I am attempting to perform taylor expansion on the following numerical differentiation formula:
$f'''(0) = \frac {−f(−3h/2) + 3f(−h/2) − 3f(h/2) + f(3h/2)) }{ h^3 }$
Over the reference interval [−3h/2, 3h/2]
After the derivation of the weights using Lagrange polynomials, the result is:
$p'''0 = f0*l_0'''(0) + f1*l_1'''(0) + f2*l_2'''(0) + f2*l_3'''(0)$
$ \frac {-f0 + 3f1 - 3f2 + f3} {h^3} $
Where $f0 = (-3h/2), f1 = (-h/2), f2 = (h/2), f3 = (3h/2)$
When trying to find the leading error term with Taylor expansion, my leading error term is $\frac {-3}{2h^2} fo'$. This gives an order of accuracy equal to -2.
The taylor expansion is done as follows:
$ \frac {-f0 + 3f1 - 3f2 + f3} {h^3} - f'''0$
$\frac {1} {h^{3}} (-f0 + 3(f0 +(-h/2)f0' + ...) - 3(f0 +(h/2)f0' + ...) + (f0 +(3h/2)f0' + ...))-f'''0$
Am I doing something wrong?
You should decide which of the formulas $$ \frac{-f(x)+3f(x+h)-3f(x+2h)+f(x+3h)}{h^3} $$ or $$ \frac{-f(x-3h/2)+3f(x-h/2)-3f(x+h/2)+f(x+3h/2)}{h^3} $$ you are actually considering and then stay with it. Your Taylor expansions mix (partial) terms from both formulas.