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Simplifying to a Geometric Series

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    I have a question with asks to solve a differential equation via power series and I've done everything up to finding the recurrence relation which is a[itex]_{n+2} = -[/itex][itex]\frac{a_{n}}{n+2}[/itex]
    Given the initial conditions a[itex]_{o}[/itex] = 1 and a[itex]_{1}[/itex] = 0 I'm trying to simplify the series into a geometric series. The series is 1,-1/2, 1/8, -1/48, 1/480, -1/5760 etc...


    3. The attempt at a solution

    So far I've gotten (-1)[itex]^{n}[/itex] in the numerator to account for the alternating negative sign, however I can't find the denominator for the life of me...
    my best attempt is [itex]\frac{(-1)^{n}}{(2^n)}[/itex] but it only works for the first 2 terms :S
     
  2. jcsd
  3. Nov 30, 2011 #2

    Mute

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    Don't just try to randomly guess what the pattern is by looking at the first few terms. You want to try to do it a little more systematically than that. Try to write [itex]a_{n+2}[/itex] in terms of [itex]a_0[/itex] (hopefully you noticed that all the odd coefficients are zero).

    i.e., start writing
    [tex]a_{2k+2} = -\frac{a_{2k}}{2k+2} = (-1)^2\frac{a_{2k-2}}{(2k+2)(2k)} = (-1)^3\frac{a_{2k-4}}{(2k+2)(2k)(2k-2)} = \dots[/tex]
    and then see if you can guess a pattern that gives [itex]a_{2k+2} = a_0/(\mbox{stuff})[/itex]. (note that I set n = 2k).

    Once you figure out the pattern, if you want to legitimately prove it's correct you can prove it using induction (i.e., prove it holds for k = 1, then prove that if you assume the form is true for k it follows that it's true for k+1).

    Also, note that the terms in a geometric series have the form [itex]ar^k[/itex]. Your terms won't look quite like this, so it's not exactly a geometric series.
     
  4. Nov 30, 2011 #3
    The best I can do is 2k!! which gives 2k(2k-2)(2k-4)...right? it works up k=3 to but it's missing the 2k+2 term and doesnt work for k=4
    I know 2k!! is expressed as [itex]2^{k}[/itex]k! but how is (2k+2)!! expressed in such a way that I can calculate it?

    edit*
    oops, 2k!! is right I believe. I miscalculated the series in my first post, instead of -1/48, 1/480 it should have been -1/48, 1/384 etc..
     
    Last edited: Nov 30, 2011
  5. Nov 30, 2011 #4
    Looks like 1/480 should be 1/384 instead, and 1/5760 be 1/3840. Did you skip a term?

    In any case, you might want to check out Sloan's Sequences: https://oeis.org/
     
  6. Nov 30, 2011 #5

    Mute

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    If you're still not sure, just take (2k)!! and replace k with k + 1: (2(k+1))!! = (2k+2)!! = 2k+1(k+1)!, such that

    [tex]a_{2k+2} = \frac{(-1)^{k+1}}{2^{k+1}(k+1)!}[/tex]
     
  7. Dec 1, 2011 #6

    HallsofIvy

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    (2k+2)!!= (2(k+1))!! so it would just be [itex]2^{k+1}(k+1)![/itex]

     
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