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nwtmaple - Lab 7 The Bisection Method and Newtons Method in...

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Lab 7 The Bisection Method and Newton’s Method in Maple (Classic Version for Windows) Author: B. A. Forrest ([email protected]) Copyrighted / r1.5 Contents 1 Objectives for this Lab i 2 Approximate Solutions of Equations ii 3 Bisection Method iii 4 Error in the Bisection Method vi 5 Newton’s Method vii 6 Lab 7 Exercises xi 1 Objectives for this Lab Understand the Bisection Method. Use Maple to aid in the calculations for the Bisection Method. Understand Newton’s Method. Use Maple to aid in the calculations for Newton’s Method. Use plots to investigate the Bisection Method and Newton’s Method in Maple. One of the most common tasks in mathematics is to find the solutions (roots) of equations. General solutions to linear and quadratic equations have existed for a long time. However, it has only been since 1545 AD (about 100 years before the “invention” of calculus) that algebraic solutions to cubic (3rd degree) equations were formalized. It is interesting to i
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Copyrighted by B.A. Forrest ([email protected]) Newton’s Method in Maple note that these solutions were discovered by Italian mathematicians through math contests! Although algebraic methods for finding solutions to quartic (4th degree) equations exist, they turn out to be very complicated expressions. In fact, it is possible to prove that there is no special formula for finding solutions to quintics (5th degree) or higher degree polynomial equations. Instead, calculus has given us a number of methods for determining approximate solutions to certain equations. Two of the methods that we will focus on in this lab are the Bisection Method and Newton’s Method . In this lab we use Maple to help with the calculations and visualization involved in applying these methods. 2 Approximate Solutions of Equations Suppose that we wanted to show that the equation e x = - 3 x 2 + 4 has a solution in the interval [0 , 1]. We could apply the Intermediate Value Theorem (IVT). To do so we first introduce the function F ( x ) = e x + 3 x 2 - 4 obtained by subtracting the function on the right-hand side of the equation from the function on the left-hand side. We then note that a point c is a solution of the equation if and only if F ( c ) = 0. However, F ( x ) is a continuous function (can you explain why?) and F (0) = e 0 + 3(0) 2 - 4 = - 3 < 0 while F (1) = e + 3 - 4 > 0 . Therefore, the IVT shows that there is at least one point c [0 , 1] such that F ( c ) = 0. That is e c + 3 c 2 - 4 = 0 . If we wanted a better idea of the location of c , we could bisect the original interval [0 , 1] at the midpoint 1 2 , and evaluate F ( 1 2 ). If F ( 1 2 ) < 0, then since F (1) > 0, we would have a solution in the new interval [ 1 2 , 1]. Otherwise, if F ( 1 2 ) > 0, c would be located in the interval [0 , 1 2 ]. We can then bisect the new interval, test the midpoint and repeat the procedure.
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