# Taylor Series Expansion only upto two terms.

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 $$\xi\in(x_r,x_n).$$
• I wouldn't say $\xi$ alone is the remainder, rather the term involving it, with $\xi$ restricted but unknown in an interval. Jun 9, 2021 at 10:45
• I of course didn't intend to convey $\xi$ itself is the remainder. I meant the term with $\xi$. Edited the answer for clarity. OP already specified where $\xi$ must live, and this is also included in the definition of Taylor remainder. Jun 9, 2021 at 10:49