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21: 3.8 Nonlinear Equations
This iterative method for solving z = ϕ ( z ) is given by … p ( z ) = z 4 1 . … p ( z ) = z 4 2 z 2 + 1 . … Consider x = 20 and j = 19 . We have p ( 20 ) = 19 ! and a 19 = 1 + 2 + + 20 = 210 . …
22: 32.8 Rational Solutions
where the Q n ( z ) are monic polynomials (coefficient of highest power of z is 1 ) satisfying …with Q 0 ( z ) = 1 , Q 1 ( z ) = z . … Next, let p m ( z ) be the polynomials defined by p m ( z ) = 0 for m < 0 , and …where τ n ( z ) is the n × n Wronskian determinant … where P m ( z ) and Q m ( z ) are polynomials of degree m , with no common zeros. …
23: 28.1 Special Notation
m , n integers.
z = x + i y complex variable.
ce ν ( z , q ) , se ν ( z , q ) , fe n ( z , q ) , ge n ( z , q ) , me ν ( z , q ) ,
The functions Mc n ( j ) ( z , h ) and Ms n ( j ) ( z , h ) are also known as the radial Mathieu functions. …
Abramowitz and Stegun (1964, Chapter 20)
The radial functions Mc n ( j ) ( z , h ) and Ms n ( j ) ( z , h ) are denoted by Mc n ( j ) ( z , q ) and Ms n ( j ) ( z , q ) , respectively.
24: 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. …
25: 6.16 Mathematical Applications
6.16.1 sin x + 1 3 sin ( 3 x ) + 1 5 sin ( 5 x ) + = { 1 4 π , 0 < x < π , 0 , x = 0 , 1 4 π , π < x < 0 .
6.16.2 S n ( x ) = k = 0 n 1 sin ( ( 2 k + 1 ) x ) 2 k + 1 = 1 2 0 x sin ( 2 n t ) sin t d t = 1 2 Si ( 2 n x ) + R n ( x ) ,
6.16.3 R n ( x ) = 1 2 0 x ( 1 sin t 1 t ) sin ( 2 n t ) d t .
6.16.5 li ( x ) π ( x ) = O ( x ln x ) , x ,
See accompanying text
Figure 6.16.2: The logarithmic integral li ( x ) , together with vertical bars indicating the value of π ( x ) for x = 10 , 20 , , 1000 . Magnify
26: 10.3 Graphics
See accompanying text
Figure 10.3.14: H 5 ( 1 ) ( x + i y ) , 20 x 10 , 4 y 4 . … Magnify 3D Help
See accompanying text
Figure 10.3.16: H 5.5 ( 1 ) ( x + i y ) , 20 x 10 , 4 y 4 . … Magnify 3D Help
27: 3.4 Differentiation
B 2 5 = 1 120 ( 6 10 t 15 t 2 + 20 t 3 5 t 4 ) ,
B 3 6 = 1 720 ( 12 8 t 45 t 2 + 20 t 3 + 15 t 4 6 t 5 ) ,
B 2 6 = 1 60 ( 9 9 t 30 t 2 + 20 t 3 + 5 t 4 3 t 5 ) ,
B 2 6 = 1 60 ( 9 + 9 t 30 t 2 20 t 3 + 5 t 4 + 3 t 5 ) ,
f ( z ) = e z , x 0 = 0 . …
28: 12.19 Tables
  • Murzewski and Sowa (1972) includes D n ( x ) ( = U ( n 1 2 , x ) ) for n = 1 ( 1 ) 20 , x = 0 ( .05 ) 3 , 7S.

  • Zhang and Jin (1996, pp. 455–473) includes U ( ± n 1 2 , x ) , V ( ± n 1 2 , x ) , U ( ± ν 1 2 , x ) , V ( ± ν 1 2 , x ) , and derivatives, ν = n + 1 2 , n = 0 ( 1 ) 10 ( 10 ) 30 , x = 0.5 , 1 , 5 , 10 , 30 , 50 , 8S; W ( a , ± x ) , W ( a , ± x ) , and derivatives, a = h ( 1 ) 5 + h , x = 0.5 , 1 and a = h ( 1 ) 5 + h , x = 5 , h = 0 , 0.5 , 8S. Also, first zeros of U ( a , x ) , V ( a , x ) , and of derivatives, a = 6 ( .5 ) 1 , 6D; first three zeros of W ( a , x ) and of derivative, a = 0 ( .5 ) 4 , 6D; first three zeros of W ( a , ± x ) and of derivative, a = 0.5 ( .5 ) 5.5 , 6D; real and imaginary parts of U ( a , z ) , a = 1.5 ( 1 ) 1.5 , z = x + i y , x = 0.5 , 1 , 5 , 10 , y = 0 ( .5 ) 10 , 8S.

  • 29: 36.4 Bifurcation Sets
    z 0 ,
    x = 9 20 z 2 .
    x = 3 20 z 2 ,
    Elliptic umbilic bifurcation set (codimension three): for fixed z , the section of the bifurcation set is a three-cusped astroid … The + sign labels the cusped sheet; the sign labels the sheet that is smooth for z 0 (see Figure 36.4.4). …
    30: 27.15 Chinese Remainder Theorem
    Their product m has 20 digits, twice the number of digits in the data. …These numbers, in turn, are combined by the Chinese remainder theorem to obtain the final result ( mod m ) , which is correct to 20 digits. …