Dear all,
This a stupid question, but how can I solve directly in, Scilab an equation
such as :
0.403*X^(-0.121) + 60.5*X^(-0.73) - 0.1839 = 0
?
Is-it necessary to code a function ? from memory : dichotomy method, secant
method, Brent one etc. .
Regards
Paul
__
Hi
I would code the function, then plot it, then use fsolve
On 05/11/2012 13:13, Paul Carrico wrote:
Dear all,
This a stupid question, but how can I solve directly in, Scilab an
equation such as :
0.403*X^(-0.121) + 60.5*X^(-0.73) -- 0.1839 = 0
?
Is-it necessary to code a function ? fr
Hi,
This is a nonlinear equation, for which we are searching a zero.
The fsolve function is designed for this purpose:
http://help.scilab.org/fsolve
Notice that there might not be a solution. This is why the algorithm is
an optimization problem, where the square norm of f(x) is minimized.
Th
part de michael.bau...@contrib.scilab.org
Envoyé : lundi 5 novembre 2012 13:59
À : International users mailing list for Scilab.
Objet : Re: [Scilab-users] root calculation
Hi,
This is a nonlinear equation, for which we are searching a zero.
The fsolve function is designed for this purpose:
http://he
part de michael.bau...@contrib.scilab.org
Envoyé : lundi 5 novembre 2012 13:59
À : International users mailing list for Scilab.
Objet : Re: [Scilab-users] root calculation
Hi,
This is a nonlinear equation, for which we are searching a zero.
The fsolve function is designed for this purpose:
http://help.scilab.org
d'origine-
De :users-boun...@lists.scilab.org [mailto:users-boun...@lists.scilab.org] De
la part demichael.bau...@contrib.scilab.org
Envoyé : lundi 5 novembre 2012 13:59
À : International users mailing list for Scilab.
Objet : Re: [Scilab-users] root calculation
Hi,
This is a nonlin
Hi Paul,
If the exponent are always negative and the coefficients are always
positive, the function is strictly decreasing (so the derivative is
non-null), you can use a Newton method.
function y=newton(fun, dfun, x0, eps)
y=fun(x0)
while abs(y) > eps
x0 = x0 - y / dfun(x0);
blocks), isnt it ?
Paul
De : users-boun...@lists.scilab.org [mailto:users-boun...@lists.scilab.org]
De la part de Calixte Denizet
Envoyé : lundi 5 novembre 2012 16:10
À : users@lists.scilab.org
Objet : Re: [Scilab-users] root calculation
Hi Paul,
If the exponent are always negative
sor
(solving by "blocks"), isn't it ?
Paul
*De :*users-boun...@lists.scilab.org
[mailto:users-boun...@lists.scilab.org] *De la part de* Calixte Denizet
*Envoyé :* lundi 5 novembre 2012 16:10
*À :* users@lists.scilab.org
*Objet :* Re: [Scilab-users] root calculation
Hi Paul,
If th
Dear all,
For my information, is there a faster method than the Newton one to
calculate roots (1 root for a non-linear equation) ?
NB : in case of "division by zero" in a function, is it possible to generate
a specific code in order to avoid program crash ?
Thanks
Paul
___
>NB : in case of “division by zero” in a function, is it possible to generate a
>specific code in order to avoid program crash ?
ieee(2)
___
users mailing list
users@lists.scilab.org
http://lists.scilab.org/mailman/listinfo/users
And how do you use it ? I've not understood
-Message d'origine-
De : users-boun...@lists.scilab.org [mailto:users-boun...@lists.scilab.org] De
la part de sgoug...@free.fr
Envoyé : jeudi 13 décembre 2012 16:58
À : International users mailing list for Scilab.
Objet : Re: [Sc
Le 13/12/2012 18:20, Paul Carrico a écrit :
And how do you use it ? I've not understood
after ieee(2), any division by zero will yield %inf or -%inf instead
of an error breaking the execution of the remainig code.
help ieee // for more info
___
13 matches
Mail list logo
