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Math Help - Limit (2)

  1. #1
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    Limit (2)

    Let f: [0,1] \longrightarrow \mathbb{R} be continuous. We all know that \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n f \left(\frac{j}{n} \right)=\int_0^1 f(x) \ dx. Now let L=\lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right). What do you think the value of L is?
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  2. #2
    Senior Member pankaj's Avatar
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    Is it Zero
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  3. #3
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    Quote Originally Posted by pankaj View Post
    Is it Zero
    that's the natural candidate and it's absolutely correct. so now we need a proof!
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  4. #4
    Super Member Random Variable's Avatar
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    Can you direct me to a proof of the limit that everyone knows except for me?
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  5. #5
    Senior Member pankaj's Avatar
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    \int_{a}^{b}f(x)dx=\lim_{h\to 0}h(f(a+h)+f(a+2h)+f(a+3h)+.....+f(a+nh))

    If a=0 and b=1,we get h=\frac{b-a}{n}=\frac{1-0}{n}=\frac{1}{n}.

    It is then that we have \int_{0}^{1}f(x)dx=\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^nf(\frac{r}{n})= \lim_{h\to 0}h(f(h)+f(2h)+f(3h)+.....+f(nh))

    What NCA has asked is
    \lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^n(-1)^{r-1}f(\frac{r}{n})
     <br />
=\lim_{n\to \infty}\frac{1}{n}(f(\frac{1}{n})-f(\frac{2}{n})+f(\frac{3}{n})-..........+(-1)^{n-1}f(\frac{n}{n}))<br />

     <br />
=\lim_{h\to 0}h(f(h)-f(2h)+f(3h)-f(4h)+f(5h)-....+(-1)^{n-1}f(nh))<br />

    = \lim_{h\to 0}h(f(h)+f(3h)+f(5h)+....)-\lim_{h\to 0}h(f(2h)+f(4h)+f(6h)+...)

     <br />
=\frac{1}{2}\lim_{h\to 0}(2h)(f(h)+f(3h)+f(5h)+....)-\frac{1}{2}\lim_{h\to 0}(2h)(f(2h)+f(4h)+f(6h)+...)<br />

     <br />
=\frac{1}{2}\int_{0}^1f(x)dx-\frac{1}{2}\int_{0}^1f(x)dx<br />

     <br />
=0<br />
    Last edited by pankaj; June 11th 2009 at 06:27 PM.
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  6. #6
    Senior Member pankaj's Avatar
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    I do not believe that I have done it right.
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Let f: [0,1] \longrightarrow \mathbb{R} be continuous. We all know that \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n f \left(\frac{j}{n} \right)=\int_0^1 f(x) \ dx. Now let L=\lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right). What do you think the value of L is?
    Write

    L_n = \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right). Then

    L_{2n} = \frac{1}{2n} \sum_{j=1}^n \Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n} \right)\Big]

    hence
    L_{2n} \leq \frac{1}{2n} n \mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big] = \frac{1}{2}\mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big]

    and similarily L_{2n} \geq  \frac{1}{2}\mbox{ inf }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big] so that

    \frac{1}{2}\mbox{ inf }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big] \leq L_{2n} \leq \frac{1}{2}\mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big]

    but both sides go to 0 as n \rightarrow \infty since f is continous!
    Last edited by Bruno J.; June 20th 2009 at 06:38 PM.
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  8. #8
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    Quote Originally Posted by Bruno J. View Post
    Write

    L_n = \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right). Then

    L_{2n} = \frac{1}{n} \sum_{j=1}^n \Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n} \right)\Big] n here should be 2n.

    hence
    L_{2n} \leq \frac{1}{n} n \mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big] = \mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big]

    and similarily L_{2n} \geq \mbox{ inf }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big] so that

    \mbox{ inf }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big] \leq L_{2n} \leq \mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big]


    but both sides go to 0 as n \rightarrow \infty since f is continous!

    even L_{2n} \to 0 will not necessarily imply that L_n \to 0 because we don't know that \lim L_n exists.
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  9. #9
    MHF Contributor Bruno J.'s Avatar
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    Thanks for the heads up! I was in a hurry when I wrote that (excuses excuses ).

    Indeed L_{2n}\rightarrow 0 does not imply the result, but a slight modification of the argument should show that L_{n}\rightarrow 0, ex. by writing

    L_{2n+1} = \frac{1}{2n+1}\Big[\sum_{j=1}^n\Big[f\Big(\frac{2j-1}{n}\Big)-f\Big(\frac{2j}{n}\Big)\Big]+f\Big(\frac{2n+1}{n}\Big)\Big]
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