Help with non linear equations.

Can someone please help me solve these equations?

I tried to use fsolver but the estimates are very bad. I expect \mu to be 8, \sigma = 4 and \alpha =2.

I will be very glad if someone helps me out on how to better my estimates and improve the results.

Thank you!

Answer 1

[Torsten]

fun = @(x)[8.07-(x(1)-x(2)+x(2)*0.9^(-1/x(3)));11.82-(x(1)-x(2)+x(2)*0.5^(-1/x(3)));284.23-(x(1)-x(2)+x(2)*0.1^(-1/x(3)))];
x = fsolve(fun,[156 45 0.8])

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.

x = 1×3

7.8119 0.8515 0.3980

fun(x)

ans = 3×1

0 0 0

[Mohammed Adjieteh]

Thank you so much!

Answer 2

[Bora Eryilmaz]

fsolve actually finds a correct solution:

x = fsolve(@(x) fcn(x),[1;1;1])

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.

x = 3×1

7.8119 0.8515 0.3980

You can see that the solution satisfies the equations (within a tolarance):

fcn(x)

ans = 1×3

1.0e-13 * 0.0022 0.0089 0.5684

Actually, your suggested values are not a solution for these equations:

fcn([8;4;2])

ans = 1×3

-0.1464 2.1631 267.5809

function F = fcn(x)
 m = x(1);
 s = x(2);
 a = x(3);

 F = [ ...
     8.07 - m + s - s*(1-0.1)^(-1/a) ...
    11.82 - m + s - s*(1-0.5)^(-1/a) ...
   284.23 - m + s - s*(1-0.9)^(-1/a) ...
   ];
end

[Mohammed Adjieteh]

Thank you so much!

Answer 3

[John D'Errico]

syms mu sigma alpha

eq(1) = 8.07 == mu - sigma + sigma*(1-0.1)^(-1/alpha)

eq = 

eq(2) = 11.82 == mu - sigma + sigma*(1-0.5)^(-1/alpha)

eq = 

eq(3) = 284.23 == mu - sigma + sigma*(1-0.9)^(-1/alpha)

eq = 

First, make the problem simpler, subtract equations 1 and 2, then 1 and 3.

eqhat(1) = eq(3) - eq(2)

eqhat = 

eqhat(2) = eq(3) - eq(1)

eqhat = 

And replace alpha by a, where a = -1/alpha

syms a
eqhat = subs(eqhat,alpha,-1/a)

eqhat = 

Next, factor out sigma from each equation. We can use that to eliminate sigma also.

sig1 = solve(eqhat(1),sigma)

sig1 = 

eqa = subs(eqhat(2),sigma,sig1)

eqa = 

That yields one equation, involving only a. It seems to have only one solution, but no analytical solution seems to drop out.

a = vpasolve(eqa)

a = 

and therefore we have alpha.

-1/a

ans = 

0.39797912021003886009252959228167

which gives us sigma. At this point, mu and sigma are linear parameters.

solve(subs(eq(1),alpha,-1/a),subs(eq(2),alpha,-1/a))

ans = struct with fields:
       mu: 7.811904694538537823283206754899
    sigma: 0.85154515981572698988870615312211

Direct enough. However, the solutions you expected are not even close to what comes out. Not my fault. The mathematics won't lie.

[Mohammed Adjieteh]

Thank you so much