1

EDIT: How to get this, if it is possible to get it: $$f^{(n)}(z_o)=\frac{n!}{2\pi}\int_0^{2\pi}f(e^{i\theta}+z_o)(e^{ni\theta})^{-1}\ d\theta$$

I was reading this question, and wanted to write it down to also find Cauchy's differentiation formula. This is what I have so far:

We have Cauchy's formula: $$f(z_o)=\frac{1}{2\pi i}\int_\Gamma \frac{f(z)}{z-z_o}dz$$ then we consider $\gamma:[0,2\pi]\to\Bbb C$ as $\theta\mapsto re^{i\theta}+z_o$, which is circle with center in $z_o$ and radius $r>0$ such that it is entirely contained in the geometric interior of $\Gamma$. We get: $$f(\gamma(\theta))=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(\gamma(\theta))}{\gamma(\theta)-z_o}\gamma'(\theta)\ d\theta \\ =\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(re^{i\theta}+z_o)}{re^{i\theta}}(ire^{i\theta})\ d\theta \\ =\frac{1}{2\pi}\int_0^{2\pi} f(re^{i\theta}+z_o)\ d\theta$$ (a): Is this what it says in the answer? I got a little confused by the notation.

To get the differentiation formulas, I did pretty much the same as above, but with $f'(z_o)$:

$$f'(\gamma(\theta))=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(\gamma(\theta))}{(\gamma(\theta)-z_o)^2}\gamma'(\theta)\ d\theta=\frac{1}{2\pi}\int_0^{2\pi}f(re^{i\theta}+z_o)(re^{i\theta})^{-1}\ d\theta \tag{1}$$

(b): Is there a way to get something prettier looking like: $$f'(z_o)=\frac{1}{2\pi}\int_0^{2\pi}f(e^{i\theta}+z_o)(e^{i\theta})^{-1}\ d\theta \tag{2}$$

In $(2)$ I feel that we are only moving the argument.

Community
  • 1
Ana Galois
  • 2,466
  • can't we just make substitution in Cauchy Integral formula? – S L May 28 '14 at 18:00
  • @SantoshLinkha That's what I thought, but I can't find a way to get rid of the $r$ – Ana Galois May 28 '14 at 18:04
  • take $r=1$, a unit circle, it will still include only the singularity since $f(z)$ is assumed to be analytic. – S L May 28 '14 at 18:05
  • @SantoshLinkha but what if the unit circle is too big for $\Gamma$? can't that happen? – Ana Galois May 28 '14 at 18:08
  • 1
    it shouldn't make difference unless $f(z)$ has pole outside of region $\Gamma$ – S L May 28 '14 at 18:10
  • @SantoshLinkha uhm... so if we had as hypothesis that $f$ is analytic in and over $\Gamma$ then we're save. What if that doesn't happen? What can we do? – Ana Galois May 28 '14 at 18:12
  • perhaps you wouldn't get this nice formula, I mean you cannot freely manipulate the values of $r$. I have't experimented with it, but in many cases, usually $f$ is nice we can evaluate integrals like $$\int_{0}^{2\pi} \log (a + b \sin (x)) dx$$ – S L May 28 '14 at 18:16
  • here is an example i had difficulty dealing with it, but IDK how integral happened to be real part of it. – S L May 28 '14 at 18:19
  • @SantoshLinkha wow, that was intense... – Ana Galois May 28 '14 at 18:26
  • 1
    haha ... i suppose no $f(z)$ has to be analytic in region $|z-z_0| < 1$ from here since what you claim in just higher order generalization. still let's see what experts has to say ,,, I am just small fry on Complex Analysis. – S L May 28 '14 at 18:29

0 Answers0