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Full-Width-Half-Maximum for difference of Gaussians

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I am trying to figure out the width-at-half-maximum for a speical case of the difference of Guaussians.  In this scenario, I start with a standard formula for a Gaussian:

f(x, x0, S, Gmax) = -Gmax*(exp(S*(x - x0)^2*(-1/2)) - 1)*heaviside(x - x0)

 

where x0, is the location of the peak, S is the spread, and Gmax is the amplitude.  However, then I take a difference of two, assuming that x0 is 0 for both, that Gmax is unity for both, and the only thing free to vary between them is the spread.

fDOG(x, S_a, S_d) = f(x, 0, S_a, 1) - f(x, 0, S_d, 1)

 

Here I also assume that S_a > S_d, and that all values (including x) are real positive numbers.  In this case, I (believe) I always get a peak function that rises from y = 0 at x = 0 to some peak, and then falls back to y = 0 at infinity.

Differentiating fDOG and solving for y = 0, I can find the time of the peak of this function:

tpeak(S_a, S_d) = (sqrt(2)*sqrt(ln(S_a/S_d))/sqrt(S_a - S_d))

 

I can then find the amplitude at the peak by substituting tpeak for x in fDOG. However, what I would like to do now is is find the (two) points on x where y = fDOG(tpeak, S_a, S_d)/2.  Is this possible?


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