Consider a polynomial . We say a number is a zero of multiplicity if
- 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 .
Consider the equation
Show that there are two values of such that the equation is satisfied.
Proof. Let . (We want then to find the zeros of this function since these will be the points that .) Then,
Since for any (since ), we have
Then, is continuous and differentiable everywhere, so we may apply Rolle’s theorem on any interval. So, by Rolle’s theorem we know has at most two zeros (if there were three or more, say and , then there must be distinct numbers and with such that , but we know there is only one such that ).
Furthermore, has at least two zeros since , , and . Thus, by Bolzano’s theorem there are zeros between each of these points. We have that the number of zeros of is at most two and at least 2. Hence, the number of zeros must be exactly two