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Problem
- Thread starter daf44
- Start date
Stirling's approximation is:
\(\displaystyle n!\sim n^{n}e^{-n}\sqrt{2\pi n}\)
But, this would actually involve the Psi function, not Stirling. I suppose it could be done with Stirling, but there is a function which defines the derivative of logs and factorials. It's called the Psi or Digamma function.
The Psi function is defined as the derivative of the Log Gamma function. Thus,
\(\displaystyle \psi(n)=\frac{d}{dn}ln\Gamma(n)=\frac{\Gamma'(n)}{\Gamma(n)}\)
So, if we have \(\displaystyle ln\left(\frac{n!}{s!(n-s)!}\right)\) and noting that \(\displaystyle n!=\Gamma(n+1)\), then differentiating in terms of s gives:
\(\displaystyle ln(\frac{n!}{(n-s)!s!})=ln(n!)-ln((n-s)!)-ln(s!)\)
\(\displaystyle =ln(\Gamma(n+1))-ln(\Gamma(n-s+1))-ln(\Gamma(s+1))\)
\(\displaystyle \frac{d}{ds}\left[ln(\Gamma(n+1))-ln(\Gamma(n-s+1))-ln(\Gamma(s+1))\right]\)
\(\displaystyle =\psi(n-s+1)-\psi(s+1)\).
\(\displaystyle n!\sim n^{n}e^{-n}\sqrt{2\pi n}\)
But, this would actually involve the Psi function, not Stirling. I suppose it could be done with Stirling, but there is a function which defines the derivative of logs and factorials. It's called the Psi or Digamma function.
The Psi function is defined as the derivative of the Log Gamma function. Thus,
\(\displaystyle \psi(n)=\frac{d}{dn}ln\Gamma(n)=\frac{\Gamma'(n)}{\Gamma(n)}\)
So, if we have \(\displaystyle ln\left(\frac{n!}{s!(n-s)!}\right)\) and noting that \(\displaystyle n!=\Gamma(n+1)\), then differentiating in terms of s gives:
\(\displaystyle ln(\frac{n!}{(n-s)!s!})=ln(n!)-ln((n-s)!)-ln(s!)\)
\(\displaystyle =ln(\Gamma(n+1))-ln(\Gamma(n-s+1))-ln(\Gamma(s+1))\)
\(\displaystyle \frac{d}{ds}\left[ln(\Gamma(n+1))-ln(\Gamma(n-s+1))-ln(\Gamma(s+1))\right]\)
\(\displaystyle =\psi(n-s+1)-\psi(s+1)\).
Last edited:
Well, the solution from my teacher was:
dln(Q)/dS=
d[(N-S)ln(N-S)+(N-S)-Sln(S)]/dS=
(N-S)/S
We've never discussed the Psi fuction, and I've never heard of it, so its seems unlikely that I should use it.
Further on, my book say's that the stirling's approximation for large N is ln(N!) = Nln(N) - N
dln(Q)/dS=
d[(N-S)ln(N-S)+(N-S)-Sln(S)]/dS=
(N-S)/S
We've never discussed the Psi fuction, and I've never heard of it, so its seems unlikely that I should use it.
Further on, my book say's that the stirling's approximation for large N is ln(N!) = Nln(N) - N