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Homework Help: Cauchy Integral Number 2 (Proof)

  1. Oct 17, 2007 #1
    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.
     
  2. jcsd
  3. Oct 17, 2007 #2

    HallsofIvy

    User Avatar
    Science Advisor

    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
    [tex]\frac{1}{\zeta-(z+h)}-\frac{1}{\zeta-z}[/tex]

     
  4. Oct 17, 2007 #3
    Oh! Duh... you get

    [tex]\frac{h}{\zeta-(z+h)(\zeta-z)}[/tex]

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

    Does anyone see where the 2! in the numerator of
    [tex]\frac{2!}{2i\pi}}\int_{C}\frac{f(\zeta)d\zeta}{(\zeta-z)^3}[/tex]
     
    Last edited: Oct 17, 2007
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