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Math Help - Arccot series and Arithmetic progressions?

  1. #1
    Super Member fardeen_gen's Avatar
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    Arccot series and Arithmetic progressions?

    If x_1,x_2,x_3,\mbox{...} are in arithmetic progression with common difference d, show that:

    \mbox{arccot}\left(\frac{1 + x_{1}x_{2}}{d}\right) + \mbox{arccot}\left(\frac{1 + x_{2}x_{3}}{d}\right) + \mbox{arccot}\left(\frac{1 + x_{3}x_{4}}{d}\right) + \mbox{...} +\ \mbox{arccot}\left(\frac{1 + x_{n}x_{n + 1}}{d}\right) = \mbox{arccot}\left(\frac{1 + x_{1}x_{n + 1}}{d}\right), where x_1 > 0 and d > 0
    Last edited by fardeen_gen; May 21st 2009 at 11:14 AM.
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  2. #2
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    This statement is false.

    Letting x=\{2,4,6,8,10,12\} ,

    \sum_{k=1}^5 \arctan(\frac{2}{1+x_kx_{k+1}}) = .381

    but \arctan(\frac{2}{1+x_1x_{6}}) = .0798
    Last edited by Media_Man; May 22nd 2009 at 12:32 PM. Reason: small error in calculation
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  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    If x_1,x_2,x_3,\mbox{...} are in arithmetic progression with common difference d, show that:

    \mbox{arccot}\left(\frac{1 + x_{1}x_{2}}{d}\right) + \mbox{arccot}\left(\frac{1 + x_{2}x_{3}}{d}\right) + \mbox{arccot}\left(\frac{1 + x_{3}x_{4}}{d}\right) + \mbox{...} +\ \mbox{arccot}\left(\frac{1 + x_{n}x_{n + 1}}{d}\right) = \mbox{arccot}\left(\frac{1 + x_{1}x_{n + 1}}{d}\right), where x_1 > 0 and d > 0
    Since they are in A.P with common difference 'd', x_2 - x_1 = x_3 - x_2 = x_4 - x_3 = \cdots = d

    \sum_{k=1}^{k=n}\mbox{arccot}\left(\frac{1 + x_{k}x_{k+1}}{d}\right) = \sum_{k=1}^{k=n}\mbox{arccot}\left(\frac{1 + x_{k}x_{k+1}}{x_{k+1} - x_k}\right) = \sum_{k=1}^{k=n}\mbox{arccot}(x_{k+1}) - \mbox{arccot}(x_{k}) = \mbox{arccot}(x_{n+1}) - \mbox{arccot}(x_1) =\mbox{arccot}\left(\frac{1 + x_{1}x_{n+1}}{nd}\right)
    Last edited by Isomorphism; May 22nd 2009 at 01:51 PM. Reason: Thanks Media_Man
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  4. #4
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    Slight Correction

    Isomorphism:

    \mbox{arccot}(x_{n+1}) - \mbox{arccot}(x_1) = \mbox{arccot}\left(\frac{1 + x_{1}x_{n+1}}{x_{n+1}-x_{1}}\right) = \mbox{arccot}\left(\frac{1 + x_{1}x_{n+1}}{nd}\right)

    This slight correction will make the theorem true.
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