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Math Help - Help Mathematical Proofs pleasee

  1. #1
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    Help Mathematical Proofs pleasee

    Heyy Forum
    Im having difficulty figuring out this question soo any help would be greatly appriciated
    Thanks so much

    Help Mathematical Proofs pleasee-1.png
    Last edited by jess0517; July 13th 2010 at 09:26 AM. Reason: SOLVED
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  2. #2
    Senior Member roninpro's Avatar
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    Did you try working out the product for different values of n?
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  3. #3
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    Quote Originally Posted by roninpro View Post
    Did you try working out the product for different values of n?
    Yes i did but i still couldnt seem to figure it out
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  4. #4
    Senior Member roninpro's Avatar
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    I computed the values for 2\leq n\leq 20:

    \frac{3}{4},\frac{2}{3},\frac{5}{8},\frac{3}{5},\f  rac{7}{12},\frac{4}{7},\frac{9}{16},\frac{5}{9},\f  rac{11}{20},\frac{6}{11},\frac{13}{24},\frac{7}{13  },\frac{15}{28},\frac{8}{15},\frac{17}{32},\frac{9  }{17},\frac{19}{36},\frac{10}{19},\frac{21}{40}

    Do you see a pattern, at least in the denominators?
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  5. #5
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    Quote Originally Posted by jess0517 View Post
    Heyy Forum
    Im having difficulty figuring out this question soo any help would be greatly appriciated
    Thanks so much

    Click image for larger version. 

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    Do some particular cases: n=2\Longrightarrow 1-\frac{1}{4}=\frac{3}{4}\,,\,\,n=3\Longrightarrow \frac{3}{4}\cdot\frac{8}{9}=\frac{2}{3} , n=4\Longrightarrow \frac{2}{3}\cdot \frac{15}{16}=\frac{5}{8} ,

    etc., and now separate in odd and even n's...can you see a pretty clear patern for each case?

    Tonio
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  6. #6
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    Hello, jess0517!

    Find a prove a formula for: . \displaystyle{\prod  ^n_{i=2}\left(1 - \frac{1}{i^2}\right) }

    I cranked out the first few partial products:

    . . \begin{array}{cccc}<br />
n & P(n) \\ \hline \\[-4mm]<br />
2 & \frac{3}{4} \\ \\[-4mm]<br />
3 & \frac{4}{6}\\ \\ [-4mm]<br />
4 & \frac{5}{8}\\ \\[-4mm]<br />
5 & \frac{6}{10}\\ \\[-4mm]<br />
6 & \frac{7}{12} \\ \\[-4mm]<br />
7 & \frac{8}{14} \\<br />
\vdots & \vdots \end{array}


    It appears that: . P(n) \:=\:\dfrac{n+1}{2n}

    I'll let you supply the proof.

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  7. #7
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    Direct computation.

    I assume "find and prove" means that induction is involved.

    But anyway you could find the value of \prod_{i=2}^{n}(1-\frac{1}{i^2}) by direct manipulation of the terms of the product (1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2}).
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  8. #8
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    Notice that 1-\dfrac1{i^2} = \dfrac{i^2-1}{i^2} = \dfrac{(i-1)(i+1)}{i^2}. Therefore

    \left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\ldots\left(1-\dfrac{1}{n^2}\right) = \dfrac{1\cdot3}{2^2} \cdot \dfrac{2\cdot4}{3^2} \cdot \dfrac{3\cdot5}{4^2} \cdots\dfrac{(n-1)(n+1)}{n^2}.

    Now start cancelling: each of the two factors in the numerators cancels with a factor in a neighbouring denominator. The only terms left uncancelled are \frac12 at the start and \frac{n+1}n at the end.
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  9. #9
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    Quote Originally Posted by Soroban View Post
    Hello, jess0517!


    I cranked out the first few partial products:

    . . \begin{array}{cccc}<br />
n & P(n) \\ \hline \\[-4mm]<br />
2 & \frac{3}{4} \\ \\[-4mm]<br />
3 & \frac{4}{6}\\ \\ [-4mm]<br />
4 & \frac{5}{8}\\ \\[-4mm]<br />
5 & \frac{6}{10}\\ \\[-4mm]<br />
6 & \frac{7}{12} \\ \\[-4mm]<br />
7 & \frac{8}{14} \\<br />
\vdots & \vdots \end{array}


    It appears that: . P(n) \:=\:\dfrac{n+1}{2n}

    I'll let you supply the proof.

    So using induction would this be correct?.. im not to sure how to write it out formaly on this site but this is how i did it:

    [(k+1)+1]/2k =[k+2]/2k = 1/2+1/k = [k+2]/2k ?
    Obviously adding a formal base step, induction hypothesis and induction step
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  10. #10
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    Thread closed due to this member deleting questions after getting help.
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