Help revise the code

Question

Jiiim Wu 2023-8-22

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2011217-help-revise-the-code

评论： Walter Roberson 2023-8-22

% Given constants
q = 0.2;
alpha = 0.7;
gamma = 0.1;
K_bar = 1;
z_hat = 1.1;
eta = 0.2;
% Define the range of epsilon values
epsilon_range = 0.1:0.01:1;
% Preallocate array to store z_star values
z_star_values = zeros(size(epsilon_range));
% Define the options for fsolve
options = optimoptions('fsolve', 'Display', 'off', 'TolFun', 1e-6);
for i = 1:length(epsilon_range)
    epsilon_val = epsilon_range(i);
    
    % Define the equation for fsolve
    equation = @(z_star) shooting_equation(z_star, epsilon_val, q, alpha, gamma, K_bar, z_hat, eta);
    
    % Solve for z_star using fsolve
    z_star_initial_guess = 0.5;  % Initial guess for z_star
    z_star_values(i) = fsolve(equation, z_star_initial_guess, options);
end
% Plot z_star as a function of epsilon
figure;
plot(epsilon_range, z_star_values, 'b', 'LineWidth', 2);
xlabel('\epsilon');
ylabel('z^*');
title('Plot of z^* as a Function of \epsilon');
grid on;
function value = shooting_equation(z_star, epsilon, q, alpha, gamma, K_bar, z_hat, eta)
value = z_star - z_star * (((1 - alpha) * epsilon) / (1 - alpha * epsilon))^((1 - epsilon) / epsilon) ...
    - q^((1 - epsilon) / epsilon) * (K_bar)^((epsilon - 1) / epsilon) ...
    * (1)^((epsilon - 1) / (alpha * epsilon)) * ((1 - alpha) / alpha)^((1 - alpha) * (epsilon - 1) / (alpha * epsilon)) ...
    * (epsilon)^(((1 - epsilon + alpha * epsilon) * (1 - epsilon)) / (alpha * epsilon^2)) ...
    * ((alpha) / (1 - alpha * epsilon))^((1 - epsilon) / epsilon) ...
    * (alpha * epsilon)^((epsilon - 1) / (alpha * epsilon^2)) ...
    * (((1 - epsilon) / (0 - 2))^((1 - epsilon) * (epsilon - 1 + alpha * epsilon) / (alpha * epsilon^2))) ...
    * ((((2)^(1 / (1 - epsilon)) - (z_hat + eta)^(1 / (1 - epsilon))) + (((z_hat - eta)^(1 / (1 - epsilon))) - (z_star)^(1 / (1 - epsilon))) ...
    + ((1 - gamma)^(epsilon / (1 - epsilon))) * (((z_hat + eta)^(1 / (1 - epsilon))) - ((z_hat - eta)^(1 / (1 - epsilon))))) ...
    ^((1 - epsilon) * (epsilon - 1 + alpha * epsilon) / (alpha * epsilon^2)) - z_star;
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
end

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2023-8-22

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2011217-help-revise-the-code#answer_1291002

在 MATLAB Online 中打开

starting from the beginning of

    * ((((2)^(1 / (1 - epsilon)) - (z_hat + eta)^(1 / (1 - epsilon))) + (((z_hat - eta)^(1 / (1 - epsilon))) - (z_star)^(1 / (1 - epsilon))) ...

you have one extra ( that does not have a matching )

value = z_star - z_star * (((1 - alpha) * epsilon) / (1 - alpha * epsilon))^((1 - epsilon) / epsilon) - q^((1 - epsilon) / epsilon) * (K_bar)^((epsilon - 1) / epsilon) * (1)^((epsilon - 1) / (alpha * epsilon)) * ((1 - alpha) / alpha)^((1 - alpha) * (epsilon - 1) / (alpha * epsilon)) * (epsilon)^(((1 - epsilon + alpha * epsilon) * (1 - epsilon)) / (alpha * epsilon^2)) * ((alpha) / (1 - alpha * epsilon))^((1 - epsilon) / epsilon) * (alpha * epsilon)^((epsilon - 1) / (alpha * epsilon^2)) * (((1 - epsilon) / (0 - 2))^((1 - epsilon) * (epsilon - 1 + alpha * epsilon) / (alpha * epsilon^2))) * ((((2)^(1 / (1 - epsilon)) - (z_hat + eta)^(1 / (1 - epsilon))) + (((z_hat - eta)^(1 / (1 - epsilon))) - (z_star)^(1 / (1 - epsilon))) + ((1 - gamma)^(epsilon / (1 - epsilon))) * (((z_hat + eta)^(1 / (1 - epsilon))) - ((z_hat - eta)^(1 / (1 - epsilon)))))^((1 - epsilon) * (epsilon - 1 + alpha * epsilon) / (alpha * epsilon^2)) - z_star;
%                         123         2          1   2                   10 12           1          0     12           1          0   1     0 12           1          0   1 0 12           1   2               10   12         1        0 12         1   2           1   2               10   1       0 123                             2   3           21   2                 10   12     1   2                   10 12           1          0   1               0 12           1   2                 10   123           2   3     21 23           2   3                             2   3                 210   1234 3 4    5           43   4           3 4    5           432   345           4 5    6           543   4      3 4    5           432   34         3 4          5           432   345           4 5    6           543   45           4 5    6           54321 23           2   3                             2   3                 21    

The number under each bracket is the number of open brackets in effect "after" the character above the number.

2 个评论
显示无隐藏无

Jiiim Wu 2023-8-22

移动：Bruno Luong 2023-8-22

Would it be possible that you can provide the correct code. It looks still confusing to me. Thanks

Walter Roberson 2023-8-22

No, I cannot do that, as I do not know what the correct equations are.

I highly recommend that you break the function up into multiple statements, instead of using one long long line of code.

That would also allow you to do optimizations such as calculating alpha * epsilon and 1-epsilon into variables so that those expressions do not need to be calculated multiple times.

Using a long line of source is harder for humans to read, and harder for MATLAB to analyze the flow of. And tends to involve calculating the same sub-expression multiple times unnecessarily.

请先登录，再进行评论。

Help revise the code

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

回答（1 个）

2 个评论
显示无隐藏无

另请参阅

类别

标签

Community Treasure Hunt

Help revise the code

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

回答（1 个）

2 个评论 显示 无隐藏 无

另请参阅

类别

标签

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

2 个评论
显示无隐藏无