Question

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].

Answer 1

n = 7

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.