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Approximating unsolvable recursion relations

  1. Jul 5, 2010 #1
    I have a complicated recursion replation, which I'm sure is unsolvable. (By "unsolvable" I mean that there is no closed form solution expressing [tex]\xi_1[/tex], [tex]\xi_2[/tex], [tex]\xi_3[/tex], etc. in terms of [tex]\xi_0[/tex].) It goes

    [tex]\frac{(k+4)!}{k!}\xi_{k+4} +K_1 (k+2)(k+1)\xi_{k+2}+ [ K_2 k(k-1) +K_3] \xi_{k} +K_4 \xi_{k-2} =0, [/tex]

    for [tex]k=0,1,2,3...[/tex], where [tex]K_1[/tex], [tex]K_2[/tex], [tex]K_3[/tex] and [tex]K_4[/tex] are constants. However, what I do know about the [tex]\xi_k[/tex]'s is that

    [tex]\sum_{k_{even}} \xi_{k}=\sum_{k_{odd}} \xi_k=1,[/tex]

    and that

    [tex]\sum_{k_{even}}k \xi_{k}=\sum_{k_{odd}}k \xi_k=0.[/tex]

    [tex]\xi_k[/tex] is also zero for all [tex]k<0[/tex]. What I want to do is produce an approximation of the sum for

    [tex]S(z) =\sum_{k=0}^\infty \xi_k z^k.[/tex]

    Does anyone have any idea how to do this?
     
    Last edited: Jul 5, 2010
  2. jcsd
  3. Jul 5, 2010 #2
    If I remember correctly that type of recursion relation is solvable. Somehow you take differences of consecutive equations until you have
    [tex]\sum_1^N a_i\xi_{k+i}=0[/tex]
    and then you use a geometric series as the trial solution. Not sure if the degree of that polynomial becomes too large.

    But maybe the contraints help.
     
  4. Jul 5, 2010 #3
    What do you mean? Isn't S=2 then?
     
  5. Jul 5, 2010 #4
    Oops, sorry. I had the wrong equation for S. I've fixed it now.
     
  6. Jul 5, 2010 #5
    Using something which I think is called z-transform you get
    [tex]S''''+(a+bx^2)S''+(c+dx^2)S=e_5x^5+\dotsb+e_0[/tex]
    I guess that's your initial problem in reverse :)
    Anyone knows how to solve this? I've found
    http://eqworld.ipmnet.ru/en/solutions/ode/ode0406.pdf
    but that only works for a special contraints on the parameters.
     
    Last edited: Jul 5, 2010
  7. Jul 5, 2010 #6
    This series will converge really quickly - this is definitely an entire function. It's probably just as good to deal with the recursive relation. Even if you had a closed form, it wouldn't help you to evaluate the function.

    Do you need a closed form, or do you just think you need it? For the purpose of it being a solution to an ODE, this is just as good.
     
  8. Jul 6, 2010 #7
    That's correct. In fact my equation is

    [tex]S''''+(a+bx^2)S''+(c+dx^2)S=0,[/tex]

    with some inhomogenious boundary conditions.
     
  9. Jul 6, 2010 #8
    Sorry for messing around and confusing people :)
    But maybe the special case from that webpage helps a bit :)
     
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