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Spivak's Calculus, 13.11

  1. Jan 16, 2013 #1
    These are self-study problems, which, unfortunately, aren't answered in the back. I'd appreciate if someone could review my answers and tell me what's weak about them, and what's good about them. And I think I need some help on D in particular.

    1. The problem statement, all variables and given/known data

    (a) Which functions have the property that every lower sum equals every upper sum?

    (b) Which functions that some upper sum equals some (other) lower sum?

    (c) Which continuous functions have the property that all lower sums are equal?

    *(d) Which integral functions have the property that all lower sums are equal? (Bear in mind that one such function is f(x)=0 for x irrational, f(x) = 1/q for x = p/q in lowest terms.) Hint: You will need the notion of a dense set, introduced in problem 8-7, as well as the results of Problem 30.

    2. Relevant equations

    Let L(f,P) be a lower sum of some function f for some partition P of the domain of f.
    Let U(f,P) be the analogous upper sum.

    3. The attempt at a solution

    a) This one is obviously constant functions, but I don't know how to prove it formally. I don't think it quite matters. Something along the lines of the remark that the mi must = Mi for all possible partitions, otherwise it would be possible to construct a partition P such that mi ≠ Mi for some i.

    b) Answer: I have a partial answer. I can think of some equations for which this is true. First of all, constant functions. Second of all functions that intersect zero at some point in their domains.

    We know first all that L(f,P) ≤ [itex]\int f[/itex] ≤ U(f,P), and thus L(f,P) = [itex]\int f[/itex] = U(f,P) whenever L(f,P) = U(f,P). Therefore we need to think of functions where L(f,P) = [itex]\int f[/itex] for some partition P.

    I don't think we can get there with any strictly positive continuous function. L(f,P) is always going to be some amount smaller than [itex]\int f[/itex]. If we consider functions that are sometimes negative and sometimes positive in the domain, then we can "make up" that difference by choosing less granular partitions across that part of the domain for which f is negative.

    We can do the same thing in reverse for U(f,P).

    For integrable, non-continous functions, I can't think of a way to "make up" the difference between any L(f,P) and [itex]\int f[/itex], again, unless there is some x such that f(x) < 0.

    This is far from proof though. I really need some assistance on this one, I think.

    c) Well, constant functions, certainly. I can't think of any others, but I can't prove that there are no others. I might go something like this: we know that we must be able to get arbitrarily close to [itex]\int f[/itex], if we think of any function that has one partition such that L(f,P) ≠ [itex]\int f[/itex], then we would not be able to make a partition arbitrarily closer to [itex]\int f[/itex] without having two partitions such that L(f,P) ≠ L(f,P'). Thus all lower sums must be equal to [itex]\int f[/itex], and for continuous functions, this is only true for constant functions.

    d) Any function f(x) such that inf(f(x)) is always the same no matter how it is partitioned. One such species would, again, be constant functions, but since we're considering discontinuous functions as well, intuitively I believe that the only kind are constant functions except point differences from the constant. In other words, no interval can fail to include inf(f(x)). I'm supposed to say the word "dense" in my answer. I think that answering this question is as much about learning how to "speak math" than it is about having some insight about lower sums or the geometric picture of the graph of f, and for that, I think I need to hear the words of mathemeticians.


    Best regards,

    Mason
     
    Last edited: Jan 16, 2013
  2. jcsd
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