How to prove that successive derivatives also have real roots?

Question

Mathematics

Good evening, please tell me how to approach.
Rolle's theorem does not work, since the roots can be multiple?

Answer 1

First, you only need to prove this for one derivative. All the rest are proved by induction.
If there are no multiple roots, then everything is simple. You have n-1 consecutive gaps between two roots. Each must have a zero derivative.
Now, let's say there is some root of degree k > 1.
Then P(x) = (x-a)^k Q(x).
If you take the derivative, you can see that it is divisible by (xa) to the power of k-1, i.e. it has k-1 roots:
P'(x) = k*(x-a)^(k-1)*Q(x)+(x-a)^k*Q'(x)
That is, if you have m different roots (with a total multiplicity of n), then the derivative, like in the first case, it has m-1 roots between the roots of P(x) and repeats the roots of the polynomial nm times more, as in the second case. In total - n-1 real roots, so all roots are real.