kronecker product by number of iteration
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
lets say A=[1 2;3 4], i want for i=1:5 times, to multiply A itself kronecker product. in this case ,manual will be kron(kron(kron(kron(kron(A,A),A),A),A),A)
0 Commenti
Risposta accettata
Geoff Hayes
il 8 Lug 2014
Unless you are looking for a way to do this in one line of code, a simple for loop would do the trick
A = [1 2;3 4];
B = A;
for k=1:5
B = kron(B,A);
end
4 Commenti
Geoff Hayes
il 9 Lug 2014
Akymrat - Please format the above code so that it is readable. Highlight the code and press the {}Code button.
There is a bug in the code with the
else
B==2;
R=F2
end
I suspect you meant
elseif B==2
Please make the correction and address the case where B is neither 1 nor 2.
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Logical 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!