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I have seen the following simplification:

$$\left|\frac{1}{(-1-\frac{1}{n})^4 - 1}\right| = \frac{1}{\left|-1-\frac{1}{n}\right|^4 - 1}$$

I really don't have a clue why this is possible...

I am sorry but I can't give any ideas because I don't have any.

Thank you for advice.


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up vote 2 down vote accepted

Note that $x \geq 0$ if and only if $|x|=x$. So we have $\frac{1}{(-1-\frac{1}{n})^4 - 1}\geq 0$.

Inequlity $(-1-\frac{1}{n})^4 - 1\geq 0$ implies that $1+\frac{1}{n}\geq 1$. Hence for every natural number $n$, we have $|\frac{1}{(-1-\frac{1}{n})^4 - 1}| = \frac{1}{|-1-\frac{1}{n}|^4 - 1}$.

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That made it clear -thank you! – FunkyPeanut Jan 25 '14 at 12:08
You mean "$1+\frac{1}{n}\geq 1$ implies that $(-1-\frac{1}{n})^4-1\leq 0$", right? – JiK Jan 25 '14 at 12:23
NO, $(-1-\frac{1}{n})^4-1\geq 0$ implies that $1+\frac{1}{n}\geq 1$ – Babak Miraftab Jan 25 '14 at 12:25
But $(-1-\frac{1}{n})^4-1\geq 0$ is what we want to prove to show the equality in the question. – JiK Jan 25 '14 at 12:27
@JiK i was intersected in for which n, we have $\left|\frac{1}{(-1-\frac{1}{n})^4 - 1}\right| = \frac{1}{\left|-1-\frac{1}{n}\right|^4 - 1}$. We can move to chat if you want more discussion. – Babak Miraftab Jan 25 '14 at 13:26

Have you tried $\left|f(x)\right|=\sqrt{f(x)^2}$?

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