Math Introduction to Numerical Methods - Winter 2011 Homework 2 Assigned: Friday, January 14, Due: Thursday, January 27,

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1 Math Introduction to Numerical Methods - Winter 2011 Homework 2 Assigned: Friday, January 14, Due: Thursday, January 27, Include a cover page. You do not need to hand in a problem sheet. Write the problems in order and indicate clearly where to look to see any code or output you have attached. Always clearly label all plots (title, x-label, y-label, and legend). Use the subplot command from MATLAB when comparing 2 or more plots to make comparisons easier and to save paper. Include all of your script. A print out of your script should be included in the write up you submit and any m files should be ed to me as well. Place a comment at the top of each function or script that you submit which includes the name of the function or script, your name, the date, and MATH 371. For this assignment you will need to write three MATLAB programs. One for the bisection method, one for Newtons Method and one for the Secant Method. You will use these programs to carry out several parts of the problems below (Bisection on 1 and 5, Newton s on 4 and 5, and Secant on 5 and 8). I will give you code for Newton s Method and a shell of a program for the Bisection Method. You may also write code other problems if you like (Problem 3 part b, Problem 10) or you may do that by hand with the assistance of a calculator. 1

2 Problem 1 Write a MATLB routine that carries out the Bisection Method for finding a root. A shell for your MATLAB code is included below. Then use that code to find the roots of the following functions using an error tolerance of Discuss briefly if the error results agree with the predicted order of convergence discussed in class. (a) x 4 x 1 = 0. Include evidence that you have found all the roots (a plot or a theoretical verification). For this problem there are 2 roots. f(x) = x 4 x 1. I know this by looking at the graph of From this plot we can see that there are two roots, one positive and one negative. To find the first root I ran the bisection method with the initial interval of [1, 1.5], 15 steps were required to reach the desired accuracy. n c n error estimate e e e e-06 2

3 To find the second root I used an interval of [-1, -.5], which again took 15 iterations to achieve the desired accuracy. n c n error estimate e e e e-06 3

4 (b) Find the smallest nonzero positive root of x = tan(x). Include a graph that demonstrates the location of the root. (This will help in choosing a and b.) For this problem we begin by looking at the graph of y = x and y = tan(x). I have included two of these graphs one near x=4 and the other near x = 100 so that you can see the roots closest to these points. For the first root a suitable interval [a, b] is [3.8, 4.5]. You must be careful to note that the the graphs may appear to have other roots due to the fact that asymptotes are drawn. However the first root is close to 4. The output is then: n c n error estimate e-05 y=x y=tan(x)

5 (c) Find the root of x = tan(x) which is closest to 100. Include a graph that demonstrates the location of the root. (This will help in choosing a and b.) For the second root a suitable interval [a, b] is [98.9, 98.96]. The output is then: n c n error estimate e-05 y=x y=tan(x)

6 Problem 2 Let α be the unique solution of x = x 4 Write a function that has a root equal to α. Find an interval [a, b] containing α and for which the bisection method will converge to α. Then estimate the number of iterates needed to find α within an accuracy of Do not actually carry out the Bisection method for this problem For this problem we will consider the function f(x) = x 3 1+x 4. First we find an interval [a, b] for which f(a) and f(b) are opposite signs. One possible choice would be [0, 2] as f(0) = 3 and f(2) = 31/17 To estimate the number of iterates needed for the bisection method to produce an approximate c n such that c n α < we use the fact that c n α b a 2 n+1. Thus we need to choose n so that: (b a) 2 n+1 = n+1 = 2 n Thus we need that 10 8 /5 2 n or 8 ln(10) ln(5) ln(2) be at least 25 for this to be satisfied. n, computing this we see that n must 6

7 Problem 3 (a) Show that Newton s Method for f(x) = x m a reduces to the iterative scheme: x n+1 = 1 ( ) a m (x n ) + (m 1)x m 1 n. (b) Using the iterative scheme and an initial guess of 1 compute m 2 for m = 3, 4, 5, 6, 7 and 8, each to within an accuracy of You may do this either writing MATLAB code or any other computing device you desire. Include the iterates for each of the six calculations. If we use the function x m a and construct the Newtons method scheme we obtain that the iteration formula is: x n+1 = x n f(x n) f (x n ) = x n (x n) m a m(x n = m(x n) m (x n ) m + a m 1 m(x n ) m 1 = 1 ( ) a m (x n ) + (m 1)x m 1 n. We now perform Newtons method several times each for a different value of m. In each we set a = 2. For each value of m I have used 1 as my starting value. m = 3 n x n error estimate e e-10 m = 5 n x n error estimate e e-09 m = 7 n x n error estimate e e-09 m = 4 n x n error estimate e e-10 m = 6 n x n error estimate e e-09 m = 8 n x n error estimate e e-09 7

8 Problem 4 Bradie, p. 105, #11:The function f(x) = 27x x 3 180x x 7 has a zero at x = 1/3. Perform ten iterations of Newton s method on this function, starting with p 0 = 0. What is the apparent order of convergence of the sequence of approximations? What is the multiplicity of the zero at x = 1/3? Would the sequence generated by the bisection method converge faster? n p(n) e(n) ln e(n)/ln e(n-1) e(n)/e(n-1) The apparent order of convergence is linear with asymptotic rate constant The Bisection method converges faster. (Why?) Problem 5 Before doing this problem you should write programs that implement Newton s Method, the Bisection Method, and the Secant Method. Your code should be properly commented and the m files should be ed to me as part of your assignment. (a) Suppose f(x) is differentiable on [a, b]. Discuss how you might use a rootfinding method to identify a local extremum of f(x) inside [a, b]. Roots of the derivative give possible locations of local extrema. (b) Let f(x) = log x sin x. Prove that f(x) has a unique maximum in the interval [4, 6]. (Note that log means natural logarithm.) f (4) > 0, f (6) < 0, and the continuity of f (x) give the existence of at least one maximum. f (x) < 0 for x [4, 6] shows the maximum is unique. (c) Approximate this local maximum using six iterations of the Bisection Method with starting interval [4, 6]. (see table) (d) Approximate this local maximum using six iterations of the two fixed-point methods (Secant and Newton). For Newton s Method, use p 0 = 4. For the Secant Method, use p 0 = 6 and p 1 = 4. 8

9 (see table) (e) What is your best estimate for p, the location of the maximum? (f) Compare the three algorithms using the following two tables. Use your best estimate for p when comparing the error for the second table. Please do not write the tables by hand. n Bisection Secant Newton Table 2: Absolute error p n p versus iteration number n n Bisection Secant Newton e e e e e (g) On the same set of axes plot the log of the absolute error against the log of n for all four methods. Use n as the independent variable and log(e n ) as the dependent variable, plot each method in a different color. What does this indicate about the convergence in this case? 9

10 5 0 Plot of the Log of the errors of each method Bisection Secant Newton 5 Log of the absolute value of the error n iterations (h) What happens if you attempt to approximate the maximum by starting Newton s Method with p 0 = 6? The method converges to a root outside [4, 6]. 10

11 Problem 6 Do Problem 1 from section 2.3 and use the results to answer Problem 4 from the same section. (a) Given that sequence {p n } converges linearly to the fixed point p, we know that g(p) = p and g(p n ) = p n+1. By the intermediate value theorem we have that: p n+1 p p n p = g(p n) g(p) p n p = g (c n ) where c n is between p n and p. Since p n p as n then c n p as n as well. Thus for n large we have that: Thus, e n+1 e n = p n+1 p p n p p n p n 1 p n 1 p n 2 = (p n p) (p n 1 p) (p n 1 p) (p n 2 p) = e n e n 1 e n 1 e n 2 e n 1 e n 1 = (e n/e n 1 1) (1 e n 2 /e n 1 ) (g (p) 1) (1 g (p) 1 ) = g (p), g (p) where we have used that e n /e n 1 g (p) and e n 2 /e n 1 g (p) 1. This proves the desired result. (b) Recall that when a fixed point iteration scheme converges linearly, it has the asymptotic error constant λ = g (p), Thus e n g (p)e n 1, or e n 1 e n /g (p). Now, e n = p p n = p n p n 1 + p n 1 p = p n p n 1 + e n 1 Substituting e n 1 e n /g (p) into this last expression, solving for e n, and taking the absolute value gives e n g (p) g (p) 1 p n p n 1 This completes the proof. 11

12 Problem 7 For each of the following decided if the indicated iteration scheme will converge to the indicated α, provided x 0 is chosen sufficiently close to α. If it does converge, determine the convergence order. (a) x n+1 = 15x2 n 24x n x n, α = 1 We consider the function g(x) = (15x 2 24x + 13)/(4x). It is easy to check that g(1) = ( )/4 = 1 So 1 is indeed a fixed point. Also, g (x) = 15/4 13/(4x 2 ) (b) and 0 < g (1) = 1/2 < 1. Thus the iteration scheme converges linearly. Remember it is important that g (1) is not only non-zero but it is also less then 1. The asymptotic rate constant is then λ = g (1) = 1/2 x n+1 = x3 n + 6x n 3x 2 n + 2, α = 2 We consider the function g(x) = x3 +6x, and g( 2) = = 2, thus 2 is 3x indeed a fixed point. Moreover, g (x) = (3x2 + 2)(3x 2 + 6) (x 3 + 6x)(6x) (3x 2 + 2) 2 = 3(x2 2) 2 (3x 2 + 2) 2, g ( 2) = 0, thus we know that the sequence will converge since g ( 2) = 0 < 1, the order of convergence is at least 2. g (x) = (3x2 + 2) 2 (6)(x 2 2)2x 3(x 2 2) 2 (2)(3x 2 + 2)(6x) (3x 2 + 2) 4 = 96x(x2 2) (3x 2 + 2) 3, thus g ( 2) = 0, and the order of convergence is at least 3. g (x) = 96(9x4 36x 2 + 4) (3x 2 + 2) 4, g ( 2) = 3/4 so the order of convergence is exactly 3. Moreover, the asymptotic error constant is λ = g ( 2) 3! =

13 Problem 8 (a) Show that if x n converges with order α, then lim n log e n log e n 1 = α. Let This implies that or or or e n e n 1 α = L. ( ) en log = log(l), e n 1 α log( e n ) α log( e n 1 ) = log(l), log( e n ) log(l) = α log( e n 1 ) log( e n ) log e n 1 log(l) log e n 1 = α. Now, by assumption lim L = λ and since lim e n = 0 then lim log e n 1 = and thus lim log(l)/ log e n 1 = 0. Consequently, we have: ( log( en ) lim n log e n 1 log(l) ) = α. log e n 1 or ( ) ( log( en ) lim n log e n 1 ) log(l) 0 lim = α. n log e n 1 (b) Carry out seven iterations of the Secant Method for the function f(x) = 3 1 and include a table with the following columns: x n x n e n log( e n ) log( e n 1 ) n x n e n = 1 x 3 n log e n / log e n NAN

14 What can you say about the order of convergence of the Secant Method in this case based on this table and the calculation from (a)? Based on the table once can conclude that at least for this example the Secant Method is converging with and order of convergence somewhere between 1 and 2. The number should be (1 + 5)/ however, what occurs here is that you achieve 1/3 to all digits of accuracy storable by the computer before the ratio reaches this value. Problem 9 Problem 11 from Section 2.3 (Problems 12 and 13 are also quite similar). Consider the function g(x) = e x2. (a) Prove that g has a unique fixed point on the interval [0, 1]. Let g(x) = e x2. We will proceed by showing that g is continuous on [0, 1], maps [0, 1] to [0, 1] and there exists a k < 1 such that g (x) k for all x [0, 1]. First, note that g is the composition of the functions e x and x 2, both of which are continuous on [0, 1]. Consequently, g(x) = e x2 is continuous on [0, 1]. Next, we see that g (x) = 2xe x2 < 0 for all x (0, 1). Hence, g is decreasing on (0, 1). Combining this fact with g(0) = 1 and g(1) = e , it then follows that for x [0, 1], g(x) [e 1, 1] [0, 1]. Finally, we find g (x) = (4x 2 2)e x2 = 0 when x = 2/2. Because g (0) = 0, g ( 2 2 ) = 2e 1/ and g (1) = 2e , we find that g (x) 2e 1/2 for all x [0, 1]. Thus, we take k = 2e 1/2. Having established that g is continuous on [0, 1], maps [0, 1] to [0, 1] and there exists a k < 1 such that g (x) k for all x [0, 1], we conclude that g has a unique fixed point on the interval [0, 1]. (b) With a starting approximation of p 0 = 0, use the iteration scheme p n = e p2 n 1 to approximate the fixed point on [0, 1] to within

15 With a starting approximation of p 0 = 0 and a convergence tolerance of ɛ = , fixed point iteration using g(x) = e x2 p 1 = 1 p 10 = p 20 = p 128 = with an error estimate of This was using the value of p = as the actual root calculated in Mathematica. I you use the error estimate p n p n 1 to approximate the error then you will reach error less than at at p 133. (c) Use the theoretical error bound p n p kn 1 k p 1 p 0 to obtain a theoretical bound on the number of iterations needed to approximate the fixed point to within How does the number of iterations performed in part (b) compare with the theoretical bound? In part (a) we found k = 2e 1/2. With p 0 = 0, it follows that p 1 = g(p 0 ) = e 0 = 1. Solving the equation k n 1 k p 1 p for n yields n 152.3, or, since n must be an integer, n 153. As these calculations were carried out using k = max x [0,1] g (x), we see that the upper bound on the number of iterations needed to guarantee an absolute error less than is n = 153. In part (b), we found that only 128 iterations were needed to achieve the prescribed level of accuracy, confirming the theoretical upper bound. (d) Accelerate the convergence of the sequence obtained in part (a) using Aitken s 2 method. By how much has Aitken s 2 -method reduced the asymptotic error constant? 15

16 When carrying out the iteration above it appears that the method converges linearly with an asymptotic rate constant of λ = If we perform Aitken s 2 -method then we get the table below where the last row corresponds to n=53. The right most column indicates that the asymptotic rate constant has decreased to approximately λ = p n e n for p n λ approximation NAN

17 Newton s Method Code function newtonsimple(x0, error_bd, max_iterate) %%%%%% x0 is the initial guess. %%%%%% error_bd, is the accuracy we seek, we stop the iterations if the %%%%%% error is < error_bd. %%%%%% max_iterate is the maximum number of times we will carry out the %%%%%% iterations. If the error bound is not reached in this many %%%%%% iterations the program will halt and display an error message. format long g error=1; d=1; lerror=1; it_count =0; %%%%%% Above we have initialized the variable "error" which will be %%%%%% updated with the error at each step of the loop. We have also %%%%%% initialized the variable lerror, which will be the error from the %%%%%% previous step. while abs(error)>error_bd & it_count < max_iterate %Stops the while loop if the error bound is met %or the number of iterates (counted by %it_count) is more then the prescribed amount. fx=f(x0); dfx=deriv_f(x0); %Sets fx and dfx equal to f and f evaluated at %the current x value. if dfx==0; disp( The derivative is zero. Stop ) %Displays an error if the derivative is 0. return end x1=x0-fx/dfx; %error=x1-x0 error=actualroot-x0 %%%%%%%% When the actual root is not known you should use the first error %%%%%%%% line. In this case the error is e_n= actual root - x(n) is %%%%%%%% approximated by x(n+1)-x(n). The second error line should be used %%%%%%%% when the actual root is known ahead of time. The error is e_n= %%%%%%%% actual root - x(n). 17

18 Table(it_count+1,1)=it_count; Table(it_count+1,2)=x0; Table(it_count+1,3)=fx; Table(it_count+1,4) =error; Table(it_count+1,5)=error/d; Table(it_count+1,6)=error/d^2; Table(it_count+1,7)=error/d^3; %%%%%%%% The first column of the matrix Table, is n. The second column is %%%%%%%% x_n the third column is f(x_n), the fourth column is e_n, the %%%%%%%% fifth is x_n/x_n-1, the sixth is x_n/(x_(n-1))^2 and the final %%%%%%%% column is x_n/(x_(n-1))^3. These columns are useful in %%%%%%%% identifying the order of convergence of the series produced. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% d=error; x0=x1; it_count=it_count+1; %%%%%% This saves the error value for use during the next iteration, %%%%%% updates x0 to x1 and updates the counter as well. end if it_count>=max_iterate disp( The number of iterates calculated exceeded ) disp( max_iterate. An accurate root was not ) disp( calculated. ) %%%%%%% This displays an error message if the error bound is not reached %%%%%%% in the maximum allowed iterations. else format long root = x1 %%%%%%%% Displays the root. end format long g Table=Table(:, :) %%%%%%%% Displays the table. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function value= f(x) value=3-x^2; %%%% Defines the function f to be used. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function actualroot=actualroot actualroot=sqrt(3); 18

19 %%%%% Defines the actual root, this part can be commented out if not using %%%%% a function where you know the root. %%%%%%%%%%%%%%%%%%%%%%%%%%% function value =deriv_f(x) value= -2.*x; %%%%% Defines the derivative of the function f(x) used above. 19

20 Secant Method Code function secant(x0,x1, error_bd, max_iterate) %%%%%% x0 and x1 are the initial iterates. %%%%%% error_bd, is the accuracy we seek, we stop the iterations if the %%%%%% error is < error_bd. %%%%%% max_iterate is the maximum number of times we will carry out the %%%%%% iterations. If the error bound is not reached in this many %%%%%% iterations the program will halt and display an error message. format long g error=1; d=1; it_count =0; fx0=f(x0); %%%%%% Above we have initialized the variable "error" which will be %%%%%% updated with the error at each step of the loop. We have also %%%%%% initialized the variable lerror, which will be the error from the %%%%%% previous step. We have also defined fx0 to be f evaluated at x0 while abs(error)>error_bd & it_count < max_iterate %Stops the while loop if the error bound is met %or the number of iterates (counted by %it_count) is more then the prescribed amount. fx1=f(x1); %Sets fx0 and fx1 equal to f evaluated at %the current x value. if (fx0-fx1)==0; disp( The derivative is zero. Stop ) %Displays an error if the derivative is 0. return end x2=x1-fx1*((x1-x0)./(fx1-fx0)); %error=x1-x0; error=actualroot-x0; %%%%%%%% When the actual root is not known you should use the first error %%%%%%%% line. In this case the error is e_n= actual root - x(n) is %%%%%%%% approximated by x(n+1)-x(n). The second error line should be used %%%%%%%% when the actual root is known ahead of time. The error is e_n= 20

21 %%%%%%%% actual root - x(n). Table(it_count+1,1)=it_count; Table(it_count+1,2)=x0; Table(it_count+1,3)=fx0; Table(it_count+1,4)=abs(error); Table(it_count+1,5)=abs(error)/abs(d); Table(it_count+1,6)=abs(error)/(abs(d)^(1.62)); Table(it_count+1,7)=abs(error)/d^2; %%%%%%%% The first column of the matrix Table, is n. The second column is %%%%%%%% x_n the third column is f(x_n), the fourth column is e_n, the %%%%%%%% fifth is x_n/x_n-1, the sixth is x_n/(x_(n-1))^(1.62) and the final %%%%%%%% column is x_n/(x_(n-1))^2. These columns are useful in %%%%%%%% identifying the order of convergence of the series produced. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% d=error; x0=x1; x1=x2; fx0=fx1; it_count=it_count+1; %%%%%% This saves the error value for use during the next iteration, %%%%%% updates x0 to x1 and updates the counter as well. Also fx0 is reassigned the %%%%%% value from fx1, this avoids recomputing f. end if it_count>=max_iterate disp( The number of iterates calculated exceeded ) disp( max_iterate. An accurate root was not ) disp( calculated. ) %%%%%%% This displays an error message if the error bound is not reached %%%%%%% in the maximum allowed iterations. else format long root = x1 %%%%%%%% Displays the root. end format long g Table=Table(:, :) %%%%%%%% Displays the table. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function value= f(x) value=3-1./x; 21

22 %%%% Defines the function f to be used. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function actualroot=actualroot actualroot=1./3; %%%%% Defines the actual root, these two lines can be deleted or commented out, if %%%%% the actual root is not known. 22

23 function A=bisection(a,b,error_bound,max_it) %%%%%%%%%%% The variables a and b describe an interval [a, b] in which f(x) %%%%%%%%%%% has a root. The function f(x) is refered to as myfun(x) and %%%%%%%%%%% is defined below and you can change it for the problem you are %%%%%%%%%%% working on. The variable error_bound is the error tolerance you %%%%%%%%%%% wish to specify for your root, and max_it is the maximum number %%%%%%%%%%% of times the bisection method will run. if (sign(myfun2(b))*sign(myfun2(a))>0) error( The sign of f(a) and f(b) are the same ); end %%%%%%%%%% The returns an error if f(a) and f(b) are not opposite signs. A(1,1)=0; A(1,2)=a; A(1,3)=b; %%%%%%%%%% This fills in the first three columns of row 1 of the matrix A %%%%%%%%%% with 0, a0, and b0. c=(a+b)./2; A(1,4)=c; A(1,5)=myfun2(c); %%%%%%%%%% This computed the c0 and inserts c0 and f(c0) er=b-c; A(1,6)=er; %%%%%%%%% This computes the error bound as the difference of b0 and c0 and %%%%%%%%% puts this value in the 6th column of the first row of the matrix %%%%%%%%% A. iter=1; %%%%%%%%% This initializes our counter for our loop. while (er>error_bound && iter<=max_it) if sign(myfun2(b))*sign(myfun2(c))<=0 a=c; else b=c; end %%%%%%%% This if loop assigns the new a and b c=(a+b)/2; %%%%%%%% Assigns the new c value er=b-c; %%%%%%%% Assigns the new error A(iter+1,1)=iter; A(iter+1,2)=a; A(iter+1,3)=b; A(iter+1,4)=c; A(iter+1,5)=myfun2(c); A(iter+1,6)=er; 23

24 iter=iter+1; end if (iter>=max_it), disp( Maximum iterations reached at or before desired accuracy ); end %%%%%%%%%%%%%function to find root of%%%%%%%%%%%%%%%%% function value= myfun2(x) %value=x.^4-x-1; value=tan(x)-x; 24

25 Ex Problem 1 1(a) Use the Secant method with an initial guess of x 0 = 0 and x 1 = 2 to find the root of f(x) = x 3 x 2 x 1, using an error tolerance of For this problem I applied the Secant Method starting with initial iterates x 0 = 0 and x 1 = 2 (I know a root of f(x) = x 3 x 2 x 1 is between these two numbers). The results to the desired accuracy of 10 6 are: f(x) = x 3 x 2 x 1 n x n error estimate e e-10 1(d) Use the Secant method with an initial guess of x 0 = 0 and x 1 = 2 to find the root of f(x) = x e x 1, using an error tolerance of For this problem I applied the Secant Method starting with initial iterates x 0 = 0 and x 1 = 1 (I know an root of f(x) = x e x 1 is between these two numbers). The results to the desired accuracy of 10 6 are: f(x) = x e x 1 n x n error estimate e e-08 25

26 Ex Problem 2 Perform Newton s Method on the function f(x)=3-1/x, to find the root x = 1/3. Use an initial guess of p 0 =.2. Carry out five iterations and include a table with the following columns: n x n e n e n /e n 1 e n /e 2 n 1 e n /e 3 n 1 n x n e n e n /e n 1 e n /e 2 n 1 e n /e 3 n e e e What can you say about the order of convergence of Newtons Method in this case based on this table? By looking at this table we can conclude that the order of convergence is 2 and that the asymptotic rate constant is approximately 3. Ex Problem 3 Write a Matlab code that implements the Secant method for the function f(x) = x 4 (5.4) x 3 + (10.56) x 2 (8.954) x + (2.7951), find the root α located in the interval [1, 1.2], to an accuracy of at least 10 6, and display all iterates. Once you have your table of iterates and error estimates, discuss in complete sentences the convergence of the algorithm and if it agrees with what we discussed in class and why. For this problem we perform the Secant method for the function f(x) = x 4 (5.4) x 3 + (10.56) x 2 (8.954) x + (2.7951). Starting with initial iterates of [1, 1.2] it takes 29 iterates to reach the desired level of accuracy. The reason that the Secant method takes so long to converge is that the root is that the root is of multiplicity 3. In fact f(x) = (x 1.1) 3 (x 2.1) 26

27 n x n error estimate e e e e e e e e e e e e e e e e-07 Ex Problem 4 Bradie, p. 125, #14: The Secant Method appears to be converging with order α 1.6. The lack of accuracy reflects the fact that the root has higher multiplicity. ============================================================= n p(n) e(n) ln e(n)/ln e(n-1) =============================================================

28 ============================================================= The Secant Method appears to be converging to the root. The data on the order of convergence are not really conclusive. ============================================================= n p(n) e(n) ln e(n)/ln e(n-1) ============================================================= ============================================================= The Secant Method appears to be converging toward the root with superlinear order α 1.6. Again, the data are not really conclusive. ============================================================= n p(n) e(n) ln e(n)/ln e(n-1) ============================================================= ============================================================= Ex Problem 5 Bradie, p. 125, #15 Choosing initial iterates p 0 = 0.3 and p 1 = 1.2, the Secant Method appears to converge to the root, but the apparent order is about α 1. This problem is very sensitive to initial guesses. ============================================================ n p(n) e(n) ln e(n)/ln e(n-1) ============================================================

29 ============================================================ The Secant Method converges to the root. The order of convergence cannot be determined from this data. ============================================================= n p(n) e(n) ln e(n)/ln e(n-1) ============================================================= ============================================================= Convergence to the root initially is superlinear, although we cannot really estimate the order. Later, the order becomes roughly linear. ============================================================= n p(n) e(n) ln e(n)/ln e(n-1) ============================================================= ============================================================= 29

30 Ex Problem 6 Consider the function f(x) = 2 cos(x) 2e x + 1 on the interval [ 2, 2]. When you plot the function, you will see that it has two roots on this interval. (a) Write down an explicit fixed point method for finding each root, including a reasonable starting point. The iteration function may be the same for both roots. The methods should have linear-order convergence. A reasonable choice for the negative root is g (x) = x 0.2f(x) with starting point p 0 = 0.5. A reasonable choice for the positive root is g + (x) = x + 0.5f(x) with starting point p 0 = 2. (b) Use the theorem on page 87 to prove that both methods converge for the starting points you specify. The function g increases on [ 0.5, 0.2] and maps the endpoints to 0.39 and On the same interval, the derivative increases from 0.15 to The theorem applies. The function g + is increasing on [1.8, 2] and maps the endpoints to 1.91 and On the same interval, the derivative increases from 0.19 to The theorem applies. (c) Write a program that uses this fixed point method to find the positive root of f. Use the stopping condition on page 92, with a tolerance of function p = FixPoint( g, p1, tol ); p(1) = p1; p(2) = g(p1); n = 2; errorest = Inf; while (errorest > tol ) n = n + 1; p(n) = g( p(n-1) ); end dg = (p(n) - p(n-1))/(p(n-1) - p(n-2)); errorest = abs( dg*(p(n) - p(n-1)) / ( dg - 1 ) ); 30

31 (d) Verify from the numerical output of your algorithm that the method converges linearly, and estimate the asymptotic error constant. The following formulae are relevant. α log e n log e n 1 and λ e n e n 1. ============================================================================== n p(n) e(n) alpha lambda ============================================================================== ============================================================================== (e) Write a second program that uses Aitken s 2 method from Section 2.6 to accelerate your fixed-point iteration. Apply the stopping criterion from page 92 to the sequence p n with tolerance function phat = AitkenDelta( g, p1, tol ) p(1) = p1; p(2) = g( p1 ); n = 2; errorest = Inf; while (errorest > tol) n = n + 1; p(n) = g( p(n-1) ); phat(n) = p(n) - (p(n) - p(n-1))^2 / (p(n) - 2*p(n-1) + p(n-2)); if (n >= 5) 31

32 end dg = (phat(n) - phat(n-1)) / (phat(n-1) - phat(n-2)); errorest = abs( dg*(phat(n) - phat(n-1)) / ( dg - 1 ) ); end (f) Use the numerical output of your second program to estimate the order of convergence and asymptotic rate constant of the output. How do they compare with the non-accelerated method? ============================================================================= n phat(n) e(n) alpha lambda ============================================================================= ============================================================================= Ex Problem 7 Problem 5 from section 2.6: The sequence listed below was obtained from fixed point iteration applied to the function g(x) = 10/(2 + x), which has a unique fixed point (a) Apply Aitkens 2 -method to the given sequence. (You may do this by hand or writing a small script that you can reuse). Give all of the p k s. 32

33 Using the first three terms in the fixed point iteration sequence, we calculate p 3 = p 3 (p 3 p 2 ) 2 p 3 + p 1 2p 2 ( )2 = ( ) = Next, using p 2, p 3 and p 4, we calculate p 4 = p 4 (p 4 p 3 ) 2 p 4 + p 2 2p 3 ( ) 2 = ( ) = Continuing in this manner, we find p 5 = , p 6 = , and p 7 =

34 (b) To ten digits, the fixed point of g is x = Use this to show that both the original sequence and the output from Aitkens 2 -method are linearly convergent and estimate the corresponding asymptotic error constant. By how much has Aitken s 2 -method reduced the asymptotic error constant? The values in columns 3 and 5 in the table below confirm that both the original Fixed Point sequence and the accelerated sequence converge linearly. The asymptotic error constant for the Fixed Point sequence is roughly 0.23, while the asymptotic error constant for the accelerated sequence is roughly Aitken s 2 -methodhas reduced the error constant by nearly 80%. Fixed Point Aitken s 2. Fixed Point Aitken s Λ 2 n p n e n / e n 1 p n e n / en Ex Problem 8 Problem 8 from section 2.6: (a) Perform ten iterations to approximate the fixed point of g(x) = ln(4 + x x 2 ) using p 0 = 2. Verify numerically that the sequence converges linearly and estimate the asymptotic error constant. To 20 digits, the fixed point is x = Let g(x) = ln(4 + x x 2 ) and take p 0 = 2. The sequence generated by fixed point iteration is given in the second column below. The values in the third column confirm the linear convergence of the sequence with an asymptotic error constant of roughly n p n e n / e n

35 (b) Accelerate the convergence of the sequence obtained in part (a) using Aitken s 2 method. By how much has Aitken s 2 -method reduced the asymptotic error constant? Applying Aitkens 2 -method to the sequence obtained in part (a) produces the sequence listed in the second column in the table below. The values in the third column confirm the linear convergence of the sequence with an asymptotic error constant of roughly 0.19, more than 50% lower than the error constant for the original sequence. n p n e n / e n (c) Apply Steffensens method to g(x) = ln(4 + x x 2 ) using the same starting approximation specified in part (a). Perform four iterations, and verify that convergence is quadratic. Let g(x) = ln(4 + x x 2 ) and take p 0 = 2. Steffensens method produces the sequence given in the second column of the table below. The values in the third column confirm quadratic convergence of the sequence. n p n e n / e n