2
$\begingroup$

I'm studying some numerical methods and especially the numerical differentiation.

I have some troubles with the central differentiation. How do you become from the Taylor series the formula for $f(x+h)$, $f(x-h)$, $f(x+2h)$ and $f(x-2h)$?

enter image description here

enter image description here

enter image description here

$\endgroup$
  • $\begingroup$ Take the ordinary formula and substitute $h$ for $-h,2h$ and $-2h$. (It even works with $0$ !) $\endgroup$ – Yves Daoust Jan 10 '18 at 11:55
2
$\begingroup$

Start from your first expression

$$ f(x) = \sum_{n=0}^{+\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n \tag{1} $$

Call $y = x - a$, so that $x = a + y$. Eqn. (1) becomes

$$ f(y + a) = \sum_{n=0}^{+\infty}\frac{f^{(n)}(a)}{n!}y^n \tag{2} $$

Now, to recover your second set of equations you can call $y \to h$ and $a\to x$, the result is

$$ f(x+h)=\sum_{n=0}^{+\infty}\frac{f^{(n)}(x)}{n!}h^n \tag{3} $$

With $y\to 2h$, $\cdots$ you will recover the other equations

$\endgroup$
  • $\begingroup$ Thnx, but how do you become the equations f(x-h) and f(x-2h)? $\endgroup$ – WinstonCherf Jan 10 '18 at 11:57
  • $\begingroup$ @LeneCoenen In Eqn. (2) replace $y\to -h$ and $a\to x$ $\endgroup$ – caverac Jan 10 '18 at 11:59
  • $\begingroup$ How do you know until which term you have to expand? And what does the O(h^4) for example mean? $\endgroup$ – WinstonCherf Jan 10 '18 at 12:13
  • $\begingroup$ @LeneCoenen The more terms you include the more accurate is your approximation. The symbol $O(h^4)$ just means that the terms that follow are all going to be of order $h^4$ or higher. $\endgroup$ – caverac Jan 10 '18 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.