The Bisection Method versus Newton s Method in Maple (Classic Version for Windows)

Transcription

1 The Bisection Method versus (Classic Version for Windows) Author: Barbara Forrest Contact: Copyrighted/NOT FOR RESALE version 1.1 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 Practice 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. 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

2 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 = 3x 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 + 3x 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 while F (0) = e 0 + 3(0) 2 4 = 3 < 0 F (1) = e > 0. Therefore, the IVT shows that there is at least one point c [0, 1] such that F (c) = 0. That is e c + 3c 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. In general, we now have an algorithm called the Bisection Method for finding approximate solutions of equations. This method is illustrated in the next section. ii

3 3 Bisection Method Suppose that we want to find an approximate solution to the equation where both f(x) and g(x) are continuous functions of x. f(x) = g(x) Step 1: Form the new continuous function. F (x) = f(x) g(x) Step 2: Find two points a 0 < b 0 such that either F (a 0 ) > 0 and F (b 0 ) < 0, or F (a 0 ) < 0 and F (b 0 ) > 0. The IVT now shows us that there is a c between a 0 and b 0 such that F (c) = 0. Step 3: Find the midpoint and evaluate F (d). d = a 0 + b 0 2 Step 4: If F (a 0 ) and F (d) have opposite signs (i.e., one negative and one positive), then let a 1 = a 0 and b 1 = d. Otherwise, let a 1 = d and b 1 = b 0 to obtain a new interval [a 1, b 1 ] which contains a solution to the equation. We also have that b 1 a 1 = 1 2 (b 0 a 0 ). Step 5: Repeat steps 3 and 4 to obtain new intervals [a 2, b 2 ], [a 3, b 3 ],...,[a n, b n ], each of which contains a solution to the equation. Then there is a c such that F (c) = 0 in each interval [a k, b k ]. Example [Bisection Method using Maple] Use the Bisection Method to approximate the root of the polynomial f(x) = x 5 + x 1. Start Maple. Begin a new worksheet by entering the restart: command. Enter the function f(x) in Maple. [> restart: [> f := x -> x 5 + x - 1; To get an idea of whether this function has a solution (root), look at plots of f(x) in Maple. Notice that we begin by using a large set of axes ranges, and then narrow the view by choosing a smaller and smaller x-range. [> plot( f(x), x= , y = ); [> plot( f(x), x=-4..4, y = ); [> plot( f(x), x=-1..1, y = ); iii

4 The plots indicate that f(x) crosses the x-axis in the interval [0, 1]. This will be our starting interval for the Bisection Method. You could also confirm that f(x) changes sign over this interval by checking: [> f(0); [> f(1); Note that Maple tells us that f(0) = 1 < 0 and f(1) = 1 > 0. This calculation confirms that the root c lies somewhere between 0 and 1. So, using the notation of the Bisection Method from before, let a = 0 and b = 1 (i.e., the endpoints of the interval). To assign values to a variable in Maple, we use the assignment operator :=. In Maple, enter the following to assign values to a and b : [> a := 0; [> b := 1.0; The midpoint d is found by calculating a+b 2. In Maple, this is: [> d := (a+b)/2; Maple tells us that the midpoint of the interval [0, 1] is d = 0.5. We must next find f(0.5) or equivalently, f(d) to continue with the algorithm and to find the side of the interval on which the midpoint c is located. [> f(d); Maple tells us f(d) = f(0.5) = Since f(a) = f(0) < 0 from above, the Bisection Method tells us to replace a by d. This means we must assign a the current value of d. [> a := d; Our current state in the Bisection Method is a = 0.5 and b = 1.0. We now know that c is located somewhere in the interval [0.5, 1.0]. To make sure we have not made any errors, let s try plotting f(x) in the interval [0.5, 1.0]. [> plot( f(x), x= ); (Indeed, we can see from the plot that c lies somewhere between 0.7 and 0.8.) To refine the search for c even further, we repeat this process with the new interval [0.5, 1.0]. The new midpoint is [> d := (a+b)/2; iv

5 Maple tells us that the new value of the midpoint d is We again find f(d). [> f(d); Maple tells us f(d) = f(0.75) = Since, from before, we have f(a) = f(0.5) = < 0, the Bisection Method tells us again to replace a by d. This means we must again assign a the current value of d. [> a := d; The current state in the Bisection Method is a = 0.75 and b = 1.0. We now know that c is located somewhere in the interval [0.75, 1.0]. We can keep repeating this process for a predetermined number of times or until the distance between the interval endpoints, a and b, is as small as we desire. However, this calculation is very tedious! Instead, we can use Maple s programming facilities to write a simple routine that contains a loop. WARNING: Use Maple CLASSIC version. Maple will give you the comment until you have finished typing all of the lines. Warning, premature end of input Try entering the following lines in Maple (you do not need to type the comments). BASIC BISECTION METHOD CODE [> restart: [> f := x -> x 5 + x -1; define the function f [> a := 0; assign the left-hand endpoint of the original interval to a [> b := 1.0; assign the right-hand endpoint of the original interval to b [> for n from 1 to 10 do do 10 iterations of the Bisection Method [> d := (a+b)/2; calculate the midpoint of the interval [> if ( evalf( f(a)*f(d) ) < 0) then b := d else a := d end if; check the sign of f at the midpoint [> end do; Maple returns the first 10 calculations of d, with the last calculated value as d = Maple tells us that c is approximately Let s check this value of c as the root by plotting: [> plot( f(x), x = ); We can also calculate f( ) to ensure its value is close to zero: [> f( ); It seems we have a very accurate estimate of the root c! v

6 4 Error in the Bisection Method Notice that in the previous example, the initial interval [a 0, b 0 ] = [0, 1] has length b 0 a 0 = 1 0 = 1. Hence, after n bisections, the current interval has length 1 2 n In general, since half of the interval length represents the maximum possible error in approximating c, after n bisections the maximum error in the approximation is b 0 a 0 2 n In the previous example, we ran the loop a predetermined number of times to find an approximation for c. However, we can modify this loop to stop when we reach a desired error level. Example [Bisection Method with Determined Error using Maple]: Suppose we wanted the approximation for c to have an error of at most Thus, we should stop the Bisection Method when n is large enough so that b 0 a 0 2 n < 10 4 To use Maple for this task, we add a new Step 6 to our previous algorithm for the Bisection Method: Step 6: Stop if b 0 a 0 2 n < error The modified loop in Maple is: BISECTION METHOD WITH DETERMINED ERROR CODE [> f := x -> x 5 + x -1; define the function [> a := 0; assign the left-hand endpoint of the original interval to a [> b := 1.0; assign the right-hand endpoint of the original interval to b [> L := b - a; L represents the initial length of the interval [> n := 0; n represents the number of Bisections executed [> e := 10 (-4); assign a value to the error represented by e [> while ( L /(2 n) > e ) do repeat Bisection Method untile error is less than e [> d := (a+b)/2; calculate the midpoint of the interval [> if ( evalf( f(a)*f(d) ) < 0) then b := d else a := d end if; check the sign of the midpoint [> n := n+1; counts the number of Bisections completed [> end do; Maple tells us that after n = 14 iterations of the Bisection Method, the approximated value d = is within 10 4 of the actual solution c. vi

7 OPTIONAL Challenge Exercise: This loop does not return any value of d if the initial interval length b 0 a 0 < desirederror. How would you modify the loop to deal with this issue? We have just seen that the Intermediate Value Theorem (Bisection Method) gives us a simple but effective method for finding approximate solutions to equations. Unfortunately, the method of bisection converges slowly. Instead, we now introduce Newton s Method, which is also very easy to describe and to program, but is much more efficient as a means of finding approximate solutions to equations. 5 Newton s Method Just as in the Bisection Method, Newton s Method allows us to solve equations of the type f(x) = 0 To see how this method works, we begin with the following simple case. Assume that f(x) = f(a) + m(x a) with slope m 0. This means that f(x) is a linear function whose graph passes through the point (a, f(a)). Suppose we wanted to find a point c such that f(c) = 0 (i.e., c is a root of f(x)). In this case, because the graph of f(x) is a line with non-zero slope, there is no need to estimate c since we can calculate it explicitly. We have This implies that 0 = f(c) = f(a) + m(c a). f(a) = m(c a). If m 0, we can divide both sides of the equation by m to get f(a) m = c a. Finally, adding a to both sides of the equation yields c = a f(a) m = a f(a) f (a) Therefore, if f(x) = f(a) + m(x a) and m 0, we can easily solve the equation f(x) = 0. Note: If m = 0, the graph is a horizontal line, so if f(a) 0, it does NOT cross the x-axis and NO such c exists. What do we do if the graph of f(x) is not a line? In this case we can use tangent line approximation (this is formally called linear approximation). The following steps outline a general method for using tangent lines to approximate the point c for f(c) = 0. Step 1: Pick a point x 1 that is reasonably close to a point c with f(c) = 0. (The IVT or a rough plot might be helpful in finding such an x 1.) Step 2: If f(x) is differentiable at x = x 1, then we can approximate f(x) by the function L x1 (x) = f(x 1 ) + f (x 1 )(x x 1 ) vii

8 The graph of L x1 (x) is the tangent line to f(x) at x = x 1. Since, for x values near x 1, f(x) = L x1 (x) it would make sense that the graphs of f(x) and L x1 (x) cross the x-axis at roughly the same place. Recall from above that L x1 (x) is a linear function. Therefore, if f (x 1 ) 0, the graph of L x1 (x) crosses the x-axis at the point x 2 where x 2 = x 1 f(x 1) f (x 1 ). Step 3: We now repeat the above procedure, replacing x 1 by x 2 and using the linear approximation at x 2, to get a new approximation for c. x 3 = x 2 f(x 2) f (x 2 ) The diagram shows that in this example, x 3 is very close to c. Continuing in this manner, we get a recursively defined sequence x n+1 = x n f(x n) f (x n ) where x n+1 is simply the point at which the tangent line to the graph of f(x) through (x n, f(x n )) crosses the x-axis. It can be shown that for most nice functions and reasonable choices of x 1, the sequence {x n } converges very rapidly to a number c with f(c) = 0. Still, it makes sense for us to ask: How do we know how good our approximation is? If we want to approximate c to k decimal places of accuracy, at least k decimal places must be carried throughout our calculations. The procedure stops when two successive terms, x n and x n+1, agree to k decimal places. In most cases, this will happen after only a few iterations because, as a rule of thumb, each iteration doubles the number of decimal places of accuracy. Indeed, Newton s Method is much more efficient than our previous Bisection Method for finding approximate solutions to equations since the Bisection Method requires roughly 4 iterations to improve the accuracy of the estimate by just 1 decimal place. viii

9 In the next example, Maple is used to demonstrate Newton s Method. Example []: Use Newton s Method to find all the roots of the equation f(x) = ln(4 x 2 ) x. Begin by drawing a graph to find the initial approximation, x 1. Start Maple. Begin a new worksheet by entering the restart: command. [> restart: Step 1: First define f(x) in Maple. [> f := x -> ln(4 - x 2) - x; Step 2: Determine an estimate for x 1, the first approximation for the solution (root) to f(x) = 0 by plotting f(x) in Maple. [> plot(f(x), x= ); The plot of f(x) indicates that f(x) = 0 has two roots one positive and one negative. Let s try to approximate the positive root which we will call c. After studying the graph, it looks like x 1 = 1 might be a good guess to approximate c. Let s check and see if f(1) is close to 0. [> f(1); Maple returns a symbolic answer. What is the numerical (decimal) approximation? [> evalf( f(1) ); Using x 1 = 1 gives us f(1) = Now is close to 0 so x 1 = 1 is a good approximation for c, but we can do better! Newton s Method shows us how. Step 3: Determine the next approximation x 2. Newton s Method tells us that the second approximation x 2 to f(x) = 0 is given by x 2 = x 1 f(x 1) f (x 1 ) Using the right-hand side of this equation, let s define a function in Maple called Newton that calculuates x n+1 = x n f(x n) f (x n ) We do this so we don t have to keep typing in the equation x - f(x)/d(f)(x);. ix

10 [> Newton := x -> x - f(x)/d(f)(x); Now using our first approximation of x 1 = 1, find the value of the next iteration of Newton s Method by using Newton. [> Newton(1); Notice that Maple returns a symbolic answer. In this case, we are looking for a numerical (decimal) approximation to the solution for f(x) = 0. So we will use Maple s evalf command. [> evalf( Newton(1) ); Maple tells us that the next approximation for c is x 2 = Let s see if f( ) is close to zero? [> f( ); Indeed, x 2 = is an even better approximation for f(x) = 0 than was x 1 = 1. Let s see if we can do better by doing another iteration of Newton s Method. This time we must calculate x 3 = x 2 f(x 2) f (x 2 ) But we already know that x 2 = and we have Newton, so we don t have to type this formula in again. [> evalf( Newton( ) ); Maple tells us that the next approximation for c is x 3 = Let s see if f( ) is even closer to zero? [> f( ); Maple tells us that when x 3 = is used as an approximation for c, we get f( ) = !!! This means that f( ) = which is very close to 0! We can stop now as we have just shown that x 3 = is a very good approximation for one solution of f(x) = 0. In real-world applications, the functions that are used are much more complicated and Newton s Method is a technique that scientists can use to quickly get solutions that have acceptable errors. As we did with the Bisection Method, we can also use Maple s programming facilities to write a simple routine that contains a loop to calculate the Newton Method iterates. Let s repeat the previous example using a Maple loop for Newton s Method. Try entering the following lines in Maple (you do not need to type the comments). Note: As before, Maple will give you the comment Warning, premature end of input until you type in all of the lines. NEWTON S METHOD CODE [> f := x -> ln(4 - x 2) - x; define the function [> x := 1; initial guess to estimate c [> k := 10; number of times you want to iterate Newton s Method [> for n from 1 to k do do k iterations of the Newton s Method [> x := evalf( x - f(x)/d(f)(x) ); calculate the next iterate [> end do; Notice that after the third iteration of Newton s Method, all of the values are the SAME! This means that three iterations of Newton s Method gives an estimate that is accurate up to 10 decimal places! You can now appreciate that Newton s x

11 Method is much more efficient (when it works!) at finding roots than the Bisection Method. WARNING: Newton s Method can FAIL under certain circumstances (see the practice exercises at the end of this lab). The Bisection Method, though inefficient, always works. 6 Practice Exercises Complete the following exercises in a Maple worksheet and write your answers in the space provided (where applicable). Save the worksheet as Newton.mws. Questions from these practice exercises may appear on your course assignments. 1. Create the following text input. At the top of a new Maple worksheet, type the following lines in a text region. Maple Lab : Bisection Method versus Newton s Method Name: Type your name here. ID: Type your ID here. Date: Type today s date here. You can enhance the fonts in any way you wish (e.g., change the font, fontsize, bold, etc.). Insert an execution group after the last line of the text region. Enter Maple s restart: command. 2. Complete the following steps to use the Bisection Method and Newton s Method to approximate the solution of f(x) = e x + 3x. You will then compare the efficiency of these two methods. a) In Maple, create a text region and enter the following sentence: Bisection Versus Newton s Method. Insert an execution group (command prompt) after the text region. b) In Maple, create the function f(x) = e x + 3x. HINT: remember that e x is exp(x) in Maple. c) Plot the graph of f(x) using the ranges x = and y = d) Study the plot you created. How many solutions (roots) does f(x) have? ANSWER: e) Using the Intermediate Value Theorem, explain why there is a root c in the interval [ 1.0, 0]. ANSWER: f) Edit the loop that was presented in this lab on page (vi) for the Bisection Method with Determined Error and enter it in Maple to find an estimate for c with an error of at most (Hint: You will have to edit the lines in the loop so that a := -1.0, b := 0 and e := 10 (-6). To have an error of at most 10 6 means to be accurate to 5 decimal places. ) ANSWER: ESTIMATE FOR C: g) How many iterates of the Bisection Method did it take to get the estimate for c with an error of at most ANSWER: NUMBER OF ITERATES: h) Using x 1 = 1 as your initial estimate, edit the loop provided on page (x) in this lab for Newton s Method and enter it in Maple, using 10 iterations to estimate the value for c. ANSWER: ESTIMATE FOR C: i) How many iterations of Newton s Method did it take for your estimate to be accurate to 6 decimal places? ANSWER: xi

12 j) Which of the two methods, Bisection Method or Newton s Method, is the most efficient? Explain your answer. ANSWER: 3. Complete the following steps to show that Newton s Method can fail if the choice of the initial estimate for c is poor. In this question, we will use the function f(x) = arctan(x) to show the importance of the initial choice for x 1 in Newton s Method. We know that the only solution to arctan(x) = 0 is x = 0 (look at the graph of arctan(x) to convince yourself!). We will show that if x 1 = 0.5, then the Newton s Method iterates converge quickly to the root c = 0. However, if x 1 = 2.0, then Newton s Method fails to find an accurate approximation for the root c. a) In Maple, create a text region and enter the following sentence: When Newton s Method Fails. Enter Maple s restart: command. Insert a new execution group (command prompt). b) In Maple, create the function f(x) = arctan(x). HINT: Use arctan(x) in Maple. c) Plot the graph of f(x) using the ranges x = Notice the only root at (0, 0). d) Using x 1 = 0.5 as your initial estimate, edit the loop provided in this lab for Newton s Method and enter it in Maple, using 10 iterations to estimate the value for c. ANSWER: ESTIMATE FOR C: e) Using x 1 = 2.0 as your initial estimate, edit the loop provided in this lab for Newton s Method and enter it in Maple, using 10 iterations to estimate the value for c. ANSWER: ESTIMATE FOR C: f) Study the output of the iterates for Newton s Method when you used x 1 = 2.0. What is happening to the successive estimates for c when you used x 1 = 2.0? ANSWER: ESTIMATE FOR C: g) By studying the graph for f(x) = arctan(x), explain why Newton s Method failed when you used x 1 = 2.0. HINT: What happens to the slope of the tangent lines to arctan(x) as x gets large? ANSWER: 4. Save the worksheet. Save the worksheet as Newton.mws. You may print the worksheet if you would like a copy for your notes. To print a copy of the worksheet, from the File menu, choose Print and click OK. xii