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Verifying Bessel's Differential Equation

  1. Jul 26, 2010 #1
    Hey everyone, I am trying to use the Bessel function:

    [tex]f(x)= \sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{2^{2k+1}(k!)(k+1)!}[/tex]

    To verify the Differential Equation:

    [tex]x^2y''+xy'+(x^2-1)y=0[/tex]

    So I thought this was gonna be really easy, but I am having some difficulties. First I calculated y,y',y'' to be:

    [tex]y= \sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{2^{2k+1}(k!)(k+1)!}[/tex]

    [tex]y'= \sum_{k=0}^\infty \frac{(-1)^{k}(2k+1)x^{2k}}{2^{2k+1}(k!)(k+1)!}[/tex]

    [tex]y''= \sum_{k=0}^\infty \frac{(-1)^{k}(2k+1)(2k)x^{2k-1}}{2^{2k+1}(k!)(k+1)!}[/tex]

    And then I just pluged those into the differential equation to yield:

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

    Then I just put the x terms in the summation since they are constant to get:

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

    And since the summations are all indexed the same and all have the same denominator, I combined them to one summation with one fraction:

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

    But now that I have one huge fraction, after all the simplification I have done, I can't get the numerator to zero out since the DE is homogeneous.

    Did I make an error in my calculation? Have I done everything correct, but there is even MORE simplification? And if so, can you help me? Or am I just approaching this problem completely wrong?

    Thanks a TON!

    Sincerely,
    Tony
     
  2. jcsd
  3. Jul 26, 2010 #2
    You should have expanded the term:

    [tex]\left(x^2-1\right)\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}[/tex]

    into 2 terms:

    [tex]x^2\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}-\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}[/tex]

    You shift the index, and combine it again.
     
  4. Jul 26, 2010 #3
    I did that at first, but I thought that would'd help at all. Expanding that, putting the x's in the summation and then combining, I got:

    [tex]\sum_{k=0}^\infty \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}+(-1)^kx^{2k+3}-(-1)^kx^{2k+1}}{2^{2k+1}(k!)(k+1)!}[/tex]

    But that still looks like it is going no where. The numerator doesn't look like it is zero, unless it is really cryptic and I am not catching it. Help?
     
  5. Jul 27, 2010 #4
    You didn't do the shifting of the summation variables. You should do the following:

    [tex]\sum _{k=0}^{\infty } \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}+(-1)^kx^{2k+3}-(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}[/tex]

    [tex]\sum _{k=0}^{\infty } \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}-(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}+\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+3}}{2^{2k+1}k!(k+1)!}[/tex]

    Do the shifting in this step:

    [tex]\sum _{k=0}^{\infty } \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}-(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}+\sum _{k=1}^{\infty } \frac{(-1)^{k-1}x^{2k+1}}{2^{2k-1}(k-1)!k!}[/tex]

    Grouping it together again:

    [tex]\sum _{k=1}^{\infty } \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}-(-1)^kx^{2k+1}+4k(k+1)(-1)^{k-1}x^{2k+1}}{2^{2k+1}k!(k+1)!}[/tex]

    You should be able to show that it is indeed zero.
     
  6. Jul 27, 2010 #5
    Ah, I see. That is indeed zero. Thanks a MILLION! However, I am a little fuzzy how you went from your 3rd to 4th step in your last post. I understand that you let j=k-1 to shift the end summation, but how were you able to recombine it back with the first summation when that one was indexed at k=0? I don't know if this is a stupid question, but that transition from step 3 to 4 just doesn't look natural to me.
     
  7. Jul 27, 2010 #6
    Since I take out the k=0 term, and it is again zero! I just skip a bit.
     
  8. Jul 27, 2010 #7
    AHH! I see. Duh. Thanks soooooo much! :D
     
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