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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.

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1 Answer 1

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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 $P(x)=a\prod_{i=1}^{n}{(x-x_i)}$, $x_1<x_2<...<x_n$. Pick any $x_0<x_1$ 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 $0<t<\min_{S \in \{-1,1\}^{n+1}}\epsilon_S$. 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.

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  • $\begingroup$ Probably one should assume $a_0\not=0$ in the induction hypothesis because otherwise $x P_S(x)$ would have $0$ as multiple root.. $\endgroup$ Dec 27, 2018 at 5:36

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