How to plot a function dependent on no. of iterations, provided it plots till nth iteration where n is the number where the function is converged?
9 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Hello All,
I am trying to plot a function vs no. of iterations. In the for loop, i have given the convergence condition, but I am unable to plot for f(x) vs no. iterations (till convergence). Please help.
Its of the form:
for m=1:n
pu(m)=(-1)^m)/((2*m)+1))*(exp(-((2*m)+1)))
e = e+pu;
if (abs(pu)/e)<(10^(-6)) %convergence check
break
end
end
hold all
plot(1:n,pu,'linestyle','none','marker','*')
0 comentarios
Respuesta aceptada
VBBV
el 28 de En. de 2022
Editada: VBBV
el 29 de En. de 2022
n= 10;
e = 0;
for m=1:n
pu(m)=(-1)^m/((2*m)+1)*(exp(-((2*m)+1)))
e = e+pu(m);
if (abs(pu(m))/e)>(10^(-6)) %convergence check
break
end
end
hold all
plot(1:n,pu,'linestyle','-','marker','.')
Check the convergence condition
7 comentarios
VBBV
el 29 de En. de 2022
legend(mrkr{cidx},legendInfo)
Put this line inside the inner for loop. After legendInfo. Everytime the inner for loop exits, it's over written with initial values for mrkr and clr index.
VBBV
el 29 de En. de 2022
Editada: VBBV
el 29 de En. de 2022
u=zeros(n,npoints); % change also This line in beginning
u(m,cidx)=pu(m);
hold all
plot(1:m,u(1:m,cidx),'linestyle','none','marker',mrkr{cidx},'Color',clr{cidx})
You also need to change these two lines inside the for loop. As you can see the u value is dependent on pu which is again varying with each outer iteration of for loop.
Más respuestas (0)
Ver también
Categorías
Más información sobre General Applications en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!


