About the Project

of imaginary argument

AdvancedHelp

(0.002 seconds)

31—40 of 73 matching pages

31: 20.11 Generalizations and Analogs
20.11.1 G ( m , n ) = k = 0 n 1 e π i k 2 m / n ;
If both m , n are positive, then G ( m , n ) allows inversion of its arguments as a modular transformation (compare (23.15.3) and (23.15.4)):
20.11.2 1 n G ( m , n ) = 1 n k = 0 n 1 e π i k 2 m / n = e π i / 4 m j = 0 m 1 e π i j 2 n / m = e π i / 4 m G ( n , m ) .
With the substitutions a = q e 2 i z , b = q e 2 i z , with q = e i π τ , we have …
32: 10.40 Asymptotic Expansions for Large Argument
§10.40 Asymptotic Expansions for Large Argument
Products
§10.40(ii) Error Bounds for Real Argument and Order
§10.40(iii) Error Bounds for Complex Argument and Order
33: 11.6 Asymptotic Expansions
§11.6(i) Large | z | , Fixed ν
11.6.7 𝐌 ν ( λ ν ) ( 1 2 λ ν ) ν 1 π Γ ( ν + 1 2 ) k = 0 k ! c k ( i λ ) ν k , | ph ν | 1 2 π δ .
34: 20.10 Integrals
20.10.1 0 x s 1 θ 2 ( 0 | i x 2 ) d x = 2 s ( 1 2 s ) π s / 2 Γ ( 1 2 s ) ζ ( s ) , s > 1 ,
20.10.2 0 x s 1 ( θ 3 ( 0 | i x 2 ) 1 ) d x = π s / 2 Γ ( 1 2 s ) ζ ( s ) , s > 1 ,
20.10.3 0 x s 1 ( 1 θ 4 ( 0 | i x 2 ) ) d x = ( 1 2 1 s ) π s / 2 Γ ( 1 2 s ) ζ ( s ) , s > 0 .
Let s , , and β be constants such that s > 0 , > 0 , and | β | + | β | . … For corresponding results for argument derivatives of the theta functions see Erdélyi et al. (1954a, pp. 224–225) or Oberhettinger and Badii (1973, p. 193). …
35: 5.3 Graphics
§5.3(i) Real Argument
§5.3(ii) Complex Argument
See accompanying text
Figure 5.3.4: | Γ ( x + i y ) | . Magnify 3D Help
See accompanying text
Figure 5.3.5: 1 / | Γ ( x + i y ) | . Magnify 3D Help
See accompanying text
Figure 5.3.6: | ψ ( x + i y ) | . Magnify 3D Help
36: 1.10 Functions of a Complex Variable
Phase (or Argument) Principle
If D = ( , 0 ] and z = r e i θ , then one branch is r e i θ / 2 , the other branch is r e i θ / 2 , with π < θ < π in both cases. Similarly if D = [ 0 , ) , then one branch is r e i θ / 2 , the other branch is r e i θ / 2 , with 0 < θ < 2 π in both cases. … Thus if F ( z ) is continued along a path that circles z = 1 m times in the positive sense and returns to z 0 without circling z = 1 , then F ( ( z 0 1 ) e 2 m π i + 1 ) = e α ln ( 1 z 0 ) e β ln ( 1 + z 0 ) e 2 π i m α . If the path also circles z = 1 n times in the clockwise or negative sense before returning to z 0 , then the value of F ( z 0 ) becomes e α ln ( 1 z 0 ) e β ln ( 1 + z 0 ) e 2 π i m α e 2 π i n β . …
37: 25.1 Special Notation
k , m , n nonnegative integers.
s = σ + i t complex variable.
z = x + i y complex variable.
primes on function symbols: derivatives with respect to argument.
38: 19.3 Graphics
In Figures 19.3.7 and 19.3.8 for complete Legendre’s elliptic integrals with complex arguments, height corresponds to the absolute value of the function and color to the phase. …
See accompanying text
Figure 19.3.10: ( K ( k ) ) as a function of complex k 2 for 2 ( k 2 ) 2 , 2 ( k 2 ) 2 . The imaginary part is 0 for k 2 < 1 , and is antisymmetric under reflection in the real axis. … Magnify 3D Help
See accompanying text
Figure 19.3.12: ( E ( k ) ) as a function of complex k 2 for 2 ( k 2 ) 2 , 2 ( k 2 ) 2 . The imaginary part is 0 for k 2 1 and is antisymmetric under reflection in the real axis. … Magnify 3D Help
39: 36.2 Catastrophes and Canonical Integrals
36.2.6 Ψ ( E ) ( 𝐱 ) = 2 π / 3 exp ( i ( 4 27 z 3 + 1 3 x z 1 4 π ) ) exp ( 7 π i / 12 ) exp ( π i / 12 ) exp ( i ( u 6 + 2 z u 4 + ( z 2 + x ) u 2 + y 2 12 u 2 ) ) d u ,
Ψ ( E ) ( 𝐱 ) = 4 π 3 1 / 3 exp ( i ( 2 27 z 3 1 3 x z ) ) ( exp ( i π 6 ) F + ( 𝐱 ) + exp ( i π 6 ) F ( 𝐱 ) ) ,
Ψ 2 ( 𝟎 ) = 1.67481 + i  0.69373
Ψ 4 ( 𝟎 ) = 1.79222 + i  0.48022 .
40: 20.2 Definitions and Periodic Properties
For fixed z , each of θ 1 ( z | τ ) / sin z , θ 2 ( z | τ ) / cos z , θ 3 ( z | τ ) , and θ 4 ( z | τ ) is an analytic function of τ for τ > 0 , with a natural boundary τ = 0 , and correspondingly, an analytic function of q for | q | < 1 with a natural boundary | q | = 1 . …
Figure 20.2.1: z -plane. …Left-hand diagram is the rectangular case ( τ purely imaginary); right-hand diagram is the general case. …
§20.2(iii) Translation of the Argument by Half-Periods
20.2.10 M M ( z | τ ) = e i z + ( i π τ / 4 ) ,
20.2.11 θ 1 ( z | τ ) = θ 2 ( z + 1 2 π | τ ) = i M θ 4 ( z + 1 2 π τ | τ ) = i M θ 3 ( z + 1 2 π + 1 2 π τ | τ ) ,