1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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?

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook