Check those approximations that can follow from Stirling's formula: log(NV) ~ Nlog(N) log(NV) - Nlog(N) = N log(NV) - Nlog(N) ≈ N + glog(N) (d) log(N!) ~ Nlog(N) - N + glog(N) + glog(2t) For each of those approximations, determine the smallest value of N for which the relative error in N is less than %.
Added by Nicole G.
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Step 1
First, let's recall Stirling's formula: $$ \log(N!) \approx N\log(N) - N + \frac{1}{2}\log(2\pi N) $$ Now, let's check each approximation: Show more…
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