D D y y y
e.g.6.2 \textbf{e.g.6.2 } e.g.6.2 ∬ D x 1 − x 2 + y 2 d x d y , \iint_Dx\sqrt{1-x^2+y^2}\text{d}x\text{d}y, ∬ D x 1 − x 2 + y 2 d x d y , D D D y = x , x = 1 , y = − 1 y=x,x=1,y=-1 y = x , x = 1 , y = − 1
Solution \textbf{Solution } Solution
O.F. = ∫ − 1 1 ( ∫ y 1 x 1 − x 2 + y 2 d x ) d y = ∫ − 1 1 ( ( − 1 2 ) ∫ x = y x = 1 1 − x 2 + y 2 d ( 1 − x 2 + y 2 ) ) d y = ∫ − 1 1 ( 1 3 ( 1 − x 2 + y 2 ) 3 2 ∣ x = 1 x = y ) d y = ∫ − 1 1 ( 1 3 ( 1 − ( y 2 ) 3 2 ) ) d y = ∫ − 1 1 ( 1 3 ( 1 − ∣ y ∣ 3 ) ) d y = 2 × ( ( 1 3 y − 1 12 y 4 ) ∣ 0 1 ) = 1 2 . \begin{align*}
\text{O.F.}&=\int_{-1}^1\Big(\int_y^1x\sqrt{1-x^2+y^2}\text{d}x\Big)\text{d}y\\
&=\int_{-1}^1\Bigg(\Big(-\frac{1}{2}\Big)\int_{x=y}^{x=1}\sqrt{1-x^2+y^2}\text{d}(1-x^2+y^2)\Bigg)\text{d}y\\
&=\int_{-1}^1\Bigg(\frac{1}{3}(1-x^2+y^2)^\frac{3}{2}\bigg|_{x=1}^{x=y}\Bigg)\text{d}y=\int_{-1}^1\Bigg(\frac{1}{3}\Big(1-(y^2)^\frac{3}{2}\Big)\Bigg)\text{d}y\\
&=\int_{-1}^1\Big(\frac{1}{3}(1-|y|^3)\Big)\text{d}y=2\times\Bigg(\Big(\frac{1}{3}y-\frac{1}{12}y^4\Big)\Big|_0^1\Bigg)\\
&=\frac{1}{2}.
\end{align*}
O.F. = ∫ − 1 1 ( ∫ y 1 x 1 − x 2 + y 2 d x ) d y = ∫ − 1 1 ( ( − 2 1 ) ∫ x = y x = 1 1 − x 2 + y 2 d ( 1 − x 2 + y 2 ) ) d y = ∫ − 1 1 ( 3 1 ( 1 − x 2 + y 2 ) 2 3 x = 1 x = y ) d y = ∫ − 1 1 ( 3 1 ( 1 − ( y 2 ) 2 3 ) ) d y = ∫ − 1 1 ( 3 1 ( 1 − ∣ y ∣ 3 ) ) d y = 2 × ( ( 3 1 y − 12 1 y 4 ) 0 1 ) = 2 1 .
极坐标系下二重积分的计算方法
我们本就知道极坐标系中原本的直角坐标( x , y ) (x,y) ( x , y ) { x = r cos θ y = r sin θ d x d y = r d r d θ , \begin{cases}x=r\cos\theta\\y=r\sin\theta\\\text{d}x\text{d}y=r\text{d}r\text{d}\theta\end{cases}, ⎩ ⎨ ⎧ x = r cos θ y = r sin θ d x d y = r d r d θ , { ( r , θ ) ∣ r 1 ≤ r ≤ r 2 , θ 1 ≤ θ ≤ θ 2 } \{(r,\theta)|r_1\leq r\leq r_2,\theta_1\leq\theta\leq\theta_2\} {( r , θ ) ∣ r 1 ≤ r ≤ r 2 , θ 1 ≤ θ ≤ θ 2 } D , D, D ,
∬ D x O y f ( x , y ) d x d y = ∬ D r O θ f ( r cos θ , r sin θ ) r d r d θ = ∫ θ 1 θ 2 d θ ∫ r 1 ( θ ) r 2 ( θ ) f ( r cos θ , r sin θ ) r d r . \iint_{D_{xOy}}f(x,y)\text{d}x\text{d}y=\iint_{D_{rO\theta}}f(r\cos\theta,r\sin\theta)r\text{d}r\text{d}\theta=\int_{\theta_1}^{\theta_2}\text{d}\theta\int_{r_1(\theta)}^{r_2(\theta)}f(r\cos\theta,r\sin\theta)r\text{d}r.
∬ D x O y f ( x , y ) d x d y = ∬ D r Oθ f ( r cos θ , r sin θ ) r d r d θ = ∫ θ 1 θ 2 d θ ∫ r 1 ( θ ) r 2 ( θ ) f ( r cos θ , r sin θ ) r d r .
e.g.6.3 \textbf{e.g.6.3 } e.g.6.3 D = { ( r , θ ) ∣ 2 ≤ r ≤ 2 ( 1 + cos θ ) , − π 2 ≤ θ ≤ π 2 } , D=\big\{(r,\theta)|2\leq r\leq 2(1+\cos\theta),-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\big\}, D = { ( r , θ ) ∣2 ≤ r ≤ 2 ( 1 + cos θ ) , − 2 π ≤ θ ≤ 2 π } , ∬ D x d x d y . \iint_Dx\text{d}x\text{d}y. ∬ D x d x d y .
Solution \textbf{Solution } Solution
O.F. = ∫ − π 2 π 2 d θ ∫ 2 2 ( 1 + cos θ ) r cos θ r d r = ∫ − π 2 π 2 ( 1 3 r 3 cos θ ) ∣ r = 2 r = 2 ( 1 + cos θ ) d θ = 8 3 ∫ − π 2 π 2 ( ( 1 + cos θ ) 3 − 1 ) cos θ d θ = 8 3 ∫ − π 2 π 2 ( cos 3 θ + 3 cos 2 θ + 3 cos θ ) cos θ d θ = 16 3 ∫ 0 π 2 ( cos 4 θ + 3 cos 3 θ + 3 cos 2 θ ) d θ = 16 3 ( 3 4 × 1 2 × π 2 + 3 × 2 3 × 1 + 3 × 1 2 × π 2 ) = 32 3 + 5 π . \begin{align*}
\text{O.F.}&=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\text{d}\theta\int_2^{2(1+\cos\theta)}r\cos\theta r\text{d}r\\
&=\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\Big(\frac{1}{3}r^3\cos\theta\Big)\Big|_{r=2}^{r=2(1+\cos\theta)}\text{d}\theta=\frac{8}{3}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\big((1+\cos\theta)^3-1\big)\cos\theta\text{d}\theta\\
&=\frac{8}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\cos^3\theta+3\cos^2\theta+3\cos\theta)\cos\theta\text{d}\theta=\frac{16}{3}\int_0^\frac{\pi}{2}(\cos^4\theta+3\cos^3\theta+3\cos^2\theta)\text{d}\theta\\
&=\frac{16}{3}\Big(\frac{3}{4}\times\frac{1}{2}\times\frac{\pi}{2}+3\times\frac{2}{3}\times1+3\times\frac{1}{2}\times\frac{\pi}{2}\Big)=\frac{32}{3}+5\pi.
\end{align*}
O.F. = ∫ − 2 π 2 π d θ ∫ 2 2 ( 1 + c o s θ ) r cos θ r d r = ∫ − 2 π 2 π ( 3 1 r 3 cos θ ) r = 2 r = 2 ( 1 + c o s θ ) d θ = 3 8 ∫ − 2 π 2 π ( ( 1 + cos θ ) 3 − 1 ) cos θ d θ = 3 8 ∫ − 2 π 2 π ( cos 3 θ + 3 cos 2 θ + 3 cos θ ) cos θ d θ = 3 16 ∫ 0 2 π ( cos 4 θ + 3 cos 3 θ + 3 cos 2 θ ) d θ = 3 16 ( 4 3 × 2 1 × 2 π + 3 × 3 2 × 1 + 3 × 2 1 × 2 π ) = 3 32 + 5 π .
上方例子中使用了余弦函数的幂的积分有关结论。
并不一定非要是题中给了极坐标系形式的函数才能使用极坐标系二重积分——当被积函数以“x 2 + y 2 x^2+y^2 x 2 + y 2 r r r θ . \theta. θ .
e.g.6.4 \textbf{e.g.6.4 } e.g.6.4 D D D x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1 y = x y=x y = x x x x ∬ D e ( x + y ) 2 ( x 2 − y 2 ) d x d y . \iint_De^{(x+y)^2}(x^2-y^2)\text{d}x\text{d}y. ∬ D e ( x + y ) 2 ( x 2 − y 2 ) d x d y .
Solution \textbf{Solution } Solution x = r cos θ , y = r sin θ . x=r\cos\theta,y=r\sin\theta. x = r cos θ , y = r sin θ . ( ( cos θ + sin θ ) 2 ) ′ = 2 ( cos θ + sin θ ) ( cos θ − sin θ ) = 2 ( cos 2 θ − sin 2 θ ) , \big((\cos\theta+\sin\theta)^2\big)'=2(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=2(\cos^2\theta-\sin^2\theta), ( ( cos θ + sin θ ) 2 ) ′ = 2 ( cos θ + sin θ ) ( cos θ − sin θ ) = 2 ( cos 2 θ − sin 2 θ ) ,
O.F. = ∫ 0 1 d r ∫ 0 π 4 e r 2 ( cos θ + sin θ ) 2 r 3 ( cos 2 θ − sin 2 θ ) d θ = ∫ 0 1 1 2 r d r ∫ θ = 0 θ = π 4 e r 2 ( cos θ + sin θ ) 2 d ( r 2 ( cos θ + sin θ ) 2 ) = 1 2 ∫ 0 1 r ( e r 2 ( cos θ + sin θ ) 2 ∣ θ = 0 θ = π 4 ) d r = 1 4 ∫ 0 1 ( e 2 r 2 − e r 2 ) d ( r 2 ) = 1 8 e 2 r 2 ∣ 0 1 + 1 4 e r 2 ∣ 0 1 = ( e − 1 ) 2 8 . \begin{align*}
\text{O.F.}&=\int_0^1\text{d}r\int_0^\frac{\pi}{4}e^{r^2(\cos\theta+\sin\theta)^2}r^3(\cos^2\theta-\sin^2\theta)\text{d}\theta\\
&=\int_0^1\frac{1}{2}r\text{d}r\int_{\theta=0}^{\theta=\frac{\pi}{4}}e^{r^2(\cos\theta+\sin\theta)^2}\text{d}\big(r^2(\cos\theta+\sin\theta)^2\big)=\frac{1}{2}\int_0^1r\Big(e^{r^2(\cos\theta+\sin\theta)^2}\Big|_{\theta=0}^{\theta=\frac{\pi}{4}}\Big)\text{d}r\\
&=\frac{1}{4}\int_0^1(e^{2r^2}-e^{r^2})\text{d}(r^2)=\frac{1}{8}e^{2r^2}\Big|_0^1+\frac{1}{4}e^{r^2}\Big|_0^1=\frac{(e-1)^2}{8}.
\end{align*}
O.F. = ∫ 0 1 d r ∫ 0 4 π e r 2 ( c o s θ + s i n θ ) 2 r 3 ( cos 2 θ − sin 2 θ ) d θ = ∫ 0 1 2 1 r d r ∫ θ = 0 θ = 4 π e r 2 ( c o s θ + s i n θ ) 2 d ( r 2 ( cos θ + sin θ ) 2 ) = 2 1 ∫ 0 1 r ( e r 2 ( c o s θ + s i n θ ) 2 θ = 0 θ = 4 π ) d r = 4 1 ∫ 0 1 ( e 2 r 2 − e r 2 ) d ( r 2 ) = 8 1 e 2 r 2 0 1 + 4 1 e r 2 0 1 = 8 ( e − 1 ) 2 .
当然，也不是所有给了极坐标系形式的函数的情况就一定要用极坐标方法求二重积分；有时也需要转化回直角坐标系。
e.g.6.5 \textbf{e.g.6.5 } e.g.6.5 ∬ D r 2 sin θ 1 − r 2 cos 2 θ d r d θ , \iint_Dr^2\sin\theta\sqrt{1-r^2\cos2\theta}\text{d}r\text{d}\theta, ∬ D r 2 sin θ 1 − r 2 cos 2 θ d r d θ , D = { ( r , θ ) ∣ 0 ≤ r ≤ sec θ , 0 ≤ θ ≤ π 4 } . D=\big\{(r,\theta)|0\leq r\leq\sec\theta,0\leq\theta\leq\frac{\pi}{4}\big\}. D = { ( r , θ ) ∣0 ≤ r ≤ sec θ , 0 ≤ θ ≤ 4 π } .
Solution \textbf{Solution } Solution D x O y = { ( x , y ) ∣ 0 ≤ x ≤ 1 , 0 ≤ y ≤ x } . D_{xOy}=\{(x,y)|0\leq x\leq 1, 0\leq y\leq x\}. D x O y = {( x , y ) ∣0 ≤ x ≤ 1 , 0 ≤ y ≤ x } .
O.F. = ∬ D r sin θ 1 − r 2 ( cos 2 θ − sin 2 θ ) r d r d θ = ∬ D r sin θ 1 − ( r cos θ ) 2 + ( r sin θ ) 2 r d r d θ = ∬ D x O y y 1 − x 2 + y 2 d x d y = ∫ 0 1 ( ∫ 0 x y 1 − x 2 + y 2 d y ) d x = ∫ 0 1 ( ∫ y = 0 y = x 1 2 1 − x 2 + y 2 d ( 1 − x 2 + y 2 ) ) d x = ∫ 0 1 ( 1 3 ( 1 − x 2 + y 2 ) 3 2 ) ∣ y = 0 y = x d x = ∫ 0 1 ( 1 3 ( 1 − ( 1 − x 2 ) 3 2 ) ) d x = ∫ 0 1 1 3 d x − 1 3 ∫ 0 1 ( 1 − x 2 ) 3 2 d x = 令 x = sin α 1 3 − 1 3 ∫ 0 1 cos 3 α d ( sin α ) = 1 3 − 1 3 ∫ 0 π 2 cos 4 α d α = 1 3 − 1 3 ∫ 0 π 2 ( 1 + cos 2 α 2 ) 2 d α = 1 3 − 1 3 ∫ 0 π 2 ( 1 4 + 1 2 cos 2 α + 1 4 cos 2 2 α ) d α = 1 3 − 1 3 × π 8 − 1 3 ∫ 0 π 2 ( 1 2 cos 2 α + 1 4 ⋅ 1 + cos 4 α 2 ) d α = 1 3 − π 24 − 1 3 × π 16 − ∫ 0 π 2 ( 1 6 cos 2 α + 1 24 cos 4 α ) d α = 1 3 − π 16 − ∫ 0 π 2 cos 2 α 6 d α − ∫ 0 π 2 cos 4 α 24 d α = 1 3 − π 16 − 0 − 0 = 1 3 − π 16 . \begin{align*}
\text{O.F.}&=\iint_Dr\sin\theta\sqrt{1-r^2(\cos^2\theta-\sin^2\theta)}r\text{d}r\text{d}\theta=\iint_Dr\sin\theta\sqrt{1-(r\cos\theta)^2+(r\sin\theta)^2}r\text{d}r\text{d}\theta\\
&=\iint_{D_{xOy}}y\sqrt{1-x^2+y^2}\text{d}x\text{d}y=\int_0^1\big(\int_0^xy\sqrt{1-x^2+y^2}\text{d}y\big)\text{d}x\\
&=\int_0^1\Big(\int_{y=0}^{y=x}\frac{1}{2}\sqrt{1-x^2+y^2}\text{d}(1-x^2+y^2)\Big)\text{d}x=\int_0^1\Big(\frac{1}{3}(1-x^2+y^2)^\frac{3}{2}\Big)\bigg|_{y=0}^{y=x}\text{d}x\\
&=\int_0^1\Big(\frac{1}{3}\big(1-(1-x^2)^\frac{3}{2}\big)\Big)\text{d}x=\int_0^1\frac{1}{3}\text{d}x-\frac{1}{3}\int_0^1(1-x^2)^\frac{3}{2}\text{d}x\xlongequal{\text{令}x=\sin\alpha}\frac{1}{3}-\frac{1}{3}\int_0^1\cos^3\alpha\text{d}(\sin\alpha)\\
&=\frac{1}{3}-\frac{1}{3}\int_0^\frac{\pi}{2}\cos^4\alpha\text{d}\alpha=\frac{1}{3}-\frac{1}{3}\int_0^\frac{\pi}{2}\Big(\frac{1+\cos2\alpha}{2}\Big)^2\text{d}\alpha=\frac{1}{3}-\frac{1}{3}\int_0^\frac{\pi}{2}\Big(\frac{1}{4}+\frac{1}{2}\cos2\alpha+\frac{1}{4}\cos^22\alpha\Big)\text{d}\alpha\\
&=\frac{1}{3}-\frac{1}{3}\times\frac{\pi}{8}-\frac{1}{3}\int_0^\frac{\pi}{2}\Big(\frac{1}{2}\cos2\alpha+\frac{1}{4}\cdot\frac{1+\cos4\alpha}{2}\Big)\text{d}\alpha\\
&=\frac{1}{3}-\frac{\pi}{24}-\frac{1}{3}\times\frac{\pi}{16}-\int_0^\frac{\pi}{2}\Big(\frac{1}{6}\cos2\alpha+\frac{1}{24}\cos4\alpha\Big)\text{d}\alpha=\frac{1}{3}-\frac{\pi}{16}-\int_0^\frac{\pi}{2}\frac{\cos 2\alpha}{6}\text{d}\alpha-\int_0^\frac{\pi}{2}\frac{\cos 4\alpha}{24}\text{d}\alpha\\
&=\frac{1}{3}-\frac{\pi}{16}-0-0=\frac{1}{3}-\frac{\pi}{16}.
\end{align*}
O.F. = ∬ D r sin θ 1 − r 2 ( cos 2 θ − sin 2 θ ) r d r d θ = ∬ D r sin θ 1 − ( r cos θ ) 2 + ( r sin θ ) 2 r d r d θ = ∬ D x O y y 1 − x 2 + y 2 d x d y = ∫ 0 1 ( ∫ 0 x y 1 − x 2 + y 2 d y ) d x = ∫ 0 1 ( ∫ y = 0 y = x 2 1 1 − x 2 + y 2 d ( 1 − x 2 + y 2 ) ) d x = ∫ 0 1 ( 3 1 ( 1 − x 2 + y 2 ) 2 3 ) y = 0 y = x d x = ∫ 0 1 ( 3 1 ( 1 − ( 1 − x 2 ) 2 3 ) ) d x = ∫ 0 1 3 1 d x − 3 1 ∫ 0 1 ( 1 − x 2 ) 2 3 d x 令 x = s i n α 3 1 − 3 1 ∫ 0 1 cos 3 α d ( sin α ) = 3 1 − 3 1 ∫ 0 2 π cos 4 α d α = 3 1 − 3 1 ∫ 0 2 π ( 2 1 + cos 2 α ) 2 d α = 3 1 − 3 1 ∫ 0 2 π ( 4 1 + 2 1 cos 2 α + 4 1 cos 2 2 α ) d α = 3 1 − 3 1 × 8 π − 3 1 ∫ 0 2 π ( 2 1 cos 2 α + 4 1 ⋅ 2 1 + cos 4 α ) d α = 3 1 − 24 π − 3 1 × 16 π − ∫ 0 2 π ( 6 1 cos 2 α + 24 1 cos 4 α ) d α = 3 1 − 16 π − ∫ 0 2 π 6 cos 2 α d α − ∫ 0 2 π 24 cos 4 α d α = 3 1 − 16 π − 0 − 0 = 3 1 − 16 π .
高中时就学过偏心圆( x ± a ) 2 + y 2 = a 2 (x\pm a)^2+y^2=a^2 ( x ± a ) 2 + y 2 = a 2 x 2 + ( y ± a ) 2 = a 2 x^2+(y\pm a)^2=a^2 x 2 + ( y ± a ) 2 = a 2 r = ( ± ) 2 a cos θ r=(\pm)2a\cos\theta r = ( ± ) 2 a cos θ r = ( ± ) 2 a sin θ . r=(\pm)2a\sin\theta. r = ( ± ) 2 a sin θ .
交换累次积分次序
将一种次序的累次积分转化为另一种次序的累次积分，从而降低运算量。
下面给出一些“交换积分次序”的例子。
对∫ − 1 0 d y ∫ 2 1 − y f ( x , y ) d x , \int_{-1}^0\text{d}y\int_2^{1-y}f(x,y)\text{d}x, ∫ − 1 0 d y ∫ 2 1 − y f ( x , y ) d x , − ∫ − 1 0 d y ∫ 1 − y 2 f ( x , y ) d x , -\int_{-1}^0\text{d}y\int_{1-y}^2f(x,y)\text{d}x, − ∫ − 1 0 d y ∫ 1 − y 2 f ( x , y ) d x , − 1 ≤ y ≤ 0 , 1 − y ≤ x ≤ 2. -1\leq y\leq 0,1-y\leq x\leq 2. − 1 ≤ y ≤ 0 , 1 − y ≤ x ≤ 2. ( x , y ) (x,y) ( x , y ) D D D x y x + y = 1 x = 2 D
因此，区域D D D { ( x , y ) ∣ 1 ≤ x ≤ 2 , 1 − x ≤ y ≤ 0 } . \big\{(x,y)|1\leq x\leq 2,1-x\leq y\leq 0\big\}. { ( x , y ) ∣1 ≤ x ≤ 2 , 1 − x ≤ y ≤ 0 } . − ∫ 1 2 d x ∫ 1 − x 0 f ( x , y ) d y . -\int_1^2\text{d}x\int_{1-x}^0f(x,y)\text{d}y. − ∫ 1 2 d x ∫ 1 − x 0 f ( x , y ) d y .
对于∫ π 2 π d x ∫ sin x 1 f ( x , y ) d y , \int_\frac{\pi}{2}^\pi\text{d}x\int_{\sin x}^1f(x,y)\text{d}y, ∫ 2 π π d x ∫ s i n x 1 f ( x , y ) d y , π 2 ≤ x ≤ π , sin x ≤ y ≤ 1. \frac{\pi}{2}\leq x\leq\pi,\sin x\leq y\leq 1. 2 π ≤ x ≤ π , sin x ≤ y ≤ 1. ( x , y ) (x,y) ( x , y ) D D D x y y = 1 y = sin x x = ¼ x = ¼ 2 D
所以D D D { 0 ≤ y ≤ 1 , π − arcsin y ≤ x ≤ π } , \big\{0\leq y\leq 1,\pi-\arcsin y\leq x\leq\pi\big\}, { 0 ≤ y ≤ 1 , π − arcsin y ≤ x ≤ π } , ∫ 0 1 d y ∫ π − arcsin y π f ( x , y ) d x . \int_0^1\text{d}y\int_{\pi-\arcsin y}^\pi f(x,y)\text{d}x. ∫ 0 1 d y ∫ π − a r c s i n y π f ( x , y ) d x .
对于积分∫ − π 4 π 2 d θ ∫ 0 2 a cos θ f ( ρ , θ ) d ρ , \int_{-\frac{\pi}{4}}^{\frac{\pi}{2}}\text{d}\theta\int_0^{2a\cos\theta}f(\rho,\theta)\text{d}\rho, ∫ − 4 π 2 π d θ ∫ 0 2 a c o s θ f ( ρ , θ ) d ρ , − π 4 ≤ θ ≤ π 2 , 0 ≤ ρ ≤ 2 a cos θ , -\frac{\pi}{4}\leq\theta\leq\frac{\pi}{2},0\leq\rho\leq 2a\cos\theta, − 4 π ≤ θ ≤ 2 π , 0 ≤ ρ ≤ 2 a cos θ , ( x , y ) (x,y) ( x , y ) D D D µ = ¡ ¼ 4 µ = ¼ 2 ½ = 2 a cos µ ½ D
所以D D D { ( ρ , θ ) ∣ 0 ≤ ρ ≤ 2 a , − π 4 ≤ θ ≤ arccos ρ 2 a } , \big\{(\rho,\theta)|0\leq\rho\leq 2a,-\frac{\pi}{4}\leq\theta\leq\arccos\frac{\rho}{2a}\big\}, { ( ρ , θ ) ∣0 ≤ ρ ≤ 2 a , − 4 π ≤ θ ≤ arccos 2 a ρ } , ∫ 0 2 a d ρ ∫ − π 4 arccos ρ 2 a f ( ρ , θ ) d θ . \int_0^{2a}\text{d}\rho\int_{-\frac{\pi}{4}}^{\arccos\frac{\rho}{2a}}f(\rho,\theta)\text{d}\theta. ∫ 0 2 a d ρ ∫ − 4 π a r c c o s 2 a ρ f ( ρ , θ ) d θ .
奇偶对称·轮换对称·质心坐标
二重积分的奇偶对称
关于y y y D D D ∬ D f ( x , y ) d δ = { 0 , f ( − x , y ) = − f ( x , y ) 2 ∬ D 1 f ( x , y ) d δ , f ( − x , y ) = f ( x , y ) , \iint_Df(x,y)\text{d}\delta=\begin{cases}0&,f(-x,y)=-f(x,y)\\2\iint_{D_1}f(x,y)\text{d}\delta&,f(-x,y)=f(x,y)\end{cases}, ∬ D f ( x , y ) d δ = { 0 2 ∬ D 1 f ( x , y ) d δ , f ( − x , y ) = − f ( x , y ) , f ( − x , y ) = f ( x , y ) , D 1 = { ( x , y ) ∣ ( x , y ) ∈ D , x ≥ 0 } . D_1=\{(x,y)|(x,y)\in D,x\geq 0\}. D 1 = {( x , y ) ∣ ( x , y ) ∈ D , x ≥ 0 } . x x x D D D ∬ D f ( x , y ) d δ = { 0 , f ( x , − y ) = − f ( x , y ) 2 ∬ D 2 f ( x , y ) d δ , f ( x , − y ) = f ( x , y ) , \iint_Df(x,y)\text{d}\delta=\begin{cases}0&,f(x,-y)=-f(x,y)\\2\iint_{D_2}f(x,y)\text{d}\delta&,f(x,-y)=f(x,y)\end{cases}, ∬ D f ( x , y ) d δ = { 0 2 ∬ D 2 f ( x , y ) d δ , f ( x , − y ) = − f ( x , y ) , f ( x , − y ) = f ( x , y ) , D 2 = { ( x , y ) ∣ ( x , y ) ∈ D , y ≥ 0 } . D_2=\{(x,y)|(x,y)\in D,y\geq 0\}. D 2 = {( x , y ) ∣ ( x , y ) ∈ D , y ≥ 0 } .
e.g.6.6 \textbf{e.g.6.6 } e.g.6.6 D = { ( x , y ) ∣ − a ≤ x ≤ a , x ≤ y ≤ a } , D 1 = { ( x , y ) ∣ 0 ≤ x ≤ a , x ≤ y ≤ a } , D=\{(x,y)|-a\leq x\leq a,x\leq y\leq a\},D_1=\{(x,y)|0\leq x\leq a,x\leq y\leq a\}, D = {( x , y ) ∣ − a ≤ x ≤ a , x ≤ y ≤ a } , D 1 = {( x , y ) ∣0 ≤ x ≤ a , x ≤ y ≤ a } , ∬ D ( x y + cos x sin y ) d x d y = \iint_D(xy+\cos x\sin y)\text{d}x\text{d}y= ∬ D ( x y + cos x sin y ) d x d y =
A. 2 ∬ D 1 cos x sin y d x d y 2\iint_{D_1}\cos x\sin y\text{d}x\text{d}y 2 ∬ D 1 cos x sin y d x d y
B. 2 ∬ D 1 x y d x d y 2\iint_{D_1}xy\text{d}x\text{d}y 2 ∬ D 1 x y d x d y
C. 4 ∬ D 1 ( x y + cos x sin y ) d x d y 4\iint_{D_1}(xy+\cos x\sin y)\text{d}x\text{d}y 4 ∬ D 1 ( x y + cos x sin y ) d x d y
Analyses \textbf{Analyses } Analyses D D D D 1 D_1 D 1
D D 1 D 2 D 3 D 4 y = a y = x x = ¡ a a x y
现在将原积分式拆为两部分考察。如果对D D D D 1 , D 2 , D 3 , D 4 D_1,D_2,D_3,D_4 D 1 , D 2 , D 3 , D 4 ∬ D 1 + D 2 x y d x d y = 0 = ∬ D 3 + D 4 x y d x d y . \iint_{D_1+D_2}xy\text{d}x\text{d}y=0=\iint_{D_3+D_4}xy\text{d}x\text{d}y. ∬ D 1 + D 2 x y d x d y = 0 = ∬ D 3 + D 4 x y d x d y .
∬ D x y d x d y = ∬ D 1 + D 2 x y d x d y + ∬ D 3 + D 4 x y d x d y = 0 , \iint_Dxy\text{d}x\text{d}y=\iint_{D_1+D_2}xy\text{d}x\text{d}y+\iint_{D_3+D_4}xy\text{d}x\text{d}y=0,
∬ D x y d x d y = ∬ D 1 + D 2 x y d x d y + ∬ D 3 + D 4 x y d x d y = 0 ,
所以∬ D cos x sin y d x d y \iint_D\cos x\sin y\text{d}x\text{d}y ∬ D cos x sin y d x d y f ( x , y ) = cos x sin y , f(x,y)=\cos x\sin y, f ( x , y ) = cos x sin y , f ( − x , y ) = f ( x , y ) , f ( x , − y ) = − f ( x , y ) , f(-x,y)=f(x,y),f(x,-y)=-f(x,y), f ( − x , y ) = f ( x , y ) , f ( x , − y ) = − f ( x , y ) , ∬ D 3 + D 4 cos x sin y d x d y = 0 , ∬ D 1 + D 2 cos x sin y d x d y = 2 ∬ D 1 cos x sin y d x d y . \iint_{D_3+D_4}\cos x\sin y\text{d}x\text{d}y=0,\iint_{D_1+D_2}\cos x\sin y\text{d}x\text{d}y=2\iint_{D_1}\cos x\sin y\text{d}x\text{d}y. ∬ D 3 + D 4 cos x sin y d x d y = 0 , ∬ D 1 + D 2 cos x sin y d x d y = 2 ∬ D 1 cos x sin y d x d y .
∬ D cos x sin y d x d y = ∬ D 1 + D 2 cos x sin y d x d y + ∬ D 3 + D 4 cos x sin y d x d y = 2 ∬ D 1 cos x sin y d x d y . \iint_D\cos x\sin y\text{d}x\text{d}y=\iint_{D_1+D_2}\cos x\sin y\text{d}x\text{d}y+\iint_{D_3+D_4}\cos x\sin y\text{d}x\text{d}y=2\iint_{D_1}\cos x\sin y\text{d}x\text{d}y.
∬ D cos x sin y d x d y = ∬ D 1 + D 2 cos x sin y d x d y + ∬ D 3 + D 4 cos x sin y d x d y = 2 ∬ D 1 cos x sin y d x d y .
Choice \textbf{Choice } Choice
二重积分的轮换对称
关于y = x y=x y = x D D D ∬ D f ( x , y ) d x d y = ∬ D f ( y , x ) d x d y . \iint_Df(x,y)\text{d}x\text{d}y=\iint_Df(y,x)\text{d}x\text{d}y. ∬ D f ( x , y ) d x d y = ∬ D f ( y , x ) d x d y .
轮换对称在使用时可能需要对被积函数进行“伸缩”处理。
e.g.6.7 \textbf{e.g.6.7 } e.g.6.7 I = ∬ D sin x 2 cos y 2 d x d y . I=\iint_D\sin x^2\cos y^2\text{d}x\text{d}y. I = ∬ D sin x 2 cos y 2 d x d y . D = { ( x , y ) ∣ x 2 + y 2 ≤ a 2 , a > 0 } . D=\{(x,y)|x^2+y^2\leq a^2,a>0\}. D = {( x , y ) ∣ x 2 + y 2 ≤ a 2 , a > 0 } .
Solution \textbf{Solution } Solution D D D
D x 2 + y 2 = a 2 y = x x y
毫无疑问，这个区域关于y = x y=x y = x ∬ D sin x 2 cos y 2 d x d y = I = ∬ D sin y 2 cos x 2 d x d y . \iint_D\sin x^2\cos y^2\text{d}x\text{d}y=I=\iint_D\sin y^2\cos x^2\text{d}x\text{d}y. ∬ D sin x 2 cos y 2 d x d y = I = ∬ D sin y 2 cos x 2 d x d y .
2 I = ∬ D sin x 2 cos y 2 d x d y + ∬ D sin y 2 cos x 2 d x d y = ∬ D ( sin x 2 cos y 2 + cos x 2 sin y 2 ) d x d y = ∬ D sin ( x 2 + y 2 ) d x d y . \begin{align*}
2I&=\iint_D\sin x^2\cos y^2\text{d}x\text{d}y+\iint_D\sin y^2\cos x^2\text{d}x\text{d}y\\
&=\iint_D(\sin x^2\cos y^2+\cos x^2\sin y^2)\text{d}x\text{d}y=\iint_D\sin(x^2+y^2)\text{d}x\text{d}y.
\end{align*}
2 I = ∬ D sin x 2 cos y 2 d x d y + ∬ D sin y 2 cos x 2 d x d y = ∬ D ( sin x 2 cos y 2 + cos x 2 sin y 2 ) d x d y = ∬ D sin ( x 2 + y 2 ) d x d y .
可令{ x = r cos θ y = r sin θ . \begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}. { x = r cos θ y = r sin θ .
2 I = ∬ D sin r 2 r d r d θ = 1 2 ∫ 0 2 π d θ ∫ r = 0 r = a sin r 2 d ( r 2 ) = 1 2 ∫ 0 2 π ( ( − cos r 2 ) ∣ r = 0 r = a ) d θ = − 1 2 ∫ 0 2 π ( cos a 2 − 1 ) d θ = − 1 2 × 2 π ( cos a 2 − 1 ) = π ( 1 − cos a 2 ) . \begin{align*}
2I&=\iint_D\sin r^2r\text{d}r\text{d}\theta=\frac{1}{2}\int_0^{2\pi}\text{d}\theta\int_{r=0}^{r=a}\sin r^2\text{d}(r^2)\\
&=\frac{1}{2}\int_0^{2\pi}\Big((-\cos r^2)\big|_{r=0}^{r=a}\Big)\text{d}\theta=-\frac{1}{2}\int_0^{2\pi}(\cos a^2-1)\text{d}\theta\\
&=-\frac{1}{2}\times2\pi(\cos a^2-1)=\pi(1-\cos a^2).
\end{align*}
2 I = ∬ D sin r 2 r d r d θ = 2 1 ∫ 0 2 π d θ ∫ r = 0 r = a sin r 2 d ( r 2 ) = 2 1 ∫ 0 2 π ( ( − cos r 2 ) r = 0 r = a ) d θ = − 2 1 ∫ 0 2 π ( cos a 2 − 1 ) d θ = − 2 1 × 2 π ( cos a 2 − 1 ) = π ( 1 − cos a 2 ) .
故所求为I = π 2 ( 1 − cos a 2 ) . I=\frac{\pi}{2}(1-\cos a^2). I = 2 π ( 1 − cos a 2 ) .
e.g.6.8 \textbf{e.g.6.8 } e.g.6.8 I = ∬ D x sin ( π x 2 + y 2 ) x + y d x d y . I=\iint_D\frac{x\sin(\pi\sqrt{x^2+y^2})}{x+y}\text{d}x\text{d}y. I = ∬ D x + y x s i n ( π x 2 + y 2 ) d x d y . D = { ( x , y ) ∣ 1 ≤ x 2 + y 2 ≤ 4 , x ≤ 0 , y ≤ 0 } . D=\{(x,y)|1\leq x^2+y^2\leq 4,x\leq 0,y\leq 0\}. D = {( x , y ) ∣1 ≤ x 2 + y 2 ≤ 4 , x ≤ 0 , y ≤ 0 } .
Solution \textbf{Solution } Solution D D D
D x y x 2 + y 2 = 4 x 2 + y 2 = 1
显然这个图线关于y = x y=x y = x ∬ D x sin ( π x 2 + y 2 ) x + y d x d y = I = ∬ D y sin ( π y 2 + x 2 ) y + x d x d y . \iint_D\frac{x\sin(\pi\sqrt{x^2+y^2})}{x+y}\text{d}x\text{d}y=I=\iint_D\frac{y\sin(\pi\sqrt{y^2+x^2})}{y+x}\text{d}x\text{d}y. ∬ D x + y x s i n ( π x 2 + y 2 ) d x d y = I = ∬ D y + x y s i n ( π y 2 + x 2 ) d x d y .
2 I = ∬ D x sin ( π x 2 + y 2 ) x + y d x d y + ∬ D y sin ( π y 2 + x 2 ) y + x d x d y = ∬ D ( x + y ) sin ( π x 2 + y 2 ) x + y d x d y = ∬ D sin ( π x 2 + y 2 ) = Polar Coordinates ∫ 0 π 2 d θ ∫ 1 2 sin π r d r = 分部积分 ∫ 0 π 2 ( ( − r π cos π r + 1 π 2 sin π r ) ∣ r = 1 r = 2 ) d θ = π 2 × ( − 3 π ) = − 3 2 . \begin{align*}
2I&=\iint_D\frac{x\sin(\pi\sqrt{x^2+y^2})}{x+y}\text{d}x\text{d}y+\iint_D\frac{y\sin(\pi\sqrt{y^2+x^2})}{y+x}\text{d}x\text{d}y\\
&=\iint_D\frac{(x+y)\sin(\pi\sqrt{x^2+y^2})}{x+y}\text{d}x\text{d}y=\iint_D\sin(\pi\sqrt{x^2+y^2})\\
&\xlongequal{\text{Polar Coordinates}}\int_0^\frac{\pi}{2}\text{d}\theta\int_1^2\sin\pi r\text{d}r\xlongequal{\text{分部积分}}\int_0^\frac{\pi}{2}\bigg(\Big(-\frac{r}{\pi}\cos\pi r+\frac{1}{\pi^2}\sin\pi r\Big)\Big|_{r=1}^{r=2}\bigg)\text{d}\theta\\
&=\frac{\pi}{2}\times\Big(-\frac{3}{\pi}\Big)=-\frac{3}{2}.
\end{align*}
2 I = ∬ D x + y x sin ( π x 2 + y 2 ) d x d y + ∬ D y + x y sin ( π y 2 + x 2 ) d x d y = ∬ D x + y ( x + y ) sin ( π x 2 + y 2 ) d x d y = ∬ D sin ( π x 2 + y 2 ) Polar Coordinates ∫ 0 2 π d θ ∫ 1 2 sin π r d r 分部积分 ∫ 0 2 π ( ( − π r cos π r + π 2 1 sin π r ) r = 1 r = 2 ) d θ = 2 π × ( − π 3 ) = − 2 3 .
因此所求为I = − 3 4 . I=-\frac{3}{4}. I = − 4 3 .
二重积分的质心坐标
对几何中心重心 坐标为( X ‾ , Y ‾ ) (\overline{X},\overline{Y}) ( X , Y ) D , D, D , ∬ D x d x d y = X ‾ S D , ∬ D y d x d y = Y ‾ S D . \iint_Dx\text{d}x\text{d}y=\overline{X}S_D,\iint_Dy\text{d}x\text{d}y=\overline{Y}S_D. ∬ D x d x d y = X S D , ∬ D y d x d y = Y S D . S D S_D S D D D D
利用质心坐标可以有效简化一部分二重积分的计算。
级数
一个无穷数列的各项用加号连接得到的式子。∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n s i = u 1 + u 2 + ⋯ + u i s_i=u_1+u_2+\cdots+u_i s i = u 1 + u 2 + ⋯ + u i { s n } \{s_n\} { s n } lim n → ∞ s n = S , \lim_{n\rightarrow\infty}s_n=S, lim n → ∞ s n = S , { u n } \{u_n\} { u n } ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n S S S S = u 1 + u 2 + ⋯ + u n + ⋯ . S=u_1+u_2+\cdots+u_n+\cdots. S = u 1 + u 2 + ⋯ + u n + ⋯ .
数项级数的收敛性定义可以得到结论：
lim n → ∞ a n 存在 ⟺ ∑ n = 1 ∞ ( a n + 1 − a n ) 收敛 . \lim_{n\rightarrow \infty}a_n\text{存在}\iff\sum_{n=1}^\infty(a_{n+1}-a_n)\text{收敛}.
n → ∞ lim a n 存在 ⟺ n = 1 ∑ ∞ ( a n + 1 − a n ) 收敛 .
级数本身有如下性质：
若k ≠ 0 , k≠0, k = 0 , ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n ∑ n = 1 ∞ k u n \sum_{n=1}^\infty ku_n ∑ n = 1 ∞ k u n ∑ n = 1 ∞ u n = S \sum_{n=1}^\infty u_n=S ∑ n = 1 ∞ u n = S ∑ n = 1 ∞ k u n = k S . \sum_{n=1}^\infty ku_n=kS. ∑ n = 1 ∞ k u n = k S .
在级数的前面增删有限个数的项，级数的收敛性不会发生变化。
收敛级数在不改变各项顺序的情况下可以任意添加括号，级数之和与收敛性都不会发生变化。
加上括号后的级数发散时原级数一定发散。
收敛级数去掉括号后不一定收敛（举例：∑ n = 1 ∞ ( ( − 1 ) 2 n − 1 + ( − 1 ) 2 n ) \sum_{n=1}^\infty\big((-1)^{2n-1}+(-1)^{2n}\big) ∑ n = 1 ∞ ( ( − 1 ) 2 n − 1 + ( − 1 ) 2 n ) ∑ n = 1 ∞ ( − 1 ) n − 1 \sum_{n=1}^\infty(-1)^{n-1} ∑ n = 1 ∞ ( − 1 ) n − 1
∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n lim n → ∞ u n = 0. \lim_{n\rightarrow\infty}u_n=0. lim n → ∞ u n = 0. lim n → ∞ u n ≠ 0 , \lim_{n\rightarrow\infty}u_n≠0, lim n → ∞ u n = 0 , ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n
e.g.7.1 \textbf{e.g.7.1 } e.g.7.1 f ( x ) f(x) f ( x ) f ( 0 ) = 1 , 0 < f ′ ( x ) < 1 2 , f(0)=1,0<f'(x)<\frac{1}{2}, f ( 0 ) = 1 , 0 < f ′ ( x ) < 2 1 , { x n } \{x_n\} { x n } x n + 1 = f ( x n ) , x_{n+1}=f(x_n), x n + 1 = f ( x n ) ,
(1)级数∑ n = 1 ∞ ( x n + 1 − x n ) \sum_{n=1}^\infty(x_{n+1}-x_n) ∑ n = 1 ∞ ( x n + 1 − x n )
(2)lim n → ∞ x n \lim_{n\rightarrow\infty}x_n lim n → ∞ x n ( 0 , 2 ) (0,2) ( 0 , 2 )
(1)Proof \textbf{Proof } Proof f ( x ) f(x) f ( x ) ∀ a , b ∈ R , ∃ ξ ∈ ( a , b ) , s.t. f ′ ( ξ ) = f ( b ) − f ( a ) b − a . \forall a,b\in\mathbb{R},\exists\xi\in(a,b),\text{ s.t. }f'(\xi)=\frac{f(b)-f(a)}{b-a}. ∀ a , b ∈ R , ∃ ξ ∈ ( a , b ) , s.t. f ′ ( ξ ) = b − a f ( b ) − f ( a ) .
∣ x n + 1 − x n ∣ = ∣ f ( x n ) − f ( x n − 1 ) ∣ = f ′ ( ξ ) ∣ x n − x n − 1 ∣ . |x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})|=f'(\xi)|x_n-x_{n-1}|.
∣ x n + 1 − x n ∣ = ∣ f ( x n ) − f ( x n − 1 ) ∣ = f ′ ( ξ ) ∣ x n − x n − 1 ∣.
这时我们又注意到0 < f ′ ( x ) < 1 2 , 0<f'(x)<\frac{1}{2}, 0 < f ′ ( x ) < 2 1 ,
∣ x n + 1 − x n ∣ = f ′ ( ξ ) ∣ x n − x n − 1 ∣ < 1 2 ∣ x n − x n − 1 ∣ < 1 4 ∣ x n − 1 − x n − 2 ∣ < ⋯ < ( 1 2 ) n − 1 ∣ x 2 − x 1 ∣ . |x_{n+1}-x_n|=f'(\xi)|x_n-x_{n-1}|<\frac{1}{2}|x_n-x_{n-1}|<\frac{1}{4}|x_{n-1}-x_{n-2}|<\cdots<\Big(\frac{1}{2}\Big)^{n-1}|x_2-x_1|.
∣ x n + 1 − x n ∣ = f ′ ( ξ ) ∣ x n − x n − 1 ∣ < 2 1 ∣ x n − x n − 1 ∣ < 4 1 ∣ x n − 1 − x n − 2 ∣ < ⋯ < ( 2 1 ) n − 1 ∣ x 2 − x 1 ∣.
显然x 1 , x 2 x_1,x_2 x 1 , x 2 ∑ n = 1 ∞ ( 1 2 ) n − 1 \sum_{n=1}^\infty\big(\frac{1}{2}\big)^{n-1} ∑ n = 1 ∞ ( 2 1 ) n − 1 ∑ n = 1 ∞ ( 1 2 ) n − 1 ∣ x 2 − x 1 ∣ \sum_{n=1}^\infty\big(\frac{1}{2}\big)^{n-1}|x_2-x_1| ∑ n = 1 ∞ ( 2 1 ) n − 1 ∣ x 2 − x 1 ∣ ∑ n = 1 ∞ ∣ x n + 1 − x n ∣ \sum_{n=1}^\infty|x_{n+1}-x_n| ∑ n = 1 ∞ ∣ x n + 1 − x n ∣ f ′ ( x ) > 0 f'(x)>0 f ′ ( x ) > 0 ∣ x n + 1 − x n ∣ = x n + 1 − x n , |x_{n+1}-x_n|=x_{n+1}-x_n, ∣ x n + 1 − x n ∣ = x n + 1 − x n ,
Q.E.D. \textbf{Q.E.D.} Q.E.D.
(2)Proof \textbf{Proof } Proof ∑ n = 1 ∞ ( x n + 1 − x n ) \sum_{n=1}^\infty(x_{n+1}-x_n) ∑ n = 1 ∞ ( x n + 1 − x n )
x n + 1 = x 1 + ( x 2 − x 1 ) + ( x 3 − x 2 ) + ⋯ + ( x n + 1 − x n ) = x 1 + ∑ n = 1 ∞ ( x n + 1 − x n ) \begin{align*}
x_{n+1}&=x_1+(x_2-x_1)+(x_3-x_2)+\cdots+(x_{n+1}-x_n)\\
&=x_1+\sum_{n=1}^\infty(x_{n+1}-x_n)
\end{align*}
x n + 1 = x 1 + ( x 2 − x 1 ) + ( x 3 − x 2 ) + ⋯ + ( x n + 1 − x n ) = x 1 + n = 1 ∑ ∞ ( x n + 1 − x n )
也就收敛；换言之，{ x n } \{x_n\} { x n } lim n → ∞ x n \lim_{n\rightarrow\infty}x_n lim n → ∞ x n c , c, c , f ( c ) = c . f(c)=c. f ( c ) = c . g ( x ) = f ( x ) − x , g(x)=f(x)-x, g ( x ) = f ( x ) − x , g ′ ( x ) = f ′ ( x ) − 1 < 0 , g'(x)=f'(x)-1<0, g ′ ( x ) = f ′ ( x ) − 1 < 0 , g ( x ) g(x) g ( x ) g ( 0 ) = f ( 0 ) − 0 = 1 > 0 , g ( 2 ) = f ( 2 ) − 2 < f ( 0 ) + 1 2 × ( 2 − 0 ) − 2 = 0 , g(0)=f(0)-0=1>0,g(2)=f(2)-2<f(0)+\frac{1}{2}\times(2-0)-2=0, g ( 0 ) = f ( 0 ) − 0 = 1 > 0 , g ( 2 ) = f ( 2 ) − 2 < f ( 0 ) + 2 1 × ( 2 − 0 ) − 2 = 0 , g ( x ) = 0 g(x)=0 g ( x ) = 0 ( 0 , 2 ) (0,2) ( 0 , 2 ) c ∈ ( 0 , 2 ) , c\in(0,2), c ∈ ( 0 , 2 ) ,
Q.E.D. \textbf{Q.E.D.} Q.E.D.
正项级数敛散性判别
我们知道对0 ≤ u n ≤ v n 0\leq u_n\leq v_n 0 ≤ u n ≤ v n ∑ n = 1 ∞ v n \sum_{n=1}^\infty v_n ∑ n = 1 ∞ v n ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n ∑ n = 1 ∞ v n \sum_{n=1}^\infty v_n ∑ n = 1 ∞ v n p p p ∑ n = 1 ∞ 1 n p \sum_{n=1}^\infty\frac{1}{n^p} ∑ n = 1 ∞ n p 1 p ≤ 1 p\leq 1 p ≤ 1 ∑ n = 1 ∞ a q n \sum_{n=1}^\infty aq^n ∑ n = 1 ∞ a q n a ≠ 0. ∣ q ∣ > 1 a≠0.|q|>1 a = 0.∣ q ∣ > 1 a 1 − q \frac{a}{1-q} 1 − q a
这种「比较判别法」对应也有比例极限形式。如果lim n → ∞ u n v n = l , \lim_{n\rightarrow\infty}\frac{u_n}{v_n}=l, lim n → ∞ v n u n = l , l ∈ ( 0 , + ∞ ) l\in(0,+\infty) l ∈ ( 0 , + ∞ ) ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n ∑ n = 1 ∞ v n \sum_{n=1}^\infty v_n ∑ n = 1 ∞ v n l = 0 l=0 l = 0 ∑ n = 1 ∞ v n \sum_{n=1}^\infty v_n ∑ n = 1 ∞ v n ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n l = + ∞ l=+\infty l = + ∞ ∑ n = 1 ∞ v n \sum_{n=1}^\infty v_n ∑ n = 1 ∞ v n ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n
选用比较判别法的情况通常是正项级数含有n n , n ! , a n , n k n^n,n!,a^n,n^k n n , n ! , a n , n k u n + 1 u n \frac{u_{n+1}}{u_n} u n u n + 1
e.g.7.2 \textbf{e.g.7.2 } e.g.7.2 ∑ n = 1 ∞ ( 1 n − ln ( 1 + 1 n ) ) \sum_{n=1}^\infty\big(\frac{1}{n}-\ln(1+\frac{1}{n})\big) ∑ n = 1 ∞ ( n 1 − ln ( 1 + n 1 ) )
Solution n → ∞ \textbf{Solution }n\rightarrow\infty Solution n → ∞ 1 n − ln ( 1 + 1 n ) ∼ 1 2 ⋅ 1 n 2 . \frac{1}{n}-\ln(1+\frac{1}{n})\sim\frac{1}{2}\cdot\frac{1}{n^2}. n 1 − ln ( 1 + n 1 ) ∼ 2 1 ⋅ n 2 1 . ∑ n = 1 ∞ 1 2 n 2 \sum_{n=1}^\infty\frac{1}{2n^2} ∑ n = 1 ∞ 2 n 2 1
e.g.7.3 \textbf{e.g.7.3 } e.g.7.3 a n = ∫ 0 π 4 tan n x d x , a_n=\int_0^\frac{\pi}{4}\tan^nx\text{d}x, a n = ∫ 0 4 π tan n x d x ,
(1)求∑ n = 1 ∞ 1 n ( a n + a n + 2 ) . \sum_{n=1}^\infty\frac{1}{n}(a_n+a_{n+2}). ∑ n = 1 ∞ n 1 ( a n + a n + 2 ) .
(2)求证：∀ λ > 0 , ∑ n = 1 ∞ a n n λ \forall\lambda>0,\sum_{n=1}^\infty\frac{a_n}{n^\lambda} ∀ λ > 0 , ∑ n = 1 ∞ n λ a n
(1)Solution \textbf{Solution } Solution
a n + a n + 2 = ∫ 0 π 4 tan n x ( 1 + tan 2 x ) d x = 1 n + 1 tan n + 1 x ∣ 0 π 4 = 1 n + 1 , a n + a n − 2 = 1 n − 1 . \begin{align*}
a_n+a_{n+2}&=\int_0^\frac{\pi}{4}\tan^nx(1+\tan^2x)\text{d}x=\frac{1}{n+1}\tan^{n+1}x\Big|_0^\frac{\pi}{4}=\frac{1}{n+1},\\
a_n+a_{n-2}&=\frac{1}{n-1}.
\end{align*}
a n + a n + 2 a n + a n − 2 = ∫ 0 4 π tan n x ( 1 + tan 2 x ) d x = n + 1 1 tan n + 1 x 0 4 π = n + 1 1 , = n − 1 1 .
因此
O.F. = ∑ n = 1 ∞ 1 n × 1 ( n + 1 ) = ∑ n = 1 ∞ 1 n ( n + 1 ) = ∑ n = 1 ∞ ( 1 n − 1 n + 1 ) = 1 − 1 n + 1 . \begin{align*}
\text{O.F.}&=\sum_{n=1}^\infty\frac{1}{n}\times\frac{1}{(n+1)}=\sum_{n=1}^\infty\frac{1}{n(n+1)}\\
&=\sum_{n=1}^\infty\Big(\frac{1}{n}-\frac{1}{n+1}\Big)=1-\frac{1}{n+1}.
\end{align*}
O.F. = n = 1 ∑ ∞ n 1 × ( n + 1 ) 1 = n = 1 ∑ ∞ n ( n + 1 ) 1 = n = 1 ∑ ∞ ( n 1 − n + 1 1 ) = 1 − n + 1 1 .
(2)Proof \textbf{Proof } Proof a n = ∫ 0 π 4 tan n x d x = t = tan x ∫ 0 1 t n 1 + t 2 d t < ∫ 0 1 t n d t = 1 n + 1 t n + 1 ∣ 0 1 = 1 n + 1 , a_n=\int_0^\frac{\pi}{4}\tan^nx\text{d}x\xlongequal{t=\tan x}\int_0^1\frac{t^n}{1+t^2}\text{d}t<\int_0^1t^n\text{d}t=\frac{1}{n+1}t^{n+1}\big|_0^1=\frac{1}{n+1}, a n = ∫ 0 4 π tan n x d x t = t a n x ∫ 0 1 1 + t 2 t n d t < ∫ 0 1 t n d t = n + 1 1 t n + 1 0 1 = n + 1 1 ,
a n n λ < 1 ( n + 1 ) n λ < 1 n ⋅ n λ . \frac{a_n}{n^\lambda}<\frac{1}{(n+1)n^\lambda}<\frac{1}{n\cdot n^\lambda}.
n λ a n < ( n + 1 ) n λ 1 < n ⋅ n λ 1 .
而级数∑ n = 1 ∞ 1 n λ + 1 \sum_{n=1}^\infty\frac{1}{n^{\lambda+1}} ∑ n = 1 ∞ n λ + 1 1 λ > 0 \lambda>0 λ > 0
Q.E.D. \textbf{Q.E.D.} Q.E.D.
（D’Alembert判别法）对正项级数∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n lim n → ∞ u n + 1 u n = ρ , \lim_{n\rightarrow\infty}\frac{u_{n+1}}{u_n}=\rho, lim n → ∞ u n u n + 1 = ρ ,
ρ < 1 ⇒ ∑ n = 1 ∞ u n \rho<1\Rightarrow\sum_{n=1}^\infty u_n ρ < 1 ⇒ ∑ n = 1 ∞ u n ρ > 1 ⇒ ∑ n = 1 ∞ u n \rho>1\Rightarrow\sum_{n=1}^\infty u_n ρ > 1 ⇒ ∑ n = 1 ∞ u n ρ = 1 \rho=1 ρ = 1
（Cauchy根值判别法）对lim n → ∞ u n n = ρ : \lim_{n\rightarrow\infty}\sqrt[n]{u_n}=\rho: lim n → ∞ n u n = ρ :
ρ < 1 ⇒ ∑ n = 1 ∞ u n \rho<1\Rightarrow\sum_{n=1}^\infty u_n ρ < 1 ⇒ ∑ n = 1 ∞ u n ρ > 1 ⇒ ∑ n = 1 ∞ u n \rho>1\Rightarrow\sum_{n=1}^\infty u_n ρ > 1 ⇒ ∑ n = 1 ∞ u n ρ = 1 \rho=1 ρ = 1
在正项级数的一般项出现幂次形式时，一般会先考虑使用Cauchy根值判别法；如果此法失效才会考虑使用比较判别法。
现在仔细考虑比较判别法。我们知道u n > 0 , ρ > 0 u_n>0,\rho>0 u n > 0 , ρ > 0 lim n → ∞ u n + 1 u n = ρ \lim_{n\rightarrow\infty}\frac{u_{n+1}}{u_n}=\rho lim n → ∞ u n u n + 1 = ρ lim n → ∞ u n n = ρ \lim_{n\rightarrow\infty}\sqrt[n]{u_n}=\rho lim n → ∞ n u n = ρ lim n → ∞ u n + 1 u n \lim_{n\rightarrow\infty}\frac{u_{n+1}}{u_n} lim n → ∞ u n u n + 1 lim n → ∞ u n n \lim_{n\rightarrow\infty}\sqrt[n]{u_n} lim n → ∞ n u n ∑ n = 1 ∞ 2 − n − ( − 1 ) n . \sum_{n=1}^\infty 2^{-n-(-1)^n}. ∑ n = 1 ∞ 2 − n − ( − 1 ) n .
e.g.7.4 \textbf{e.g.7.4 } e.g.7.4 ∑ n = 1 ∞ n cos 2 n π 3 2 n \sum_{n=1}^\infty\frac{n\cos^2\frac{n\pi}{3}}{2^n} ∑ n = 1 ∞ 2 n n c o s 2 3 nπ
Solution \textbf{Solution } Solution n cos 2 n π 3 2 n ≤ n 2 n . \frac{n\cos^2\frac{n\pi}{3}}{2^n}\leq\frac{n}{2^n}. 2 n n c o s 2 3 nπ ≤ 2 n n . lim n → ∞ n 2 n n = 1 2 < 1 \lim_{n\rightarrow\infty}\sqrt[n]{\frac{n}{2^n}}=\frac{1}{2}<1 lim n → ∞ n 2 n n = 2 1 < 1 ∑ n = 1 ∞ n 2 n \sum_{n=1}^\infty\frac{n}{2^n} ∑ n = 1 ∞ 2 n n
e.g.7.5 \textbf{e.g.7.5 } e.g.7.5 ∑ n = 1 ∞ a n ( a n ≥ 0 ) \sum_{n=1}^\infty a_n(a_n\geq0) ∑ n = 1 ∞ a n ( a n ≥ 0 ) ∑ n = 1 ∞ a n 2 , ∑ n = 1 ∞ a n n , ∑ n = 1 ∞ a n 1 + a n \sum_{n=1}^\infty{a_n}^2,\sum_{n=1}^\infty\frac{\sqrt{a_n}}{n},\sum_{n=1}^\infty\frac{a_n}{1+a_n} ∑ n = 1 ∞ a n 2 , ∑ n = 1 ∞ n a n , ∑ n = 1 ∞ 1 + a n a n
Proof \textbf{Proof } Proof
lim n → ∞ a n + 1 a n < 1. \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}<1.
n → ∞ lim a n a n + 1 < 1.
因此就有
lim n → ∞ ( a n + 1 a n ) 2 < 1 , lim n → ∞ a n + 1 n 2 a n n 2 < 1. \begin{align}
\lim_{n\rightarrow\infty}\Big(\frac{a_{n+1}}{a_n}\Big)^2&<1,\tag{1}\\
\lim_{n\rightarrow\infty}\frac{\frac{a_{n+1}}{n^2}}{\frac{a_n}{n^2}}&<1.\tag{2}
\end{align}
n → ∞ lim ( a n a n + 1 ) 2 n → ∞ lim n 2 a n n 2 a n + 1 < 1 , < 1. ( 1 ) ( 2 )
由式( 1 ) (1) ( 1 )
lim n → ∞ a n + 1 2 a n 2 < 1 , \lim_{n\rightarrow\infty}\frac{a_{n+1}^2}{a_n^2}<1,
n → ∞ lim a n 2 a n + 1 2 < 1 ,
故∑ n = 1 ∞ a n 2 \sum_{n=1}^\infty{a_n}^2 ∑ n = 1 ∞ a n 2
由式( 2 ) (2) ( 2 )
lim n → ∞ a n + 1 ( n + 1 ) 2 a n n 2 < n 2 ( n + 1 ) 2 < 1 , \lim_{n\rightarrow\infty}\frac{\frac{a_{n+1}}{(n+1)^2}}{\frac{a_n}{n^2}}<\frac{n^2}{(n+1)^2}<1,
n → ∞ lim n 2 a n ( n + 1 ) 2 a n + 1 < ( n + 1 ) 2 n 2 < 1 ,
进而
lim n → ∞ a n + 1 ( n + 1 ) 2 a n n 2 < 1 ⇒ lim n → ∞ a n + 1 n + 1 a n n < 1 , \lim_{n\rightarrow\infty}\sqrt{\frac{\frac{a_{n+1}}{(n+1)^2}}{\frac{a_n}{n^2}}}<1\Rightarrow\lim_{n\rightarrow\infty}\frac{\frac{\sqrt{a_{n+1}}}{n+1}}{\frac{\sqrt{a_n}}{n}}<1,
n → ∞ lim n 2 a n ( n + 1 ) 2 a n + 1 < 1 ⇒ n → ∞ lim n a n n + 1 a n + 1 < 1 ,
故∑ n = 1 ∞ a n n \sum_{n=1}^\infty\frac{\sqrt{a_n}}{n} ∑ n = 1 ∞ n a n
而a n ≥ 0 a_n\geq 0 a n ≥ 0 a n 1 + a n < a n , \frac{a_n}{1+a_n}<a_n, 1 + a n a n < a n , ∑ n = 1 ∞ a n \sum_{n=1}^\infty a_n ∑ n = 1 ∞ a n ∑ n = 1 ∞ a n 1 + a n \sum_{n=1}^\infty\frac{a_n}{1+a_n} ∑ n = 1 ∞ 1 + a n a n
Q.E.D. \textbf{Q.E.D.} Q.E.D.
（积分判别法）若单调减少的非负函数f ( x ) ( 1 ≤ x ≤ + ∞ ) f(x)(1\leq x\leq +\infty) f ( x ) ( 1 ≤ x ≤ + ∞ ) u n = f ( n ) , u_n=f(n), u n = f ( n ) , ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n ∫ 1 + ∞ f ( x ) d x \int_1^{+\infty}f(x)\text{d}x ∫ 1 + ∞ f ( x ) d x
e.g.7.6 \textbf{e.g.7.6 } e.g.7.6 ∑ n = 2 ∞ 1 n ln p n \sum_{n=2}^\infty\frac{1}{n\ln^pn} ∑ n = 2 ∞ n l n p n 1
Solution \textbf{Solution } Solution f ( x ) = 1 x ln p x . f(x)=\frac{1}{x\ln^px}. f ( x ) = x l n p x 1 . ∫ 2 ∞ f ( x ) d x , \int_2^\infty f(x)\text{d}x, ∫ 2 ∞ f ( x ) d x ,
∫ 2 ∞ 1 x ln p x d x = ∫ 2 ∞ ln − p x d ( ln x ) = ln 1 − p x 1 − p ∣ 2 ∞ . \begin{align*}
\int_2^\infty\frac{1}{x\ln^px}\text{d}x&=\int_2^\infty\ln^{-p}x\text{d}(\ln x)=\frac{\ln^{1-p}x}{1-p}\Big|_2^\infty.
\end{align*}
∫ 2 ∞ x ln p x 1 d x = ∫ 2 ∞ ln − p x d ( ln x ) = 1 − p ln 1 − p x 2 ∞ .
因此当p > 1 p>1 p > 1
交错级数｜任意项级数
交错级数
形如∑ n = 1 ∞ ( − 1 ) n u n \sum_{n=1}^\infty(-1)^nu_n ∑ n = 1 ∞ ( − 1 ) n u n ∑ n = 1 ∞ ( − 1 ) n − 1 u n \sum_{n=1}^\infty(-1)^{n-1}u_n ∑ n = 1 ∞ ( − 1 ) n − 1 u n u n > 0. u_n>0. u n > 0.
通俗地说，交错级数就是每个项正负交错。
（Leibniz判别法）如果交错级数∑ n = 1 ∞ ( − 1 ) n u n \sum_{n=1}^\infty(-1)^nu_n ∑ n = 1 ∞ ( − 1 ) n u n u n + 1 ≤ u n u_{n+1}\leq u_n u n + 1 ≤ u n lim n → ∞ u n = 0 , \lim_{n\rightarrow\infty u_n}=0, lim n → ∞ u n = 0 ,
Leibniz判别法是交错级数收敛的充分不必要 条件；这是因为，如果u n + 1 > u n u_{n+1}>u_n u n + 1 > u n
绝对收敛｜条件收敛（任意项级数）
毫无规律的级数即任意项级数。对于任意项级数∑ n = 1 ∞ u n , \sum_{n=1}^\infty u_n, ∑ n = 1 ∞ u n , ∑ n = 1 ∞ ∣ u n ∣ \sum_{n=1}^\infty |u_n| ∑ n = 1 ∞ ∣ u n ∣ 绝对收敛 。如果∑ n = 1 ∞ ∣ u n ∣ \sum_{n=1}^\infty |u_n| ∑ n = 1 ∞ ∣ u n ∣ ∑ n = 1 ∞ u n , \sum_{n=1}^\infty u_n, ∑ n = 1 ∞ u n , ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n 条件收敛 。
根据绝对收敛和条件收敛的定义，不难得到一般项级数判断敛散性的通用流程。要求∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n ∑ n = 1 ∞ ∣ u n ∣ \sum_{n=1}^\infty |u_n| ∑ n = 1 ∞ ∣ u n ∣ ∑ n = 1 ∞ ∣ u n ∣ \sum_{n=1}^\infty |u_n| ∑ n = 1 ∞ ∣ u n ∣
e.g.7.7 \textbf{e.g.7.7 } e.g.7.7 k > 0 , k>0, k > 0 , ∑ n = 1 ∞ ( − 1 ) n k + n n 2 \sum_{n=1}^\infty(-1)^n\frac{k+n}{n^2} ∑ n = 1 ∞ ( − 1 ) n n 2 k + n
A. 发散
B. 绝对收敛
C. 条件收敛
D. 敛散性仅与k有关
Analyses \textbf{Analyses } Analyses u n = ( − 1 ) n k + n n 2 u_n=(-1)^n\frac{k+n}{n^2} u n = ( − 1 ) n n 2 k + n ∑ n = 1 ∞ ∣ u n ∣ \sum_{n=1}^\infty|u_n| ∑ n = 1 ∞ ∣ u n ∣ { ∣ u n ∣ } \{|u_n|\} { ∣ u n ∣ } { u n } \{u_n\} { u n }
Choice \textbf{Choice } Choice
e.g.7.8 \textbf{e.g.7.8 } e.g.7.8 λ > 0 \lambda>0 λ > 0 ∑ n = 1 ∞ a n 2 \sum_{n=1}^\infty a_n^2 ∑ n = 1 ∞ a n 2 ∑ n = 1 ∞ ( − 1 ) n ∣ a n ∣ n 2 + λ \sum_{n=1}^\infty\frac{(-1)^n|a_n|}{\sqrt{n^2+\lambda}} ∑ n = 1 ∞ n 2 + λ ( − 1 ) n ∣ a n ∣
A. 发散
B. 绝对收敛
C. 条件收敛
D. 敛散性仅与λ有关
Analyses \textbf{Analyses } Analyses ∣ a n ∣ n 2 + λ ≤ 1 2 ( a n 2 + 1 n 2 + λ ) , \frac{|a_n|}{\sqrt{n^2+\lambda}}\leq\frac{1}{2}\big(a_n^2+\frac{1}{n^2+\lambda}\big), n 2 + λ ∣ a n ∣ ≤ 2 1 ( a n 2 + n 2 + λ 1 ) , a n 2 a_n^2 a n 2 1 n 2 + λ \frac{1}{n^2+\lambda} n 2 + λ 1 ∣ a n ∣ n 2 + λ \frac{|a_n|}{\sqrt{n^2+\lambda}} n 2 + λ ∣ a n ∣ u n = ( − 1 ) n ∣ a n ∣ n 2 + λ u_n=\frac{(-1)^n|a_n|}{\sqrt{n^2+\lambda}} u n = n 2 + λ ( − 1 ) n ∣ a n ∣ ∑ n = 1 ∞ ∣ u n ∣ \sum_{n=1}^\infty|u_n| ∑ n = 1 ∞ ∣ u n ∣ ∑ n = 1 ∞ u n \sum_{n=1}^\infty u_n ∑ n = 1 ∞ u n
Choice \textbf{Choice } Choice
e.g.7.9 \textbf{e.g.7.9 } e.g.7.9 a n > 0 , λ ∈ ( 0 , π 2 ) a_n>0,\lambda\in\big(0,\frac{\pi}{2}\big) a n > 0 , λ ∈ ( 0 , 2 π ) ∑ n = 1 ∞ a n \sum_{n=1}^\infty a_n ∑ n = 1 ∞ a n ∑ n = 1 ∞ ( − 1 ) n ( n tan λ n ) a 2 n \sum_{n=1}^\infty(-1)^n\big(n\tan\frac{\lambda}{n}\big)a_{2n} ∑ n = 1 ∞ ( − 1 ) n ( n tan n λ ) a 2 n
A. 发散
B. 绝对收敛
C. 条件收敛
D. 敛散性仅与λ有关
Analyses \textbf{Analyses } Analyses u n = ( − 1 ) n ( n tan λ n ) a 2 n , u_n=(-1)^n\big(n\tan\frac{\lambda}{n}\big)a_{2n}, u n = ( − 1 ) n ( n tan n λ ) a 2 n , n → ∞ n\rightarrow\infty n → ∞ n tan λ n ∼ λ , n\tan\frac{\lambda}{n}\sim\lambda, n tan n λ ∼ λ , ∑ n = 1 a n \sum_{n=1}a_n ∑ n = 1 a n ∑ n = 1 a 2 n \sum_{n=1}a_{2n} ∑ n = 1 a 2 n ∑ n = 1 ∣ u n ∣ \sum_{n=1}|u_n| ∑ n = 1 ∣ u n ∣
Choice \textbf{Choice } Choice
特别地，对选择题判断收敛性的情景下，常用下面几个反例：
条件收敛
绝对收敛
∑ n = 1 ∞ ( − 1 ) n 1 n \sum_{n=1}^\infty (-1)^n\frac{1}{n} ∑ n = 1 ∞ ( − 1 ) n n 1
∑ n = 1 ∞ ( − 1 ) n 1 n \sum_{n=1}^\infty (-1)^n\frac{1}{\sqrt{n}} ∑ n = 1 ∞ ( − 1 ) n n 1
∑ n = 1 ∞ ( − 1 ) n 1 n 2 \sum_{n=1}^\infty (-1)^n\frac{1}{n^2} ∑ n = 1 ∞ ( − 1 ) n n 2 1
一般地，如果∑ n = 1 ∞ a n \sum_{n=1}^\infty a_n ∑ n = 1 ∞ a n ∑ n = 1 ∞ ( 1 2 ( a n + ∣ a n ∣ ) ) \sum_{n=1}^\infty \big(\frac{1}{2}(a_n+|a_n|)\big) ∑ n = 1 ∞ ( 2 1 ( a n + ∣ a n ∣ ) ) ∑ n = 1 ∞ ( 1 2 ( a n − ∣ a n ∣ ) ) \sum_{n=1}^\infty \big(\frac{1}{2}(a_n-|a_n|)\big) ∑ n = 1 ∞ ( 2 1 ( a n − ∣ a n ∣ ) )
幂级数
形如∑ n = 0 ∞ a n x n \sum_{n=0}^\infty a_nx^n ∑ n = 0 ∞ a n x n ∑ n = 0 ∞ a n ( x − x 0 ) n \sum_{n=0}^\infty a_n(x-x_0)^n ∑ n = 0 ∞ a n ( x − x 0 ) n
收敛半径｜收敛域
若幂级数∑ n = 0 ∞ a n x n \sum_{n=0}^\infty a_nx^n ∑ n = 0 ∞ a n x n lim n → ∞ a n + 1 a n = ρ \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\rho lim n → ∞ a n a n + 1 = ρ lim n → ∞ a n n = ρ , \lim_{n\rightarrow\infty}\sqrt[n]{a_n}=\rho, lim n → ∞ n a n = ρ ,
R = { 1 ρ , ρ ≠ 0 + ∞ , ρ = 0 0 , ρ = + ∞ R=\begin{cases}\frac{1}{\rho}&,\rho≠0\\+\infty&,\rho=0\\0&,\rho=+\infty\end{cases}
R = ⎩ ⎨ ⎧ ρ 1 + ∞ 0 , ρ = 0 , ρ = 0 , ρ = + ∞
就是这个幂级数的收敛半径。
e.g.7.10 \textbf{e.g.7.10 } e.g.7.10 ∑ n = 1 ∞ ( x − 3 ) n 2 n n 2 \sum_{n=1}^\infty\frac{(x-3)^n}{2^nn^2} ∑ n = 1 ∞ 2 n n 2 ( x − 3 ) n
Solution \textbf{Solution } Solution a n = 1 2 n n 2 , x n = ( x − 3 ) n . a_n=\frac{1}{2^nn^2},x_n=(x-3)^n. a n = 2 n n 2 1 , x n = ( x − 3 ) n . lim n → ∞ a n + 1 a n = lim n → ∞ 1 2 n + 1 ⋅ 1 ( n + 1 ) 2 1 2 n ⋅ 1 n 2 = n 2 2 ( n + 1 ) 2 = 1 2 . \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty}\frac{\frac{1}{2^{n+1}}\cdot\frac{1}{(n+1)^2}}{\frac{1}{2^n}\cdot\frac{1}{n^2}}=\frac{n^2}{2(n+1)^2}=\frac{1}{2}. lim n → ∞ a n a n + 1 = lim n → ∞ 2 n 1 ⋅ n 2 1 2 n + 1 1 ⋅ ( n + 1 ) 2 1 = 2 ( n + 1 ) 2 n 2 = 2 1 . R = 2 , R=2, R = 2 ,
另一方面，x = 1 x=1 x = 1 x = 5 x=5 x = 5 [ 1 , 5 ] . [1,5]. [ 1 , 5 ] .
在利用上方定义求收敛半径公式时一定要注意级数是否“缺项”。如果缺项，这个级数就是一般项级数而不是幂级数，需要用lim n → ∞ ∣ u n + 1 ( x ) u n ( x ) ∣ < 1 \lim_{n\rightarrow\infty}\big|\frac{u_{n+1}(x)}{u_n(x)}\big|<1 lim n → ∞ u n ( x ) u n + 1 ( x ) < 1 x x x
e.g.7.11 \textbf{e.g.7.11 } e.g.7.11 ∑ n = 1 ∞ 1 2 n + ( − 3 ) n x 2 n − 1 \sum_{n=1}^\infty\frac{1}{2^n+(-3)^n}x^{2n-1} ∑ n = 1 ∞ 2 n + ( − 3 ) n 1 x 2 n − 1
Solution \textbf{Solution } Solution a n ( x ) = 1 2 n + ( − 3 ) n x 2 n − 1 . a_n(x)=\frac{1}{2^n+(-3)^n}x^{2n-1}. a n ( x ) = 2 n + ( − 3 ) n 1 x 2 n − 1 . lim n → ∞ ∣ a n + 1 ( x ) a n ( x ) ∣ < 1 \lim_{n\rightarrow\infty}\big|\frac{a_{n+1}(x)}{a_n(x)}\big|<1 lim n → ∞ a n ( x ) a n + 1 ( x ) < 1
lim n → ∞ ∣ x 2 ( 2 n + ( − 3 ) n ) 2 n + 1 + ( − 3 ) n + 1 ∣ = 1 3 x 2 < 1. \lim_{n\rightarrow\infty}\Big|\frac{x^2\big(2^n+(-3)^n\big)}{2^{n+1}+(-3)^{n+1}}\Big|=\frac{1}{3}x^2<1.
n → ∞ lim 2 n + 1 + ( − 3 ) n + 1 x 2 ( 2 n + ( − 3 ) n ) = 3 1 x 2 < 1.
解之得− 3 < x < 3 . -\sqrt{3}<x<\sqrt{3}. − 3 < x < 3 . 3 . \sqrt{3}. 3 .
一般地，如果∑ n = 0 a n x n \sum_{n=0}a_nx^n ∑ n = 0 a n x n ∑ n = 0 b n x n \sum_{n=0}b_nx^n ∑ n = 0 b n x n R 1 , R 2 , R_1,R_2, R 1 , R 2 , ∑ n = 0 ( a n + b n ) x n \sum_{n=0}(a_n+b_n)x^n ∑ n = 0 ( a n + b n ) x n R = min ( R 1 , R 2 ) . R=\min(R_1,R_2). R = min ( R 1 , R 2 ) .
（Abel定理）若幂级数∑ n = 0 a n x n \sum_{n=0}a_nx^n ∑ n = 0 a n x n x = x 0 ≠ 0 x=x_0≠0 x = x 0 = 0 ∣ x ∣ < ∣ x 0 ∣ |x|<|x_0| ∣ x ∣ < ∣ x 0 ∣ x = x 0 ≠ 0 x=x_0≠0 x = x 0 = 0 ∣ x ∣ > ∣ x 0 ∣ |x|>|x_0| ∣ x ∣ > ∣ x 0 ∣
e.g.7.12 \textbf{e.g.7.12 } e.g.7.12 ∑ n = 1 ∞ a n ( x − 1 ) n \sum_{n=1}^\infty a_n(x-1)^n ∑ n = 1 ∞ a n ( x − 1 ) n x = − 1 x=-1 x = − 1 x = 2 x=2 x = 2
A. 发散
B. 绝对收敛
C. 条件收敛
D. 敛散性不确定
Analyses \textbf{Analyses } Analyses x = − 1 x=-1 x = − 1 ( x − 1 ) n = ( − 2 ) n . (x-1)^n=(-2)^n. ( x − 1 ) n = ( − 2 ) n . x = 2 x=2 x = 2 ∣ x − 1 ∣ = 1 < 2 = ∣ − 2 ∣ , |x-1|=1<2=|-2|, ∣ x − 1∣ = 1 < 2 = ∣ − 2∣ ,
Choice \textbf{Choice } Choice
幂级数在收敛区间内逐项积分或求导后收敛半径不会发生变化。然而，收敛域在端点处的敛散性可能会变。
幂级数求和与展开
我们知道幂级数s ( x ) = ∑ n = 0 ∞ a n x n s(x)=\sum_{n=0}^\infty a_nx^n s ( x ) = ∑ n = 0 ∞ a n x n
s ′ ( x ) = ∑ n = 1 ∞ n a n x n − 1 , ∫ 0 x s ( x ) d x = ∑ n = 0 ∞ a n n + 1 x n + 1 , \begin{align*}
s'(x)&=\sum_{n=1}^\infty na_nx^{n-1},\\
\int_0^xs(x)\text{d}x&=\sum_{n=0}^\infty\frac{a_n}{n+1}x^{n+1},
\end{align*}
s ′ ( x ) ∫ 0 x s ( x ) d x = n = 1 ∑ ∞ n a n x n − 1 , = n = 0 ∑ ∞ n + 1 a n x n + 1 ,
因此可以通过逐项求导、求积分把幂级数转化为等比级数，从而求幂级数。这个方法有重要公式如下：
∑ n = 0 ∞ x n = 1 1 − x , ∑ n = 1 ∞ x n x 1 − x . ( ∣ x ∣ < 1 ) \sum_{n=0}^\infty x^n=\frac{1}{1-x},\sum_{n=1}^\infty x^n\frac{x}{1-x}.(|x|<1) ∑ n = 0 ∞ x n = 1 − x 1 , ∑ n = 1 ∞ x n 1 − x x . ( ∣ x ∣ < 1 ) ∑ n = 0 ∞ ( − 1 ) n x n = ∑ n = 0 ∞ ( − x ) n = 1 1 + x , ∑ n = 1 ∞ ( − 1 ) n x n = − x 1 + x . ( ∣ x ∣ < 1 ) \sum_{n=0}^\infty(-1)^nx^n=\sum_{n=0}^\infty(-x)^n=\frac{1}{1+x},\sum_{n=1}^\infty(-1)^nx^n=-\frac{x}{1+x}.(|x|<1) ∑ n = 0 ∞ ( − 1 ) n x n = ∑ n = 0 ∞ ( − x ) n = 1 + x 1 , ∑ n = 1 ∞ ( − 1 ) n x n = − 1 + x x . ( ∣ x ∣ < 1 ) ∑ n = 1 ∞ n x n = x ( 1 − x ) 2 , ∣ x ∣ < 1. \sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2},|x|<1. ∑ n = 1 ∞ n x n = ( 1 − x ) 2 x , ∣ x ∣ < 1.
e.g.7.13 \textbf{e.g.7.13 } e.g.7.13 ∑ n = 1 ∞ n x n \sum_{n=1}^\infty nx^n ∑ n = 1 ∞ n x n
Solution \textbf{Solution } Solution a n = n , x n = x n , a_n=n,x_n=x^n, a n = n , x n = x n , R = lim n → ∞ 1 a n + 1 a n = n n + 1 = 1 , R=\lim_{n\rightarrow\infty}\frac{1}{\frac{a_{n+1}}{a_n}}=\frac{n}{n+1}=1, R = lim n → ∞ a n a n + 1 1 = n + 1 n = 1 , ( − 1 , 1 ) . (-1,1). ( − 1 , 1 ) . s ( x ) , s(x), s ( x ) ,
s ( x ) = ∑ n = 1 ∞ n x n = x ∑ n = 1 ∞ n x n − 1 = x s 1 ( x ) , ∫ s 1 ( x ) d x = ∫ ∑ n = 1 ∞ n x n − 1 d x = ∑ n = 1 ∞ ∫ n x n − 1 d x = ∑ n = 1 ∞ x n = x 1 − x . \begin{align*}
s(x)&=\sum_{n=1}^\infty nx^n=x\sum_{n=1}^\infty nx^{n-1}=xs_1(x),\\
\int s_1(x)\text{d}x&=\int\sum_{n=1}^\infty nx^{n-1}\text{d}x=\sum_{n=1}^\infty\int nx^{n-1}\text{d}x=\sum_{n=1}^\infty x^n=\frac{x}{1-x}.
\end{align*}
s ( x ) ∫ s 1 ( x ) d x = n = 1 ∑ ∞ n x n = x n = 1 ∑ ∞ n x n − 1 = x s 1 ( x ) , = ∫ n = 1 ∑ ∞ n x n − 1 d x = n = 1 ∑ ∞ ∫ n x n − 1 d x = n = 1 ∑ ∞ x n = 1 − x x .
因此s 1 ( x ) = ( ∫ s 1 ( x ) d x ) ′ = 1 ( 1 − x ) 2 . s_1(x)=\big(\int s_1(x)\text{d}x\big)'=\frac{1}{(1-x)^2}. s 1 ( x ) = ( ∫ s 1 ( x ) d x ) ′ = ( 1 − x ) 2 1 .
s ( x ) = x ( 1 − x ) 2 . s(x)=\frac{x}{(1-x)^2}.
s ( x ) = ( 1 − x ) 2 x .
e.g.7.14 \textbf{e.g.7.14 } e.g.7.14 ∑ n = 1 ∞ n 2 x n \sum_{n=1}^\infty n^2x^n ∑ n = 1 ∞ n 2 x n
Solution \textbf{Solution } Solution a n = n 2 , x n = x n , a_n=n^2,x_n=x^n, a n = n 2 , x n = x n , R = lim n → ∞ 1 a n + 1 a n = n 2 ( n + 1 ) 2 = 1 , R=\lim_{n\rightarrow\infty}\frac{1}{\frac{a_{n+1}}{a_n}}=\frac{n^2}{(n+1)^2}=1, R = lim n → ∞ a n a n + 1 1 = ( n + 1 ) 2 n 2 = 1 , ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) s ( x ) , s(x), s ( x ) ,
s ( x ) = ∑ n = 1 ∞ n 2 x n = x ∑ n = 1 ∞ n 2 x n − 1 = x s 1 ( x ) ∫ s 1 ( x ) d x = ∑ n = 1 ∞ ∫ n 2 x n − 1 d x = ∑ n = 1 ∞ n x n . \begin{align*}
s(x)&=\sum_{n=1}^\infty n^2x^n=x\sum_{n=1}^\infty n^2x^{n-1}=xs_1(x)\\
\int s_1(x)\text{d}x=\sum_{n=1}^\infty\int{n^2}x^{n-1}\text{d}x=\sum_{n=1}^\infty nx^n.
\end{align*}
s ( x ) ∫ s 1 ( x ) d x = n = 1 ∑ ∞ ∫ n 2 x n − 1 d x = n = 1 ∑ ∞ n x n . = n = 1 ∑ ∞ n 2 x n = x n = 1 ∑ ∞ n 2 x n − 1 = x s 1 ( x )
由例7.13可得∫ s 1 ( x ) d x = x ( 1 − x ) 2 , \int s_1(x)\text{d}x=\frac{x}{(1-x)^2}, ∫ s 1 ( x ) d x = ( 1 − x ) 2 x , s 1 ( x ) = ( x ( 1 − x ) 2 ) ′ = 1 + x ( 1 − x ) 3 . s_1(x)=\big(\frac{x}{(1-x)^2}\big)'=\frac{1+x}{(1-x)^3}. s 1 ( x ) = ( ( 1 − x ) 2 x ) ′ = ( 1 − x ) 3 1 + x .
s ( x ) = x ( x + 1 ) ( 1 − x ) 3 . s(x)=\frac{x(x+1)}{(1-x)^3}.
s ( x ) = ( 1 − x ) 3 x ( x + 1 ) .
e.g.7.15 \textbf{e.g.7.15 } e.g.7.15 ∑ n = 2 ∞ n ( n − 1 ) x n − 2 \sum_{n=2}^\infty n(n-1)x^{n-2} ∑ n = 2 ∞ n ( n − 1 ) x n − 2
Solution \textbf{Solution } Solution a n = n ( n − 1 ) , x n = x n − 2 , a_n=n(n-1),x_n=x^{n-2}, a n = n ( n − 1 ) , x n = x n − 2 , R = lim n → ∞ 1 a n + 1 a n = 1 , R=\lim_{n\rightarrow\infty}\frac{1}{\frac{a_{n+1}}{a_n}}=1, R = lim n → ∞ a n a n + 1 1 = 1 , ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) s ( x ) , s(x), s ( x ) ,
s ( x ) = ∑ n = 2 ∞ n ( n − 1 ) x n − 2 = ∑ n = 2 ∞ ( x n ) ′ ′ = ( ∑ n = 2 ∞ x n ) ′ ′ = ( 1 1 − x ) ′ ′ = 2 ( 1 − x ) 3 . \begin{align*}
s(x)&=\sum_{n=2}^\infty n(n-1)x^{n-2}=\sum_{n=2}^\infty(x^n)''=\Big(\sum_{n=2}^\infty x^n\Big)''\\
&=\Big(\frac{1}{1-x}\Big)''=\frac{2}{(1-x)^3}.
\end{align*}
s ( x ) = n = 2 ∑ ∞ n ( n − 1 ) x n − 2 = n = 2 ∑ ∞ ( x n ) ′′ = ( n = 2 ∑ ∞ x n ) ′′ = ( 1 − x 1 ) ′′ = ( 1 − x ) 3 2 .
一般级数求和的思路也就是构造一个相应的幂级数后对后者求和——只要把待求原级数的常数项级数看作幂级数在x x x x 0 x_0 x 0
e.g.7.16 \textbf{e.g.7.16 } e.g.7.16 ∑ n = 0 ∞ 4 n 2 + 4 n + 3 2 n + 1 x 2 n \sum_{n=0}^\infty\frac{4n^2+4n+3}{2n+1}x^{2n} ∑ n = 0 ∞ 2 n + 1 4 n 2 + 4 n + 3 x 2 n
Solution \textbf{Solution } Solution a n ( x ) = 4 n 2 + 4 n + 3 2 n + 1 x 2 n , a_n(x)=\frac{4n^2+4n+3}{2n+1}x^{2n}, a n ( x ) = 2 n + 1 4 n 2 + 4 n + 3 x 2 n , lim n → ∞ ∣ a n + 1 ( x ) a n ( x ) ∣ < 1 , \lim_{n\rightarrow\infty}\big|\frac{a_{n+1}(x)}{a_n(x)}\big|<1, lim n → ∞ a n ( x ) a n + 1 ( x ) < 1 , − 1 < x < 1 , -1<x<1, − 1 < x < 1 , ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) s ( x ) , s(x), s ( x ) ,
s ( x ) = ∑ n = 0 ∞ 4 n 2 + 4 n + 3 2 n + 1 x 2 n = ∑ n = 0 ∞ 4 n 2 + 4 n + 1 2 n + 1 x 2 n + ∑ n = 0 ∞ 2 2 n + 1 x 2 n = ∑ n = 0 ∞ ( 2 n + 1 ) x 2 n + 2 ∑ n = 0 ∞ x 2 n 2 n + 1 . \begin{align*}
s(x)&=\sum_{n=0}^\infty\frac{4n^2+4n+3}{2n+1}x^{2n}=\sum_{n=0}^\infty\frac{4n^2+4n+1}{2n+1}x^{2n}+\sum_{n=0}^\infty\frac{2}{2n+1}x^{2n}\\
&=\sum_{n=0}^\infty(2n+1)x^{2n}+2\sum_{n=0}^\infty\frac{x^{2n}}{2n+1}.
\end{align*}
s ( x ) = n = 0 ∑ ∞ 2 n + 1 4 n 2 + 4 n + 3 x 2 n = n = 0 ∑ ∞ 2 n + 1 4 n 2 + 4 n + 1 x 2 n + n = 0 ∑ ∞ 2 n + 1 2 x 2 n = n = 0 ∑ ∞ ( 2 n + 1 ) x 2 n + 2 n = 0 ∑ ∞ 2 n + 1 x 2 n .
现在记∑ n = 0 ∞ ( 2 n + 1 ) x 2 n = s 1 ( x ) , ∑ n = 0 ∞ x 2 n 2 n + 1 = s 2 ( x ) . \sum_{n=0}^\infty(2n+1)x^{2n}=s_1(x),\sum_{n=0}^\infty\frac{x^{2n}}{2n+1}=s_2(x). ∑ n = 0 ∞ ( 2 n + 1 ) x 2 n = s 1 ( x ) , ∑ n = 0 ∞ 2 n + 1 x 2 n = s 2 ( x ) .
∫ s 1 ( x ) d x = ∑ n = 0 ∞ x 2 n + 1 = x 1 − x 2 , s 2 ( x ) = ∑ n = 0 ∞ x 2 n 2 n + 1 = 1 x ∑ n = 0 ∞ x 2 n + 1 2 n + 1 . \begin{align*}
\int s_1(x)\text{d}x&=\sum_{n=0}^\infty x^{2n+1}=\frac{x}{1-x^2},\\
s_2(x)&=\sum_{n=0}^\infty\frac{x^{2n}}{2n+1}=\frac{1}{x}\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}.
\end{align*}
∫ s 1 ( x ) d x s 2 ( x ) = n = 0 ∑ ∞ x 2 n + 1 = 1 − x 2 x , = n = 0 ∑ ∞ 2 n + 1 x 2 n = x 1 n = 0 ∑ ∞ 2 n + 1 x 2 n + 1 .
再记s 3 ( x ) = ∑ n = 0 ∞ x 2 n + 1 2 n + 1 , s_3(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}, s 3 ( x ) = ∑ n = 0 ∞ 2 n + 1 x 2 n + 1 ,
s 3 ′ ( x ) = ∑ n = 0 ∞ x 2 n = 1 1 − x 2 . s_3'(x)=\sum_{n=0}^\infty x^2n=\frac{1}{1-x^2}.
s 3 ′ ( x ) = n = 0 ∑ ∞ x 2 n = 1 − x 2 1 .
因此
s 2 ( x ) = 1 x ∫ s 3 ′ ( x ) d x = 1 2 x ( ln ( 1 + x ) − ln ( 1 − x ) ) ⇒ s ( x ) = s 1 ( x ) + 2 s 2 ( x ) = x 1 − x 2 + ln 1 + x 1 − x x . \begin{align*}
&s_2(x)=\frac{1}{x}\int s'_3(x)\text{d}x=\frac{1}{2x}\big(\ln(1+x)-\ln(1-x)\big)
\Rightarrow&s(x)=s_1(x)+2s_2(x)=\frac{x}{1-x^2}+\frac{\ln\frac{1+x}{1-x}}{x}.
\end{align*}
s 2 ( x ) = x 1 ∫ s 3 ′ ( x ) d x = 2 x 1 ( ln ( 1 + x ) − ln ( 1 − x ) ) ⇒ s ( x ) = s 1 ( x ) + 2 s 2 ( x ) = 1 − x 2 x + x ln 1 − x 1 + x .
而注意到x = 0 x=0 x = 0 a 0 ( 0 ) = 3 , a_0(0)=3, a 0 ( 0 ) = 3 ,
s ( x ) = { 3 , x = 0 x 1 − x 2 + ln 1 + x 1 − x x , x ∈ ( − 1 , 0 ) ∪ ( 0 , 1 ) . s(x)=\begin{cases}
3&,x=0\\
\frac{x}{1-x^2}+\frac{\ln\frac{1+x}{1-x}}{x}&,x\in(-1,0)\cup(0,1)
\end{cases}.
s ( x ) = { 3 1 − x 2 x + x l n 1 − x 1 + x , x = 0 , x ∈ ( − 1 , 0 ) ∪ ( 0 , 1 ) .
e.g.7.17 \textbf{e.g.7.17 } e.g.7.17 a 0 = 4 , a 1 = 1 , a n = a n − 2 n ( n − 1 ) , a_0=4,a_1=1,a_n=\frac{a_{n-2}}{n(n-1)}, a 0 = 4 , a 1 = 1 , a n = n ( n − 1 ) a n − 2 , S ( x ) = ∑ n = 0 ∞ a n x n . S(x)=\sum_{n=0}^\infty a_nx^n. S ( x ) = ∑ n = 0 ∞ a n x n .
Solution \textbf{Solution } Solution a n ⋅ n ( n − 1 ) = a n − 2 . a_n\cdot n(n-1)=a_{n-2}. a n ⋅ n ( n − 1 ) = a n − 2 .
S ′ ( x ) = ∑ n = 1 ∞ n a n x n − 1 ⇒ S ′ ′ ( x ) = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 = ∑ n = 2 ∞ a n − 2 x n − 2 = ∑ n = 0 ∞ a n x n = S ( x ) . \begin{align*}
&S'(x)=\sum_{n=1}^\infty na_nx^{n-1}\\
\Rightarrow&S''(x)=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}=\sum_{n=2}^\infty a_{n-2}x^{n-2}=\sum_{n=0}^\infty a_nx^n=S(x).
\end{align*}
⇒ S ′ ( x ) = n = 1 ∑ ∞ n a n x n − 1 S ′′ ( x ) = n = 2 ∑ ∞ n ( n − 1 ) a n x n − 2 = n = 2 ∑ ∞ a n − 2 x n − 2 = n = 0 ∑ ∞ a n x n = S ( x ) .
换言之，y = S ( x ) y=S(x) y = S ( x ) y ′ ′ = y , y''=y, y ′′ = y , y ′ ′ − y = 0 ; y''-y=0; y ′′ − y = 0 ; S ( x ) = y = c 1 e x + c 2 e − x . S(x)=y=c_1e^x+c_2e^{-x}. S ( x ) = y = c 1 e x + c 2 e − x . a 0 = 4 , a 1 = 1 , a_0=4,a_1=1, a 0 = 4 , a 1 = 1 , S ( 0 ) = a 0 = 4 , S ′ ( 0 ) = a 1 = 1. S(0)=a_0=4,S'(0)=a_1=1. S ( 0 ) = a 0 = 4 , S ′ ( 0 ) = a 1 = 1.
{ c 1 + c 2 = 4 c 1 − c 2 = 1 ⇒ { c 1 = 5 2 c 2 = 3 2 ⇒ S ( x ) = 5 2 e x + 3 2 e − x . \begin{cases}
c_1+c_2&=4\\
c_1-c_2&=1
\end{cases}
\Rightarrow\begin{cases}
c_1&=\frac{5}{2}\\
c_2&=\frac{3}{2}
\end{cases}
\Rightarrow
S(x)=\frac{5}{2}e^x+\frac{3}{2}e^{-x}.
{ c 1 + c 2 c 1 − c 2 = 4 = 1 ⇒ { c 1 c 2 = 2 5 = 2 3 ⇒ S ( x ) = 2 5 e x + 2 3 e − x .
将函数展开为幂级数的思路是通过变量代换、恒等变形或微/积分把函数化为1 1 ± x , e x , sin x , cos x , ln ( 1 + x ) \frac{1}{1\pm x},e^x,\sin x,\cos x,\ln(1+x) 1 ± x 1 , e x , sin x , cos x , ln ( 1 + x )
e.g.7.18 \textbf{e.g.7.18 } e.g.7.18 f ( x ) = x 2 + x − x 2 f(x)=\frac{x}{2+x-x^2} f ( x ) = 2 + x − x 2 x x x x
Solution \textbf{Solution } Solution
O.F. = x ( 2 − x ) ( 1 + x ) = 1 3 ( 2 2 − x − 1 1 + x ) = 1 3 ( 1 1 − x 2 − 1 1 + x ) = 1 3 ( ∑ n = 0 ∞ ( x 2 ) n − ∑ n = 0 ∞ ( − 1 ) n x n ) = ∑ n = 0 ∞ ( − 1 ) n + 1 + ( 1 2 ) n 3 x n . \begin{align*}
\text{O.F.}&=\frac{x}{(2-x)(1+x)}=\frac{1}{3}\Big(\frac{2}{2-x}-\frac{1}{1+x}\Big)\\
&=\frac{1}{3}\Big(\frac{1}{1-\frac{x}{2}}-\frac{1}{1+x}\Big)=\frac{1}{3}\bigg(\sum_{n=0}^\infty\Big(\frac{x}{2}\Big)^n-\sum_{n=0}^\infty(-1)^nx^n\bigg)\\
&=\sum_{n=0}^\infty\frac{(-1)^{n+1}+\big(\frac{1}{2}\big)^n}{3}x^n.
\end{align*}
O.F. = ( 2 − x ) ( 1 + x ) x = 3 1 ( 2 − x 2 − 1 + x 1 ) = 3 1 ( 1 − 2 x 1 − 1 + x 1 ) = 3 1 ( n = 0 ∑ ∞ ( 2 x ) n − n = 0 ∑ ∞ ( − 1 ) n x n ) = n = 0 ∑ ∞ 3 ( − 1 ) n + 1 + ( 2 1 ) n x n .
e.g.7.19 \textbf{e.g.7.19 } e.g.7.19 f ( x ) = arctan 1 + x 1 − x f(x)=\arctan\frac{1+x}{1-x} f ( x ) = arctan 1 − x 1 + x x x x
Solution \textbf{Solution } Solution
f ′ ( x ) = 1 1 + x 2 = ∑ n = 0 ∞ ( − 1 ) n x 2 n ⇒ f ( x ) = ∫ ∑ n = 0 ∞ ( − 1 ) n x 2 n d x = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 2 n + 1 + C . \begin{align*}
&f'(x)=\frac{1}{1+x^2}=\sum_{n=0}^\infty(-1)^nx^{2n}\\
\Rightarrow&f(x)=\int\sum_{n=0}^\infty(-1)^nx^{2n}\text{d}x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}+C.
\end{align*}
⇒ f ′ ( x ) = 1 + x 2 1 = n = 0 ∑ ∞ ( − 1 ) n x 2 n f ( x ) = ∫ n = 0 ∑ ∞ ( − 1 ) n x 2 n d x = n = 0 ∑ ∞ ( − 1 ) n 2 n + 1 x 2 n + 1 + C .
注意到f ( 0 ) = π 4 , f(0)=\frac{\pi}{4}, f ( 0 ) = 4 π , C = π 4 . C=\frac{\pi}{4}. C = 4 π .
f ( x ) = π 4 + ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 2 n + 1 . f(x)=\frac{\pi}{4}+\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}.
f ( x ) = 4 π + n = 0 ∑ ∞ ( − 1 ) n 2 n + 1 x 2 n + 1 .
sin x , cos x , e x \sin x,\cos x,e^x sin x , cos x , e x ( 2 n + 1 ) ! , ( 2 n ) ! , n ! (2n+1)!,(2n)!,n! ( 2 n + 1 )! , ( 2 n )! , n !
e.g.7.20 \textbf{e.g.7.20 } e.g.7.20 ∑ n = 1 ∞ ( − 1 ) n n ( 2 n + 1 ) ! \sum_{n=1}^\infty\frac{(-1)^nn}{(2n+1)!} ∑ n = 1 ∞ ( 2 n + 1 )! ( − 1 ) n n
Solution \textbf{Solution } Solution s ( x ) = ∑ n = 1 ∞ ( − 1 ) n n x 2 n − 1 ( 2 n + 1 ) ! ; s(x)=\sum_{n=1}^\infty\frac{(-1)^nnx^{2n-1}}{(2n+1)!}; s ( x ) = ∑ n = 1 ∞ ( 2 n + 1 )! ( − 1 ) n n x 2 n − 1 ; s ( 1 ) . s(1). s ( 1 ) .
∫ s ( x ) d x = 1 2 ∑ n = 1 ∞ ( − 1 ) n x 2 n ( 2 n + 1 ) ! = 1 2 x ∑ n = 1 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! = 1 2 x ( sin x − x ) = sin x 2 x − 1 2 , \begin{align*}
\int s(x)\text{d}x&=\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\\
&=\frac{1}{2x}\sum_{n=1}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}=\frac{1}{2x}(\sin x-x)\\
&=\frac{\sin x}{2x}-\frac{1}{2},
\end{align*}
∫ s ( x ) d x = 2 1 n = 1 ∑ ∞ ( 2 n + 1 )! ( − 1 ) n x 2 n = 2 x 1 n = 1 ∑ ∞ ( 2 n + 1 )! ( − 1 ) n x 2 n + 1 = 2 x 1 ( sin x − x ) = 2 x sin x − 2 1 ,
故
s ( x ) = ( sin x 2 x − 1 2 ) ′ = x cos x − sin x 2 x 2 . s(x)=\Big(\frac{\sin x}{2x}-\frac{1}{2}\Big)'=\frac{x\cos x-\sin x}{2x^2}.
s ( x ) = ( 2 x sin x − 2 1 ) ′ = 2 x 2 x cos x − sin x .
这时再取s ( 1 ) , s(1), s ( 1 ) ,
O.F. = cos 1 − sin 1 2 . \text{O.F.}=\frac{\cos 1-\sin 1}{2}.
O.F. = 2 cos 1 − sin 1 .
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