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.
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3 Answers
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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.
answered May 21, 2018 at 21:02
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$\begingroup$ Is the idea then to take the limit as $h \to \infty$? Maybe the h's cancel in some cases. $\endgroup$– John DoeCommented Feb 19, 2022 at 13:22
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$\begingroup$ The idea is to establish the approximations to $f'(x)$ and $f''(x)$ from the OP. Not sure what you mean by "the $h$'s cancel". $\endgroup$ Commented Feb 19, 2022 at 13:32
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$\begingroup$ The approximation of $f'(x)$ and $f''(x)$ depends on some small $h>0$ so I was confirming that you could take the limit as $h \to 0$ since you have expressions in terms of $h$ on the RHS. $\endgroup$– John DoeCommented Feb 19, 2022 at 13:47
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$\begingroup$ I meant that for $f'(x)$ the $h$ in the numerator may cancel with the $h$ in the denominator hence no need for taking the limit. $\endgroup$– John DoeCommented Feb 19, 2022 at 14:04
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1$\begingroup$ Cancelling in that sense would be extremely unusual, it would happen for just a handful of functions. The $h$ does not have to be positive. And yes, these are approximations, so the smaller the $h$, the better the approximation. As mentioned in the question and in the answer, these are $o(h^4)$, which means that they are much better than the naive approach. $\endgroup$ Commented Feb 19, 2022 at 14:29
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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, $f^{(0)}=f$).
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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.
answered May 21, 2018 at 21:04