Jenny asked in Science & MathematicsMathematics · 2 months ago

# Approximate the function at the given value of x, using the Taylor polynomial of degree n = 3, centered at c = 8.?

Approximate the function at the given value of x, using the Taylor polynomial of degree

n = 3,

centered at

c = 8.

Relevance
• 2 months ago

f(8) + f'(8) * (x - 8) / 1! + f''(8) * (x - 8)^2 / 2! + f'''(8) * (x - 8)^3 / 3!

f(x) = x^(1/3)

f'(x) = (1/3) * x^(-2/3)

f''(x) = (-2/9) * x^(-5/3)

f'''(x) = (10/27) * x^(-8/3)

f(8) = 8^(1/3) = 2

f'(8) = (1/3) * 8^(-2/3) = (1/3) * 2^(-2) = (1/3) * (1/4) = 1/12

f''(8) = (-2/9) * 8^(-5/3) = (-2/9) * 2^(-5) = (-2/9) * (1/32) = -1/144

f'''(8) = (10/27) * 8^(-8/3) = (10/27) * 2^(-8) = (10/27) * (1/512) = 5/(27 * 256) = 5/6912

2 + (1/12) * (x - 8) / 1! + (-1/144) * (x - 8)^2 / 2 + (5/6912) * (x - 8)^3 / 6 =>

2 + (1/12) * (x - 8) - (1/288) * (x - 8)^2 + (5/41472) * (x - 8)^3

x = 8.07

2 + (1/12) * (8.07 - 8) - (1/288) * (8.07 - 8)^2 + (5/41472) * (8.07 - 8)^3 =>

2 + (1/12) * 0.07 - (1/288) * 0.07^2 + (5/41472) * 0.07^3 =>

(2 * 41472 + 0.07 * 3456 - 0.07^2 * 144 + 5 * 0.07^3) / 41472 =>

82944 + 0.07 * (3456 - 0.07 * 144 + 5 * 0.07^2)) / 41472 =>

(82944 + 241.216115) / 41472 =>

83185.216115 / 41472 =>

2.005 816 360797646604938271604938....

8.07^(1/3) = 2.005 816 401671469566765690923571.....

So we're good until 6 decimal places, and we're off by 4 * 10^(-8).  Not bad.

• 2 months ago

Given that polynomial, P~3 = 2 + .07/12 - .0049/288 + 5(.000343)/20736

= 2 + 0.00583… - 0.000017013 + 8.2×(10^-8) ≈ 2.006