Answer to Question #130273 in Complex Analysis for tom

Question

𝑓 ^(𝑛) (𝑧) = 𝑛! /2𝜋𝑖 ∫_𝜕𝐷 𝑓(𝑤)𝑑𝑤 / (𝑤−𝑧)^𝑛+1 . using the generalized Cauchy integral formula (GCIF) expressed in

Prove it with the Mathematical Induction Method.(As is known, the formula for = 1 is Cauchy Integral

It becomes the Cauchy Integral Formula (CIF), which is the result of the theorem and it is correct.)

Answer 1

Proof by induction:

For all $n\in\N_{>0},$ let $P(n)$ be the proposition:

$$f^{(n)}(z)=\dfrac{n!}{2\pi i}\int_{\partial D}\dfrac{f(w)}{(w-z)^{n+1}}dw$$

Basis for the Induction

$P(0)$ holds, as this is:

$$f^{(0)}(z)=\dfrac{0!}{2\pi i}\int_{\partial D}\dfrac{f(w)}{(w-z)^{0+1}}dw$$

$$f(z)=\dfrac{1}{2\pi i}\int_{\partial D}\dfrac{f(w)}{w-z}dw$$

which is Cauchy's Integral Formula.

This is our basis for the induxtion. 

Induction Hypothesis

Now we need to show that, if $P(k)$ is true, where $k\geq0,$ then it logically follows that $P(k+1)$ is true.

So this is our induction hypothesis:

$$f^{(k)}(z)=\dfrac{k!}{2\pi i}\int_{\partial D}\dfrac{f(w)}{(w-z)^{k+1}}dw$$

Then we need to show:

$$f^{(k+1)}(z)=\dfrac{(k+1)!}{2\pi i}\int_{\partial D}\dfrac{f(w)}{(w-z)^{(k+1)+1}}dw$$

Induction Step

This is our induction step:

$$\dfrac{d}{dz}f^{(k)}(z)=\dfrac{k!}{2\pi i}\int_{\partial D}\dfrac{d}{dz}\dfrac{f(w)}{(w-z)^{k+1}}dw=$$

$$=\dfrac{k!}{2\pi i}\int_{\partial D}(k+1)\dfrac{f(w)}{(w-z)^{k+2}}dw=$$

$$=\dfrac{(k+1)!}{2\pi i}\int_{\partial D}\dfrac{f(w)}{(w-z)^{(k+1)+1}}dw=P(k+1)$$

So $P(k)=>P(k+1)$ and the result follows by the Principle of Mathematical Induction.

There $\forall n\in\N:$

$$f^{(n)}(z)=\dfrac{n!}{2\pi i}\int_{\partial D}\dfrac{f(w)}{(w-z)^{n+1}}dw$$