How to feed additional variables into fsolve function
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Hello,
I have a function F = quad(x,REF)
function F = quad(x,REF)
F(1) = x(3)/((T(1)/x(1))-1) + log(x(2)) - log(dv(1))
F(2) = x(3)/((T(2)/x(1))-1) + log(x(2)) - log(dv(2))
F(3) = x(3)/((T(3)/x(1))-1) + log(x(2)) - log(dv(3))
F(4) = x(3)/((T(4)/x(1))-1) + log(x(2)) - log(dv(4))
F(5) = x(3)/((T(5)/x(1))-1) + log(x(2)) - log(dv(5))
F(6) = x(3)/((T(6)/x(1))-1) + log(x(2)) - log(dv(6))
F(7) = x(3)/((T(7)/x(1))-1) + log(x(2)) - log(dv(7))
F(8) = x(3)/((T(8)/x(1))-1) + log(x(2)) - log(dv(8))
end
I'd like to feed different dv and T values from the main program and solve. I've tried the below code from another post (modified and played around a bit to no avail). How can I feed dv and T values into the function quad and then use fsolve?
REF = [x,dv,T];
f =@(x) quad(x,REF);
options = optimoptions('fsolve','Display','iter');
% fun(dv,T) = @quad;
x0 = [150,0.045,7.5];
[out,fval] = fsolve(f,x0,options)
This works fine when the T and dv are already in the function quad, but I don't want to manually copy and paste new values of dv and T in.
function F = quad(x)
dv = [3610.30 2040.80 1203.70 483.04 106.03 15.54 3.39 1.97]
T = [-40 -35 -30 -20 0 40 100 135] + 273.15
F(1) = x(3)/((T(1)/x(1))-1) + log(x(2)) - log(dv(1))
F(2) = x(3)/((T(2)/x(1))-1) + log(x(2)) - log(dv(2))
F(3) = x(3)/((T(3)/x(1))-1) + log(x(2)) - log(dv(3))
F(4) = x(3)/((T(4)/x(1))-1) + log(x(2)) - log(dv(4))
F(5) = x(3)/((T(5)/x(1))-1) + log(x(2)) - log(dv(5))
F(6) = x(3)/((T(6)/x(1))-1) + log(x(2)) - log(dv(6))
F(7) = x(3)/((T(7)/x(1))-1) + log(x(2)) - log(dv(7))
F(8) = x(3)/((T(8)/x(1))-1) + log(x(2)) - log(dv(8))
end
(CALLED FROM THE MAIN AS:)
fun = @quad;
x0 = [150,0.045,7.5];
[out,fval] = fsolve(fun,x0,options)
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채택된 답변
Stephen23
2021년 4월 3일
dv = [3610.30,2040.80,1203.70,483.04,106.03,15.54,3.39,1.97];
T = [-40,-35,-30,-20,0,40,100,135] + 273.1;
f = @(x) myquad(x,dv,T);
x0 = [150,0.045,7.5];
[out,fval] = fsolve(f,x0)
function F = myquad(x,dv,T)
F(1) = x(3)/((T(1)/x(1))-1) + log(x(2)) - log(dv(1));
F(2) = x(3)/((T(2)/x(1))-1) + log(x(2)) - log(dv(2));
F(3) = x(3)/((T(3)/x(1))-1) + log(x(2)) - log(dv(3));
F(4) = x(3)/((T(4)/x(1))-1) + log(x(2)) - log(dv(4));
F(5) = x(3)/((T(5)/x(1))-1) + log(x(2)) - log(dv(5));
F(6) = x(3)/((T(6)/x(1))-1) + log(x(2)) - log(dv(6));
F(7) = x(3)/((T(7)/x(1))-1) + log(x(2)) - log(dv(7));
F(8) = x(3)/((T(8)/x(1))-1) + log(x(2)) - log(dv(8));
end
추가 답변 (1개)
Sulaymon Eshkabilov
2021년 4월 2일
편집: Sulaymon Eshkabilov
2021년 4월 2일
Here is a combined and a bit simplified solution:
options = optimoptions('fsolve','Display','iter');
x0 = [150,0.045,7.5];
dv = [3610.30 2040.80 1203.70 483.04 106.03 15.54 3.39 1.97];
T = [-40 -35 -30 -20 0 40 100 135] + 273.15;
[out,fval] = fsolve(@(x)([ x(3)/((T(1)/x(1))-1) + log(x(2)) - log(dv(1));
x(3)/((T(2)/x(1))-1) + log(x(2)) - log(dv(2));
x(3)/((T(3)/x(1))-1) + log(x(2)) - log(dv(3));
x(3)/((T(4)/x(1))-1) + log(x(2)) - log(dv(4));
x(3)/((T(5)/x(1))-1) + log(x(2)) - log(dv(5));
x(3)/((T(6)/x(1))-1) + log(x(2)) - log(dv(6));
x(3)/((T(7)/x(1))-1) + log(x(2)) - log(dv(7));
x(3)/((T(8)/x(1))-1) + log(x(2)) - log(dv(8));]), x0,options)
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