Here is the original problem
Gradient descent iteratively moves closer to a desired value (the minimum) of a function. Another algorithm in this spirit is known as Newton’s method, which is an iterative algorithm that finds the root of a function. Here, we consider Newton’s method which, given a function $f : ℝ → ℝ$, finds a value $x^*$ such that $f(x^* ) = 0$. The algorithm moves through a series of points $x^{(0)}, x^{(1)}, …$. If the algorithm is currently at a point $x^{(t)}$, then to find point $x^{(t+1)}$, it first takes the equation of the line tangent to the curve at $x=x^{(t)}$,
$y = f′(x^{(t)})(x – x^{(t)}) + f(x^{(t)})$.
It then uses the $x$-intercept of this line as the next point $x^{(t+1)}$.
a. Show that the algorithm described above can be summarized by the update rule
$$x^{(t+1)}=x^{(t)}-\dfrac{f(x^{(t)})}{f'(x^{(t)})}$$
We restrict our attention to some domain $I$ and assume that $f′(x) ≠ 0$ for all $x ∈ I$ and that $f''(x)$ is continuous. We also assume that the starting point $x^{(0)}$ is sufficiently close to $x^{*}$, where “sufficiently close” means that we can use only the first two terms of the Taylor expansion of $f(x^{*})$ about $x^{(0)}$, namely
$$f(x^{\ast})=f(x_0)+f'(x^{(0)})(x^{\ast}-x^{(0)})+\frac{1}{2}f''(\gamma^{(0)})(x^{\ast}-x^{(0)})^2\qquad(33.34)$$
where $γ^{(0)}$ is some value between $x^{(0)}$ and $x^*$. If the approximation in equation (33.34) holds for $x^{(0)}$, it also holds for any point closer to $x^*$.
b. Assume that the function $f$ has exactly one point $x^*$ for which $f(x^*) = 0$. Let $ε^{(t)} = |x^{(t)} – x^*|$. Using the Taylor expansion in equation (33.34), show that
$\epsilon^{(t+1)}=\dfrac{|f''(\gamma^{(t)})|}{|2f'(\gamma^{(t)})|}\epsilon^{(t)}$
where $γ^{(t)}$ is some value between $x^{(t)}$ and $x^*$.
Here is where I'm stuck
I now hope to solve the subproblem b. I hope to prove the case when $t=0$. Then I calculate $\epsilon^{(1)}$, I can get
$\begin{aligned} \epsilon^{(1)}&=|x^{(1)}-x^{\ast}|\\ &=\left|x^{(0)}-\dfrac{f(x^{(0)})}{f'(x^{(0)})}-x^{\ast}\right|\\ &=\left|x^{(0)}+\dfrac{f'(x^{(0)})(x^{\ast}-x^{(0)})+\frac{1}{2}f''(\gamma^{(0)})(x^{\ast}-x^{(0)})^2-f(x^{\ast})}{f'(x^{(0)})}-x^{\ast}\right|\\ &=\left|\dfrac{\frac{1}{2}f''(\gamma^{(0)})(x^{\ast}-x^{(0)})^2-f(x^{\ast})}{f'(x^{(0)})}\right|\\ &=\left|\dfrac{f''(\gamma^{(0)})(x^{\ast}-x^{(0)})^2}{2f'(x^{(0)})}\right|\\ &=\left|\dfrac{f''(\gamma^{(0)})}{2f'(x^{(0)})}\right|(\epsilon^{(0)})^2\\ &=\left|\dfrac{f''(\gamma^{(0)})}{2f'(\gamma^{(0)})}\right|\epsilon^{(0)}&\qquad(?)\\ \end{aligned}$
However, the step $(?)$ I don't know how to get. This means I need to prove that $f'(x^{(0)})=f'(\gamma^{(0)})\epsilon^{(0)}$, can anyone please prove this?