The definition of a derivative states that the derivative quantifies the relationship between two points. We prove that higher order derivatives quantify the relationship among several points.

Theorem

\begin{align} f^{(n)}(a) = \lim_{h\rightarrow 0}\frac{1}{h^l}\sum_{k=0}^l\binom{l}{k} (-1)^kf^{(n-l)}(a + (l-k)h) \label{eqn:binder} \end{align}

Proof

Consider

\[f^{(n)}(a) = \lim_{h\rightarrow 0}\frac{f^{(n-1)}(a+h)-f^{(n-1)}(a)}{h}\]

then

\[\begin{align*} f^{(n)}(a) &= \lim_{h\rightarrow 0}\frac{f^{(n-1)}(a+h)-f^{(n-1)}(a)}{h}\\ &= \lim_{h\rightarrow 0}\frac{\frac{f^{(n-2)}(a+2h)-f^{(n-2)}(a+h)}{h}-\frac{f^{(n-2)}(a+h)-f^{(n-2)}(a)}{h}}{h}\\ &= \lim_{h\rightarrow 0}\frac{f^{(n-2)}(a+2h)-2f^{(n-2)}(a+h)+f^{(n-2)}(a)}{h^2}\\ &= \lim_{h\rightarrow 0}\frac{\frac{f^{(n-3)}(a+3h)-f^{(n-3)}(a+2h)}{h}-2\frac{f^{(n-3)}(a+2h)-f^{(n-3)}(a+h)}{h}+\frac{f^{(n-3)}(a+h)-f^{(n-3)}(a)}{h}}{h^2}\\ &= \lim_{h\rightarrow 0}\frac{f^{(n-3)}(a+3h)-3f^{(n-3)}(a+2h)+3f^{(n-3)}(a+h) -f^{(n-1)}(a)}{h^3}\\ & = \cdots \nonumber \\ & = \lim_{h\rightarrow 0}\frac{1}{h^l}\sum_{k=0}^l\binom{l}{k} (-1)^kf^{(n-l)}(a + (l-k)h) \end{align*}\]

QED

where $n-l$ denotes which lower order derivative we decide to use to represent the higher order derivative $n$. If we choose to evaluate on the underived function, the zero derivative, then $l=n$, and the result above reveals that we need $n$ evaluations of $f$ at $n$ different points, at a distance $h$ of each other, to approximate the $n$-th derivative. Therefore, an analytic function has the information about all the function in its derivatives in one point.