If you really want to use Newton's method you need to
calculate the derivative dy/dx first.
If I am not wrong, differentiating both sides...
dy/dx= 1/(a*(1-1/b-ln(y/b))
Use this derivative at your own risk (bla bla, GPL stuff ;-)
Please check it!!
Lorenzo
Il 16/04/2003 6.03.55, Heber Farnsworth
<address@hidden> ha scritto:
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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