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Power series

  1. Aug 21, 2005 #1
    I really need help with this exercise. Consider the power series

    [tex]\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{2n+1}.[/tex]​

    for [tex]z\in\mathbb{C}.[/tex]

    I need to answer the following questions:

    a) Is the series convergent for z = 1?

    This is easy; just plug in z = 1 and observe that the alternating series obtained is convergent using some basic theorems and stuff.

    b) Is the series convergent for z = i?

    Here I'm in trouble; the absolute series (series of absolute values) diverges, but that tells me nothing... Any hints?

    c) Show that the radius of convergence is R=1.

    I have done this, but in a complicated way that isn't the right way, for sure. What's confusing me is that this is not a power series in the standard form, [tex]\sum a_n z^n[/tex] - if you write this series in this way every second term is 0 (corresponding to even n's), so the standard formulas in my book for finding radius of convergence are not applicable (at least, I'm not able to apply them).

    Thanks.
     
  2. jcsd
  3. Aug 21, 2005 #2

    Hurkyl

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    b) This should be rather straightforward... what did you get when you made the substitution?

    c) They don't have to be in standard form to apply the various tests you know (like ratio test or n-th root test... BTW, the latter should apply to the standard form one as well).

    But you're right, this is not the way you're supposed to do it: you're supposed to do it by using basic facts about radius of convergence!
     
  4. Aug 21, 2005 #3
    For z = i I get:

    [tex]\sum_{n=0}^{\infty}\frac{i}{2n+1}.[/tex]

    Most of the series theorems in my book involves only series with real, not complex, terms. The only theorem I can think of is the one that states that for a series to be convergent, the individual terms must tend to 0 - they do in this case, but it is not a sufficient condition, so that's a worhtless observation (if they didn't tend to 0, I could conclude that it diverges, but they do unfortunately). Similarly, if I could show the series to be absolutely convergent, then it would surely be convergent too - but it is not absolutely convergent, so this approach also yields nothing.

    What particular facts do you have in mind? I can't think of any...:biggrin:
     
    Last edited: Aug 21, 2005
  5. Aug 21, 2005 #4

    Hurkyl

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    Can you think of any manipulations you can do with it so you can get your hands on a real summation?


    Something along the lines of... it's definition. :biggrin:
     
  6. Aug 21, 2005 #5
    I can of course throw the i outside the summation and conclude that the series diverges? Is that a valid step when considering infinite sums? Remember, it has to be rigorous (this is a course, where we started out proving that 1+1=2...)


    Hehe, my book defines radius of convergence as:

    [tex]R=\frac{1}{\lim~\sup_n |a_n|^{\frac1{n}}}.[/tex]

    What the **** is that??
     
  7. Aug 21, 2005 #6

    Hurkyl

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    If you're not sure, then prove it. :smile:


    You sure that formula is the definition of "Radius of Convergence", and not just a formula for computing it?

    Anyways, in your own words, what does it mean for the radius of convergence of a power series to be equal to R?
     
  8. Aug 21, 2005 #7
    Now, I've done the b-question in a rigorous and satisfying way. Thanks for helping.

    That is actually the definition in my book (pretty non-intuitive, grrr...). The radius of convergence is just the radius of the circle in the complex plane inside which the power series converges - I don't know what else to say about it??
     
  9. Aug 21, 2005 #8

    Hurkyl

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    Yep. You can say more than that, though...

    Anyways, have the previous parts shed any light on the radius of the circle in the complex plane inside which the power series converges?
     
  10. Aug 22, 2005 #9
    Yes, since it converges at z=1 and diverges at z=i, R must be 1! :) You're a genius, thanks!
     
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