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I was wondering how to show entropy of the zeta distribution.

It is: $ H_\mathrm{zeta}(X) = \sum_{k=1}^\infty \frac{1/k^s}{\zeta(s)} \log(k^s \zeta(s))$

The entropy of the zipf distribution is:

$ H_\mathrm{zipf}(X) = \frac{s}{H_{N,s}} \sum_{k=1}^N \frac{\ln(k)}{k^s} + \ln(H_{N,s})$

The zipf distribution with parameter $N = \infty$ is the zeta distribution.

Then, taking the limit $H_\mathrm{zipf}(X)$ as $N$ goes to $\infty$,

$ \lim_{N \to \infty} \frac{s}{H_{N,s}} \sum_{k=1}^N \frac{\ln(k)}{k^s} + \ln(H_{N,s}) $

$ = \frac{s}{\zeta(s)} \sum_{k=1}^\infty (1/k^s) \ln(k) + \ln(\zeta(s))$

$ = \sum_{k=1}^\infty (1/k^s) \ln(k^s) (1/\zeta(s)) + \ln(\zeta(s))$

$ = \sum_{k=1}^\infty \frac{1/k^s}{\zeta(s)} \ln(k^s) + \ln(\zeta(s))\cdots\text{(1)}$

Note that I have used the identity: $\lim_{N \to \infty} H_{N,s} = \zeta(s)$

In the last equation, equation (1), I was unable to combine the natural log terms because it seems $\ln(\zeta(s))$ is outside of the summation. How do I proceed?

Thanks.

1 Answers 1

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Since $\sum\limits_{k=1}^\infty \frac{1/k^s}{\zeta(s)} =1$, (1) is $\left(\sum_{k=1}^\infty \frac{1/k^s}{\zeta(s)} \ln(k^s)\right) + \ln(\zeta(s))=\left(\sum_{k=1}^\infty \frac{1/k^s}{\zeta(s)} \ln(k^s)\right) + \left(\sum_{k=1}^\infty \frac{1/k^s}{\zeta(s)}\right) \ln(\zeta(s)), $ that is, as desired, $ \sum_{k=1}^\infty \frac{1/k^s}{\zeta(s)} \ln(k^s\zeta(s)). $

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    Thank you. The definition of the zeta function times (1 divided by the zeta function) is equal to 1. Also, $\sum_{i=1}^\infty k_i g_i + \sum_{i=1}^\infty k_i A = \sum_{i=1}^\infty k_i(g_i + A)$2012-08-27