Using taylor series expansion to approximate the derivative of a function

Question

I know that $f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+\frac{h^3}{3!}f'''(x)$ but I do not know what to do to reach what is shown above, I would appreciate any collaboration.

Answer 1

The equality you have is $$\tag1 f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+\frac{h^3}{3!}f'''(x)+o(h^4). $$ Also $$\tag2 f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(x)-\frac{h^3}{3!}f'''(x)+o(h^4), $$ $$\tag3 f(x+2h)=f(x)+2hf'(x)+2h^2f''(x)+\frac{4h^3}{3}f'''(x)+o(h^4), $$ $$\tag4 f(x-2h)=f(x)-2hf'(x)+2h^2f''(x)-\frac{4h^3}{3}f'''(x)+o(h^4). $$ If you use the above, you get $$ 8f(x+h)-8f(x-h)-f(x+2h)+f(x-2h)=12hf'(x)+o(h^4), $$ which gives the first formula. The second one is similar.

Answer 2

Just repeatedly use $\sum_i c_i f(x+ih)=\sum_{n\ge 0}\dfrac{h^n\sum_i c_i^n}{n!}f^{(n)}(x)$, where $f^{(n)}$ denotes the $n$th derivative of $f$ (in particular, $f^{(0)}=f$).

Answer 3

Yes, $f(x+ h)= f(x)+ hf'(x)+ \frac{h^2}{2}f''(x)+ \frac{h^3}{6}f'''(x)$ so $f(x+ 2h)= f(x)+ 2hf'(x)+ \frac{4h^2}{2}f''(x)+ \frac{8h^3}{6}f'''(x)$, $f(x- h)= f(x)- hf'(x)+ \frac{h^2}{2}f''(x)- \frac{h^3}{6}f'''(x)$, and $f(x- 2h)= f(x)- 2hf'(x)+ \frac{4h^2}{2}x^2- \frac{8h^3}{6}x^3$.

Add those.