Approximating $\sin 2$ with an error of less than $0.001$

Question

How many terms of a Maclaurin Polynomial are needed to approximate $\sin 2$ with an error of less than $0.001$?

I'm pretty new to this stuff, but anyway, since the nth term of the polynomial $\sin x$ would be $\frac{2^{2n-1}}{(2n-1)!}$ (assuming n=1,2,3...), the n+1 th term would be $\frac{2^{2n+1}}{(2n+1)!}$, and that is less than $\frac{1}{1000}$. So cross multiplying, $2^{2n+1} \cdot 1000 < (2n+1)!$. Then it turns out that when $n=5$ the inequality first becomes true. So I thought the answer would be $4$ terms, but it is $5$ and I'm not sure why $n=5$ has to be included.

Answer 1

To estimate the error you need Taylor's expansion with Lagrange's remainder

$$\sin(x) = \sum\limits_{k=0}^n (-1)^k\dfrac{x^{2k+1}}{(2k+1)!} + R_n(x)$$

with

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

Answer 2

Let me give some example why you need the 5th term as well.

Say the 5th term is something like 0.00999, which is smaller than 0.001. After the 4th term you're off by some value (the error), which is then corrected by the 5th term. But still the result after the 5th term isn't entirely correct. Some $\epsilon$, which is the sum of the remaining terms, is missing.

This means that the error after the 4th term is $0.00999 + \epsilon$, which is likely to be greater than 0.001.

This means, looking at the size of a single term is not sufficient. You need to look at the sum of the remaining terms (or an upper bound of those).