用泰勒展开证明fx=fx-2h-8fx-h+8fx+h-fx+2h12h+oh^4结果不要出现其他符号

根据泰勒展开公式：

$$f(x \pm h) = f(x) \pm hf'(x) + \frac{h^2}{2!}f''(x) \pm \frac{h^3}{3!}f'''(x) + \frac{h^4}{4!}f^{(4)}(x) + O(h^5)$$

将其代入式子：

$$f'(x) = \frac{f(x+h)-f(x)}{h} - \frac{h}{2}f''(x) + \frac{h^2}{6}f'''(x) - \frac{h^3}{24}f^{(4)}(x) + O(h^4)$$

$$f'(x) = \frac{f(x+2h)-f(x)}{2h} - \frac{h}{2}f''(x) + \frac{2h^2}{3!}f'''(x) - \frac{2h^3}{4!}f^{(4)}(x) + O(h^4)$$

$$f'(x) = \frac{8f(x+h)-8f(x-h)}{4h} + O(h^3)$$

将第二个式子乘以3和第三个式子乘以-1，然后相加可得：

$$3f'(x) = \frac{3f(x+2h)-3f(x)}{2h} - 3hf''(x) + 3\cdot\frac{2h^2}{3!}f'''(x) - 3\cdot\frac{2h^3}{4!}f^{(4)}(x) + O(h^4)$$

$$-f'(x) = \frac{-f(x-2h)+f(x)}{2h} + hf''(x) + \frac{h^2}{3!}f'''(x) + \frac{h^3}{4!}f^{(4)}(x) + O(h^4)$$

将两式相加可得：

$$2f'(x) = \frac{f(x+2h)-f(x-2h)}{2h} - 8f'(x) + 8\cdot\frac{2h^2}{3!}f'''(x) - 2\cdot\frac{2h^3}{4!}f^{(4)}(x) + O(h^4)$$

整理得：

$$f'(x) = \frac{f(x+2h)-8f(x+h)+8f(x-h)-f(x-2h)}{12h} + O(h^4)$$

即：

$$f'(x) = \frac{f(x-2h)-8f(x-h)+8f(x+h)-f(x+2h)}{12h} + O(h^4)$$

证毕