# I Limit of spherical bessel function of the second kind

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1. Mar 1, 2016

### Mr. Rho

I know that the limit for the spherical bessel function of the first kind when $x<<1$ is:

$$j_{n}(x<<1)=\frac{x^n}{(2n+1)!!}$$

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

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

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):

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

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

$$y_n(x<<1)=\frac{(-1)^{n+1}(-n)!}{2^{n}(-2n)!}\frac{1}{x^{n+1}}$$

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:

$$y_{n}(x<<1)=-\frac{(2n-1)!}{x^{n+1}}$$

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

2. Mar 2, 2016

### alivedude

If i'm not way off here I think you can rewrite those negative factorials as gamma function, maybe that will help? That would give you $(-n)!=\Gamma(1-n)$ and $(-2n)! = \Gamma(1-2n)$ and then you could use some suitable notation for these and plug that in. I'm not 100% sure tho, its been a while since I did something similar but it might be worth a shot.

3. Mar 2, 2016

4. Mar 2, 2016

### TeethWhitener

Alright, this ended up really bugging me all day, but I think it's because Mathworld gives a bad power series for the spherical Bessel function (how is k supposed to start at 0 when negative factorials are undefined?). Here's an alternative expression that gives the correct asymptotic behavior:

http://dlmf.nist.gov/10.53

(BTW, @Mr. Rho in your first post, I believe the asymptotic expression you wanted is $$y_n(x\rightarrow 0) \approx -\frac{(2n-1)!!}{x^{n+1}}$$
with a double factorial, instead of a single factorial.)

5. Mar 2, 2016

### Mr. Rho

yes, sorry for that, it is double factorial, thank you for the correction and thank you for those relations you give, the asymptotic limit it's very clear with them!

why would wolfram give a wrong relation?

Last edited: Mar 2, 2016
6. Mar 2, 2016

### TeethWhitener

It happens occasionally. There used to be a way to submit errors and corrections, but I don't see the link anymore.

7. Mar 3, 2016

### alivedude

Haha yeah, that was throwing me off too.