How can i solve function with an dependent variable?

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Ceren Öztürk 2021-6-10

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评论： Ceren Öztürk 2021-6-11

采纳的回答： Rik

There is a g(u) function and u variable's values are dependent to g(u) function, its derivative and smaller u.

I do not know how to write this as a matlab code.

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Rik 2021-6-10

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Below I show the structure of a possible solution. The rest is up to you. Feel free to comment if you have a specific question.

g=@(mu,B1,B2,B3,a1,a2,a3) B1./a1;
g_=@(mu,B1,B2,B3,a1,a2,a3) 2./(a1-mu);
mu_k=0;
while true
    mu_kp1=mu_k-g(mu_k,B1,B2,B3,a1,a2,a3)./g_(mu_k,B1,B2,B3,a1,a2,a3);
    if mu_kp1/mu_k < 1
        break
    end
end

After a minor edit, the code runs fine. And you only need pre-allocation if you expect many iterations, in which case I would suggest avoiding storing all intermediate values, unless the explicit goal is to find the trend in iterations of mu for different values of your betas and alphas.

%generate random values, but make sure they are the same every run
rng(0);[B1,al1,B2,al2,B3,al3]=deal(rand);
g = @(u) (B1./(al1 - u)).^2 + (B2./(al2 - u)).^2 + (B3./(al3 - u)).^2 -1;
derivg = @(u) 2/(al1-u).*(B1./(al1 - u)).^2 + 2/(al2-u).*(B2./(al2 - u)).^2 + 2/(al3-u).*(B3./(al3 - u)).^2 ;
u=0; k=1;
while true
    u(k+1)= u(k)-g(u(k))./derivg(u(k));
    k=k+1;
    if abs((u(k)- u(k-1))./u(k)) < 0.000001
        %  ^            ^
        % you forgot these
        break
    end    
end
disp(u)
         0   -0.2716   -0.4929   -0.5853   -0.5963   -0.5964   -0.5964