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reciprocity law

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1: 27.9 Quadratic Characters
27.9.2 ( 2 | p ) = ( 1 ) ( p 2 1 ) / 8 .
If p , q are distinct odd primes, then the quadratic reciprocity law states that … Both (27.9.1) and (27.9.2) are valid with p replaced by P ; the reciprocity law (27.9.3) holds if p , q are replaced by any two relatively prime odd integers P , Q .
2: 25.16 Mathematical Applications
which satisfies the reciprocity law
3: 36.6 Scaling Relations
§36.6 Scaling Relations
4: 5.3 Graphics
See accompanying text
Figure 5.3.1: Γ ( x ) and 1 / Γ ( x ) . … Magnify
See accompanying text
Figure 5.3.5: 1 / | Γ ( x + i y ) | . Magnify 3D Help
5: Bibliography F
  • H. E. Fettis (1970) On the reciprocal modulus relation for elliptic integrals. SIAM J. Math. Anal. 1 (4), pp. 524–526.
  • G. Freud (1976) On the coefficients in the recursion formulae of orthogonal polynomials. Proc. Roy. Irish Acad. Sect. A 76 (1), pp. 1–6.
  • 6: 8.23 Statistical Applications
    In queueing theory the Erlang loss function is used, which can be expressed in terms of the reciprocal of Q ( a , x ) ; see Jagerman (1974) and Cooper (1981, pp. 80, 316–319). …
    7: 8.24 Physical Applications
    The function γ ( a , x ) appears in: discussions of power-law relaxation times in complex physical systems (Sornette (1998)); logarithmic oscillations in relaxation times for proteins (Metzler et al. (1999)); Gaussian orbitals and exponential (Slater) orbitals in quantum chemistry (Shavitt (1963), Shavitt and Karplus (1965)); population biology and ecological systems (Camacho et al. (2002)). …
    8: 5.2 Definitions
    5.2.1 Γ ( z ) = 0 e t t z 1 d t , z > 0 .
    9: 27.16 Cryptography
    To do this, let s denote the reciprocal of r modulo ϕ ( n ) , so that r s = 1 + t ϕ ( n ) for some integer t . …
    10: 27.14 Unrestricted Partitions
    Euler introduced the reciprocal of the infinite product
    27.14.2 f ( x ) = m = 1 ( 1 x m ) = ( x ; x ) , | x | < 1 ,
    27.14.3 1 f ( x ) = n = 0 p ( n ) x n ,
    27.14.4 f ( x ) = 1 x x 2 + x 5 + x 7 x 12 x 15 + = 1 + k = 1 ( 1 ) k ( x ω ( k ) + x ω ( k ) ) ,
    27.14.15 5 ( f ( x 5 ) ) 5 ( f ( x ) ) 6 = n = 0 p ( 5 n + 4 ) x n