Find T 5 (x): Taylor polynomial of degree 5 of the function f(x) = cos(x) at a =

Question

Find T5(x): Taylor polynomial of degree 5 of the function f(x) = cos(x) at a = 0.
(You need to enter function.)
T5(x) =
Find the largest value of d for which Taylor's inequality (on page 754) can be used to show that, if |x| ? d, then | cos(x) ? T5(x) | ? 0.001744.
d =  (Give your answer to at least 4 significant digits.) Find T5(x): Taylor polynomial of degree 5 of the function f(x) = cos(x) at a = 0.
(You need to enter function.)
T5(x) =
Find the largest value of d for which Taylor's inequality (on page 754) can be used to show that, if |x| ? d, then | cos(x) ? T5(x) | ? 0.001744.
d =  (Give your answer to at least 4 significant digits.)

Explanation / Answer

A) Via cos t = 1 - t^2/2 + t^4/4! - t^6/6! + ..., letting t = 0.6 x yields
cos(0.6 x) = 1 - (0.6x)^2/2 + (0.6x)^4/4! - ...

==> T5(x) = 1 - 0.18x^2 + 0.0054x^4.

B) We apply the Taylor remainder formula:
Error = f^(6)(c) (x - 0)^6/6! for some c between 0 and x

However, f^(6)(c) = -(0.6)^6 cos(0.6c).

Thus, |Error|
= |-(0.6)^6 cos(0.6c) x^6 / 6!|
<= (0.6)^6 * 1 * x^6 / 6!, since |cos(0.6x)| <= 1 for all x.
<= (0.6)^6 * 1 * 1^6 / 6!, since the maximum of x^6 on [-1,1] is 1^6.
= (0.6)^6 / 6!.

Thus, we need (0.6)^6 / 6! < 1/10^k
==> 6! / (0.6)^6 > 10^k
==> k < log [6! / (0.6)^6] = 4.188

So, k can be no bigger than 4.

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