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
___
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
>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
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
___
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
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
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);
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
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
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
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
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
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
__
13 matches
Mail list logo