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Bessel function of second kind with integer order.

  1. Sep 28, 2013 #1
    I have a question about deriving the Bessel function of the second kind with integer order. I understand that the Bessel function and the second independent variable is defined as:
    [tex]L(y)=x^2y''+xy'+(x^{2}-n^{2})y=0[/tex]
    [tex]y_{2}(x)=aJ_m(x) ln(x)+\sum_{u=0}^{\infty} C_{u} x^{u+n}[/tex]
    and Bessel first kind for integer order is
    [tex]J_{n}(x)=\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}[/tex]

    Without going through the series manipulations and factoring out, let me jump to the grouping with terms containing ##a ln(x)##
    [tex]a ln(x)\left[ \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)(2k+n-1)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n+2}}{k!(k+n)!2^{2k+n}} - n^{2}\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}\right][/tex]

    You can see this is in form of
    [tex]L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}[/tex]
    Where
    [tex] y_{1}=\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}[/tex]
    My question is in the next step, the derivation claimed
    [tex]L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}=0[/tex]
    [tex]\Rightarrow\;a ln(x)\left[ \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)(2k+n-1)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n+2}}{k!(k+n)!2^{2k+n}} + n^{2}\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}\right]=0[/tex]
    And all these disappeared!!!

    I understand the definition for the Bessel function is
    [tex]L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}=0[/tex]
    But that does not imply when you see anything like [tex]L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}[/tex] it is automatically equal to zero. Please explain.


    Thanks
     
    Last edited: Sep 28, 2013
  2. jcsd
  3. Sep 29, 2013 #2
    I resolve it already. It is very simple

    It is given already that
    [tex]L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}=0[/tex]
    Where
    [tex] y_{1}=\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}[/tex]

    [tex]a ln(x)\left[ \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)(2k+n-1)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n+2}}{k!(k+n)!2^{2k+n}} - n^{2}\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}\right]=a ln(x)\left[x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}\right]=0[/tex]
     
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