# 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?

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 ?

• 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. – arts Nov 30 '17 at 17:59
• The first few terms you can choose almost at random, say $a_0=-1, a_1=2, a_2=1$. – arts Nov 30 '17 at 18:05
• @arts why not post an answer along those lines? – zhw. Nov 30 '17 at 18:57

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 $$p_n(x)$$.

• I wonder if someone could give a more explicit construction/example of such a sequence (e.g. a closed form expression for $a_n$)... – John Don Nov 30 '17 at 22:22
• Thank you. I will mark your answer, just waiting if someone has an example as you said. – user371663 Dec 1 '17 at 4:49
• @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 $a_n = \frac{1}{(2n)!}$). – John Don Dec 2 '17 at 3:10