MATLAB: Nonlinear system of equations

equationnonlinearsystem

I am trying to solve a system of nonlinear equations. I am not sure what method is the best to use for my problem but using syms does not seem to work. How do I get a numerical solution to the following system of equations?
    clear
    b1=1000;
    b2=100;
    c=50;
    t=0.76;
    T=950;
    n=5;
        
    syms sum R1 R2 R3 R4 R5
    eq1=(R1-R2)-((1+t)*T*(R2^t-R1^t)+50)/b2 ==0;
    eq2=(R2-R3)-((1+t)*T*(R3^t-R2^t)+50)/b2 ==0;
    eq3=(R3-R4)-((1+t)*T*(R4^t-R3^t)+50)/b2 ==0;
    eq4=(R4-R5)-((1+t)*T*(R5^t-R4^t)+50)/b2 ==0;
    eq5=sum-(n*R5+(R4-R5)+2*(R3-R4)+3*(R2-R3)+4*(R1-R2)) ==0;
    eq6=R1-(b1-b2*sum-(1+t)*T*R1^t-c)/b2 ==0;
    eq7=R2-(b1-b2*sum-(1+t)*T*R2^t-2*c)/b2 ==0;
    eq8=R3-(b1-b2*sum-(1+t)*T*R3^t-3*c)/b2 ==0;
    eq9=R4-(b1-b2*sum-(1+t)*T*R4^t-4*c)/b2 ==0;
    eq10=R5-(b1-b2*sum-(1+t)*T*R5^t-5*c)/b2 ==0;
    
    sol=solve(eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9, eq10);
    sol;

Best Answer

https://de.mathworks.com/help/optim/ug/fsolve.html
function myEq
b1=1000;
b2=100;
c=50;
t=0.76;
T=950;
n=5;
x0 = [1 1 1 1 1 1];
opts = optimoptions('fsolve', 'Algorithm', 'levenberg-marquardt');
sol = fsolve(@fun, x0, opts)
test = fun(sol)
function eq = fun(x)
Sum = x(1);
R1 = x(2);
R2 = x(3);
R3 = x(4);
R4 = x(5);
R5 = x(6);
eq(1)=(R1-R2)-((1+t)*T*(R2^t-R1^t)+50)/b2;
eq(2)=(R2-R3)-((1+t)*T*(R3^t-R2^t)+50)/b2;
eq(3)=(R3-R4)-((1+t)*T*(R4^t-R3^t)+50)/b2;
eq(4)=(R4-R5)-((1+t)*T*(R5^t-R4^t)+50)/b2;
eq(5)=Sum-(n*R5+(R4-R5)+2*(R3-R4)+3*(R2-R3)+4*(R1-R2));
eq(6)=R1-(b1-b2*Sum-(1+t)*T*R1^t-c)/b2;
eq(7)=R2-(b1-b2*Sum-(1+t)*T*R2^t-2*c)/b2;
eq(8)=R3-(b1-b2*Sum-(1+t)*T*R3^t-3*c)/b2;
eq(9)=R4-(b1-b2*Sum-(1+t)*T*R4^t-4*c)/b2;
eq(10)=R5-(b1-b2*Sum-(1+t)*T*R5^t-5*c)/b2;
end
end
gives acceptabe results:
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
sol =
    1.5076    0.3566    0.3279    0.2998    0.2722    0.2453
test =
   1.0e-09 *
  Columns 1 through 9
   -0.0108   -0.0043    0.0059    0.1264   -0.0998   -0.0157   -0.0049   -0.0006   -0.0065
  Column 10
   -0.1329
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