How to solve for a b c d ?? If i have 4 equations like these in matlab.

i have a set of equations like these. i am trying to solve for a b c d. matlab returns numerical approximation instead and the values are way too big. is there anything wrong with the equation??

syms a b c d;

eqn1 = (a+(b/(1+(exp(-((47.71-c)/d))))))==11.1111;

eqn2 = (a+(b/(1+(exp(-((72.17-c)/d))))))==18.5185;

eqn3 = (a+(b/(1+(exp(-((85.38-c)/d))))))==33.3333;

eqn4 = (a+(b/(1+(exp(-((94.79-c)/d))))))==55.5556;

sol = solve(eqn1,eqn2,eqn3,eqn4,a,b,c,d);

[Walter Roberson]

If you work iterative, you can easily solve eqn1 for a, substitute that into eqn2, solve for b, substitute the a and b into eqn3. You can then solve that for c; I have not tried in MATLAB, but in Maple I get a closed form expression with thousands of terms.

You can then substitute all of those into eqn4, but you can be sure there will be no practical way of getting a closed form solution for d, and even looking for a numeric solution is going to be very time consuming.

[Walter Roberson]

11.1111 appears to be standing in for 100/9

33.3333 appears to be standing in for 100/3

55.5556 appears to be standing in for 500/9

But is 18.5185 intended to be standing in for 18 + 14/27 ?

Are there intended fractions for the other numbers such as the 47.71 ?

[Walter Roberson]

My investigations so far suggest that to within floating point precision, there are many solutions. If you solve the first three equations for a, b, c, then there are multiple thousand terms of exp() adding together, and there is a lot of round-off involved in that process.

Answer 1

[Walter Roberson]

One of the solutions is

a = 9.82047978530749
b = 320.687014884004
c = 117.506137526386
d = 12.6641844232045

this gives a residue on the order of 1E-22, which is in the noise level compared to the uncertainty on the input coefficients such as 72.17

At the moment I do not seem to find any other solution.

Answer 2

[Rik]

If you take a look at the documentation, you will notice that you need to enter your equations as a vector: sol = solve([eqn1,eqn2,eqn3,eqn4],[a,b,c,d]);

However, this still gives the same answer, so that can't be the problem. What is, I couldn't say, but if you use fit, you get much smaller values.

x=[47.71 72.17 85.38 94.79]';
y=[11.1111 18.5185 33.3333 55.5556]';
sol=fit(x,y,'(a+(b/(1+(exp(-((x-c)/d))))))');

[Walter Roberson]

Entering equations as individual terms is the older form and still supported.

[Rik]

Thanks for the clarification, I couldn't find anywhere in the new doc why this even worked.

[Walter Roberson]

You can see the form in the R2014b documentation http://www.mathworks.com/help/releases/R2014b/symbolic/solve.html . Search for example for "Solve Multivariate Equations"

[Walter Roberson]

When I use the above method, fit(), then the fit is not a good one. I get

 a = 26.9987073814621
 b = 2.63091761853793
 c = 0.412024384777503
 d = 0.308022637282378

If you back-substitute those into the equations then you get

29.629625000000006984635092521785 == 11.111100000000000420641299569979,
29.629625000000006984635092521785 == 18.5185,
29.629625000000006984635092521785 == 33.333300000000001261923898709938,
29.629625000000006984635092521785 == 55.555599999999998317434801720083

which is not good at all.

The fit gets stuck in a local minima with residue approximately 1152.