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21: 1.12 Continued Fractions
when p k 0 , k = 1 , 2 , 3 , . …when c k 0 , k = 1 , 2 , 3 , . … Define … The continued fraction a 1 b 1 + a 2 b 2 + converges when … Then the convergents C n satisfy …
22: 19.32 Conformal Map onto a Rectangle
with x 1 , x 2 , x 3 real constants, has differential …
19.32.3 x 1 > x 2 > x 3 ,
z ( x 2 ) = z ( x 1 ) + z ( x 3 ) ,
z ( x 3 ) = R F ( x 3 x 1 , x 3 x 2 , 0 ) = i R F ( 0 , x 1 x 3 , x 2 x 3 ) .
As p proceeds along the entire real axis with the upper half-plane on the right, z describes the rectangle in the clockwise direction; hence z ( x 3 ) is negative imaginary. …
23: 34.10 Zeros
§34.10 Zeros
In a 3 j symbol, if the three angular momenta j 1 , j 2 , j 3 do not satisfy the triangle conditions (34.2.1), or if the projective quantum numbers do not satisfy (34.2.3), then the 3 j symbol is zero. Similarly the 6 j symbol (34.4.1) vanishes when the triangle conditions are not satisfied by any of the four 3 j symbols in the summation. …However, the 3 j and 6 j symbols may vanish for certain combinations of the angular momenta and projective quantum numbers even when the triangle conditions are fulfilled. …
24: 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
25: 35.8 Generalized Hypergeometric Functions of Matrix Argument
§35.8(iii) F 2 3 Case
Kummer Transformation
Let c = b 1 + b 2 a 1 a 2 a 3 . … Let a 1 + a 2 + a 3 + 1 2 ( m + 1 ) = b 1 + b 2 ; one of the a j be a negative integer; ( b 1 a 1 ) , ( b 1 a 2 ) , ( b 1 a 3 ) , ( b 1 a 1 a 2 a 3 ) > 1 2 ( m 1 ) . … Again, let c = b 1 + b 2 a 1 a 2 a 3 . …
26: 23.2 Definitions and Periodic Properties
If ω 1 and ω 3 are nonzero real or complex numbers such that ( ω 3 / ω 1 ) > 0 , then the set of points 2 m ω 1 + 2 n ω 3 , with m , n , constitutes a lattice 𝕃 with 2 ω 1 and 2 ω 3 lattice generators. … then 2 ω 2 , 2 ω 3 are generators, as are 2 ω 2 , 2 ω 1 . …where a , b , c , d are integers, then 2 χ 1 , 2 χ 3 are generators of 𝕃 iff … If 2 ω 1 , 2 ω 3 is any pair of generators of 𝕃 , and ω 2 is defined by (23.2.1), then … where …
27: 3.4 Differentiation
The B k n are the differentiated Lagrangian interpolation coefficients: … where ξ 0 and ξ 1 I . For the values of n 0 and n 1 used in the formulas below … For partial derivatives we use the notation u t , s = u ( x 0 + t h , y 0 + s h ) . …
28: 24.3 Graphs
See accompanying text
Figure 24.3.1: Bernoulli polynomials B n ( x ) , n = 2 , 3 , , 6 . Magnify
See accompanying text
Figure 24.3.2: Euler polynomials E n ( x ) , n = 2 , 3 , , 6 . Magnify
29: 1.11 Zeros of Polynomials
f ( z ) = z 3 6 z 2 + 6 z 2 , g ( w ) = w 3 6 w 6 , A = 3 4 3 , B = 3 2 3 . Roots of f ( z ) = 0 are 2 + 4 3 + 2 3 , 2 + 4 3 ρ + 2 3 ρ 2 , 2 + 4 3 ρ 2 + 2 3 ρ . … For the roots α 1 , α 2 , α 3 , α 4 of g ( w ) = 0 and the roots θ 1 , θ 2 , θ 3 of the resolvent cubic equationResolvent cubic is z 3 + 12 z 2 + 20 z + 9 = 0 with roots θ 1 = 1 , θ 2 = 1 2 ( 11 + 85 ) , θ 3 = 1 2 ( 11 85 ) , and θ 1 = 1 , θ 2 = 1 2 ( 17 + 5 ) , θ 3 = 1 2 ( 17 5 ) . So 2 α 1 = 1 + 17 , 2 α 2 = 1 17 , 2 α 3 = 1 + 5 , 2 α 4 = 1 5 , and the roots of f ( z ) = 0 are 1 2 ( 3 ± 17 ) , 1 2 ( 1 ± 5 ) . …
30: 16.12 Products
16.12.3 ( F 1 2 ( a , b c ; z ) ) 2 = k = 0 ( 2 a ) k ( 2 b ) k ( c 1 2 ) k ( c ) k ( 2 c 1 ) k k ! F 3 4 ( 1 2 k , 1 2 ( 1 k ) , a + b c + 1 2 , 1 2 a + 1 2 , b + 1 2 , 3 2 k c ; 1 ) z k , | z | < 1 .