Here's an ideal solution to HW1. Note the following * I used ''format compact'' to minimize empty lines and thus save electrons and trees. * I included as much of the problem statement as possible (dropping hard-to-type mathematical expressions * I edited out errors, calls to ''help'' and so on. * In some cases I showed several ways to accomplish the same task. I don't expect that of you. My homework set prints in just three pages, if I show just one way of solving each problem! **Please be sure to make your homework solutions as compact and readable as possible, using ''format compact'' and editing out extraneous material!** % John F. Gibson % 2014-09-12 % Math 445 HW1 format compact % Problem 1: Given two numeric variables x and y, write a Matlab expression that % evaluates to true (1) if and have opposite signs and false (0) otherwise. % Test the expression by evaluating it with the following pairs of numbers % (x,y) = (-5, 4), (5,4), (5,-4), (0,-2), and (3,0). % There are many ways to answer this question, ranging from the most obvious and % literal to the most readable and efficiently executed. Here's the obvious and % literal way x=-5; y=4; ((x<0) && (y>0)) || ((x>0) && (y<0)) ans = 1 x=5; y=4; ((x<0) && (y>0)) || ((x>0) && (y<0)) ans = 0 x=5; y=-4; ((x<0) && (y>0)) || ((x>0) && (y<0)) ans = 1 x=0; y=-2; ((x<0) && (y>0)) || ((x>0) && (y<0)) ans = 0 x=3; y=0; ((x<0) && (y>0)) || ((x>0) && (y<0)) ans = 0 % Here's a less complicated expression that accomplishes the same x=-5; y=4; x*y < 0 ans = 1 % and another x=-5; y=4; sign(x)*sign(y) == -1 ans = 1 % Problem 2: Find a matlab expression to computer the combined resistance RT of % three resistors R1,R2, R3. % The simple, straightforward way R1=5; R2=3; R3=4; RT = 1/(1/R1 + 1/R2 + 1/R3) RT = 1.2766 % Here's a slicker way to do the same R = [5 3 4]; RT = 1/sum(R.^(-1)) RT = 1.2766 % or even better RT = 1/sum([5 3 4].^(-1)) RT = 1.2766 % Problem 3: % (a) Create a row vector x whose elements are the numbers 5, 7, 10, 1. x = [5 7 10 1] x = 5 7 10 1 % (b) Create a column vector x whose elements are the numbers 5, 7, 10, 1. x = [5 ; 7 ; 10 ; 1] x = 5 7 10 1 x = [5 7 10 1]' x = 5 7 10 1 % (c) Use colon syntax to create a row vector x whose elements start at 0, end at 1, % and increase in steps of 0.1. x = 0:0.1:1 x = Columns 1 through 10 0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 Column 11 1.0000 % (d) Determine the dimension of x from (c) and assign the value to the variable d % (using Matlab, not by counting!). d = length(x) d = 11 d = size(x,2) d = 11 % (e) Use the linspace function to create a 10-dimensional vector of numbers evenly spaced between 0 and 1. x = linspace(0,1,10) x = 0 0.1111 0.2222 0.3333 0.4444 0.5556 0.6667 0.7778 0.8889 1.0000 % Problem 4 % (a) Create the given matrix A = [ 3 9 2 ; -1 4 6 ; 5 -2 0] A = 3 9 2 -1 4 6 5 -2 0 % (b) Change A(2,3) to 7 A(2,3) = 7 A = 3 9 2 -1 4 7 5 -2 0 % (c) Assign the third column of A to the variable v v = A(:,3) v = 2 7 0 % (d) Change the first row of A to 8,1,4 A(1,:) = [8 1 4] A = 8 1 4 -1 4 7 5 -2 0 % Problem 5. Create the given A matrix and b vector and solve Ax=b A = [4 2 ; -1 5] A = 4 2 -1 5 b = [3 4]' b = 3 4 x = A\b x = 0.3182 0.8636 % verify solution satisfies Ax=b A*x ans = 3 4 A*x-b ans = 0 0 norm(A*x-b) ans = 0 % Problem 6: Use Matlab to solve the problem. Nilanjana has 40 coins worth $6.40. % They're all quarters and nickels. How many nickels and how many quarters does % she have? Verify that your answer solves the problem. A = [1 1 ; 0.10 0.25] A = 1.0000 1.0000 0.1000 0.2500 b = [40 ; 6.40] b = 40.0000 6.4000 x = A\b x = 24.0000 16.0000 % She has 24 dimes and 16 quarters. That sums to 40 coins, and 24 * $0.10 + 16 * $0.25 = $6.40 A*x ans = 40.0000 6.4000 % Problem 7: Suhasini has 44 coins worth $7.50. They're all quarter, dimes, and nickels. % She has twice as many dimes as nickels. How many of each type of coin does she have? % Find the answer, and then verify that the solution satisfies the problem. The equations % are % % n + d + q = 44 % 0.05 n + 0.10 d + 0.25 q = 7.50 % d - 2n = 0 A = [1 1 1 ; 0.05 0.10 0.25 ; -2 1 0] A = 1.0000 1.0000 1.0000 0.0500 0.1000 0.2500 -2.0000 1.0000 0 b = [44 7.50 0]' b = 44.0000 7.5000 0 x = A\b x = 7 14 23 % She has 7 nickels, 14 dimes, and 23 quarters. Verifying... n = x(1); d = x(2); q = x(3); n + d + q ans = 44 0.05*n + 0.10*d + 0.25*q ans = 7.5000 d - 2*n ans = 0 norm(A*x-b) ans = 0