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23 April, 18:48

Let f (x) = (1 - x) - 1 and x0 = 0.

Find the nth Taylor polynomial Pn (x) for f (x) about x0. Find a value of n necessary for Pn (x) to approximate f (x) to within 10-6 on [0, 0.5].

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  1. 23 April, 22:04
    0
    n = 7

    Step-by-step explanation:

    Recalling Taylor nth polynimial, we have,

    Pₙ (x) = f (a) f ¹ (a) (x - a) + f ¹¹ (a) (x - a) ²/2 + f ¹¹¹ (a) (x - a) ³/3x2 + ... + f ⁽ⁿ⁾ (a) (x-a) ⁿ / n (n - 1) (n - 2) ... 3x2

    Therefore,

    If f (x) = (1 - x) ⁻¹,

    Then the Lagrange form of the remainder hold which sates that,

    (1 - x) ⁻¹ - Pₙ (x) = f ⁿ⁺¹ (c) x ⁿ⁺¹ / (n + 1) !

    Noting that,

    where C is between 0 and x,

    Hence therefore, the error is bounded by,

    / (0.5) ⁿ⁺¹ (1 - x) ⁻¹ ⁽⁰⁵⁾ : (n + 1) ! / ≤ 10⁻⁶

    These then equates to the fact that, that inequality first equates to 7.

    Therefore, n = 7.
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