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Derive the following difference approximation for the first derivative:

$f'(x_0) = (f'(x_0 + 2h) - f(x_0 - h))/3h$

I really just need some pointers in how to start this out. If I were to guess, it looks like it starts out with the Lagrange form of the interpolating polynomial, differentiate f with respect to x, and when we get to

$f'(x_0) = (f'(x_1) - f(x_0))/(x_1 - x_0)$

We substitute $x_1 = x_0 + 2h$

But I'm not really sure if this is correct, and if it is, I'm not sure how to go about starting the derivation... If somebody could give me some pointers that'd be great. I'm also not sure which tags should be attached to this, so if anybody feels like editing it to add some better ones, that'd be great.

Thank you

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Your difference quotient is really centered not at $x_0$, but rather at $x_0+h/2$ (the average of places where you evaluate $f(x)$ in getting the difference quotient. I guess if $f'(x)$ is assumed continuous that should be OK... – coffeemath Oct 29 '12 at 0:24

Hello person most likely in M348 or otherwise stuck with Bradie's book!

I've been racking my brains with this problem for the better part of an hour, and the only way I can think of to solve this problem is by taking advantage of a Taylor Series expansion for $f^{\prime}(x)$.

So we have:

$$f^{\prime}(x+h)=f^{\prime}(x)+h f^{\prime\prime}(x)+\frac{h^2}{2}f^{\prime\prime\prime}(x)+O(h^3)$$

which fiddles around to

$$f^{\prime}(x)=f^{\prime}(x+h)-h f^{\prime\prime}(x)-\frac{h^2}{2}f^{\prime\prime\prime}(x)-O(h^3)$$

which is basically


with $e(x)$ being the associated error term for the difference approximation needed for part b. Then we can say that $f^{\prime}(x_0) \approx f^{\prime}(x_0+h)$ for an appropriate $h$, and calculate a central difference formula for $f^{\prime}(x_0 + h)$.

Hope that gives you an idea of where to start, but I'd give anything for someone with a better understanding to either back me up or tell me where I went wrong.

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Assuming that you have made a small typo in your question and actually meant $$f'(x_{0}) = \frac{f(x_{0} + 2h) - f{(x_{0} - h)}}{3h} $$ we can derive this finite difference approximation easily by using the Taylor series expansion. Notice that: $$f(x_{0} + 2h) = f(x_{0}) + \frac{(2h)^{1}}{1!}f'(x_{0}) + \text{higher order terms} $$ and $$f(x_{0} -h) = f(x_{0}) + \frac{(-h)^{1}}{1!}f'(x_{0}) + \text{higher order terms}. $$ Subtract the second expansion from the first: $$f(x_{0} + 2h) - f(x_{0} -h) = f(x_{0}) - f(x_{0}) + 2hf'(x_{0}) + hf'(x_{0}) + \text{higher order terms}.$$ Rearrange and you will recover the approximation you are looking for.

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Assume that $f(x)$ has a continuous derivative in some open interval $O$ containing $x_0$. Then for sufficiently small $h$ we may apply the mean value theorem to the interval $[x_0-h,x_0+2h]$ and conclude that there is a real number $c_h$ with $x_0-h<c_h<x_0+2h$ for which $$ \frac{f(x_0+2h)-f(x_0-h)}{(x_0+2h)-(x_0-h)}=f'(c_h).$$ Then since $c_h$ approaches $x_0$ as $h\to 0$ (it's trapped between two things which both approach $x_0$), and using continuity of $f'(x)$, we see that as $h \to 0$ the difference quotient $$ \frac{f(x_0+2h)-f(x_0-h)}{3h} \to f'(x_0).$$

The statement of the question might not hold for functions $f$ which are only assumed to be differentiable at $x_0$, but I can't think of an example of this immediately.

Re-statement of a previous approach: break three times the given difference quotient into three parts: $$\frac{f(x_0+2h)-f(x_0+h)}{h}+\frac{f(x_0+h)-f(x_0)}{h}+\frac{f(x_0)-f(x_0-h)}{h}$$ and note that each of these difference quotients approaches $f'(x_0)$ as $h \to 0$, under the assumption that $f'(x)$ is continuous in an open interval about $x_0$. This may be in a way more intuitive to some than the above version using the mean value theorem.

EDIT: I just noticed that this "previous approach" shows one only need assume that $f(x)$ is differentiable at $x_0$. To see this, note that the second and third terms approach respectively the right hand and left hand derivatives of $f(x)$ at $x_0$. And note that the first term may be broken into $$2\frac{f(x_0+2h)-f(x_0)}{2h} + \frac{f(x_0)-f(x_0+h)}{h}.$$ As $h \to 0$ first term here approches $2f'(0)$ and the second approaches $-f'(x_0)$. So combining all three terms, or all four terms given the splitting of the earlier first term into two parts, we get a net of $$2f'(x_0)-f'(x_0)+f'(x_0)+f'(x_0)=3f'(x_0)$$ for the limit, which as before on division by 3 gives the desired $f'(x_0).$ So no need of any extra assumptions, beyond that $f'(x_0)$ exists.

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