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$\begingroup$

I was reading page 12 of this note on the internet, but it seems unclear to me why the following inequality holds:

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from the 2nd to 3rd line of the above image.

All I could argue is that $e^x-1 = (e^{x/2}-1)(e^{x/2}+1) $, hence, we have that $$ (e^x-1)^{-1} \le (e^{x/2}+1)^{-1} .$$

where $x$ is such that $e^{x/2}-1 \ge 1 \Leftrightarrow x \ge \log 4. $ But this is no where as good as the inequality provided... I guess we split into cases? As this is part of a proof for DCT, so we need a function that dominates $g_N$.

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    Your problem is to understand why $$|1-(-e^{-x})^{N+1}|\leqslant2$$ right? Well, what is the set of values of $(-e^{-x})$? What is the set of values of $(-e^{-x})^{N+1}$? Hence... – 2017-02-26
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    I understand that part, but I don't understsand why we turn from $e^x-1$ to $e^x+1$. – 2017-02-26
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    Well spotted! $1-e^{-x}$ in the denominator is a typo, which should read $1-(-e^{-x})=1+e^{-x}$. – 2017-02-26
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    $e^x-1\leq e^x+1$ should be true right? And $-e^{-x}\leq -1$ for positive x, I believe. So I think what is done is finding the max of the numerator, and divide by something bigger than the denominator, assuming x is positive. (One has moved the left factor into the denominator I believe) – 2017-02-26
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    @Emil Dividing by something larger than the denominator would allow to prove *anything*. Please rethink. – 2017-02-26
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    @Did nah I won't rethink. I think the inequality on line 3 is the wrong way though. And the equality on line 2 is probably wrong. – 2017-02-26
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    @Emil "nah I won't rethink" Your call. Then be aware that your comments do not make any sense. No big deal. – 2017-02-26
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    Also, it is a pretty useless predicate unless one fixes the typo you found so it is a chain of $\leq$. $1<5>-2<0$ is also useless. – 2017-02-26

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