18
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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