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Trivial power series question

  1. Feb 23, 2012 #1
    How do I find the power series for [itex]z^7[/itex]?

    I can't remember.
     
  2. jcsd
  3. Feb 23, 2012 #2

    Mark44

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    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?
    .
    .
    .
     
  4. Feb 23, 2012 #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.
     
  5. Feb 23, 2012 #4
    Around zero. I have been looking through Rudin and Rosenlicht but I don't see an example of what I am looking for.
     
  6. Feb 23, 2012 #5

    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.
     
  7. Feb 23, 2012 #6
    $$
    f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_0^{2\pi}\frac{u^7}{u^{n + 1}}du z^n\right] = z^7.
    $$

    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?
     
  8. Feb 23, 2012 #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.
     
    Last edited: Feb 23, 2012
  9. Feb 23, 2012 #8
    u is a complex number and u^7 is analytic and continuous.



    Only difference I am dealing with u^7 not the conjugate.
     
  10. Feb 23, 2012 #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.
     
  11. Feb 23, 2012 #10
    Ok, I understand now, thanks.
     
  12. Feb 23, 2012 #11
    Ok so if the function was 1/u, we would have u^{-n-2}. This one would always be 0 then?
     
  13. Feb 23, 2012 #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.
     
  14. Feb 23, 2012 #13
    If we haven't done Laurent Series yet, how should I handle it then?
     
  15. Feb 23, 2012 #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.
     
  16. Feb 23, 2012 #15
    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. Feb 23, 2012 #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.
     
  18. Feb 23, 2012 #17
    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. Feb 23, 2012 #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.
     
  20. Feb 23, 2012 #19
    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. Feb 23, 2012 #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.
     
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