Lyapunov Exponent diagram of one variable out of 3D chaotic map

I am using the code following code to find the lyapunov exponent of variable x:

clear all; clc
rStart = 3.3;
rStep = 0.001;
rEnd = 3.8;
rValues = rStart:rStep:rEnd;  
nPoints = length( rValues );
nIterations = 1000; % number of iterations
LE      = zeros( 1, nPoints );
x       = zeros( 1, nIterations + 1 );
x(1) = 0.1;
y(1) = 0.002;
z(1) = 0.022;
lamda = 3.79;
beta = 0.01;
alpha = 0.0112;

for k = 1:nPoints
  sum = 0;
  for i = 1:nIterations
      
        x(i+1)=k*(lamda*x(i)*(1-x(i))+beta*((y(i)).^2)*x(i)+alpha*((z(i)).^3)); % compute and save next x
        y(i+1)=(lamda*y(i)*(1-y(i))+beta*((z(i)).^2)*y(i)+alpha*((x(i)).^3)); % compute and save next x
        z(i+1)=(lamda*z(i)*(1-z(i))+beta*((z(i)).^2)*z(i)+alpha*((y(i)).^3)); % compute and save next x
        sum = sum + log( abs( rValues( k ) - 2*rValues( k )*x( i ) ) );   
  end
  LE( k ) = sum / nIterations;     
end
figure; plot( rStart:rStep:rEnd, LE, 'k.-' ); 

But I get a single dot plot. any help!!!

[KSSV]

Check your code...your LE going to inf becuase (x,y,z) are going to inf. YOu have only one value, so it shows up as dot.

[tanveer haq]

Yes. you are right KSSV. But when i run only the middle for loop to generate points it work fine.

[KSSV]

Did you check the iteration update of x,y,z....it is correct?

[tanveer haq]

yes. it is working fine. if just the 1st for loop and k is removed.