# Solve equations in a loop with fsolve

8 views (last 30 days)
Mepe on 18 Feb 2020
Commented: Mepe on 19 Feb 2020
Hi there,
I have two problems solving an equation:
Problem 1:
The equation below is to be solved component by component and the results are to be stored line by line in the vector F1. So far so good, how do I teach the loop to use the correct column for the calculations (e.g. f1 (f ,:) or f8 (f ,:) without integrating the function into fsolve?
tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
eq = @(s) s*tau-(0.1.*s^2+3.54.*s-9.53).*f4.^2-f8;
for f = 1:1:length (f4)
F1 (f,:) = fsolve (eq, 0)
end
Problem 2:
The eq described above actually consists of two equations:
eq1 = 0.01*s.^2+3.54.*s-y*9.53
eq2 = y.*f4.^2-f8-s.*tau
It would be desirable to be able to insert both equations separately. Here is the variable y, which disappears after summarizing. Is there a way to combine this with the "problem" above?
Thanks a lot!
##### 2 CommentsShowHide 1 older comment
Mepe on 18 Feb 2020
s is the variable we are looking for. f4 and f8 are given by the vectors. Do these still have to be specified as you have declared?

Matt J on 18 Feb 2020
Edited: Matt J on 18 Feb 2020
Your equations are quadratic and therefore generally have two solutions, s. Fsolve cannot find them both for you. Why aren't you using roots()? Regardless, here are the code changes pertaining to your question:
Problem 1
tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
for i = 1:1:length (f4)
eq = @(s) s*tau-(0.1.*s^2+3.54.*s-9.53).*f4(i).^2-f8(i);
F1 (i,:) = fsolve (eq, 0);
end
Problem 2
for i = 1:1:length (f4)
eq1 = @(sy) [0.01*sy(1).^2+3.54.*sy(1)-sy(2)*9.53 ; ...
sy(2).*f4(i).^2-f8(i)-sy(1).*tau];
F2 (i,:) = fsolve (eq, [0,0]);
end
Mepe on 19 Feb 2020
I still have a question.
Now if I needed both solutions to the quadratic equation, how exactly would that work with the roots () command? according to help, a vector with numerical values must be used. How can I apply this to my problem?

darova on 18 Feb 2020
This is the correct form
tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
eq = @(s,f4,f8) s*tau-(0.1.*s^2+3.54.*s-9.53).*f4.^2-f8;
for f = 1:1:length (f4)
F1 (f,:) = fsolve (@(s)eq(s,f4(f),f8(f), 0);
end
Mepe on 19 Feb 2020
No problem. Thanks a lot for this solution!!!