Bessel function of second kind with integer order.

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Main Question or Discussion Point

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
 
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Answers and Replies

  • #2
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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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