- 5.7.1: Use the recursion formulas to calculate (a) T4, T5 and (b) H4, H5.
- 5.7.2: Let p0(x), p1(x), and p2(x) be orthogonal with respect to the inner...
- 5.7.3: Show that the Chebyshev polynomials have the following properties: ...
- 5.7.4: Find the best quadratic least squares approximation to ex on [1, 1]...
- 5.7.5: Let p0, p1, . . . be a sequence of orthogonal polynomials and let a...
- 5.7.6: Let Tn(x) denote the Chebyshev polynomial of degree n, and define U...
- 5.7.7: Let Un1(x) be defined as in Exercise 6 for n 1, and define U1(x) = ...
- 5.7.8: Show that the Ui s defined in Exercise 6 are orthogonal with respec...
- 5.7.9: Verify that the Legendre polynomial Pn(x) satisfies the second-orde...
- 5.7.10: Prove each of the following: (a) H_ n(x) = 2nHn1(x), n = 0, 1, . . ...
- 5.7.11: Given a function f (x) that passes through the points (1, 2), (2,1)...
- 5.7.12: Show that if f (x) is a polynomial of degree less than n, then f (x...
- 5.7.13: Use the zeros of the Legendre polynomial P2(x) to obtain a two-poin...
- 5.7.14: (a) For what degree polynomials will the quadrature formula in Exer...
- 5.7.15: Let x1, x2, . . . , xn be distinct points in the interval [1, 1] an...
- 5.7.16: Let x1, x2, . . . , xn be the roots of the Legendre polynomial Pn. ...
- 5.7.17: Let Q0(x), Q1(x), . . . be an orthonormal sequence of polynomials; ...
Solutions for Chapter 5.7: Orthogonal Polynomials
Full solutions for Linear Algebra with Applications | 8th Edition
Adjacency matrix of a graph.
Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected). Adjacency matrix of a graph. Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected).
Augmented matrix [A b].
Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.
Basis for V.
Independent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!
Put CI, ... ,Cn in row n and put n - 1 ones just above the main diagonal. Then det(A - AI) = ±(CI + c2A + C3A 2 + .•. + cnA n-l - An).
z = a - ib for any complex number z = a + ib. Then zz = Iz12.
Diagonal matrix D.
dij = 0 if i #- j. Block-diagonal: zero outside square blocks Du.
Echelon matrix U.
The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.
0,1,1,2,3,5, ... satisfy Fn = Fn-l + Fn- 2 = (A7 -A~)I()q -A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n - r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.
Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.
Hypercube matrix pl.
Row n + 1 counts corners, edges, faces, ... of a cube in Rn.
A symmetric matrix with eigenvalues of both signs (+ and - ).
Krylov subspace Kj(A, b).
The subspace spanned by b, Ab, ... , Aj-Ib. Numerical methods approximate A -I b by x j with residual b - Ax j in this subspace. A good basis for K j requires only multiplication by A at each step.
Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
IIA II. The ".e 2 norm" of A is the maximum ratio II Ax II/l1x II = O"max· Then II Ax II < IIAllllxll and IIABII < IIAIIIIBII and IIA + BII < IIAII + IIBII. Frobenius norm IIAII} = L La~. The.e 1 and.e oo norms are largest column and row sums of laij I.
The diagonal entry (first nonzero) at the time when a row is used in elimination.
Spectral Theorem A = QAQT.
Real symmetric A has real A'S and orthonormal q's.
Constant down each diagonal = time-invariant (shift-invariant) filter.
Trace of A
= sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.
v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.