# Absolute value squared - Meaning

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.

• 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? – Benjamin Feb 20 at 18:56
• Your second paragraph is mysterious. – Yves Daoust Feb 20 at 19:00
• "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. – fleablood Feb 20 at 19:04
• Hello @YvesDaoust, I wrote a wrong sentence. – Gennaro Arguzzi Feb 20 at 19:04

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".

• excellent explanation; thank you a lot. – Gennaro Arguzzi Feb 20 at 19:01

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.