Number of roots: _exponential_ quadratic equation. Let $q_1(x), q_2(x), \ldots, q_n(x)$ be polynomials of degree 2 (quadratics) with real coefficients and $a_n$ be real constants.
How can I prove that
$$\sum_{i=1}^n a_n\cdot \exp(q_n(x))=0$$
cannot have more than two roots?
PS: By root I mean the function has a change of sign either side of the root value of x
It seems intuitively correct, I have played about on my graphical calc and never exceed $2$ roots. I have tried logs and differentiation but nothing solid enough to call a proof.
Thanks in advance for hints, pointers and solutions.
 A: Your claim is incorrect, the equation can have an arbitrarily high number of roots.
Consider $a_1=1, q_1(x)=1-10(x-1)^2$ and $a_2=-1, q_2(x)=1-10(x-2)^2.$
We have $a_1e^{q_1(1)}=1e^1=e$ and for $|x-1| \ge 1$ that $q_1(x) \le 1 - 10 = -9$ and hence $0  < a_1e^{q_1(x)} \le 1e^{-9} < 2\times 10^{-4}$ under that condition.
Similarly, $a_2e^{q_2(2)}=-1e^1=-e$ and $0 > a_2e^{q_2(x)} \ge -1e^{-9} > -2\times 10^{-4}$ if $|x-2| \ge 1$.
In other words, for $i=1,2$ the term $a_ie^{q_i(i)}$ is $(-1)^{i+1}e$, but for $|x-i| \ge 1$ we have $|a_ie^{q_i(x)}| \le e^{-9}$, which is "small".
The sum $a_1e^{q_i(x)} + a_2e^{q_2(x)}$ is thus positive at $x=1$ (first summand is $e$, second summand is "small" so that the sum is positive), while it is negative for $x=2$ (second summmand is $-e$, first summand is small). Due the the continuity of $a_1e^{q_i(x)} + a_2e^{q_2(x)}$, this means there is a zero of that function in the interval $(1,2)$.
But we can continue this idea, we set, for $i=3,4,\ldots$
$$a_i =(-1)^{i+1}, q_i(x)=1-10(x-i)^2.$$
We get the same result as above, but now for all $i=1,2,\ldots$
$$a_ie^{q_i(i)}=(-1)^{i+1}e, \;|a_ie^{q_i(x)}| \le e^{-9} \text{ if } |x-i| \ge 1$$
If we consider the for some $n$ the sum
$$\sum_{i=1}^na_ie^{q_i(x)}$$
for $x=1,2,\ldots,n$, we find that one summand (namely for $i=x$) is either $e$ or $-e$, while the others are "small".
Even with $n=5001$, the "small" values add up (in absolute values) to less then $5000\times2\times10^{-4}=1$, so less then $e$, so they cannot change the sign dictated by the big value of absolute value $e$.
That means that sum has alternating signs at $x=1,2,\ldots, n$, so has at least $n-1$ zeros.
Sure, at some higher $n$ the argument breaks down because the small values can add up to more than $e$, but if you want more zeros, just change the 10 in the definition of the $q_i(x)$ to a higher value, which guarantees that the "small" contributions get even smaller, so you can choose $n$ as high as you want.
