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Math Help - a number logic probem

  1. #1
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    Question a number logic probem

    It is a major theorem that

    a number logic probem-untitled.jpg

    Thanks!
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  2. #2
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    Hello, soyeahiknow!

    It is a major theorem that:

    . . \displaystyle{S \;=\; \sum^{\infty}_{n=1}\frac{1}{n^2} \;=\;\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots \;=\;\frac{\pi^2}{6} }

    Use this to calculate: . [1]\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots

    We have: . X \;=\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots

    . . . . . . . . X \;=\;\dfrac{1}{2^2}\underbrace{\bigg[\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \hdots \bigg]}_{\text{This is }S}

    . . . . . . . . X \;=\;\dfrac{1}{4}\cdot\dfrac{\pi^2}{6}

    . . . . . . . . X \;=\;\dfrac{\pi^2}{24}




    [2]\;\dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \hdots

    Are the denominators primes? . 2, 3, 5, 7, 11, 13,\:\hdots

    or a Fibonacci-type sequence? . 2, 3, 5, 8, 13, 21,\:\hdots

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  3. #3
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    I believe they are primes
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  4. #4
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    can u tell me why you took out the 1/(2^2) ?
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, soyeahiknow!


    We have: . X \;=\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots

    . . . . . . . . X \;=\;\dfrac{1}{2^2}\underbrace{\bigg[\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \hdots \bigg]}_{\text{This is }S}

    . . . . . . . . X \;=\;\dfrac{1}{4}\cdot\dfrac{\pi^2}{6}

    . . . . . . . . X \;=\;\dfrac{\pi^2}{24}





    Are the denominators primes? . 2, 3, 5, 7, 11, 13,\:\hdots

    or a Fibonacci-type sequence? . 2, 3, 5, 8, 13, 21,\:\hdots



    I think the second one should be the sum of the squared inverses of the odd natural numbers, otherwise it seems to be pretty hard to calculate the sum over the primes...

    Tonio
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  6. #6
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    Quote Originally Posted by tonio View Post
    I think the second one should be the sum of the squared inverses of the odd natural numbers, otherwise it seems to be pretty hard to calculate the sum over the primes...

    Tonio
    Prime Zeta Function -- from Wolfram MathWorld

    id:A085548 - OEIS Search Results

    EDIT: I should have added some comment to be clear; it's likely that tonio is right that the problem is misstated; nevertheless, if it is correctly stated then the above links are relevant.
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  7. #7
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    Thanx, I knew that, but I highly doubt an obviously beginner student can come up with a research about the prime zeta function...

    Tonio
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  8. #8
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    Quote Originally Posted by tonio View Post
    Thanx, I knew that, but I highly doubt an obviously beginner student can come up with a research about the prime zeta function...

    Tonio
    Yes, I edited my post above with a comment, but not quick enough.
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  9. #9
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    wait wait,

    i think there may have been a mistake.

    The 2nd problem is just simply:

    1/(1^2) + 1(3^2) + 1/(5^2)

    So it is simply the odd numbers. So would I just do 1/(3^2) * S= 1/9 * (Pi^2)/6 = (Pi^2)/ 54? Thanks
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  10. #10
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    Quote Originally Posted by soyeahiknow View Post
    wait wait,

    i think there may have been a mistake.

    The 2nd problem is just simply:

    1/(1^2) + 1(3^2) + 1/(5^2)

    So it is simply the odd numbers. So would I just do 1/(3^2) * S= 1/9 * (Pi^2)/6 = (Pi^2)/ 54? Thanks
    In post #2, Soroban factored out (1/2)^2 because it evenly divides each term, and because doing so allows us to express the sum as a product of already known quantities.

    For this problem, think about what happens when you add this series to the series that Soroban solved.
    Last edited by undefined; July 25th 2010 at 01:29 PM. Reason: clarity
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  11. #11
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    if I add this series to the one Soroban solves, it would just be back to the original series right?


    OOOOO so all I have to do is ( original series - series with all the even numbers)= series for problem 2 ( the odd numbers). So (pi^2)/6-(pi^2)/24= (pi^2)/8

    and what kind of problems are these called? Thanks so much and sorry for the typo in the beginning! I posted it at like 3 am.
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  12. #12
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    Quote Originally Posted by soyeahiknow View Post
    if I add this series to the one Soroban solves, it would just be back to the original series right?


    OOOOO so all I have to do is ( original series - series with all the even numbers)= series for problem 2 ( the odd numbers). So (pi^2)/6-(pi^2)/24= (pi^2)/8

    and what kind of problems are these called? Thanks so much and sorry for the typo in the beginning! I posted it at like 3 am.

    You're right about the sum of the squared inverses of the odd naturals, and these series could be classified as ones related to the Riemann zeta function, a truly fascinating and intriguing series.

    Tonio
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