Exercise [07.03] b

vasco · 13 Jul 2008, 16:18

This is what I think Penrose wants you to do:

Taylor's expansion for f(z)

about the point

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

is an integer other than

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.

Exercise [07.03] b

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