Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Minimization of the sum of integrals with unknown bounds Date: Fri, 8 Jul 2011 21:39:12 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 20 Message-ID: <iv7te0$ovp$1@newscl01ah.mathworks.com> References: <iv6ob2$b6t$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-02-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1310161152 25593 172.30.248.47 (8 Jul 2011 21:39:12 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Fri, 8 Jul 2011 21:39:12 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:735611 "Ita Atz" wrote in message <iv6ob2$b6t$1@newscl01ah.mathworks.com>... > Hi all, > I have to solve a pretty messed up problem such: > > min_y sum_{h=1}^H int(-inf,y(h)) fun(y) > > where y is the H-dimensional vector of the elements y(h). > Does anyone know how to solve such a problem? > Thank you in advance! > Ita - - - - - - - - - - - You haven't made your problem at all clear to me either, Ita. I think you better keep on trying to improve your description. You can do better than you have up to this point. However, the following fact may (or may not) be of use to you. The minimum of a single integral with respect to variation of its upper limit of integration, assuming nothing else varies but the limit, must occur either at a point where its integrand is zero or else at an extreme value in the range allowed for that limit. You can show this by taking the derivative of the integral which would be its integrand evaluated at the upper limit. Of course having a zero integrand does not guarantee a minimum. It might be a maximum. It depends on whether the integrand is descending or ascending at that point. In your sum of integrals if you are able to vary each upper limit independently of the others, then the minimum must occur in such circumstances in each integral independently of the others. Roger Stafford