Cauchy Integral Number 2 (Proof)

[Milky]

Homework Statement

If f(z) is analytic interior to and on a simple closed contour C, then all the derivatives $$f^k(z)$$ , k=1,2,3... exist in the domain D interior to C, and

$$f^k(z)=\frac{k!}{2i\pi}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^{k+1}}$$
Prove for second derivative.

The Attempt at a Solution

For the first derivative, I would start with
$$\frac{f(z+h)-f(z)}{h}=\frac{1}{2i\pi}\frac{1}{h}\int_{C}{f(\zeta)(\frac{1}{\zeta-(z+h)}-\frac{1}{\zeta-z})d\zeta$$

=$$\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)(\zeta-(z+h)}$$
(HERE, I DON'T UNDERSTAND WHERE THE H DISAPPEARED TO)
=$$\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2}+R$$

where R= $$\frac{h}{2i\pi}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2(\zeta-z-h)}$$

By the ML Theorem, we end up with the R term, and therefore the whole term going to zero, so that all that is left is:

$$\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2}$$

For the second derivative, I would start with
$$\frac{f'(z+h)-f'(z)}{h}=\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2(\zeta-z-h)}$$

Again, I get a term R that goes to zero, and am suppose to be left with

$$\frac{2}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^3}$$

Where does the two in the numerator come from? I'm confused by the calculations.

 

[HallsofIvy]

Homework Statement

If f(z) is analytic interior to and on a simple closed contour C, then all the derivatives $$f^k(z)$$ , k=1,2,3... exist in the domain D interior to C, and

$$f^k(z)=\frac{k!}{2i\pi}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^{k+1}}$$
Prove for second derivative.

The Attempt at a Solution

For the first derivative, I would start with
$$\frac{f(z+h)-f(z)}{h}=\frac{1}{2i\pi}\frac{1}{h}\int_{C}{f(\zeta)(\frac{1}{\zeta-(z+h)}-\frac{1}{\zeta-z})d\zeta$$

=$$\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)(\zeta-(z+h)}$$
(HERE, I DON'T UNDERSTAND WHERE THE H DISAPPEARED TO)

First, there was no "H" to disappear! You mean, of course, "h", but you need to be careful about that: in general "H" and "h" may be very different things.

What do you get if you actually do the subtraction? What is
$$\frac{1}{\zeta-(z+h)}-\frac{1}{\zeta-z}$$

=$$\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2}+R$$

where R= $$\frac{h}{2i\pi}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2(\zeta-z-h)}$$

By the ML Theorem, we end up with the R term, and therefore the whole term going to zero, so that all that is left is:

$$\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2}$$

For the second derivative, I would start with
$$\frac{f'(z+h)-f'(z)}{h}=\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2(\zeta-z-h)}$$

Again, I get a term R that goes to zero, and am suppose to be left with

$$\frac{2}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^3}$$

Where does the two in the numerator come from? I'm confused by the calculations.

 

[Milky]

Oh! Duh... you get

$$\frac{h}{\zeta-(z+h)(\zeta-z)}$$

And that h cancels with the 1/h on the outside.
Thanks!

Does anyone see where the 2! in the numerator of
$$\frac{2!}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^3}$$