Does there exist a sequence $(a_n)_{n≥0}$ such that, for all $n$, $a_0 +a_1 X +\cdots+a_nX^n$ has exactly $n$ distinct real roots ?

I wonder if such a sequence exists. Maybe something with algebraically independent real numbers ?

Is it possible to give an example of such a sequence ?

  • 2
    $\begingroup$ At each step (for $n>2$) make $a_{n+1}$ so small that the absolute value of $a_{n+1}x^{n+1}$ is smaller than the maximum of the absolute value of the previous polynomial in the interval from $r_1$ to $r_n$, where $r_1$ and $r_n$ are the smallest and largest root of the previous polynomial. That's all. $\endgroup$
    – arts
    Nov 30, 2017 at 17:59
  • $\begingroup$ The first few terms you can choose almost at random, say $a_0=-1, a_1=2, a_2=1$. $\endgroup$
    – arts
    Nov 30, 2017 at 18:05
  • 1
    $\begingroup$ @arts why not post an answer along those lines? $\endgroup$
    – zhw.
    Nov 30, 2017 at 18:57

1 Answer 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 $s_n$).


  • $p_n(x)$ 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 $p_n(x)$.

  • $\begingroup$ I wonder if someone could give a more explicit construction/example of such a sequence (e.g. a closed form expression for $a_n$)... $\endgroup$
    – John Don
    Nov 30, 2017 at 22:22
  • $\begingroup$ Thank you. I will mark your answer, just waiting if someone has an example as you said. $\endgroup$
    – user371663
    Dec 1, 2017 at 4:49
  • $\begingroup$ @Lucas I have yet to verify it, but I suspect that something like $$\cos(-\sqrt{x}) = \sum_{n=0}^{\infty} \frac{x^n}{(2n)!}$$ might do the trick (i.e. taking $a_n = \frac{1}{(2n)!}$). $\endgroup$
    – John Don
    Dec 2, 2017 at 3:10

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