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Thread: Proof by Induction?

  1. #1
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    Proof by Induction?

    Prove that

    $\displaystyle {\prod}_{i=2}^n \left(1-\frac{1}{i^2}\right) = \frac{n+1}{2n} $ for integers $\displaystyle n\ge2 $

    I want to know - what does the $\displaystyle \prod $ mean? Can I do this question through induction? And how does the $\displaystyle i = 2 $ affect my working - as opposed to $\displaystyle i=0 $ or $\displaystyle i=1 $?

    I did attempt this, but I haven't got anywhere near close enough. I, sort of, treated the $\displaystyle \prod $ sign as a $\displaystyle \sum $ - so the term to add on to both sides was $\displaystyle \left(1 - \frac{1}{(k+1)^2}\right) $. So no doubt I was completely wrong there.

    Help would be much appreciated. Thanks.
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  2. #2
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    Sigma, $\displaystyle \sum {} $, stands for SUM where as Pi, $\displaystyle \Pi $ stands for product.
    $\displaystyle \prod\limits_{i = 2}^4 {\left( {1 - \frac{1}{{i^2 }}} \right)} = \left( {1 - \frac{1}{{2^2 }}} \right)\left( {1 - \frac{1}{{3^2 }}} \right)\left( {1 - \frac{1}{{4^2 }}} \right)$
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  3. #3
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    Quote Originally Posted by Plato View Post
    Sigma, $\displaystyle \sum {} $, stands for SUM where as Pi, $\displaystyle \Pi $ stands for product.
    $\displaystyle \prod\limits_{i = 2}^4 {\left( {1 - \frac{1}{{i^2 }}} \right)} = \left( {1 - \frac{1}{{2^2 }}} \right)\left( {1 - \frac{1}{{3^2 }}} \right)\left( {1 - \frac{1}{{4^2 }}} \right)$
    Oh ok thanks!

    But what's the difference if it was

    $\displaystyle \prod\limits_{i = 1}^4 {\left( {1 - \frac{1}{{i^2 }}} \right)} $

    i.e. what difference does the i=2 make as opposed to i=1 or i=0 ?

    I really should know this
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  4. #4
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    Firstly, the problem states that the index begins i=2.
    If i=1 then the product would equal zero, would it not?
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  5. #5
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    Quote Originally Posted by Plato View Post
    Firstly, the problem states that the index begins i=2.
    Yes, of course. But I was confused what the i=2 meant.

    Quote Originally Posted by Plato View Post
    If i=1 then the product would equal zero, would it not?
    Yes. So under the pi sign, if it's i=3, does that mean that n=3 is the first
    term? (Sorry, I know it's probably very simple to understand, but I'm not quite grasping it).

    As for the question, I was able to prove it. Thanks.


    __________________________________________________ _______________

    $\displaystyle P_2 $

    $\displaystyle \prod\limits_{i = 2}^2 {\left( {1 - \frac{1}{4}} \right)} = \frac{3}{4}$

    Therefore true for $\displaystyle P_2$, assume it's true for $\displaystyle P_k$

    $\displaystyle P_{k+1}$

    $\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k+1}{2k} \left(1-\frac{1}{(k+1)^2}\right)$

    $\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k+1}{2k} - \frac{k+1}{2k(k+1)^2}$

    $\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{(k+1)^3-(k+1)}{2k(k+1)^2} $

    $\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k^3+3k^2+2k}{2k(k+1)^2} $

    $\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k(k+1)(k+2)}{2k(k+1)^2} $

    $\displaystyle \prod\limits_{i = 2}^{k+1} {\left( {1 - \frac{1}{k^2}} \right)} = \frac{k+2}{2(k+1)} $

    If $\displaystyle P_k$ is true, then $\displaystyle P_{k+1}$ is also true. But since $\displaystyle P_2$ is true, $\displaystyle \rightarrow$ $\displaystyle P_3$ is true, $\displaystyle \rightarrow$ $\displaystyle P_4$ is true...etc. Hence proved.
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  6. #6
    Senior Member DivideBy0's Avatar
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    You can't have k as both the upper bound and the index.

    It should be

    $\displaystyle \prod_{i=2}^{k+1} \left(1-\frac{1}{i^2} \right)=\frac{(k+1)+1}{2(k+1)}$

    with $\displaystyle i$ as the index.
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  7. #7
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    Quite right. Rookie mistake

    Thanks.
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