Can we make a sequence of real numbers such that polynomial of any degree with co-efficients of the sequence has all its roots real and distinct ?

Question

Does there exist a sequence of real numbers $(a_n)$ such that $\forall n \in \mathbb N$ , the polynomial $a_nx^n+a_{n-1}x^{n-1}+...+a_o$ has all $n$ real roots ? Can we make a sequence so that all the $n$ real roots are distinct ?

Answer 1

Yes.

Suppose $p_n(x) = a_0 + \ldots + a_n x^n$ has $n$ distinct real roots $r_1 < \ldots < r_n$. Since these are roots of multiplicity $1$, $p_n(x)$ changes sign at each of these roots. Take $s_1 < r_1 < s_2 < r_2 < \ldots < r_n < s_{n+1}$. Thus $p_n(s_j)$ are nonzero, with alternating signs. Now take $a_{n+1}$ nonzero but so small that each $p_{n+1}(s_j) = a_{n+1} s_j^{n+1} + p_n(s_j)$ has the same sign as $p_n(s_j)$. Therefore the Intermediate Value Theorem says $p_{n+1}$ has a real root in each of the $n$ intervals $(s_j, s_{j+1})$. In addition I claim it has another real root outside $[s_1, \ldots, s_{n+1}]$: if $n$ is odd, $p_{n+1}(s_1)$ and $p_{n+1}(s_{n+1})$ have opposite signs, while $p_{n+1}(x)$ has the same sign as $x \to +\infty$ and as $x \to -\infty$. If $n$ is even, $p_{n+1}(s_1)$ and $p_{n+1}(s_{n+1})$ have the same sign, while $p_{n+1}(x)$ has opposite signs as $x \to +\infty$ and $x \to -\infty$. In either case, the Intermediate Value Theorem gives us either a root in $(-\infty, s_1)$ or a root in $(s_1, + \infty)$.

Answer 2

Here is an inductive, semi-non-constructive approach: assume I have a polynomial $p(x) = a_0 + \dots + a_n x^n$, with $n$ real roots and which goes to infinity as $x$ goes to infinity. Let M be large enough that all the zeros of $p$ are contained in $[-M,M]$.

Now, pick $a_{n+1}$ positive but so small that $a_{n+1}x^{n+1}$ is essentially zero on $[-2M,2M]$. Then $p(x) + a_{n+1}x^{n+1}$ looks almost the same as $p(x)$ on the interval $[-M,M]$ (in particular it has all the old zeros, slightly perturbed), but it also has a new real root, somewhere to the right of $-2M$.