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Re: Nonlinear equation


From: Heber Farnsworth
Subject: Re: Nonlinear equation
Date: Tue, 15 Apr 2003 23:03:55 -0500

Newton's method finds (hopefully) zeros of functions. So if you move the x0 to the right hand side you have a function which is zero at the correct value of y. If a, b, and x0 are constants then just define your function that way. If they are parameters that you may have to change on subsequent runs you may want to make them global variables as I've done below. Define a function

function f = myfunc(y)
global a b x0
f = a*y*log(b) - a*y*log(y) + a*y - x0;
endfunction

Then at the octave prompt (after you have set your global variables) type

fsolve("myfunc",1)

where 1 is a starting value for y. You may want to pick a better one if you have an idea where y is.

Heber

On Tuesday, April 15, 2003, at 09:43 PM, address@hidden wrote:



Hello,

Can you help me how to solve a nonlinear equation given by:

x = a*y*ln(b) - a*y*ln(y) + a*y

where a and b are constant parameters.
I know the inital value of x0 and I want to know the value of y.

Can I use Newton's method? How?

Regards,
Paulo



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Octave is freely available under the terms of the GNU GPL.

Octave's home on the web:  http://www.octave.org
How to fund new projects:  http://www.octave.org/funding.html
Subscription information:  http://www.octave.org/archive.html
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