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Bessel Equations help

  1. Dec 2, 2007 #1
    I have been working on this for some hours now. How can I prove
    Jsub2(x) = (2/x)*Jsub1(x) - Jsub0(x)??
  2. jcsd
  3. Dec 2, 2007 #2


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    Staff: Mentor

    Well, the thing to do would be to prove the general recursive relationship,


    using the identity

    and Bessel's equation.

    With what identities or properties of Bessel's function of the first kind is one familiar?
  4. Dec 2, 2007 #3
    We are using Bessel Equations of the first kind.

    So I did the recursion relation for 2, 1, and zero.
    Like for 2nd i got
    n=0 to inf
    i started doing n=0 though 3 to see a pattern but my TA told us not to do that and just to just the general part which it what i wrote about for Jsub2

    I dont know if I am explaining myself right.
    Sorry if I am not, and thank you for replying :)
  5. Dec 3, 2007 #4


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    Staff: Mentor

    It seems that one was trying to do it inductively, but the TA indicated that one should start with the infinite series for J2(x) and manipulate the terms and show that they can be rearranged to give (2/x)J1(x) - J0(x), which is fine.

    I would recommend that at some point try to prove the general recursive relationship, i.e.

    Jn+1(x) = (2n/x)Jn(x) - Jn-1(x), using the method proposed by the TA.
  6. Dec 3, 2007 #5


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    Staff: Mentor

    It occurred to me that one could use the definition of Jp(x)

    [tex]J_p(x)\,=\,\sum_{m=0}^\infty \frac{(-1)^m(\frac{x}{2})^{2m+p}} {m!\,{\Gamma(1+m+p)}} [/tex]

    then using p = n+1 and p = n-1, show


    which is just a reordering of terms.

    This book might prove useful: Special Functions
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