# Quod Erat Demonstrandum

## 2011/04/15

### 又談MVT

Filed under: Fun,HKALE,Pure Mathematics — johnmayhk @ 11:05 下午
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（from MathUpdate）

$f'(\gamma)=0$

$f'(\gamma)=0$

$g'(\gamma)=0$

$f'(\gamma)-m=0\Rightarrow f'(\gamma)=\frac{f(b)-f(a)}{b-a} \Rightarrow f(b)-f(a)=f'(\gamma)(b-a)$

Let $\alpha,\beta$ ($\alpha<\beta$) be distinct real roots of $P(x)=0$.

It is given that $P(x)$ is differentiable on $(\alpha,\beta)$.

Prove that $P'(\gamma)+2011P(\gamma)=0$ for some $\gamma \in (\alpha,\beta)$.

$f(x)=e^{2011x}P(x)$

$f(\alpha)=f(\beta)=0$

$f'(\gamma)=0$

$f(x)=e^{g(x)}P(x)$，再配合 MVT，或許可以產生不太直觀的結果。

P.S.

http://momath.org/

https://simonsfoundation.org/mathematics-physical-sciences/news/-/asset_publisher/bo1E/content/kindling-the-mathematical-muse?redirect=%2Fmathematics-physical-sciences%2Fnews