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So while I was waking up I started thinking about parabolas and absolute value functions and I suddenly realised that if you first square any function and then take the square root of it you will end up with the absolute value of that, since you're taking out the negatives by squaring them and putting them back to place by rooting them. Does anypne know if this method has been used anywhere?

Otomeram
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  • Method of what exactly? Are you just trying to take the absolute value of a function? – Dave Jan 04 '17 at 20:57
  • Exactly, but instead of just multiplying the negative chunks by -1 I do it this way – Otomeram Jan 04 '17 at 20:58
  • It works for real numbers. But if you have complex numbers you have to use geometric mean value with conjugate instead of square: $z\overline z = |z|^2 \Leftrightarrow \sqrt{z\overline z} = |z|$ – mathreadler Jan 04 '17 at 20:58
  • Uh, didn't think of the complex guys. Either way it then should be a valid way to do it for reals right? – Otomeram Jan 04 '17 at 20:59
  • For a function (let's say it is real valued) $f(x)$, instead of doing $\sqrt{(f(x))^2}$ you could just do $|f(x)|$, and take the absolute value directly. – Dave Jan 04 '17 at 20:59
  • Yes it is valid for all reals. – mathreadler Jan 04 '17 at 21:00
  • @Dave I know, but one of the most important parts of math is not staying with one and only one solution – Otomeram Jan 04 '17 at 21:01
  • @mathreadler nice – Otomeram Jan 04 '17 at 21:01
  • One place where something like this can be useful: if you want to approximate $|f(x)|$ by something smooth (where $f$ itself is smooth), take $\sqrt{\epsilon + f(x)^2}$ where $\epsilon > 0$ is small. – Robert Israel Jan 04 '17 at 21:04
  • Well, for the same real valued function, $\sqrt{(f(x))^2}=|f(x)|$, so the two methods are essentially the same. If the goal is to take the absolute value of a function, I am not aware of any application of the method you're referring to, rather most people would just take the absolute value directly. However your method is valid for such functions. – Dave Jan 04 '17 at 21:04
  • For some popular representations of numbers the sign is easily accessible and then just throwing the sign away would be cheaper computationally speaking, but for some other, squaring and square rooting may be easier than calculating the sign and turning it positive. – mathreadler Jan 04 '17 at 21:06
  • Yeah, like when roots are indeterminate right? – Otomeram Jan 04 '17 at 21:08
  • There was this question somewhere,... wait I'll see if I can find it. http://math.stackexchange.com/q/1731070/213607 – mathreadler Jan 04 '17 at 21:10

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This is a common fact, and without more details on what you want to use it for, I can't think of any place where it is critically useful.

A fun way of using this is in the "false proofs," i.e. $$ -2=-2\\ 4-6=1-3\\ 4-6+9/4=1-3+9/4\\ (2-3/2)^2=(1-3/2)^2\\ 2-3/2=1-3/2\\ 2=1 $$

TomGrubb
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