Results 1 to 3 of 3

Thread: Approximation and error formula

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    72

    Approximation and error formula

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

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

    Attemp:

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

    But I don't understand how to obtain the big O bound of $\displaystyle O(n^p)$ using the right rectangle rule. I cannot see how the rule is relevant. Can someone please help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    May be this helps.
    We have

    $\displaystyle
    \displaystyle
    \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}
    $

    Inserting

    $\displaystyle
    \displaystyle
    f(x)=x^p
    $

    we get

    $\displaystyle
    \displaystyle
    \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}
    $

    Making substitution

    $\displaystyle
    \displaystyle
    x=\frac{y}{n} \ \ \xi=\frac{\eta}{n}
    $

    we get

    $\displaystyle
    \displaystyle
    \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}
    $

    $\displaystyle
    \displaystyle
    \int_0^n y^pdy=\sum_{k=1}^n \ k^p-\frac{1}{2}pn \ \eta_n^{p-1}
    $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2011
    Posts
    72
    Thank you. You have solved my problem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Oct 17th 2011, 12:21 PM
  2. Approximation and error
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Oct 9th 2011, 04:14 AM
  3. Replies: 0
    Last Post: Apr 1st 2009, 04:22 AM
  4. Help! Finding error of approximation
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Mar 29th 2009, 09:00 PM
  5. [SOLVED] error propagation: the approximation formula
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Aug 3rd 2008, 09:12 AM

Search Tags


/mathhelpforum @mathhelpforum