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21: 22.19 Physical Applications
The periodicity and symmetry of the pendulum imply that the motion in each four intervals θ ( 0 , ± α ) and θ ( ± α , 0 ) have the same “quarter periods” K = K ( sin ( 1 2 α ) ) . …
See accompanying text
Figure 22.19.1: Jacobi’s amplitude function am ( x , k ) for 0 x 10 π and k = 0.5 , 0.9999 , 1.0001 , 2 . …As k 1 , plateaus are seen as the motion approaches the separatrix where θ = n π , n = ± 1 , ± 2 , , at which points the motion is time independent for k = 1 . … Magnify
As a 1 / β from below the period diverges since a = ± 1 / β are points of unstable equilibrium. … For an initial displacement with 1 / β | a | < 2 / β , bounded oscillations take place near one of the two points of stable equilibrium x = ± 1 / β . …For initial displacement with | a | 2 / β the motion extends over the full range a x a : …
22: 14.27 Zeros
P ν μ ( x ± i 0 ) (either side of the cut) has exactly one zero in the interval ( , 1 ) if either of the following sets of conditions holds: …For all other values of the parameters P ν μ ( x ± i 0 ) has no zeros in the interval ( , 1 ) . …
23: 4.37 Inverse Hyperbolic Functions
In (4.37.1) the integration path may not pass through either of the points t = ± i , and the function ( 1 + t 2 ) 1 / 2 assumes its principal value when t is real. In (4.37.2) the integration path may not pass through either of the points ± 1 , and the function ( t 2 1 ) 1 / 2 assumes its principal value when t ( 1 , ) . …In (4.37.3) the integration path may not intersect ± 1 . … Arcsinh z and Arccsch z have branch points at z = ± i ; the other four functions have branch points at z = ± 1 . …
24: 8.21 Generalized Sine and Cosine Integrals
8.21.1 ci ( a , z ) ± i si ( a , z ) = e ± 1 2 π i a Γ ( a , z e 1 2 π i ) ,
8.21.2 Ci ( a , z ) ± i Si ( a , z ) = e ± 1 2 π i a γ ( a , z e 1 2 π i ) .
8.21.3 0 t a 1 e ± i t d t = e ± 1 2 π i a Γ ( a ) , 0 < a < 1 ,
25: 33.6 Power-Series Expansions in ρ
33.6.5 H ± ( η , ρ ) = e ± i θ ( η , ρ ) ( 2 + 1 ) ! Γ ( ± i η ) ( k = 0 ( a ) k ( 2 + 2 ) k k ! ( 2 i ρ ) a + k ( ln ( 2 i ρ ) + ψ ( a + k ) ψ ( 1 + k ) ψ ( 2 + 2 + k ) ) k = 1 2 + 1 ( 2 + 1 ) ! ( k 1 ) ! ( 2 + 1 k ) ! ( 1 a ) k ( 2 i ρ ) a k ) ,
where a = 1 + ± i η and ψ ( x ) = Γ ( x ) / Γ ( x ) 5.2(i)). … Corresponding expansions for H ± ( η , ρ ) can be obtained by combining (33.6.5) with (33.4.3) or (33.4.4).
26: 4.21 Identities
4.21.1 sin u ± cos u = 2 sin ( u ± 1 4 π ) = ± 2 cos ( u 1 4 π ) .
4.21.2 sin ( u ± v ) = sin u cos v ± cos u sin v ,
4.21.3 cos ( u ± v ) = cos u cos v sin u sin v ,
4.21.4 tan ( u ± v ) = tan u ± tan v 1 tan u tan v ,
4.21.5 cot ( u ± v ) = ± cot u cot v 1 cot u ± cot v .
27: 32.7 Bäcklund Transformations
and …furnish solutions of P II , provided that α 1 2 . … The solutions w α = w ( z ; α ) , w α ± 1 = w ( z ; α ± 1 ) , satisfy the nonlinear recurrence relation … Let w 0 = w ( z ; α 0 , β 0 ) and w j ± = w ( z ; α j ± , β j ± ) , j = 1 , 2 , 3 , 4 , be solutions of P IV  with … with ε = ± 1 . …
28: 9.12 Scorer Functions
e 2 π i / 3 Hi ( z e 2 π i / 3 ) ,
9.12.10 e 2 π i / 3 Hi ( z e 2 π i / 3 ) , Ai ( z ) , Ai ( z e ± 2 π i / 3 ) , π ± ph z 1 3 π .
9.12.13 Gi ( z ) = e π i / 3 Hi ( z e ± 2 π i / 3 ) ± i Ai ( z ) ,
9.12.14 Hi ( z ) = e ± 2 π i / 3 Hi ( z e ± 2 π i / 3 ) + 2 e π i / 6 Ai ( z e 2 π i / 3 ) .
If ζ = 2 3 z 3 / 2 or 2 3 x 3 / 2 , and K 1 / 3 is the modified Bessel function (§10.25(ii)), then …
29: 10.19 Asymptotic Expansions for Large Order
10.19.9 H ν ( 1 ) ( ν + a ν 1 3 ) H ν ( 2 ) ( ν + a ν 1 3 ) } 2 4 3 ν 1 3 e π i / 3 Ai ( e π i / 3 2 1 3 a ) k = 0 P k ( a ) ν 2 k / 3 + 2 5 3 ν e ± π i / 3 Ai ( e π i / 3 2 1 3 a ) k = 0 Q k ( a ) ν 2 k / 3 ,
with sectors of validity 1 2 π + δ ± ph ν 3 2 π δ . …
P 2 ( a ) = 9 100 a 5 + 3 35 a 2 ,
10.19.13 H ν ( 1 ) ( ν + a ν 1 3 ) H ν ( 2 ) ( ν + a ν 1 3 ) } 2 5 3 ν 2 3 e ± π i / 3 Ai ( e π i / 3 2 1 3 a ) k = 0 R k ( a ) ν 2 k / 3 + 2 4 3 ν 4 3 e π i / 3 Ai ( e π i / 3 2 1 3 a ) k = 0 S k ( a ) ν 2 k / 3 ,
R 2 ( a ) = 9 100 a 5 + 57 70 a 2 ,
30: 19.34 Mutual Inductance of Coaxial Circles
19.34.3 2 a b I ( 𝐞 5 ) = a 3 I ( 𝟎 ) I ( 𝐞 3 ) = a 3 I ( 𝟎 ) r + 2 r 2 I ( 𝐞 3 ) = 2 a b ( I ( 𝟎 ) r 2 I ( 𝐞 1 𝐞 3 ) ) ,
where a 1 + b 1 t = 1 + t and
19.34.4 r ± 2 = a 3 ± 2 a b = h 2 + ( a ± b ) 2
19.34.5 3 c 2 8 π a b M = 3 R F ( 0 , r + 2 , r 2 ) 2 r 2 R D ( 0 , r + 2 , r 2 ) ,
19.34.6 c 2 2 π M = ( r + 2 + r 2 ) R F ( 0 , r + 2 , r 2 ) 4 R G ( 0 , r + 2 , r 2 ) .