Solve equations in a loop with fsolve

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Mepe
Mepe 2020 年 2 月 18 日
コメント済み: Mepe 2020 年 2 月 19 日
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 件のコメント
Mepe
Mepe 2020 年 2 月 18 日
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?

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採用された回答

Matt J
Matt J 2020 年 2 月 18 日
編集済み: Matt J 2020 年 2 月 18 日
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 件のコメント
Mepe
Mepe 2020 年 2 月 19 日
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?

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その他の回答 (1 件)

darova
darova 2020 年 2 月 18 日
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
  3 件のコメント
Mepe
Mepe 2020 年 2 月 19 日
No problem. Thanks a lot for this solution!!!

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