How to calculate the inverse function of $y=-\frac{1}{2} \ln(1-x^2) \times \text{sign}(x)$?

How can the continuous random variable $$x$$ by isolated by itself on one side of the following equation

$$y = -\frac{1}{2} \ln(1-x^2) \times \text{sign}(x)$$

without resorting to a piece-wise equation?

$$x = ?$$

Below is my initial, incomplete and probably wrong attempt since I don't know the exponential of a product or the exponential of $$\text{sign}()$$:

$$-2 y = \ln(1-x^2) \times \text{sign}(x)$$ $$\exp(-2y) = (1-x^2) \times \exp(\text{sign}(x))$$

• The function is an odd function, so you can do it just for $x>0$, whence we just have $y = -\frac 12\ln(1-x^2)$.(The square root , when eventually taken to determine $x$ will be positive). Sep 7, 2020 at 7:00
• the derivation without the sign multiplier is easy and well-known, so I would like to learn how to do it for the function shown instead ($x\in \mathbb{R}$ not just $x>0$) Sep 7, 2020 at 7:09
• The domain of $y$ is $x \in (-1,1)$,since otherwise $\ln (1-x^2)$ would be undefined. My comment still stands : break $y$ into a piecewise function by breaking the $\mbox{sign}$ into one, then find the piecewise inverses and put them together. In a follow up question you have asked for the derivative : you can get it everywhere except at $0$ where you need to create the differential quotient. The sign multiplier is handled by breaking the function into pieces, so that on each piece we know what to do. Sep 7, 2020 at 14:18
• is it possible to isolate $x$ without resorting to piece-wise functions, I meant to add that to the question and will do so now Sep 7, 2020 at 14:23
• Thanks for the edits, I now see what you want to do. Here's something : note that $sign(x) = sign(y(x))$ for all $x$ (proved by noting that $\frac{sign(x)}{x} = \frac 1{|x|}$ is always positive for $x \neq 0$, so the quotient $\frac{y}{x}$ is also always positive for $x \neq 0$), so you can replace $sign(x)$ with $sign(y)$. This should help. Let us discuss this. Sep 7, 2020 at 16:23

There are 2 cases:

1. $$x \geq 0$$: \begin{align*} y&=-\frac{1}{2}ln(1-x^2)\\ -2y &= ln(1-x^2)\\ e^{-2y} &= 1-x^2\\ x^2 &= 1-e^{-2y}\\ x &= +\sqrt{1-e^{-2y}} \qquad :\text{since } x \geq 0 \end{align*}
2. $$x < 0$$: \begin{align*} y&=\frac{1}{2}ln(1-x^2)\\ 2y &= ln(1-x^2)\\ e^{2y} &= 1-x^2\\ x^2 &= 1-e^{2y}\\ x &= -\sqrt{1-e^{2y}} \qquad :\text{since } x < 0 \end{align*}

Now, as stated in the comments, you notice that $$y(x)$$ has the same sign as $$x$$, i.e: $$\text{sign}(x) = \text{sign}(y(x))$$ So, the different formulas for $$x$$ can be unified using $$\text{sign}(y)$$, as follows: $$x=\text{sign}(y)\sqrt{1-e^{-2y.\text{sign}(y)}}$$ Also, since $$y.\text{sign}(y)=|y|$$, we can write $$x$$ as: $$x=\text{sign}(y)\sqrt{1-e^{-2|y|}}$$

We can use chain rule noting that for $$x\neq 0$$

$$(\text{sign}(x))'=0$$

therefore

$$(f(x)\cdot \text{sign}(x))'=f'(x)\cdot \text{sign}(x)$$

• this omits all the actual content of the function and is incomplete since $x$ is not isolated, could you show the line-by-line operations without masking with chain rule Sep 7, 2020 at 7:25
• what's the problem? The $sign$ function is $1$ for x >0 and $-1$ for $x <0$ Sep 7, 2020 at 15:47
• @Tortar what does that mean for the first derivative of $\text{sign}(x)$? Sep 7, 2020 at 16:43

From

$$y=-\frac12\ln(1-x^2)\text{ sgn}(x)$$ we can draw

$$x=\text{sgn}(x)\sqrt{1-e^{-2y\text{ sgn(x)}}}$$ because the square root is a positive number. But the function is odd, $$\text{sgn}(x)=\text{sgn}(y)$$, and

$$x=\text{sgn}(y)\sqrt{1-e^{-2|y|}}.$$

• doesn't differenting the equation cause an unwanted called $y'$? I would like to retain $y$ as is while isolating $x$ Sep 7, 2020 at 16:44
• you singled out a single component of the equation while ignoring two others: $y$ and $\text{sign}(x)$. if i take the derivative of the entire equation, i have to deal with a new quantity called $y'$, which is the first derivative of variable $y$. I would like to retain $y$ as is, while isolating $x$ Sep 7, 2020 at 16:49
• read the question again, especially the first two formulas Sep 7, 2020 at 16:51
• @develarist: oops, sorry, because of the other answer, I was on the way of a differentiation question. That was silly !
– user65203
Sep 7, 2020 at 16:52