“Teach A Level Maths” Vol. 2: A2 Core Modules 17: Iteration using x g( x ) © Christine Crisp Iteration Module C3 "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" Iteration There are some equations that we can’t solve. e.g. x 1 x 3 However, we can find an approximate solution to some of these equations. The approximation can be very accurate, say to 6 or more decimal places. There are several methods of finding approximate solutions and in this presentation we will study one of them. Iteration There are 2 stages to getting a solution: Stage 1. Find a 1st estimate Stage 2. Find a formula to improve the estimate. Sometimes we can just spot an approximate solution to an equation. Can you spot the approximate value of the solution to x 2x 1 ? 3 Ans: It’s quite close to 0 as the l.h.s. is then 0 and the r.h.s. is 1. If we can’t quickly spot an approximation, we can use a method involving finding bounds for the solution. Iteration Bounds are numbers which lie on either side of the solution. If these are integers we call them integer bounds. e.g. To find integer bounds for x 1 x 3 we can sketch y x and y 1 x 3 y 1 x 3 y x This is the point . . . where y x and y 1 x 3 so the x coordinate gives the solution to x 1 x 3 Iteration Bounds are numbers which lie on either side of the solution. If these are integers we call them integer bounds. e.g. To find integer bounds for x 1 x 3 we can sketch y x and y 1 x 3 y 1 x 3 y a x The solution lies between 0 and 1. 0 and 1 are the integer bounds. We usually call the solution a, so 0 a 1 . Iteration Bounds are numbers which lie on either side of the solution. If these are integers we call them integer bounds. e.g. To find integer bounds for x 1 x 3 we can sketch y x and y 1 x 3 y 1 x 3 y a x The solution lies between 0 and 1. 0 and 1 are the integer bounds. We usually call the solution a, so 0 a 1 . Our first approximation to a, is any number between the bounds, say 0 5. Iteration If we can use Autograph or a graphical calculator, sketching is a good method of finding the integer bounds. Even without a graph plotter you may have been able to sketch these 2 graphs. However, if we can spot likely bounds, or if we are given values and want to show they are bounds, we can use the algebraic method that follows and avoid sketching. Iteration Rearrange the equation to get zero on one side. e.g. For x 1 x 3 , get x 3 1 x 0 Call this f ( x) 0 To show how the method works I’m going to sketch y f ( x ) ( but you won’t usually have to do this ). 3 The solution, a, is f ( x) x 1 x now where f ( x) 0 a At a, f ( x ) 0 To the left of a, e.g. at x = 0, f ( x ) 1 0 To the right of a, e.g. at x = 1, f ( x ) 1 0 Iteration Rearrange the equation to get zero on one side. e.g. For x 1 x 3 , get x 3 1 x 0 Call this f ( x) 0 To show how the method works I’m going to sketch y f ( x ) ( but you won’t usually have to do this ). f ( x) x 1 3 a x f ( x ) has opposite signs on the left and right of a. To the left of a, e.g. at x = 0, f ( x ) 1 0 To the right of a, e.g. at x = 1, f ( x ) 1 0 Iteration The Algebraic Method: If we want to show that 0 and 1 are integer bounds, we show that f ( x ) has different signs at 0 and 1. • Rearrange the equation to the form f ( x ) 0 x 1 x x 1 x 0 Define f ( x ): Let f ( x ) x 3 1 x Find f ( 0) : e.g. • • 3 3 Iteration The Algebraic Method: If we want to show that 0 and 1 are integer bounds, we show that f ( x ) has different signs at 0 and 1. • Rearrange the equation to the form f ( x ) 0 x 1 x x 1 x 0 Define f ( x ): Let f ( x ) x 3 1 x Find f ( 0) : f (0 ) 0 3 1 0 1 Change of sign 3 Find f ( 1 ) : f (1 ) 1 1 1 1 e.g. • • • • 3 The change of sign 3 Iteration The Algebraic Method: If we want to show that 0 and 1 are integer bounds, we show that f ( x ) has different signs at 0 and 1. • Rearrange the equation to the form f ( x ) 0 x 1 x x 1 x 0 Define f ( x ): Let f ( x ) x 3 1 x Find f ( 0) : f (0 ) 0 3 1 0 1 Change of sign 3 Find f ( 1 ) : f (1 ) 1 1 1 1 e.g. • • • • 3 3 The change of sign 0 a 1 You must of always line. Our 1st estimate a is include betweenthis these values, say 0 5 Iteration It is possible to find bounds that are closer than the nearest integers. For example, to find bounds accurate to 1 decimal place, we could use a decimal search. So if we had 0 a 1 , a decimal search would calculate f ( 0 1 ), f ( 0 2 ), f ( 0 3 ) . . . looking for a change of sign. However, integer bounds are good enough for the method of iteration we are studying. Iteration e.g. 1 (a) Using a graphical calculator, or otherwise, sketch, on the same axes, the graphs of 3 y x 2 and y 5 x (b) From the sketch, find integer bounds for the 3 solution, a, of the equation x 2 5 x (c) Use an algebraic method to confirm these are correct and give a 1st approximate solution. Solution: y 5x (a) (b) a is the x-value at the point of 3 intersection, so 0 yx 2 and 1 are integer bounds. a Iteration e.g. 1 (a) Using a graphical calculator, or otherwise, sketch, on the same axes, the graphs of 3 y x 2 and y 5 x (b) From the sketch, find integer bounds for the 3 solution, a, of the equation x 2 5 x (c) Use an algebraic method to confirm these are correct and give a 1st approximate solution. (c) ( Confirm bounds are 0 and 1 ) • Rearrange equation to f ( x ) 0 : x 3 5 x 2 0 • • • Define f ( x ) : So, f ( x) x 5 x 2 f ( 0) 2 0 f (1) 1 5 2 2 0 The change of sign 0 a 1 3 A 1st approximation is any number between 0 and 1. Iteration Not all functions f ( x ) that have a sign change between 2 numbers have a solution to f ( x ) 0 between the numbers. Look at this function: f ( 0 ) 2 , y f ( x) x 1 2 f (1) 2 There is a change of sign . . . but no solution between 0 and 1. Do you notice anything that explains this? Ans: The function has an asymptote between 0 and 1. We couldn’t draw the curve without lifting the pencil off the paper. We say it is discontinuous. Iteration Not all functions f ( x ) that have a sign change between 2 numbers have a solution to f ( x ) 0 between the numbers. Look at this function: f ( 0 ) 2 , y f ( x) x 1 2 f (1) 2 There is a change of sign . . . but no solution between 0 and 1. Do you notice anything that explains this? You are unlikely to meet discontinuous functions in this work so just remember the effect on solutions. Iteration Exercise 1. Using a graphical calculator or otherwise, sketch suitable graphs to find integer bounds for the solution to x 1 e 2 x . Give a 1st approximation to the solution. 2. Use the algebraic method to show that ln x 4 x has a root between 2 and 3. Give an approximation to this root. An equation has a solution which may consist of one or more roots. Iteration 1. x 2 1 e x Solution: Sketch y x 2 1 and y e x y x 1 2 ye x a The integer bounds for a are 1 and 2. So, 1a 2 Any number between 1 and 2 could be used as the 1st approximation. Iteration 2. Use the algebraic method to show that ln x 4 x has a root between 2 and 3. Solution: • Rearrange to • ln x 4 x f ( x) 0: ln x x 4 0 Define f ( x ) : f ( x ) ln x x 4 f ( 2) ln 2 2 4 1 3 f ( 3) ln 3 3 4 0 1 • Change of sign (continuous function) 2 a 3 Any number between 2 and 3 could be used as the 1st approximation. Iteration The 1st approximate solution lies anywhere between the bounds. The next stage is to improve this estimate. • Rearrange the equation to the form x g( x ) . You may spot lots of ways of doing this. e.g. e.g. For the equation 3 x 1 x : Iteration The 1st approximate solution lies anywhere between the bounds. The next stage is to improve this estimate. • Rearrange the equation to the form x g( x ) . You may spot lots of ways of doing this. e.g. e.g. For the equation (i) Square: 3 x 1 x : 3 2 3 x 1 x x (1 x ) or (ii) Rearrange: x 1 x Cube root: 3 x 1 3 x x 1 x or (iii) Rearrange: x 1 x 3 x 3 1 x 1 x Divide by x : 2 x x 2 1 3 Iteration Let’s take the 2nd arrangement: x 1 x 1 3 Our 1st estimate of a we will call x0. We substitute into the of is the formula and ) ( Some peoplex0start withr.h.s. x1 which just as good. the result gives the new estimate x1. We now have x1 1 x0 1 3 We will then keep repeating the process so we write the formula as 1 x n 1 1 x n 3 This is called an iterative formula. to iterate means to repeat Iteration So, x n 1 1 xn Starting with x 0 0 5 x1 1 x0 1 3 1 3 we get x1 1 05 1 3 x1 0 664105 ( 6 d.p. ) Because we are going to repeat the calculation, we use the ANS function on the calculator. • Type 0 5 and press ENTER • Type the r.h.s. of the equation, replacing x with ANS, using the ANS button, giving 1 ANS • • Press ENTER and you get 0 664105 ( 6 d.p. ) Pressing ENTER again replaces 0 5 with 0 664105 and gives the next estimate and so on. 1 3 We get Iteration x0 0 5 x1 0 664105 x 2 0 569878 Although I’ve only written down 6 decimal places, the calculator is using the greatest possible accuracy. If we continue to iterate we eventually get a 0 605423 ( to 6 d.p. ) Error Bounds Since the answer is correct to 6 decimal places, the exact value of a must be within 6 of our answer. 0 0000005 or 0 5 10 Tip: The index equals the number of d.ps. in the answer. Iteration SUMMARY To find an approximation to a solution ( or root ) of an equation: Find a 1st approximation, often by finding integer bounds for the solution. Let this be x0 . Rearrange the equation into the form x g ( x ) Write the arrangement as an iterative formula: x n1 g( x n ) Key x0 into a calculator and ENTER. Key the r.h.s. of the formula into the calculator, replacing x with ANS. Press ENTER as many times as required to get the solution to the specified accuracy. Iteration e.g. 1(a) Show that the equation 2 x x 3 has a root a in the interval 1 3 a 1 4 . (b) Using the arrangement x 3 2 x write down an iterative formula and use it to find the root correct to 4 decimal places. Solution: (a) Let f ( x ) 2 x x 3 0 Then, f (1 3) 2 f (1 4) 2 1.3 1 3 0 27 1.4 1 4 0 10 3 3 Change of sign 1 3 a 1 4 Iteration e.g. 1(a) Show that the equation 2 x x 3 has a root a in the interval 1 a 2. (b) Using the arrangement x 3 2 x write down an iterative formula and use it to find the root correct to 4 decimal places. Solution: (b) Let x 3 2 gives x x 0 1 5 x1 2 0 1 414 3 x x n1 3 2 3 2 xn 15 x 2 1 3865 x 3 1 3776 It takes about 7 iterations to reach a 1 3735 (4 d.p.) Iteration Exercise 1. (a) Show that the equation x 3 x 2 25 has a solution a between 2 and 3. 2 (b) Use the iterative formula x n1 3 25 x n with x 0 2 5 to find the solution, giving your answer correct to 4 d.p. 2. (a) Show that the equation ln x 2 x 3 has a solution between 1 and 2. (b) Use the iterative formula x n1 ln x n x n 3 with x 0 1 5 to find the solution, giving your answer correct to 4 d.p. Iteration Solutions 1. (a) Show that the equation x 3 x 2 25 has a solution between 2 and 3. Solution: (a) Let f ( x ) x 3 x 2 25 0 f ( 2) 2 2 25 13 0 3 2 f ( 3) 3 3 25 11 0 3 2 Change of sign 2 a 3 (b) Use the iterative formula x n1 3 25 x n with x 0 2 5 to find the solution, giving your answer correct to 4 d.p. 2 x 0 2 5, x1 2 6566 , . . . a 2 6258 ( 4 d.p. ) Iteration Solutions 2. (a) Show that the equation ln x 2 x 3 has a solution between 1 and 2. Solution: Let f ( x ) ln x 2 x 3 f (1) ln 1 2 3 1 0 f ( 2) ln 2 4 3 0 3 0 Change of sign 1 α 2 (b) Use the iterative formula x n1 ln x n x n 3 with x 0 1 5 to find the solution, giving your answer correct to 4 d.p. Solution: We need ln( ANS) ANS 3 x 0 1 5, x1 1 9055 , . . . a 1 7915 ( 4 d.p. ) Iteration Some arrangements of an equation give formulae which do not give a solution. We earlier met 3 arrangements of x 1 x 3 1 1 x (i) x (1 x 3 ) 2 (ii) x 1 x 3 (iii) x 2 x We used (ii) with x 0 0 5 to find the solution a 0 605423 ( to 6 d.p. ) Now try (i) with x 0 0 5 We get x1 0 77 , x 2 0 30 , x 3 0 94 , . . . and after a while the sequence just oscillates between 1 and 0. This iterative sequence does not converge. Iteration Now try the formula x n1 1 We get x1 1 17 , x 2 0 06 , xn xn 2 with x 0 0 5 . The iteration then fails because we are trying to square root a negative number. Some arrangements of an equation give an iterative sequence which converges to a solution; others do not converge. The next presentation investigates convergence of iterative sequences. Iteration Iteration The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet. Iteration There are some equations that we can’t solve. e.g. x 1 x 3 However, we can find an approximate solution to some of these equations. The approximation can be very accurate, say to 6 or more decimal places. There are several methods of finding approximate solutions and in this presentation we will study one of them. Iteration There are 2 stages to getting a solution: Stage 1. Find a 1st estimate Stage 2. Find a formula to improve the estimate. Sometimes we can just spot an approximate solution to an equation. The solution to x 2x 1 3 is quite close to 0 as the l.h.s. is then 0 and the r.h.s. is 1. If we can’t quickly spot an approximation, we can use a method involving finding bounds for the solution. Iteration We often start by finding numbers ( bounds ) that lie on either side of the solution. If these are integers we call them integer bounds. e.g. To find integer bounds for x 1 x 3 we can sketch y x and y 1 x 3 y 1 x 3 y a x The solution lies between 0 and 1. 0 and 1 are the integer bounds. We usually call the solution a, so 0 a 1 . Our first approximation to a, is any number between the bounds, say 0 5. Iteration The Algebraic Method: If we want to show that 0 and 1 are integer bounds, we show that f ( x ) has different signs at 0 and 1. • Rearrange the equation to the form f ( x ) 0 x 1 x x 1 x 0 Define f ( x ): Let f ( x ) x 3 1 x Find f ( 0) : f (0 ) 0 3 1 0 1 Change of sign Find f ( 1 ) : f (1 ) 13 1 1 1 e.g. • • • • 3 3 The change of sign 0 a 1 Our 1st estimate of a is between these values, say 0 5 Iteration e.g. 1 (a) Using a graphical calculator, or otherwise, sketch, on the same axes, the graphs of 3 y x 2 and y 5 x (b) From the sketch, find integer bounds for the 3 solution, a, of the equation x 2 5 x (c) Use an algebraic method to confirm these are correct and give a 1st approximate solution. Solution: (c) ( Confirm bounds are 0 and 1 ) • Rearrange equation to f ( x ) 0 : x 3 5 x 2 0 • • • Define f ( x ) : So, f ( x) x 5 x 2 f ( 0) 2 0 f (1) 1 5 2 2 0 The change of sign 0 a 1 3 A 1st approximation is any number between 0 and 1. Iteration The 1st approximate solution lies anywhere between the bounds. The next stage is to improve this estimate. • Rearrange the equation to the form x g( x ) . You may spot lots of ways of doing this. e.g. e.g. For the equation (i) Square: x 1 x : 3 2 3 x 1 x x (1 x ) (ii) Rearrange: x 1 x 3 3 Cube root: (iii) Rearrange: Divide by x 2: x 1 3 x 1 x 3 x x 1 x x3 1 x 1 x x x 2 1 3 Iteration Let’s take the 2nd arrangement: x 1 x 1 3 Our 1st estimate of a we will call x0. ( Some people start with x1 which is just as good. ) We substitute x0 into the r.h.s. of the formula and the result gives the new estimate x1. We now have x1 1 x0 1 3 We will then keep repeating the process so we write the formula as 1 x n 1 1 x n 3 This is called an iterative formula. to iterate means to repeat Iteration SUMMARY To find an approximation to a solution ( or root ) of an equation: Find a 1st approximation, often by finding integer bounds for the solution. Let this be x0 . Rearrange the equation into the form x g ( x ) Write the arrangement as an iterative formula: x n1 g( x n ) Key x0 into a calculator and ENTER. Key the r.h.s. of the formula into the calculator, replacing x with ANS. Press ENTER as many times as required to get the solution to the specified accuracy. Iteration e.g. 1(a) Show that the equation 2 x x 3 has a root a in the interval 1 3 a 1 4 . (b) Using the arrangement x 3 2 x write down an iterative formula and use it to find the root correct to 4 decimal places. Solution: (a) Let f ( x) 2 x 0 x f (1 3) 2 f (1 4) 2 3 1.3 1 3 0 27 1.4 1 4 0 10 3 3 Change of sign 1 3 a 1 4 Iteration Solution: (b) Let x 3 2 gives x x 0 1 5 x1 2 0 1 414 3 x x n1 3 2 3 2 xn 15 x 2 1 3865 x 3 1 3776 It takes about 7 iterations to reach a 1 3735 (4 d.p.) Iteration Some arrangements of an equation give formulae which do not give a solution. We earlier met 3 arrangements of x 1 x 3 x 1 x (i) x (1 x ) 1 1 x 3 x x 1 x 2 x We used (ii) with x 0 0 5 to find the solution 3 2 (ii) x 1 3 (iii) x 3 a 0 605423 ( to 6 d.p. ) Trying (i) with x 0 0 5 gives x1 0 77 , x 2 0 30 , x 3 0 94 , . . . and after a while the sequence just oscillates between 1 and 0. The iterative sequence does not converge. Iteration Trying the arrangement x 1 x x 2 with x 0 0 5 gives x1 1 17 , x 2 0 06 , The iteration then fails because we are trying to square root a negative number. Some arrangements of an equation give an iterative sequence which converges to a solution; others do not converge.