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## Re: Nonlinear equation

 From: Lorenzo Fiorentini Subject: Re: Nonlinear equation Date: Wed, 16 Apr 2003 12:36:36 +0200

```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 ;-)

Lorenzo

Il 16/04/2003 6.03.55, Heber Farnsworth

>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,
>
>>
>>
>> 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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