# Thread: Finding the Taylor polynomial and remainder

1. ## Finding the Taylor polynomial and remainder

I want to make sure I'm doing this right.

Find the third degree Taylor polynomial of $f(x) = \cos x$ centered at c $=$ $\frac {\pi}{6}$. Write the remainder in Lagrange form and find an upper bound for the remainder.

So $n = 3$ and $c = \frac{\pi}{6}$ right?

$f(x) = \cos x$; $f{(\frac{\pi}{6}}) = \frac{\sqrt{3}}{2}$
$f'(x) = -\sin x$; $f'{(\frac{\pi}{6}}) = -\frac{1}{2}$
$f''(x) = -\cos x$; $f''{(\frac{\pi}{6}}) = -\frac{\sqrt{3}}{2}$
$f'''(x) = \sin x$; $f'''{(\frac{\pi}{6}}) = \frac{1}{2}$
(error) $f^{(4)}(x) = \cos x$;

$P_3(\frac{\pi}{6}) = (\frac{\sqrt{3}}{2}) - (\frac{1}{2 * 1!} (x - \frac{\pi}{6})) - (\frac{\sqrt{3}}{2 * 2!}(x - \frac{\pi}{6})^2) + (\frac{1}{2 * 3!}(x - \frac{\pi}{6})^3)$

$R_3(x) = \frac{cos (c)}{4!}(x - \frac{\pi}{6})^4$, where $c \exists {[\frac{\pi}{6}, x]}$

I was confused about the part where it says to find an upper bound for the remainder. Does $c \exists {(\frac{\pi}{6}, x)}$ show the upper bound or do I need more?

2. ## Re: Finding the Taylor polynomial and remainder

$|Error| \le \frac{\left(x - \frac{\pi}{6}\right)^4}{4!}$

max possible value of $\cos(c) = 1$