Patricia Bean

Answered

2022-07-20

Let $a\u03f5{\mathbb{R}}_{+}$ and $b=\mathrm{exp}(-a).$

What is the range of $\{-{b}^{2}\mathrm{log}(b)\}$? Does it range $(-\mathrm{\infty},+\mathrm{\infty})$?

How can I show $\frac{-b\mathrm{log}(b)}{1-b}\le 1$ ?

What is the range of $\{-{b}^{2}\mathrm{log}(b)\}$? Does it range $(-\mathrm{\infty},+\mathrm{\infty})$?

How can I show $\frac{-b\mathrm{log}(b)}{1-b}\le 1$ ?

Answer & Explanation

Sandra Randall

Expert

2022-07-21Added 17 answers

Given that a>0

$-{b}^{2}\mathrm{log}(b)=-{\left({e}^{-a}\right)}^{2}\mathrm{log}({e}^{-a})=-{\left({e}^{-a}\right)}^{2}(-a)=a{e}^{-2a}$

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 $(0,\mathrm{\infty})$

Second question:

$\frac{-b\mathrm{log}(b)}{1-b}\le 1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{-{e}^{-a}\mathrm{log}({e}^{-a})}{1-{e}^{-a}}\le 1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{a{e}^{-a}}{1-{e}^{-a}}\le 1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{a}{{e}^{a}-1}\le 1$

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 $

$-{b}^{2}\mathrm{log}(b)=-{\left({e}^{-a}\right)}^{2}\mathrm{log}({e}^{-a})=-{\left({e}^{-a}\right)}^{2}(-a)=a{e}^{-2a}$

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 $(0,\mathrm{\infty})$

Second question:

$\frac{-b\mathrm{log}(b)}{1-b}\le 1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{-{e}^{-a}\mathrm{log}({e}^{-a})}{1-{e}^{-a}}\le 1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{a{e}^{-a}}{1-{e}^{-a}}\le 1\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{a}{{e}^{a}-1}\le 1$

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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