If you need the source code from the sample solutions, copy and paste from the PDF should work. If you re having trouble, send me an .

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1 AM117 Sample Solutions, HW#1 Hi, I m Andreas Kloeckner, your grader for this course. Feel free to me at kloeckner@dam.brown.edu. The code for the sample solutions is written so that it will run in both Matlab and Octave. (I test it in both.) Octave is a language for scientific computing that is generally very similar to Matlab. Unlike Matlab, Octave is freely downloadable. You may grab a copy at If you need the source code from the sample solutions, copy and paste from the PDF should work. If you re having trouble, s me an . When grading your homework, I use four symbols to denote how right you got that particular problem: perfect ( ) mostly OK you made a small mistake, or there s a little piece missing what you wrote is somewhat like a solution, but requires substantial work X either you wrote nothing, or what you wrote is completely wrong Please follow these guidelines when turning in homework: Please staple your homework. In particular, paperclips and folded corners are evil they won t keep your homework together long enough to survive the tangle of the dropoff/pickup box. Please don t use red pen in your answers. When you turn in results of computational exercises, include something that makes it credible that you actually did write code for the problem include the code itself or a (reasonably short) printout of the results. Use your own best judgement to turn in the least amount of material that you think will convince me that you did the problem and did it correctly. If you include computer output, try to keep it concise. Norms (especially norms of errors) are much shorter and more meaningful than full vector output. I am not interested in intermediate results of your code. When you find a result using facts from the book, then always mention what you are using. For example, it will not be enough to say This matrix is so-and-so even if that is true. You need to give a reason why you believe so. If you are proving something, write down what and how. It doesn t take much time, but it helps you and me understand what you re writing and tells me that you know what you re talking about. A happy grader is a thorough and exact grader, you know. ;-) On this homework set, many people encountered these issues: When estimating the order of convergence in problem 1, don t guess or use trial-and-error. Instead, solve for α using the logarithm. 1

2 When looking for an alternative expression in problem 2, using Taylor series only counts if you can put an explicit bound on the error in that approximation and argue that it doesn t change the final graph visibly. When finding roots in the last two problems, you need to do two things before running your code: Make sure you re finding all the roots of the functions. A plot is not sufficient! You need to show that there can t be any more roots other than the ones you re trying to find. Write down how you obtained your guesses for the initial interval (bisection) or starting value (Newton). A plot is a valid help here, by the intermediate value theorem. See below for examples. 1. Problem 6, p.28 We are attempting to find α and λ such that Taking the logarithm of this, we obtain e n+1 λ e n α. log e n+1? y log λ? b + α? a log e n? x (1) If we look closely, we can see that this reduces the problem to finding a line (given by slope a and y-intercept b). We need to find two parameters, and since we have three data points for each method, we may write two equations of the form (1) to find them: log e 2 = log λ +α log e 1, log e 3 = log λ +α log e 2. Subtracting the two, we find Knowing α, we solve the first one for λ: log e 2 log e 3 log e 1 log e 2 = α. log λ = log e 2 α log e 1, λ = exp(log e 2 α log e 1 ). This yields: From these calculations it seems the orders of convergence are roughly 2, 4 and 3, respectively. Here s some code to perform this computation: function p1 verify_order_of_convergence( Method 1, [4.0e-2, 9.1e-4, 4.8e-7], 2) verify_order_of_convergence( Method 2, [3.7e-4, 1.2e-15, 1.5e-60], 3.9) 2

3 verify_order_of_convergence( Method 3, [4.3e-3, 1.8e-8, 1.4e-24], 3) function verify_order_of_convergence(label, seq, order) disp(label) alpha = (log(seq(2))-log(seq(3))) / (log(seq(1))-log(seq(2))) lambda = exp(log(seq(2))-alpha*log(seq(1))) 2. Problem 9, p.52 a) 1.4e-15 "1-cos.dat" 1.2e-15 1e-15 8e-16 6e-16 4e-16 2e e-08-4e-08-3e-08-2e-08-1e e-08 2e-08 3e-08 4e-08 5e-08 Code: function p3 p3_do_plot(-5e-8, 5e-8, 1000) pause function do_plot(a, b, n) x = linspace(a, b, n); y = 1-cos(x); plot(x,1-cos(x)); b) Plot sin 2 x and note that it looks a bit like the above. Remember or, for our case, cos(α + β) = cos(α)cos(β) sin(α)sin(β) cos(2x)=cos 2 x sin 2 x. (2) 3

4 Also remember 1=cos 2 x + sin 2 x. (3) Now compute (3) (2): 1 cos(2x) =2sin 2 (x), which, after substituting ξ = 2x, yields 1 cos(ξ) =2sin 2 (ξ/2) and a much cleaner plot: 1.4e-15 "1-cos.dat" 2*sin(x/2)**2 1.2e-15 1e-15 8e-16 6e-16 4e-16 2e e-08-4e-08-3e-08-2e-08-1e e-08 2e-08 3e-08 4e-08 5e Problem 13, p.52 a) g(x) f(x)= cos2 x 1+sin x (1 sin x)= cos2 x (1 sin x)(1 + sin x) = cos2 x (1 sin 2 x) = sin x 1 + sin x In preparation for b) and c), we create this table: x sin(x) cos 2 (x) π/ π/2-1 0 b) Near π/2, sin(x) 1, so f(x) = 1 sin(x) subtracts two values of order 1 to obtain a result very close to zero, and so it will encounter significant cancellation error. The logical choice is g(x) = cos 2 (x)/(1 + sin(x)), for which we observe no such phenomena. c) Near 3π/2, sin(x) 1, so the denominator of g (which is 1+sin x) will encounter cancellation error. 4. Problem 6, p.69 The length of the initial interval is b a. After each iteration, the length of the interval is halved, so we have b a 2 1 after the first, b a 2 2 after the second,, b a 2 n after the nth. 4

5 Once we know that the length of our candidate interval is shorter than ε, our tolerance, i.e. b a 2 n <ε, we are done. We can now compute the number of iterations needed to achieve this precision: 5. Problem 16, p.70 b a 2 n < ε b a ε < 2 n b a log ε < n log 2 b a log ε log 2 < n. a) f(x)=e x +x 2 x 4. Consider this function in two parts. f 1 (x) = e x f 2 (x) = x 2 +x+4=(x 1/2) 2 17/4 f(x) = f 1 (x) f 2 (x), so that the zeros we seek are the intersections of the graphs of f 1 and f 2, as shown below: 10 -x**2+x+4 exp(x) Using our knowledge of the behavior of the exponential function and of parabolas, it is not hard to see that there will be two zeros, one each in the intervals [ 2, 1] and [1, 2]. Bisection search results are listed at the of the problem in one big batch. b) f(x)=x 3 x 2 10x+7. Theory says that this polynomial cannot have more than three zeros. Plot it: 5

6 20 x**3-x**2-10*x It s not hard to see that we have three candidate intervals, [ 4, 3], [0, 1] and [3, 4]. c) f(x)= x + log x. Once again consider this function as composed of two parts: f 1 (x) = ( x), f 2 (x) = log x, f(x) = f 2 (x) f 1 (x). Again, we can view the sought zeros as intersections of the graphs of two well-understood functions: 0.6 -( *x) log(x) We are led to believe that we should look in [0.8, 1] and [1, 1.2]. 6

7 Code: function p6 disp( ) disp( Part a) ) disp( ) bisect(-2, bisect(1, disp( ) disp( Part b) ) disp( ) bisect(-4, bisect(0, bisect(3, disp( ) disp( Part c) ) disp( ) bisect(0.8, bisect(1, function bisect(a, b, f) tol = 1e-6 fa = f(a); fb = f(b); disp( ) a b % intermediate value theorem must guarantee existence of a % zero in (a,b). if (fa*fb > 0) error( Potentially no zero in search interval. ) while (abs(b-a) > tol) c = (a+b)/2.; fc = f(c); if (fc*fa <= 0) b = c; fb = fc; else a = c; fa = fc; 7

8 disp( ) a b function y = f_part_a(x) y = exp(x)+x^2-x-4; function y = f_part_b(x) y = x^3-x^2-10*x+7; function y = f_part_c(x) y = *x+log(x); Results: Part a) a = -2 b = -1 a = b = a = 1 b = 2 a = b = Part b) a = -4 b = -3 a = b = a = 0 b = 1 a = b = a = 3 8

9 b = 4 a = b = Part c) a = b = 1 a = b = a = 1 b = a = b = Problem 14, p. 105 First note that the functions in this problem are the same as above, so we may recycle our arguments for how many roots there are and in which interval we expect them to lie. a) f(x) = e x + x 2 x 4, f (x) = e x + 2x 1. As above, we need one root each in the intervals [ 2, 1] and [1, 2]. Each time, we choose the left boundary of the interval as a starting point. Results for Newton s method are listed in one big batch at the. b) f(x) = x 3 x 2 10x + 7, f (x) = 3x 2 2x 10. The candidate intervals are [ 4, 3], [0, 1] and [3, 4]. Again, we choose the left boundary of the interval as a starting point. c) f(x) = x + log x, f (x) = /x. [0.8, 1] and [1, 1.2] are the candidate intervals, and again we choose the left boundary of each interval as a starting point. Results: Part a) Iteration:0 x:-2 f(x): e+00 Iteration:1 x: f(x): e-01 Iteration:2 x: f(x): e-03 Iteration:3 x: f(x): e-07 Iteration:0 x:1 f(x): e+00 Iteration:1 x: f(x): e-01 Iteration:2 x: f(x): e-03 Iteration:3 x: f(x): e-06 Iteration:4 x: f(x): e-12 Part b) Iteration:0 x:-4 f(x): e+01 9

10 Iteration:1 x: f(x): e+00 Iteration:2 x: f(x): e-01 Iteration:3 x: f(x): e-03 Iteration:4 x: f(x): e-07 Iteration:0 x:0 f(x): e+00 Iteration:1 x:0.7 f(x): e-01 Iteration:2 x: f(x): e-04 Iteration:3 x: f(x): e-10 Iteration:0 x:3 f(x): e+00 Iteration:1 x: f(x): e+00 Iteration:2 x: f(x): e-02 Iteration:3 x: f(x): e-04 Iteration:4 x: f(x): e-09 Part c) Iteration:0 x:0.8 f(x): e-03 Iteration:1 x: f(x): e-04 Iteration:2 x: f(x): e-06 Iteration:3 x: f(x): e-10 Iteration:0 x:1 f(x): e-02 Iteration:1 x:1.25 f(x): e-02 Iteration:2 x: f(x): e-03 Iteration:3 x: f(x): e-04 Iteration:4 x: f(x): e-06 Iteration:5 x: f(x): e-11 Code: function p6 disp( ) disp( Part a) ) disp( 1e-6); disp( 1e-6); disp( ) disp( Part b) ) disp( 1e-6); disp( 1e-6); disp( 1e-6); disp( ) disp( Part c) ) disp( 1e-6); disp( 1e-6); 10

11 % function newton(x, f, fprime, tol) i = 0; while abs(f(x)/fprime(x)) > tol disp(sprintf( Iteration:%d x:%.10g f(x):%.10e, i, x, f(x))) x = x - f(x)/fprime(x); i = i + 1; disp(sprintf( Iteration:%d x:%.10g f(x):%.10e, i, x, f(x))) % function y = f_part_a(x) y = exp(x)+x^2-x-4; function y = fp_part_a(x) y = exp(x)+2*x-1; % function y = f_part_b(x) y = x^3-x^2-10*x+7; function y = fp_part_b(x) y = 3*x^2-2*x - 10 ; % function y = f_part_c(x) y = *x+log(x); function y = fp_part_c(x) y = /x; 11