Section01a_中值定理 (1)

Th1 Rolle中值定理

定理

• 若 $\displaystyle{\begin{cases}f(x)\in C[a,b],\\(a,b)\text{内可导}\\ f(a)=f(b) \end{cases}\Rightarrow \exists\ \xi\in(a,b)}$，使得 $$f'(\xi)=0$$

Proof $$\begin{array}{ll} & f \in C[a,b] \Rightarrow \exists\ m,M \\ & \text{When } m=M \\ \because & f(x)\equiv c_{0} \\ \therefore & f'(x) = 0 \\ \therefore & \exists\ \xi \in (a,b), f'(\xi)=0 \\ & \text{When } m \ne M \\ \because & f(a) = f(b) \\ \therefore & m,M\text{至少有一个存在于}(a,b) \\ & \text{设}\xi\in(a,b), f(\xi) = M \\ \therefore & f'(\xi) = 0 \text{ or } \nexists\ f'(\xi) \\ \because & f\in C[a,b] \\ \therefore & f'(\xi) = 0 \end{array}$$

例题

• 例1 $f(x)\in C[0,2]$ 且在 $(0,2)$ 内可导且 $f(0)=\frac{1}{2}, f(1)=0, f(2)=2$ 证明 $\exists\ \xi\in(0,2),\ f'(\xi)=0$
• 解 $$\begin{array}{ll} & \text{令}\phi(x) = f(x)-\frac{1}{2} \\ \therefore & \phi(1) = -\frac{1}{2}, \phi(2) = \frac{3}{2} \\ \therefore & \phi(1)\phi(2) < 0 \\ \therefore & \exists\ c\in(1,2), \phi(c) = 0 \\ & \text{而}f(c) = \frac{1}{2} = f(0) \\ \therefore & \exists\ \xi\in(0,c)\subset(0,2),\ f'(\xi)= 0\\ \end{array}$$

Th2 Lagrange中值定理

定理

• 若 $\displaystyle{\begin{cases}f\in C[a,b]\\(a,b)\text{内可导}\end{cases}}\Rightarrow \exists\ \xi\in(a,b)$ 使 $$\displaystyle{f'(\xi) = \frac{f(b)-f(a)}{b-a}}$$

Proof $$\begin{array}{ll} & \text{经}A,B\text{的两条曲线} \begin{cases} \text{L}: y=f(x) \\ \text{L}_{AB}: y-f(a) = \frac{f(b)-f(a)}{b-a}(x-a) \end{cases} \\ &\text{令}\phi(x) = \text{L}_{AB}-\text{L} = f(a)+\frac{f(b)-f(a)}{b-a}(x-a)-f(x) \\ \because & \phi(a) = \phi(b) = 0 \\ \therefore & \exists\ \xi\in(a,b),\ \phi'(\xi) = 0 \\ & \text{而}\phi'(x) = \frac{f(b)-f(a)}{b-a}-f'(x) \\ \therefore & \frac{f(b)-f(a)}{b-a}-f'(\xi) = 0 \\ & f'(\xi) = \frac{f(b)-f(a)}{b-a} \end{array}$$

Notes

1. 若 $f(a)=f(b) \Rightarrow \text{Rolle中值定理}$
2. $$\begin{split} & f'(\xi)=\frac{f(b)-f(a)}{b-a}\\ \Leftrightarrow & f(b)-f(a)=f'(\xi)(b-a)\\ \Leftrightarrow & f(b)-f(a) = f'[a + (b-a)\theta](b-a) \quad \theta\in(0,1) \end{split}$$

Th3 Cauchy中值定理

定理

• 若 $\begin{cases} f,g\in C[a,b] \\ f,g \text{在}(a,b)\text{内可导} \\ g'(x) \neq 0 & x\in(a,b)\end{cases}$ 则 $\exists\ \xi\in(a,b)$ 使得 $$\frac{f'(\xi)}{g'(\xi)} = \frac{f(b)-f(a)}{g(b)-g(a)}$$

  Analysis

  1. $g'(x)\neq 0 \Rightarrow \begin{cases}g'(\xi)\neq 0 \\ g(b)-g(a) \neq 0\end{cases}$

  2. Lagrange: $\phi(x) = f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$

     Cauchy: $\phi(x)= f(x) - f(a) - \frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)]$

  Proof $$\begin{array}{ll} & \text{令}\phi(x) = f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)]\\ \therefore & \phi(a)=\phi(b) \\ \therefore & \exists\ \xi\in(a,b), \phi'(\xi)=0 \\ & \text{而}\phi'(x) = f'(x)-\frac{f(b)-f(a)}{g(b)-g(a)}g'(x) \\ \because & g'(x) \neq 0 \\ \therefore & \frac{f'(\xi)}{g'(\xi)}=\frac{f(b)-f(a)}{g(b)-g(a)} \\ \end{array}$$ Notes

  • 若$g(x)=x$，则$\text{Cauchy}\Rightarrow\text{Lagrange}$