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Math Help - Taylor polynomial question

  1. #1
    Super Member
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    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
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Paymemoney View Post
    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.
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