redefine array in for loop

Hi Guys,
for n = 3:length(x)
y1 = x(2:n) ;
t1 = ((n-2):-1:0)*dt ;
% y2 = x(1:n-1) ;
y2 = y1 - 1 ;
t2 = t1 +dt;
int(n) = sum (t1.^(k-1)/factorial(k-1).*y1 + t2.^(k-1)/factorial(k-1).*y2)*dt/2 ;
end
I want to define y2= y1 -1 instead y2= x (1:n-1) which I have commented above. By y2 = y1 - 1, occurred some errors in MATLAB. Why? It is not correct?

1 commentaire

Walter Roberson
Walter Roberson le 6 Juil 2012
What error did you encounter with the code you tried?

Connectez-vous pour commenter.

Réponses (1)

John Petersen
John Petersen le 5 Juil 2012

0 votes

y1-1 means the vector y1 with the quantity 1 subtracted from it. The commented code shifts the elements by one element. Which do you want to do?

5 commentaires

Sara
Sara le 5 Juil 2012
I intend instead of defining the variable y2 with x(1:n-1) , define it by y1 -1 . Maybe I can omit the y2 variable.
F.
F. le 5 Juil 2012
could you give us an exemple of x ??? I'm ok with John y2 = x(1:n-1) and y2 = y1 - 1 , it's not the same thing !!!
John Petersen
John Petersen le 5 Juil 2012
I think you want to shift the elements by 1, but you want to define it your own way. If you want to use Matlab, you need to use it's syntax.
Sara
Sara le 11 Juil 2012
int = zeros(1,length(x));
if k <= 0
int = x ;
return
end
for n = 3:length(x)
y1 = x(2:n) ;
t1 = ((n-2):-1:0)*dt ;
y2 = x(1:n-1) ;
% y2 = -1 +y1 ;
t2 = t1 +dt;
int(n) = sum (t1.^(k-1)/factorial(k-1).*y1 + t2.^(k-1)/factorial(k-1).*y2)*dt/2 ;
end
I mean, is there anyway to define y2 from y1? And not from variable x? Could I omit the variable y2 from the code ?
F.
F. le 11 Juil 2012
If you don't use y1 and y2 after...
if k <= 0
int = x ;
return
else
int = zeros(1,numel(x));
end
for n = 3:numel(x)
t1 = ((n-2):-1:0)*dt ;
t2 = t1 +dt;
int(n) = sum ( ...
t1.^(k-1)/factorial(k-1) .* x( 2 : n ) + ...
t2.^(k-1)/factorial(k-1) .* x( 1 : n-1 ) ...
)*dt/2 ;
end

Connectez-vous pour commenter.

Catégories

En savoir plus sur Loops and Conditional Statements dans Centre d'aide et File Exchange

Question posée :

le 5 Juil 2012

Community Treasure Hunt

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

Start Hunting!

Translated by