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Math Help - Function proof

  1. #1
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    Function proof

    f(n) = [1-(1/2^2)]*[1-(1/3^2)]*...*[1-(1/n^2)] where f: N − {1}→Q

    I need to use some small values of n to come up with a conjecture of F(n) and the prove it using induction.

    So far I have:

    Once I calculated some values I conjectured that f(n)=(n+1)/2n

    So my basis is n=2, ([1-(1/2^2)] = 3/4 = (2+1)/2*2

    Hypothesis is n=k, then [1-(1/2^2)]*[1-(1/3^2)]*...*[1-(1/k^2)] = (k+1)/2k

    Induction Step: n=k+1,
    [1-(1/2^2)]*[1-(1/3^2)]*...*[1-(1/k^2)]*[1-(1/(k+1)^2)] = [(k+1)+1]/2(k+1)

    So the right side becomes k+2/(2k+2)

    And the left side becomes k+1/2k*[1-(1/(k+1)2)]

    I tried simplifying the left side but I am definitely not getting anything similar to the right side. Any advice?
    Last edited by mr fantastic; October 20th 2009 at 03:06 AM.
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  2. #2
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    Hello CoraGB
    Quote Originally Posted by CoraGB View Post
    f(n) = [1-(1/2^2)]*[1-(1/3^2)]*...*[1-(1/n^2)] where f: N − {1}→Q

    I need to use some small values of n to come up with a conjecture of F(n) and the prove it using induction.

    So far I have:

    Once I calculated some values I conjectured that f(n)=(n+1)/2n

    So my basis is n=2, ([1-(1/2^2)] = 3/4 = (2+1)/2*2

    Hypothesis is n=k, then [1-(1/2^2)]*[1-(1/3^2)]*...*[1-(1/k^2)] = (k+1)/2k

    Induction Step: n=k+1,
    [1-(1/2^2)]*[1-(1/3^2)]*...*[1-(1/k^2)]*[1-(1/(k+1)^2)] = [(k+1)+1]/2(k+1)

    So the right side becomes k+2/(2k+2)

    And the left side becomes k+1/2k*[1-(1/(k+1)2)]

    I tried simplifying the left side but I am definitely not getting anything similar to the right side. Any advice?
    I agree with your conjecture. The induction proof goes like this:

    P(n) is the propositional function: f(n) = \frac{n+1}{2n}

    So P(k) \Rightarrow f(k) = \frac{k+1}{2k}

    \Rightarrow f(k+1) = f(k)\Big(1-\frac{1}{(k+1)^2}\Big)

    = \Big(\frac{k+1}{2k}\Big)\Big(1-\frac{1}{(k+1)^2}\Big)

    = \Big(\frac{k+1}{2k}\Big)\Big(\frac{(k+1)^2-1}{(k+1)^2}\Big)

    = \Big(\frac{1}{2k}\Big)\Big(\frac{k^2+2k}{k+1}\Big)

    = \Big(\frac{1}{2k}\Big)\Big(\frac{k(k+2)}{k+1}\Big)

    = \Big(\frac{1}{2}\Big)\Big(\frac{k+2}{k+1}\Big)

    = \frac{(k+1)+1}{2(k+1)}

    \Rightarrow P(k+1)

    You've already proved that P(2) is true. Hence, by Induction, P(n) is true for all n \ge 2.

    Grandad
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