- 220.127.116.11.44: Verify that the polynomials in (11') satisfy Legendre's equation.
- 18.104.22.168.45: Derive (11 ') from (11).
- 22.214.171.124.46: Obtain P6 and P7 from (11).
- 126.96.36.199.47: Convergence) Show that for any 11 for which (6) or (7) does nol red...
- 188.8.131.52.48: Legendre function Qo(x) for n = 0) Show that (6)with 11 = 0 gives Y...
- 184.108.40.206.49: (Legendre function -Ql(X) for II = 1) Show that (7) with 11 = I giv...
- 220.127.116.11.50: (ODE) Find a solution of (a 2 - x 2 )y" - 2xy' + n(/1 + l)y = 0. a ...
- 18.104.22.168.51: [Rodrigues's formula (12)]2 Applying the binomial theorem to (X2 - ...
- 22.214.171.124.52: (Rodrigues's formula) Obtain (II ') from (12).
- 126.96.36.199.53: Graph P2 (x) ... PIO(.\) on common axes. For what \" (approximately...
- 188.8.131.52.54: From what /I on will your CAS no longer produce faithful graphs of ...
- 184.108.40.206.55: Graph Qo(x), QI (x), and some further Legendre functions.
- 220.127.116.11.56: Substitute asxs + a s+ IXs + 1 + as+2xs+2 into Legendre's equation ...
- 18.104.22.168.57: TEAM PROJECT. Generating Functions. Generating functions playa sign...
Solutions for Chapter 5.3: Legendre's Equation. Legendre Polynomials P nex)
Full solutions for Advanced Engineering Mathematics | 9th Edition
Tv = Av + Vo = linear transformation plus shift.
Circulant matrix C.
Constant diagonals wrap around as in cyclic shift S. Every circulant is Col + CIS + ... + Cn_lSn - l . Cx = convolution c * x. Eigenvectors in F.
Cross product u xv in R3:
Vector perpendicular to u and v, length Ilullllvlll sin el = area of parallelogram, u x v = "determinant" of [i j k; UI U2 U3; VI V2 V3].
Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.
Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).
Fast Fourier Transform (FFT).
A factorization of the Fourier matrix Fn into e = log2 n matrices Si times a permutation. Each Si needs only nl2 multiplications, so Fnx and Fn-1c can be computed with ne/2 multiplications. Revolutionary.
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].
Gram-Schmidt orthogonalization A = QR.
Independent columns in A, orthonormal columns in Q. Each column q j of Q is a combination of the first j columns of A (and conversely, so R is upper triangular). Convention: diag(R) > o.
Independent vectors VI, .. " vk.
No combination cl VI + ... + qVk = zero vector unless all ci = O. If the v's are the columns of A, the only solution to Ax = 0 is x = o.
Inverse matrix A-I.
Square matrix with A-I A = I and AA-l = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B-1 A-I and (A-I)T. Cofactor formula (A-l)ij = Cji! detA.
Jordan form 1 = M- 1 AM.
If A has s independent eigenvectors, its "generalized" eigenvector matrix M gives 1 = diag(lt, ... , 1s). The block his Akh +Nk where Nk has 1 's on diagonall. Each block has one eigenvalue Ak and one eigenvector.
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.
Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.
= Xl (column 1) + ... + xn(column n) = combination of columns.
Reduced row echelon form R = rref(A).
Pivots = 1; zeros above and below pivots; the r nonzero rows of R give a basis for the row space of A.
Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)-l has AA+ = 1m.
Schur complement S, D - C A -} B.
Appears in block elimination on [~ g ].
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.
Singular matrix A.
A square matrix that has no inverse: det(A) = o.
Skew-symmetric matrix K.
The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.
v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.