Finding roots of a two variable function

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Chaitanya
Chaitanya 2022-12-11
移动：Walter Roberson 2022-12-14
Basically I have x which is a column vector with 350 rows of random data. I need to model this as a Birnbaum Sauders distribution and estimate its parameters. I can do this directly using the mle function, but I wonder if its possible to use fsolve. Below is my function file.
function f = BirnbaumSaunders(alpha,lamb)
F = (alpha.^2)-((1/350)*sum((lamb.*x)+(1./(lamb.*x))-2));
G = (-175.*lamb)+((1./(2.*(alpha.^2)))*sum(((lamb.*lamb.*x))-(1./x))+sum(lamb./((lamb.*x)+1)));
end
Here, I have to find the value of alpha and lamb for which both my functions F and G are zero. I have tried using the fsolve with initial alpha and lamb values as 0.65 and 2.13 (I got these values after using the mle function). But I'm getting an error using fsolve.
fun = @BirnbaumSaunders;
x = fsolve(fun,[0.65,2.13])
This is giving me an error. Is there any other way to find the values of alpha and lamb when F and G are set to 0?

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回答（2 个）

Torsten
Torsten 2022-12-12
Ran in:
x = rand(10,1);
fun = @(z)BirnbaumSaunders(z(1),z(2),x);
z = fsolve(fun,[0.65,2.13])
Solver stopped prematurely. fsolve stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 2.000000e+02.
z = 1×2
0.1961 -1.1427
norm(fun(z))
ans = 0.1632
function f = BirnbaumSaunders(alpha,lamb,x)
F = (alpha.^2)-((1/350)*sum((lamb.*x)+(1./(lamb.*x))-2));
G = (-175.*lamb)+((1./(2.*(alpha.^2)))*sum(((lamb.*lamb.*x))-(1./x))+sum(lamb./((lamb.*x)+1)));
f = [F;G];
end
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Chaitanya
Chaitanya 2022-12-12
编辑：Chaitanya 2022-12-12
I got a correct value using this data and your code!!! Thank you very much for your time!
Torsten
Torsten 2022-12-12
You want to estimate two parameters of a distribution given seven realizations ? That's ... courageous.

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Walter Roberson
Walter Roberson 2022-12-14
移动：Walter Roberson 2022-12-14
Ran in:
format long g
x = [
0.260594
0.50998
0.609437
0.365165
0.949793
0.590385
0.902765];
fun = @(z)BirnbaumSaunders(z(1),z(2),sym(x));
syms z [1 2]
F = simplify(fun(z))
F = 
sol = vpasolve(F,[0.65;2.13])
sol = struct with fields:
z1: [16×1 sym] z2: [16×1 sym]
vpa(subs(z, sol))
ans = 
subs(F, sol)
ans = 
fun = @(z)BirnbaumSaunders(z(1),z(2),x);
opt = optimoptions('fsolve', 'MaxFunctionEvaluations', 1e4);
Z = fsolve(fun,[0.06,2], opt)
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.
Z = 1×2
0.0630815494828271 2.00076854993031
function f = BirnbaumSaunders(alpha,lamb,x)
F = (alpha.^2)-((1/350)*sum((lamb.*x)+(1./(lamb.*x))-2));
G = (-175.*lamb)+((1./(2.*(alpha.^2)))*sum(((lamb.*lamb.*x))-(1./x))+sum(lamb./((lamb.*x)+1)));
f = [F;G];
end

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