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11: 24.2 Definitions and Generating Functions
B 2 n + 1 = 0 ,
24.2.4 B n = B n ( 0 ) ,
Table 24.2.4: Euler numbers E n .
n E n
Table 24.2.5: Coefficients b n , k of the Bernoulli polynomials B n ( x ) = k = 0 n b n , k x k .
k
Table 24.2.6: Coefficients e n , k of the Euler polynomials E n ( x ) = k = 0 n e n , k x k .
k
12: 3.6 Linear Difference Equations
Given numerical values of w 0 and w 1 , the solution w n of the equation …These errors have the effect of perturbing the solution by unwanted small multiples of w n and of an independent solution g n , say. … The unwanted multiples of g n now decay in comparison with w n , hence are of little consequence. … The latter method is usually superior when the true value of w 0 is zero or pathologically small. … beginning with e 0 = w 0 . …
13: 10.75 Tables
  • Wills et al. (1982) tabulates j 0 , m , j 1 , m , y 0 , m , y 1 , m for m = 1 ( 1 ) 30 , 35D.

  • MacDonald (1989) tabulates the first 30 zeros, in ascending order of absolute value in the fourth quadrant, of the function J 0 ( z ) i J 1 ( z ) , 6D. (Other zeros of this function can be obtained by reflection in the imaginary axis).

  • Abramowitz and Stegun (1964, Chapter 11) tabulates 0 x J 0 ( t ) d t , 0 x Y 0 ( t ) d t , x = 0 ( .1 ) 10 , 10D; 0 x t 1 ( 1 J 0 ( t ) ) d t , x t 1 Y 0 ( t ) d t , x = 0 ( .1 ) 5 , 8D.

  • Abramowitz and Stegun (1964, Chapter 11) tabulates e x 0 x I 0 ( t ) d t , e x x K 0 ( t ) d t , x = 0 ( .1 ) 10 , 7D; e x 0 x t 1 ( I 0 ( t ) 1 ) d t , x e x x t 1 K 0 ( t ) d t , x = 0 ( .1 ) 5 , 6D.

  • Zhang and Jin (1996, pp. 296–305) tabulates 𝗃 n ( x ) , 𝗃 n ( x ) , 𝗒 n ( x ) , 𝗒 n ( x ) , 𝗂 n ( 1 ) ( x ) , 𝗂 n ( 1 ) ( x ) , 𝗄 n ( x ) , 𝗄 n ( x ) , n = 0 ( 1 ) 10 ( 10 ) 30 , 50, 100, x = 1 , 5, 10, 25, 50, 100, 8S; x 𝗃 n ( x ) , ( x 𝗃 n ( x ) ) , x 𝗒 n ( x ) , ( x 𝗒 n ( x ) ) (Riccati–Bessel functions and their derivatives), n = 0 ( 1 ) 10 ( 10 ) 30 , 50, 100, x = 1 , 5, 10, 25, 50, 100, 8S; real and imaginary parts of 𝗃 n ( z ) , 𝗃 n ( z ) , 𝗒 n ( z ) , 𝗒 n ( z ) , 𝗂 n ( 1 ) ( z ) , 𝗂 n ( 1 ) ( z ) , 𝗄 n ( z ) , 𝗄 n ( z ) , n = 0 ( 1 ) 15 , 20(10)50, 100, z = 4 + 2 i , 20 + 10 i , 8S. (For the notation replace j , y , i , k by 𝗃 , 𝗒 , 𝗂 ( 1 ) , 𝗄 , respectively.)

  • 14: 27.2 Functions
    where p 1 , p 2 , , p ν ( n ) are the distinct prime factors of n , each exponent a r is positive, and ν ( n ) is the number of distinct primes dividing n . … Note that σ 0 ( n ) = d ( n ) . …Note that J 1 ( n ) = ϕ ( n ) . In the following examples, a 1 , , a ν ( n ) are the exponents in the factorization of n in (27.2.1). … Table 27.2.1 lists the first 100 prime numbers p n . …
    15: 3.11 Approximation Techniques
    Beginning with u n + 1 = 0 , u n = c n , we apply … With b 0 = 1 , the last q equations give b 1 , , b q as the solution of a system of linear equations. … (3.11.29) is a system of n + 1 linear equations for the coefficients a 0 , a 1 , , a n . … With this choice of a k and f j = f ( x j ) , the corresponding sum (3.11.32) vanishes. … Two are endpoints: ( x 0 , y 0 ) and ( x 3 , y 3 ) ; the other points ( x 1 , y 1 ) and ( x 2 , y 2 ) are control points. …
    16: 3.7 Ordinary Differential Equations
    The path is partitioned at P + 1 points labeled successively z 0 , z 1 , , z P , with z 0 = a , z P = b . … Write τ j = z j + 1 z j , j = 0 , 1 , , P , expand w ( z ) and w ( z ) in Taylor series (§1.10(i)) centered at z = z j , and apply (3.7.2). … If, for example, β 0 = β 1 = 0 , then on moving the contributions of w ( z 0 ) and w ( z P ) to the right-hand side of (3.7.13) the resulting system of equations is not tridiagonal, but can readily be made tridiagonal by annihilating the elements of 𝐀 P that lie below the main diagonal and its two adjacent diagonals. … The values λ k are the eigenvalues and the corresponding solutions w k of the differential equation are the eigenfunctions. … where h = z n + 1 z n and …
    17: 3.2 Linear Algebra
    where u j = c j , j = 1 , 2 , , n 1 , d 1 = b 1 , and …Forward elimination for solving 𝐀 𝐱 = 𝐟 then becomes y 1 = f 1 , …and back substitution is x n = y n / d n , followed by … Define the Lanczos vectors 𝐯 j and coefficients α j and β j by 𝐯 0 = 𝟎 , a normalized vector 𝐯 1 (perhaps chosen randomly), α 1 = 𝐯 1 T 𝐀 𝐯 1 , β 1 = 0 , and for j = 1 , 2 , , n 1 by the recursive scheme … Start with 𝐯 0 = 𝟎 , vector 𝐯 1 such that 𝐯 1 T 𝐒 𝐯 1 = 1 , α 1 = 𝐯 1 T 𝐀 𝐯 1 , β 1 = 0 . …
    18: 19.29 Reduction of General Elliptic Integrals
    Let …where … Next, for j = 1 , 2 , define Q j ( t ) = f j + g j t + h j t 2 , and assume both Q ’s are positive for y < t < x . …where …If Q 1 ( t ) = ( a 1 + b 1 t ) ( a 2 + b 2 t ) , where both linear factors are positive for y < t < x , and Q 2 ( t ) = f 2 + g 2 t + h 2 t 2 , then (19.29.25) is modified so that …
    19: 3.9 Acceleration of Convergence
    A transformation of a convergent sequence { s n } with limit σ into a sequence { t n } is called limit-preserving if { t n } converges to the same limit σ . … This transformation is accelerating if { s n } is a linearly convergent sequence, i. … Then the transformation of the sequence { s n } into a sequence { t n , 2 k } is given by … Then t n , 2 k = ε 2 k ( n ) . … We give a special form of Levin’s transformation in which the sequence s = { s n } of partial sums s n = j = 0 n a j is transformed into: …
    20: 18.2 General Orthogonal Polynomials
    The h n are defined as: … Assume that the p n ( x ) are monic, so A n = 1 = a n . … Let x 1 , , x n be the zeros of the OP p n , so …The discriminant of p n is defined by … where x 1 , x 2 , , x n are the zeros of p n ( x ) and …