We are told that the starting point is the difference $ f(x+h) - f(x-h) $ (each of which represents it's own series; the last an alternating series) for getting the Richardson extrapolation.
When I do this subtraction the even exponential terms cancel out because of the difference sign while the odd exponential terms ones sum up yielding the following:
$ f(x+h) - f(x-h) = 2hf'(x)+{2h^{3}\over 3!}f'''(x)+... $
So the expression for the derivative yields if i understand it correctly, where now the h's get even exponentials:
$f'(x) = \frac{f(x+h) - f(x-h)}{2h} = f'(x) + {h^{2}\over 3!}f'''(x)+...$
Making the stepsize h half as big by replacing every $h$ with $h/2$ in the last expression yields (where $4 = 2^2$ and $16 = 2^4$) :
$f'(x) = \frac{f(x+h/2) - f(x-h/2)}{h} + {h^{2}\over 4*3!}f'''(x)+{h^{4}\over 16*5!}f'''''(x)...$
Now we are supposed to get a series on the form to get an approximation for the derivative: $f'(x) = \frac{4(f(x+h/2) - f(x-h/2))}{3*h} - \frac{(f(x+h) - f(x-h))}{3*2h} -\frac{1}{4}a_4 h^4 + \frac{5}{16}a_6 h^6 + ...$
Here the a's are constants.
I'm wondering how the numerical values before each term are achieved. The deviation from my result as far as I can see are the numbers above the fractions, and the 2 first terms. I don't really see how to get the last expression, any suggestions?
Since $$ \frac{f(x+h) - f(x-h)}{2h} = f'(x) + {h^{2}\over 3!}f'''(x)+ {h^{4}\over 5!}f'''''(x)...$$ and $$ \frac{f(x+h/2) - f(x-h/2)}{h}= f'(x) + {h^{2}\over 4*3!}f'''(x)+{h^{4}\over 16*5!}f'''''(x)...$$ One write $$\frac{4(f(x+h/2) - f(x-h/2))}{3*h} - \frac{(f(x+h) - f(x-h))}{3*2h} = \frac{4}{3} \left(f'(x) + {h^{2}\over 4*3!}f'''(x)+{h^{4}\over 16*5!}f'''''(x) ...\right)- \frac{1}{3} \left(f'(x) + {h^{2}\over 3!}f'''(x) + {h^{4}\over 5!}f'''''(x) ...\right) = f'(x) + \frac{4}{3}{h^{4}\over 16*5!}f'''''(x) - \frac{1}{3}{h^{4}\over 5!}f'''''(x)... = f'(x) - \frac{1}{4}{h^{4}\over 5!}f'''''(x)... $$
Hope this helps.
Connection to Richardson Extrapolation
Richardson Extrapolation for a method defined as $$A(h)=A_0+Ch^n+o(h^{n+1})$$ is defined as $$R(h,k)=\frac{k^nA(h)-A(k h)}{k^n-1}$$ In the question above we have $k=2$, $n=2$ and $$A(h)=\frac{f(x+h/2) - f(x-h/2)}{h} + Ch^2...$$