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Recurrence relation for Bessel Functions

  1. Aug 12, 2016 #1
    1. The problem statement, all variables and given/known data
    I want to prove this relation
    ##J_{n-1}(x) + J_{n+1}(x)=\frac{2n}{x}J_{n}(x))##
    from the generating function. The same question was asked in this page with solution.
    http://www.edaboard.com/thread47250.html
    My problem is the part with comparing the coefficient. I don't understand how to compare them. They all have different powers. All pages that i visited have the same solution. Just said "compare the coefficient." They don't explain it.
    Do you have a homepage that contains a detailed solution?
     
  2. jcsd
  3. Aug 12, 2016 #2

    fresh_42

    Staff: Mentor

    If two polynomials ##x^2+2x+3## and ##a_3x^3+a_2x^2+a_1x+a_0## are equal, what can you say about the ##a_i##?
     
  4. Aug 12, 2016 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    If you have two power series in ##t## that are supposed to be equal for all ##t##, then for each ##n## the coefficients of ##t^n## must be the same on both sides. So, look at the terms in ##t^0## on both sides; what will that give you? Then look at the terms in ##t^1## on both sides, and see what you get. Now try to extract the terms in ##t^n## on both sides.

    Note that if you have something like ##\sum_n a_n t^n + \sum b_n t^{n-1} + \sum_n c_n t^{n+1}## then the terms in ## t^n## are ##(a_n + b_{n+1} + c_{n-1}) t^n##. Do you see why?
     
  5. Aug 13, 2016 #4
    I think i got it.
    Also
    ##a_0=3##
    ##a_1=2##
    ##a_2=1##
    ##a_3=0##


    Sorry but i never had to use this method. I dont know why i can change the index "n". Is it because of the summation?
     
  6. Aug 13, 2016 #5

    Ray Vickson

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    Science Advisor
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    For the summation I presented, write out ALL the terms for n = 1, 2 and 3; there are 9 of them. Now pick out all the terms in ##t^2##.
     
  7. Aug 13, 2016 #6
    Thank you. I understand it now.

    I got this
    ##(a_1t^1+a_2t^2+a_3t^3+b_1t^0+ b_2t^1+b3t^2+c_1t^2+c_2t^3+c_3t^4##.
    The terms for ##t^2## are
    ##a_2+b_3+c_1##
    Which is the same as

    ##(a_n + b_{n+1} + c_{n-1}) t^n##.
     
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