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Math Help - Riemann sum.

  1. #1
    Member integral's Avatar
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    Riemann sum.

    \displaystyle \lim_{x\rightarrow \infty}\sum_{k=1}^{x}\frac{(k-1)^4}{x^5}

    I'm sure this is a integral of some sort but I can not get it into the correct format to figure out the limits of integration (or the integrand! ):

    \displaystyle \int_{a}^{b}f(t)dt=\lim_{x\rightarrow \infty}\sum_{k=1}^{x} \left ( \frac{b-a}{x}\right )\cdot f\left ( a+k\frac{b-a}{x} \right )

    edit: sorry, typo, I changed it.
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  2. #2
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    We may write

    <br />
\displaystyle<br />
\sum \frac{1}{x} \; \frac{(k-1)^4}{x^4}=\sum \frac{1}{x} \; (\frac{k-1}{x})^4<br />
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  3. #3
    Member integral's Avatar
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    That is where I got originally. I simply do not know what to do from there.
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  4. #4
    Super Member girdav's Avatar
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    We have \displaystyle \sum_{k=1}^x\frac 1x\left(\frac{k-1}x\right)^4=\sum_{k=2}^x\frac 1x\left(\frac{k-1}x\right)^4=\sum_{j=1}^{x-1}\frac 1x\left(\frac jx\right)^4 =\sum_{j=1}^x\frac 1x\left(\frac jx\right)^4-\frac 1x and now you can conclude.
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  5. #5
    Member integral's Avatar
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    So:

    \displaystyle \sum_{i=1}^{n}\frac{(i-1)^4}{n^5}

    =\displaystyle \sum_{i=1}^{n}\left (\frac{i-1}{n} \right )^4 \frac{1}{n}

    =\displaystyle \sum_{J=2}^{n-1}\left ( J\frac{1}{n} \right )^4 \frac{1}{n}

    \therefore a=0, b=1

    \displaystyle \sum_{i=1}^{n}\frac{(i-1)^4}{n^5}= \sum_{J=2}^{n-1}\left ( J\frac{1}{n} \right )^4 \frac{1}{n}=\int_{0}^{1}x^4dx=\frac{1}{5}

    ?
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  6. #6
    Super Member girdav's Avatar
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    Quote Originally Posted by integral View Post
    So

    =\displaystyle \sum_{J=2}^{n-1}\left ( J\frac{1}{n} \right )^4 \frac{1}{n}
    I think you have to start the sum from J=1 (but it doesn't change anything because the result you found is correct).
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