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 Exercise [07.03] 
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Joined: 30 Jun 2008, 22:14
Posts: 25
Post Exercise [07.03]
This is an interesting exercise though the wording is vague (perhaps deliberately in order to test your ability?). It asks to prove f(z) analytic at the origin if it is complex smooth. In other words, if the Cauchy formula (and hence the origin-shifted Cauchy formula) is valid for f(z), prove it has nth derivative. At the level of formal expressions, one could simply prove that the first derivative exists.

Derivative is defined by the limit: f'(z) = \lim\limits_{h \to 0}\frac{f(z + h) - f(z)}{h}
At the origin: f'(0) = \lim\limits_{h \to 0}\frac{f(h) - f(0)}{h}. Substitute Cauchy and origin-shifted cauchy formulas to the fraction, and get immediately the expression for the first derivative.


08 Jul 2008, 23:42
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Joined: 07 Jun 2008, 08:21
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Post Re: Exercise [07.03]
I agree that the wording in the book about this Exercise is vague. 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^{(k)}(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, 15:02

Joined: 30 Jun 2008, 22:14
Posts: 25
Post Re: Exercise [07.03]
Nah. I don't think that's what the excercise meant. The stuffs you've put in are fine, but they are more or less covered the previous excercise. In fact, they're just the reverse of exercise 7.2.

Your starting point is wrong and you applied the wrong origin-shifted Cauchy because you haven't got the nth derivative established at this point yet. You need to prove its existence first; and the way forward is to derive the 1st derivative. All higher derivatives are then analogically derived.

As I said the text is vague and appears in reversed order.


14 Jul 2008, 17:14

Joined: 25 May 2009, 03:16
Posts: 2
Post Re: Exercise [07.03]
As I undersand the excercise, we are supposed to show that the McLaurin series actually converges to f(z) when f^(n) (0) is defined by the Cauchy formula.

I am new on complex analysis, and do not know how to do that, but I think that you both misinterpreted what is being asked. I agree that the wording is pretty vague though.


29 May 2009, 16:54

Joined: 30 Jun 2008, 22:14
Posts: 25
Post Re: Exercise [07.03]
A bit vague (and confused) around the exercise there. Er.. R. Penrose?
See your point. Like my History teacher used to say: exercising alot only turns one into an exercise slave rather than a master (who actually came up with the exercise). He's right, but useless. Unfortunately, unless one were one of the A's (Arnold/Albert) at birth, exercise's the only way to get there. Sad, eh?


18 Jul 2009, 19:05

Joined: 28 May 2010, 07:42
Posts: 1
Post Re: Exercise [07.03]
This is really an exercise to our brain. I am trying to understand these equations. Anyone could help me? Hope I will get the point of it soon.

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28 May 2010, 08:25

Joined: 03 Jul 2011, 14:43
Posts: 8
Post Re: Exercise [07.03]
http://en.wikipedia.org/wiki/Proof_that ... e_analytic

This Wikipedia article more or less answers the question. There is a step in the proof (the interchange of the summation and the integral) that require some rigorous justification, but Penrose states in the question that it is not necessary. As stated in his hint, it also makes use of the origin-shifted Cauchy integral formula. Also of note is that in the exercise, a is the origin.


Last edited by cwjian on 04 Jul 2011, 18:44, edited 4 times in total.

04 Jul 2011, 15:07
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Joined: 07 Jun 2008, 08:21
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Post Re: Exercise [07.03]
cwjian wrote:
variounes wrote:
This is an interesting exercise though the wording is vague (perhaps deliberately in order to test your ability?). It asks to prove f(z) analytic at the origin if it is complex smooth. In other words, if the Cauchy formula (and hence the origin-shifted Cauchy formula) is valid for f(z), prove it has nth derivative. At the level of formal expressions, one could simply prove that the first derivative exists.

Derivative is defined by the limit: f'(z) = \lim\limits_{h \to 0}\frac{f(z + h) - f(z)}{h}
At the origin: f'(0) = \lim\limits_{h \to 0}\frac{f(h) - f(0)}{h}. Substitute Cauchy and origin-shifted cauchy formulas to the fraction, and get immediately the expression for the first derivative.


I think the part that doesn't require rigorous justification is proving the limit
f'(z) = \lim\limits_{h \to 0}\frac{f(z + h) - f(z)}{h}

because you are dealing with 2 variables in complex analysis, and this is thus the domain of multivariable calculus, which the book hasn't explored yet at this point.

Complex analysis is NOT about 2 variables. It is about ONE variable z. This is the whole point about complex analysis. If you think Complex Analysis is about 2 variables then you have completely missed the point.
The calculus of functions of many variables is completely different and is NOT part of Complex Analysis.


04 Jul 2011, 15:38

Joined: 03 Jul 2011, 14:43
Posts: 8
Post Re: Exercise [07.03]
Actually yes, you are correct, Vasco. I'm sorry, I was confusing it with the multivariable calculus, since the z is of course splittable into 2 components but, of course, complex analysis only involves one variable, so the definition of the derivative applies just as well. But the previous answers don't answer Penrose's question. The proof in the Wikipedia article is actually the correct one, I feel.


04 Jul 2011, 18:07
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Joined: 07 Jun 2008, 08:21
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Post Re: Exercise [07.03]
Yes, I agree that my solution is incorrect. I'll have a look at the wikipedia link when I get some free time.


04 Jul 2011, 18:42
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