Solve equations in a loop with fsolve
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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!
采纳的回答
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
4 个评论
Mepe
2020-2-18
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
2020-2-19
Mepe
2020-2-19
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