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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.

  1. 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 kth derivative, f^{(k)} has at least r-k zeros in the interval.
  2. Assume the kth 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?

  1. 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 kth derivative f^{(k)}(x) has at least r-k zeros.

  2. If the kth 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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