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Homework Help: Is there a differnt way to write this expression?

  1. Mar 7, 2010 #1
    f(z) = SUM [(-1)n

    Where SUM = sum from n=0 to infinity

    Looks to me like there isn't but there must be a simpler way to write that.

    Thanks
     
    Last edited: Mar 7, 2010
  2. jcsd
  3. Mar 7, 2010 #2

    Dick

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    It's divergent, isn't? The nth term doesn't approach 0.
     
  4. Mar 7, 2010 #3
    The question is to write this power series as a function

    2hn01gp.jpg
     
    Last edited: Mar 7, 2010
  5. Mar 7, 2010 #4

    Dick

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    z^(2n)/n! isn't equal to e^(z^2). The SUM of the series z^(2n)/n! is e^(z^2). That would be fine if the (-1)^n weren't there. Can't you think of a function that's a lot like e^(z^2) that will give you the alternating sign in the series?
     
  6. Mar 7, 2010 #5
    I just keep getting into the routine of trying to simply it like before. Do you mean replace e^(z^2) or multiply in some function f(z)/f(z)?
     
  7. Mar 7, 2010 #6

    Dick

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    (-1)^n*z^(2n). Can you think of a way to write that as r^n for some number r? What would r be?
     
  8. Mar 7, 2010 #7
    ahh yes
     
    Last edited: Mar 7, 2010
  9. Mar 7, 2010 #8
    For the 2nd part I have it down to

    z-2 SUM (-1)n.(z2/4)n+1

    Is there anyway I can progress from here or should I go back and try a different way?
     
  10. Mar 7, 2010 #9

    Dick

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    It's a geometric series.
     
  11. Mar 7, 2010 #10
    (z2/4)n+1 starting at n=0

    = 1/(1-z2/4) - 1

    => Series = z-2 SUM (-1)n/(1-z2/4) + (-1)n+1
     
  12. Mar 7, 2010 #11

    Dick

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    You just did the same wrong thing you did in the first problem. You replaced a term is the series with the sum of a series. And you missed the cue to do the same thing you did with the (-1)^n in the first problem. You series has the form a*r^n. r just might have a (-1) in it.
     
  13. Mar 7, 2010 #12
    ah yes, i have my r now thanks, was simple after that!

    Thanks alot
     
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