MATLAB: Infinite or Not-a-Number function value encountered.

quad

Originally I tried to calculate

int(int((sin(q*A*sin(phi)/2)*sin(q*B*sin(phi)/2)*4/(q^2*A*B*sin(phi)*cos(phi)))^2*sin(q*C*cos(theta)/2)*2/(C*q))^2*sin(theta),theta,0,pi/2),phi,0.pi/2)

however MATLAB fail to show proper equation. So I change to use 'quad' function instead of 'int'

I just put A=B=C=7 and q=0.1 and tried to calculate…but failed. How can I calculate this equation?

-(183750365601059840*sin((7/20*sin(phi))).^4)./(8477503880968187*sin(phi).^2.*(sin(phi).^2. – 1) –> this equation comes from int((sin(q*A*sin(phi)/2)*sin(q*B*sin(phi)/2)*4/(q^2*A*B*sin(phi)*cos(phi)))^2*(sin(q*C*cos(theta)/2)*2/(C*q))^2*sin(theta),theta,0,pi/2) and put A, B, C, q values.

——————————————————————————————

quad(@(phi) -(183750365601059840*sin((7/20*sin(phi))).^4)./(8477503880968187*sin(phi).^2.*(sin(phi).^2. – 1)),0.0001,pi/2)

Warning: Infinite or Not-a-Number function value encountered.

> In quad at 109

ans =

NaN

Best Answer

Besides the parenthesis unbalanced, you are dividing by zero.

for phi = pi/2

(8477503880968187*sin(phi)).^2.*(sin(phi).^2. - 1) = 0

Then:

r=-(183750365601059840*sin((7/20*sin(phi))).^4)/...
    (8477503880968187*sin(phi)).^2.*(sin(phi).^2. - 1)

will yield NaN.

Related Solutions

MATLAB: How to define a function that evaluates in input

function y=T(theta_Degree)
theta=theta_Degree*pi/180;
y=[cos(theta)^2 sin(theta)^2 2*cos(theta)*sin(theta);sin(theta)^2 cos(theta)^2 -2*cos(theta)*sin(theta);-cos(theta)*sin(theta) cos(theta)*sin(theta) cos(theta)^2-sin(theta)^2];
end

Then type in command line

T(40)

and see the results.

MATLAB: Solving non linear equation using fsolve

You are not passing theta_n and other parameters to root2d(). One fix is to make root2d a Nested Function. Also, you must choose a different initial point. phi0 = [10,10] worked for me.

function solve_it()
    beta_a = 10;
    alpha_n = 10 ;
    eta= 18;
    i = 18;
    theta_i = 10;
    theta_n = 10 ;
    
    fun = @root2d;
    phi0 = [10,10];
    phi = fsolve(fun,phi0);
    phi_n = phi(1),
    phi_i = phi(2),
    function F = root2d(phi)
    F = zeros(2,1);
    F(1) = (pi*sin((pi*(phi(1) + theta_n))/180)*cos((pi*phi(2))/180)*(cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180)))/(180*sin((pi*phi(1))/180)*(cos((pi*(phi(1) + theta_n))/180)*cos((phi(2)*pi)/180) + sin((phi(2)*pi)/180)*tan((theta_i*pi)/180))^2) - (pi*cos((pi*phi(1))/180)*(cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180)))/(180*sin((phi(1)*pi)/180)^2*(cos((pi*(phi(1) + theta_n))/180)*cos((pi*phi(2))/180) + sin((pi*phi(2))/180)*tan((pi*theta_i)/180)));
    F(2) = ((cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180))*((pi*cos((pi*(phi(1) + theta_n))/180)*sin((pi*phi(2))/180))/180 - (pi*cos((pi*phi(2))/180)*tan((pi*theta_i)/180))/180))/(sin((pi*phi(1))/180)*(cos((pi*(phi(1) + theta_n))/180)*cos((phi(2)*pi)/180) + sin((phi(2)*pi)/180)*tan((theta_i*pi)/180))^2);
    end
end

Best Answer

Related Solutions

MATLAB: How to define a function that evaluates in input

MATLAB: Solving non linear equation using fsolve

Related Question