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in a proof I'm reading, the author infers from the following expression: $$\left(1-\frac{1}{x}\right)^{i}\le\frac{1}{2^{2n}}$$ that: $i\ge\ln{(2)}2nx$

It is not clear to me how to get from the first expression to the result. Can you point out to me the algebraic tricks that were used here?

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  • $\begingroup$ I assume $\frac{1}{x} < 1$? $\endgroup$
    – Alex Botev
    Apr 20, 2013 at 19:55
  • $\begingroup$ of course... :) $\endgroup$
    – gilad hoch
    Apr 20, 2013 at 20:20

2 Answers 2

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Take the base 2 log of both sides and use the expansion of $\ln(1-\frac 1x)^i$ to show it is greater than $-ix$

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  • $\begingroup$ instead of $i$ i just wrote $x\cdot{\frac{i}{x}}$ which got me to $\frac{1}{e^{\frac{i}{x}}}$ and using ln, solved it. thanks. $\endgroup$
    – gilad hoch
    Apr 20, 2013 at 22:03
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$$(1-\frac{1}{x})^i \leq \frac{1}{2^{2n}}$$
$$ \Rightarrow i\ln(1-\frac{1}{x}) \leq ln(2^{-2n})$$ $$ i \geq \frac{-2nln(2)}{ln(1-\frac{1}{x})}$$ What is left is to proove that $$\frac{-1}{\ln(1-\frac{1}{x})} \geq x$$ This is equivalent to $$-1 \leq x\ln(1-\frac{1}{x}) \Leftrightarrow \frac{1}{e} \leq (1-\frac{1}{x})^x$$ The last one is well known limit, so it holds. Note that you must have $1-\frac{1}{x} \geq 0$.

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