$\sigma(x) = \sigma^{-1}(x)$ I want to find the solution to $\sigma(x) =\sigma^{-1}(x)$ Where $\sigma(x)=\frac{1}{1+e^{-x}}$ and so $\sigma^{-1}(x)=\ln(x)-\ln(1-x)$. I've got it down to $\frac{e^x}{x}=e^x+1$ but I can't get any further. Desmos tells me that the solution is around $0.659$ but I want an exact expression for it... Is this possible?
 A: I don't believe your equation has an analytic solution. You can, however, get close by trying some approximations. Let's rewrite our equation as
$$e^x=\frac{x}{1-x}$$
Say we expand the functions into a Taylor series
$$e^x=1+x+\frac{x^2}{2}+...$$
$$\frac{x}{1-x}=x-x^2+...$$
Dropping higher order terms, we have
$$1+x+\frac{x^2}{2}=x+x^2$$
which, when simplified, gives us
$$x^2-2=0$$
Which tells us $x=\pm\sqrt{2}\approx1.414$
Okay not great... but adding another higher order term gives us
$$5x^3+3x^2-6=0$$
Which has one real root at $x\approx.895$. Adding the fourth order term gives $x\approx.767$. A fifth gives $x\approx.716$.
By the time you've reached seventh order terms, your approximation is at $x\approx.679$ which is within 3% of the actual answer. This is just one way to get at the solution by using functions we know the solutions to.
A: Use the second order approximation of $\textbf{exp}$ i.e $e^x \approx 1+x+\frac{x^2}{2!}$. We would then have,
$$\frac{e^x}{x} = e^x + 1 \Rightarrow \frac{1+x+\frac{x^2}{2!}}{x} \approx \left(1+ x+ \frac{x^2}{2!}\right)+1$$
The above implies,
$$1+ x + \frac{x^2}{2!}  \approx 2x + x^2 + \frac{x^3}{2!} $$
And so $x$ will be approximately the root of,
$$ p(u) =u^3 + u^2 + 2u-2 $$
Now you can use intermediate value theorem first noting that $p(.5)<0$ and $p(1)>0$ i.e your zero lies in the interval $[.5,1]$. If you want to approach this from a qualitative point of view. I would just keep increasing $.5$ until $p(.5+ \epsilon)>0$. In any case, at least now you can an expression which approximates the root.  
A: At first, the value approximated by your computation system is correct. Using algebraic adjustments we can determine an equality: $$-\log\left({\frac{1}{x}-1}\right)= \frac{e^x}{e^x+1}$$ 
The only possible way to evaluate the solution to this equation is explicitly numeric, therefore indescribable as an analytical expression. The closest thing to get an explicit number I can think of is a convergence of Fourier or Taylor series, which can be described as an infinite sum of functions. There should not exist any way to do this precisely and nothing should be more precise considering absolute convergence than those series.
A: To find a root of $g(x)=(x-1)e^x+x$ in $[0,1]$ is numerically pretty simple since such function is convex and increasing over $[0,1]$. By initializing Newton's method
$$ z_{n+1} = z_n - \frac{g(z_n)}{g'(z_n)} $$
at $z_0=1$ we get $z_2<\frac{2}{3}$ and $z_4=\color{green}{0.659046}\ldots$
You might have an explicit representation of such constant in terms of Lambert $W$ function, but the computation of $W$ still is performed through Newton's method, raising an interesting debate about what explicit really means.
