Proving Rnl(r) Has (n-l-1) Zeros - Interesting Proof

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SUMMARY

The discussion centers on proving that the function Rnl(r) has (n-l-1) zeros, excluding those at r=0 and r=infinity. The equation for Rnl(r) is given as Rnl(r) = Anlunl, where Anl = sqrt((n-l-1)!/(2n((n+l)!)^3). Participants suggest focusing on the properties of Laguerre Polynomials and applying the fundamental theorem of algebra to establish the number of zeros. The polynomial's degree, n-l-1, guarantees the existence of n-l-1 complex roots, which may include distinct zeros depending on the coefficients of the polynomial.

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

I think you can build the proof really easily.

Good luck! :smile:

Daniel.
 
Ed Quanta said:
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?

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.
 
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:
dextercioby said:
I'm sorry for not sharing your enthusiasm,but i don't find that proof to be interesting... :-p

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

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