# Solve equations in a loop with fsolve

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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!

darova on 18 Feb 2020
Shouldn't the function be dependent? 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

Show 1 older comment
Matt J on 18 Feb 2020
What exactly do sy(1) or sy(2) mean in the equation definition?
In Problem 2, you have two unknowns, s and y. Fsolve requires that they be bundled into a vector sy=[s,y].
in F2 the solution is filled with a 2x4 matrix. How does that come about?
Each row is a solution [s,y] corresponding to i=1,2,3,4.
Mepe on 19 Feb 2020
Now I see. Thanks a lot for your help!!!
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 18 Feb 2020
Thank you very much for your support. I really like your idea of how the fsolve expression is built!
Unfortunately I get an error message (The input to FSOLVE should be either a structure with valid fields or consist of at least two arguments).
Can you help again here?
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
darova on 18 Feb 2020 