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 (an)n0 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 ?


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,


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,


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).

Source : Link , Question Author : Community , Answer Author : John Don

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