Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: How to solve transecedental equation using matlab Date: Sat, 14 Feb 2009 20:07:02 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 10 Message-ID: <gn7895$o9t$1@fred.mathworks.com> References: <3e609bfa-176d-4203-883a-6e51a6dca80d@g1g2000pra.googlegroups.com> Reply-To: <HIDDEN> NNTP-Posting-Host: webapp-02-blr.mathworks.com Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1234642022 24893 172.30.248.37 (14 Feb 2009 20:07:02 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Sat, 14 Feb 2009 20:07:02 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:518278 vinneelaero@gmail.com wrote in message <3e609bfa-176d-4203-883a-6e51a6dca80d@g1g2000pra.googlegroups.com>... > Hi friends..... > when i was doing a vibration problem i got a equation tanh(ax)+tan > (ax)=0, which is a transecedental equation....i want the roots of this > equation..... could any body help me to find out the roots of this > equation by using matlab It should be pointed out that for this particular equation there is no need to resort to matlab for roots. Ordinary calculus should suffice. It is clear that there is one root at x = 0. The derivative of the left side with respect to x is a*(sech(a*x)^2+sec(a*x)^2), which can never be zero assuming a is non-zero. This means there could never be any other root than at x=0 without violating the famous law of the mean of calculus. Roger Stafford