# Quod Erat Demonstrandum

## 2010/07/26

### MVT 某推廣

Filed under: HKALE,Pure Mathematics — johnmayhk @ 11:54 上午
Tags: ,

$F(x) = -3x + \frac{\pi}{2}\cos(\frac{\pi}{2}x) + \frac{e^x}{e - 1} + \frac{1}{(x + 2)\ln 2}$ 在區間 (0,1) 存在零點 ………. (*)

$F(0) , F(1)$ 皆大於零，故不宜用介值定理（Intermediate Value Theorem）。

$F(1) \approx$ -0.696675773，即可用介值定理一步 KO 的。

$\alpha_1, \alpha_2, \dots , \alpha_n$ 為 n 個實數，且有 $\alpha_1 + \alpha_2 + \dots + \alpha_n = 0$

$f_1(x), f_2(x), \dots , f_n(x)$ 為 n 個在 [a , b] 上連續並在 (a , b) 上可導之函數，

$\displaystyle\sum_{i = 1}^n \frac{\alpha_i f_i'(c)}{f_i(b) - f_i(a)} = 0$

$n = 4$
$a = 0$
$b = 1$
$f_1(x) = x^2$
$f_2(x) = \sin(\frac{\pi}{2}x)$
$f_3(x) = e^x$
$f_4(x) = \ln(x + 1)$
$\alpha_1 = -3, \alpha_2 = \alpha_3 = \alpha_4 = 1$

$g(x)$
$= \alpha_1(f_2(b) - f_2(a))(f_3(b) - f_3(a))(f_1(x) - f_1(a))$
$+ \alpha_2(f_1(b) - f_1(a))(f_3(b) - f_3(a))(f_2(x) - f_2(a))$
$+ \alpha_3(f_1(b) - f_1(a))(f_2(b) - f_2(a))(f_3(x) - f_3(a))$

$g(a) = 0$

$g(b) = (f_1(b) - f_1(a))(f_2(b) - f_2(a))(f_3(b) - f_3(a))(\alpha_1 + \alpha_2 + \alpha_3) = 0$

result follows.