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Upper & Lower Partitions

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f(x) = x, x [tex]\in[/tex] [0,1], [tex]P_{n}[/tex] = {0, [tex]\frac{1}{n}[/tex], [tex]\frac{2}{n}[/tex],..., [tex]\frac{n}{n}[/tex] = 1}.

    Calculate [tex]U_{P_{n}}[/tex](f) and [tex]L_{P_{n}}[/tex](f).

    2. Relevant equations

    [tex]U_{P_{n}}[/tex](f) is the sum of the upper partitions and [tex]L_{P_{n}}[/tex](f) is the sum of the lower partitions.

    A hint was [tex]\sum^{n}_{k=1}[/tex] = [tex]\frac{n(n+1)}{2}[/tex].

    3. The attempt at a solution

    I know that:

    [tex]U_{P_{n}}[/tex](f) = 1/2 + 1/2n
    [tex]L_{P_{n}}[/tex](f) = 1/2 - 1/2n

    just can't figure out how to get them.
  2. jcsd
  3. Feb 17, 2009 #2


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    Let's first work through some examples. If n = 3, then what are the upper and lower partition sums? Draw a picture and indicate the partition sums.
    If you make n = 4, or n = 5, then what happens.

    I don't want to hear the answer (for n = 3, U = 2/3 - I can read that off from the final answer which you already seem to have) but a description of how you find the answer, which can be generalized to any n
  4. Feb 17, 2009 #3


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    And I'll do n= 2 for you. Pn= {0, 1/2, 1}. That is, the x-axis is divided in to 2 intervals from 0 to 1: 0 to 1/2 and 1/2 to 1. If you draw y= x on coordinate system you see that it is increasing so the lowest value on each interval is on the left highest value is on the right. Draw vertical lines at x= 0, x= 1/2, x= 1. For L, the lower interval using the lowest value, draw horizontal lines from (0, 0) to (1/2, 0) and from (1/2,1/2) to (1, 1/2). You have divided the area into two "rectangles". I put "rectangles" in quotes because the first one is just the line from (0,0) to (1/2, 0). It has base 1/2 and height 0 so has area 0. The second rectangle has corners (1/2, 0), (1/2, 1/2), (1, 1/2) and (1, 0). Its base is 1/2 and its height is 1/2 so it has area 1/4. The total area of the two rectangles is 0+ 1/4 and so the total area L2= 1/4.

    To get the "upper area" use the highest value in each interval, which is on the right.
    Draw horizontal lines from (0, 1/2) to (1/2, 1/2) and from (1/2, 1) to (1, 1). Now we really have two rectangles, one with corners at (0,0), (0, 1/2), (1/2, 1/2) and (1/2, 0) and the other with corners at (1/2, 0), (1/2, 1), (1, 1) and (1, 0). The first rectangle has base 1/2 and height 1/2 and so has area 1/4. The second rectangle has base 1/2 and height 1 and so has area 1/2. The total area is 1/4+ 1/2= 3/4 and so U[sub2[/sub]= 3/4.

    NOw you do n= 3 where you divide the interval from 0 to 1 into three parts.
  5. Feb 17, 2009 #4
    got it, thanks guys. where n=3 the lower partition would be 0+1/9+2/9 = 1/3 and the upper partitions would be 1/9+2/9+1/3 = 2/3. then for the n case, in my question, the sum of the upper partitions would be 1/n^2 + . . . + n/n^2 where the sum of 1 to n = n(n+1)/2, and plugging in you get n+1/2n, and then similarily for the sum of the lower partitions. thanks again for the help.
  6. Feb 18, 2009 #5


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    Yes, you get Un= (n+1)/2n and Ln= (n-1)/2n as you say. As n goes to infinity, they both converge to 1/2 proving that the integral of x dx from 0 to 1 exists and is 1/2. Also notice that the "area under the curve" is inside the rectangles forming every upper sum, and so is always less than (n+1)/2n but always contains the rectangles forming every lower sum and so is always larger than (n+1)/2n. Since the sums for both upper and lower rectangles go to 1/2, and the area is always between them, the area of the triangle is 1/2. Of course, here, you could have calculated that from the formula for area of a triangle but for more complicated curves, that proves the "area beneath the curve" is the integral.
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