# can't do this

2 visualizzazioni (ultimi 30 giorni)
Tom il 8 Gen 2012
can't do this
Hi, I'm attempting nested for loops and failing miserably.
Here's my code: -
close all;
clear all;
l=50;R=100;T=68;m=0.000125;w0=0.1;t=0;
mu=m/l;c=T/mu;
r=2;x0=l/r;
iter=1;
for x=1:33;
iter=1;
for n=1:5000;
wxt=((4*w0)/pi)*exp(-R*t)*((l/(pi*n^2*x0))*sin((n*pi*x0)/l)-(1/n)*cos((n*pi*x0)/l)+(l/(2*n*x0))*cos((n*pi*x0)/l))*sin((n*pi*x)/l)*cos((c*n*pi*t)/l);
wxtrec(iter)=wxt;
iter=iter+1;
end
w=sum(wxtrec)
wrec(iter)=w;
iter=iter+1;
end
plot(x,wrec)
For the first outer loop run (x=1), I want the inner loop to do 5000 circuits, then I want the sum of that result. Then I want it to go around again 5000 times for x=2, taking a sum of that result etc.
Once I have 33 summed results, I want to plot those sums against x
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Image Analyst il 3 Nov 2012
Next time, could you pick a more meaningful subject line than "can't do this"?

Accedi per commentare.

### Risposta accettata

Walter Roberson il 8 Gen 2012
With minimal changes. Improvements are possible.
close all;
clear all;
l=50;R=100;T=68;m=0.000125;w0=0.1;t=0;
mu=m/l;c=T/mu;
r=2;x0=l/r;
iter=1;
for x=1:33;
iter=1;
for n=1:5000;
wxt=((4*w0)/pi)*exp(-R*t)*((l/(pi*n^2*x0))*sin((n*pi*x0)/l)-(1/n)*cos((n*pi*x0)/l)+(l/(2*n*x0))*cos((n*pi*x0)/l))*sin((n*pi*x)/l)*cos((c*n*pi*t)/l);
%wxtrec(iter)=wxt;
wxtrec(n)=wxt; %NEW
iter=iter+1;
end
w=sum(wxtrec)
%wrec(iter)=w;
wrec(x)=w; %NEW AND IMPROVED
iter=iter+1;
end
plot(x,wrec)
##### 2 CommentiMostra NessunoNascondi Nessuno
Tom il 9 Gen 2012
Hi thanks for answering. The code returns a message about subscript indices having to be either real positive integers or logicals. Should I be changing something?
Walter Roberson il 9 Gen 2012
Sorry, I had a typo which I have now corrected.

Accedi per commentare.

### Categorie

Scopri di più su Introduction to Installation and Licensing in Help Center e File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by