for loop not working????
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fidelities value does not change with different values of r.
i'm trying to get the different values of fidelity with different r values,and plot(fidelity,r).
1 comentario
Atsushi Ueno
el 17 de Mzo. de 2023
Editada: Atsushi Ueno
el 17 de Mzo. de 2023
No, for loop is not working because the number sequence is not recognized as a column vector.
for r=0.01;0.02;0.03;0.04;0.05;0.06;0.07;0.08;0.09;0.10;
r
end
But this fixing does not work as you expected:
for r = [0.01;0.02;0.03;0.04;0.05;0.06;0.07;0.08;0.09;0.10;]
r
end
It can be fixed as following:
for r = 0.01:0.01:0.1 % or r = [0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10]
r
end
Respuestas (1)
Bhanu Prakash
el 17 de Mzo. de 2023
Editada: Bhanu Prakash
el 17 de Mzo. de 2023
Hi Abu,
As per my understanding, you are facing some errors with the "for" loop.
Consider the line below:
plot(fidelity,r);
For the "plot" function to work, "r" must be of the same size as "fidelity", i.e., 1x15. If not, it throws an error.
To make "r" to be of size 1x15, the function "linspace" can be used.
I have made some changes in the code and the edited part of the code is attached below:
% Define the commutator of E and F
R_1 = E-F ;
r=linspace(0.01,0.1,15)
% Compute the exponential of R_1
for r_m=r
U = exp(0.01*(R_1));
With the help of "linspace", 15 equally spaced values are assigned to the matrix "r" and the "for" loop runs for 15 times, which is the size of "r".
You can refer to the documentation of "linspace" and "plot" functions, for more info:
Hope this answer helps you.
Thanks,
Bhanu Prakash.
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