How does the map $X_{n}^3=7X_{n-1}+6$ reach the fixed point -1?

Question

The map $X_{n}^3=7X_{n-1}+6 ....(1)$ gives rise to bounded sequence(s): $X_1, X_2, X_3, X_4,....X_{\infty}$. For large values of $n$ let $X_{n-1}=X_n=x \implies x^3-7x-6=0 \implies x=-1,-2,3.$ By numerics, We find that the three branches: $X_{n}=\sqrt[3]{7X_{n-1}+6}....(2), X_{n}= \omega \sqrt[3]{7X_{n-1}+6}....(3), X_{n}=\omega^2 \sqrt[3]{7X_{n-1}+6}....(4)$; where $\omega$ is cube root of unity, which for real or nonreal (complex) initial values for $X_1$ give rise to $X_{\infty}$ as $3,-2$ and $-2$, respectively avoiding -1, as a fixed point.

So the question is how do we reach the fixed point -1 for this map (1).

Answer 1

you chose one way to iterate finding a root of $x^3 - 7x - 6.$ Use Newton's instead

parisize = 4000000, primelimit = 500000
? x = -0.7
%1 = -0.7000000000000000000000000000
? x = ( 2 * x^3 + 6 ) / ( 3 * x^2 - 7 )
%2 = -0.9609403254972875226039783002
? x = ( 2 * x^3 + 6 ) / ( 3 * x^2 - 7 )
%3 = -0.9989460941317897304240340199
? x = ( 2 * x^3 + 6 ) / ( 3 * x^2 - 7 )
%4 = -0.9999991688603336875749119385
? x = ( 2 * x^3 + 6 ) / ( 3 * x^2 - 7 )
%5 = -0.9999999999994819060742966954
? x = ( 2 * x^3 + 6 ) / ( 3 * x^2 - 7 )
%6 = -0.9999999999999999999999997987
? x = ( 2 * x^3 + 6 ) / ( 3 * x^2 - 7 )
%7 = -1.000000000000000000000000000
?