Problem with numerator in a series expansion

  1. 1. The problem statement, all variables and given/known data
    The problem I am having has to do with part (d) in the picture which I have attached. I have managed to get as far as to determine that the coefficients in the series expansion have the recurrence relation shown below in part (2). From this I think that I have been able to determine that the general form of the coefficients must what is shown in part (3) below. The issue is I am unsure of how to get the proper form of the numerator. Any assistance would be greatly appreciated, thanks!


    2. Relevant equations
    [itex]a_{n+2}=\frac{n(n+3)-\lambda}{R^{2}(n+2)(n+3)}[/itex] where [itex]a_{o}=1[/itex]

    [itex]λ=\frac{2m^{2}}{\omega_{o}^{2}}[/itex] where m is the separation constant

    3. The attempt at a solution
    [itex]a_{2n}=\frac{something}{(R^{2})^{n}(2n+1)!}[/itex]
     

    Attached Files:

  2. jcsd
  3. SammyS

    SammyS 8,887
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    Let's a least make the image more accessible.
    [​IMG]
     
  4. Much better, so any ideas?
     
  5. vela

    vela 12,767
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    Part (d) only asks for the recurrence relation. You don't need to get a closed form for an.
     
  6. ugh now I feel dumb. Guess I should have read it most closely. Thanks though. I don't suppose you have any ideas about part (e). I realize that if r=R then the series is simply an expansion of terms that go to infinity but I am unsure of how I would truncate the series.
     
  7. vela

    vela 12,767
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    You need the numerator in the recurrence relation to vanish at some point so that all the subsequent coefficients will be 0. That requirement tells you the possible values of λ.
     
  8. is there a way to determine at what specific point it should vanish?
     
  9. vela

    vela 12,767
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    Nope, because you're obtaining a family of solutions. If the series terminates after one term, that's one mode of oscillation. If it terminates after two terms, that's another mode of oscillations, and so on.
     
  10. Oh ok, I will work at this and see what I can do. Thank you for your help!
     
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