某值定理杂烩
XantC
极值定理
f(x) ∈ C[a, b], ∃c, d ∈ [a, b], s.t. f(c) = f(x)min, f(d) = f(x)max
介值定理
f(x) ∈ C[a, b], f(a) ≠ f(b)
∀N ∈ (f(a), f(b)), ∃c ∈ (a, b), s.t. f(c) = N
罗尔中值定理
f(x) ∈ C[a, b], f(x)在(a, b)可导, f(a) = f(b)
∃c ∈ (a, b), s.t. f′(c) = 0
证明
Case 1 f(x) = f(a)
Case 1: f(x) = f(a)
显然 ∃c ∈ (a, b), s.t. f′(c) = 0
Case 2 f(x) > f(a)
Case 2: f(x) > f(a)
由极值定理: ∃f(x)max
∴ ∃c ∈ (a, b), s.t. f′(c) = 0
Case 3 f(x) < f(a)
Case 3: f(x) < f(a)
由极值定理: ∃f(x)min
∴ ∃c ∈ (a, b), s.t. f′(c) = 0
拉格朗日中值定理
f(x) ∈ 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
$$ 令g(x)=f(x)-\frac{f(b)-f(a)}{b-a}x $$
$$ g(a)=f(a)-\frac{f(b)-f(a)}{b-a}a=\frac{bf(a)-af(b)}{b-a} $$
$$ g(b)=f(b)-\frac{f(b)-f(a)}{b-a}b=\frac{bf(a)-af(b)}{b-a} $$
⇒ g(a) = g(b)
⇒ ∃c ∈ (a, b), 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) ∈ 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
Rn(x0) = Rn′(x0) = … = Rn(n)(x0) = 0
$$ \begin{align*} \frac{R_n(x)}{(x-x_0)^{n+1}}&=\frac{R_n(x)-R_n(x_0)}{(x-x_0)^{n+1}-(x_0-x_0)^{n+1}} \\ &=\frac{R_n^{'}(\xi)}{(n+1)(x-x_0)^n} \\ &\dots \\ &=\frac{R_n^{(n+1)}(\xi)}{(n+1)!} \\ &=\frac{f^{(n+1)}(\xi)}{(n+1)!} \end{align*} $$
$$ \therefore R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1} $$