Math 502 Fall 2005 Solutions to Homework 3

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1 Math 502 Fall 2005 Solutions to Homework 3 (1) As shown in class, the relative distance between adjacent binary floating points numbers is 2 1 t, where t is the number of digits in the mantissa. Since every single precision number also has a double precision representation and distances are being given in a relative sense we can assume the adjacent numbers are of unit size. Then length of the interval between them is s The length of an interval between adjacent double precision numbers is d Since there is s / d 2 29 subintervals of length d in an interval of length s, there are double precision numbers strictly between two adjacent single precision numbers. (2) Here s a simple MATLAB script that fulfills the requirements of problem 2. The graph it produces follows the solution of problem 6. Script File: polyplot This scripts compares two different ways of generating data for plotting the polynomial p(x) (x-2)^9 a(10)*x^ a(2)x + a(1) on the interval 1.920,2.080]. This is a solution to problem 13.3 in Trefethen s book the coefficients in the expansion of p(x) are: a -512;2304;-4608;5376;-4032;2016;-672;144;-18;1]; x 1.920:0.001:2.080; pts spaced evenly in 1.920,2.080] p1 a(10)*ones(size(x)); for k 9:-1:1 p1 a(k) + x.*p1; end p2 (x - 2).^9; Horner s algorithm with vector processing to calculate all values at once the alternate method of evaluation plot(x,p1,x,p2, r-. ); xlabel( x ) ylabel( p ) title( Graphs of p(x) (x-2)^9 computed two different ways. ) (3) By definition, there are positive constants C 1, C 2 and numbers ɛ 1, ɛ 2 > 0 such that φ(ɛ) C 1 ɛ, for all ɛ ɛ 1, ψ(ɛ) C 2 ɛ, for all ɛ ɛ 2.

2 (a) Clearly f(ɛ) 1 + φ(ɛ) + ψ(ɛ) + φ(ɛ)ψ(ɛ). Let ɛ 0 min{1, ɛ 1, ɛ 2 }. From the discussion above it follows that f(ɛ) 1 C 1 ɛ + C 2 ɛ + C 1 C 2 ɛ 2 (C 1 + C 2 + C 1 C 2 ɛ ) ɛ C ɛ, for all ɛ ɛ 0, where C C 1 + C 2 + C 1 C 2. Hence f(ɛ) 1 + O(ɛ), as ɛ 0. (b) By a simple algebraic trick we have g(ɛ) φ(ɛ) φ(ɛ) 1 φ(ɛ) 1 + φ(ɛ) 1 + φ(ɛ) 1 + φ(ɛ). Let ɛ 0 min{ 1 2C 1, ɛ 1 }. Then for all ɛ ɛ 0 we have φ(ɛ) C 1 ɛ and φ(ɛ) 1 2. The second inequality implies 1 + φ(ɛ) 1 + φ(ɛ) 1 2. Therefore g(ɛ) 1 φ(ɛ) 2 φ(ɛ) 2C1 ɛ, 1 + φ(ɛ) for all ɛ ɛ 0. Thus g(ɛ) 1 + O(ɛ), as ɛ 0. (4) ( ) Suppose that A has an LU factorization. Since A is nonsingular det(a) det(l)det(u) u 11 u 22 u mm 0. In particular u ii 0 for i 1,..., m. For each k 1,..., m there are conformable partitions of A, L, U such that ] ] ] A11 A 12 L11 0 U11 U 12 A 21 A 22 L 21 L 22 0 U 22 with L 11 being a (k k) unit lower triangular matrix, U 11 being a (k k) upper triangular matrix, and A 11 L 11 U 11. Since det(a 11 ) det(u 11 ) u 11 u 0, A 11 is nonsingular. ( ) Let A denote the upper left (k k) block of A and suppose A is nonsingular for each k 1,..., m. We use induction to show that a zero pivot element is never encountered during the Gaussian elimination process. Since a 11 det(a 11 ) 0 this is true for the first step of Gaussian elimination. After k 1 steps have been performed we have ] Uk 1 V L k 1 L 2 L 1 A 0 W where U k 1 is (k 1) (k 1)] upper triangular with nonzero diagonal entries. The upper left (k k) block of this matrix is Uk 1 v ] 0 a (k 1) where v is a column vector and a (k 1) is the diagonal entry that is obtained from a after k 1 steps, and is the pivot element for step k. This matrix is identical to the matrix that would be obtained from A by Gaussian elimination. Since the elementary rows operations used in Gaussian elimination do not alter determinants, we have det(a ) a (k 1) det(u k 1 ). Hence a (k 1) is a nonzero pivot. (Uniqueness) Suppose that A L 1 U 1 L 2 U 2 where L 1, L 2 are unit lower triangular matrices and U 1, U 2 are upper triangular matrices. Since

3 A is nonsingular so are L 1, L 2, U 1, U 2. Hence the identity L 1 U 1 L 2 U 2 implies L 1 2 L 1 U 2 U1 1. It is easily verified that the inverse of a unit lower triangular matrix is also a unit lower triangular matrix, and that the product of two unit lower triangular matrices is again a unit lower triangular matrix. Analogous statements hold for upper triangular matrices. Thus L 1 2 L 1 is unit lower triangular and U 2 U 1 1 is upper triangular. Since they are equal they must be diagonal, with ones along the diagonal. That is, L 1 2 L 1 I U 2 U 1 1, which implies L 1 L 2, U 1 U 2. (5) (a) This is obvious once it is observed that A 21 A 1 11 A 11 + A 21 A 21 + A (b) After n steps of Gaussian elimination we have L11 0 I L 21 ] ] A11 A 12 A 21 A 22 U11 V 0 W where L 11, A 11 and U 11 are all (n n) and the block structures are conformable. Comparing the lower left blocks shows L 21 A 11 +A Since A 11 is invertible by assumption, L 21 A 21 A Comparing the lower right blocks and using this fact we have ], W L 21 A 12 + A 22 A 22 A 21 A 1 11 A 12. (6) We use induction on the size of the matrix. Obviously it is true if m 2. Suppose it is true for matrices of size m 1, and consider a matrix A of size m. Since a 11 > i>1 a i1 a k1, for each k 2,..., m there is no row interchange made in the first step. Let A (1) a (1) ij ] L 1A be the matrix obtained after the the first elimination step. The rest of the process only involves the subblock a (1) ij ]m i,j2 which is a matrix of size (m 1) (m 1)]. We show that this matrix is strictly column diagonally dominant, and thus conclude from the induction hypothesis that no row interchanges will take place during the entire process. The entries of the subblock are a (1) ij a ij (a i1 /a 11 )a 1j. For each j 2,..., m, we have m i2,i j ij i2,i j i2,i j i1,i j a ij a i1 a 11 a 1j a ij + a 1j a 11 i2,i j a ij a 1j + a 1j a 11 a i1 ( m i2 < a jj a 1j + a 1j a 11 ( a 11 a j1 ) a jj a 1j a 11 a j1. ) a i1 a j1 We also know ij a jj a 1j a j1 a jj a 1j a 11 a 11 a j1 a jj a 1j a 11 a j1,

4 since a 1j a 11 a j1 a 1j a jj. Combining these inequalities shows ij < a(1) jj, j 2,..., m. i2,i j Thus, the subblock is strictly column diagonally dominant.

5 1.5 x Graphs of p(x) (x 2) 9 computed two different ways p x