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By Ilijas Farah

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Extra info for Analytic Quotients: Theory of Liftings for Quotients over Analytic Ideals on the Integers (Memoirs of the American Mathematical Society)

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0 0 0 0 a1n a11 a2n a22 .. an−1n an−1n−1 0        The Gauss-Seidel process converges if and only if all the eigenvalues of the matrix [I + L]−1 U have modulus less than one. Successive over-relaxation (SOR) The SOR iterative scheme is x(r+1) = x(r) + ω(b − Lx(r+1) − x(r) − Ux(r) ), where 1 < ω < 2 for over-relaxation, and ω = 1 for Gauss-Seidel. The SOR process converges if and only if all the eigenvalues of the matrix [I + ωL]−1 [(1 − ω)I − ωU] have modulus less than one. 7 Chebyshev polynomials Recurrence relation Tn+1 (x) = 2xTn (x) − Tn−1 (x); T0 (x) = 1, T1 (x) = x Tn (x) = cos(n cos−1 x) (coefficient of xn is 2n−1 ) T2 (x) = 2x2 − 1, T3 (x) = 4x3 − 3x, T4 (x) = 8x4 − 8x2 + 1, T5 (x) = 16x5 − 20x3 + 5x The polynomial (1/2n−1 )Tn (x) has a smaller upper bound to its magnitude over [−1, 1] than that of any other polynomial with leading term xn .

If any off-diagonal element is non-zero, return to 1. LR method (for all the eigenvalues) Form the sequence A0 = A, Ar+1 = Ur Lr , where Ar = Lr Ur with Lr lower triangular, all diagonal elements are equal to 1 and Ur is upper triangular. For suitable A the sequence converges to an upper triangular matrix whose diagonal elements are the eigenvalues of A, arranged in order of decreasing modulus. If the method fails to converge try applying it again to A + pI where p is a suitable real number. Convergence in this case will be to λi + p.

The Fourier series for f (x) is given by: ∞ ∞ 1 nπx nπx F (x) = a0 + an cos + bn sin 2 l l n=1 n=1 where a0 = an = 1 l l f (x) cos −l 1 l l f (x)dx, −l nπx dx; l bn = l 1 l f (x) sin −l nπx dx l F (x) extends f (x) periodically with period 2l. F (x) = f (x) at all points, −l < x < l, at which f is continuous. If f is discontinuous at x0 , then F (x0 ) = {f (x0 +) + f (x0 −)}/2. Half-range series f (x) is defined for 0 < x < l. The Fourier half-range cosine series is: 1 Fc (x) = a0 + 2 ∞ an cos n=1 nπx ; l an = 2 l l f (x) cos 0 nπx dx.

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