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ploting in for loop

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Ahmet Oguz
Ahmet Oguz el 13 de Nov. de 2016
Editada: Ahmet Oguz el 15 de Nov. de 2016
I wrote sum which is xs for below calculation There is something wrong with my code below that i can not figure it out?
clc
t=cputime;
k=0;
for p=1:1:7
dt=10^-p;
k=0:1.35/dt;
xs = 2/(sqrt(pi))*sum(exp(-(k*dt).^2)*dt)
e=cputime-t;
semilogx(e,dt)
end

Respuesta aceptada

Walter Roberson
Walter Roberson el 14 de Nov. de 2016
Try this:
k=0;
for p=1:1:7
t=cputime;
dt(p)=10^-p;
k=0:1.35/dt(p);
xs = 2/(sqrt(pi))*sum(exp(-(k*dt(p)).^2)*dt(p))
e(p)=cputime-t;
end
semilogx(1./dt, e)
You were plotting timestep as if it were a consequence of execution time. Also, as p increases, dt decreases, so if you plot dt then it is getting smaller and smaller and so your datapoints were getting further left, which is more difficult for people to understand. If you plot against 1./dt then you are plotting time as a consequence of number of data samples used, which is much more natural for people.
  1 comentario
Ahmet Oguz
Ahmet Oguz el 14 de Nov. de 2016
Thanks i figure it out now.

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Más respuestas (1)

Daniel kiracofe
Daniel kiracofe el 13 de Nov. de 2016
Editada: Daniel kiracofe el 13 de Nov. de 2016
"there is something wrong" is pretty vague, so I'm totally guessing at what your problem is. But this seems like a reasonable guess. If this doesn't answer you question then you need to post more detail about what is your specific problem.
clc
t=cputime;
k=0;
for p=1:1:7
dt(p)=10^-p;
k=0:1.35/dt(p);
xs = 2/(sqrt(pi))*sum(exp(-(k*dt(p)).^2)*dt(p))
e(p)=cputime-t;
end
semilogx(e,dt)
  2 comentarios
Ahmet Oguz
Ahmet Oguz el 14 de Nov. de 2016
same mistake accured there is no graph
Walter Roberson
Walter Roberson el 14 de Nov. de 2016
I would make a small change, and move the
t=cputime;
to inside the for p loop. Otherwise you are getting cumulative time since you started, instead of time for that particular refinement.

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