Originally Posted by

**Paymemoney** Hi

Can someone tell me if this is correct?

1) Find (exactly) the Taylor polynomial of degree four for:

$\displaystyle f(x) = cos(\frac{\pi(x)}{6})$ about x = 2

f(x) = $\displaystyle cos(\frac{\pi(x)}{6})$

f'(x) = $\displaystyle \frac{-\pi}{6}sin(\frac{\pi(x)}{6})$

f''(x) = $\displaystyle \frac{-\pi}{6}(\frac{\pi}{6}cos(\frac{\pi(x)}{6}))$

f'''(x) = $\displaystyle \frac{-\pi^2}{36}(\frac{-\pi}{6}sin(\frac{\pi(x)}{6}))$

f''''(x) = $\displaystyle \frac{\pi^3}{216}(\frac{\pi}{6}cos(\frac{\pi(x)}{6 }))$

f(2) = $\displaystyle cos(\frac{\pi(2)}{6}) = 0.5 $

f'(2) = $\displaystyle \frac{-\pi}{6}sin(\frac{\pi(2)}{6}) = \frac{-\sqrt{3}\pi}{12}$

f''(2) = $\displaystyle \frac{-\pi}{6}(\frac{\pi}{6}cos(\frac{\pi(2)}{6})) = \frac{-\pi^2}{72}$

f'''(2) = $\displaystyle \frac{-\pi^2}{36}(\frac{-\pi}{6}sin(\frac{\pi(2)}{6})) = \frac{\sqrt{3}\pi^3}{432}$

f''''(2) = $\displaystyle \frac{\pi^3}{216}(\frac{\pi}{6}cos(\frac{\pi(2)}{6 })) = \frac{\pi^4}{2592}$

Therefore this is the Taylor series i got:

$\displaystyle 0.5 - 0.45(x-2) - \frac{0.14(x-2)^2}{2} + \frac{0.12(x-2)^3}{6} - \frac{0.038(x-2)^4}{24}$

Note i rounded off the values from my derivatives to 2 decimal places.

P.S