New answers tagged

2

The first thing you should understand is why central differencing gives you a more precise solution. Consider the Taylor expansion of $f$ around $x$: $$f(x + h) = f(x) + h f'(x) + \frac{1}{2} h^2 f''(x) + \frac{1}{3!} h^3 f'''(x) \cdots$$ Then: $$\frac{f(x+h) - f(x)}{h} = f'(x) + \frac{1}{2} h f''(x) + \frac{1}{3!} h^2 f'''(x)\cdots$$ That is: $$f'(x) = \...


Top 50 recent answers are included