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ODE power series method

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data
    using the fact that [tex]\frac{sint}{t}[/tex] = [tex]\sum^{}_{n=0}[/tex][tex]\frac{ (-1)[tex]^{n}[/tex] t[tex]^{2n}[/tex]}{(2n+1)!}[/tex]

    i wanted to write:

    using that sint/t = ∑ (-1)ⁿ·t²ⁿ/(2n+1)!

    find a power series solution of the equation tx'' +sint x = 0
    2. Relevant equations
    sorry i pushed the wrong button,


    3. The attempt at a solution
    I'm a beginner at LaTeX, a disaster at LaTeX

    i get something like ∑n(n-1)a_nt^(n-2) +sint∑a_n = 0
     
    Last edited: Nov 4, 2009
  2. jcsd
  3. Nov 5, 2009 #2


    You only need to use the tex-brackets at the beginning and at the end; don't clutter the formula with them. Also, you should use \sin (or\mathrm{sin}) instead of sin, makes the font right.

    [tex]\frac{\sin t}{t} = \sum_{n=0}\frac{ (-1)^{n} t^{2n}}{(2n+1)!}[/tex]

    Okay, so how do find a power series solution? You write the DE as a sum over powers of t and then require that the coefficient is zero for all n. That gives you a recursion relation between the coefficients a_n.

    You're missing a t and a t^n there. Then write out the series expansion for sint/t and you can find a formal solution of the form

    [tex] \sum_{n=0}^{\infty} A_n t^n = 0 [/tex] where now A_n is something quite complicated, but it contains the recursion relation for the coefficients a_n. I am not sure if you can find an answer in terms of elementary functions; the form of the equation suggests to me that you can't.
     
  4. Nov 5, 2009 #3
    Well this is how far i get:
    Using that

    [tex]\frac{\sin t}{t} = \sum_{n=0}^{\infty}\frac{(-1)^{n} t^{2n} }{(2n+1)!}[/tex]

    for t≠0 i rewrite tx''+sint x =0 as

    [tex](\aleph); x'' + \frac{\sin t}{t}x = 0 [/tex]

    [tex] Ansats: x(t) = \sum_{n=0}^{\infty}a_{n}t^{n} [/tex]

    then

    [tex] x''(t) = \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2}[/tex]

    pluggin this and the ansats for x(t) into [tex](\aleph) [/tex]

    i get

    [tex] \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sum_{n=0}^{\infty}\frac{(-1)^{n}t^{2n} }{(2n+1)!}\sum_{n=0}^{\infty}a_{n}t^{n} [/tex]

    here is where i'm not 100% sure.

    i multiply the last two series getting

    [tex] \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sum_{n=0}^{\infty}\frac{(-1)^{n}t^{2n+1} }{(2n+1)!}\sum_{n=0}^{\infty}a_{n} [/tex]

    which is equivalent to

    [tex]\sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sin t\sum_{n=0}^{\infty}a_{n}[/tex]

    now what's next?
     
  5. Nov 5, 2009 #4
    You cannot use the same indices with the two sums. Use for example m and n. What you should have then is something proportional to [tex] t^{2m+n} [/tex]. Now you need to rescale the sums so that you will have something of the form [tex] \sum A_n t^n [/tex] where then A_n will be zero for all n. So for example for the first sum you'd write
    [tex] \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} t^{n} [/tex].
     
  6. Nov 5, 2009 #5
    [tex]
    \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sum_{m=0}^{\infty}\frac{(-1)^{m}t^{2m} }{(2m+1)!}\sum_{n=0}^{\infty}a_{n}t^{n}
    [/tex]

    [tex]\Leftrightarrow[/tex]

    [tex] \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}t^{n} + \sum_{^{m=0}_{n=0}}^{\infty}\frac{(-1)^{m}t^{2m+n} }{(2m+1)!}a_{n} [/tex]

    i know how to shift series and add them but now what?

    sincerely
     
  7. Nov 5, 2009 #6
    You shift the term on the right as well. Just notice when you do it: then you only sum over a finite number of m:s for each n.
     
  8. Nov 5, 2009 #7
    what's the substitution in the second series?

    give me a hint
     
  9. Nov 5, 2009 #8
    Since you want to have t^n there, you have to do n+2m -> n. Then a_n -> a_(n-2m), and the summation over m will be cut off by the fact that a_(-1) = 0. That's all there is to it.
     
  10. Nov 5, 2009 #9
    yes that is exactly what i tried but got confused.

    i used 2m+n = k then m = 1/2·(k-n) and (2m+1)! --> (k+1-n)! and [tex]a_{n} \rightarrow a_{k-2m}[/tex]
     
    Last edited: Nov 5, 2009
  11. Nov 5, 2009 #10
    You don't want to replace any m:s, only n:s. So the only things you need to touch are an and tn+2m. You should end up with

    [tex] \sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2} + \sum_{m=0}^{n/2} a_{n-2m} \frac{(-1)^m}{(2m+1)!}\right] t^n, [/tex]

    where the sum over m goes to m=n/2 for even values of n and to m=(n-1)/2 for odd values.
     
  12. Nov 5, 2009 #11
    Danke Clamtrox
     
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