how to find the root of transcedental equation while loop
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k0=(2*pi/0.6328)*1e6;
t2=1.5e-6;
n1=1.512;n2=1.521;n3=4.1-1i*0.211;
n4=1;
m=0;
t3=1e-9;
k1=k0*sqrt(n1^2-x^2);
k2=k0*sqrt(n2^2-x^2);
k3=k0*sqrt(n3^2-x^2);
k4=k0*sqrt(n4^2-x^2);
tol = 1e-12;
n = 1;
y=-(k2)*t2+atan(k1/1i*k2)+atan((k3/k2)*tan(atan(k4/1i*k2)-k3*t3))+m*pi;
how to calculate the root of equation i.e the vaue of x numerically by while loop
Answers (1)
Walter Roberson
on 17 Feb 2022
format long g
k0=(2*pi/0.6328)*1e6;
t2=1.5e-6;
n1=1.512;n2=1.521;n3=4.1-1i*0.211;
n4=1;
m=0;
t3=1e-9;
tol = 1e-12;
n = 1;
x = rand
y = inf;
while n < 5 & abs(y) > tol
k1=k0*sqrt(n1^2-x^2);
k2=k0*sqrt(n2^2-x^2);
k3=k0*sqrt(n3^2-x^2);
k4=k0*sqrt(n4^2-x^2);
y=-(k2)*t2+atan(k1/1i*k2)+atan((k3/k2)*tan(atan(k4/1i*k2)-k3*t3))+m*pi;
[n, x, y]
n = n + 1;
end
All of the results are the same because you have not defined any way for x to change.
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