how to find the root of transcedental equation while loop

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

[shiv gaur]

what will be root of this equation using while loop

[shiv gaur]

any one pl iterate the value of x is the root help

Answer 1

[Walter Roberson]

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

x = 
         0.599324733046148

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

ans = 
                          1 +                     0i          0.599324733046148 +                     0i          -17.6929697847717 +  7.90103155130006e-07i

ans = 
                          2 +                     0i          0.599324733046148 +                     0i          -17.6929697847717 +  7.90103155130006e-07i

ans = 
                          3 +                     0i          0.599324733046148 +                     0i          -17.6929697847717 +  7.90103155130006e-07i

ans = 
                          4 +                     0i          0.599324733046148 +                     0i          -17.6929697847717 +  7.90103155130006e-07i

All of the results are the same because you have not defined any way for x to change.

[shiv gaur]

this is not the root of the equation i.e y pl any one to help find the root

[Torsten]

You search for x such that y=0, or what's your aim ?

[Walter Roberson]

Why are you writing the root-finding logic by hand, instead of using fzero() or fsolve() ?

[shiv gaur]

from muller method ans is different why is so

here is program

function kps3

T3 = 1e-9:1e-9:1e-6;

for j=1:numel(T3)

t3 = T3(j);

p0 = 0.5;

p1 = 1;

p2 = 1.5;

TOL = 10^-8;

N0 = 100; format long

h1 = p1 - p0;

h2 = p2 - p1;

DELTA1 = (f(p1,t3) - f(p0,t3))/h1;

DELTA2 = (f(p2,t3) - f(p1,t3))/h2;

d = (DELTA2 - DELTA1)/(h2 + h1);

i=3;

while i <= N0

b = DELTA2 + h2*d;

D = (b^2 - 4*f(p2,t3)*d)^(1/2);

if abs(b-D) < abs(b+D)

E = b + D;

else

E = b - D;

end

h = -2*f(p2,t3)/E;

p = p2 + h;

if abs(h) < TOL

%disp(p)

break

end

p0 = p1;

p1 = p2;

p2 = p;

h1 = p1 - p0;

h2 = p2 - p1;

DELTA1 = (f(p1,t3) - f(p0,t3))/h1;

DELTA2 = (f(p2,t3) - f(p1,t3))/h2;

d = (DELTA2 - DELTA1)/(h2 + h1);

i=i+1;

end

if i > N0

formatSpec = string('The method failed after N0 iterations,N0= %d \n');

fprintf(formatSpec,N0);

end

P(j)=real(p);

end

plot(T3,P)

end

function y=f(x, t3)

k0=(2*pi/0.6328)*1e6;

t2=1.5e-6;

n1=1.512;n2=1.521;n3=4.1-%i*0.211;

n4=1;

m=0;

m=0;

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;

end

why results are so different using while loop pl sort out the problem using this while loop

[shiv gaur]

I think root will be same for both program use while loop or muller