with CLEAR WRITING AND EXPLAINATION! Consider the following function. f(x) = x^5
ID: 2883833 • Letter: W
Question
with CLEAR WRITING AND EXPLAINATION!
Consider the following function. f(x) = x^5/7, a= 1, n = 3, 0.7 lessthanorequalto x lessthanorequalto 1.3 Approximate f by a Taylor polynomial with degree n at the number a. T_3(x) = 1 + 5/7 middot (x - 1) - (10/49 middot (x - 1)^2)/2! + (90/343(x - 1)^3)/3! Use Taylor's Inequality to estimate the accuracy of the approximation f(x) TildeTilde T_n(x) when x lies in the given interval. (Round your answer to eight decimal places.) |R_3(x)| lessthanorequalto _______Explanation / Answer
Solution:
Using the definition of Taylor Series:
f(x) = x^(5/7) ==> f(1) = 1
f '(x) = (5/7)x^(-2/7) ==> f '(1) = 5/7
f ''(x) = (5/7)*(-2/7)x^(-9/7) ==> f ''(1) = (5/7)*(-2/7) = -10/49
f '''(x) = (5/7)*(-2/7)*(-9/7)x^(-16/7) ==> f ''(1) = (5/7)*(-2/7)*(-9/7) = 90/343
So, T3(x) = 1 + (5/7) (x - 1) - ((10/49)/2!) (x - 1)^2 + ((90/343)/3!) (x - 1)^3
The error is given by
|(1/4!) f ''''(c) (x - 1)^4| for some c between x and 1.
However,
|f '''(c)| = |(90/343)(-16/7) c^(-23/7)|
.........= (1440/(2401*c^(23/7)))
......... 1440/(2401*(0.7)^(23/7)), since c is bounded below by x, which is itself bounded below by 0.7
Hence, the error bound is
(1/4!) * (1440)/(2401*(0.7)^(23/7)) * |x - 1|^4
60/(2401*(0.7)^(23/7)) * (1.3-1)^4
60/(2401*(0.7)^(23/7)) * (0.3)^4
0.00065344
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