Use the Remainder Estimation Theorem to find an interval containing x=0 over which f(x) can be approximated by p(x) to three decimal-place accuracy throughout the interval.
f(x)=cosx p(x)=1 - x^2/2! + x^4/4!
Not when using this theorem.
Theorem:
if $\displaystyle S_n=\sum (-1)^nb_n$ is a sum of an alternating series
that satisfies
$\displaystyle 0 \le b_{n+1} \le b_n$ and $\displaystyle \lim_{n \to \infty}b_n =0$
Then the remainder $\displaystyle |R_n| \le b_{n+1}$
The question saidIt is not unique.to find an interval containing x=0
we could use any decimal that doesn't change in the thousandths column
The Macclaurin Series for cosine converges for all real x
$\displaystyle \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$
So by the theorm above we use the forth term (n=3)
$\displaystyle b_3=-\frac{x^6}{6!}$ we set this less than the error
$\displaystyle |b_3| < .0001$ and solve for x.
I hope this clears it up.
Good luck.
We set the term $\displaystyle b_3$ less then the error (x and all)
and solve the inequality for x. This gives us the distance x can be from the center of the series and get the accuracy we want. The Series is centered at zero.
since $\displaystyle x<.65$ we get the set $\displaystyle (-.65,.65)$
lets do a quick check
p(.65)=.796187
cos(.65)=.79608
The error is within what we wan't yeah!!
$\displaystyle \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=\underbrace{1-\frac{x^2}{2}+\frac{x^4}{4!}}_{p(x)}-\frac{x^6}{6!}+\frac{x^8}{8!}-...$
according to the AST remainder theorem we use the next term in the series
$\displaystyle -\frac{x^6}{6!}$ this is when n=3 in the sum we set that term less than the error.