Midterm Exam with solutions

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1 Midterm Exam with solutions CS227-Introduction to Scientific Computation November 8, The following is a transcript of a MATLAB session. >> x=1/62.5 x = >> y=(1+x)-1 y = >> y==x ans = 0 >> xx=1/64 xx = >> yy=(1+xx)-1 yy = >> yy==xx ans =1 There are three things for you to explain here: (a) (5 points) The expressions x and (1+x)-1 are mathematically equivalent, but in the comparison y==x, MATLAB reports that they are unequal. Explain carefully and succinctly why this is the case. Solution. 1/62.5 is not represented exactly in binary so has an infinite binary expansion. It s representation by x is accurate to within 52 bits, but when we add 1, some of those bits are lost. Thus when we subtract 1, we get the value of 1/62.5 with some of the bits removed. (This is the same principle as catastrophic cancellation, but not quite as big a catastrophe.) (b) (5 points) Nonetheless, when asked to print the values of these two expressions in long decimal format, MATLAB gives the same result Why did this happen? 1

2 Solution. MATLAB is only displaying the results rounded to fourteen significant decimal digits, which is about 46 bits. So even though the values are different, they round the same. (c) (5 points) When the experiment is repeated with xx=1/64 in place of 1/62.5, MATLAB now reports that xx and (1+xx)-xx are equal. What is different in this case? Solution. 1/64 = 2 6 is exactly representable in binary (and most of the bits of the representation are zero). So no precision is lost when passing to or upon subtraction. 2

3 2. This problem concerns the MATLAB function ht displayed below. function vec = ht( c,t ) vec=zeros(c+1,1); for j=1:t result=floor(2*rand(1,c)); sumresult = sum(result); vec(sumresult+1)=vec(sumresult+1)+1; end Suppose you call this function with the command v=ht(5,30); (a) (5 points) What kind of object is v? That is, what are its dimensions, and what is its possible range of values? (To answer this second question, you might want to look at (b) below). Solution. v is a vector with 6 rows (you can see this from the initialization of the variable vec in the code). The answer below shows that every entry of v is an integer between 0 and 30, inclusive. (b) (5 points) Suppose you then typed the command sum(v) What would the response be? (This can be answered without understanding what the first line within the for loop is actually doing!) Solution. The response is 30. At each pass through the loop, one component of vec is incremented. (c) (5 points) Suppose you then typed the command vec What would you expect to see as the response? Solution. Unless you have already used a variable called vec in the Command Window, you should expect an error message about an undefined variable. This is because the name vec is private to the function. 3

4 (d) (5 points) Finally, let s address what this function is actually doing. Yes, it calls the function rand(), so it is performing some kind of random simulation. What is it simulating? What is the significance of the entries in v in terms of this simulation? Solution. It is simulating 30 tosses of 5 coins and counting, for each j between 0 and 5, the number of tosses that resulted in exactly j heads. (More generally, t tosses of c coins.) In fact, the j th entry of v is the number of tosses that resulted in exactly j 1 heads this is because we are obliged to begin the indexing at 1 rather than (5 points) What is the result of typing the commands below? >> m = [3 1-1;6 4 8; 0 0 1] >> [l u p]=lu(m) Just to be clear here, the function lu called in this fashion with a matrix M as an input argument and with three output arguments [L U P] returns three matrices satisfying P M = LU. Solution. Here is what MATLAB prints. l = u = p = 4

5 A single row interchange, switching the first two rows, is performed because of the partial pivoting policy: always bring the largest element in a column to the upper left. This is reflected in the matrix p. A single row operation, subtracting 1/2 times the first row from the second, is performed, which already gives an upper-triangular matrix. The matrix l is the inverse of this row operation. 4. You are working with a collection of student exam grades. The grades are on a scale between 0 and 100, and you want to estimate the grade distribution given the first, second, and third quartile grades. The first quartile is the grade q 1 such that 25% of the students scored q 1 or below (i.e., it is the 25 th percentile grade); the second quartile is the grade q 2 such that 50% of the students scored q 2 or below (i.e., q 2 is the median grade), and, likewise, q 3 is the 75 th percentile grade. You try to estimate the number of students who score any grade x or below by interpolating the points (0, 0), (q 1, 0.25), (q 2, 0.5), (q 3, 0.75), (100, 1) with a spline curve. The nice result for q 1 = 50, q 2 = 73, q 3 = 89, is shown in Figure 1. (a)(10 points) Write a sequence of MATLAB commands (not a function) that takes values q 1, q 2, q 3 and plots this spline distribution, highlighting the quartile points, as shown. Solution. This sequence of commands works, although as (b) indicates, the choice of the interpolation method is misguided. x=[0 q1 q2 q3 100]; y=linspace(0,1,5); xx=linspace(0,100,200); yy=interp1(x,y,xx, spline ); plot(x,y, ro,xx,yy); plot(xx,yy2); (b)(5 points) Unfortunately, certain choices of quartile values lead to pictures like Figure 2. This was created using q 1 = 40, q 2 = 60, and q 3 = 70. It makes no sense as a grade distribution, since it suggests, among other things, that 120% of the students received a grade of 90 or less! What is a very simple thing we can do to fix this problem? 5

6 Solution. Choosing linear or pchip interpolating gets rid of the problem. Using pchip gives smoother interpolation, but both of them give strictly increasing distributions. 5. If we use Newton s Method to compute the root of x 3 a, we get the following iteration: x k+1 = 2 3 x k + a, given some appropriate choice of x 0. We try to implement this method with the following function, which is supposed to return both the computed cube root and the number of iterations performed. 3x 2 k function [x numiter]=cuberoot(a) x=a; numiter =0; while xˆ3 = a x=2*x/3+a/(3*xˆ2); end numiter=numiter+1; end (a) (10 points) The function contains several errors, one of which causes it to never terminate for some values, even when the sequence of iterates really does converge, and the other of which causes it to return an incorrect number of iterations. Describe how to correct these errors. Solution. One problem is trying to find the root exactly. Typically this will not be possible. One correction is to look for only an approximate match, say to get the residual to within a. We could do this with while abs(xˆ3-a)> (1e-10)* a The second problem is that numiter is incremented outside the loop, so that it always will return with the value 1. It should be moved before the end of the while statement. (b) (5 points) To answer this question, you might want to consult the graph of x 3 a, which is pictured below. Assume a > 0. Does the function, as written, 6

7 always converge? What if we used a different initial guess? For how many initial guesses does the algorithm fail to converge, and where are these guesses located? Solution. For a > 0 Newton s method always converges. If a > 1 then the tangent lines approach the cube root. If a < 1, the first tangent line lands to the right of the root, and then again approaches from the left. This is clear from the direction of concavity of the function. If we use an initial guess of x 0 = 0, then the tangent line is horizontal, so the method does not converge. But it also fails to converge if we choose x 0 = b 1 so that the tangent line passes through (0, 0). You can tell from the graph that there must be such a point (it actually occurs at b 1 = (a/2) 1 3. As a result, it will fails to converge if we choose x 0 = b 2 < b 1 so that the tangent line at b 2 passes through (b 1, 0). Proceeding in this fashion we find infinitely many negative values of the initial guess for which the method fails to converge. Figure 1: Grade Distribution from Splines 7

8 Figure 2: Ridiculous Grade Distribution from Splines 8

9 Figure 3: Graph of f(x) = x