# Math Help - Taylor polynomial question

1. ## Taylor polynomial question

Hi

Can someone tell me if this is correct?

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

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

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

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

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

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

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

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

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

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

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

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

Therefore this is the Taylor series i got:

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

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

P.S

2. Originally Posted by Paymemoney
Hi

Can someone tell me if this is correct?

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

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

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

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

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

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

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

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

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

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

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

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

Therefore this is the Taylor series i got:

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

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

P.S
What you have done, until the last line, is correct but why did you round off? Why did you change to decimal form at all?

$\frac{1}{2}- \frac{\sqrt{3}\pi}{24}(x- 2)- \frac{\pi^2}{144}(x-2)^2+ \frac{\sqrt{3}\pi^2}{2592}(x- 2)^3+ \frac{\pi^4}{62208}(x- 2)^4$ is exact whereas your answer is only approximate.