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Thread: Rate of convergence

  1. #1 Rate of convergence 
    Moderator Moderator AlexP's Avatar
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    When dealing with series, is there any sense of 'rate of change' or 'rate of convergence/divergence'? I know different series converge or diverge at different rates (for example, it can be said that the harmonic series converges very very slowly), so I'm just wondering if there's a way to kind of put a number on how quickly it does so.


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  3. #2  
    Forum Professor serpicojr's Avatar
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    If you have a convergent series, s is the sum of the series, and s<sub>N</sub> is the Nth partial sum, then you can try to find a good bound for the error |s-s<sub>N</sub>|. The techniques for doing so include comparing the error to an integral, using patterns in the signs of the terms of the series, or exactly evaluating the series. So, for example, you could ask how fast the series



    converges. You can bound the error by the integral



    So the function 1/N gives you an idea how fast this converges.

    This sort of reasoning also works for series that diverge. Here, however, you hope to find a function asymptotic to your series--namely, you want to find a nice function f(N) so that |s<sub>N</sub>-f(N)| grows smaller than f(N) does. For example, you spoke of the harmonic series



    We can approximate the partial sums by certain integrals. For example, we have the inequalities



    and



    So we see that ln(N) ≤ ln(N+1) ≤ s<sub>N</sub> ≤ 1+ln(N), and we have |s<sub>N</sub>-ln(N)| ≤ 1 for all values of N. Thus s<sub>N</sub> grows like ln(N).


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  4. #3  
    Moderator Moderator AlexP's Avatar
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    Interesting... I don't think I understand the work in your last example, but I think I get the idea of s<sub>N</sub> growing like ln(N), and in your first example, that 1/N describes how the series converges. So considering the graph of 1/x, can it be said that the series converges more slowly as n approaches infinity?

    Would the the sum from 1 to infinity of n (1+2+3+...+) grow like x<sup>2</sup>/2 by any chance?
    "There is a kind of lazy pleasure in useless and out-of-the-way erudition." -Jorge Luis Borges
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  5. #4  
    Forum Professor serpicojr's Avatar
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    Quote Originally Posted by Chemboy
    Interesting... I don't think I understand the work in your last example
    Yeah, I left out details. The basic idea is this. For the lower bound on 1+...+1/N, look at the function f(x) = 1/x on the interval from 1 to N+1. You can break it up into N disjoint subintervals of length 1, each having integer endpoints. On the interval from m to m+1, 1 ≤ m ≤ N, we have that 1/m is bigger than f(x), and so the integral from m to m+1 of 1/x is less than 1/m. Adding up these inequalities, we obtain the lower bound. The upper bound is similar--we break up the interval from 1 to N into N-1 disjoint subintervals of length 1. On the interval from m-1 to m, 2 ≤ m ≤ N, we have that 1/x is greater than 1/m. Adding up the inequalities, we have that 1/2+1/3+...+1/N is less than the integral from 1 to N. Adding one to both sides gives us the upper bound.

    in your first example, that 1/N describes how the series converges. So considering the graph of 1/x, can it be said that the series converges more slowly as n approaches infinity?
    Yes, the series does converge more slowly as n approach infinity, but this will have to be true for any convergent series. So this isn't really a useful mathematical fact. What is useful is comparing rates of convergence. So I can think of 1/n as being a certain rate of convergence, 1/n<sup>2</sup> as being faster, 1/ln(n) as being slower, etc.

    Would the the sum from 1 to infinity of n (1+2+3+...+) grow like x<sup>2</sup>/2 by any chance?
    Indeed. You can show this via integrals as I do for the harmonic series, or you can in fact show this by an exact formula. Indeed, we have:

    1+2+3+...+N = N(N+1)/2 = (N<sup>2</sup>+N)/2

    This differs from your estimate by N/2, but N/2 gets really small with respect to N<sup>2</sup>/2 as N goes to infinity, so we may as well ignore it if we're looking for the "main term" of this sequence's growth.

    To see the above equality, we can use induction. It's true for N = 1. If it's true for some N, then:

    1+2+...+N+(N+1) = N(N+1)/2+(N+1) = (N+1)(N/2+1) = (N+1)(N+2)/2

    And this is precisely the statement for N+1.
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  6. #5  
    Forum Professor river_rat's Avatar
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    There is a whole host of stuff you can talk about w.r.t the convergence rate of sequences - do you know anything about big O notation Chem?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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