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Show this sequence converges and find its limit

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose f is continuously differentiable, and a < b. The sequence is defined as follows:

    [itex]a_{n} = n\int_a^b \! f(x) \, \mathrm{d} x - n(\frac{b-a}{n})\displaystyle\sum\limits_{i=1}^n f(a + \frac{b-a}{n}i)[/itex]

    3. The attempt at a solution
    I've been busting my *** with this one for well over 3 hours now, and thus far the only thing I was able to is to use the theorem that states that for some y, [itex]f(y)(b-a)=\int_a^b \! f(x) \, \mathrm{d} x[/itex].

    I have then expanded the sequence, took apart the summation to group the same values together, and ultimately got:

    [itex]a_{n} = (b - a)((nf(y) - \displaystyle\sum\limits_{i=1}^n f(a + \frac{b-a}{n}i)) = ... = (b - a)((f(y) - f(a + \frac{b-a}{n}) + ... + (f(y) - f(b)).[/itex]

    I'm unable to go anywhere from here :frown: I'm also not sure whether I should've shown that the sequence converges first, and then use a different approach to find its limit. I've been trying to show it converges, but I don't think it's monotone (even though it's bounded), and I also can't show that it's Cauchy. I figured the approach I took would take care of both the fact that it converges and the limit, but so far I'm not getting any of them.

    Thanks in advance for any help!
     
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  3. Dec 5, 2011 #2

    Dick

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    You should recognize the summation as an approximation to the integral, like a Riemann sum. What's the limit of just the summation without the extra n in front?
     
  4. Dec 5, 2011 #3

    Dick

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    Just a warning not to jump to the conclusion that the limit is zero just because the summation term is an approximation of n times the integral. It's not a particularly good approximation. You'll need to approximate the difference to get the limit. Didn't we used to have a 'Edit' button on posts?
     
  5. Dec 5, 2011 #4
    Yeah, I do recognize it, and that was actually a part of the question (the limit is the integral). It's just that I didn't post it, as I knew how to answer it. And I've tried inserting various functions into WolframAlpha, and I can see that the limit of this sequence is not zero. But I can't seem to find a pattern as to what it is, because some functions throw me off. I also figured maybe the approach I was contemplating last, i.e. grouping individual terms taken out of the summation, isn't the best one, but I don't really know what to do to approximate the difference between the whole summation and f(y).

    Intuitively, it seems to me that f(y) is the average value of the continuous function, whereas the summation, divided by n, is the approximation of that average value that gets better with increasing n.

    I'm thinking it might have something to do with the maximum and minimum of the function, but I can't quite figure out whether that's actually true, or whether it only "works" for some functions.

    Could you offer any additional help here perhaps?
     
  6. Dec 5, 2011 #5

    Dick

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    The integral of f(x) between a and b is the area under the graph of f(x). The summation is an approximation to that integral by bunch of rectangles. Draw a graph of this (pick any function you like). Now you want to estimate the difference between the integral and the sum of the area of the rectangles. The difference is the sum of a bunch of small "triangular" regions. Estimate the area of each triangular region. Use f'(x) to do it. It's the slope, yes? You can use that to estimate the height of the triangle. You should see another Riemann sum type integral emerging.
     
    Last edited: Dec 5, 2011
  7. Dec 6, 2011 #6
    Alright, so thinking about this, this is what I came up with. As n increases, wouldn't the difference between the actual area under the curve and the rectangle be [itex](f(a + \frac{b-a}{n}(i+1)) - f(a + \frac{b-a}{n}i))(\frac{b-a}{2n})[/itex], since (b-a)/n is the length of each interval and the height of the triangle is the difference between the two function values at the end of the interval. If this were the case, then after summing up the differences, everything but would cancel out, and we'd only be left with the [itex]\frac{(f(b)-f(a))(b-a)}{2n}[/itex] term. Since there's n terms in total, this would leave us with the limit being [itex]\frac{(f(b)-f(a))(b-a)}{2}.[/itex]

    Is my reasoning correct here? I'm namely unsure, because you also said this ...

    I don't think I quite get what you mean by using f'(x) to get the height of the triangle. In the method described above there doesn't seem to be a need for it, so I was wondering how exactly you envisioned this. Because if I'm wrong with the above approach, I'm in trouble, as I don't see what else I could do, even with the f'(x) hint :frown:
     
  8. Dec 6, 2011 #7

    Dick

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    That's fine. It's actually nicer than my way. I was approximating the height of the triangles by multiplying f'(x) times the width of the rectangle and winding up with a sum that converged to the integral of f'(x) from a to b. Which is f(b)-f(a). You might want to check the sign on your result. You are subtracting the rectangles from the integral, not the integral from the rectangles.
     
  9. Dec 6, 2011 #8
    Great to hear that this is the right approach then! The only thing I'm concerned is whether this is rigorous enough. I guess by doing this, we would establish a limit for each of the n terms in the sum, and we'd do this by using a geometric argument. Supposing this flies (is there another, non-geometric way, perhaps?), we'd still have to take n of those terms, and multiplying n, which by itself tends to infinity, by something that tends to a certain limit doesn't necessarily result in the product converging. So I'm still a bit unsure of how to explain this, since the product rule for the limits couldn't be applied here. Or is it that we're actually saying that as n gets larger, the difference between the integral and the summation actually is [itex]\frac{(f(b)-f(a))(b-a)}{2n}[/itex]?
     
  10. Dec 6, 2011 #9
    Well, the limit for each individual term of the sum is clearly zero, so you're right, you can't take the limit of each individual term in the sum first and then multiply everything by n.

    As for a non-geometric approach, the way I would do it is first show that a_n is equivalent to the expression:

    [tex]n\sum_{i=1}^{n} \int_{a+\frac{b-a}{n}(i-1)}^{a+\frac{b-a}{n}i}f(x) - f(a+\frac{b-a}{n}i)\ dx[/tex]

    Then add and subtract from each integral the term [itex](x-(a+\frac{b-a}{n}i))f'(a+\frac{b-a}{n}i)[/itex] to obtain:

    [tex]n\sum_{i=1}^{n} \int_{a+\frac{b-a}{n}(i-1)}^{a+\frac{b-a}{n}i}f(x) - f(a+\frac{b-a}{n}i) - (x-(a+\frac{b-a}{n}i))f'(a+\frac{b-a}{n}i)\ dx + n\sum_{i=1}^{n} \int_{a+\frac{b-a}{n}(i-1)}^{a+\frac{b-a}{n}i}(x-(a+\frac{b-a}{n}i))f'(a+\frac{b-a}{n}i)\ dx[/tex]

    It can be shown that the first term goes to zero and the second term (once you actually compute the integrals) is [itex]-\frac{b-a}{2}[/itex] times a Riemann sum for f'(x) on the integral [a,b], and hence converges to [itex]-\frac{(f(b)-f(a))(b-a)}{2}[/itex]
     
  11. Dec 6, 2011 #10

    Dick

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    You are right to be concerned that it's rigorous enough. The remaining thing to check is that the sum of the triangles approximates the error well enough. The remaining error is the same as the error you would get by doing a trapezoidal approximation of the integral. There's an easy error bound for that involving the second derivative of the function and it's proportional to 1/n^2. So that would fix it up. Except that you seem to only be given that the function has a continuous derivative. Not a second derivative.
     
  12. Dec 6, 2011 #11
    Thanks for the alternative approach, although I wish I could wrap my head around it. I just don't see how the first integral would go to zero, or how I'd prove that, and I don't know what to do with the second one, either. I mean, I'm trying, but I just can't make it work. Do you need to further separate the first integral? I guess what I don't see is how that extra term helps. And this is not a knock on your approach, I don't doubt it, and if anything, I feel really stupid for not getting it.
    Hmm, yeah, we are only given that it's C1. I really want to make this approach work, since, like I said, I can't wrap my head around that alternative one, at least not yet, and I don't think I could convince anyone. I also don't just want to go with something that I don't fully understand.
     
  13. Dec 6, 2011 #12
    The extra term helps by making the integrand smaller. Basically, you're subtracting off the tangent line to the function at the right endpoint of each integral, so the integrand is o(a+(b-a)i/n - x). Now, if you integrate a+(b-a)i/n - x over the interval [a+(b-a)(i-1)/n - x, a+(b-a)i/n - x], you get a constant times 1/n², so this tells you that the integrals are each o(1/n²). But you have n integrals in the sum on the left, so that sum is o(1/n), and multiplying it by n gives you o(1). Since this is a little o, that means that the left term goes to 0 as n → ∞.

    As for actually proving that it goes to 0, probably the best way is to do direct ε-δ work. Choose δ so that |x-y|<δ implies that |f'(x) - f'(y)|<Cε (where C is a constant that will be specified later), and suppose that n is large enough so that (b-a)/n < δ. Then you can show using the mean value theorem that:

    [tex]|f(x) - f(a + \frac{b-a}{n}i) - (x-(a+\frac{b-a}{n}i))f'(a+\frac{b-a}{n}i)| \leq |x-(a+\frac{b-a}{n}i)| C \varepsilon[/tex]

    And then you can compute directly each integral of the form:

    [tex]\int_{a+\frac{b-a}{n}(i-1)}^{a+\frac{b-a}{n}i} |x-(a+\frac{b-a}{n}i)|\ dx[/tex]

    You'll get the same result for each integral, so then computing the sum of all those integrals will be easy. At then end you'll get that the term on the left is bounded by come constant times Cε. Then go back to the beginning of the proof and set C to be less than the reciprocal of that constant, and then you'll have a nice ε-δ proof that the terms on the left really do approach 0 as n→∞.

    For the term on the right, just literally compute each integral (the integrands are polynomials in x, so it should be easy). Factor out everything that doesn't belong in a Riemann sum.
     
  14. Dec 7, 2011 #13
    Hey, sorry for the late response, but I appreciate responses from both of you guys. Unfortunately, as I got stuck even with the help you offered, with limited time I had to skip to other questions on the assignment. I hope to get back to this one later on, and since it's due today, I have to finish it in a few hours anyway.

    Citan Uzuki, quickly glancing at your further post, I may have trouble with that mean value theorem thing, but I'll look into it more thoroughly later on. Hopefully I can do the rest, although I've noticed that when I get into a certain mode of thinking, it's hard to snap out of it, and that's why I'm sometimes having trouble seeing what you mean, and seem as if I had never seen the stuff before.

    Dick, so basically without the second derivative, I can't really prove convergence with your approach?
     
  15. Dec 7, 2011 #14
    The mean value theorem should be fairly easy to apply with the hints I've given you, but here's a little nudge further -- you know that:

    [tex]\frac{f(x)-f(a+\frac{b-a}{n}i)}{x-(a+\frac{b-a}{n}i)}=f'(c)[/tex]

    For some c between x and [itex]a + \frac{b-a}{n}i[/itex]. How far can c be from [itex]a + \frac{b-a}{n}i[/itex]? Hence how far apart can f'(c) and [itex]f'(a + \frac{b-a}{n}i)[/itex] be?
     
  16. Dec 7, 2011 #15

    Dick

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    Not unless you can think of another way to prove the difference between a trapezoidal integral approximation and the integral shrinks faster than 1/n with the assumption of only one continuous derivative.
     
  17. Dec 9, 2011 #16
    Alright, thanks again to both of you. In the end, I unfortunately didn't get it. I went with what Citan Uzuki suggested, but got stuck on proving the first term goes to 0. I tried to handwave my way through, but I doubt it will count for anything. Meh, it's fine, although I always get frustrated when I think of how much time I spend with some problems only to get them wrong or not at all.
     
  18. Dec 9, 2011 #17

    Dick

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    You made a good try. I think the limit is correct. At least on C^2 functions. That ought to count for something. Post back if you figure out what I was missing.
     
  19. Dec 20, 2011 #18
    This response is way overdue, but I did want to give some feedback on this. I can't really post the solution as envisioned by our professor, and to be honest, since there was so much other stuff to do, I didn't go through it carefully enough to fully understand it. But basically, yeah, the limit that Citan Uzuki posted (i.e. with the corrected minus sign when compared to my post) is correct, and his approach is similar to what our professor used. That is, at least as far as splitting the interval goes, but it does start to diverge after that. At first I had my doubts about his approach, not in terms of correctness, but it seemed very complicated. Well, it seems to me now his is actually the simpler of the two :smile: So thanks again, Citan Uzuki, for the help.

    Dick, as for what the geometric approach might be missing, I don't really know. It's clear it leads to the same result, but I have no idea whether it can be made to work with enough rigour given our assumptions. But it did give me a new way to think about it, so thanks to you, as well :wink:

    I really like how in the end three different approaches can get you to the same point, though. Ah, the beauty in mathematics, eh? :biggrin:
     
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