Find the maxima of the sum of two Gaussian curves.

Question

$$y=A_1e^{-\frac{(x-d)^2}{2{\sigma _1}^2}}+A_2e^{-\frac{(x+d)^2}{2{\sigma _2}^2}}$$

To find the maxima I should solve $\frac{dy}{dx}=0$.

$$\frac{dy}{dx}=-\frac{A_1}{{\sigma _1}^2}e^{-\frac{(x-d)^2}{2{\sigma _1}^2}}(x-d)-\frac{A_2}{{\sigma _2}^2}e^{-\frac{(x+d)^2}{2{\sigma _2}^2}}(x+d)=0$$

But I have no idea how to solve this equation. Suggestions are welcome.

By imagining adding the two Gaussians it's clear that there are local maxima somewhere to the left of d and somewhere to the right of -d. The values of y at these two points is a little greater than A1 and a little greater than A2 respectively. Which is bigger will depend on the values of the A's and the sigma's. — berkeleychocolate, Jun 01 '14 at 22:14

score 2 · Answer 1 · answered Jun 01 '14 at 22:13

how to solve this equation

Numerically, that's the only way. The most you can expect from theoretical consideration is whether or not the sum has a single maximum or two maxima, and their rough locations.

To begin with: the second derivative of the Gaussian is negative when $|x-d|<\sigma$ and positive when $|x-d|>\sigma$. Also, the first derivative of the sum is negative when $x>d$ and positive when $x<-d$. Therefore, any maximum must lie in the set $$A= [-d,-d+\sigma_1] \cup [d-\sigma_2,d]$$ If $2d\le \min(\sigma_1,\sigma_2)$, then both summands are concave on $A$. Therefore the maximum is unique in this case. One can go on to finer analysis, but since it is not clear if you need it, I'll stop here.

I don't really understand your answer. Why is the second derivative of a Gaussian relevant here? Don't you usually want the second derivative of the sum? I'm also interested in for what values of the $\sigma _1$, $\sigma _2$, $A_1$, $A_2$ and $d$ there are one or two maxima. — m009, Jun 04 '14 at 08:55
@m009 The second derivative of the sum is the sum of second derivatives. — , Jun 04 '14 at 12:14

Find the maxima of the sum of two Gaussian curves.

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