# Second order Taylor expansion

1. Feb 19, 2012

### AxiomOfChoice

1. The problem statement, all variables and given/known data

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

$$F(x+h) = F(x) + h F'(x) + \frac{h^2}{2} F''(x) + h^2 \varphi(h),$$

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

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

$$F(x+h) = F(x) + h F'(x) + \frac{h^2}{2} F''(x) + \int_0^h w \psi(w) dw,$$

where

$$\psi(h) = \frac{F'(x+h) - F'(x)}{h} - F''(x),$$

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

$$\int_0^h w \psi(w) dw = h^2 \varphi(h).$$

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

2. Feb 19, 2012

### fauboca

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 $\mathit{O}(h^2)$.

3. Feb 20, 2012

### Ray Vickson

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

RGV

4. Feb 21, 2012

### AxiomOfChoice

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

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

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 $F'''(t)$ makes sense. Second of all, I'm not sure how one can finagle the formula I quoted to somehow turn $\psi(t)$ into $F'''(t)$; 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...

5. Feb 21, 2012

### Ray Vickson

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
$$F(x+h) = F(x) + h F'(x) + \int_{x}^{x+h} (x+h-t)F''(t) dt,$$
or
$$F(x+h) = F(x) + h F'(x) + \frac{1}{2} h^2 F''(x + \theta h), \; (0 < \theta < 1).$$

RGV

6. Feb 23, 2012

### AxiomOfChoice

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

$$F(x+h) = F(x) + h F'(x) + \int_x^{x+h} (x+h - t)F''(t)dt$$

and

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

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

$$F(x+h) = F(x) + h F'(x) + \frac{h^2}{2} F''(x) + h^2 \varphi(h),$$

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

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

...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?

Last edited: Feb 23, 2012
7. Feb 23, 2012

### Ray Vickson

If you take the $(h^2/2)F''(x + \theta h)$ form, and use continuity of F'' you get $(h^2/2)F''(x) + r(h),$ where r(h) goes to zero faster than h^2. (In fact, $r(h) = (h^2/2)[F''(x + \theta h) - F''(x)]$ 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