polynomial series and root multiplicity

Question

Excuse me, because I know this is a double post but I can't for the life of me find the original post.

Given a sequence $(a_n)$, one can construct a polynomial of the form $\sum\limits_{n=0}^j(x^na_n)$. Is it possible to find a sequence for which this sum/polynomial never has a root of multiplicity $> 1$ for every value of $j$ ($j\in \mathbb{N}$) in the real numbers?

I tried proving this and the answer I get is 'no', but the proof is not really trivial and involves a rather questionable $z$ transform. Can someone verify this result or give me a more elegant proof?

Excuse me if this is trivial, I'm not studying maths ...

Answer 1

Let $\alpha$ be a transcendental number, for instance $\alpha=\pi$. Then for any sequence $(a_n)_{n>0}$ of nonzero rational numbers (or even algebraic numbers), the sequence $(a_0=\alpha,a_1,a_2,\dots)$ has the property you desire.

Indeed, if $P_N=\sum_{n=0}^Na_nX^n$ had a root in common with its derivative, say $r$, then $r$ would be algebraic for being a root of $P_N'$, and $\alpha$ would be a rational expression in powers of $r$ (for being a root of $P_N$), contradicting its transcendance.

Answer 2

What happens if you take the sequence $a_n=1$, for all $n$?

Then the polynomials are of the form $$ p_j(x)=\sum_{n=0}^j x^n = \frac{x^{j+1} -1}{x-1}. $$

The roots of this polynomial are the $(j+1)$-th roots of unity $$ e^{i 2 \pi k/ (j+1)}= \cos (\frac{2 \pi k}{j+1}) + i \sin (\frac{2 \pi k}{j+1}), $$ for $k=1, \cdots j$. So there are $j$ distinct (complex) roots for each value of $j$, each of multiplicity 1.

And if you only care about real roots, then note that (depending on whether $j+1$ is even or odd) there will either be $1$ real root of multiplicity 1 (namely $-1$) or no real roots.