Recurrence relations in asymptotic regime

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Homework Help Overview

The discussion revolves around solving the Schrödinger equation for the quantum harmonic oscillator using a power series method. A recurrence relation has been derived, and participants are exploring its asymptotic behavior as n approaches infinity.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how the recurrence relation leads to an asymptotic law. Some participants suggest simplifying the relation by ignoring terms that become negligible as n increases.

Discussion Status

Participants are actively discussing the implications of ignoring certain terms in the recurrence relation as n becomes large. There is a general agreement on the approach of simplifying the expression, though no consensus on the specific reasoning has been reached.

Contextual Notes

There is a focus on the behavior of the recurrence relation in the asymptotic regime, with specific attention to the limits and simplifications involved. The original poster expresses confusion regarding the derivation of the asymptotic law from the recurrence relation.

dingo_d
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Homework Statement



I'm solving the quantum harmonic oscillator. And I'm solving Schrödinger 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...
 
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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 ^^
 

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