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Math Help - unusual integral involving ceiling and floor functions

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    Super Member Random Variable's Avatar
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    unusual integral involving ceiling and floor functions



    where b is an integer greater than 1.

    If I was confident in my answer 1/(b-1) I would have posted this in the Math Challenge forum. Wolfram Alpha and Maple have no idea what to do with it, even for specific values of b and for finite intervals.
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    Quote Originally Posted by Random Variable View Post


    where b is an integer greater than 1.

    If I was confident in my answer 1/(b-1) I would have posted this in the Math Challenge forum. Wolfram Alpha and Maple have no idea what to do with it, even for specific values of b and for finite intervals.
    Plugging into mathematica with specific values for b you get

    b = 1.1, integral = 8.04619
    b = 2, integral = 0.989315
    b = 10, integral = 0.101846
    b = 100, integral = 0.00999388

    This seems close to 1/(b-1) and I wouldn't be too suprised if the discrepancies are due to the fact this integral is poorly behaved.
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  3. #3
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    Quote Originally Posted by Random Variable View Post


    where b is an integer greater than 1.

    If I was confident in my answer 1/(b-1) I would have posted this in the Math Challenge forum. Wolfram Alpha and Maple have no idea what to do with it, even for specific values of b and for finite intervals.
    I agree with the answer 1/(b-1) in the case where b is an integer. For non-integer values of b I'm not so happy with it. I think that if no power of b is an integer then the answer will be \sum_{k=1}^{\infty} \frac{1}{\lceil b^k \rceil}. If it were not for the ceiling function in the denominator, that would agree with your answer.

    If b is a number like \sqrt 2 with some powers integral and some powers non-integral, the answer would be very hard to compute.

    Without LaTeX being available at present, I couldn't face trying to give a detailed solution. Basically, you need to notice that the integrand is zero when x>1, so you only need to look at the intervals 1/(n+1) < x < 1/n, where the integrand is equal to the integer part of \log_{b}n. The difficulties arise from handling the cases where \log_{b}n is already an integer.

    Edit. As usual, I didn't read the problem carefully enough, where it stipulates that b is an integer. So yes, I agree with the given solution!
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  4. #4
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    Quote Originally Posted by Opalg View Post
    Without LaTeX being available at present, I couldn't face trying to give a detailed solution.
    You can use http://www.quicklatex.com/ to render latex and then imbed them as images. It's very quick and simple.
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