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With apologies for promoting my own work, there's a whole book on the mathematics of the exponentials of various entropies: Tom Leinster, Entropy and Diversity: The Axiomatic Approach. Cambridge University Press, 2021. You can download a free copy by clicking, although persons of taste will naturally want to grace their bookshelves with the bound work. The ...


6

If I may broaden the question somewhat to include other interpretations of the exponential of entropy, it is commonly used in ecology to measure how many different species there are in a community. In that context one considers the generalized diversity index (or Hill number) $${}^q\!D=\left ( {\sum_{i} p_i^q} \right )^{1/(1-q)},$$ with the exponential of ...


3

It looks like the inequality is true (at least for the function with the cotangent). The proof (I hope it is correct but, please, check the details: I'm not in my top shape today) is as follows. Claim 1: For every $y>0$, we have $\left|\int_0^y t^3e^{it}\,dt\right|\le y^3$. Proof: One trivial bound for the LHS is $\int_0^y t^3=\frac {y^4}4\le y^3$ when $y\...


2

If $\min(a,b)\ge100$, then the mean of the beta distribution will be $$m=\frac a{a+b},$$ and for its standard deviation we will have \begin{equation} s\approx\sqrt{\frac{ab}{(a+b)^3}}=m\sqrt{\frac b{a(a+b)}}<m\sqrt{\frac1a}\le\frac m{10} \tag{0} \end{equation} and similarly $$s\approx\sqrt{\frac{ab}{(a+b)^3}}=(1-m)\sqrt{\frac a{b(a+b)}}<(1-m)\sqrt{\...


1

Here is another solution: Noting the identity $$\sum_{n=-\infty}^{\infty} \frac{1}{(t+2\pi n)^2} = \frac{1}{2(1-\cos t)}$$ and taking advantage of the fact that the integrand is non-negative, we may apply Tonelli's theorem to get \begin{align*} f(x) &= \frac{1}{2\sqrt{x}} \int_{-\pi}^{\pi} \left( \sum_{n=-\infty}^{\infty} \frac{1}{(t+2\pi n)^2} \right) \...


1

Here is a fairly simple condition. It uses the following notion: A family of functions $f_t\in L^1$ depending on a parameter $t$ in a measure space $X$ is said to tend to $f$ somewhere if the essential infimum of $\lVert f_t-f\rVert_1$ over $X$ is $0$. Put $g_s(t)=g(t-s)$. The condition is that $afg_s\to\lvert f\rvert\lVert g\rVert_\infty$ somewhere, for ...


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