Use taylor series to arrive at the expression f'(x)=1/h[-3*f(x)/2+2f(x+h)-f(x+2h)/2]
I'm not really sure how to go about this.. any help is appreciated.
Solution 1:
Apply the extended mean value formula several times to $$ \frac{-3f(x)+4f(x+h)-f(x+2h)-2hf'(x)}{h^3} $$
$$ =\frac{4f'(x+h_1)-2f'(x+2h_1)-2f'(x)}{3h_1^2}\\ =\frac{4f''(x+h_2)-4f''(x+2h_2)}{6h_2}\\ =\frac{4f'''(x+h_2+h_3)}{6}\\ $$ where $0<h_3<h_2<h_1<h$.
Of course one can also do this as Taylor series. Consider $$g(h)=-3f(x)+4f(x+h)-f(x+2h).$$ Then $g(0)=0$, $g'(0)=2f'(x)$, $g''(0)=0$, $g'''(0)=-4f'''(x)$, thus $$g(h)=2h·f'(x)-\frac23h^3·(2f'''(x+2h_1)-f'''(x+h_1)).$$