About the Project

Visit (-- RXLARA.COM --) pharmacy buy Penegra over counter. Sildenafil Citrate Penegra female 100mg pills price prescription onli

AdvancedHelp

(0.009 seconds)

11—20 of 380 matching pages

11: 36.7 Zeros
x m , n ± = 2 y m ( 2 n + 1 2 + ( 1 ) m 1 2 ± 1 4 ) π , m = 1 , 2 , 3 , , n = 0 , ± 1 , ± 2 , .
Table 36.7.1: Zeros of cusp diffraction catastrophe to 5D. …
Zeros { x y } inside, and zeros [ x y ] outside, the cusp x 2 = 8 27 | y | 3 .
{ ± 0.52768 4.37804 } [ ± 2.35218 1.74360 ]
{ ± 1.41101 5.55470 } { ± 2.36094 5.52321 } [ ± 4.42707 3.05791 ]
{ ± 0.38488 8.31916 } { ± 2.71193 8.22315 } { ± 3.49286 8.20326 } { ± 5.96669 7.85723 } { ± 6.79538 7.80456 } [ ± 9.17308 5.55831 ]
x n = ± ( 8 27 ) 1 / 2 | y n | 3 / 2 ( 1 + ξ n ) ,
12: 7.23 Tables
  • Zhang and Jin (1996, pp. 637, 639) includes ( 2 / π ) e x 2 , erf x , x = 0 ( .02 ) 1 ( .04 ) 3 , 8D; C ( x ) , S ( x ) , x = 0 ( .2 ) 10 ( 2 ) 100 ( 100 ) 500 , 8D.

  • Zhang and Jin (1996, pp. 638, 640–641) includes the real and imaginary parts of erf z , x [ 0 , 5 ] , y = 0.5 ( .5 ) 3 , 7D and 8D, respectively; the real and imaginary parts of x e ± i t 2 d t , ( 1 / π ) e i ( x 2 + ( π / 4 ) ) x e ± i t 2 d t , x = 0 ( .5 ) 20 ( 1 ) 25 , 8D, together with the corresponding modulus and phase to 8D and 6D (degrees), respectively.

  • Fettis et al. (1973) gives the first 100 zeros of erf z and w ( z ) (the table on page 406 of this reference is for w ( z ) , not for erfc z ), 11S.

  • 13: Charles W. Clark
    He has been a Visiting Fellow at the Australian National University, a Dr. Lee Fellow at Christ Church College of the University of Oxford, and Visiting Professor at the National University of Singapore. Clark received the R&D 100 Award, Distinguished Presidential Rank Award of the U. …
    14: 6.4 Analytic Continuation
    6.4.4 Ci ( z e ± π i ) = ± π i + Ci ( z ) ,
    6.4.5 Chi ( z e ± π i ) = ± π i + Chi ( z ) ,
    6.4.6 f ( z e ± π i ) = π e i z f ( z ) ,
    6.4.7 g ( z e ± π i ) = π i e i z + g ( z ) .
    15: 14.2 Differential Equations
    Standard solutions: 𝖯 ν ( ± x ) , 𝖰 ν ( ± x ) , 𝖰 ν 1 ( ± x ) , P ν ( ± x ) , Q ν ( ± x ) , Q ν 1 ( ± x ) . … Standard solutions: 𝖯 ν μ ( ± x ) , 𝖯 ν μ ( ± x ) , 𝖰 ν μ ( ± x ) , 𝖰 ν 1 μ ( ± x ) , P ν μ ( ± x ) , P ν μ ( ± x ) , 𝑸 ν μ ( ± x ) , 𝑸 ν 1 μ ( ± x ) . …
    16: 26.15 Permutations: Matrix Notation
    where the sum is over 1 g < k n and n h > 1 . … For ( j , k ) B , B [ j , k ] denotes B after removal of all elements of the form ( j , t ) or ( t , k ) , t = 1 , 2 , , n . B ( j , k ) denotes B with the element ( j , k ) removed.
    26.15.5 R ( x , B ) = x R ( x , B [ j , k ] ) + R ( x , B ( j , k ) ) .
    17: 33.8 Continued Fractions
    33.8.2 H ± H ± = c ± i ρ a b 2 ( ρ η ± i ) + ( a + 1 ) ( b + 1 ) 2 ( ρ η ± 2 i ) + ,
    a = 1 + ± i η ,
    b = ± i η ,
    c = ± i ( 1 ( η / ρ ) ) .
    F = ± ( q 1 ( u p ) 2 + q ) 1 / 2 ,
    18: 4.24 Inverse Trigonometric Functions: Further Properties
    4.24.13 Arcsin u ± Arcsin v = Arcsin ( u ( 1 v 2 ) 1 / 2 ± v ( 1 u 2 ) 1 / 2 ) ,
    4.24.14 Arccos u ± Arccos v = Arccos ( u v ( ( 1 u 2 ) ( 1 v 2 ) ) 1 / 2 ) ,
    4.24.15 Arctan u ± Arctan v = Arctan ( u ± v 1 u v ) ,
    4.24.16 Arcsin u ± Arccos v = Arcsin ( u v ± ( ( 1 u 2 ) ( 1 v 2 ) ) 1 / 2 ) = Arccos ( v ( 1 u 2 ) 1 / 2 u ( 1 v 2 ) 1 / 2 ) ,
    4.24.17 Arctan u ± Arccot v = Arctan ( u v ± 1 v u ) = Arccot ( v u u v ± 1 ) .
    19: 28.20 Definitions and Basic Properties
    When z is replaced by ± i z , (28.2.1) becomes the modified Mathieu’s equation: … Then from §2.7(ii) it is seen that equation (28.20.2) has independent and unique solutions that are asymptotic to ζ 1 / 2 e ± 2 i h ζ as ζ in the respective sectors | ph ( i ζ ) | 3 2 π δ , δ being an arbitrary small positive constant. …
    Mc 2 n ( j ) ( z ± 1 2 π i , h ) = Mc 2 n ( j ) ( z , ± i h ) ,
    Ms 2 n + 1 ( j ) ( z ± 1 2 π i , h ) = Mc 2 n + 1 ( j ) ( z , ± i h ) ,
    Mc 2 n + 1 ( j ) ( z ± 1 2 π i , h ) = Ms 2 n + 1 ( j ) ( z , ± i h ) ,
    20: 10.34 Analytic Continuation
    10.34.3 I ν ( z e m π i ) = ( i / π ) ( ± e m ν π i K ν ( z e ± π i ) e ( m 1 ) ν π i K ν ( z ) ) ,
    10.34.4 K ν ( z e m π i ) = csc ( ν π ) ( ± sin ( m ν π ) K ν ( z e ± π i ) sin ( ( m 1 ) ν π ) K ν ( z ) ) .
    10.34.6 K n ( z e m π i ) = ± ( 1 ) n ( m 1 ) m K n ( z e ± π i ) ( 1 ) n m ( m 1 ) K n ( z ) .