Consider a polynomial . We say a number is a zero of multiplicity if
where .
- Prove that if the polynomial has zeros in , then its derivative has at least zeros in . More generally, prove that the th derivative, has at least zeros in the interval.
- Assume the th derivative has exactly zeros in the interval . What can we say about the number of zeros of in the interval?
- Proof. Let denote the distinct zeros of in and their multiplicities, respectively. Thus, the total number of zeros is given by,
By the definition given in the problem, if is a zero of of multiplicity then
Taking the derivative (using the product rule), we have
Thus, again using the definition given in the problem, is a zero of of multiplicity .
Next, we know from the mean-value theorem for derivatives, that for distinct zeros and of there exists a number (assuming, without loss of generality, that ) such that . Hence, if has distinct zeros, then the mean value theorem guarantees numbers such that . Thus, has at least:By induction then, the th derivative has at least zeros.
- If the th derivative has exactly zeros in , then we can conclude that has at most zeros in .
For part (a), could we just use Rolle’s theorem to say if f has r zeros, we can divide f into r-1 partitions where f’ must equal zero somewhere in the partition, and thus f’ must have at least r-1 partitions?
and then f’ must have at least r-1 zeros*