Patricia Bean

2022-07-20

Let $aϵ{\mathbb{R}}_{+}$ and $b=\mathrm{exp}\left(-a\right).$
What is the range of $\left\{-{b}^{2}\mathrm{log}\left(b\right)\right\}$? Does it range $\left(-\mathrm{\infty },+\mathrm{\infty }\right)$?
How can I show $\frac{-b\mathrm{log}\left(b\right)}{1-b}\le 1$ ?

Sandra Randall

Expert

Given that a>0
$-{b}^{2}\mathrm{log}\left(b\right)=-{\left({e}^{-a}\right)}^{2}\mathrm{log}\left({e}^{-a}\right)=-{\left({e}^{-a}\right)}^{2}\left(-a\right)=a{e}^{-2a}$
therefore you need to find the range of $a{e}^{-2a}$. Since a>0, you can think of this as a positive constant times ${e}^{-2a}$. This produces the range of $\left(0,\mathrm{\infty }\right)$
Second question:
$\frac{-b\mathrm{log}\left(b\right)}{1-b}\le 1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{-{e}^{-a}\mathrm{log}\left({e}^{-a}\right)}{1-{e}^{-a}}\le 1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{a{e}^{-a}}{1-{e}^{-a}}\le 1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{a}{{e}^{a}-1}\le 1$
which simplifies to
$a\le {e}^{a}-1$
or
${e}^{a}\ge a+1$
I would recommend applying Bernoulli's inequality. Alternatively, you could observe that
$0<{e}^{a}=1+a+\frac{{a}^{2}}{2}+\frac{{a}^{3}}{3!}+\frac{{a}^{4}}{4!}+\frac{{a}^{5}}{5!}+\cdots$

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