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Math Help - Finding the Taylor polynomial and remainder

  1. #1
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    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?
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  2. #2
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    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
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