Summation of logs

Question

Are there any useful identities for quickly calculating the sum of consecutive logs? For example $\sum_{k=1}^{N} log(k)$ or something to this effect. I should add that I am writing code to do this (as opposed to doing this on a calculator) so N can be very large.

$\log a+\log b=\log (a\cdot b)$ – lab bhattacharjee May 08 '13 at 14:57 — lab bhattacharjee, May 08 '13 at 14:57

score 10 · Accepted Answer · answered May 08 '13 at 15:15

10

For large $N$, we have $N!\approx N^Ne^{-N}\sqrt {2\pi N}$ (Stirling formula) and hence $$\sum_{k=1}^N\ln k\approx\left( N+\frac12\right)\ln N-N+\frac12\ln(2\pi).$$

answered May 08 '13 at 15:15

Hagen von Eitzen

374,180

What expression would you get for $\sum_{k=1}^N \log^2 k$? – EGME May 09 '22 at 13:22
It’s interesting to note that $\frac{1}{2}\log(2\pi)$ is the divergent sum of the logarithms of all natural numbers – Sidharth Ghoshal Jul 19 '23 at 02:23

Inceptio · Answer 2 · 2013-05-08T15:03:55.847

7

Hint:

Use the fact that $\log(a)+\log(b)=\log(ab)$

Your expression simply becomes $\sum_1^N \log k=\log(N!)$, and now you can have Stirling Approximation to approximate $N!$

edited May 08 '13 at 15:03

answered May 08 '13 at 14:58

Inceptio

7,881

I am trying to do this with a computer, so for large $N$, calculating N! may not be possible – user74255 May 08 '13 at 14:58
@Inceptio, it's better to show the means than end – lab bhattacharjee May 08 '13 at 14:59
@labbhattacharjee: Considered it. Edited.:) – Inceptio May 08 '13 at 15:03
Will the approximation of N! interfere with the log calculation? I am not sure what O(ln(n)) is – user74255 May 08 '13 at 15:04
@user74255: No. You can calculate the approximate value of it, and then take $\log$ of it. – Inceptio May 08 '13 at 15:07
Perhaps instead of calculating $N!$ and then taking the $log$, you could in one step calculate the log-gamma function, i.e., $ln Γ (N+1)$. You didn't mention what language you are coding in; this would be very easy in, say, Mathematica. – Ray May 08 '13 at 15:10
@Ray I am using Python – user74255 May 08 '13 at 15:11
@Inceptio, what you suggest is a very numerically unstable approach – Ray May 08 '13 at 15:13
@Ray: Stirling Approximation can reduce the load. – Inceptio May 08 '13 at 15:18
I'm not talking about computational complexity--you can read about numerical stability here: http://en.wikipedia.org/wiki/Numerical_stability – Ray May 08 '13 at 15:55

score 7 · Answer 3 · edited Apr 13 '17 at 12:21

The Euler-Maclaurin Sum Formula, can also be used to get an asymptotic expansion. It gives $$ \sum_{k=1}^n\log(k)=\overbrace{\vphantom{\frac12}C}^{\frac12\log(2\pi)}+\overbrace{\vphantom{\frac12}n\log(n)-n}^{\int f(n)\,\mathrm{d}n}+\overbrace{\frac12\log(n)}^{\frac12f(n)}+\overbrace{\frac1{12n}}^{\frac1{12}f'(n)}-\overbrace{\frac1{360n^3}}^{\frac1{720}f'''(n)}+\dots $$ The constant $\frac12\log(2\pi)$ is derived as in the proof of Stirling's Formula.

Charles · Answer 4 · 2013-05-08T20:55:17.027

In the particular case of $$ \sum_{k=1}^N\log k=\log N!=\log\Gamma(N+1) $$ this is just the loggama function, which is implemented in many software systems. This is fast and accurate.

For example, it is lngamma in GP, lgamma in C (math.h), LogGamma in Mathematica, lnGAMMA in Maple, LogGamma in Magma, gammaln in MatLab, lnGamma in Mathcad, log_gamma in Sage, math.lgamma in Python, and gammaln in Perl (Math::SpecFun::Gamma).

Summation of logs

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