Timeline for Effect of boundary condition on the local error
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 15, 2012 at 5:41 | comment | added | David Ketcheson | In the example you give, you are not "at the boundary" but rather a distance $h$ from it. Then $f(x+h)$ still has the usual Taylor expansion; nothing prohibits you from using it. | |
| Oct 15, 2012 at 1:25 | comment | added | Kamil | Assume it is a right boundary and $f(x+h):=g$. By Taylor $$f(x-h)=f(x)-f'(x)h+0.5f^{''}(x)h^2+O(h^2)$$ However, before I could do the same for $f(x+h)$. Now it is fixed and $$\frac{f(x+h)-f(x-h)}{2h}=\frac{g-f(x)+f'(x)h-0.5f^{''}(x)h^2}{2h}+O(h^2)$$ and, therefore, it doesn't look as before and I can't say it is an approximation for the derivative. What am I missing? | |
| Oct 14, 2012 at 21:03 | history | edited | vanCompute | CC BY-SA 3.0 |
added 65 characters in body
|
| Oct 14, 2012 at 20:58 | history | answered | vanCompute | CC BY-SA 3.0 |