Approximating feature advantage of the Symmetric Difference Quotient over the Divided Difference

In numerical differentiation, symmetric Newton’s quotient (a.k.a. symmetric difference quotient) is known to more effectively approximate the first derivative of given function $f\in C^3(\mathbb{R})$ , in comparison with the (first-order) divided difference. This statement can be summarized as the following proposition.

Proposition 1.
$|f'(a)-\frac{f(a+h)-f(a-h)}{2h}|\leq |f'(a)-\frac{f(a+h)-f(a)}{h}|$

for sufficiently small $h$ and $\forall a\in \mathbb{R}$ .

This sounds intuitively correct, since taking average of left and right divided differences at the point $a$ seem to smoothen the drastic slope change that can occasionally occur only one side. This is shown by somewhat technical manner.

Proof. Note that $f^{(k)}\quad (k\leq 3)$ is bounded around $a$ since $f$ is assumed to be of class $C^3$ as a function on $\mathbb{R}$ . By the Taylor series expression of $f(a+h)$ and $f(a-h)$ at $a$ , we get:

$\begin{array}{l}\displaystyle{ f(a+h)=f(a)+f'(a)h+\frac{f''(a)}{2}h^2+\frac{f'''(\xi_h)}{6}} \quad (\xi_h\in (a,a+h)), \\\\\displaystyle{f(a-h)=f(a)-f'(a)h+\frac{f''(a)}{2}h^2-\frac{f'''(\xi_{-h})}{6}} \quad (\xi_{-h}\in (a-h,a)).\end{array}$

Subtracting the second row from the first, and divided by $2h$ yealds:

$f'(a)-\frac{f(a+h)-f(a-h)}{2h} = -\frac{h^2}{12}(f'''(\xi_h) - f'''(\xi_{-h})) \sim o(h)(h\to 0).$

Similarly we have the lower order infinitesimal equation:

$f'(a)-\frac{f(a+h)-f(a)}{h} = -\frac{h}{2}f''(\eta_h) \sim o(1)(h\to 0).$

By taking $M=\displaystyle{\sup_{x,y\in (a-h,a+h)}|f'''(x)-f'''(y)|}$ and $m=\displaystyle{\inf_{x,y\in (a,a+h)}|f''(x)-f''(y)|}$ , we can evaluate the equation only by the order of infinitesimal with regard to $M$ and $m$ being constant when $h$ is sufficiently small, which finish the proof ■