TITLE: Uses of $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$
QUESTION [35 upvotes]: I have been wondering whether the following limit is being used somehow, as a variation of the derivative:
$$\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h} .$$
Edit:
I know that this limit is defined in some places where the derivative is not defined, but it gives us some useful information.
The question is not whether this limit is similar to the derivative, but whether it is useful somehow.
Thanks.

REPLY [23 votes]: The "symmetric difference" form of the derivative is quite convenient for the purposes of numerical computation; to wit, note that the symmetric difference can be expanded in this way:
$$D_h f(x)=\frac{f(x+h)-f(x-h)}{2h}=f^\prime(x)+\frac{f^{\prime\prime\prime}(x)}{3!}h^2+\frac{f^{(5)}(x)}{5!}h^4+\dots$$
and one thing that should be noted here is that in this series expansion, only even powers of $h$ show up.
Consider the corresponding expansion when $h$ is halved:
$$D_{h/2} f(x)=\frac{f(x+h/2)-f(x-h/2)}{h}=f^\prime(x)+\frac{f^{\prime\prime\prime}(x)}{3!}\left(\frac{h}{2}\right)^2+\frac{f^{(5)}(x)}{5!}\left(\frac{h}{2}\right)^4+\dots$$
One could take a particular linear combination of this half-$h$ expansion and the previous expansion in $h$ such that the term with $h^2$ zeroes out:
$$4D_{h/2} f(x)-D_h f(x)=3f^\prime(x)-\frac{f^{(5)}(x)}{160}h^4+\dots$$
and we have after a division by $3$:
$$\frac{4D_{h/2} f(x)-D_h f(x)}{3}=f^\prime(x)-\frac{f^{(5)}(x)}{480}h^4+\dots$$
Note that the surviving terms after $f^\prime(x)$ are (supposed to be) much smaller than either of the terms after $f^\prime(x)$ in the expansions for $D_h f(x)$ and $D_{h/2} f(x)$. Numerically speaking, one could obtain a slightly more accurate estimate of the derivative by evaluating the symmetric difference at a certain (well-chosen) step size $h$ and at half of the given $h$, and computing the linear combination $\dfrac{4D_{h/2} f(x)-D_h f(x)}{3}$. (This is akin to deriving Simpson's rule from the trapezoidal rule). The procedure generalizes, as one keeps taking appropriate linear combinations of a symmetric difference for some $h$ and the symmetric difference at half $h$ to zero out successive powers of $h^2$; this is the famous Richardson extrapolation.