# Homework Help: Trivial power series question

1. Feb 23, 2012

### fauboca

How do I find the power series for $z^7$?

I can't remember.

2. Feb 23, 2012

### Staff: Mentor

What would be the coefficient for the constant term?
What would be the coefficient for the z term?
What would be the coefficient for the z2 term?
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3. Feb 23, 2012

### Dick

If you mean the power series expansion around a point that's not 0, say around z=1, then write z^7=((z-1)+1)^7 and expand that.

4. Feb 23, 2012

### fauboca

Around zero. I have been looking through Rudin and Rosenlicht but I don't see an example of what I am looking for.

5. Feb 23, 2012

### Dick

Then what are you looking for? z^7 IS the power series for the function f(z)=z^7 around z=0.

6. Feb 23, 2012

### fauboca

$$f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_0^{2\pi}\frac{u^7}{u^{n + 1}}du z^n\right] = z^7.$$

I want to show the above equality. I know since u^7 is analytic in the unit disk, g will be the same as f. But is there a way to show that without stating this?

7. Feb 23, 2012

### Dick

Well, the integral is zero unless n=7, isn't it? The integral of u^k is zero unless k=(-1). And I'm assuming you actually meant a contour integral around the origin, not u=0 to u=2pi.

Last edited: Feb 23, 2012
8. Feb 23, 2012

### fauboca

u is a complex number and u^7 is analytic and continuous.

Only difference I am dealing with u^7 not the conjugate.

9. Feb 23, 2012

### Dick

$\int_C \frac{u^7}{u^{n + 1}}du=\int_C u^{7 - n - 1}du$. That's zero unless the exponent is -1 which happens when n=7.

10. Feb 23, 2012

### fauboca

Ok, I understand now, thanks.

11. Feb 23, 2012

### fauboca

Ok so if the function was 1/u, we would have u^{-n-2}. This one would always be 0 then?

12. Feb 23, 2012

### Dick

Sure, but f(u)=1/u doesn't have a power series expansion around 0 with only positive powers. You'd need a Laurent series instead of a power series to represent it.

13. Feb 23, 2012

### fauboca

If we haven't done Laurent Series yet, how should I handle it then?

14. Feb 23, 2012

### Dick

You don't. f(z)=1/z doesn't have a power series expansion around z=0. f(z) has to be analytic at z=0 to have a power series expansion. 1/z isn't analytic at z=0.

15. Feb 23, 2012

### fauboca

Evaluating the sum and the integral for it yields f(z) = 0. Is that correct to put down after evaluating f(z) since all the terms are 0?

16. Feb 23, 2012

### Dick

All of the terms in your series are zero, yes. But that still doesn't make f(z)=1/z=0. I'm not sure you are paying attention here.

17. Feb 23, 2012

### fauboca

I understand what you are saying but I am trying to solve for
$$f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_0^{2\pi}\frac{\frac{1}{u}}{u^{n + 1}}du z^n\right].$$
Since Laurent series are out and all the terms are 0, what else could f(z) be?

18. Feb 23, 2012

### Dick

I am trying to tell you that the series you are quoting is NOT valid for all functions f(z). f(z) has to be analytic at z=0 to apply that. f(z)=1/z is NOT analytic at z=0. I've already told you this.

19. Feb 23, 2012

### fauboca

By the integral transform theorem, if you put a continuous function g(u) into f(z), you get out an analytic function. If g is analytic, you get the same function. So f(z) has to equal something.

20. Feb 23, 2012

### Dick

I'm not quite sure why this is so difficult. 1/u is continuous on the contour. And yes, you get an analytic function out. It's f(z)=0. Now you say "If g is analytic, you get the same function.". g(u)=1/u ISN'T analytic at u=0. So the function you get out f(z)=0, ISN'T the same as the function you put in f(z)=1/z.