So I'm having trouble figuring this problem out so if someone can help me out that'd be great.
Find all the fixed points for the associated Newton iteration function for $$ f(x) = \frac{x}{(x-1)^n} $$ when $n = 1, 2, 3, .....$ Which are attracting and which are repelling?
$$f(x)={{x} \over {(x-1)^n}}$$ $$f^{'}(x)=\left({{-n} \over {x-1}}+1 \right) \cdot (x-1)^{-n}$$ Implementing Newton's method recurrence... $$x_{n+1}=x_n-{{f(x)} \over {f^{'}(x)}}$$ $$x_{n+1}=x_n+{{x_n \cdot (x_n-1)} \over {(n-1) \cdot x_n+1}}$$ A fixed point is defined as a value $v$ such that $x_1=v$ So solve for x... $$x=x+{{x \cdot (x-1)} \over {(n-1) \cdot x+1}}$$ Clearly, $x=1$, and $x=0$ are the fixed points
It might be enlightening to now investigate the periodic points of the dynamical system. A periodic point of period n is an $x_n=v$, so the solution oscillates between n values. An period "1" solution is just a typical fixed point...