# Existence of $n$ distinct roots of a set of polynomials

The question is

Let $$n$$ be a positive integer. Show that there are positive real numbers $$a_0,a_1,\dots, a_n$$ such that for each choice of signs the polynomial $$\pm a_{n} x^{n} \pm a_{n-1} x^{n-1}\pm a_{n-2} x^{n-2 }+\cdots \pm a_{1} x^{1}\pm a_{0} x^{0}$$ has $$n$$ distinct real roots.

I cannot determine a condition to guarantee $$n$$ distinct real roots for a polynomial.

One way for getting multiple roots is to use intermediate value theorem to produce at least one root in $$(0,1),(1,2),(2,3), \dots, (n-1,n)$$, but we are changing signs of coefficient for the rest of polynomials so I am not able to proceed from here.

## 1 Answer

Lemma: Suppose $$P(x)$$ is a polynomial with distinct real roots. Then $$\exists \epsilon>0$$ such that the polynomial $$P(x)+c$$ has distinct real roots for all $$|c|<\epsilon$$.

Proof: Write $$P(x)=a\prod_{i=1}^{n}{(x-x_i)}$$, $$x_1. Pick any $$x_0 and $$x_{n+1}>x_n$$. Then $$P(\frac{x_i+x_{i+1}}{2})$$ is nonzero and alternates sign for $$i=0,1,...,n$$. Take $$\epsilon=\min_{0 \leq i \leq n} |P(\frac{x_i+x_{i+1}}{2})|$$ Consider any $$c$$ with $$|c|<\epsilon$$. Then by intermediate value theorem, for $$i=1,...,n$$, exists $$s_i \in (\frac{x_{i-1}+x_i}{2},\frac{x_i+x_{i+1}}{2})$$, with $$P(s_i)=-c$$. Thus we get $$n$$ distinct real roots of $$P(x)+c$$. As $$P(x)+c$$ has order $$n$$, it has distinct real roots.

Now if $$a_0,a_1,...,a_n$$ work, for each choice of sign $$S \in \{-1,1\}^{n+1}$$ we get a polynomial $$P_S(x)$$ with distinct nonzero real roots, so $$xP_S(x)$$ has distinct real roots, so $$\exists \epsilon_S>0$$ such that the polynomial $$xP_S(x)+c$$ has distinct real roots for all $$|c|<\epsilon_S$$. Take $$0. Then for each choice of signs S, $$xP_S(x) \pm t$$ has distinct real roots. Therefore $$t, a_0,a_1,...,a_n$$ work for $$n+1$$. As the base case $$n=1$$ is obvious, we are done by induction.

• Probably one should assume $a_0\not=0$ in the induction hypothesis because otherwise $x P_S(x)$ would have $0$ as multiple root.. Dec 27 '18 at 5:36