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].

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.