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.


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    $\begingroup$ Are you using IEEE 754 double floating point calculations by any chance? $\endgroup$
    – WimC
    Oct 2, 2017 at 5:54
  • $\begingroup$ Just to state that there is a chat room on Dynamical Systems and Chaos theory , in order to gain attention of users interested in the area of Dynamical Systems,Non linear Dynamics,Chaos theory! $\endgroup$
    – BAYMAX
    Oct 2, 2017 at 6:16
  • $\begingroup$ By default, MATLAB® uses 16 digits of precision. $\endgroup$
    – Adam
    Jun 28, 2019 at 17:23
  • $\begingroup$ commons.wikimedia.org/wiki/… $\endgroup$
    – Adam
    Jun 28, 2019 at 18:47
  • $\begingroup$ It should be noted that here the parameter of the tent map is 2 $\endgroup$
    – Adam
    Jun 28, 2019 at 19:29

2 Answers 2


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 $(0,1)$.

  • $\begingroup$ What is the name of this error ? Is it loss of significance ? $\endgroup$
    – Adam
    Jun 28, 2019 at 19:06
  • $\begingroup$ sciencedirect.com/science/article/pii/S1007570418300716 $\endgroup$
    – Adam
    Jun 28, 2019 at 19:23
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    $\begingroup$ @Adam I don't think it has a specific name. It is not really loss of significance, but rather a problem of (the binary) representation. When choosing your initial point "randomly" in fact it is very "non-random" for a dynamical system that multiplies by 2 but fine if you multiply by e.g. 3. The citation you give, (artificially) circumvene this problem by adding a noise on the last digits. Probably the cheapest way to get something which behaves as expected (but a mathematical rigorous treatment is not easy). $\endgroup$
    – H. H. Rugh
    Jun 30, 2019 at 9:26
  • $\begingroup$ iopscience.iop.org/article/10.1088/1742-6596/955/1/012025 $\endgroup$
    – Adam
    Jul 7, 2019 at 19:27

The question was answered, but I would like to give brute force solution based on above answer: increasing precision ( and digits of initial value) moves the problem to higher iterations.

One needs:

  • iMax binary digits
  • 5+iMax/3.6 decimal digits

Here iMax is maximal number of iterations

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