某值定理杂烩
XantC
极值定理
\[ f(x)\in\mathrm{C}[a, b], \] \[ \exists c,d\in[a, b], \text{s.t.}\, f(c)=f(x)_{min},\,f(d)=f(x)_{max} \]
介值定理
\[ f(x)\in \mathrm{C}[a, b],\, f(a)\neq f(b) \]
\[ \forall N\in (f(a), f(b)), \exists c\in (a, b), \text{s.t.}\,f(c)=N \]
罗尔中值定理
\[ f(x)\in\mathrm{C}[a, b],\, f(x)在(a,b)可导,\,f(a)=f(b) \]
\[ \exists c\in(a,b), \text{s.t.}\, f^{'}(c)=0 \]
证明
Case 1 f(x) = f(a)
\[ \text{Case 1}\colon f(x)=f(a) \]
\[ 显然\,\exists c\in(a,b), \text{s.t.}\, f^{'}(c)=0 \]
Case 2 f(x) > f(a)
\[ \text{Case 2}\colon f(x)>f(a) \]
\[ 由极值定理\colon\exists f(x)_{max} \]
\[ \therefore \exists c\in(a,b), \text{s.t.}\, f^{'}(c)=0 \]
Case 3 f(x) < f(a)
\[ \text{Case 3}\colon f(x)<f(a) \]
\[ 由极值定理\colon\exists f(x)_{min} \]
\[ \therefore \exists c\in(a,b), \text{s.t.}\, f^{'}(c)=0 \]
拉格朗日中值定理
\[ f(x)\in\mathrm{C}[a, b],\, f(x)在(a,b)可导 \]
\[ \exists c\in(a,b), \text{s.t.}\, f^{'}(c)=\frac{f(b)-f(a)}{b-a} \]
证明
\[ 令g(x)=f(x)-\frac{f(b)-f(a)}{b-a}x \]
\[ g(a)=f(a)-\frac{f(b)-f(a)}{b-a}a=\frac{bf(a)-af(b)}{b-a} \]
\[ g(b)=f(b)-\frac{f(b)-f(a)}{b-a}b=\frac{bf(a)-af(b)}{b-a} \]
\[ \Rightarrow g(a)=g(b) \]
\[ \Rightarrow \exists c\in(a,b), \text{s.t.}\, g^{'}(c)=0 \]
\[ \Rightarrow f^{'}(c)-\frac{f(b)-f(a)}{b-a}=0 \]
\[ \therefore \exists c\in(a,b), \text{s.t.}\, f^{'}(c)=\frac{f(b)-f(a)}{b-a} \]
柯西中值定理
\[ f(x),g(x)\in\mathrm{C}[a, b],\, f(x)与g(x)在(a,b)可导 \]
\[ \exists \xi\in(a,b), \text{s.t.}\, \frac{f^{'}(\xi)}{g^{'}(\xi)}=\frac{f(b)-f(a)}{g(b)-g(a)} \,(g^{'}(\xi)\neq 0)\]
证明
\[ f^{'}(\xi)=\frac{f(b)-f(a)}{b-a} \]
\[ g^{'}(\xi)=\frac{g(a)-g(b)}{b-a} \]
\[ \therefore \frac{f^{'}(\xi)}{g^{'}(\xi)}=\frac{f(b)-f(a)}{g(b)-g(a)} \]
泰勒中值定理
\[f(x)=\sum_{n=0}^{+\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+R_n(x) \]
\[ R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1} \;(\xi \in (x,x_0)) \]
证明
\[ R_n(x_0)=R_n^{'}(x_0)=\dots =R_n^{(n)}(x_0)=0 \]
\[ \begin{align*} \frac{R_n(x)}{(x-x_0)^{n+1}}&=\frac{R_n(x)-R_n(x_0)}{(x-x_0)^{n+1}-(x_0-x_0)^{n+1}} \\ &=\frac{R_n^{'}(\xi)}{(n+1)(x-x_0)^n} \\ &\dots \\ &=\frac{R_n^{(n+1)}(\xi)}{(n+1)!} \\ &=\frac{f^{(n+1)}(\xi)}{(n+1)!} \end{align*} \]
\[ \therefore R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1} \]