Finding roots using recurrence relations Everyone knows the Fibonacci sequence:
$s[ ] = 1, 1, 2, 3, 5, 8, ...$
where 
$s_{n+2} = s_n + s_{n+1}$
This represents a single solution to the polynomial, $x^2 = x + 1$. 
Recurrence relations can be applied to find roots of other polynomials. For example, from the relation, $x^4 = 3 + 2x + x^2$ , a sequence $s$ can be constructed, where $s_{n+4} = 3\cdot s_n + 2\cdot s_{n+1} + s_{n+2}$. For example, $$s = 1, 1, 1, 6, 6, 11, 21, 41, 61, 116, 206, 361, 621, 1121, 1961, 3446, 6066, 10731, \ldots$$ The ratio of secessive terms approaches 1.76137782854929251, which is one of the roots (the largest one) of the given relation. 
How can this method be used to find more roots than 1?
 A: I am not sure regarding its relation to the Newton-Raphson method, but consider the function $f(x)=1+\frac 1 x$, and the recursively defined sequence
$$a_1=1,\qquad a_{n+1}=f(a_n), \quad \forall n\in \mathbb N $$.
Clearly, such sequences (i.e. defined recursively by a continuous function) can converge only to a fixed point of $f$.
That is, since if the series converges to $L\in \mathbb R$ then $$L=\lim_{n\to\infty} a_{n+1}=\lim_{n\to\infty}f(a_n)=f(\lim_{n\to\infty}a_n)=f(L).$$
An easy induction will show that $a_n=\frac{F_{n+1}}{F_n}$ where $F_n$ is the $n$'th Fibonacci number.
Because $f([1.1,2])\subset [1.1,2]$ and for $x>y\in [1.1,2]$
$$|f(y)-f(x)|=\bigg|\frac{x-y}{xy}\bigg|\leq\frac{1}{1.21}|x-y| $$
we see that $f$ is shrinking, and therefore every such iterative sequence (since $a_2=2\in[1.1,2]$) is Cauchy, and therefore converges to a value $L\in [1.1,2]$. This $L$ must be the unique fixed point of $f|_{[1.1,2]}$.
Obviously, this fixed point is the golden ratio $\phi=\frac{\sqrt5+1}{2}$!
