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Homework Help: 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

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    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

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    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

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    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
     
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