Taylor & Maclaurin Series...?

Question

let g(x)=tanx

a.) write the second degree Taylor polynomial for g(x) about x=pie/4

b.) estimate g(0.75) using the quadratic Taylor polynomial in part (a.)

Answer 1

f(x)

= (x-a)^0 / (0!) + f'(a) * (x-a)^1 / (1!) + f''(a) * (x-a)^2 / (2!)

= tan(pi/4) * (x-pi/4)^0 / 1+ sec^2(pi/4) * (x-pi/4)^1 / 1 + 2 * sec^2(pi/4) * tan^2(pi/4) * (x-pi/4)^2 / 2

= tan(pi/4) + sec^2(pi/4) * (x-pi/4) + sec^2(pi/4) * tan^2(pi/4) * (x-pi/4)^2

= 1 + 2 (x-pi/4) + 2 (x-pi/4)^2

And then just plug in 0.75.

Answer 2

tanx = 1+(x-pi/4)*2 +(x-pi/4)^2 /2! *4
tan(0.75) =1-0.070796 +0.002506