Trivial power series question

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

I can't remember.
 

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  • #2
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How do I find the power series for [itex]z^7[/itex]?

I can't remember.
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?
.
.
.
 
  • #3
Dick
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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
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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.
Around zero. I have been looking through Rudin and Rosenlicht but I don't see an example of what I am looking for.
 
  • #5
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.
Then what are you looking for? z^7 IS the power series for the function f(z)=z^7 around z=0.
 
  • #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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$$
f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_0^{2\pi}\frac{u^7}{u^{n + 1}}du z^n\right] = z^7.
$$

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?
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.
 
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  • #8
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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).
u is a complex number and u^7 is analytic and continuous.


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.
 
  • #9
Dick
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[itex]\int_C \frac{u^7}{u^{n + 1}}du=\int_C u^{7 - n - 1}du[/itex]. That's zero unless the exponent is -1 which happens when n=7.
 
  • #10
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[itex]\int_C \frac{u^7}{u^{n + 1}}du=\int_C u^{7 - n - 1}du[/itex]. That's zero unless the exponent is -1 which happens when n=7.
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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Ok so if the function was 1/u, we would have u^{-n-2}. This one would always be 0 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.
 
  • #13
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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.
If we haven't done Laurent Series yet, how should I handle it then?
 
  • #14
Dick
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If we haven't done Laurent Series yet, how should I handle it then?
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
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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.
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
Dick
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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?
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
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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.
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
Dick
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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?
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
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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.
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
Dick
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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'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.
 
  • #21
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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.
I understand that. I was verifying that f(z) = 0 is correct. You kept saying you don't understand what I telling you.
 
  • #22
Dick
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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.
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.
 
  • #23
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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.
No problem. I know f(z) = 1/z\neq 0 but when I said f(z) I meant,

$$
f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_0^{2\pi}\frac{\frac{1}{u}}{u^{n + 1}}du z^n\right].
$$
 
  • #24
Dick
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No problem. I know f(z) = 1/z\neq 0 but when I said f(z) I meant,

$$
f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_0^{2\pi}\frac{\frac{1}{u}}{u^{n + 1}}du z^n\right].
$$
Yeah, of course it's 0. Sorry again for being a bit thick about what you were asking.
 

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