Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Question on the derivate /calculus of a 2-norm matrix . Thanks a lot Date: Mon, 26 Jul 2010 03:01:04 +0000 (UTC) Organization: Anhui University Lines: 14 Message-ID: <i2itpg$h3j$1@fred.mathworks.com> References: <i2he90$21a$1@fred.mathworks.com> <i2i2st$n4j$1@fred.mathworks.com> <i2irns$94i$1@fred.mathworks.com> <i2it4t$73b$1@fred.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: webapp-05-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1280113264 17523 172.30.248.35 (26 Jul 2010 03:01:04 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Mon, 26 Jul 2010 03:01:04 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 2419503 Xref: news.mathworks.com comp.soft-sys.matlab:656060 "Matt J " <mattjacREMOVE@THISieee.spam> wrote in message <i2it4t$73b$1@fred.mathworks.com>... > "Antony " <mutang.bing@gmail.com> wrote in message <i2irns$94i$1@fred.mathworks.com>... > > > I have another problem. Maybe we can not directly solve it and I think the result might be more complxe than my former problem. The problem is: > > if g(x) = ||KX-B||^0.6 with all the other settings as the former problem, what is \partial{g}/\partial{x}? > ======== > > This is equivalent to (||KX-B||^2)^0.3 > > So you can use your original result, with one more step of the chain rule leading to > > Gradient = 0.3*(||KX-B||^2)^(-.7) * 2*K'*(K*X-B) Why not write it as Gradient = 0.3 *2*K'*(K*X-B)*(||KX-B||^2)^(-.7) according to the chain rule? It is because K'*(K*X-B) is a scalar and there is no difference between them? Thank you!