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Math Help - closest integer value

  1. #1
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    closest integer value

    Hello!
    Let {x} be the distance of x to the nearest integer value and let x=0.\overline{b_{1}b_{2} \cdots b_{m}} be a binary expansion with repetition.

    I want to prove that for all integers k and k_{0} the following is valid:
    \{2^{mk + k_{0} } x\} = \{2^{k_{0}  } x \}

    Well, in my opinion it is only valid for positive integers k and k_{0}.
    Anyway, here is what I did:

    \{2^{mk + k_{0} } x\}  = \{2^{mk + k_{0} } \cdot  0.\overline{b_{1}b_{2} \cdots b_{m}}\} = \{ 2^{k_{0}} \cdot 2^{mk} \cdot  } 0.\overline{b_{1}b_{2} \cdots b_{m}} \} =  \{ 2^{k_{0}} \cdot \cdot \sum_{i=1}^{m} \overline{b_{i}} 2^{-i + mk} \}

    Now some of these values in the sum must be integer values.
    Or do I just have to say that there is going to be a k-times shift of m digits and because of the repetition of x it will be always the same value. I just donīt know how to represent that integer value in the sum which will then disappear because of the definition of {}.

    Does anyone have any advice? Thanks!

    Regards
    Last edited by Tahoe; July 12th 2011 at 01:04 PM.
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  2. #2
    Junior Member
    Joined
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    Re: closest integer value

    Here is my solution: If there are mistakes could you please let me know? Thank you so much

    If we look at x=0.b_{1} b_{2} \cdots b_{m} then
    2^{m} \cdot x = 2^{m} \sum_{i=1}^{m} b_{i} 2^{-i} = \sum_{i=1}^{m} b_{i} 2^{m-i} = b_{1} 2^{m-1} + b_{2} 2^{m-2} \cdots + b_{m} 2^{m-m}
    will be an integer.

    When we look at x=\overline{0.b_{1} b_{2} \cdots b_{m}} we get
    2^{m} \cdot \overline{0.b_{1} b_{2} \cdots b_{m}} = 2^{m} \cdot \left(\sum_{i=1}^{m} b_{i} 2^{-i}  + \underbrace{0.0 \cdots 0}_{m \ zeros} \overline{b_{1} \cdots b_{m}}\right) = \\  2^{m} \cdot \sum_{i=1}^{m} b_{i} 2^{-i} + 2^{m} \cdot \sum_{i=1}^{m} \overline{b_{i}} \cdot 2^{-i-m} = 2^{m} \cdot \sum_{i=1}^{m} b_{i} 2^{-i} + \sum_{i=1}^{m} \overline{b_{i}} \cdot 2^{-i}

    Thus
    \{ 2^{m} \cdot \sum_{i=1}^{m} b_{i} 2^{-i} + \sum_{i=1}^{m} \overline{b_{i}} \} = \{\sum_{i=1}^{m} \overline{b_{i}} \}

    We can repeat such a procedure k times and that is why we get the result.

    I want to ask: What do you think of that solution? Thanks!
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  3. #3
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    Re: closest integer value

    Hello guys!

    I donīt want to bother but I really wonder what you think of my solution? I just want to make sure I donīt have any big mistakes in there

    Thank you!
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