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Math Help - Find polynomial if n is odd

  1. #1
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    Find polynomial if n is odd

    Let p(x) be a polynomial of degree n and p(k) = \frac{k}{k+1}, k = 0,1,2,3,..., n. Find p(n+1) if n is odd.
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  2. #2
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    Re: Find polynomial if n is odd

    We claim the degree n polynomial
    p(x) = \sum_{i=1}^n (-1)^{i+1} \frac{1}{i+1} \binom{x}{i}
    where
    \binom{x}{i} = \frac{x(x-1)(x-2) \cdots (x-i+1)}{i!}
    (a generalized binomial coefficient) satisfies
    p(k) = \frac{k}{k+1} for k = 0, 1, 2, \dots n
    and that
    p(n+1) = 1 for n odd.

    The key to these claims is the identity
    (*) ... \sum_{i=1}^k (-1)^{i+1} \frac{1}{i+1} \binom{k}{i} = \frac{k}{k+1}
    To prove this, start with the binomial theorem
    (1+k)^k = \sum_{i=0}^k \binom{k}{i} x^i
    Integrate both sides from 0 to x and then set x=-1, with the result
    \frac{-1}{k+1} = \sum_{i=0}^k (-1)^{i+1} \frac{1}{i+1} \binom{k}{i} = -1 + \sum_{i=1}^k (-1)^{i+1} \frac{1}{i+1} \binom{k}{i}
    so
    \sum_{i=1}^k (-1)^{i+1} \frac{1}{i+1}\binom{k}{i} = 1 - \frac{1}{k+1} = \frac{k}{k+1}
    which shows that p(k) = \frac{k}{k+1} for k = 0, 1, 2, \dots n, as claimed.

    To show that p(n+1) = 1, start with (*) for k = n+1, i.e.
    \sum_{i=1}^{n+1} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} = \frac{n+1}{n+2}
    so
    \sum_{i=1}^{n} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} + (-1)^{n+2} \frac{1}{n+2}= \frac{n+1}{n+2}
    so for n odd
    \sum_{i=1}^{n} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} - \frac{1}{n+2}= \frac{n+1}{n+2}
    hence
    \sum_{i=1}^{n} (-1)^{i+1} \frac{1}{i+1} \binom{n+1}{i} = \frac{n+1}{n+2} + \frac{1}{n+2} = 1
    Last edited by awkward; August 8th 2012 at 07:41 PM.
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