Azzera filtri
Azzera filtri

Help with For-Loop question?

1 visualizzazione (ultimi 30 giorni)
Abdullah Tamimi
Abdullah Tamimi il 5 Ott 2011
I keep getting the same error, about not being any real or logical answer.
here is the question; Create a time vector x = 0:2*pi. Using a for-loop, plot ten different curves on the same graph where the first curve is x1 = sin(x), the second curve is x2 = sin(x1), the third curve is x3 = sin(x2) and so on. Note that by using a for-loop it is not necessary to create ten separate variables to solve this problem.
Here is what I've been entering
x=0
>> for i=[pi/5:pi/5:2*pi]
x(i)=sin(x(i-(pi/5)))
end
??? Attempted to access x(0); index must be a positive integer or logical.
Any help would be appreciated

Risposta accettata

Robert
Robert il 5 Ott 2011
N=10;
x=zeros(N,numel(pi/5:pi/5:2*pi));
x(1,:)=pi/5:pi/5:2*pi;
for j=2:N
x(j,:)=sin(x(j-1,:));
end
plot(x(1,:),x')

Più risposte (1)

Walter Roberson
Walter Roberson il 5 Ott 2011
t = 0:pi/5:2*pi;
curvenums = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
for K = 1 : numel(curvenums)./size(curvenums,1)
if ~curvenum(K)
x(:,curvenum(K+1)) = sin(t);
else
x(:,curvenum(K+1)) = sin(x(:,curvenum(K)));
end
end
plot(t,x)
  2 Commenti
Robert
Robert il 5 Ott 2011
That seems needlessly complicated. Why do you keep track of a "curvenum" array that is just 0:9? See my reply below.
Walter Roberson
Walter Roberson il 5 Ott 2011
The posting is clearly a homework assignment. The code I provided is functional but would get no marks from any marginally alert marker as it would obviously not be written by the student. It is code that can be learned from but not profitably cloned as the assignment answer.
Straight forward and readily understood solutions, on the other hand, are fodder for student copying without any real understanding.

Accedi per commentare.

Categorie

Scopri di più su Loops and Conditional Statements 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