Math-3315 & CSE-3365, exam 2 answer sheet

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1 Math-3315 & CSE-3365, exam 2 answer sheet Oct. 23, 2012 Problem 1. (20 points) Given a vector z that contains all distinct elements z 1,z 2,,z n. The first task is to generate an n n Vandermonde-type matrix V that has the following form z1 2 z2 2 z3 2 z4 2 zn 2 z 2 n V = z1 4 z2 4 z3 4 z4 4 zn 4 zn z1 2n 2 z2 2n 2 z3 2n 2 z4 2n 2 zn 2n 2 zn 2n 2 That is, the i-th row of V is z 2(i), elementwise, for i = 1,2,,n. The second task is to solve a linear equation Vx = b for x, assuming that b is a given vector with length n. Complete the following Matlab code for these two tasks. You are allowed to use only one explicit loop when generating V, without using the Matlab power function. Your code should strive for as less computational cost as possible. function [V, x] = vandsquare(z, b) n = length(z); V = ones(n); %make sure z is a row vector, transpose z if it is a column vector if (size(z,1) > size(z,2)), z=z ; zsqr = z.* z; %should reuse this computed vector in the loop generating V %generate V using just one loop (by constructing V row by row) for i = 2 : n V(i, :) = V(i-1, :).* zsqr; %make sure b is a column vector, transpose b if it is a row vector %(hint: you can get some hint from a similar line for vector z above) if (size(b,1) < size(b,2)), b=b ; %solve the linear equation Vx =b for x x = V \ b; 1

2 Problem 2. (20 points) Given a vector z that contains all distinct elements z 1,z 2,,z n. The task is to generate the n n Newton matrix N that has the following lower triangular form (here the upper triangular 0 elements are not listed) 1 1 z 2 z 1 1 z 3 z 2 1 l=1 N = (z 3 z l ) 1 z 4 z 2 1 l=1 (z 4 z l )... 1 z n z 2 1 l=1 (z n z l ) 3 l=1 (z 4 z l ) l=1 (z n z l ). n l=1 (z n z l ) That is, the first column contains all 1 s, and for j > 1 the (i,j)-th element of N is (z i z l ). Complete the following Matlab code to generate the N matrix. You are allowed to use only one explicit loop when generating N. Your code should strive for as less computational cost as possible. j l=1 function [ N ] = newtonmat(z); % set n as the length of z n = length(z); % assign N as an nxn zero matrix N = zeros(n); % assign the first column of N as all 1 s N(:,1) = ones(n,1); % make sure z is a column vector (this is needed when constructing N) if (size(z,1) < size(z,2)), z = z ; % apply one loop to generate N column by column for j = 1 : n-1 N(:, j+1) = N(:, j).* (z - z(j)); 2

3 Problem 3. (20 points) Evaluation of polynomials using Horner s method. (i) Complete the Matlab code such that for a given coefficient vector coef and a given x (which could be a scalar or a vector), it will return the polynomial value coef(1)+coef(2) x+coef(3) x.ˆ2+ +coef(n+1) x.ˆn. in pval (which could be a scalar or a vector, deping on x). function [pval] = hornerval(coef, x) % evaluate polynomial in standard form using Horner s method n = length(coef)-1; pval = coef(n+1); for i = n : -1 : 1 pval = pval.*x + coef(i); Download and experiment with this code to see the Horner s method in action: (ii) Complete the Matlab code: For a given coefficient vector coef, another vector pt storing interpolation points, and a given x (which could be a scalar or a vector), the code will return the Newton polynomial value coef(1)+coef(2) (x pt(1))+coef(3) (x pt(1)). (x pt(2))+ n +coef(n+1) (x pt(j)) in pval (which could be a scalar or a vector, deping on x). j=1 function [pval]=newtonpval(coef, pt, x) % evaluate polynomial in Newton form using Horner s method n = length(coef)-1; pval = coef(n+1); for i = n : -1 : 1 pval = pval.*(x - pt(i)) + coef(i); Download and experiment with this code to see the Horner s method in action: 3

4 Problem 4. (20 points, each line 2 points) Complete the following Matlab code that implements the composite midpoint, trapezoidal, and 1/3-Simpson rules. (There are certainly different ways to implement the rules, but your code should fit in the structure that is already in place here.) function [quadf] = myquad(f, a, b, n, method) % compute integral of f(x) on [a,b] using n+1 equi-distant % grid points with three composite quadrature rules. % the input variable method controls which rule to use: % method = composite midpoint rule % method = composite trapezoidal rule % method = composite 1/3 simpson rule % the integral value is output as quadf. n = max(ceil(n), 3); %make sure n is a positive integer > 2 h = (b-a)/n; if (method==1) %the composite midpoint rule quadf = 0; x = a+h/2; for i = 1 : n quadf = quadf + f(x); x = x + h; quadf = quadf*h; elseif (method==2) %the composite trapezoidal rule quadf = (f(a)+f(b))/2 ; x = a; for i = 1 : n-1 quadf = quadf + f(x); quadf = quadf*h; elseif (method==3) %the composite 1/3 simpson rule if (mod(n,2)~=0), n=n+1; h=(b-a)/n; %if n odd, make n even and recompute h quadf = f(a)+f(b); q4 = 0; %q4 used to store the term that is to be multiplied by 4 q2 = 0; %q2 used to store the term that is to be multiplied by 2 x = a; for i = 1 : n/2-1 q4 = q4 + f(x); q2 = q2 + f(x); q4 = q4 + f(x+h); quadf = (quadf + 4*q4 + 2*q2)*(h/3); Download and experiment with this code about the quadrature rules: 4

5 Problem 5. (20 points) Find the coefficients A, B, C for the quadrature rule f(x)dx Af( 2 )+Bf(0)+Cf(1 2 ) such that it is accurate for 1,x,x 2. Is the quadrature rule you find accurate for x 2k+1 for any integer k 1? Is it also accurate for x 4? Answer: Plugging f(x) = 1,x,x 2 respectively into the quadracture rule we get 3 equntions 1dx = 2 = A+B +C xdx = 0 = 2 A+0B C x 2 dx = 2 3 = 1 4 A+0B C Solving the equations we get the quadrature rule is A = C = 4 3, B = 2 3, f(x)dx 4 3 f( 2 ) 2 3 f(0)+ 4 3 f(1 2 ). Plugging in f(x) = x 2k+1 into the rule we get x 2k+1 dx = 0 = 4 3 ( 2 )2k (1 2 )2k+1, which is always true. So the quadrature rule is exact for x 2k+1 for any integer k 1. Plugging in f(x) = x 4 into the rule we get x 4 dx = 2 5 = 4 3 ( 2 ) (1 2 )4 = 1 6, which is not true, therefore the quadrature rule is not exact for x 4. 5