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11: 22.5 Special Values
For example, at z = K + i K , sn ( z , k ) = 1 / k , d sn ( z , k ) / d z = 0 . …
Table 22.5.1: Jacobian elliptic function values, together with derivatives or residues, for special values of the variable.
z
sn z 0 , 1 1 , 0 1 / k , 0 , 1 / k 0 , 1 0 , 1 0 , 1
sd z 0 , 1 k 1 , 0 , i ( k k ) 1 i k 1 , 0 0 , 1 0 , 1 0 , 1
dc z 1 , 0 , 1 0 , k k , 0 1 , 0 1 , 0 1 , 0
sc z 0 , 1 , k 1 i k 1 , 0 i , 0 0 , 1 0 , 1 0 , 1
12: 16.8 Differential Equations
is a value z 0 of z at which all the coefficients f j ( z ) , j = 0 , 1 , , n 1 , are analytic. If z 0 is not an ordinary point but ( z z 0 ) n j f j ( z ) , j = 0 , 1 , , n 1 , are analytic at z = z 0 , then z 0 is a regular singularity. … Equation (16.8.4) has a regular singularity at z = 0 , and an irregular singularity at z = , whereas (16.8.5) has regular singularities at z = 0 , 1 , and . … More generally if z 0 ( ) is an arbitrary constant, | z z 0 | > max ( | z 0 | , | z 0 1 | ) , and | ph ( z 0 z ) | < π , then …(Note that the generalized hypergeometric functions on the right-hand side are polynomials in z 0 .) …
13: 12.8 Recurrence Relations and Derivatives
12.8.1 z U ( a , z ) U ( a 1 , z ) + ( a + 1 2 ) U ( a + 1 , z ) = 0 ,
12.8.2 U ( a , z ) + 1 2 z U ( a , z ) + ( a + 1 2 ) U ( a + 1 , z ) = 0 ,
12.8.3 U ( a , z ) 1 2 z U ( a , z ) + U ( a 1 , z ) = 0 ,
12.8.6 V ( a , z ) 1 2 z V ( a , z ) ( a 1 2 ) V ( a 1 , z ) = 0 ,
12.8.7 V ( a , z ) + 1 2 z V ( a , z ) V ( a + 1 , z ) = 0 ,
14: 36.5 Stokes Sets
The Stokes set takes different forms for z = 0 , z < 0 , and z > 0 . For z = 0 , the set consists of the two curves … For z 0 , the Stokes set is expressed in terms of scaled coordinates … For z < 0 , there are two solutions u , provided that | Y | > ( 2 5 ) 1 / 2 . … For z > 0 the Stokes set has two sheets. …
15: 10.33 Continued Fractions
Assume I ν 1 ( z ) 0 . …
10.33.1 I ν ( z ) I ν 1 ( z ) = 1 2 ν z 1 + 1 2 ( ν + 1 ) z 1 + 1 2 ( ν + 2 ) z 1 + , z 0 ,
10.33.2 I ν ( z ) I ν 1 ( z ) = 1 2 z / ν 1 + 1 4 z 2 / ( ν ( ν + 1 ) ) 1 + 1 4 z 2 / ( ( ν + 1 ) ( ν + 2 ) ) 1 + , ν 0 , 1 , 2 , .
16: 10.10 Continued Fractions
Assume J ν 1 ( z ) 0 . …
10.10.1 J ν ( z ) J ν 1 ( z ) = 1 2 ν z 1 1 2 ( ν + 1 ) z 1 1 2 ( ν + 2 ) z 1 , z 0 ,
10.10.2 J ν ( z ) J ν 1 ( z ) = 1 2 z / ν 1 1 4 z 2 / ( ν ( ν + 1 ) ) 1 1 4 z 2 / ( ( ν + 1 ) ( ν + 2 ) ) 1 , ν 0 , 1 , 2 , .
17: 3.7 Ordinary Differential Equations
The path is partitioned at P + 1 points labeled successively z 0 , z 1 , , z P , with z 0 = a , z P = b . …
f 0 ( z ) = 1 ,
g 0 ( z ) = 0 ,
h 0 ( z ) = 0 ,
If the solution w ( z ) that we are seeking grows in magnitude at least as fast as all other solutions of (3.7.1) as we pass along 𝒫 from a to b , then w ( z ) and w ( z ) may be computed in a stable manner for z = z 0 , z 1 , , z P by successive application of (3.7.5) for j = 0 , 1 , , P 1 , beginning with initial values w ( a ) and w ( a ) . …
18: 3.3 Interpolation
The divided differences of f relative to a sequence of distinct points z 0 , z 1 , z 2 , are defined by … Newton’s formula has the advantage of allowing easy updating: incorporation of a new point z n + 1 requires only addition of the term with [ z 0 , z 1 , , z n + 1 ] f to (3.3.38), plus the computation of this divided difference. Another advantage is its robustness with respect to confluence of the set of points z 0 , z 1 , , z n . For example, for k + 1 coincident points the limiting form is given by [ z 0 , z 0 , , z 0 ] f = f ( k ) ( z 0 ) / k ! . … It can be used for solving a nonlinear scalar equation f ( z ) = 0 approximately. …
19: 10.35 Generating Function and Associated Series
For z and t { 0 } , …
10.35.2 e z cos θ = I 0 ( z ) + 2 k = 1 I k ( z ) cos ( k θ ) ,
10.35.4 1 = I 0 ( z ) 2 I 2 ( z ) + 2 I 4 ( z ) 2 I 6 ( z ) + ,
10.35.5 e ± z = I 0 ( z ) ± 2 I 1 ( z ) + 2 I 2 ( z ) ± 2 I 3 ( z ) + ,
cosh z = I 0 ( z ) + 2 I 2 ( z ) + 2 I 4 ( z ) + 2 I 6 ( z ) + ,
20: 32.7 Bäcklund Transformations
with ζ = 2 1 / 3 z and ε = ± 1 , where W ( ζ ; 1 2 ε ) satisfies P II  with z = ζ , α = 1 2 ε , and w ( z ; 0 ) satisfies P II  with α = 0 . … 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 …
z 1 = z 0 ,
z 2 = z 0 ,
Let W 0 = W ( z ; α 0 , β 0 , γ 0 , 1 2 ) and W 1 = W ( z ; α 1 , β 1 , γ 1 , 1 2 ) be solutions of P V , where …