# Does there exist a sequence (an)(a_n) such that, for all nn, a0+a1X+⋯+anXna_0 +a_1 X +\cdots+a_nX^n has exactly nn 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 ?

We can give inductive method of constructing such a sequence.

We can begin with the base case $$a0=1,a1=−1a_0 = 1, a_1 = -1$$.

Suppose we have the first $$nn$$ coefficients, so that $$pn(x)p_n(x)$$ has exactly $$nn$$ distinct roots, $$r1. Now we select $$(n+1)(n + 1)$$ points,

s0rn\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 $$(ri,ri+1)(r_i, r_{i+1})$$ contains no roots, $$pn(x)p_n(x)$$ has the same sign on the whole interval and that adjacent intervals have different signs, as all of the roots have multiplicity $$11$$. $$†^{\dagger}$$ Hence, we have that $$pn(si)p_n(s_i)$$ and $$pn(si+1)p_n(s_{i+1})$$ always have different sign.

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

$$|an+1|

then we will retain the property that $$p_{n+1}(s_i)p_{n+1}(s_i)$$ and $$p_{n+1}(s_{i+1})p_{n+1}(s_{i+1})$$ always have different sign. By the intermediate value theorem, this gives us exactly $$nn$$ distinct roots lying between $$s_0s_0$$ and $$s_ns_n$$.

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

$$\dagger\dagger$$:

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

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