What is the simplification of the second order Taylor expansion for F(x+h)?

[AxiomOfChoice]

Homework Statement

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + h^2 \varphi(h),<br />

where \varphi(h) \to 0 as h\to 0.

Homework Equations

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + \int_0^h w \psi(w) dw,<br />

where

<br /> \psi(h) = \frac{F&#039;(x+h) - F&#039;(x)}{h} - F&#039;&#039;(x),<br />

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

<br /> \int_0^h w \psi(w) dw = h^2 \varphi(h).<br />

What do you think the \varphi(h) they're wanting here is?

 

[fauboca]

Homework Statement

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + h^2 \varphi(h),<br />

where \varphi(h) \to 0 as h\to 0.

Homework Equations

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + \int_0^h w \psi(w) dw,<br />

where

<br /> \psi(h) = \frac{F&#039;(x+h) - F&#039;(x)}{h} - F&#039;&#039;(x),<br />

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

<br /> \int_0^h w \psi(w) dw = h^2 \varphi(h).<br />

What do you think the \varphi(h) they're wanting here is?

I am guessing \varphi(h) are the higher order terms from the expansion. However, I have never seen it written as h^2\varphi(h) but instead as \mathit{O}(h^2).

 

[Ray Vickson]

Homework Statement

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + h^2 \varphi(h),<br />

where \varphi(h) \to 0 as h\to 0.

Homework Equations

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + \int_0^h w \psi(w) dw,<br />

where

<br /> \psi(h) = \frac{F&#039;(x+h) - F&#039;(x)}{h} - F&#039;&#039;(x),<br />

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

<br /> \int_0^h w \psi(w) dw = h^2 \varphi(h).<br />

What do you think the \varphi(h) they're wanting here is?

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

RGV

 

[AxiomOfChoice]

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

RGV

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + \frac 12 \int_x^{x+h} (x+h-t)^2 F&#039;&#039;&#039;(t) dt.<br />

This is problematic for two reasons: First, in the problem I'm trying to solve, I know only that F is C^2, so I'm not even sure F&#039;&#039;&#039;(t) makes sense. Second of all, I'm not sure how one can finagle the formula I quoted to somehow turn \psi(t) into F&#039;&#039;&#039;(t); 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 \psi. Defining it to be the difference between the difference quotient and the derivative just seems so obvious...

 

[Ray Vickson]

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + \frac 12 \int_x^{x+h} (x+h-t)^2 F&#039;&#039;&#039;(t) dt.<br />

This is problematic for two reasons: First, in the problem I'm trying to solve, I know only that F is C^2, so I'm not even sure F&#039;&#039;&#039;(t) makes sense. Second of all, I'm not sure how one can finagle the formula I quoted to somehow turn \psi(t) into F&#039;&#039;&#039;(t); 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 \psi. Defining it to be the difference between the difference quotient and the derivative just seems so obvious...

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
F(x+h) = F(x) + h F&#039;(x) + \int_{x}^{x+h} (x+h-t)F&#039;&#039;(t) dt,
or
F(x+h) = F(x) + h F&#039;(x) + \frac{1}{2} h^2 F&#039;&#039;(x + \theta h), \; (0 &lt; \theta &lt; 1).

RGV

 

[AxiomOfChoice]

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
F(x+h) = F(x) + h F&#039;(x) + \int_{x}^{x+h} (x+h-t)F&#039;&#039;(t) dt,
or
F(x+h) = F(x) + h F&#039;(x) + \frac{1}{2} h^2 F&#039;&#039;(x + \theta h), \; (0 &lt; \theta &lt; 1).

RGV

Thanks very much, again, for your help. I've managed to verify that both of the following are true:

<br /> F(x+h) = F(x) + h F&#039;(x) + \int_x^{x+h} (x+h - t)F&#039;&#039;(t)dt<br />

and

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + \frac 12 \int_x^{x+h} (x+h - t)^2F&#039;&#039;&#039;(t)dt.<br />

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 F(x+h) = F(x+h).

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + h^2 \varphi(h),<br />

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

<br /> \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).<br />

...and this goes to zero as h\to 0! So why is it necessary to just throw in an extra h^2 like they want? IS it necessary?

 

[Ray Vickson]

Thanks very much, again, for your help. I've managed to verify that both of the following are true:

<br /> F(x+h) = F(x) + h F&#039;(x) + \int_x^{x+h} (x+h - t)F&#039;&#039;(t)dt<br />

and

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + \frac 12 \int_x^{x+h} (x+h - t)^2F&#039;&#039;&#039;(t)dt.<br />

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 F(x+h) = F(x+h).

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

<br /> F(x+h) = F(x) + h F&#039;(x) + \frac{h^2}{2} F&#039;&#039;(x) + h^2 \varphi(h),<br />

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

<br /> \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).<br />

...and this goes to zero as h\to 0! So why is it necessary to just throw in an extra h^2 like they want? IS it necessary?

If you take the (h^2/2)F&#039;&#039;(x + \theta h) form, and use continuity of F'' you get (h^2/2)F&#039;&#039;(x) + r(h), where r(h) goes to zero faster than h^2. (In fact, r(h) = (h^2/2)[F&#039;&#039;(x + \theta h) - F&#039;&#039;(x)] does have the form h^2 \varphi(h), where \varphi(h) \rightarrow 0 as h --> 0.) I think that is all that is involved.

RGV