Calculus: Approximating Error / Taylor Series?

Question

The problem asks us to do the following:

Find the exact error in approximating ln[sec(0.2)] by its fourth degree Taylor polynomial at a = 0

Please show how to do this problem step-by-step.

Answer 1

ln sec x = ln(1/cosx) = -ln cosx
cos x = 1-x^2/2!+x^4/4!.. and
ln(1+z)= z-z^2/2 +z^3/3
so
ln cos = (-x^2/2+x^4/4!)-
(-x^2/2+x^4/4!)^2/2 and no more terms of 4rth degree
=-x^/2+x^4(1/24-1/8) and stop terms of 4rth degree
so -ln cos(0.2) = 0.04/2 +(0.2)^4 (1/12)=0.0201

Answer 2

? ? (cos a million)^ok ok=a million The sum of an limitless geometric sequence is a/(a million-r) a = the 1st term r = ratio between term and the previous term cos1/(a million-cos1) ? ? n^2 / e^n ok=3 This sequence is convergent. it fairly is shown with the shrink returned attempt to the l'hospitals rule. n^2 will derive to a persevering with on a similar time as e^2 will proceed to be. you're maximum stunning on the main suitable undertaking. The sequence will continually save getting extra effective effective via a minimum of a million and so it won't converge.