# Interesting proof

1. Feb 6, 2005

### Ed Quanta

How can it be proved that Rnl(r) has (n-l-1) zeros (not counting those at r-0 and r= infinity?

I tried doing this inductively but found it hard to get anywhere since the equations for Rnl(r) aren't too pretty. For instance, Rnl(r)=Anlunl

where Anl= sqrt((n-l-1)!/(2n((n+l)!)^3)

this last equation seems to bear a similarity with what I am supposed to prove but I am having trouble with this. Help anyone?

2. Feb 6, 2005

### dextercioby

I'm sorry for not sharing your enthusiasm,but i don't find that proof to be interesting... :tongue2:
Yu can find the general formula for $R_{nl}(r)$ here

I think you can build the proof really easily.

Good luck!!

Daniel.

3. Feb 6, 2005

### kanato

I don't think you need to worry about $$A_{nl}$$ in your proof, since that is just a constant for a given n and l, and not dependent on r. I think what you want to look at is the Laguerre Polynomials, (specifically their order), and remember what the fundamental theorem of algebra says about how many zeros a polynomial will have, and somehow come up with an argument that those zeros will be real.

4. Feb 6, 2005

### mikeu

The link Daniel provided states that
$$R_{n\ell}(r) = r^\ell \exp\left(\frac{zr}{na}\right) \sum_{j=0}^{n-\ell-1} b_jr^j.$$
The sum is simply the definition of a polynomial of degree $$n-\ell-1$$, which the fundamental theorem of algebra guarantees will have exactly $$n-\ell-1$$ complex roots (not necessarily distinct). The coefficients in front of the sum provide the zeros at $$r=0$$ and $$z\to-\infty$$, and the others are provided by the polynomial.

If you need to show that there are $$n-\ell-1$$ distinct zeros then there will obviously be something more to the proof (I would guess it involves looking at the definition of the $$b_j$$), but this might be a good starting point at least....

Last edited: Feb 6, 2005
5. Feb 6, 2005

### Ed Quanta

Ha ha, I just didn't know what to title it. Thanks for help.