Solution of an Equation I have this equation, and I want to find solution for x.
$\begin{align*}&(-2 x+2 α+1/(2 σ^2))\exp[(-(x-α)^2+(x-μ)/(2 σ^2))]+\\&(-2 x+2 β+1/(2σ^2))\exp[(-(x-β)^2+(x-μ)/(2 σ^2))]=0\end{align*}$
I have already used Wolfram, but it calculates the solution for $\sigma$, and other solvers say that they "Can not solve for x". Does anyone have an idea? Thank you all, in advance, for your concern.
P.S. Can I say that if this equation is equal to $0$, then only $(-2x+2 α+1/(2 σ^2))=0$ and $(-2 x+2 β+1/(2σ^2))=0$ at the same time, since exp is always $>0$.
 A: First a trivial algebraic simplification:
$$\begin{align}&{}\qquad (-2 x+2 α+1/(2 σ^2))\exp[(-(x-α)^2+(x-μ)/(2 σ^2))]\\
&{}\quad+(-2 x+2 β+1/(2σ^2))\exp[(-(x-β)^2+(x-μ)/(2 σ^2))] \\  \\
& = (-2 x+2 α+1/(2 σ^2))\exp[-(x-α)^2] \cdot \exp[(x-μ)/(2 σ^2)]\\
&{}\quad+(-2 x+2 β+1/(2σ^2))\exp[-(x-β)^2]\cdot\exp(x-μ)/(2 σ^2)],
\end{align}$$
so if this is $0$, then you can divide both sides by $\exp[(x-\mu)/(2\sigma^2)]$, since, as a value of the exponential function, that can never be $0$.
You have
$$
\left(-2(x-\alpha) + \frac{1}{2\sigma^2}\right) \exp(-(x-\alpha)^2) + \left(-2(x-\beta) + \frac{1}{2\sigma^2}\right) \exp(-(x-\beta)^2) = 0.
$$
A trivial substitution moves some complications into one place rather than two:
$$
\left(-2w + \frac{1}{2\sigma^2}\right) \exp(-w^2) + \left(-2(w-\gamma) + \frac{1}{2\sigma^2}\right) \exp(-(w-\gamma)^2) = 0.
$$
At this point I ponder whether some other trivial simplifications might help.  E.g. you could cancel $\exp(-w^2)$ from both sides, and a few other things like that.  But those don't seem potentially fruitful.
So, I'm thinking maybe Newton--Raphson or the like.
But: This looks like something that might have come from trying to find some MLEs.  So are you sure you shouldn't be trying to solve for $\alpha$, $\beta$, and $\sigma$ instead?
