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$$ \frac{f^{(n)}(z_0)}{n!} = \lim_{z \rightarrow z_0} \frac{f(z) -f(z_0)}{(z-z_0)^n} $$

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.

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Is this perhaps from complex analysis? The expression does look vaguely familiar. –  EuYu Oct 2 '12 at 2:53
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2 Answers

up vote 4 down vote accepted

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 $$\lim_{z\rightarrow z_0}\frac{f(z)-f(z_0)}{(z-z_0)^n} = \frac{f^{(n)}(z_0)}{n!}$$

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Thank you very much!!!! –  HCCHUNG Oct 2 '12 at 3:11
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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 $n>1$.

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Thank you! but I changed the problem. Is the above image true? –  HCCHUNG Oct 2 '12 at 3:09
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