My book writes that the Taylor series expansion for Newton's method is $$f(x^*) = f(x^{(n)}) + f'(x^{(n)})(x^*-x^{(n)})+\frac{1}{2!}f''(\xi^{(n)})(x^*-x^{(n)})^2 = 0,$$ where $x^*$ is the root, $x^{(n)}$ is the current guess, and $\xi^{(n)}$ is a value lying between $x^{n}$ and $x^*$.
What is the reason why we have the $f''(\xi^{(n)})$ term in the formula, as opposed to $f''(x^{(n)})$, as is the case for a general two-degree Taylor series expansion?
Usually, in a truncated Taylor series you have a remainder in the implicit form $R_n(x)=o((x-x^*)^n)$. Your book uses the explicit form: $$ R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x^*)^{n+1} $$
That's because, I guess, you're going to need some kind of estimate of the approximation error.