Does there exist a sequence $(a_n)$ such that, for all $n$, $a_0 +a_1 X +\cdots+a_nX^n$ has exactly $n$ distinct real roots?

Solution 1:

We can give inductive method of constructing such a sequence.

We can begin with the base case $a_0 = 1, a_1 = -1$.

Suppose we have the first $n$ coefficients, so that $p_n(x)$ has exactly $n$ distinct roots, $r_1 \lt r_2 \lt \, ... \lt r_n$. Now we select $(n + 1)$ points,

\begin{align} s_0 &\lt r_1\\ s_1 &\in (r_1, r_2)\\ s_2 &\in (r_2, r_3)\\ &...\\ s_{n-1} &\in (r_{n-1}, r_n)\\ s_n &\gt r_n \end{align}

Notice that, since $(r_i, r_{i+1})$ contains no roots, $p_n(x)$ has the same sign on the whole interval and that adjacent intervals have different signs, as all of the roots have multiplicity $1$. $^{\dagger}$ Hence, we have that $p_n(s_i)$ and $p_n(s_{i+1})$ always have different sign.

If we now choose $a_{n+1}$ to be small enough - specifically,

$$|a_{n+1}| \lt \min_{0 \le i \le n} \left|\frac{p_n(s_i)}{s_i^{\, n+1}}\right|$$

then we will retain the property that $p_{n+1}(s_i)$ and $p_{n+1}(s_{i+1})$ always have different sign. By the intermediate value theorem, this gives us exactly $n$ distinct roots lying between $s_0$ and $s_n$.

Now, to get the final root, consider the sign of $p_{n+1}(s_n)\, $; if we choose the sign of $a_{n+1}$ to be the opposite, then, for sufficiently large $x \gg s_n$, we will have that $$sign(p_{n+1}(x)) = sign(a_{n+1}) = sign(-p_{n+1}(s_n))$$ so that there must be a root lying between the two points (which is necessarily distinct from the other $n$ roots which are less than $s_n$).

$\dagger$:

• $p_n(x)$ must take the same sign on the whole interval, else by the intermediate value theorem, there would be another root in the interval.

• If $p_n(x)$ took the same sign on two adjacent intervals, $(r_{i-1}, r_i), (r_i, r_{i+1})$, the root $r_i$ would be a local extremum, so, by Fermat's theorem, we would have that ${p_n}'(r_i) = 0$ which would necessarily mean that $(x - r_i)^2$ was a factor of $p_n(x)$.