- #1

Mr. Rho

- 15

- 1

[tex]j_{n}(x<<1)=\frac{x^n}{(2n+1)!}[/tex]

I can see this from the formula for $j_{n}(x)$ (taken from wolfram's webpage):

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

and applying the relation: $(2n+1)!=(2n+1)!/2^{n}n!$ . So, the formula for the spherical bessel function of the second kind is (also taken from wolfram):

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

The result I get for the limit $x<<1$ in this case is:

[tex]y_n(x<<1)=\frac{(-1)^{n+1}(-n)!}{2^{n}(-2n)!}\frac{1}{x^{n+1}}[/tex]

But I don't know how to deal with these negative factorials. I think maybe I would need the relation $(2n-1)!=(2n)!/2^{n}n!$ because in Jackson's Classical Electrodynamics 2nd edition book they give a result for this in chapter 16, it is:

[tex]y_{n}(x<<1)=-\frac{(2n-1)!}{x^{n+1}}[/tex]

The question is: how to obtain this Jackson's result?