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Math Help - Compute N...

  1. #1
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    Compute N...

    Compute N if:

    N = 1 + 1/(2+1/(2+1/(2+...)))



    P.S. This will probably be the last problem I will post for a long time, if again. Thank you all for all of your help. Without you, these problems would remain unsolved and I'd be in trouble.
    Last edited by ceasar_19134; May 10th 2007 at 01:02 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ceasar_19134 View Post
    Compute N if:

    N = 1 + 1/2+1/2+1/2+...
    You're kidding right?

    Let me try this. The question you REALLY want answered is:
    N = 1 + 1/(2 + 1/(2 + 1/(2 + ...)))

    Either way, PLEASE use parenthesis!!

    -Dan
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  3. #3
    Member Rimas's Avatar
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    Yes what Dan said was right but could you answer the question

    The problem should read

    N = 1 + 1/(2 + 1/(2 + 1/(2 + ...)))
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  4. #4
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    Whoops. Sorry.

    My paper has it as a picture and I forgot I needed parenthesis.

    "Either way"?
    Last edited by ceasar_19134; May 10th 2007 at 01:29 PM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Rimas View Post
    Yes what Dan said was right but could you answer the question

    The problem should read

    N = 1 + 1/(2 + 1/(2 + 1/(2 + ...)))
    Assume this converges (in fact we know it does but I wont go into that now,
    Put:

    X = 1/(2 + 1/(2 + 1/(2 + ...)))

    Then:

    X= 1/(2+X),

    or:

    X^2+2X - 1=0

    or X = [-2 +/- sqrt(4+4)]/2 = +/-sqrt(2)-1

    but as we know X>0, we have X=sqrt(2)-1, so

    N = 1 + 1/(2 + 1/(2 + 1/(2 + ...))) = 1 + X = sqrt(2)

    RonL
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  6. #6
    Forum Admin topsquark's Avatar
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    I knew I had seen how to do this. I had just forgotten.

    -Dan
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  7. #7
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    The harder thing in this problem besides for the sum is to show it actually converges. But since this is a high school question we can ignore that.
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