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How do I find the power series for [itex]z^7[/itex]?

I can't remember.

I can't remember.

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- #1

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How do I find the power series for [itex]z^7[/itex]?

I can't remember.

I can't remember.

- #2

Mark44

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What would be the coefficient for the constant term?How do I find the power series for [itex]z^7[/itex]?

I can't remember.

What would be the coefficient for the z term?

What would be the coefficient for the z

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.

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- #3

Dick

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

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Dick

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Then what are you looking for? z^7 IS the power series for the function f(z)=z^7 around z=0.Around zero. I have been looking through Rudin and Rosenlicht but I don't see an example of what I am looking for.

- #6

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

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

Dick

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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.$$

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?

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u is a complex number and u^7 is analytic and continuous.Well, the integral is zero unless n=7, isn't it? The integral of u^k is zero unless k=(-1).

For all z inside of C (C the unit circle oriented counterclockwise),

[tex]

f(z) = \frac{1}{2\pi i}\int_C \frac{g(u)}{u-z} du

[/tex]

where [itex]g(u) = \bar{u}[/itex] is a continuous function and [itex]f[/itex] is analytic in C. Describe [itex]f[/itex]in C in terms of a power series.

[itex]\displaystyle f(z) = \frac{1}{2\pi i}\int_C \frac{\bar{u}}{u-z} du[/itex] I am confused with what I am supposed to do. I know it says describe [itex]f[/itex] in terms of a power series.

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

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Dick

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Ok, I understand now, thanks.

- #11

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

- #12

Dick

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

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

- #14

Dick

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

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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?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.

- #16

Dick

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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.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?

- #17

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I understand what you are saying but I am trying to solve forAll 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.

$$

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

Dick

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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.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?

- #19

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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.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.

- #20

Dick

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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.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.

- #21

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I understand that. I was verifying that f(z) = 0 is correct. You kept saying you don't understand what I telling you.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.

- #22

Dick

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Apologies if I'm misunderstanding. But I'm just saying 0 isn't the power series representation of 1/z because it doesn't have one. And f(z)=1/z is not equal to 0. That's all. So, ok. Yes f(z)=0. I somehow thought you were trying to represent 1/z as a power series. Sorry.I understand that. I was verifying that f(z) = 0 is correct. You kept saying you don't understand what I telling you.

- #23

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No problem. I know f(z) = 1/z\neq 0 but when I said f(z) I meant,Apologies if I'm misunderstanding. But I'm just saying 0 isn't the power series representation of 1/z because it doesn't have one. And f(z)=1/z is not equal to 0. That's all. So, ok. Yes f(z)=0. I somehow thought you were trying to represent 1/z as a power series. Sorry.

$$

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].

$$

- #24

Dick

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Yeah, of course it's 0. Sorry again for being a bit thick about what you were asking.No problem. I know f(z) = 1/z\neq 0 but when I said f(z) I meant,

$$

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].

$$

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