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Theoretical calculus

  1. Jul 25, 2012 #1
    1. The problem statement, all variables and given/known data
    comparing two partitions: Let f be defined on the interval [0,2] by f(x)=0, if 0≤x≤1
    and f(x)=1 otherwise. Consider two partitions P={0,1,2} and
    Q={0,1-[itex]\epsilon[/itex], 1+[itex]\epsilon[/itex],2}, and calculate the difference of the upper and the lower sum with respect to each partition. By using partitions like Q prove directly from the definition of integrability that the function f is integrable and calculate the integral.


    2. Relevant equations
    spf=ƩJ1mj(xj-xj-1)
    and replace m with M for Spf

    the definition of integrability I use in my textbook is:
    If f is a bounded function on [a,b], the following conditions are equivalent:
    a. f is integrable on [a,b]
    b. For every ε>0 there is a partition P of [a,b] such that SPf-sPf<ε

    3. The attempt at a solution
    I have made an attempt on solving for the upper and lower sum of P and I got 1+1+0 since the points {1,2} in P is 1 and {0} is 0. For Q I got the upper and lower sum as 2 as well as {0,1-ε} is in 0 and {1+ε,2} is in 1. However I don't know if I'm right or not. Neither do I know how to proceed with this question. Thank you very much in advance for anyone that's going to help me!
     
  2. jcsd
  3. Jul 25, 2012 #2

    LCKurtz

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    This is confusingly written, not to mention incorrect. What does "{0,1-ε} is in 0" mean??

    First, the sum I have highlighted in red is incorrectly copied. You wouldn't have the index of summation j also in the limits of the summation. The j at the top of the sum would probably be n, which is the number of intervals in the partition.

    Let's start with P. There are two intervals in P so you won't get three terms in either the upper or lower sum. The partition P is ##x_0=0,\, x_1 = 1,\, x_2=2##. Start by literally writing the sum (after you correct it) for this problem. By this, I mean figure out ##n## and ##m_j## and write out the terms for this problem. Once you do that correctly we can look at Q.
     
  4. Jul 26, 2012 #3
    Oh so i found out that the upper sum of P is 2 and the lower sum of P is 1, so the difference of upper sum and lower sum for P is 1.
    Upper sum of P: Lower sum of P:
    1(1-0)+1(2-1) = 2 0(1-0)+1(2-1) = 1
    Difference:
    2-1=1

    For upper sum of Q: Lower sum of Q:
    0+ε+1-ε=1 0+0+1-ε=1-ε
    Difference:
    1-(1-ε)=ε

    Then Since the definition of integrability says that if it is bounded ( which it is)
    and that upper sum - lower sum < ε, then its integrable (for Q)

    is everything right here? i have some doubts to how simple this answer looks as my professor tend to make really difficult questions...

    Thanks!
     
  5. Jul 26, 2012 #4

    LCKurtz

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    No, they aren't correct and the way you have combined things, it is very hard to follow what you have done. You aren't going to get them correct until you go step by step. I suggested before that you begin by writing out the sums for P. Start like this:##x_0=0,\, x_1=1,\, x_2=2##. Then$$

    L_p = \sum_{j=1}^2 m_j(x_j-x_{j-1}) =m_1(x_1-x_0)+m_2(x_2-x_1)=...$$
    Now put the numbers in unsimplified and then simplify what you get. That way it will be possible to follow your work and you have some chance of getting it correct. Then do the same thing for ##U_p##:$$
    U_p = ...$$ Continue with the same steps for Q, beginning with listing the ##x_j##.
     
  6. Jul 26, 2012 #5
    I have actually done them in the way you described it.
    for lower sum of P,
    Lp=∑mj(xj−xj−1)=m1(x1−x0)+m2(x2−x1)=...
    m1 = 0 , m2 = 1, then subsequently i got
    0(1-0)+1(2-1) = 1

    As for higher sum of P,
    M1=1 and M2=1
    so
    1(1-0)+1(2-1) = 2

    is it correct?
     
  7. Jul 26, 2012 #6

    LCKurtz

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    Not quite. How do you get and m2 = 1 and M1=1? Look carefully at your definition of f(x). Fix that then let's see the steps for partition Q.
     
  8. Jul 26, 2012 #7
    OH im so sorry, my question is actually wrong with the definition of f(x), its supposed to be
    0≤x<1 not less than or equal to 1.....
    then will my answers be correct?
     
  9. Jul 26, 2012 #8

    LCKurtz

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    So ##f(x) = 0## on ##0\le x < 1## and ##f(x) = 1## on ##1\le x \le 2##. Then your calculations for P are correct. Now let's see the full details for Q.
     
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