1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Second order Taylor expansion

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that if [itex]F[/itex] is twice continuously differentiable on [itex](a,b)[/itex], then one can write

    [tex]
    F(x+h) = F(x) + h F'(x) + \frac{h^2}{2} F''(x) + h^2 \varphi(h),
    [/tex]

    where [itex]\varphi(h) \to 0[/itex] as [itex]h\to 0[/itex].

    2. Relevant equations



    3. The attempt at a solution
    I'm posting this here because it's a problem in Stein-Shakarchi's "Fourier Analysis". I'm working through this book on my own (so this problem is not homework), but I thought it'd look suspicious if I posted it in the regular forums.

    I believe I've managed to show that

    [tex]
    F(x+h) = F(x) + h F'(x) + \frac{h^2}{2} F''(x) + \int_0^h w \psi(w) dw,
    [/tex]

    where

    [tex]
    \psi(h) = \frac{F'(x+h) - F'(x)}{h} - F''(x),
    [/tex]

    but I'm not sure how I'm supposed to go about showing that

    [tex]
    \int_0^h w \psi(w) dw = h^2 \varphi(h).
    [/tex]

    What do you think the [itex]\varphi(h)[/itex] they're wanting here is?
     
  2. jcsd
  3. Feb 19, 2012 #2
    I am guessing [itex]\varphi(h)[/itex] are the higher order terms from the expansion. However, I have never seen it written as [itex]h^2\varphi(h)[/itex] but instead as [itex]\mathit{O}(h^2)[/itex].
     
  4. Feb 20, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The results you seek are proved in http://en.wikipedia.org/wiki/Taylor's_theorem . Google is your friend.

    RGV
     
  5. Feb 21, 2012 #4
    Well, after having consulted that website and changing variables a little bit in the "Taylor's theorem with integral remainder" formula, I've got that

    [tex]
    F(x+h) = F(x) + h F'(x) + \frac{h^2}{2} F''(x) + \frac 12 \int_x^{x+h} (x+h-t)^2 F'''(t) dt.
    [/tex]

    This is problematic for two reasons: First, in the problem I'm trying to solve, I know only that [itex]F[/itex] is [itex]C^2[/itex], so I'm not even sure [itex]F'''(t)[/itex] makes sense. Second of all, I'm not sure how one can finagle the formula I quoted to somehow turn [itex]\psi(t)[/itex] into [itex]F'''(t)[/itex]; I simply don't see how this is possible. Can someone provide some hints? Perhaps I'm on the wrong track with what I was trying to do, but I can't see how else one can conveniently define the function [itex]\psi[/itex]. Defining it to be the difference between the difference quotient and the derivative just seems so obvious...
     
  6. Feb 21, 2012 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You have gone one term too far: you don't know anything about F'''(t) because all you assumed was "twice continuously differentiable". Instead, try
    [tex] F(x+h) = F(x) + h F'(x) + \int_{x}^{x+h} (x+h-t)F''(t) dt, [/tex]
    or
    [tex] F(x+h) = F(x) + h F'(x) + \frac{1}{2} h^2 F''(x + \theta h), \; (0 < \theta < 1). [/tex]

    RGV
     
  7. Feb 23, 2012 #6
    Thanks very much, again, for your help. I've managed to verify that both of the following are true:

    [tex]
    F(x+h) = F(x) + h F'(x) + \int_x^{x+h} (x+h - t)F''(t)dt
    [/tex]

    and

    [tex]
    F(x+h) = F(x) + h F'(x) + \frac{h^2}{2} F''(x) + \frac 12 \int_x^{x+h} (x+h - t)^2F'''(t)dt.
    [/tex]

    All one needs to do is perform the integrals and write them out, and then everything cancels on the righthand side to just leave you with the equation [itex]F(x+h) = F(x+h)[/itex].

    I guess what confuses me is the form the authors of the text want you to supply. They want you to show that

    [tex]
    F(x+h) = F(x) + h F'(x) + \frac{h^2}{2} F''(x) + h^2 \varphi(h),
    [/tex]

    where [itex]\varphi(h) \to 0[/itex] as [itex]h \to 0[/itex]. But what I have above for [itex]\varphi(h)[/itex] is, by monotonicity of the integral,

    [tex]
    \int_0^h t \psi(t) dt \leq \int_0^h |t| |\psi(t)| dt \leq h^2 \sup_{t\in [0,h]} \psi(t).
    [/tex]

    ...and this goes to zero as [itex]h\to 0[/itex]! So why is it necessary to just throw in an extra [itex]h^2[/itex] like they want? IS it necessary?
     
    Last edited: Feb 23, 2012
  8. Feb 23, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you take the [itex](h^2/2)F''(x + \theta h)[/itex] form, and use continuity of F'' you get [itex](h^2/2)F''(x) + r(h),[/itex] where r(h) goes to zero faster than h^2. (In fact, [itex] r(h) = (h^2/2)[F''(x + \theta h) - F''(x)][/itex] does have the form [itex]h^2 \varphi(h), [/itex] where [itex] \varphi(h) \rightarrow 0 [/itex] as h --> 0.) I think that is all that is involved.

    RGV
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook