(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

[tex]f^k(z)=\frac{k!}{2i\pi}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^{k+1}}[/tex]

Prove for second derivative.

3. The attempt at a solution

For the first derivative, I would start with

[tex]\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[/tex]

=[tex]\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)(\zeta-(z+h)}[/tex]

(HERE, I DON'T UNDERSTAND WHERE THE H DISAPPEARED TO)

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

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

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:

[tex]\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2}[/tex]

For the second derivative, I would start with

[tex]\frac{f'(z+h)-f'(z)}{h}=\frac{1}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^2(\zeta-z-h)}[/tex]

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

[tex]\frac{2}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^3}[/tex]

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

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# Homework Help: Cauchy Integral Number 2 (Proof)

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