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:
Method of Answering Question
\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: $\frac{.1^4}{4!}\approx. 4.1666666…*10^{-6}$
Note these results are from a TI-83 Plus, and its in float mode.
Is my answer right?
Best Answer
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.