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## Re: Nonlinear equation

 From: Heber Farnsworth Subject: Re: Nonlinear equation Date: Wed, 16 Apr 2003 07:42:00 -0500

I think you may be confused. Fsolve implements Newton's method (with a dogleg search as I recall) and it does not need you to calculate an analytic derivative.
```
Heber

On Wednesday, April 16, 2003, at 05:36 AM, Lorenzo Fiorentini wrote:

```
```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,
```
```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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