Trial and error problem
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Dear Matlab Community Members
My objective is to find the values a, b, and c. of the following equation:
log(m)=log(a)+(b/(T-c))
I have three values of m & T
m=0.0701 & T=293.65
m=0.0262 & T=313.15
m=0.00433 & T=373.15
I want to know what is the best code or technique to find these values.
Thank you very much
8 Kommentare
the cyclist
am 26 Jan. 2023
Bearbeitet: the cyclist
am 26 Jan. 2023
How did you choose that function for the fit? It seems like a poor one, because of the singularity that will happen when T==c.
c = 280;
T = 260 : 380;
f = @(t) 1./(t -c);
plot(T,f(T))
Even if your values of T are strictly greater than c, (the "critical" temperature), this behavior will make it difficult for any algorithm to fit it.
One can try (e.g. with fitnlm), but MATLAB spits out warnings about overparameterization, and the fit is not good.
Antworten (1)
Torsten
am 26 Jan. 2023
Bearbeitet: Torsten
am 26 Jan. 2023
The results are not convincing, but that's the way to go with three data points.
fun1 = @(x)[0.0701-x(1)*exp(x(2)/(293.65-x(3)));0.0262-x(1)*exp(x(2)/(313.15-x(3)));0.00433-x(1)*exp(x(2)/(373.15-x(3)))];
fun2 = @(x)[log(0.0701)-(log(x(1))+x(2)/(293.65-x(3)));log(0.0262)-(log(x(1))+x(2)/(313.15-x(3)));log(0.00433)-(log(x(1))+x(2)/(373.15-x(3)))];
x0 = [2 1 450];
x = fsolve(fun1,x0)
fun1(x)
x = fsolve(fun2,x0)
fun2(x)
5 Kommentare
Torsten
am 27 Jan. 2023
Bearbeitet: Torsten
am 27 Jan. 2023
Isn't the form of the equation usually
m = 10^(A-B/(C+T))
where T is in degreeC ?
Thus
log10(m) = A - B/(T+C)
where log10 is not the natural, but the decadic logarithm and T is temperature in degreeC ?
m1=0.0701;
T1=293.65-273.15;
m2=0.0262;
T2=313.15-273.15;
m3=0.00433;
T3=373.15-273.15;
syms a b c
eqns = [a-b/(T1+c)== log10(m1), a-b/(T2+c)== log10(m2), a-b/(T3+c)==log10(m3)];
S = vpasolve(eqns,[a b c])
double(subs([a-b/(T1+c)- log10(m1), a-b/(T2+c)- log10(m2), a-b/(T3+c)-log10(m3)],[a,b,c],[S.a,S.b,S.c]))
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