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

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.

• $\sin (y + \pi) = - \sin y$ Nov 24, 2016 at 21:10
• It's pretty easy to see that $\frac{d^4}{dx^4}\sin(x) = \sin(x)$, so you only need to calculate the four cases that you have and then continue that pattern on.
– Tom
Nov 24, 2016 at 21:10
• @Tom Did you even read my post? Serious question. Daniel, I found that online as well, but I'm not sure how/why that applies here. Any other way to do it? Nov 24, 2016 at 21:12
• @AleksandrH It applies because the Maclaurin series of $\sin$ is easy (should be memorized, if you're a student). Nov 24, 2016 at 21:31

Hint

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

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

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

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

=

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

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.