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 $f(x-2h)$?
Asked
Active
Viewed 5,869 times
2
-
Take the ordinary formula and substitute $h$ for $-h,2h$ and $-2h$. (It even works with $0$ !) – Jan 10 '18 at 11:55
1 Answers
2
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
caverac
- 19,345
-
-
-
How do you know until which term you have to expand? And what does the O(h^4) for example mean? – WinstonCherf Jan 10 '18 at 12:13
-
@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. – caverac Jan 10 '18 at 12:17