How to show that the polynomilal $\sum_{k=0}^n \dfrac 1{3^{k^2}}x^k$ has $n$ distinct real roots for any positive integer $n$ ?

Question

From this Rational roots of polynomials ; How might we show that the polynomilal $\sum_{k=0}^n \dfrac 1{3^{k^2}}x^k$ has $n$ distinct real roots $\forall n \in \mathbb N$ ?

Answer 1

Evaluating the polynomial at $x=-3^{2m}$ for $0\le m\le n$ yields $$ \sum_{k=0}^n \frac1{3^{k^2}} (-3^{2m})^k = (-1)^m 3^{m^2} \sum_{k=0}^n (-1)^{m-k} 3^{-(m-k)^2}. $$ In the sum on the right-hand side, the $k=m$ term equals $1$, while the sum of all the other terms is less than $1$ in absolute value (one can show). Therefore the sign of the polynomial at $-3^{2m}$ is equal to $(-1)^m$. Since this is true for the $n+1$ points $m=0,1,\dots,n$, the polynomial must vanish at least $n$ times.