Timeline for Error of midpoint method for functions that are not twice-differentiable
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 20, 2014 at 23:39 | vote | accept | James Propp | ||
| Feb 20, 2014 at 19:38 | comment | added | Linda Brown Westrick | I meant with the checkbox, but in any case I'm glad it helped. | |
| Feb 18, 2014 at 22:47 | comment | added | James Propp | I gratefully accept it. I'll turn my follow-up question into a separate post; see mathoverflow.net/questions/157988/… . (Instead of absolutely continuity, I'll go for differentiability, which not only is stronger but also is probably easier to work with if one is looking for positive results as opposed to counterexamples.) | |
| Feb 18, 2014 at 5:45 | comment | added | Linda Brown Westrick | There are certainly some hypotheses one could add to get rid of the kind of bad behavior above. For a very weak example, I don't know how to make an $O(1/n)$ convergence with an absolutely continuity restriction. If you're satisfied with my answer, though, I hope you will accept it :-) | |
| Feb 18, 2014 at 3:41 | comment | added | James Propp | I still can't help thinking that there's a big gap between $O(1/n^3)$ (the behavior of the error that's guaranteed as long as the second derivative of $f$ is bounded) and $O(1/n)$, and suspecting that some hypothesis on $f$ stronger than mere continuity but considerably weaker than boundedness of the second derivative would imply that the error falls faster than $1/n$. But if nobody knows of results along these lines, I guess that's just how it is. | |
| Feb 17, 2014 at 9:04 | comment | added | Linda Brown Westrick | Now I think I understand your previous comments -- you are just as interested in the "$o$" as the "$1/n$". I added a nasty function for which the error fails to be $o(1/n)$. | |
| Feb 17, 2014 at 8:56 | history | edited | Linda Brown Westrick | CC BY-SA 3.0 |
Added an example which illustrates that $O(1/n)$ is an exact bound.
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| Feb 16, 2014 at 5:24 | comment | added | James Propp | Can Westrick (or anyone else) come up with a function $f$ for which the error of the midpoint method fails to be $o(1/n)$? If $f$ is a continuous piecewise linear function on $[0,1]$ with $k$ "breakpoints" (points at which the second derivative is undefined), and one divides $[0,1]$ in to $n$ subintervals of width $1/n$, then I can show that the error of the midpoint method is $o(1/n)$. Note that $k$ and $n$ play very different roles here. I agree that if $k$ is allowed to increase as $n$ does, e.g. if we take $k=n$, bad things happen, but then we're not talking about a fixed $f$ anymore. | |
| Feb 15, 2014 at 18:45 | comment | added | Linda Brown Westrick | Hi James. I improved the notation and added some details. (The function $f$ does not depend on $n$.) Let me know if that helps. | |
| Feb 15, 2014 at 18:39 | history | edited | Linda Brown Westrick | CC BY-SA 3.0 |
added details and clearer notation
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| Feb 15, 2014 at 17:35 | comment | added | James Propp | This addresses my first question ("What bounds are available?") but it's not clear to me that it resolves the question of whether, for each particular $f$, the error of the midpoint method is $o(1/n)$. Note that in Westrick's example, the function $f$ depends on $n$. | |
| Feb 15, 2014 at 0:56 | history | answered | Linda Brown Westrick | CC BY-SA 3.0 |