Finding the Taylor polynomial for f(x) = sin(x) at x = $\pi$

Question

I set up a little table for $n$, $f^{n}(x)$, $f^{n}(a)$, and $c_n$

n = 0: sin(x) ---- 0 ---- 0

n = 1: cos(x) ---- -1 ---- -1

n = 2: -sin(x) ---- 0 ---- 0

n = 3: -cos(x) ---- 1 ---- 1/6

n = 4: sin(x) ---- 0 ---- 0

I'm not sure how to generalize the pattern for the $f^{n}(x)$. It's clear that the even terms will all disappear because sin($\pi$) is always 0. For the odd terms, the sign alternates from positive to negative. Any hints and/or guidance in the right direction would be greatly appreciated.

Answer 1

Didn't really like the other answers here, but apparently there was an answer key/guide explaining how to solve this problem. It's pretty straightforward once you expand the sigma notation. We have:

n = 0: sin(x) ---- 0 ---- 0

n = 1: cos(x) ---- -1 ---- -1

n = 2: -sin(x) ---- 0 ---- 0

n = 3: -cos(x) ---- 1 ---- 1/3!

n = 4: sin(x) ---- 0 ---- 0

n = 5: cos(x) ---- -1 ---- -1/5!

f(x) = $\sum_{0}^{\infty} \frac{f^{n}(\pi)}{n!}(x-\pi)^{n}$

= $\frac{-1}{n!}(x-\pi)^1 + \frac{1}{3!}(x-\pi)^3 - \frac{1}{5!}(x-\pi)^5...$

=

$\sum_{0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!}(x-\pi)^{2n+1}$

Answer 2

Hint

with $f(x)=\sin(x)$

$$f^{(n)}(x)=\sin(x+n\frac{\pi}{2})$$

Answer 3

Instead of finding derivative of order n, you should write it as derivative of order (2n+1), then, in the taylor's formula, you should change other n's into 2n+1.