Taylor Approximation of $\cos(0.02)$

Question

Use a Maclaurin $(a=0)$ polynomial for $\cos{(x)}$ with $3$ nonzero terms to approximate $\cos{(0.02)}$. Also, use the Taylor Remainder Theorem to find a bound on the error $\left(\displaystyle R_n=\frac{M|x-a|^{n+1}}{(n+1)!}: M\leq|f^{(n+1)}(x)|\right)$ in the approximation.

Since,

$f^{(0)}(0)=1 ,\ f^{(1)}(0)=0 ,\ f^{(2)}(0)=-1 ,\ f^{(3)}(0)=0 ,\ f^{(4)}(0)=1 ,\ f^{(5)}(0)=0 ,\ f^{(6)}(0)=1$

Then,

$\displaystyle \cos{(x)} \approx \left[T_5(x)=\frac{1\cdot x^0}{0!}+\frac{0\cdot x^1}{1!}+\frac{-1\cdot x^2}{2!}+\frac{0\cdot x^3}{3!}+\frac{1\cdot x^4}{4!}+\frac{0\cdot x^5}{5!}\right] \implies$

Hence,

$\cos{(0.02)}\approx \left[T_5(0.02)+R_5=1-\frac{(0.02)^2}{2!}+\frac{(0.02)^4}{4!}+\frac{1\cdot|0.02|^{6}}{6!}=0.9998000066667\overline{5}\right]$

Correct?

Answer 1

O.K. That's right you did exactly .