Jump to content

Wikipedia:Reference desk/Archives/Mathematics/2010 January 23

From Wikipedia, the free encyclopedia
Mathematics desk
< January 22 << Dec | January | Feb >> January 24 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


January 23

[edit]

Finding a surface on which integrals of multiple functions are 0

[edit]

Given a set of N linearly independent functions on some bounded simply-conected 2D surface , under what conditions does there exist another surface on which the integrals of all are 0?

does not need to be connected.

I wrote a simple program to find given , and the results indicate that it is sufficient that all are continuous functions that change sign somewhere on . I know this is not a necessary condition, but I am most interested in knowing if this condition is indeed sufficient. 83.134.167.153 (talk) 08:32, 23 January 2010 (UTC)[reply]

Not sure what you're up to but it sounds like you want to look at Orthogonal functions. Dmcq (talk) 15:57, 23 January 2010 (UTC)[reply]
I don't think just changing sign is sufficient. For example if f1 is strictly greater than f2, then it won't work. Rckrone (talk) 17:30, 23 January 2010 (UTC)[reply]


By the Lyapunov convexity theorem the set is a closed convex set in , and (in fact, as a consequence) the same holds if you also prescribe or for a given number c. Therefore, if you are ok with a measurable subset s of S instead of an open set s, you have the following necessary and sufficient condition: there exists such a measurable subset s, say with if and only if there are measurable sets with such that some convex combination of the vectors in vanishes. If you really need s to be a surface (that is, an open subset of S since S itself is a surface) then you need the analogous Lyapunov convexity theorem, that I think is still true and existing somewhere there out, especially if are continuous. pma. --84.220.118.69 (talk) 18:17, 24 January 2010 (UTC)[reply]