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Math Help - Approximation and error formula

  1. #1
    Junior Member
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    Mar 2011
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    Approximation and error formula

    Show for p=0,1,2,\ldots that \displaystyle \sum^n_{k=1} k^p = \frac{n^{p+1}}{p+1} + O(n^p).

    You may assume the error formula for the right rectangle rule: if f' exists and is continuous on [0,1], then for each n there exists \xi_n \in [0,1,] such that \displaystyle \int^1_0 f(x) dx = h \sum^n_{k=1} f(kh) - \frac{1}{2} f'(\xi_n) h, \qquad h = \frac{1}{h}, which is greater than \displaystyle \sum^n_{k=1} k^p supposedly by error bound O(n^p).

    Attemp:

    Let f(x) = x^p, then \displaystyle \sum^n_{k=1} k^p can be approximated by \displaystyle \int^n_0 f(x) dx = \int^n_0 x^p dx = \left[\frac{1}{p + 1} x^{p+1} \right]^n_0 = \frac{1}{p+1} n^{p+1}.

    But I don't understand how to obtain the big O bound of O(n^p) using the right rectangle rule. I cannot see how the rule is relevant. Can someone please help?
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  2. #2
    Senior Member
    Joined
    Mar 2010
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    280
    May be this helps.
    We have

    <br />
\displaystyle<br />
\int_0^1 f(x)dx=\frac{1}{n} \sum_{k=1}^n(\frac{k}{n})^p-\frac{1}{2}f'(\xi_n)\frac{1}{n}<br />

    Inserting

    <br />
\displaystyle<br />
f(x)=x^p<br />

    we get

    <br />
\displaystyle<br />
\int_0^1 x^pdx=\frac{1}{n} \sum_{k=1}^n(\frac{k}{n})^p-\frac{1}{2}p(\xi_n)^{p-1}\frac{1}{n}<br />

    Making substitution

    <br />
\displaystyle<br />
x=\frac{y}{n} \ \ \xi=\frac{\eta}{n}<br />

    we get

    <br />
\displaystyle<br />
\int_0^n (\frac{y}{n})^p\frac{dy}{n}=\frac{1}{n} \sum_{k=1}^n(\frac{k}{n})^p-\frac{1}{2}p(\frac{\eta_n}{n})^{p-1}\frac{1}{n}<br />

    <br />
\displaystyle<br />
\int_0^n y^pdy=\sum_{k=1}^n \ k^p-\frac{1}{2}pn \ \eta_n^{p-1}<br />
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  3. #3
    Junior Member
    Joined
    Mar 2011
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    Thank you. You have solved my problem.
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