Using linear stability analysis, I would like to compute the range of stability of the fixed points and the 2-cycles of the following iterative map: $x_n = x_{n-1}^{2} - 3\mu$.

Setting $x = x^{2} - 3\mu$, using hands and Mathematica I calculated the fixed points: $x_1 = \frac{\sqrt{1+12\mu}}{2}$ and $x_2 = \frac{\sqrt{1-12\mu}}{2}$.
And the two cycles are the fixed points and two other additional points: $x_3 = \frac{-1-\sqrt{3(-1+4\mu)}}{2}$ and $x_4 = \frac{-1+\sqrt{3(-1+4\mu)}}{2}$.

Now I am struggling to find the range of stability of the fixed points and 2-cycles using hands and Mathematica.
What I know is we need to compute the slope of the map at the points. So if we differentiate $x^{2} - 3\mu$ and calculate the derivative at the fixed point $x_1 = \frac{\sqrt{1+12\mu}}{2}$, we get $2(x_1) = 1+\sqrt{1+12\mu}$. And similarly for $x_2$, $2(x_2) = 1-\sqrt{1+12\mu}$. Then we set the absolute values < 1. So we get $|1+\sqrt{1+12\mu}| < 1$ and $|1-\sqrt{1+12\mu}| < 1$. Now here is the problem, I can only find the range of stability for $x_2$ but not for $x_1$. Here is my (hands) work:
$$|1-\sqrt{1+12\mu}| < 1$$ $$-1<1-\sqrt{1+12\mu} < 1$$ $$-2<-\sqrt{1+12\mu} < 0$$ $$-\frac{1}{3} < \mu < \frac{3}{13}$$

But now I try to find the range of stability of $x_1$:
$$|1+\sqrt{1+12\mu}| < 1$$ $$-1<1+\sqrt{1+12\mu} < 1$$ $$-2<\sqrt{1+12\mu} < 0$$ and there is no solution for this inequality.

So what can we conclude for the range of stability? And how can I solve this problem using Mathematica? One way of doing it using Mathematica that I can think of is by drawing the cobweb diagram and checking for some values of $\mu$, but this method is exhaustive and time consuming. Is there any more efficient method?

Any helps on methods and code are greatly appreciated. Many thanks in advance.

  • 1
    $\begingroup$ Assuming your math is correct, doesn't this mean that the fixed point $x_1$ is unstable? $\endgroup$
    – bill s
    Commented May 18, 2013 at 7:37
  • $\begingroup$ @bills I am not sure, that is also one of my doubts and questions. $\endgroup$
    – user71346
    Commented May 18, 2013 at 7:46

1 Answer 1


You can find the roots of your system:

Roots[x^2 - x - 3 μ == 0, x]

which gives

x == 1/2 (1 - Sqrt[1 + 12 μ]) || x == 1/2 (1 + Sqrt[1 + 12 μ])

Indeed these are the two values you found. For stability, you are asking when the derivative of $x^2-3 \mu$ evaluated at the roots is less than 1. Since the derivative is $2 x$, you are asking when |2 x| < 1 with x taking on each of the values. As you see, one of these has a solution and the other does not. This means that one root is stable and the other is an unstable root.

  • $\begingroup$ If your polynomial has numeric (as opposed to symbolic) coefficients, CountRoots[] is a useful thing. $\endgroup$ Commented May 18, 2013 at 13:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.