3
$\begingroup$

I have that

$$ x |x| = -y, $$ Clearly the sign of x is determined by the sign of y. But then how to write what x is?What happens if y is discontinuous at 0?

Is it $$ x = sgn(x) \sqrt{ |y|} ?$$ or doesn't it make any sense?there is something I probably don't understand.

silly example $$ 5 |5| = - (-25)$$ x is positive when y negative and $$ - 5 |-5| = - (25)$$ x is negative when y positive.

$\endgroup$
2
  • 1
    $\begingroup$ We can go at this by analyzing two cases (if we exclude $x=0$ as we just get the origin there): either $x>0$ or $x<0$. Try graphing it by plotting each individual case and seeing how it all comes together. $\endgroup$ May 19, 2021 at 0:27
  • $\begingroup$ When $x\geq 0$, $x^2=-y$, and when $x<0$, $-x^2=-y$, i.e. $x^2=y$. So it looks like $x^2=-\text{sgn}(x)y$. $\endgroup$
    – klein4
    May 19, 2021 at 0:31

3 Answers 3

5
$\begingroup$

I'm assuming you're asking this question in the case where $x$ and $y$ are real numbers rather than complex numbers.

It looks to me as if $$ x = -\operatorname{sgn}(y) \sqrt{|y|} $$ where you might have to do something when $y = 0$, depending on the definition of "sgn".

In other words, your conjectured answer was almost correct.

$\endgroup$
1
$\begingroup$

There are four solutions:

$x = \pm \sqrt{y}$ and $\pm i \sqrt{y}$.

$\endgroup$
2
  • 3
    $\begingroup$ But some of those are spurious: if $y=+4$ then $x=+2$ is not a solution $\endgroup$
    – Henry
    May 19, 2021 at 0:26
  • $\begingroup$ thanks for the answer, I should have mentioned x and y are real. $\endgroup$
    – Geo
    May 19, 2021 at 0:38
1
$\begingroup$

$y = \begin{cases}x^2 & x\le 0\\-x^2 & x>0 \end{cases}$

and to invert this.

$x = \begin{cases} \sqrt {-y} & y\le 0\\ -\sqrt{y} & y> 0\end {cases}$

or $x = -\text{sgn}(y)\sqrt{|y|}$

$\endgroup$
1
  • 1
    $\begingroup$ I guess(typo) in the first is $$x >0.$$ Shouldn't be minus sign in the end as in the other comment? The signs were from beginning what was confusing to me. $\endgroup$
    – Geo
    May 19, 2021 at 0:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.