The Newton-Raphson method for finding the root $z$ of a function $f(x)$ is given by the iterative formula:

$$x_{n+1} = x_n – \frac{f(x_n)}{f'(x_n)},$$

where $f'(x_n)$ denotes the derivative of $f(x)$ evaluated at $x_n$. Suppose that $z$ is a fixed point of the iteration, i.e., $z=x_{n+1}=f(x_n)$.

Then, using the mean value theorem, we can write:

$$f(x_n) – f(z) = f'(\xi_n)(x_n – z),$$

for some $\xi_n$ between $x_n$ and $z$. Rearranging this expression and substituting $z$ for $x_{n+1}$, we get:

$$x_{n+1} – z = \frac{f(x_n) – f(z)}{f'(x_n)} = \frac{f'(\xi_n)}{f'(x_n)}(x_n – z).$$

Taking absolute values on both sides and using the triangle inequality, we have:

$$\left|x_{n+1} – z\right| \leq \frac{\left|f'(\xi_n)\right|}{\left|f'(x_n)\right|} \left|x_n – z\right|.$$

Since $f'(z)=0$, we have $\left|f'(z)\right|=0$, so $\left|f'(x_n)\right|$ must be bounded away from zero for $n$ large enough, since the iterations approach $z$. Thus, we can assume that $\left|f'(\xi_n)\right|$ is bounded by some constant $M$ for all $n$. Therefore, we have:

$$\left|x_{n+1} – z\right| \leq \frac{M}{\left|f'(x_n)\right|} \left|x_n – z\right|.$$

Now, suppose that $\left|x_n – z\right| = h$, where $h$ is small enough so that the inequality above is valid. Then, we have:

$$\left|x_{n+1} – z\right| \leq \frac{M}{\left|f'(x_n)\right|} h.$$

If we can show that $\left|f'(x_n)\right|$ is bounded away from zero for all $n$ sufficiently large, then we have:

$$\left|x_{n+1} – z\right| \leq K h^2,$$

where $K$ is some constant. This would imply that the Newton-Raphson method converges quadratically.

To show that $\left|f'(x_n)\right|$ is bounded away from zero for all $n$ sufficiently large, we note that $f'(z) = 0$, so $f'(x_n)$ must approach zero as $n$ approaches infinity. However, since $z$ is a fixed point of the iteration, we have $x_n \rightarrow z$ as $n \rightarrow \infty$. Therefore, $f'(x_n)$ must change sign infinitely many times as $n$ increases, and hence it must cross zero infinitely many times. This implies that there exist $n$ sufficiently large such that $\left|f'(x_n)\right|$ is bounded away from zero.

Therefore, the Newton-Raphson method converges quadratically, i.e., $\left|x_{n+1} – z\right| \approx h^2$ for $h$ small enough.

Let $f_\mu(x) = \mu x(1-x)$ for $0 \leq x \leq 1$. Note that $f_\mu(x)$ is a continuous function on the interval $[0,1]$. We want to show that for $0 \leq x \leq 1$ and $0 \leq \mu \leq 1$, the sequence $(f_\mu^n(x))_{n=1}^\infty$ converges to $0$ as $n$ goes to infinity.

We will prove this result by induction on $n$. For the base case, note that $f_\mu(x) \leq x$ for $0 \leq x \leq 1$ and $0 \leq \mu \leq 1$. Therefore, $f_\mu(x)^n \leq x^n$ for all $n\ge 1$, and so $\lim_{n\rightarrow \infty} f_\mu^n(x) \leq \lim_{n\rightarrow \infty} x^n = 0$.

Now assume that $f_\mu^{n-1}(x) \rightarrow 0$ as $n\rightarrow \infty$ for $0 \leq x \leq 1$ and $0 \leq \mu \leq 1$. We want to show that $f_\mu^n(x) \rightarrow 0$ as $n\rightarrow \infty$.

Since $f_\mu(x)$ is continuous on $[0,1]$, it attains a maximum value $M$ on this interval. Note that $M \leq \frac{1}{4}$ for $0 \leq \mu \leq 1$, since $f_\mu\left(\frac{1}{2}\right) = \frac{\mu}{4}$ and $f_\mu(x)$ is symmetric around $x = \frac{1}{2}$. Therefore, we have:

\begin{align*} |f_\mu^n(x)| &= |f_\mu(f_\mu^{n-1}(x))| \ &\leq M|f_\mu^{n-1}(x)| \ &\leq M^2|f_\mu^{n-2}(x)| \ &\leq \cdots \ &\leq M^n|f_\mu(x)| \ &\leq M^n \cdot \frac{1}{4}. \end{align*}

Since $M < 1$, we have $\lim_{n\rightarrow \infty} M^n = 0$. Therefore, $\lim_{n\rightarrow \infty} f_\mu^n(x) = 0$ as well, completing the induction step.

Thus, we have shown that for $0 \leq x \leq 1$ and $0 \leq \mu \leq 1$, the sequence $(f_\mu^n(x))_{n=1}^\infty$ converges to $0$ as $n$ goes to infinity.

We will prove the statement by induction on $n$.

Base Case: $n=1$

If $n=1$, then any point $z$ in $[a,b]$ satisfies $f^1(z)=z$, since $f^1$ is just the identity function.

Inductive Step: Assume the statement holds for $n=k$, and we will show that it holds for $n=k+1$.

Let $z_0$ be a point in $[a,b]$ with period three under $f$. That is, $f^3(z_0)=z_0$ and $f(z_0) \neq z_0 \neq f^2(z_0)$.

We will construct a point $z_1$ with period $k+1$ under $f$ as follows:

Let $z_1=f^{k}(z_0)$. By the induction hypothesis, there exists a point $y$ in $[a,b]$ with period $k$ under $f$. That is, $f^k(y)=y$. Let $z_1=y$.

Then, we have $f^{k+1}(z_1)=f(z_1)$ since $f^{k+1}(z_1)=f(f^{k}(z_0))=f^{k+1}(y)=f(y)=z_1$. Therefore, $z_1$ has period $k+1$ under $f$.

Therefore, by induction, we have shown that for any $n$, there exists a point $z$ in $[a,b]$ with period $n$ under $f$.