# Range of stability for iterative map

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.

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Assuming your math is correct, doesn't this mean that the fixed point $x_1$ is unstable? – bill s May 18 '13 at 7:37
@bills I am not sure, that is also one of my doubts and questions. – user71346 May 18 '13 at 7:46

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.

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If your polynomial has numeric (as opposed to symbolic) coefficients, CountRoots[] is a useful thing. – J. M. May 18 '13 at 13:17