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-1- (Positive Definite Matrices.) A principal submatrix A1 of an nxn matrix A is obtained by picking a set IC {1, 2, . . . , n} and crossing out all rows and columns whose indices are not in J. For example, if a12 a13 a14 ] a22 a23 a24 a32 a33 a34 a42 a43 a44 then the principal submatrix corresponding to the set {2, 4} is Show that every principal submatrix of a positive definite matrix is positive definite. (1) (2) -2- (The UL factorization.) Show how to compute the factorization A = UL where U is upper triangular with ls along the diagonal and Lis lower triangular. Show how this relates to a way of solving Ax = b by transforming the system into an equivalent system with a lower triangular matrix. (In other words, show that what we did for the LU factorization also works for a UL factorization.) Note: For the purposes of this exercise you may assume that no pivoting is required. This is of course unrealistic but pivoting would only distract from the point of this exercise ( which is that conceptually there is no difference between an LU and a UL factorization). -3- (Spectral Radius.) We saw that the spectral radius of a (square) matrix A never exceeds an induced matrix norm of IIAII. It can be shown that for any particular matrix A one 1 can find a vector norm such that the induced matrix norm of A is arbitrarily close to the spectral radius of A. Does the spectral radius itself define a norm? \i\Thy, or why not? -4- (Inequalities are sharp.) Explain the meaning of 1 llrll < llell < -1 llrll IIAII IIA-1111fij - TI;if - IIAII IIA lllfij and show how we derived these inequalities in class. (3) a. For a general matrix A, and II · II = II · 1'2, show that there are non-trivial examples (i.e., x =I= 0 =I= e) where the right hand inequality is satisfied with equality in (3). Do the same for the left hand i neq uali ty in ( 3). b. Repeat part a. for II· II = 11 · lloo -5- (Backward Error Analysis.) This problem explores the effects of a perturbation in the coefficient matrix (rather than the right hand side) of the linear system Ax= b Suppose we solve instead of ( 4) the system ( A - E) ( x - e) = b where E is a perturbation of A that causes an error e in the solution x. Show that llell llx -ell - < IIAII IIA-1IIIIAII -6- (First Order Systems.) \\Trite the second order initial value problem y" = xy2 ' y(O) = 1, y'(O) = 2 as an autonomous first order system y' = f(y), y(a) = Yo• (4) (5) (6) (7) (8) (In other words, specify y, f, a, and y0 such that the two problems are equivalent. Of course, y will have different meanings for the two problems.) 2 -7 - (Improper Integrals.) Let I= f l ln(x 2 + 1) dx = 21rln [ 1 + )2] · -1 J1 - x2 2 (9) This integral is improper because the integrand approaches infinity as x approaches ±1. Attempt to approximate this integral by using Simpson's Rule (or any other Newton-Cotes formula) on the interval [-1 + t, 1 - t] for small E > 0. Describe the results of your efforts. Then for n = 4, 5, 6, 7, 8, use the Gaussian Quadrature formula f l f ( x ) dx = t f (cos ( (2i - l)1r)) _1 J1 - x2 n i=l 2n which we discussed in class. Compute the error and discuss your results. (10) -8- (Numerical Comparison.) For Euler's Method, the Trapezoidal Rule, Simpson's Rule, and the standard 4-th order Runge-Kutta Method* solve the initial value problem y' = y , y ( 0) = 1, x E [ 0, 1] with step-sizes h = 2-s where s = 3, 4, · · ·, 15. Plot the error at the right end-point, i.e., y( 1) - Yn ( where n = l / h), against h. You may wish to superimpose the plots. If you like use appropriate logarithmic graph paper. Comment on your results. (For Simpson's rule use the exact starting value y1 = y(h).) Send me your program by e-mail. * The "standard" Runge-Kutta methods is: h Yn+1 = Yn + 6 (K1 + 2K2 + 2I<3 + K4) where I<1 = f (xn· Yn ) K2 = f (xn + %,Yn + %K1) (11) I<3 = f (xn + %,Yn + K2) J<4 = f (xn + h. Yn + hK3) . 3

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%y'=y; y(0)=1; y(1)=e

clear all

ss=3:15;
errorE=zeros(size(ss));
errorT=zeros(size(ss));
errorS=zeros(size(ss));
errorR=zeros(size(ss));
for k=1:length(ss)
    s=ss(k);
    h=2^(-s);
    %yE: Euler method
    %yT: Trapezoid method
    %yS: Simpsons's method
    %yR: Runge Kutta
    n=2^s;
    for j=1:n
       if j==1
            yE(j)=1+h*1;
            yT(j)=(1+h/2)/(1-h/2)*1;
            yS(j)=exp(h);%exact solution
            K1=1;
            K2=1+h/2*K1;
            K3=1+h/2*K2;
            K4=1+h*K3;
            yR(j)=1+h/6*(K1+2*K2+2*K3+K4);
       else
            yE(j)=yE(j-1)+h*yE(j-1);
            yT(j)=(1+h/2)/(1-h/2)*yT(j-1);
            if j==2
                yS(j)=(1+h/3*(1+4*yS(j-1)))/(1-h/3);
            else
                yS(j)=(yS(j-2)+h/3*(yS(j-2)+4*yS(j-1)))/(1-h/3);
            end
            K1=yR(j-1);
            K2=yR(j-1)+h/2*K1;
            K3=yR(j-1)+h/2*K2;
            K4=yR(j-1)+h*K3;
            yR(j)=yR(j-1)+h/6*(K1+2*K2+2*K3+K4);
       end
    end...

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