Approximation of a function’s derivative

Question

I came across this result when I was tinkering with some summations and integrals. Any ideas if it could be useful?

$\frac{d}{dx}f(x)\approx f(x+\frac{1}{2})-f(x-\frac{1}{2})$

Answer 1

This centered difference is often used to estimate $f'(x)$. More generally, $f'(x)$ can be approximated as $$ f'(x) \approx \frac{f(x + \frac{\Delta x}{2}) - f(x - \frac{\Delta x}{2})}{\Delta x}, $$ where $\Delta x$ is a small number. In your case, you're taking $\Delta x = 1$.

Visually, we are approximating $f'(x)$ by the slope of the line through the points $(x - \Delta x/2, f(x - \Delta x/2))$ and $(x + \Delta x/2, f(x + \Delta x/2))$

Answer 2

The standard definition of differentiability is: $f$ is differentiable at $a$, denoted $f'(a)$, if the following limit exists: $$ f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h} $$Sometimes it is convenient to take a symmetric version: $$ f'(a) = \frac{1}{2}\left( \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}- \frac{f(a-h)-f(a)}{h}\right) $$ $$ f'(a) = \frac{1}{2}\lim_{h\to 0} \frac{f(a+h)-f(a-h)}{h} $$If you put $h=1/2$, you recover your result. As to its utility...probably nothing too useful unless $f'$ is very small and unknown or hard to calculate.