I am stuck on the following question:
Develop a second-order method for approximating $f'(x)$ that uses the data $f(x-h), f(x)$, and $f(x+3h)$ only.
Do you have any hints or tips? Thanks in advance.
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user2574431
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Here is the technique. – Mhenni Benghorbal Oct 08 '13 at 20:52
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By Taylor, we have \begin{align*} f(x-h) &= f(x) - hf'(x) + \frac 12 h^2 f''(x) + O(h^3)\\ f(x+3h) &= f(x) + 3hf'(x) + \frac 92 h^2 f''(x) + O(h^3) \end{align*} So $$ \frac 1h \bigl(\alpha f(x-h) + \beta f(x) + \gamma f(x+3h)\bigr) = \frac 1h (\alpha + \beta + \gamma)f(x) + (3\gamma - \alpha)f'(x) + \frac h2(\alpha + 9\gamma)f''(x) + O(h^2) $$ We must therefore have $\alpha+\beta+\gamma = 0$, $3\gamma - \alpha = 1$ and $\alpha + 9\gamma = 0$, the second and third equations give $12\gamma = 1$, so $\gamma = \frac 1{12}$, hence $\alpha = -\frac 34$, and $\beta = \frac 23$. That is $$ f'(x) = \frac{-9f(x-h) + 8f(x) + f(x+3h)}{12 h} + O(h^2)$$
martini
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You beat me by 25 minutes, but mine was easier to enter ($a$ instead of $\alpha$, ...).:) – marty cohen Oct 08 '13 at 21:26
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Expanding on Mhenni Benghorbal's link:
$f(x+h) =f(x)+hf'(x)+h^2f''(x)/2+h^3f'''(x)/6+... $, and $f(x-h) =f(x)-hf'(x)+h^2f''(x)/2-h^3f'''(x)/6+... $, so $f(x+3h) =f(x)+3hf'(x)+9h^2f''(x)/2+27h^3f'''(x)/6+... $.
To combine $f(x), f(h-x)$, and $f(x+3h)$ to get $f'(x)$, let $g(x) =af(x)+bf(x-h)+cf(x+3h) $.
Then $g(x) =(a+b+c)f(x) +(-b+3c)hf'(x) +(b+9c)h^2f''(x)/2 +(-b+27c)h^3f'''(x)/6 +... $.
To make this a second order approximation, we need $a+b+c=0$, $-b+3c = 1$, and $b+9c=0$.
This will make $g(x) =hf'(x)+O(h^3)$, so $\frac{g(x)}{h} = f'(x) + O(h^2) $.
The solution to these is $c = 1/12$, $b = -3/4$, and $a = 2/3$.
Since $-b+27c =-3/4+27/12 =-9/12+27/12 =18/12 =3/2 $, this makes $\begin{align} \dfrac{\frac{2}{3}f(x)-\frac{3}{4}f(x-h)+\frac1{12}f(x+3h)}{h} &=\dfrac{g(x)}{h}\\ &=f'(x)+h^2f''(x)(3/2)/6+O(h^3)\\ &=f'(x)+h^2f''(x)/4+O(h^3)\\ \end{align} $.
marty cohen
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