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 zerosand
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*