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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 $>0$.

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It looks to me that you'd be hard-pressed to find a closed-form solution, given that you have polynomials of different degrees inside and outside the exponential... –  J. M. Nov 20 '11 at 15:25
    
@J.M. What do you think about my idea at (P.S.)? It's completely wrong? –  johan paul Nov 20 '11 at 15:33
    
According to Maple solution for $x$ is composed of Lambert W functions –  pedja Nov 20 '11 at 15:34
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If you're sure everything is always real, then the P.S. ought to nail it, yes. –  Henning Makholm Nov 20 '11 at 15:45
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In fact, if we assume $\alpha \lt \beta$, \alpha+1/(4\sigma^2) \lt x \lt \beta+1/(4\sigma^2)$ (Otherwise just flip the inequalities). This will allow numeric solution to proceed easily. You could check out chapter 9 of apps.nrbook.com/c/index.html or any numerical analysis text. Like the others, I don't think you will find an algebraic solution. –  Ross Millikan Nov 20 '11 at 16:33

1 Answer 1

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?

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i have to solve this for x, in any way, because this part is only a calculating part of my whole exercise, in order to find a rejection algorithm. I want the solution that verifies my equation, in order to use it to find an upper bound. –  johan paul Nov 20 '11 at 17:09

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