Estimate the error if $P_2 = 1-\frac{x^2}{2}$ is used to estimate the value of $cos(x)$ at $x = 0.6$

Question

This is how I tried to work through the problem:

$ |Error| < $

$ P_2 = 1-x^2/2 $

$n=0$ ␣␣ $ ƒ(x) = cox(x) $

$n=1$ ␣␣ $ ƒ^{'}(x) = -sin(x) $

$n=2$ ␣␣ $ ƒ^{''}(x) = -cox(x) $

$n=3$ ␣␣ $ ƒ^{'''}(x) = sin(x) $

Find M such that $ |ƒ^{(3)}| ≤ M, OnI [0,0.6]$

$cox(x)$ can't be more than 1, $M=1$

Plug everything into the Remainder Estimation Theorem

$ | Rn(x)| ≤ M \frac{|x-a|^{n+1}}{n+1!} $

$ a = 0 $, since centered at 0

$ R_2(0.6) ≤ (1.0) \frac{|0.6-0|^3}{3!} ≈ 0.036 $

$ ≈ 0.036 $ is what I'm getting as my answer but the platform I'm entering the answer into says it's wrong.

Answer 1

Hint: The term of degree three in the Taylor expansion of $\ \cos x\ $ is zero.

Answer 2

Note $\cos x = 1-\frac{1}{2!}x^2+ \frac{1}{4!}x^4-\>...$. So, the error estimation should be

$$R_3(0.6) \approx \frac{0.6^{3+1}}{(3+1)!}= 0.0054$$

Answer 3

This is an alternating series, so the error is less than the absolute value of the next term of the series which in this case is $$x^4 /{4!}=0.0054$$

The exact error is $0.005335616..$