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25 April, 22:15

Use an appropriate Taylor polynomial about 0 and the Lagrange Remainder Formula to approximate sin (4/7) with an error less than 0.0001.

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  1. 25 April, 22:29
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    For a taylor polynomial

    f (x) = Sum a_n x^ (n)

    = Pn + epsilon

    epsilon = Sum a_n+1 (x-x_0) ^ (n+1)

    Lagrange remainder

    a_n = f^ (n) (x_0) / n!

    |epsilon| < ∫ [x_0 to x] f^ (n+1) (t) / (t - x_0) ^n / n! dt

    < max (|f^ (n+1) |) (x-x_0) ^n+1 / n+1!

    In the Taylor expansion of sin x

    sin x = x - x^3/3! ...

    find n such that epsilon < 0.0001

    epsilon < max (|f^ (n+1) |) (x-x_0) ^n+1 / n+1! < 0.0001

    for sin x, max (|f^ (n+1) |) < 1

    (6/7) < 1 so (6/7) ^n <1

    1 / (n+1) ! < 0.0001

    n+1! > 0.0001

    (6/7) ^7 / (n+1) ! < 0.0001

    the coefficient of x^6 = 0 in the expansion of sin x

    sin x = x - x^3 + x^5 + - 0.0001
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