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 Exercise [07.03] b 
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Joined: 07 Jun 2008, 08:21
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Post Exercise [07.03] b
This is what I think Penrose wants you to do:

Taylor's expansion for f(z) about the point p is:

f(z)=\sum^\infty _{k=0}a_k(z-p)^k,\ p\ and\ z\ complex\ and\ a_k=\frac{f^{(n)}(p)}{k!}

Also Cauchy's formula in the origin shifted form is:

f^{(n)}(p)=\frac{n!}{2\pi i} \oint\frac{f(z)}{{(z-p)}^{n+1}}dz

Show that a_n is indeed given by:

\frac{f^{(n)}(p)}{n!}=\frac{1}{2\pi i} \oint\frac{f(z)}{{(z-p)}^{n+1}}dz

I am about to substitute the Taylor series for f(z) given above into the contour integral without worrying about the rigorous justification, which is what Penrose suggests you do in the exercise:

=\frac{1}{2\pi i} \oint\frac{\sum^\infty _{k=0}a_k(z-p)^k}{{(z-p)}^{n+1}}dz

and now a further step without rigorous justification

=\frac{1}{2\pi i} \oint\sum^\infty _{k=0}\frac{a_k(z-p)^k}{{(z-p)}^{n+1}}dz

and then expanding the summation

=\frac{1}{2\pi i} \oint\{\frac{a_0}{(z-p)^{n+1}}+\frac{a_1}{(z-p)^{n}}+...+\frac{a_n}{(z-p)}+a_{n+1}+a_{n+2}(z-p)+a_{n+3}(z-p)^2+...\}dz

Using the result of exercise 7.1 that \oint z^ndz=0 when n is an integer other than -1 we can see that the above simplifies to

\frac{1}{2\pi i} \oint\frac{a_n}{(z-p)}dz=\frac{1}{2\pi i}\cdot2\pi ia_n since \oint\frac{dz}{z-p}=2\pi i see RTR book section 7.2

=a_n which was to be shown.


13 Jul 2008, 16:18
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