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Thread: Limit (2)

  1. #1
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    Limit (2)

    Let $\displaystyle f: [0,1] \longrightarrow \mathbb{R}$ be continuous. We all know that $\displaystyle \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n f \left(\frac{j}{n} \right)=\int_0^1 f(x) \ dx.$ Now let $\displaystyle L=\lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right).$ What do you think the value of $\displaystyle L$ is?
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  2. #2
    Senior Member pankaj's Avatar
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    Is it Zero
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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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    $\displaystyle \int_{a}^{b}f(x)dx=\lim_{h\to 0}h(f(a+h)+f(a+2h)+f(a+3h)+.....+f(a+nh))$

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

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

    What NCA has asked is
    $\displaystyle \lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^n(-1)^{r-1}f(\frac{r}{n})$
    $\displaystyle
    =\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}))
    $

    $\displaystyle
    =\lim_{h\to 0}h(f(h)-f(2h)+f(3h)-f(4h)+f(5h)-....+(-1)^{n-1}f(nh))
    $

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

    $\displaystyle
    =\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)+...)
    $

    $\displaystyle
    =\frac{1}{2}\int_{0}^1f(x)dx-\frac{1}{2}\int_{0}^1f(x)dx
    $

    $\displaystyle
    =0
    $
    Last edited by pankaj; Jun 11th 2009 at 05:27 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 $\displaystyle f: [0,1] \longrightarrow \mathbb{R}$ be continuous. We all know that $\displaystyle \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n f \left(\frac{j}{n} \right)=\int_0^1 f(x) \ dx.$ Now let $\displaystyle L=\lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right).$ What do you think the value of $\displaystyle L$ is?
    Write

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

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

    hence
    $\displaystyle 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 $\displaystyle 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

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

    but both sides go to 0 as $\displaystyle n \rightarrow \infty$ since $\displaystyle f$ is continous!
    Last edited by Bruno J.; Jun 20th 2009 at 05:38 PM.
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  8. #8
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    Quote Originally Posted by Bruno J. View Post
    Write

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

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

    hence
    $\displaystyle 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 $\displaystyle 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

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


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

    even $\displaystyle L_{2n} \to 0$ will not necessarily imply that $\displaystyle L_n \to 0$ because we don't know that $\displaystyle \lim L_n$ 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 $\displaystyle L_{2n}\rightarrow 0$ does not imply the result, but a slight modification of the argument should show that $\displaystyle L_{n}\rightarrow 0$, ex. by writing

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