Suppose that $f$ is an infinitely differentiable function on $\mathbb{R}$. Assume that there exist constants $0<C_{1}, C_{2}<\infty$ so that $\left|f^{\prime \prime}(x)\right| \geqslant C_{1}$ and $\left|f^{\prime \prime \prime}(x)\right| \leqslant C_{2}$ for all $x \in \mathbb{R}$. Fix $x_{0} \in \mathbb{R}$ and for each $n \in \mathbb{N}$ set

$$x_{n}=x_{n-1}-\frac{f^{\prime}\left(x_{n-1}\right)}{f^{\prime \prime}\left(x_{n-1}\right)} .$$

Let $x^{*}$ be the unique value of $x$ where $f$ attains its minimum. Prove that

$$\left|x^{*}-x_{n+1}\right| \leqslant \frac{C_{2}}{2 C_{1}}\left|x^{*}-x_{n}\right|^{2} \quad \text { for all } n \in \mathbb{N} .$$

[Hint: Express $f^{\prime}\left(x^{*}\right)$ in terms of the Taylor series for $f^{\prime}$ at $x_{n}$ using the Lagrange form of the remainder: $f^{\prime}\left(x^{*}\right)=f^{\prime}\left(x_{n}\right)+f^{\prime \prime}\left(x_{n}\right)\left(x^{*}-x_{n}\right)+\frac{1}{2} f^{\prime \prime \prime}\left(y_{n}\right)\left(x^{*}-x_{n}\right)^{2}$ where $y_{n}$ is between $x_{n}$ and $\left.x^{*} .\right]$