Results 1 to 5 of 5

Math Help - Taylor series and remainder

  1. #1
    Newbie
    Joined
    Dec 2007
    Posts
    9

    Arrow Taylor series and remainder

    Let f be a continuous function and its third derivative exists. I need to show that [ -f(x+2h) + 4f(x+h) - 3f(x)] / 2h = f(x) + O[h^2].

    I know how to show that -f(x+2h) + 4f(x+h) - 3f(x) = 2h f(x) by using the Taylor expansions:

    f(x+2h) = f(x) + 2h f(x) + [(2h)^2] f(x) / 2! + O[(2h)^3],

    f(x+h) = f(x) + h f(x) + [h^2] f(x) / 2! + O[h^3].

    Note that O[h^(n+1)] is the remainder of the Taylor expansion, and O[h^(n+1)] = f^(n+1)(z_n) h^(n+1) / (n+1)!, where z_n is some number between x and c (since h = x-c).

    How can I prove that we get O[h^2] in the first equation above? I cannot prove it by directly using the definition above. Please help me answer this question. If you know any links that may help me, please give me.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    It seems you are looked at divided differences.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2007
    Posts
    9
    This link is similar to the problem. However, it does not deal with remainders (a.k.a. orders) O[h^(n+1)]. I'm struggling with proving that there is O[h^2]... Any help?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    In general if f is defined in the neighborhood of a then n-th divided difference is \sum_{k=0}^n \frac{{n\choose k}f(a+kh)(-1)^{n-k}}{h^n}. So for n=2 this expression is (in numerator) f(a+2h) - 2f(a+h)+f(a). This means if f is (infinitly) differenciable in the neighborhood of a then f(a+2h) = f(a) + 2hf'(a)+2h^2f''(a)+E_1(h) where E_1(h)/h^3\to 0 as h\to 0 by Taylor's theorem. Furthermore, 2f(a+h) = 2f(a)+2hf'(a)+h^2f''(a)+E_2(h) where E_2(h)/h^3\to 0 as h\to 0 by Taylor's theorem. This means this difference is simply d(h) = f''(a)h^2 + E_1(h) - E_2(h). Thus d(h)/h^2 \to f''(a).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2007
    Posts
    9
    I figured the problem out. Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Test functions as a taylor series with integral remainder
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 13th 2011, 09:53 AM
  2. Taylor remainder
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 24th 2010, 02:00 PM
  3. Remainder of a Taylor Series.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 4th 2009, 06:05 PM
  4. Taylor series remainder
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 8th 2008, 01:29 PM
  5. Replies: 1
    Last Post: July 4th 2008, 07:52 AM

Search Tags


/mathhelpforum @mathhelpforum