Hi Guys currently doing some Numerical Differentiation and I was looking at the different formulas such as 1) forward difference method 2) backward difference method 3) central difference method
$$\frac{f(x+h)-f(x)}{h}$$
$$\frac{f(x)-f(x-h)}{h}$$
$$\frac{f(x+h)-f(x-h)}{2}$$
Now based on using taylor's series, I know we can derive different or higher orders for the derivatives for example if i required $$f''(x)$$
using Taylor's series I know i can add f(x+h) + f(x-h) and obtain the the second derivative using taylor series.
I am trying to derive the third derivative
$$f^3(x)$$ using the taylor's series but
f(x+h)
$$f(x+h) = f(x) + hf^1(x)+h^2f^2(x)+h^3f^3(x)$$
f(x-h) $$f(x-h) = f(x)-hf^1(x)+h^2f^2(x)-h^3f^3(x)$$
This is my confusion to derive the thrid derivative for forward diff $$f^3(x)$$
$$f(x+h) - f(x)-hf^1(x)-\frac{h^2f(x)}{2!} = \frac{h^3f^3(x)}{3!}$$
$$\frac{6}{h^3}[f(x+h) - f(x)-hf^1(x)-\frac{h^2f(x)}{2!}] = f^3(x)$$
where
$$f^1(x) = \frac{f(x+h)-f(x)}{h}$$
and
$$f^2(x) = \frac{f(x+h)-2f(x)+f(x-h)}{h^2}$$
when i substitute i get the following
$$[\frac{6f(x+h)}{h^3} - \frac{6f(x)}{h^3} -\frac{6}{h^2}[\frac{f(x+h)-f(x)}{h}]-\frac{3}{h}[\frac{f(x+h)-2f(x)+f(x-h)}{h^2}]] = f^3(x)$$
And then when i attempt to break this down
$$\frac{6f(x+h)}{h^3}-\frac{6f(x)}{h^3} + \frac{-6f(x+h)+6f(x)}{h^3}+\frac{-3f(x+h)+6f(x)-3f(x-h)}{h^3}$$
I am not sure in term sof evaluating the third derivative if this is correct as i am attempting to do it for the Forward difference method and then for the backward and central difference method.Can anyone help me or clarify please.