Does there exist a sequence (an)n≥0 such that, for all n, a0+a1X+⋯+anXn 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 ?
Answer
We can give inductive method of constructing such a sequence.
We can begin with the base case a0=1,a1=−1.
Suppose we have the first n coefficients, so that pn(x) has exactly n distinct roots, r1<r2<...<rn. Now we select (n+1) points,
s0<r1s1∈(r1,r2)s2∈(r2,r3)...sn−1∈(rn−1,rn)sn>rn
Notice that, since (ri,ri+1) contains no roots, pn(x) has the same sign on the whole interval and that adjacent intervals have different signs, as all of the roots have multiplicity 1. † Hence, we have that pn(si) and pn(si+1) always have different sign.
If we now choose an+1 to be small enough – specifically,
an+1<min
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_{i1}, 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).
Attribution
Source : Link , Question Author : Community , Answer Author : John Don