I have observed when simulating a Tent map that its chaotic dynamics and attractor is not dense as that of Logistic map. Moreover, when simulating the Tent map, the plot of time series (plot of $x$ variable with respect to number of iterations) stops fluctuating and no values are observed after n=50. An example is shown in the figure. The image shows the time series plot of the state variable x and the graph is obtained in Matlab. This is only seen for the Tent Map. I have simulated other 1 D and other chaotic systems, but only for Tent map there is no value after a certain number of iterations. I have not found any information regarding such a behavior. Can somebody please tell me why this is happening? The initial value is randomly generated between 0 and 1.
You are experiencing an unfortunate interplay between computer numerics and dynamical systems.
In a binary representation $x$ the signicand part comes with a certain number of bits. Multiplying by two (and mapping the result back into $[0,1]$) shifts those bits, and the last bit becomes zero. When the significand at the starts is represented by 50 bits it becomes identically zero after 50 iterations.
This happens for a couple of dynamical systems where chaos comes from multiplication by 2, 4 ,8,...
The tent map as defined by $f(x) = 1 - |2x-1|$, on $[0,1]$ also works this way by successively putting least significant digit to zero and shifting to the left. Other examples are: $f(x) = 4x \; {\rm mod} \; 1$ (which removes two digits at the time, so it gets constant within 25 iterations only), or a more complicated one: $$ f(x) = \left\{ \begin{matrix} 4x & & x\in [0,1/4)\\ 2-4x & & x\in [1/4,1/2) \\ 2x-1 & & x \in [1/2,1] \end{matrix} \right. $$ Again at each iteration at least one more least significant binary digit becomes zero.
On the other hand, $f(x) = 3x \; {\rm mod} \; 1 $ is chaotic in binary representation (at least sufficiently for you to notice the computer errors). Similarly $f(x) = 4x(1-x)$ on $[0,1]$ is chaotic also in binary arithmetics. Indeed multiplying by 4 sets the two last digits to zero, but the product $x(1-x)$ will make these last two digits 'random' again. So apart from the last two digits (being zero) the numbers obtained on computer from $x_{n+1}=4 x_n(1-x_n)$ will appear random (with respect to a certain calculable distribution) when starting from a random initial number in $(0,1)$.