Let $f(x)$ be the investigating function. Let's say that we have the values of $f(x)$ at equidistant locations $x_0, x_1, x_2$ and $x_3$ (consecutive points are separated by $h$ apart) as $f_0, f_1, f_2$ and $f_3$ respectively. We know that we can use the two-point formula to approximate the first derivative of $f(x)$ at $x_0$, $x_1$ and $x_2$ by:
$$
f'_2 \approx \frac{f_3 – f_2}{h} \\
f'_1 \approx \frac{f_2 – f_1}{h} \\
f'_0 \approx \frac{f_1 – f_0}{h}
$$
Then, if one use the central difference scheme to get the second derivative of $f$ at $x_1$ by:
$$
(f'_1)'' \approx \frac{f'_2 – 2 f'_1 + f'_0}{h^2},
$$
then is it possible to combine these approximations into a formula for third-order approximation of $f(x)$ at $x_1$ as:
$$
f'''_1 \approx \frac{1}{h^3} \left( (f_3 – f_2) – 2(f_2 – f_1) + (f_1 – f_0) \right)
$$
If so, what order of convergence are we looking at?
Best Answer
By symmetry, $$ \frac{f(x+3h)-3f(x+2h)+3f(x+h)-f(x)}{h^3}=f'''(x+\tfrac32h)+O(h^2) $$ As approximation of the third derivative in close-by points, the error order goes down to $1$, $$f'''(x+h)=f'''(x+\tfrac32h)-\tfrac12hf^{(4)}(x+\tfrac32h)+...$$