Hi Daniel I did the test and the difference decreased. interesting, I had not noticed.
Em domingo, 12 de junho de 2016 09:53:40 UTC-3, Daniel Carrera escreveu: > > I don't know what you are trying to say, but did you try running the two > programs with the same relative and absolute tolerances? Do the results > look more similar now? > > On 12 June 2016 at 02:01, <[email protected] <javascript:>> wrote: > >> Hi daniel >> in fact, there is a relationship with the tolerance between the methods. >> >> Em sábado, 11 de junho de 2016 11:51:53 UTC-3, Daniel Carrera escreveu: >>> >>> Probably because the Julia module has different default values for the >>> absolute and relative tolerance. For Julia, the defaults are reltol=1e-5 >>> and abstol=1e-8 but in Matlab the defaults are RelTol=1e-3 and AbsTol=1e-6. >>> >>> https://github.com/JuliaLang/ODE.jl >>> http://se.mathworks.com/help/matlab/ref/odeset.html >>> >>> That would also explain why the Julia program took so much longer. Try >>> again with the same tolerance and see if the results are more similar. Oh, >>> and like Erik said, you probably meant to have parenthesis around "1/8". If >>> the results still look difference, look for more options that might have >>> different default values. >>> >>> Cheers, >>> Daniel. >>> >>> >>> >>> On Saturday, 11 June 2016 04:29:47 UTC+2, [email protected] wrote: >>>> >>>> this is the test for equation differential using runge-kutta45: f(x,y)= >>>> (-5*x - y/5)^1/8 + 10 >>>> >>>> >>>> <https://lh3.googleusercontent.com/-D6U4d5pN0pw/V1t3MaV3JpI/AAAAAAAAACU/-E_ceVhrTxkT3SAgmLUy5nHDhRJTIikSgCLcB/s1600/rk45.png> >>>> >>>> why the numerical result is different? I used : >>>> >>>> function Rk_JL() >>>> f(x,y)= (-5*x - y/5)^1/8 + 10 >>>> tspan = 0:0.001:n >>>> y0 = [0.0, 1.0] >>>> return ODE.ode45(f, y0,tspan);end >>>> >>>> >>>> and >>>> >>>> >>>> function [X1,Y1] = RK_M() >>>> f = @(x,y) (-5*x - y/5)^1/8 + 10; >>>> tspan = 0:0.001:n; >>>> y0 = 1 >>>> [X1,Y1]= ode45(f,tspan,1);end >>>> >>>> >
