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I found this information on the Web from the book:

Ecosystem Ecology: A New Systhesis, by David G. Christopher, L.J. Frid, on page 46

He calculated the Shannon entropy of a dirac delta function to be zero. Actually, he seems to be calculating for the discrete Kronecker delta. I wonder how one would go to the continuous case of the Dirac delta. Thanks.

Ecosystem Ecology: A New Systhesis, by David G. Christopher, L.J. Frid, on page 46

He calculated the Shannon entropy of a dirac delta function to be zero. Actually, he seems to be calculating for the discrete Kronecker delta. I wonder how one would go to the continuous case of the Dirac delta. Thanks.

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... However, you could define its entropy as the limit of the entropies of successive approximations to the Dirac delta. These can be taken to be increasingly narrower and higher bell curves or rectangles of constant unit area. It's easy to show that the limit is zero - the limit is in effect x*log(x) as x goes to 0.

So if we take the limit of the delta function after we calculate its entropy, then it goes to zero, right? I'm wondering if its too much trouble you to show me the math of this situation? Thanks.

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[tex]H_n = - \int_{-1/2n}^{1/2n} R_n(x) \cdot \log (R_n(x)) \textrm{d}x = - \frac{1}{n} \cdot n \cdot \log(n) = -\log(n)[/tex]

So actually you get [tex]-\infty[/tex] rather than 0. (But that's okay, because differential entropy can be negative.)

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