# Reminder Estimation Theorem - $\cos(.1)$ Approximation using MacLaurin Series

I have to do the following approximation using the Reminder Estimation Theorem, and my question is my answer right?:

## $$\cos(.1) \approx \mathbf{some \ value}$$

(I have a calculator, and google, I know such value, and the units are in radians.)

Using the MacLaurin Series:

## $$\cos(x)=\sum_\limits{n=0}^\infty\dfrac{(-1)^n x^{2n}}{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}…\frac{(-1)^n x^{2n}}{2n}$$

I then decided to substitute in the $$.1$$ for the $$x$$, and got the following.

## $$\cos(.1)=\sum_\limits{n=0}^\infty\dfrac{(-1)^n .1^{2n}}{2n}=1-\frac{.1^2}{2!}+\frac{.1^4}{4!}…\frac{(-1)^n .1^{2n}}{2n}$$

The textbook that I am using gives the Reminder Estimation Theorem in the following format:

$$\vert R_n(x)\vert \le \frac{M}{(n+1)!}\vert x-x_o\vert^{n+1}$$

I then got it to the following method:

\begin{align} 0\le\vert R_n(x)\vert&\le\frac{(.1)^{n+1}}{(n+1)!}\le.000005 \end{align} Using the first two terms I then processed the answer as the following: $$1-\frac{.1^2}{2!}=.995$$ Because the third term gave me the following value: $$\frac{.1^4}{4!}\approx. 4.1666666...*10^{-6}$$
You are (partially) right. Either use the remainder, where $$M$$ is a bound on a derivative of $$\cos x$$ (thus 1), or see that it is a series of alternating sign terms, and the error is less than the first omited term.