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0 = $f(x_{r}) = f(x_{n}) + f'(x_{n})(x_{r}-x_{n}) + \frac{f''(\xi)}{2}(x_{r}-x_{n})^{2}$
I came across a Taylor series expansion as shown in above equation while studying proof for convergence of Newton-Raphson method. I am not able to understand why the series has been terminated after writing 3 terms only and what exactly has $\xi$ to do in the third term. $\xi$ is a number between $x_{r}$ and $x_{n}$.
The last term is the remainder term of the Taylor series truncated to first order. Say you have a function $f(x)$. $\textit{Formally,}$ you can Taylor expand and write $f(x)=T_k(x)+R_k(x),$ where $T_k(x)$ is the $k$-th degree Taylor polynomial and $R_k(x)$ is the $k$-th remainder term. In your example, $k=1$, and you have a linear Taylor polynomial with the Taylor remainder being the last term involving the derivative evaluated at $\xi$, i.e.
$$R_1(x)=\frac{f''(\xi)}{2}(x_r-x_n)^2$$
for some $\xi\in(x_r,x_n).$
