Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Solve non-linear functions rapidly? Date: Sun, 3 Apr 2011 18:19:05 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 14 Message-ID: <inadmp$8u$1@fred.mathworks.com> References: <inac45$6lr$1@fred.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: fred.mathworks.com 1301854745 286 172.30.248.47 (3 Apr 2011 18:19:05 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Sun, 3 Apr 2011 18:19:05 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:719796 "Jani Korhonen" <eagle_nokia@hotmail.com> wrote in message <inac45$6lr$1@fred.mathworks.com>... > Dear all, > > I want to solve this function C1 * x^2 - exp(C2 * x) = 0 in Matlab, in which C1 and C2 are constants. I know it is easy to use fsolve to solve it, e.g. x=fsolve((x)C1*x^2-exp(C2*x), 0). My question is that x is a matrix, and I want to solve this equation at each point, i.e. obtaining an individual x value at each point. C1 and C2 might be different at different points, but they can be obtained in advance. > > However, if I use a loop to find out the answer at each point, it takes a lot of time. But if I use the following: X = fsolve((X), C1*X^2-exp(C2*X), zeros(size(X, 1), size(X, 2)), where X is a matrix. Matlab says that x must be square. It seems the function was not treated at individual points. So, how can I solve this equation rapidly? Thank you very much. > > Best regards, > > Jani - - - - - - - - - - Substituting w = -C2/2*x puts your equation in the form of Lambert's W equation. Solutions for that can be obtained from the 'lambertw' function in the Symbolic Toolbox which accepts numeric matrices. That might be faster than using 'fsolve'. Roger Stafford