Problem regarding convergence of second order

Question

Show $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with start value $x_0=1$ and $f:x\to x^5$ does not converge quadratically.

So by definition I have to show that $\mid x_{n+1}-0\mid\le c \mid x_n -0\mid^2$ for quadratic convergence, I guess. Plugging in leads to $\mid x -\frac{f(x)}{f'(x)}\mid=\mid x-\frac x5\mid=\frac45 \mid x\mid\le c \mid x \mid^2$ And since $x$ will get never negative in this iteration we can write above as $\frac45 x\le c x^2 \Rightarrow \frac45 \le cx$, but with this reasing I could set $c=\frac 1x$ and therefore this convergence would be quadratic. I am oviously on the wrong track. Could someone help me and guide me, what exactly I have to do here?

Answer 1

$$x_{n+1}=\dfrac{4}{5}x_n$$so obviously $$\lim_{n\to\infty}x_n=0$$starting from any initial condition, but $x_{n+1}$ is not proportional to $x_n^2$ so the convergence is not quadratic.

Answer 2

Suppose $f(x) = x^m$ with $m \gt 1$. Then $f'(x) = mx^{m-1} $.

If $x_{n+1} =x_n-\frac{f(x_n)}{f'(x_n)} $ then $x_{n+1} =x_n-\frac{x_n^m}{mx_n^{m-1}} =x_n-\frac{x_n}{m} =x_n(1-\frac{1}{m}) =x_n(\frac{m-1}{m}) $ so that $\dfrac{x_{n+1}}{x_n} =\frac{m-1}{m} \gt 0 $ and the convergence is linear.

The fact that $\dfrac{x_{n+1}}{x_n} =\frac{m-1}{m} \lt 1 $ shows that the iteration converges from any starting point.