[Scilab-users] root calculation

[Calixte Denizet]

If only the last constant changes, you can adapt the newton algo to your 
particular case:

function y=newton3(coeffs, expo, cste, x0, eps)
     y = zeros(cste);
     x = x0;
     dcoeffs = coeffs .* expo;
     dexpo = expo - 1;
     for i = 1:size(cste, "*")
         z = coeffs * (x .^ expo)' - cste(i);
         while abs(z) > eps
             x = x - z / (dcoeffs * (x .^ dexpo)');
             z = coeffs * (x .^ expo)' - cste(i);
         end
         y(i) = x;
      end
endfunction

in supposing that cste is a decreasing row of values (last found root is 
used as starting value in Newton's algo).

cste=10:-0.0001:0.1839;
rs=newton3([0.403 60.5], [-0.121 -0.73],cste,1,1e-10);

(around 2 secs on my computer)

to check:
deff('y=foo(x)','y=0.403*x.^(-0.121)+60.5*x.^(-0.73)')
max(abs(foo(rs)-cste))

C

On 05/11/2012 16:35, Paul Carrico wrote:
>
> Hi Calixte,
>
> Your suggestion looks encouraging ... I'll have a deeper look on that 
> specific issue  when all the aspects of my new project will be 
> bracketed ! Many thanks ...
>
> And Yes:
>
> -The exponents are always negative
>
> -The coefficients are always positive
>
> To go ahead, in one aspect of the project, only the last constant 
> change (it won(t be the case in a second stage), I mean :
>
> 0.403*X^(-0.121) + 60.5*X^(-0.73) èalways the same == constant !
>
> *-- 0.1839*èit changes ... may  be either positive or negative in a 
> first understanding
>
> Imagine I've a matrix (n x1) where reach line is independent from the 
> others, and where it's necessary to solve such equation ... I've been 
> thinking using a loop:
>
> For i=1:n
>
>    Solve my non-linear equation
>
> End
>
> (n from hundred thousand's to million's)
>
> It may be interesting to slip that loop onto several processor 
> (solving by "blocks"), isn't it ?
>
> Paul
>
> *De :*users-bounces at lists.scilab.org 
> [mailto:users-bounces at lists.scilab.org] *De la part de* Calixte Denizet
> *Envoyé :* lundi 5 novembre 2012 16:10
> *À :* users at lists.scilab.org
> *Objet :* Re: [Scilab-users] root calculation
>
> 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);
>         y = fun(x0)
>     end
>     y = x0;
> endfunction
>
> deff('y=foo(x)','y=0.403*x.^(-0.121)+60.5*x.^(-0.73)-0.1839')
> deff('y=dfoo(x)','y=0.403*-0.121*x.^(-1.121)+60.5*-0.73*x.^(-1.73)')
>
> To find a "good" starting point:
> If x is a solution then 0.403*x^(-0.121)<=0.1839 and 
> 60.5*x^(-0.73)<=0.1839, so x >=max((0.1839/0.403)^(-1/0.121), 
> (0.1839/60.5)^(-1/0.73))
>
> newton(foo,dfoo,2806,1e-10)
>
> A faster way:
> function y=newton2(coeffs, expo, cste, x0, eps)
>     y = coeffs * (x0 .^ expo)' - cste;
>     dcoeffs = coeffs .* expo;
>     dexpo = expo - 1;
>     while abs(y) > eps
>         x0 = x0 - y / (dcoeffs * (x0 .^ dexpo)');
>         y = coeffs * (x0 .^ expo)' - cste;
>     end
>     y = x0;
> endfunction
>
> newton2([0.403 60.5], [-0.121 -0.73],0.1839,2806,1e-10)
>
> Calixte
>
> 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 ? from memory : dichotomy method, 
> secant method, Brent one etc. ...
>
> Regards
>
> Paul
>
>
>
>
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>
>
> -- 
> Calixte Denizet
> Software Development Engineer
> -----------------------------------------------------------
> Scilab Enterprises
> 143bis rue Yves Le Coz - 78000 Versailles, France
> http://www.scilab-enterprises.com
>
>
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-- 
Calixte Denizet
Software Development Engineer
-----------------------------------------------------------
Scilab Enterprises
143bis rue Yves Le Coz - 78000 Versailles, France
http://www.scilab-enterprises.com

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