Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By romsek

Math Help - Numerical integration

  1. #1
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    433
    Thanks
    12

    Numerical integration

    Find \alpha such that the following differentiation formula holds:

    f'(x) \approx \alpha \int_{-h}^h t \, f(x+t) dt

    So first thing I did was tried to integrate using integration by parts

    \int_{-h}^h t \, f(x+t) dt = [t \, F(x + t) ] ^h _{-h} - \int_{-h}^h F(x+t) dt


    but I don't think that's right. I'm trying to get it to look like \lim_{h \rightarrow \infty} \dfrac{f(x + h) - f(x)}{h} so I'm guessing \alpha has a limit in it?

    Can anyone help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,789
    Thanks
    1149

    Re: Numerical integration

    I assume that h is supposed to be getting pretty small. If that's the case we can expand f as a taylor series

    $f(x+t)\approx f(x)+f'(x)t$ so

    $\displaystyle{\int_{-h}^h}t f(x+t)~dt \approx
    \displaystyle{\int_{-h}^h}t \left(f(x)+f'(x)t\right)~dt =
    \displaystyle{\int_{-h}^h}t f(x)+f'(x)t^2~dt$

    $\displaystyle{\int_{-h}^h}t f(x)+f'(x)t^2~dt= 0 + \dfrac 2 3 h^3 f'(x) = \dfrac 2 3 h^3 f'(x) $


    $f'(x) \approx \alpha \dfrac 2 3 h^3 f'(x)$

    so $\alpha = \dfrac 3 {2 h^3}$

    probably not a bad idea to get another set of eyeballs to check this.
    Thanks from Educated
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    433
    Thanks
    12

    Re: Numerical integration

    Quote Originally Posted by romsek View Post
    I assume that h is supposed to be getting pretty small. If that's the case we can expand f as a taylor series

    $f(x+t)\approx f(x)+f'(x)t$ so

    $\displaystyle{\int_{-h}^h}t f(x+t)~dt \approx
    \displaystyle{\int_{-h}^h}t \left(f(x)+f'(x)t\right)~dt =
    \displaystyle{\int_{-h}^h}t f(x)+f'(x)t^2~dt$

    $\displaystyle{\int_{-h}^h}t f(x)+f'(x)t^2~dt= 0 + \dfrac 2 3 h^3 f'(x) = \dfrac 2 3 h^3 f'(x) $


    $f'(x) \approx \alpha \dfrac 2 3 h^3 f'(x)$

    so $\alpha = \dfrac 3 {2 h^3}$

    probably not a bad idea to get another set of eyeballs to check this.
    Yea I think that's right, if we expand about the taylor series then we'd also get a f''(x) and a f'''(v) term which if you integrate the f'''(v) will give you the error of this approximation.

    I just didn't know how to start this, thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Numerical integration?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 23rd 2011, 06:54 AM
  2. numerical integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 9th 2010, 03:49 AM
  3. Example of Numerical Integration?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 7th 2009, 05:22 PM
  4. Numerical Integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 6th 2008, 05:29 PM
  5. Numerical integration!!!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2007, 03:59 PM

Search Tags


/mathhelpforum @mathhelpforum