MY NOTES ASK YOUR TEACHER EXAMPLE 1 (a) Approximate the function F(x) = V by a Taylor polynomial of degree 2 at a = 3. (

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answerhappygod
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MY NOTES ASK YOUR TEACHER EXAMPLE 1 (a) Approximate the function F(x) = V by a Taylor polynomial of degree 2 at a = 3. (

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My Notes Ask Your Teacher Example 1 A Approximate The Function F X V By A Taylor Polynomial Of Degree 2 At A 3 1
My Notes Ask Your Teacher Example 1 A Approximate The Function F X V By A Taylor Polynomial Of Degree 2 At A 3 1 (35.02 KiB) Viewed 15 times
MY NOTES ASK YOUR TEACHER EXAMPLE 1 (a) Approximate the function F(x) = V by a Taylor polynomial of degree 2 at a = 3. (b) How accurate is this approximation when 75xs 9? SOLUTION (a) f(x) F(x) == x1/3 = F(8) = 2 1 f'(x) = " F"(8) = = 3x F"(x) = -x-5/3 = F"(8) = 1 288 X 10 F"(x) = 27x = 1! Thus the second-degree Taylor polynomial is Tz(x) = f(8) + F(8) (x – 3) + F"(8) (x – 3)2 2 I + 1*- 8) The desired approximation is 2 (x-3)? T2(x) = S (b) The Taylor series is not alternating when x<8, so we can't use the Alternating Series Estimation Theorem in this example. But we can use Taylor's Inequality with n = 2 and a = 8: M 1R3(x) = x-812 3! where IF"(x) = M. Because x 27, we have x8/ 378/3, and so F"(x) = s (rounded to seven decimal places). Therefore we can take M = 0.0020656. Also 7 5X59, so -1 sx-851 and 1x - 8 s 1. Then Taylor's Inequality gives R2(x) = 13 0.0020656 3! 6 < (rounded to seven decimal places). Thus, if 75x59, the approximation in part (a) is accurate to within (rounded to seven decimal places).
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