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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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