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Cauchy's integral formula says $$ f^{(n)}(z)=\frac{n!}{2\pi i}\int_C\frac{f(\zeta)d\zeta}{(\zeta-z)^{n+1}}. $$

If we let $C$ be the circle of radius $r$, such that $|f(\zeta)|\leq M$ on $C$, then taking $z=a$, one obtains Cauchy's estimate that $$ |f^{(n)}(a)|\leq Mn!r^{-n}. $$ How is this derived? I see instead $$ |f^{(n)}(a)|\leq\frac{n!}{2\pi}\int_C \frac{|f(\zeta)||d\zeta|}{|\zeta-a|^{n+1}}\leq Mn!\int_C\frac{|d\zeta|}{|\zeta-a|^{n+1}} $$ but I don't see how this eventually gets to Cauchy's estimate.

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  • $\begingroup$ You lost a factor of $\frac{1}{2\pi}$. The last integral is equal to $2\pi r^{-n}$, assuming $C$ is centered at $a$. $\endgroup$ Feb 28, 2012 at 7:19
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    $\begingroup$ @JonasMeyer Oh, I just got rid of it since the inequality is still true. But now that you mention it, the last integral has that value since $$\int_{|\zeta-a|=r}\frac{|d\zeta|}{|r|^{n+1}}=\frac{2\pi r}{r^{n+1}}=2\pi r^{-n},$$ since the line integral is just the circumference of $C$? $\endgroup$
    – estimate
    Feb 28, 2012 at 7:26

1 Answer 1

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By Cauchy's integral formula you have given, we have $$f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(\zeta)d\zeta}{(\zeta-a)^{n+1}}$$ where $C$ is a circle of radius $r$ centered at $a$. Therefore, $C$ can be parametrized as $\zeta=a+re^{i\theta}$, $0\leq \theta\leq 2\pi$, which implies $$|f^{(n)}(a)|=\left|\frac{n!}{2\pi i}\int_0^{2\pi}\frac{f(a+re^{i\theta})rie^{i\theta}d\theta}{(re^{i\theta})^{n+1}}\right|\leq\frac{n!}{2\pi }\int_0^{2\pi}\left|\frac{f(a+re^{i\theta})rie^{i\theta}}{(re^{i\theta})^{n+1}}\right|d\theta$$ $$=\frac{n!}{2\pi }\int_0^{2\pi}\frac{|f(a+re^{i\theta})|}{r^n}d\theta\leq \frac{n!}{2\pi }\int_0^{2\pi}\frac{M}{r^n}d\theta=\frac{Mn!}{r^n}$$ where the last equality follows from $|e^{i\theta}|=1$ and $|i|=1$, and the last inequality follows from the assumption that $|f|\leq M$ on $C$.

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