# Bessel function of second kind with integer order.

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:
$$L(y)=x^2y''+xy'+(x^{2}-n^{2})y=0$$
$$y_{2}(x)=aJ_m(x) ln(x)+\sum_{u=0}^{\infty} C_{u} x^{u+n}$$
and Bessel first kind for integer order is
$$J_{n}(x)=\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}$$

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

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

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

Thanks

Last edited:

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