Question

Taylor Series: To approximate the value of cos(7pi/12), what point would you expand the Taylor series about and why?
I know the series for cos(x) is summation from n=0 to infinity of ((-1)^n(x^2n))/(2n)! but I'm not quite sure how to solve this part of the question.

Answer 1

How far you expand it is dependent upon how accurate you want to be, like, to what decimal place of certainty you want.

Answer 2

It doesn't say the error I should be estimating by.. But I'm assuming I want the sums to be close to the value of cos (7pi/12)? How would I go about this?
I know with a given error I would use Taylor's inequality or alternating series approximation..

Answer 3

both \[\large 6 \pi /12=\pi/2 \text{ and } 8\pi/12=2\pi/3\] are equidistant from \[\large 7 \pi/12\]
however \[\large f^{(n)}(\pi /2)=\pm 1, 0\]\[\large f^{(n)}(2\pi /3)=\pm \frac{1}{2}, \pm \frac{\sqrt 3}{2}\] so the convenient choice is a taylor series about \[\large a = \pi/2\]

Answer 4

now, \[\large f^{(n)}(\pi/2)=0,\quad n \text{ even}\]and\[\large f^{(n)}(\pi/2)=(-1)^n,\quad n \text{ odd}\]therefore the taylos series expansion about pi/2 is \[\large f(x)=\cos x =\sum_{n=0}^{+\infty}(-1)^{n+1}\frac{ (x-\pi/2)^{2n+1} }{ (2n+1)! }\]

Answer 5

the maclaurin series expansion \[\large \cos x=\sum_{n=0}^{+\infty}(-1)^n \frac{ x^{2n} }{ (2n)! }\] is convergent for all real x, the convergence is faster when \[\large x \approx 0\] and \[\large 7\pi/12 >> 0\] making the maclaurin series a bad choice.

Answer 6

@atogo The problem is a test on deciding which special angle to use nearest 7pi/12.

Answer 7

Ah! Thank you very much! I suppose I was over thinking it and forgetting all about the unit circle. Thank you very much :)