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In the context of the simple quantum harmonic oscillator, the asymptotic behaviour of the power series expansion of $x^ke^{x^2}$ for some natural number $k$ comes up. From my own work I find that the asymptotic behaviour of the ratio of the coefficients $a_{n+2}/a_n$ should be $$\frac{n!}{(n+2)!} = \frac{1}{n^2 + 3n + 2} $$ which is very different than the correct $\frac{2}{n}$. Where am I going wrong?

EE18
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  • I don't get either of those answers, even after replacing $x^2$ by $\frac{x^2}{2}$. To clarify, you are asking about the asymptotic behavior of the coefficient of $\frac{x^n}{n!}$, as a function of $n$? – Qiaochu Yuan Feb 05 '23 at 21:54
  • The asymptotic behaviour of the coefficients of the Taylor series of the function first given @QiaochuYuan – EE18 Feb 05 '23 at 22:13
  • @QiaochuYuan When I put it into Mathematica for terms $n=98$ and $n+2=100$ I get 1/50 as expected from the result I quoted, but I can't seem to get it. – EE18 Feb 06 '23 at 00:24
  • Why doesn't your answer depend on $k$? For $k = 0$ the coefficient of $\frac{x^{100}}{100!}$ is $\frac{100!}{50!}$ which is much larger than $\frac{1}{50}$. If you meant $e^{\frac{x^2}{2}}$ then the coefficient is $\frac{100!}{2^{50} 50!}$ which is still much larger than $\frac{1}{50}$. – Qiaochu Yuan Feb 06 '23 at 01:40
  • @QiaochuYuan The series expansion is $x^k(1+x^2/1 + x^4/2 + x^8/3! + x^{16}/4!+ \dots)$. I'm not sure where you're getting those expressions for $a_{n+2}/a_n$? – EE18 Feb 06 '23 at 02:23
  • You didn't say anything about $\frac{a_{n+2}}{a_n}$! I asked whether you were interested in the asymptotic behavior of the coefficients which is a quite different question. Try writing down what you think $a_n$ is first. – Qiaochu Yuan Feb 06 '23 at 02:49
  • @QiaochuYuan I'm so sorry, I realize now I missed including that as I asked the question on my phone! Will edit now. – EE18 Feb 06 '23 at 02:50

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