20170928.2

题 2:

设 y=f(x) 在(-1,1)内有二阶连续导数,且f’’(x)≠0. 证明

  • (1) 对于任意非零x∈(-1,1),存在唯一 θ(x)∈(0,1), 使得 f(x)=f(0)+xf’[ θ(x),x] 成立;
  • (2) $\lim _{x\rightarrow 0^{+}}\theta \left( x\right) =\dfrac {1} {2}$.

证明:

(1) 对于f(x)在[0,1]上使用拉格朗日定理,得:f(x)-f(0)=xf’[0+θ(x)*(x-0)],

f(x)=f(0)+xf’(θ(x)*x).

又f’’(x)≠在[0,1]内连续且f’’(x)≠0,故保号,于是不妨设f’’(x)>0.所以f’(x)严格单调,所以θ(x)唯一.

(2) 由泰勒公式有 $f\left( x\right) =f\left( 0\right) +f’\left( 0\right)x +\dfrac {1} {2}f’’(\varepsilon )x ^{2}$

又由(1)中的拉格朗日定理可知:f(x)=f(0)+xf’[θ(x)].

所以:

$f(0)+xf’[θ(x)*x] =f\left( 0\right) +f’\left( 0\right)x +\dfrac {1} {2}f’’(\varepsilon )x ^{2}$ =>

$xf’[θ(x)*x] =f’\left( 0\right)x +\dfrac {1} {2}f’’(\varepsilon )x ^{2}$ =>

$\dfrac {f’[θ(x)*x]-f’(0)}{x} =\dfrac {1} {2}f’’(\varepsilon )$ =>

$\lim _{x\rightarrow 0^{+}} \dfrac {f’[0+θ(x)*x]-f’(0)} {θ(x)x} θ(x)$

$=\lim _{x\rightarrow 0^{+}}f’’(0)*θ(x)$

$=\lim _{x\rightarrow 0^{+}}\dfrac {1} {2}f’’(\varepsilon ) =\dfrac {1} {2}f’’(0)$ =>

$\lim _{x\rightarrow 0^{+}}f’’(0)*θ(x)=\dfrac {1} {2}f’’(0)$

因此 $\lim _{x\rightarrow 0^{+}}θ(x)=\dfrac {1} {2}$


THE END.

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