# Central-Difference Formula

• Feb 25th 2009, 05:42 PM
zeugma
Central-Difference Formula
Good day everyone.

I'm so stuck with this homework in numerical methods. I'm supposed to derive an equation using Taylor expansion. The details are below:

Use the Taylor expansions for $f(x+h), f(x-h), f(x+2h)$ and $f(x-2h)$ and derive the central-difference formula:

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

$f(x+h)=f(x)+f'(x)h+\frac{f^{(2)}(x)h^2}{2!} + \frac{f^{(3)}(x)h^3}{3!} + \frac{f^{(4)}(x)h^4}{4!} + \frac{f^{(5)}(x)h^5}{5!}$
$f(x+h)=f(x)-f'(x)h+\frac{f^{(2)}(x)h^2}{2!} -\frac{f^{(3)}(x)h^3}{3!} + \frac{f^{(4)}(x)h^4}{4!} - \frac{f^{(5)}(x)h^5}{5!}$
$f(x+2h)=f(x)+f'(x)2h+\frac{f^{(2)}(x)4h^2}{2!} + \frac{f^{(3)}(x)8h^3}{3!} + \frac{f^{(4)}(x)16h^4}{4!} + \frac{f^{(5)}(x)32h^5}{5!}$
$f(x-2h)=f(x)-f'(x)2h+\frac{f^{(2)}(x)4h^2}{2!} - \frac{f^{(3)}(x)8h^3}{3!} + \frac{f^{(4)}(x)16h^4}{4!} - \frac{f^{(5)}(x)32h^5}{5!}$
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