How to prove that the sum of a squared exponential functions has at most three critical points?

Question

I have two functions $f_i(x)$ and $g_i(x)$ which are:

$f_i(x)=(exp(-a_i*x^2))^2$

$g_i(x)=(1-exp(-a_i*x^2))^2$

and I know that all $a_i$ s are positive. Then, functions $F$ and $G$ are defined as:

$F(x)=\sum_{i=1}^k f_i(x)$

$G(x)=\sum_{i=k+1}^N g_i(x)$

And the function $H$ is define as:

$H(x) = F(x) + G(x)$

A sample shape of H is

https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=%3D(exp(-0.05x%5E2))%5E2%2B(exp(-0.2x%5E2))%5E2%2B(exp(-0.1x%5E2))%5E2%2B(1-exp(-0.3x%5E2))%5E2%2B(1-exp(-0.01x%5E2))%5E2

By several experiments, I found out that H has at most three critical points (points in which derivation of H is zero, or equivalently local extremums) including zero and sometimes two other points which have same absolute value but different signs (Except a special case in which H has infinite critical points). I tried too hard to prove this but I could not do that. I also searched and again found nothing. Can you please help me to prove it?

Answer 1

If the $a_i$ are small compared with $x$, then $f_i(x)=(exp(-a_i*x^2))^2 \approx (1-a_i*x^2)^2 \approx 1-2a_i*x^2+a_i^2*x^4 $ and $g_i(x)=(1-\exp(-a_i*x^2))^2 \approx (a_i*x^2)^2 \approx a_i^2*x^4 $ so

$\begin{array}\\ H(x) &\approx \sum_{i=1}^k (1-2a_i*x^2+a_i^2*x^4) +\sum_{i=k+1}^N a_i^2*x^4\\ &=k+x^2( -2\sum_{i=1}^k a_i+\sum_{i=k+1}^N a_i^2) +x^4\sum_{i=1}^N a_i^2\\ \end{array} $

and since this is a quartic, its derivative can have at most three zeros.

So this supports your conjecture.