ploting in for loop

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Ahmet Oguz
Ahmet Oguz 2016 年 11 月 13 日
編集済み: Ahmet Oguz 2016 年 11 月 15 日
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

採用された回答

Walter Roberson
Walter Roberson 2016 年 11 月 14 日
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 件のコメント
Ahmet Oguz
Ahmet Oguz 2016 年 11 月 14 日
Thanks i figure it out now.

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その他の回答 (1 件)

Daniel kiracofe
Daniel kiracofe 2016 年 11 月 13 日
編集済み: Daniel kiracofe 2016 年 11 月 13 日
"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 件のコメント
Walter Roberson
Walter Roberson 2016 年 11 月 14 日
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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