a) Use base b = 10, precision k = 4, idealized, chopping floating-point arithmetic to show that fl(g(1.015)) is inaccurate, where g(x)=x14−1x−1 b) Derive the second order (n = 2) quadratic Taylor polynomial approximation for f(x)=x14 expanded about a = 1, and use it to get an accurate approximation to g(x) in part (a). c) Verify that your approximation in (b) is more accurate.

Question

avortarF · Accepted Answer

(a) Compute g(x) for x=1.015.g(1.015)=(1.015)1/4−11.015−1=0.248605929This is the exact value of g(x). In the ideal situation of chopping floating-point arithmetic, the operations are performed one at a time and the results are chopped after each operation. The floating-point number of the result with precision k=4 is fl(g(1.015))=fl((1.015)1/4)−1fl(1.015−1)=1.004−10.015=fl(1.004−1)0.015=fl(0.0040.015)=0.2667The absolute value of g(x) is 0.248605929 and the floating-point value is 0.2667. The machine precision for b=10 and k=4 is ∈mach=12b1−k=12101−4=5×10−4 The absolute error is |fl(g(1.015))−g(1.015)||g(1.015)|=|0.2667−0.248605929||0.248605929|≈0.073≈7.3%&amp;gt;5×10−4 Hence, fl(g(1.015)) is inaccurate. (b) The Taylor&#039;s polynomial of order 2 of a function f(x) around a point x = a is f(x)=f(a)+f′(a)(x−a)+f″(a)2!(x−a)2 Given f(x)=x1/4  ⇒f′(x)=14x3/4 and f′(x)=−316x74 We shall take an expansion around the point x = 1. After substituting and simplifying, we get the Taylor&#039;s quadratic polynomial as follows: f(x)=x14=1+14(x−1)−332(x−1)2 Now, approximate g(x) by using the Taylor`s polynomial f(x). fl(g(x))=fl(f(x)−1x−1)=fl(1+14(x−1)−332(x−1)2−1x−1)=fl(14(x−1)−332(x−1)2x−1)Substitute values and use idealized, chopping floating-point arithmetic.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

a) Use base b = 10, precision k = 4, idealized, chopping floating-point arithmetic to show that fl(g(1.015)) is inaccurate, where

Answered question

Answer & Explanation

New Questions in Algebra II