一道大一数分题，做不出来

wwdwwd117

设 $f(x)$ 在 $[-1,1]$ 上连续，在 $(-1,1)$ 上有三阶导数。证明：存在 $\xi\in(-1,1)$ 使得
\[ f(1)=f(-1)+2f'(0)+\frac{f'''(\xi)}{3}+8\xi. \]

青青子衿

回复 1# wwdwwd117
PS: 怎么连你也不用LaTeX码题了？（怎么喜欢上传图片了？）（2019.4.17,14:10 图片已经去掉了）
方法是“泰勒公式”
关键在于构造\(\color{red}{G(x)=f(x)+x^4}\)

\begin{align*}
G(x)&=f(x)+x^4\\
G(\xi)&=f(\xi)+\xi^4\\
G\,'(\xi)&=f'(\xi)+4\,\xi^3\\
G\,''(\xi)&=f''(\xi)+12\,\xi^2\\
G\,'''(\xi)&=f'''(\xi)+24\,\xi\\
\end{align*}

带“拉格朗日（Lagrange）余项”的泰勒公式
\begin{align*}
G(x)&=G(x_0)+G\,'(x_0)(x-x_0)+\dfrac{G\,''(x_0)}{2}\big(x-x_0\big)^2+\cdots+\dfrac{G^{(n)}(x_0)}{n!}\big(x-x_0\big)^n+R_n(x)\\
G(x)&=G(x_0)+G\,'(x_0)(x-x_0)+\dfrac{G\,''(x_0)}{2}\big(x-x_0\big)^2+\cdots+\dfrac{G^{(n)}(x_0)}{n!}\big(x-x_0\big)^n+\dfrac{G^{(n+1)}(\xi)}{(n+1)!}\big(x-x_0\big)^{n+1}\\
G(x)&=G(x_0)+G\,'(x_0)(x-x_0)+\dfrac{G\,''(x_0)}{2}\big(x-x_0\big)^2+\dfrac{G\,'''(\xi)}{6}\big(x-x_0\big)^3\\
G(1)&=G(0)+G\,'(0)(1-0)+\dfrac{G\,''(0)}{2}\big(1-0\big)^2+\dfrac{G\,'''(\xi_{\overset{\,}{1}})}{6}\big(1-0\big)^3\\
G(-1)&=G(0)+G\,'(0)(-1-0)+\dfrac{G\,''(0)}{2}\big(-1-0\big)^2+\dfrac{G\,'''(\xi_{\overset{\,}{2}})}{6}\big(-1-0\big)^3\\
\end{align*}
\begin{align*}
&&&\left\{
\begin{split}
G(1)&=G(0)+G\,'(0)+\dfrac{G\,''(0)}{2}+\dfrac{G\,'''(\xi_{\overset{\,}{1}})}{6}\\
G(-1)&=G(0)-G\,'(0)+\dfrac{G\,''(0)}{2}-\dfrac{G\,'''(\xi_{\overset{\,}{2}})}{6}\\
\end{split}
\right.\\
\Rightarrow
&&&G(1)-G(-1)=2G\,'(0)+\dfrac{1}{3}\left(\dfrac{G\,'''(\xi_{\overset{\,}{1}})+G\,'''(\xi_{\overset{\,}{2}})}{2}\right)
\end{align*}

\begin{align*}
\min\left\{G\,'''(\xi_{\overset{\,}{2}}),\,G\,'''(\xi_{\overset{\,}{2}})\right\}\leqslant
\dfrac{G\,'''(\xi_{\overset{\,}{1}})+G\,'''(\xi_{\overset{\,}{2}})}{2}
\leqslant\max\left\{G\,'''(\xi_{\overset{\,}{2}}),\,G\,'''(\xi_{\overset{\,}{2}})\right\}
\end{align*}

\[ \therefore\exists\xi\in(\xi_{\overset{\,}{1}},\,\xi_{\overset{\,}{2}}),\quad\,G\,'''(\xi)=\dfrac{G\,'''(\xi_{\overset{\,}{1}})+G\,'''(\xi_{\overset{\,}{2}})}{2} \]

\begin{align*}
&&G(1)-G(-1)&=2G\,'(0)+\dfrac{1}{3}\left(\dfrac{G\,'''(\xi_{\overset{\,}{1}})+G\,'''(\xi_{\overset{\,}{2}})}{2}\right)\\
\Rightarrow&&f(1)-f(-1)&=2f'(0)+\dfrac{1}{3}\bigg(f'''(\xi)+24\xi\bigg)\\
\Rightarrow&&f(1)&=f(-1)+2f'(0)+\dfrac{1}{3}f'''(\xi)+8\xi\\
\end{align*}

kuing

回复 2# 青青子衿

图片是我去掉的，代码也是我打的。

wwdwwd117

回复 2# 青青子衿


    赞！

wwdwwd117

这里G’’’（x）要是连续函数才可以吧？但是不一定连续吧？

wwdwwd117

还有一个细节：应该是ξ属于闭区间【ξ2，ξ1】吧？？

kuing

回复 5# wwdwwd117

这应该是没问题的，导数虽未必连续，却有介值性，“达布定理”了解一下。
至于细节，是需要改的，事关如果两个 G''' 相等，则 `\xi` 就要取 `\xi_1` 或 `\xi_2`。

wwdwwd117

长知识了，无需连续的导函数介值定理。

青青子衿

对偶的题型
\begin{align*}
f(b)+f(a)&=2f\left(\dfrac{a+b}{2}\right)+\dfrac{(b-a)^2}{4}f''(\eta)\\
f(b)-f(a)&=2(b-a)f'\left(\dfrac{a+b}{2}\right)+\dfrac{(b-a)^3}{24}f'''(\xi)\\
\end{align*}

hbghlyj

math.stackexchange.com/questions/2716902
Example 3.3.11: An interesting fact is that there do exist discontinuous functions that have the intermediate value property. The function
\[
f(x):= \begin{cases}\sin (1 / x) & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases}
\]
is not continuous at 0 , however, it has the intermediate value property. That is, for any $a<b$, and any $y$ such that $f(a)<y<f(b)$ or $f(a)>y>f(b)$, there exists a $c$ such that $f(y)=c$. Proof is left as an exercise.

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$f$ is bounded and continuous on $(0,x)$ so is integrable. $f=g'$ where $g(x) = \int_0^x \sin\frac1tdt$
By Darboux's theorem, $f$ has the intermediate value property.

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An example of a Darboux function that is discontinuous at one point is the topologist's sine curve function:
$$x \mapsto \begin{cases}\sin(1/x) & \text{for } x\ne 0, \\ 0 &\text{for } x=0. \end{cases}$$
By Darboux's theorem, the derivative of any differentiable function is a Darboux function. In particular, the derivative of the function $x \mapsto x^2\sin(1/x)$ is a Darboux function even though it is not continuous at one point.