Problem Detail: I am going through Normal Subgroup Reconstruction and Quantum Computation Using Group Representations by Hallgren et al. In the proof of the theorem $6$ of the paper on page 632, the authors go on proving the difference between the probabilities of sampling all irreps, $|p – q|_1$ of a subgroup inside the symmetric group $S_n$. $$ begin{align} |p – q|_1 &= Sigma_{rho} mid p_{rho} – q_{rho} mid &le Sigma_{rho} frac{d_{rho}}{n!} 2^{O(n)} sqrt{n}^{n / 2} &le Sigma_{rho} frac{sqrt{n!}}{n!} 2^{O(n)} sqrt{n}^{n / 2} &le frac{2^{O(n)} sqrt{n}^{n/2}}{sqrt{n!}} &= 2^{O(n)} frac{sqrt{sqrt{n}^n}} {sqrt{n!}} &le 2^{O(n)} frac{1}{sqrt{left( n / 2 right)!}} lll 2^{-Omega(n)} end{align} $$ How is $2^{O(n)} frac{1}{sqrt{left( n / 2 right)!}} lll 2^{-Omega(n)}$?
Asked By : Omar Shehab
Answered By : Raphael
Throwing away gutter, this is the claim: $qquadfrac{c^n}{sqrt{(n/2)!}} to 0$ with at least exponential rate as $n to infty$. That is, the sqare root of $(n/2)!$ grow (at least) exponentially faster than exponential functions. You can prove this by showing that $qquadfrac{c^n}{sqrt{(n/2)!}} sim 2^{-g(n)}$ for some $ g in Omega(n)$. Use Stirling’s approximation.
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Question Source : http://cs.stackexchange.com/questions/56318
