Recurrence relations in asymptotic regime

dingo_d
Messages
199
Reaction score
0

Homework Statement



I'm solving the quantum harmonic oscillator. And I'm solving Schrodinger equation. So I came up to one part where I have to use power series method of solving DE (that or Frobenius would probably work just fine). Now I have the recurrence relation:

a_{n+2}=\frac{\lambda(2n+1)-k^2}{(n+2)(n+1)}a_n

And the text in which this is solved says that for n\ton\infty that leads to asymptotic law

a_{n+2}=\frac{2\lambda}{n}a_n corresponding to the series expansion of e^{\lambda x^2}.

Now, I tried looking at the limit, via L'Hospitals rule and I really can't see how they got that! :\

So can someone explain to me how they got that? Thanks...
 
Physics news on Phys.org
Expand and divide top and bottom by n:
<br /> a_{n+1}=\frac{\lambda\left( 2+\frac{1}{n}\right)-\frac{k^{2}}{n}}{n+3+\frac{1}{n}}a_{n}<br />
For n large, the terms in 1/n can be ignored
 
And I just ignore the 3 in the denominator can be ignored because when n goes to infinity that is negligible, right?
 
Yep, sound good to me.
 
Thanks ^^
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top