Trial and error problem

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Ahmad Al-Issa
Ahmad Al-Issa 2023-1-26
编辑： Torsten 2023-1-27
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
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the cyclist
the cyclist 2023-1-26
编辑：the cyclist 2023-1-26
Ran in:
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.
Ahmad Al-Issa
Ahmad Al-Issa 2023-1-26
编辑：Ahmad Al-Issa 2023-1-27
@the cyclist
thank you for your answer, but this is not right.
I found the soution by solve a system of equations.
m1=0.0701;
T1=293.65;
m2=0.0262;
T2=313.15;
m3=0.00433;
T3=373.15;
syms a b c
eqns = [log(a)+(b/(T1-c))== log(m1), log(a)+(b/(T2-c))== log(m2), log(a)+(b/(T3-c))== log(m3)];
S = (vpasolve(eqns,[a b c]));

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Torsten
Torsten 2023-1-26
编辑：Torsten 2023-1-26
Ran in:
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)
No solution found. fsolve stopped because the problem appears regular as measured by the gradient, but the vector of function values is not near zero as measured by the value of the function tolerance.
x = 1×3
0.0345 8.2224 633.4040
fun1(x)
ans = 3×1
0.0364 -0.0075 -0.0291
x = fsolve(fun2,x0)
Solver stopped prematurely. fsolve stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 3.000000e+02.
x = 1×3
0.0217 53.9522 974.0568
fun2(x)
ans = 3×1
1.2519 0.2701 -1.5220
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Ahmad Al-Issa
Ahmad Al-Issa 2023-1-27
@Torsten
@the cyclist
thanks for your try.
The right answer, as it is in the refrences, is:
a=0.000073
b=800
c=177
because a, b, and c are constants and they need 3 equations to solve.
Torsten
Torsten 2023-1-27
编辑：Torsten 2023-1-27
Ran in:
@AHMAD AL-ISSA
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])
S = struct with fields:
a: -4.1361836318056620904194888300452 b: -347.51996229463221164872269115385 c: 96.04306650964382392391269916326
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]))
ans = 1×3
1.0e-38 * 0.0735 0.0735 0.1469

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