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Math Help - [SOLVED] Double integral of floor function

  1. #1
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    [SOLVED] Double integral of floor function

    Let denote the greatest integer that is smaller than or equal to x. Evaluate where D = {}.

    Hints: The trick is to bring up the underlying region (which is a simple square) into two parts. D = D1 + D2

    where

    • D1 is the triangle with corners at (0,0), (1,0), and (0,1), excluding the diagonal from (0,1) to (1,0).
    • D2 is the rest of the remaining region of D.

    Therefore we have
    = + .
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  2. #2
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    The whole point of dividing D into two regions is that you can simplify the integrand \lfloor{x+y}\rfloor. In D_1, 0\le{x+y}<1, while in D_2, 1\le{x+y}<2 (except at the single point (1,1), which we can safely ignore), so \lfloor{x+y}\rfloor = 0 in D_1 and \lfloor{x+y}\rfloor = 1 in D_2. I think you can probably handle it from there.

    Post again if you're still having trouble.
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  3. #3
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    Thanks. I'm not understanding the integral of a floor function, though. We had a single integral of floor function x a few weeks ago. You guys around here helped me with that. I'm looking at those notes and I'm not able to work that out to the double integral of x+y. Something about adding that y in there is confusing me. I have no idea how to integrate this function.
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  4. #4
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    \int_{y}^{1+y}{\left\lfloor t \right\rfloor \,dt}=\sum\limits_{j=y}^{y}{\int_{j}^{j+1}{\left\l  floor t \right\rfloor \,dt}}=y, now in the double integral put t=x+y so its value is \frac12.
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  5. #5
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    Think of the floor function as a piecewise constant function:

    \lfloor{x}\rfloor = 0 when 0\le{x}<1, 1 when 1\le{x}<2, 2 when 2\le{x}<3, etc.

    When you integrate, you have to divide up the pieces:

    \int_{0}^{3}\lfloor{x}\rfloor dx = \int_{0}^{1}\lfloor{x}\rfloor dx + \int_{1}^{2}\lfloor{x}\rfloor dx + \int_{2}^{3}\lfloor{x}\rfloor dx = \int_{0}^{1}0 dx + \int_{1}^{2}1 dx + \int_{2}^{3}2 dx

    The double integral is the same idea. Now we're integrating over the square with vertices (0,0), (0,1), (1,1), (1,0) instead of a line, but it's the same thing - you divide the region into pieces. In this case, the triangle with vertices (0,0), (0,1), (1,0) where \lfloor{x+y}\rfloor = 0 is one region, and the triangle with vertices (1,1), (0,1), (1,0) where \lfloor{x+y}\rfloor = 1 is the other.

    So in your last equation, you can substitute 0 for \lfloor{x+y}\rfloor in the integral over D_1 and 1 for \lfloor{x+y}\rfloor in the integral over D_2. The first integral will be zero, and the second integral is just the area of the triangle with vertices (1,1), (0,1), (1,0), which should come out to 1/2.

    Don't be afraid to post again if you're still having trouble.
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  6. #6
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    Hollywood, your last post made it click for me. I was writing down what you typed and literally said "oooohhh" out loud. It's not nearly as hard as I thought. I must have just been having trouble relating to the definition of this function because I'm not familiar with it. Once I drew the square instead of just looking at my professor's diagram, it all sunk in. Thank you!!!
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