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I'm wondering if it's correct to say that the absolute value squared means that the values above $1$ are magnified, whereas the values below $1$ are damped (I'm considering only positive values because the codomain of the absolute value is $R^+$).

In general the above consideration should be valid for all the functions which have $R^+$ as codomain.

Thank you in advance.

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    $\begingroup$ Are you asking whether $|x|^2$ makes the output larger than the input for $|x|>1$ and smaller (or the same size) for $|x|≤1$? This is true. If that isn't your question, then could you perhaps rephrase it? $\endgroup$ – Benjamin Feb 20 at 18:56
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    $\begingroup$ Your second paragraph is mysterious. $\endgroup$ – Yves Daoust Feb 20 at 19:00
  • $\begingroup$ "In general the above consideration should be valid for all the functions which have R+ as codomain" I'm not sure what this means but I don't think it follows. Just because something is true for $f(x) = x^2$ is no reason it should be true for any other function. That'd be like saying I should jump off a bridge, because my friend jumped off a bridge. $\endgroup$ – fleablood Feb 20 at 19:04
  • $\begingroup$ Hello @YvesDaoust, I wrote a wrong sentence. $\endgroup$ – Gennaro Arguzzi Feb 20 at 19:04
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well, yes. $|x| \ge 0$ for all $x \in \mathbb C$

And if $|x| = 0$ then $|x|^2 = 0^2 = 0 = |x|$ and magnitude remains unchanged.

If $0 < |x| < 1$ then, via the axiom $a > 0; x < y \implies ax < ay$, we have $|x|^2 = |x||x| < |x|\cdot 1 = |x|$ so, yes $|x|^2 < |x|$. so it is "damped".

If $|x| = 1$ then $|x|^2 = 1^2 =1 = |x|$ and magnitude remains unchanged.

If $|x| > 1$ then by the same axiom cited above. $|x|^2 = |x||x| > |x|\cdot 1 = |x|$ and magnitude is "magnified".

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  • $\begingroup$ excellent explanation; thank you a lot. $\endgroup$ – Gennaro Arguzzi Feb 20 at 19:01
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As you multiply something (namely $x$) by $x$, you will indeed obtain a smaller or larger result if $x$ is smaller or larger than $1$ respectively.

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