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Homework Help: Test question we don't know who is right

  1. Apr 7, 2009 #1
    Suppose [tex]\alpha(x) = [x][/tex] is the floor function, then what is the value of [tex]\int_{0}^{n}f d \alpha[/tex]
    for [tex]f(x) \in R_{\alpha}[0,n][/tex]

    where n is an integer?

    This was a question on my exam. I want to know if I got it right. Some say the answer is zero, but I think it is:

    [tex]\sum_{i=1}^{n-1} \max \{ f(x) : i < x \leq i+1\}[/tex]

    Because [tex]M_i = \sup \{ f(x) | x_{i-1} \leq x \leq x_i \}[/tex] gets multiplied with the difference of the endpoints of every possible partition and, with a fine enough partition, we will get the value 0 most of the time and we will get the vale 1 n-1 times.


    Let me add some more info:

    To take the Stietjes integral you write [tex]P = \{a= x_0 < x_1 < \cdots < x_n = b \}[/tex] a partition of the interval [a ,b]. Then [tex] \Delta \alpha_i = \alpha( x_i) - \alpha (x_{i-1})[/tex] for i= 1, ..., n. Next for each i = 1, ..., n we define:

    [tex]m_i = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \}[/tex]
    [tex]M_i = \sup \{ f(x) : x_{i-1} \leq x \leq x_i \}[/tex]

    Now we can define the lower and upper Stietjes sums:

    [tex]L(f, P) = \sum_{i=1}^{n}m_i\Delta \alpha_i [/tex]
    [tex]U(f, P) = \sum_{i=1}^{n}M_i\Delta \alpha_i [/tex]

    Now we can define the lower and upper Stietjes integrals, which are equal for any Stietjes integrable function over a given [tex]\alpha[/tex].

    [tex]\bar{\int_{a}^{b}}f d \alpha = \inf_P U(f,P) [/tex]

    [tex]\int_{a}^{b}f d \alpha = \sup_P L(f,P) [/tex]

    So that's what we are talking about with this problem.
    Last edited: Apr 8, 2009
  2. jcsd
  3. Apr 8, 2009 #2

    This is driving me crazy.
  4. Apr 8, 2009 #3


    User Avatar
    Science Advisor

    What you have is basically right. The best way to get the details straight is to look at simple examples. [itex]\alpha(x)[/itex] is constant except where x is an integer so [itex]d\alpha[/itex] is 0 except at integers. If n= 1, [tex]\int_0^1 f(x)d\alpha= 0[/itex]. If n= 2, the integral is f(0)+ f(1). If n= 3, the integral is f(0)+ f(1)+ f(3), etc.
  5. Apr 8, 2009 #4
    futurebird, the n in the Stieltjes sum is not the same n as in the statement of your problem.

    Your intuition should be able to guide you quickly the correct answer. Review the simpler case where [tex]\alpha(x)=0[/tex] for [tex]x<c[/tex] and [tex]\alpha(x)=1[/tex] for [tex]x\ge c[/tex]. What then is [tex]\int_a^b f(x)\,d\alpha(x)[/tex] when [tex]a<c<b[/tex]?
  6. Apr 8, 2009 #5
    Well for [tex]\int_a^b f(x)\,d\alpha(x)[/tex] the change in alpha is 1 at least once for all partitions...
  7. Apr 8, 2009 #6
    m_i = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \}


    L(f, P) = \inf \{ f(x) : x_{i-1} \leq x \leq x_i \ni c \in (x_{i-1},x_i) \}


    \int_{a}^{b}f d \alpha = \sup_P L(f,P) = \inf \{ f(x) : a \leq x \leq b \}

    It seems like the same idea as the sum I described above.
  8. Apr 8, 2009 #7
    But now it looks like:


    \inf \{ f(x) : a \leq x \leq b \} = \sup \{ f(x) : a \leq x \leq b \}


    making f(x) a constant function. I'm so lost.
  9. Apr 8, 2009 #8
    This is [tex]L(f, P)[/tex] where [tex]P[/tex] depends on the choice of [tex] x_{i-1}[/tex] and [tex]x_i [/tex], with the only requirement being that [tex] c \in (x_{i-1},x_i)[/tex].

    Take [tex] x_{i-1}[/tex] and [tex]x_i [/tex] very close to [tex] c[/tex].
  10. Apr 8, 2009 #9
    Why would I want to make the partition so snug around c? I mean I know that can be done... but why? I want to say that the vale is f(c).

    But I don't think this is justified. (or I don't understand why it is justified.)
  11. Apr 8, 2009 #10
    You have to find [tex]\sup_P L(f,P)[/tex], and you get closer and closer to the sup when the partition is more and more snug.
  12. Apr 8, 2009 #11
    OK. I just thought "sup" meant whatever partition gave the greatest value and had nothing to do with "snugness" -- so if a big messy partition gives the largest value it's the "sup" -- but now I'm thinking that this might not ever happen I think we had a theorem that said refinements are bigger... OK...

    This is making more sense now.

  13. Apr 8, 2009 #12
    Right! In fact, why not take [tex]P_j [/tex] to be the simple partition [tex]a<c-\frac1j<c+\frac1j<b [/tex]?

    Just three subintervals: two fat and one skinny!
  14. Apr 8, 2009 #13
    Thanks so much for all of your help!!
  15. Apr 8, 2009 #14


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    Staff Emeritus
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    Gold Member

    I've lost your train of thought, but I think you may have overlooked the fact that the lower integral is the supremum of an infimum -- and you forgot about the infimum part.
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