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Cauchy sum

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11: 19.21 Connection Formulas
19.21.11 6 R G ( x , y , z ) = 3 ( x + y + z ) R F ( x , y , z ) x 2 R D ( y , z , x ) = x ( y + z ) R D ( y , z , x ) ,
The latter case allows evaluation of Cauchy principal values (see (19.20.14)). …
12: 19.22 Quadratic Transformations
19.22.9 4 π R G ( 0 , a 0 2 , g 0 2 ) = 1 M ( a 0 , g 0 ) ( a 0 2 n = 0 2 n 1 c n 2 ) = 1 M ( a 0 , g 0 ) ( a 1 2 n = 2 2 n 1 c n 2 ) ,
19.22.10 R D ( 0 , g 0 2 , a 0 2 ) = 3 π 4 M ( a 0 , g 0 ) a 0 2 n = 0 Q n ,
19.22.12 R J ( 0 , g 0 2 , a 0 2 , p 0 2 ) = 3 π 4 M ( a 0 , g 0 ) p 0 2 n = 0 Q n ,
If the last variable of R J is negative, then the Cauchy principal value is
19.22.14 R J ( 0 , g 0 2 , a 0 2 , q 0 2 ) = 3 π 4 M ( a 0 , g 0 ) ( q 0 2 + a 0 2 ) ( 2 + a 0 2 g 0 2 q 0 2 + g 0 2 n = 0 Q n ) ,
13: 1.3 Determinants, Linear Operators, and Spectral Expansions
1.3.4 det [ a j k ] = = 1 n a j A j .
1.3.9 det [ a j k ] 2 ( k = 1 n a 1 k 2 ) ( k = 1 n a 2 k 2 ) ( k = 1 n a n k 2 ) .
for every distinct pair of j , k , or when one of the factors k = 1 n a j k 2 vanishes. …
Cauchy Determinant
1.3.19 j , k = | a j , k δ j , k |
14: 3.5 Quadrature
3.5.5 f ( t ) d t = h k = f ( k h ) + E h ( f ) ,
3.5.12 G 0 ( 1 2 h ) = 1 2 G 0 ( h ) + 1 2 h k = 0 n 1 f ( x 0 + ( k + 1 2 ) h ) ,
3.5.15 a b f ( x ) w ( x ) d x = k = 1 n w k f ( x k ) + E n ( f ) ,
3.5.46 f ( x ) = 1 π f ( t ) t x d t , x ,
where the integral is the Cauchy principal value. …
15: 1.2 Elementary Algebra
which for p = q = 2 is the Cauchy-Schwartz inequalityIf det ( 𝐀 ) = 0 then, depending on 𝐜 , there is either no solution or there are infinitely many solutions, being the sum of a particular solution of (1.2.61) and any solution of 𝐀 𝐛 = 𝟎 . … Nonzero vectors 𝐯 1 , , 𝐯 n are linearly independent if i = 1 n c i 𝐯 i = 𝟎 implies that all coefficients c i are zero. …The sum of all multiplicities is n . … Thus tr ( 𝐀 ) is the sum of the (counted according to their multiplicities) eigenvalues of 𝐀 . …
16: 9.12 Scorer Functions
9.12.15 Gi ( z ) = 3 2 / 3 π k = 0 cos ( 2 k 1 3 π ) Γ ( k + 1 3 ) ( 3 1 / 3 z ) k k ! ,
9.12.16 Gi ( z ) = 3 1 / 3 π k = 0 cos ( 2 k + 1 3 π ) Γ ( k + 2 3 ) ( 3 1 / 3 z ) k k ! .
9.12.23 Gi ( x ) = 4 x 2 3 3 / 2 π 2 0 K 1 / 3 ( t ) ζ 2 t 2 d t , x > 0 ,
where the last integral is a Cauchy principal value (§1.4(v)). …
17: Bibliography G
  • L. Gårding (1947) The solution of Cauchy’s problem for two totally hyperbolic linear differential equations by means of Riesz integrals. Ann. of Math. (2) 48 (4), pp. 785–826.
  • G. Gasper (1977) Positive sums of the classical orthogonal polynomials. SIAM J. Math. Anal. 8 (3), pp. 423–447.
  • E. Grosswald (1985) Representations of Integers as Sums of Squares. Springer-Verlag, New York.
  • 18: 3.4 Differentiation
    3.4.1 h f t = h f ( x 0 + t h ) = k = n 0 n 1 B k n f k + h R n , t , n 0 < t < n 1 ,
    3.4.7 h f t = k = 1 2 B k 3 f k + h R 3 , t , 1 < t < 2 ,
    3.4.11 h f t = k = 2 3 B k 5 f k + h R 5 , t , 2 < t < 3 ,
    3.4.15 h f t = k = 3 4 B k 7 f k + h R 7 , t , 3 < t < 4 ,
    If f can be extended analytically into the complex plane, then from Cauchy’s integral formula (§1.9(iii)) …
    19: 19.29 Reduction of General Elliptic Integrals
    The Cauchy principal value is taken when U α 5 2 or Q α 5 2 is real and negative. …
    19.29.12 𝐦 = ( m 1 , , m n ) = j = 1 n m j 𝐞 j ,
    The only cases that are integrals of the third kind are those in which at least one m j with j > h is a negative integer and those in which h = 4 and j = 1 n m j is a positive integer. …
    19.29.18 b j q I ( q 𝐞 l ) = r = 0 q ( q r ) b l r d l j q r I ( r 𝐞 j ) , j , l = 1 , 2 , , n ;
    20: 19.36 Methods of Computation
    When the differences are moderately small, the iteration is stopped, the elementary symmetric functions of certain differences are calculated, and a polynomial consisting of a fixed number of terms of the sum in (19.19.7) is evaluated. … All cases of R F , R C , R J , and R D are computed by essentially the same procedure (after transforming Cauchy principal values by means of (19.20.14) and (19.2.20)). …
    19.36.13 2 R G ( t 0 2 , t 0 2 + θ c 0 2 , t 0 2 + θ a 0 2 ) = ( t 0 2 + θ m = 0 2 m 1 c m 2 ) R C ( T 2 + θ M 2 , T 2 ) + h 0 + m = 1 2 m ( h m h m 1 ) .