Find $\sigma_1$ that minimizes $(e^{x^2/\sigma_1^2}e^{x^2/\sigma_2^2}-e^{x^2/\sigma_3^2})^2$

Question

I'm trying to find

\begin{equation} \arg\min_{\sigma_1} \left(\exp\left(-\frac{x^2}{\sigma_1^2}\right)\cdot \exp\left(-\frac{x^2}{\sigma_2^2}\right)-\exp\left(-\frac{x^2}{\sigma_3^2}\right)\right)^2 \end{equation}

for real $x,\sigma_i$. In other words, multiplying one Gaussian by another in order to best approximate a third.

Mathematica reports that there is no global minimum for this expression. I'm wondering if there's some way of conditioning quantities, some trick with logs I don't know about, etc. to arrive at a useful characterization.

Answer 1

Hint.

If

$$ e^{x^2/\sigma_1^2}e^{x^2/\sigma_2^2}=e^{x^2/\sigma_3^2} $$

then

$$ e^{x^2/\sigma_1^2} = e^{x^2/\sigma_3^2-x^2/\sigma_2^2} $$

or

$$ \frac{1}{\sigma_1^2} = \frac{1}{\sigma_3^2}-\frac{1}{\sigma_2^2} $$

of course if $\sigma_3 < \sigma_2$

Answer 2

Thanks Cesareo, that's simple enough. So the solution is then

\begin{equation} \sigma_1 = \left\{\begin{array}{lr} \infty, & \text{for } \sigma_3\geq \sigma_2\\ \frac{\sigma_3\sigma_2}{\sqrt{\sigma_2^2-\sigma_3^2}}, & \text{for } \sigma_3< \sigma_2\\ \end{array}\right\} \end{equation}