Given a Taylor approximation and an interval, how do we find the maximum error possible?

Question

Given an approximation of sin($x$) with a Maclaurin polynomial with degree $5$, ($x-\frac{x^3}{3!}+\frac{x^5}{5!}$), and given and interval $-0.3 < x < 0.3$, how do we find the maximum error?

Answer 1

We can refer to Taylor's series with Lagrange remainder, that is

$$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+\cdot\cdot\cdot+f^{(n)}(0)\frac{x^n}{n!}+R_n$$

with

$$R_n=\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}$$

Answer 2

The series is alternating, i.e.

i.e. $s = \sum_\limits{k=0}^\infty (-1)^k a_k$ and $0<a_{k+1}<a_k$

Each successive term in the series takes you from the positive side of s to the negative side.

That is, if $s_n$ is the partial sum of the first $n$ terms and $s_n > s$ then $s_{n+1} < s$

This implies $\epsilon = |s-s_n| < a_{n+1}$

It is a special case, and is generally more precise than the typical result from the remainder theorem.