Question

你能告诉我在哪里我必须把我的起始值放在下面提到的代码中假设我想解决它的初始值-2p为theta和phi

function myFcn    

    X = fsolve(@F, 2*pi*rand(2,1))

end

function val = F(X)

    th = X(1);
    ph = X(2);

    val =  [
        cos(2*th) + sin(2*th)
        cos(ph).^2.*cos(th).^2 + cos(ph).^2.*cos(2*th).*sin(2*th) + cos(ph).^2.*cos(2*th) + sin(2*ph).*cos(2*ph).*cos(2*th).*sin(2*th) + sin(2*ph).*cos(2*ph).*sin(th).^2 + sin(2*ph).*cos(2*ph).*sin(2*th) + 0.4837
        cos(2*ph).*cos(th).^2.*sin(2*ph) + cos(2*th).*sin(ph).^2.*sin(2*th) + cos(2*ph).*cos(2*th).*sin(2*ph).*sin(2*th) + cos(2*ph).*cos(2*th).*sin(2*ph) + sin(ph).^2.*sin(2*th) - 1
        cos(th).^2.*sin(2*ph) - cos(2*ph).*sin(th).^2 - cos(2*ph).*sin(2*th) + cos(2*th).*sin(2*ph) - cos(2*ph).*cos(2*th).*sin(2*th) + cos(2*th).*sin(2*ph).*sin(2*th)
];
end

thankx

Answer 1

fsolve的第二个参数是X的初始值。你的意思是-2 * pi作为theta和phi的初始值吗？如果是这样，您将执行以下操作：

function myFcn    

    X = fsolve(@F, [-2*pi, -2*pi])

end

您可以创建myFcn，以便在函数调用中指定起始值，如下所示：

function myFcn(initial_theta, initial_phi)

    X = fsolve(@F, [initial_theta, initial_phi])

end

然后当你调用这个函数时，你可以调用

myFcn(-2*pi, -2*pi)，

或者如果你以后决定要尝试theta = pi和phi = 0的初步猜测，你可以打电话

myFcn(pi, 0)

4个eq，用于Matlab中的两个变量

1 个答案: