# Help needed for the below logic

1 view (last 30 days)
Amy Topaz on 15 Mar 2022
Commented: Jan on 16 Mar 2022
I have implemented a simple logic as shown below. Can I implement the below logic without using for loop? Is it possible to do that without looping by using some algorithm logic? I need to find all values of x for all values of range from 1:1000 due to which we get different values of various quantities
All are constants except x.
for q = 1:1000
v = a*(((b*t(q))/(pi))^(3/2));
k = 2*(((b*t(q))/(pi))^(3/2));
h1 = (t(q));
j2 = 1.17 - (4.73e-4*((t(q))^2))/(t(q) + 636);
eq2 = @(x) ((v)*exp(-(j3-x)/h1)) + ((ff1)/(1+4*exp(-(x-j1)/h1))) - (((k)*exp(-(x-j1)/h1)) + ((ff2)/(1+2*exp(-(j2-x)/h1))));
x2 = [0 1.73];
kk(q) = fzero(eq2, x2);
end
Jan on 16 Mar 2022
"So in the last line why is it written as x2 = kk(i)?" - this is an educated guess of a good initial point for the next search by fzero: The solution for a specific parameter can be assumed to be near to the one of the former iteration.

Jan on 15 Mar 2022
Why do you want to avoid a loop? If it is running, everything is fine.
The calculation of the parameters v, k, h1, ... could be done in vectoprized form before the loop and ll and mm could be determined after the loop. Maybe this saves some time, but as long as fzero() is the most time consuming part, this is not serious. The actual call of fzero cannot be vectorized. Here only using a better initial guess is usful for an acceleration.
##### 2 CommentsShow 1 older commentHide 1 older comment
Jan on 16 Mar 2022
You can replace:
for q = 1:1000
v = a*(((b*t(q))/(pi))^(3/2));
k = 2*(((b*t(q))/(pi))^(3/2));
h1 = (t(q));
j2 = 1.17 - (4.73e-4*((t(q))^2))/(t(q) + 636);
...
end
by:
v = a * (b * t / pi) .^ (3/2);
k = 2 * (b * t / pi) .^ (3/2);
h1 = t; % Is this useful?!
j2 = 1.17 - 4.73e-4 * t .^ 2 ./ (t + 636);
This assumes, that t has the 1000 elements which are accessed in the loop. If the operations , / or ^ are applied to arrays, you need the elementwise forms ., ./ and .^ instead.