1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Riemann Sums

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Let f(x) be defined on [0,1] by

    f(x) = 1 if x is rational
    f(x) = 0 if x is irrational.

    Is f integrable on [0,1]? You may use the fact that between any two rational numbers
    there exists an irrational number, and between any two irrational numbers there exists
    a rational number.

    2. Relevant equations


    3. The attempt at a solution

    Divide into n sub-intervals.

    Δxi=1/n

    U(f,Pn) = Ʃ(f(Ui)Δxi) = [itex]\sum(1)(1/n)[/itex] = 1/n
    L(f,Pn) = Ʃ[f(li)](Δxi) = [itex]\sum(0)(1/n)[/itex] = 0

    As n[itex]\rightarrow[/itex] [itex]\infty[/itex] both U(f,Pn) and L(f,Pn) [itex]\rightarrow[/itex] 0

    Therefore [itex]\int f(x)dx[/itex] = 0

    Is this correct?
     
    Last edited: Mar 12, 2013
  2. jcsd
  3. Mar 12, 2013 #2

    Zondrina

    User Avatar
    Homework Helper

    ##f## is most certainly not integrable on any interval ##[0,b]##.

    For each sub interval ##[x_{i-1}, x_{i}]## you can always find a rational number ##r \in [x_{i-1}, x_{i}]## with ##f(r) = 0##.

    So this tells you something about the lower sum.

    We can also find an irrational number ##q \in [x_{i-1}, x_{i}]## such that ##f(q) = 1##.

    This tells you something about the upper sum.

    Using this you can deduce ##f## is not integrable on any interval ##[0,b]##.
     
  4. Mar 12, 2013 #3
    Shouldn't it be f(r) = 1 and f(q) = 0?
     
  5. Mar 12, 2013 #4

    Zondrina

    User Avatar
    Homework Helper

    Oh whoops! Yes you're right.

    My bad about that, but you see what I was getting at right?
     
  6. Mar 12, 2013 #5
    No I can't say I do.
     
  7. Mar 12, 2013 #6

    Zondrina

    User Avatar
    Homework Helper

    Well... lets do the upper sum then as an example.

    Since ##f(r) = 1##, you know ##M_i = 1## no matter what sub-interval you choose.

    The upper sum is defined as such :

    ##S_p = \sum M_i Δx_i = \sum 1 Δx_i = b##

    Now try doing the lower sum ##s_p##, what do you get?

    EDIT : Just for clearness ##M_i = sup\{ f(x) \space | \space x \in [x_{i-1}, x_i] \}##
     
  8. Mar 12, 2013 #7
    f(q) = 0
    mi = 0

    sp = ƩmiΔxi = 0
     
  9. Mar 12, 2013 #8

    Zondrina

    User Avatar
    Homework Helper

    Exactly. So sup(sp) = 0 and inf(Sp) = b.

    Since 0 ≠ b, we know that f can't be Riemann integrable on any [0,b].

    You can actually generalize it to any [a,b].
     
  10. Mar 12, 2013 #9
    Ok, Thank you very much. But would you mind telling me what I did wrong in the 1st post? What assumption of mine is incorrect?
     
  11. Mar 12, 2013 #10

    Zondrina

    User Avatar
    Homework Helper

    You took n to infinity which wasn't what you were supposed to do.

    A function is Riemann Integrable on an interval [a,b] if sup(sp) = inf(Sp).
     
  12. Mar 13, 2013 #11

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    [itex]\displaystyle U(f,P_n)=\sum_{i=1}^{n}(1)(1/n)=n(1/n)=1[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Riemann Sums
  1. Riemann sums (Replies: 2)

  2. Riemann Sums (Replies: 2)

  3. Riemann Sums (Replies: 5)

  4. Riemann sum (Replies: 1)

  5. Riemann sums (Replies: 3)

Loading...