## Transcribed Text

-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...