Sharp central-secant derivative error ($|f^{\prime }(x)-(f(x+h)-f(x-h))/(2h)|\le h^2\Vert f^{\prime \prime \prime }{\Vert }_{\infty }/6$)

For $f\in C^3[x-h,x+h]$, the Peano kernel for the error of the central secant slope has one sign. Integrating its absolute value gives the displayed constant $h^2/6$, and a cubic polynomial attains it. At $x=1$, $h=1$, the unnormalized order-two kernel is $-{\theta }^2/2$ on $[0,1]$ and $-(2-\theta {)}^2/2$ on $[1,2]$; the integral representation includes the factor $1/2!$.

Suppose $|f^{\prime \prime \prime }|\le M$ and each sampled value of $f$ has absolute error at most $\delta$. For a central finite difference with $h>0$, the triangle inequality and the sharp central-secant derivative error giveFor $M,\delta >0$, the derivative of this bound is $Mh/3-\delta /h^2$. It changes sign at $h_{\ast }=(3\delta /M{)}^{1/3}$, proving the unique minimum, of value $(3\delta {)}^{2/3}{M}^{1/3}/2$. This balance explains why reducing the step indefinitely amplifies sample absolute error even as truncation error decreases.