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1: 24.1 Special Notation
Euler Numbers and Polynomials
The notations E n , E n ( x ) , as defined in §24.2(ii), were used in Lucas (1891) and Nörlund (1924). …
2: 24.3 Graphs
See accompanying text
Figure 24.3.2: Euler polynomials E n ( x ) , n = 2 , 3 , , 6 . Magnify
3: 24.18 Physical Applications
§24.18 Physical Applications
Bernoulli polynomials appear in statistical physics (Ordóñez and Driebe (1996)), in discussions of Casimir forces (Li et al. (1991)), and in a study of quark-gluon plasma (Meisinger et al. (2002)). Euler polynomials also appear in statistical physics as well as in semi-classical approximations to quantum probability distributions (Ballentine and McRae (1998)).
4: 24.4 Basic Properties
24.4.2 E n ( x + 1 ) + E n ( x ) = 2 x n .
§24.4(ii) Symmetry
24.4.4 E n ( 1 x ) = ( 1 ) n E n ( x ) .
24.4.6 ( 1 ) n + 1 E n ( x ) = E n ( x ) 2 x n .
Next, …
5: 24.21 Software
§24.21(ii) B n , B n ( x ) , E n , and E n ( x )
6: 24.13 Integrals
24.13.3 x x + ( 1 / 2 ) B n ( t ) d t = E n ( 2 x ) 2 n + 1 ,
§24.13(ii) Euler Polynomials
24.13.7 E n ( t ) d t = E n + 1 ( t ) n + 1 + const. ,
24.13.8 0 1 E n ( t ) d t = 2 E n + 1 ( 0 ) n + 1 = 4 ( 2 n + 2 1 ) ( n + 1 ) ( n + 2 ) B n + 2 ,
§24.13(iii) Compendia
7: 24.16 Generalizations
§24.16 Generalizations
For = 0 , 1 , 2 , , Bernoulli and Euler polynomials of order are defined respectively by …When x = 0 they reduce to the Bernoulli and Euler numbers of order : …
E n ( ) = E n ( ) ( 0 ) .
§24.16(iii) Other Generalizations
8: 24.2 Definitions and Generating Functions
E ~ n ( x ) = E n ( x ) , 0 x < 1 ,
E ~ n ( x + 1 ) = E ~ n ( x ) , x .
Table 24.2.2: Bernoulli and Euler polynomials.
n B n ( x ) E n ( x )
9: 24.17 Mathematical Applications
§24.17 Mathematical Applications
24.17.2 R m ( n ) = 1 2 ( m 1 ) ! a n f ( m ) ( x ) E ~ m 1 ( h x ) d x .
24.17.3 S n ( x ) = E ~ n ( x + 1 2 n + 1 2 ) E ~ n ( 1 2 n + 1 2 ) , n = 0 , 1 , ,
§24.17(iii) Number Theory
10: 24.14 Sums
§24.14 Sums
§24.14(i) Quadratic Recurrence Relations
24.14.3 k = 0 n ( n k ) E k ( h ) E n k ( x ) = 2 ( E n + 1 ( x + h ) ( x + h 1 ) E n ( x + h ) ) ,
24.14.4 k = 0 n ( n k ) E k E n k = 2 n + 1 E n + 1 ( 0 ) = 2 n + 2 ( 1 2 n + 2 ) B n + 2 n + 2 .
24.14.5 k = 0 n ( n k ) E k ( h ) B n k ( x ) = 2 n B n ( 1 2 ( x + h ) ) ,