Prove properties about the zeros of a polynomial and its derivatives

Consider a polynomial $f$ . We say a number $\alpha$ is a zero of multiplicity $m$ if

$f(x) = (x - \alpha)^m g(x),$

where $g(\alpha) \neq 0$ .

Prove that if the polynomial $f$ has $r$ zeros in $[a,b]$ , then its derivative $f'$ has at least $r-1$ zeros in $[a,b]$ . More generally, prove that the $k$ th derivative, $f^{(k)}$ has at least $r-k$ zeros in the interval.
Assume the $k$ th derivative $f^{(k)}$ has exactly $r$ zeros in the interval $[a,b]$ . What can we say about the number of zeros of $f$ in the interval?

Proof. Let $\alpha_1, \ldots, \alpha_k$ denote the $k$ distinct zeros of $f$ in $[a,b]$ and $m_1, \ldots, m_k$ their multiplicities, respectively. Thus, the total number of zeros is given by,
$r = \sum_{i=1}^k m_i.$

By the definition given in the problem, if $\alpha_i$ is a zero of $f$ of multiplicity $m_i$ then

$f(x) = (x - \alpha_i)^{m_i} g(x) \qquad \text{where } g(x) \neq 0.$

Taking the derivative (using the product rule), we have

$\begin{align*} f'(x) &= m_i (x - \alpha_i)^{m_i - 1} g(x) + (x- \alpha_i)^{m_i} g'(x)\\ &= (x - \alpha_i)^{m_i - 1} (m_i g(x) + (x - \alpha_i) g'(x)) \end{align*}$

Thus, again using the definition given in the problem, $\alpha_i$ is a zero of $f'(x)$ of multiplicity $m_i -1$ .
Next, we know from the mean-value theorem for derivatives, that for distinct zeros $\alpha_i$ and $\alpha_j$ of $f$ there exists a number $c \in [\alpha_i, \alpha_j]$ (assuming, without loss of generality, that $\alpha_i < \alpha_j$ ) such that $f'(c) = 0$ . Hence, if $f$ has $k$ distinct zeros, then the mean value theorem guarantees $k-1$ numbers $c$ such that $f'(c) = 0$ . Thus, $f'$ has at least:

$\left(\sum_{i=1}^k (m_i -1)\right) + k - 1 = \left( \sum_{i=1}^k m_i \right) - 1 = r-1 \ \ \text{zeros}.$

By induction then, the $k$ th derivative $f^{(k)}(x)$ has at least $r-k$ zeros.
If the $k$ th derivative $f^{(k)}$ has exactly $r$ zeros in $[a,b]$ , then we can conclude that $f$ has at most $r+k$ zeros in $[a,b]$ .

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