Nth derivative can be expressed like that? $$
\frac{f^{(n)}(z_0)}{n!} = \lim_{z \rightarrow z_0} \frac{f(z) -f(z_0)}{(z-z_0)^n}
$$
Why is that nth derivative can be expressed like that limit of quotient?
I can understand the meaning but I couldnt get closed form equation.
Thanks.
 A: This looks familiar to complex analysis, so let me take a vague stab in the dark. If $f$ is holomorphic of the form
$$f(z) = f(z_0) + \sum_{k=n}^{\infty}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$$
where $n$ is the first non-zero order derivative, then this expression will hold
$$\frac{f(z)-f(z_0)}{(z-z_0)^n} = \sum_{k=n}^{\infty}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^{k-n}$$
In the above expression, we have a product of $(z-z_0)$ in every term except for $k=n$ so as $z\rightarrow z_0$ we will have the remainder
$$\lim_{z\rightarrow z_0}\frac{f(z)-f(z_0)}{(z-z_0)^n} = \frac{f^{(n)}(z_0)}{n!}$$
A: Let $f(z) = z^n$, and let $z_0 = 1$, then clearly $f^{(n)}(z) = n!$, thus $\frac{f^{(n)}(z_0)}{n!} = 1 \not= 0$. However, the right-hand-side of the you definition is infinite:
$$
   \lim_{z \to 1} \frac{z^n - 1^n}{(z-1)^n} = \lim_{z \to 1} \left( \frac{z - 1}{(z-1)^n}\cdot \sum_{m=0}^{n-1} z^m \right) = \lim_{z \to 1} \left( \frac{n}{(z-1)^{n-1}}\right) = \cases{ 1 & n=1 \cr \infty & n >1}
$$
Thus you definition can not be correct for $n>1$.
