Bisection Problem Statement. Use bisection to solve the same problem approached graphically in Example 5.1. Solution. The first step in bisection is to guess two values of the unknown (in the present problem, c) that give values for f(c) with different signs. From Fig. 5.1, we can see that the function changes sign between values of 12 and 16. Therefore, the initial estimate of the root x, lies at the midpoint of the interval Xr 12 + 16 2 This estimate represents a true percent relative error of ɛ, = 5.3% (note that the true value of the root is 14.7802). Next we compute the product of the function value at the lower bound and at the midpoint: Xr = = = 14 f(12) f(14) = 6.067(1.569) = 9.517 which is greater than zero, and hence no sign change occurs between the lower bound and the midpoint. Consequently, the root must be located between 14 and 16. Therefore, we create a new interval by redefining the lower bound as 14 and determining a revised root estimate as 14 + 16 2 = 15 which represents a true percent error of & = 1.5%. The process can be repeated to obtain refined estimates. For example, f(14) f(15) = 1.569(-0.425) = -0.666

Bisection Problem Statement. Use bisection to solve the same problem approached graphically in Example 5.1. Solution. The first step in bisection is to guess two values of the unknown (in the present problem, c) that give values for f(c) with different signs. From Fig. 5.1, we can see that the function changes sign between values of 12 and 16. Therefore, the initial estimate of the root x, lies at the midpoint of the interval Xr 12 + 16 2 This estimate represents a true percent relative error of ɛ, = 5.3% (note that the true value of the root is 14.7802). Next we compute the product of the function value at the lower bound and at the midpoint: Xr = = = 14 f(12) f(14) = 6.067(1.569) = 9.517 which is greater than zero, and hence no sign change occurs between the lower bound and the midpoint. Consequently, the root must be located between 14 and 16. Therefore, we create a new interval by redefining the lower bound as 14 and determining a revised root estimate as 14 + 16 2 = 15 which represents a true percent error of & = 1.5%. The process can be repeated to obtain refined estimates. For example, f(14) f(15) = 1.569(-0.425) = -0.666

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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