FUNCTIONS WITH DERIVATIVES GIVEN BY POLYNOMIALS IN THE FUNCTION ITSELF OR A RELATED FUNCTION GHISLAIN R. FRANSSENS
Abstract. We construct the set of holomorphic functions S1 = f : f C ! C whose members have n-th order derivatives which are given by a polynomial of degree n + 1 in the function itself. We also construct the set of holomorphic functions S2 = fg : g C ! Cg whose members have n-th order derivatives which are given by the product of the function itself with a polynomial of degree n in an element of S1 . The union S1 [ S2 contains all the hyperbolic and trigonometric functions. We study the properties of the polynomials involved and derive explicit expressions for them. As particular results, we obtain explicit and elegant formulas for the n-th order derivatives of the hyperbolic functions tanh, sech, coth and csch and the trigonometric functions tan, sec, cot and csc.
Keywords: Polynomial derivatives, sigmoid functions, hyperbolic functions, trigonometric functions. AMS 2000 MSC: 00A05, 00A22, 26A09, 33B10, 05A15. 1. Introduction We consider the set of holomorphic functions S1 = ff : f C ! Cg such that the n-th order derivative Dzn f (z) is given by a polynomial in f of degree n + 1. In addition, we also consider the related set of holomorphic functions S2 = fg : g C ! Cg such that the n-th order derivative Dzn g(z) is given by the product of g with a polynomial in f 2 S1 of degree n. The main topic of this paper is to investigate the two polynomial sequences involved. Although this problem is interesting in its own right, the motivation for studying exactly these two function sets lies in the well-known fact that the union S1 [S2 contains all the hyperbolic and trigonometric functions. As a particular and practical result of our study, we obtain explicit expressions for the (nontrivial !) n-th order derivatives of the hyperbolic functions tanh, sech, coth and csch and the trigonometric functions tan, sec, cot and csc. It is remarkable that explicit formulas for general order derivatives of such basic and simple functions are still be missing today, both in common function reference books as well as in symbolic mathematics software packages. The problem of calculating derivatives of the particular functions sec (x) and tan (x) was considered by Ho¤man in [5], but the explicit construction of the polynomials was avoided in his paper. The aim of this paper is to generalize his results and to explicitly construct expressions for the polynomials involved. The simpler problem of how to e¢ ciently compute the coe¢ cients in the Maclaurin series of sec (x) and tan (x) was addressed by Atkinson in [2]. In section 2, we explicitly construct the sets S1 and S2 and obtain the generating functions for both sequences of polynomials. In section 3, several properties for these generating functions are proved. From this, we derive the properties of the generated polynomials and give two explicit and equivalent expressions for these polynomials in section 4. In section 5, four di¤erent function subsets are considered in more detail, one containing (a generalization of) the hyperbolic functions and another containing (a generalization of) the trigonometric functions. We use here some of the notations and de…nitions as introduced in [4]. 2. Derivative polynomials 2.1. The function set S1 . Consider holomorphic functions f : for which holds, (2.1)
f
C ! C, such that z 7 ! u = f (z),
Dzn f (z) = ( 1)n Pn+1 (f (z)); 8n 2 N;
with Pn+1 (u) a polynomial in u with constant coe¢ cients and of degree n + 1. For any f satisfying (2.1), it is necessary and su¢ cient that f satis…es the …rst order di¤erential equation (2.2)
Dz f (z) =
P2 (f (z)):
Without lack of generality, we may restrict ourselves to normed polynomials P2 (u) (having a leading coe¢ cient equal to 1), as we can always absorb this coe¢ cient by a multiplicative rescaling of z. With 1
2
P2 (u) = (u (2.3)
GHISLAIN R. FRANSSENS 1 ) (u
f (z) =
8 < :
2 ),
1; 2
1 (f (z0 )
(f (z0 )
2 C, the general solution of (2.2) is
(+ 2 ) exp(+
2 ) exp
1
2 2
1
(z z0 ))
2 (f (z0 )
(
1
1 ) exp
(z z0 )) (f (z0 ) 1 ) exp( + 1+(f (zf 0(z)0 ) )(z z0 ) ; 2
2
2 2 2
1 2
(z z0 ))
(z z0 ))
We will need below also the compositional inverse function f 1 : f 1 f 1 (u), given by 8 1 < f 1 (u0 ) ln uu 1 ln uu00 1 ; if 1 2 2 2 (2.4) f 1 (u) = 1 : f 1 (u0 ) + u 1 ; if u0 In (2.3) and (2.4) is z0 2 C arbitrary, but such that u0 , f (z0 ) 6= (2.5)
Du f
1
(u) =
1 : P2 (u)
2.2. The function set S2 . Consider holomorphic functions g : which satis…es the …rst order di¤erential equation, (2.6)
Dz g(z) =
i,
g
; if
1
6=
2;
if
1
=
2
, .
C ! C, such that u 7 ! z = 1
6=
2;
1
=
2
, .
8i 2 f1; 2g. There holds
C ! C, such that z 7 ! v = g(z),
f (z)g(z);
and wherein f is given by (2.3). The n-th derivative of g then satis…es, Dzn g(z) = ( 1)n Qn (f (z))g(z); 8n 2 N;
(2.7)
with Qn (u) a polynomial in u of degree n. The general solution of (2.6) is, with c 2 Cn f0g, 8 1 + 2 (z z ) ( 1 0 ) 2 ) exp( 2 < ; if 1 = 6 2; 1 2 (z z ) 1 2 (z z ) (f (z ) ) exp + (f (z0 ) 1 ) exp( ( 0 0 ) 0 ) 2 2 2 (2.8) g(z) = c exp( (z z0 )) : if 1 = 2 , . 1+(f (z0 ) )(z z0 ) ;
2.3. Generating functions for Pn+1 (u) and Qn (u). The two di¤erentiation rules (2.1) and (2.7) generate polynomials Pn+1 (u) and Qn (u), which will be the main focus of this paper. Theorem 1. The polynomial sequences Pn+1 (u) and Qn (u) have as generating functions F (t; u) ,
(2.9)
G(t; u) ,
(2.10)
+1 X
n=0 +1 X
n=0
Pn+1 (u) Qn (u)
tn =f f n!
1
(u)
t ;
g f 1 (u) t tn = ; n! g (f 1 (u))
valid, for any …xed u, in a su¢ ciently small neighbourhood of t = 0. Proof 1. (i) Using (2.1), the generating function F (t; u) becomes F (t; u) =
+1 +1 n n X X ( t) t Pn+1 (u) = Dzn f (z) = f (z n! n! n=0 n=0
t);
and substituting z = f 1 (u) yields (2.9). (ii) Using (2.7), the generating function G(t; u) becomes G(t; u) = and substituting z = f
1
+1 +1 n n X X t ( t) Dzn g(z) g(z t) Qn (u) = = ; n! n! g(z) g(z) n=0 n=0
(u) yields (2.10).
In [5, Theorem 2.1], an alternative proof of (2.9)–(2.10) is given, based on the properties (3.13)–(3.14) below. Substituting (2.3) and (2.4) in (2.9) yields explicitly, 8 < 2 (u 1 ) exp(+ 1 2 2 t) 1 (u 2 ) exp( 1 2 2 t) ; if 1 6= 2 ; 1 2t (u 1 ) exp(+ 1 2 2 t) (u 2 ) exp( ) 2 (2.11) F (t; u) = u (u )t : if 1 = 2 , . 1 (u )t ;
FUNCTIONS W ITH DERIVATIVES GIVEN BY POLYNOM IALS IN THE FUNCTION ITSELF OR A RELATED FUNCTION3
Also, substituting (2.8) and (2.4) in (2.10) gives, 8 1+ 2 t ( 2 ) 1 ) exp( 2 < (u 1 ) exp(+ 1 2 2 t) (u 2 ) exp( (2.12) G(t; u) = : e t ; 1 (u
1
2 2
t)
; if
1
6=
2;
if
1
=
2
)t
, .
From (2.11)–(2.12) follows that the power series (2.9) and (2.10) converge absolutely and uniformly
for
(2.13)
jtj
0 P2 (u)An 1 (u
1; u
2)
1; u
2 );
are given by (2.20).
Proof 3. Combining (3.9) with (3.5), we get Du F (t; u) = (B(u
2 2 ; t))
1; u
:
Substitution of the Maclaurin series (2.9) and expression (2.18) gives +1 X
Du Pn+1 (u)
n=0
or, because t is arbitrary, 8n 2 N,
+1 X tn = An (u n! n=0
Du Pn+1 (u) = An (u
1; u
1; u
2)
2 ):
Using the recurrence relation (4.5) we get Pn+2 (u) = P2 (u)An (u Substituting n ! n
1; u
1 and because P1 (u) = u, we obtain (4.15).
2 ):
tn ; n!
6
GHISLAIN R. FRANSSENS
For u = 0 follows from (4.15) that (4.16)
Pn+1 (0) =
In particular, for
1
=
(4.17) Let
1
(4.18)
6=
2
and
1 2
=
1)n+1
n>0 (
1 2 An 1 ( 1 ; 2 ):
and due to (2.22), (4.16) reduces to Bn+1 n+1 n+1 Pn+1 (0) = 2n+1 1 : n>0 2 n+1 6 0. If we choose z0 such that u0 = f (z0 ) = 0, combining (2.1) and (4.16) gives = 2
lim Dzn f (z) =
n>0 1 2 An 1 ( 1 ; 2 );
z!z0
which yields the Taylor series of f (z) about z0 , f (z)
1
2
exp +
1
+1 X
=
An
2
2
=
1 2
(4.19)
exp +
2
2
(z
z0 )
exp
(z
z0 )
1
1( 1; 2)
(4.20)
n
Qn (u) = 2
n X
n k
(
z0 ) : n!
+
n k 2)
1
2
2
exp
(z
z0 ) ;
1
2
2
(z
z0 )
n
(z
n=1
Theorem 4. There holds, 8n 2 N,
1
Bk (u
1; u
2) ;
k=0
wherein the bivariate polynomials Bk (u
1; u
2)
are given by (2.19).
Proof 4. Substitution of the Maclaurin series (2.10) in the left-hand side of (2.16) and the Maclaurin 2 series for exp 1 + 2 t and (2.17) in the right-hand side of (2.16), gives +1 X
Qn (u)
n=0
tn n!
+1 X
=
+1
k=0 +1 X
=
Bk (u
1; u
tk X k!
2)
1
l=0
n X
n=0
n k
2
n k
(
k
Bk (u
1; u
n X
n
1
n k 2)
+
tl ; l!
2
n k
1+ 2
2)
Bk (u
l
+ 2
k=0
Because t is arbitrary holds, 8n 2 N, Qn (u) = 2
k
2
1; u
2
!
tn : n!
2 ):
k=0
In particular, for
1
=
=
2,
(4.20) reduces to
(4.21)
n
Qn (u) = 2
Bn (u
;u + ):
For u = 0 follows from (4.20) that (4.22)
Qn (0) = 2
n
n X
( 1)k
n k
(
1
n k 2)
+
Bk ( 1 ;
2 ):
k=0
In particular, for
1
=
=
2
and due to (2.21), (4.22) reduces to
(4.23) Let (4.24)
Qn (0) = En 1
6=
2.
n
:
If we choose z0 such that u0 = f (z0 ) = 0 and g(z0 ) 6= 0, combining (2.7) and (4.22) gives ! n X n k n lim Dzn g(z) = 2 n Bk ( 1 ; 2 ) g(z0 ); k ( ( 1 + 2 )) z!z0
k=0
which yields the Taylor series of g(z) about z0 , g(z) g(z0 ) (4.25)
( = 2 exp +
=
+1 X
n=0
2
n
1 ) exp
2
1
2
2
n X
k=0
n k
(z
z0 )
( (
1+
1+ 2
2
(z
1 exp n k
2 ))
z0 ) ; 1
2
2
Bk ( 1 ;
(z !
2)
z0 ) (z
n
z0 ) : n!
FUNCTIONS W ITH DERIVATIVES GIVEN BY POLYNOM IALS IN THE FUNCTION ITSELF OR A RELATED FUNCTION7
4.3.2. Second form. Theorem 5. There holds, 8n 2 N, (4.26)
n+1 X
Pn+1 (u) =
pn+1 uk ; k
k=0
wherein (4.27)
pn+1 = k
+
n=0 k=1
1)n+1
n>0 (
k
n X
n+1 l;l 1; 2
An;l Vk
;
l=1
An;l are the Eulerian numbers [7, Sloane’s A008292] and Vk n+1; l;l are the coe¢ cients of the normed 1 2 polynomial with roots 1 and 2 having respective multiplicities n + 1 l and l. Proof 5. From (4.15) and (2.20) follows Pn+1 (u)
= u
n=0
+
n>0 P2 (u)
n X1
n 1 k (u 1)
An;k+1 (u
k 2) ;
k=0
= u
n=0
+
n X1
n>0
n+1 (k+1) (u 1)
An;k+1 (u
k+1 ; 2)
k=0
= u
n=0
+
n X
n>0
n+1 l (u 1)
An;l (u
l 2) :
l=1
Using n+1 l
(u
1)
l
(u
2)
=
n+1 X
n+1 l;l 1; 2
Vk
uk ;
k=0
we get Pn+1 (u)
= u
n=0
+
n X
n>0
An;l ( 1)n+1
k
l=1
= u
n=0
+
( 1)n+1
k
k=0
n X
An;l Vk
uk ;
n+1 l;l 1; 2
l=1
Pn
From (4.26) and (4.27) with Vn+1 …nd that
n+1 l;l 1; 2
(4.28)
Pn+1 (u) = n!un+1 + O(un ):
= 1 and because
Theorem 6. There holds, 8n 2 N, (4.29)
n+1 l;l 1; 2
Vk
k=0
n+1 X
n>0
n+1 X
Qn (u) =
n X
l=1
!
uk :
An;l = n! (see [4, eq. (3.26)]), we
qkn uk ;
k=0
wherein (4.30)
qkn = 2
n
n X
n m
(
1+
n m
2)
m k
( 1)
m=k
m X
m l;l 1; 2
Bm+1;l+1 Vk
;
l=0
Bm;l are the MacMahon numbers [7, Sloane’s A060187] and Vk m ; l;l are the coe¢ cients of the normed 1 2 polynomial with roots 1 and 2 having respective multiplicities m l and l. Proof 6. Substituting (2.19) in (4.20) gives Qn (u) = 2
n
n X
n m
(
1+
n m
2)
m=0
n
n X
m=0
n m
Bm+1;l+1 (u
m l 1)
(u
l 2)
:
l=0
Expanding the second sum in powers of u, we get Qn (u) = 2
m X
(
1+
n m
2)
m X l=0
Bm+1;l+1
m X
k=0
m k
( 1)
Vk
m l;l 1; 2
uk ;
8
GHISLAIN R. FRANSSENS
Exchanging the summation order yields Qn (u) =
n X
2
n
n X
n m
(
1
+
n m
2)
m k
( 1)
m=k
k=0
Pn
l=0
=
=
(4.30) reduces to n X n k qkn = ( 1) 2 n Bn+1;l+1
(4.32)
!
uk :
Bn+1;l+1 = 2n n! (see [4, eq. (3.21)]), we …nd that
Qn (u) = n!un + O(un
(4.31) 1
Bm+1;l+1 Vk m ; l;l 1 2
l=0
From (4.29) and (4.30) and because In particular, for
m X
1
):
2,
n l;l k
(
; ):
l=0
5. Special subsets of S1 [ S2
Put f (z0 ) , u0 and let z0 = 0. We consider in this section a number of important subsets of functions, having polynomial derivative expressions. 5.1. The case 1 = 2 = . Without loss of generality, we can assume of this subset are of the form 1 (5.1) ; f (z) = 1=u0 + z 1 g(z) = c (5.2) ; 1=u0 + z and the generating functions for the polynomials are u (5.3) F (t; u) = ; 1 ut 1 (5.4) : G(t; u) = 1 ut The polynomials are easily found to be
= 0. Let u0 6= 0. The functions
Pn+1 (u) = n!un+1 ; Qn (u) = n!un ;
(5.5) (5.6)
and the di¤erentiation formulas (2.1) and (2.7) are trivial. 5.2. The case 1 = 0 and 2 6= 0. Without loss of generality, we can assume put a , (1 1=u0 ) 6= 0. The functions of this subset are of the form 1 f (z) = (5.7) ; 1 + ae z e z (5.8) g(z) = c ; 1 + ae z and the generating functions for the polynomials are 1 (5.9) F (t; u) = ; 1 (1 1=u) et et 1 (5.10) G(t; u) = : u 1 (1 1=u) et The polynomials are in this case, (5.11)
Pn+1 (u)
=
(5.12)
Qn (u)
=
n=0 u + n>0 u(u n X n 2 n k Bk (u; u k=0
1)An
1 (u; u
2
= 1. Let u0 6= 1 and
1);
1) :
The di¤erentiation formulas (2.1) and (2.7) become, (5.13)
Dzn
1 1 + ae
(5.14)
Dzn
ae z 1 + ae z
z
= =
1 n=0 1 + ae n
( 1) 2
n
z
n X
k=0
+ n k
n>0 (
Pk
l=0
n
1)
Pn
1 l=0
An;l+1 ( ae
(1 + ae
Bk+1;l+1 ( ae (1 + ae
z )k+1
z )n+1
z l+1
)
:
z l+1
)
;
FUNCTIONS W ITH DERIVATIVES GIVEN BY POLYNOM IALS IN THE FUNCTION ITSELF OR A RELATED FUNCTION9 z
ae 1 Because 1+ae 1, the right-hand sides of the expressions (5.13) and (5.14) must be equal z = 1+ae z for n > 0. This implies the following relation between Eulerian and MacMahon polynomials, holding 8n 2 Z+ and 8y 2 C, n X n k n (5.15) 2n An 1 (1; y) = y) Bk (1; y): k (1 k=0
For y = 1, (5.15) is an identity due to [4, eqs. (3.21) and (3.26)], while for y = 1 we …nd, due to [4, eqs. (3.22) and (3.27)], a well-known identity between Euler and Bernoulli numbers. For y = 0, (5.15) reduces to a well-known binomial identity. Substitution of the expression [4, eq. (4.16)] for An 1 (u; u 1) in (5.11) and using the recursion relation for the Stirling numbers of the second kind S (n; k), [1, p. 824 24.1.4 II. A.], gives (5.16)
n+1 X
Pn+1 (u) =
n+1 q
( 1)
(q
1)!S (n + 1; q) uq :
q=1
This standard form for Pn+1 (u) then yields the following equivalent expression for (5.13), 1 1 + ae
Dzn
(5.17)
=
z
n X
1 1 + ae
k
( 1) k!S (n + 1; k + 1)
k=0
k+1
:
z
Functions like (5.7)–(5.8), when regarded as function on the real line, are sometimes called sigmoid functions, because of their “S-shape”. They are often used in neural networks, to introduce nonlinearity and/or to make sure that certain signals remain bounded, and are also typical solutions of logistics models. 5.3. The case
1
=
1 and
2
= 1. Let u0 6=
1. The functions of this subset are of the form
sinh (z) + u0 cosh (z) ; cosh (z) + u0 sinh (z) u0 (5.19) g(z) = c ; cosh (z) + u0 sinh (z) and the generating functions for the polynomials are sinh (t) + u cosh (t) (5.20) F (t; u) = ; cosh (t) u sinh (t) 1 (5.21) G(t; u) = : cosh (t) u sinh (t) f (z)
(5.18)
=
5.3.1. Di¤ erentiation formulas. First form. The polynomials Pn+1 (f (z)) and Qn (f (z)) become Pn+1 (f (z))
=
(5.22) (5.23)
n=0 f (z)
+ Qn (f (z))
= 2
2 n>0 u0
n
Pn
k=0
1
Pn
1 k=0
n 1 k
An;k+1 (u0 + 1)
(u0
k
1) e+(n n+1
(cosh (z) + u0 sinh (z)) n k
k
Bn+1;k+1 (u0 + 1) (u0 1) e+(n n (cosh (z) + u0 sinh (z))
1 2k)z
;
2k)z
:
(i) Taking the limit u0 ! 0 in the expressions (5.18)–(5.19) and (5.22)–(5.23) gives for the di¤erentiation formulas (2.1) and (2.7), (i.1) for n even, Pn 1 k 1 2k) z) k=0 ( 1) An;k+1 sinh ((n (5.24) Dzn tanh (z) = n=0 tanh (z) ; n>0 coshn+1 (z) Pn k 2k) z) n n k=0 ( 1) Bn+1;k+1 cosh ((n Dz sech (z) = 2 (5.25) : n+1 cosh (z) (i.2) for n odd, (5.26)
Dzn
(5.27)
Dzn
tanh (z) sech (z)
= =
Pn
1 k=0 (
2
n
1)k An;k+1 cosh ((n coshn+1 (z)
Pn
k=0
k
1
2k) z)
( 1) Bn+1;k+1 sinh ((n coshn+1 (z)
;
2k) z)
:
10
GHISLAIN R. FRANSSENS
(ii) Taking the limit u0 ! 1 in the expressions (5.18)–(5.19) and (5.22)–(5.23) gives for the di¤erentiation formulas (2.1) and (2.7), Pn 1 An;k+1 cosh ((n 1 2k) z) (5.28) Dzn coth (z) = n=0 coth (z) + n>0 ( 1)n k=0 ; sinhn+1 (z) Pn Bn+1;k+1 cosh ((n 2k) z) (5.29) : Dzn csch (z) = ( 1)n 2 n k=0 sinhn+1 (z) Formulas (5.24)–(5.29) express the n-th derivative of the hyperbolic functions tanh, sech, coth and csch as a simple rational function of sinh and cosh functions, involving the Eulerian and MacMahon number triangles. Second form. From (2.1), (2.7) and (4.26), (4.29) follows Dzn tanh (z)
(5.30)
( 1)n
=
n+1 X
pn+1 tanhk (z) ; k
k=0
Dzn sech (z)
(5.31)
n
=
( 1) (sech (z))
n X
qkn tanhk (z) ;
k=0
and Dzn coth (z)
(5.32)
( 1)n
=
n+1 X
pn+1 cothk (z) ; k
k=0
Dzn csch (z)
(5.33)
n
=
( 1) (csch (z))
n X
qkn cothk (z) :
k=0
pn+1 k
qkn
The constants and satisfy the recursion relations (4.13)–(4.14), with in this case = 0 and = 1. Explicit expressions for pn+1 are given by (4.27), with 1 = 1 and 2 = 1, and for qkn by (4.32), k with = 1. 5.4. The case
1
=
i and
2
= i. Let u0 6=
(5.34)
f (z)
(5.35)
g(z)
i. The functions of this subset are of the form
sin (z) u0 cos (z) ; cos (z) + u0 sin (z) u0 = c ; cos (z) + u0 sin (z) =
and the generating functions for the polynomials are (5.36)
F (t; u)
=
(5.37)
G(t; u)
=
sin (t) + u cos (t) ; cos (t) u sin (t) 1 : cos (t) u sin (t)
5.4.1. Di¤ erentiation formulas. First form. The polynomials Pn+1 (f (z)) and Qn (f (z)) become Pn+1 (f (z)) (5.38) (5.39)
=
n=0 f (z)
+ Qn (f (z))
= 2
2 n>0 u0
n
Pn
k=0
+1
Pn
1 k=0
n 1 k
An;k+1 (u0 + i)
k
i) e+i(n
(u0
n+1
(cos (z) + u0 sin (z)) n k
k
Bn+1;k+1 (u0 + i) (u0 i) e+i(n n (cos (z) + u0 sin (z))
1 2k)z
;
2k)z
:
(i) Taking the limit u0 ! 0 in the expressions (5.34)–(5.35) and (5.38)–(5.39) gives for the di¤erentiation formulas (2.1) and (2.7), (i.1) for n even, Pn 1 k 1 2k) z) n=2 k=0 ( 1) An;k+1 sin ((n (5.40) ; Dzn tan (z) = n=0 tan (z) ( 1) n>0 cosn+1 (z) Pn k 2k) z) n=2 n n k=0 ( 1) Bn+1;k+1 cos ((n Dz sec (z) = 2 ( 1) (5.41) : n+1 cos (z)
FUNCTIONS W ITH DERIVATIVES GIVEN BY POLYNOM IALS IN THE FUNCTION ITSELF OR A RELATED FUNCTION 11
(i.2) for n odd, (n 1)=2
(5.42)
Dzn
tan (z)
=
( 1)
(5.43)
Dzn sec (z)
=
2
n
Pn
1 k=0
k
( 1) An;k+1 cos ((n 1 2k) z) ; cosn+1 (z) Pn k 2k) z) k=0 ( 1) Bn+1;k+1 sin ((n : n+1 cos (z)
(n 1)=2
( 1)
(ii) Taking the limit u0 ! 1 in the expressions (5.34)–(5.35) and (5.38)–(5.39) gives for the di¤erentiation formulas (2.1) and (2.7), Pn 1 1 2k) z) n n k=0 An;k+1 cos ((n (5.44) ; Dz cot (z) = n=0 cot (z) + n>0 ( 1) n+1 sin (z) Pn Bn+1;k+1 cos ((n 2k)z) (5.45) : Dzn csc (z) = ( 1)n 2 n k=0 sinn+1 (z) Formulas (5.40)–(5.45) express the n-th derivative of the hyperbolic functions tan, sec, cot and csc as a rational function of sin and cos functions, with the numerator in the form of a Fourier series with amplitudes given by the Eulerian and MacMahon number triangles. Second form. From (2.1), (2.7) and (4.26), (4.29) follows Dzn tan (z)
(5.46)
=
n+1 X
k n+1 pk
( 1)n+1
tank (z) ;
k=0
Dzn sec (z)
(5.47)
=
n X
(sec (z))
n k
( 1)
qkn tank (z) ;
k=0
and Dzn cot (z)
(5.48)
( 1)n
=
n+1 X
pn+1 cotk (z) ; k
k=0
Dzn csc (z)
(5.49)
n
=
( 1) (csc (z))
n X
qkn cotk (z) :
k=0
pn+1 k
qkn
The constants and satisfy the recursion relations (4.13)–(4.14), with in this case = 0 and = +1. Explicit expressions for pn+1 are given by (4.27), with 1 = i and 2 = i, and for qkn by (4.32), k with = i. Appendix A. Some examples of Pn+1 (u) and Qn (u) De…ne
,
1
+
2
and
,
1 2.
A.1. Pn+1 (u). P1 (u)
= u;
P2 (u)
=
1!u2
u+ ;
P3 (u)
=
2!u3
3 u2 +
=
3!u
4
12 u + 7
4!u
5
4
P4 (u) P5 (u)
=
3
+2 2
2
+8
2
+5
36 z + 14
2
+2
11
u
+8
60 u + 10 5 4
+
2
2
; u
2
+4
2
u
3
15 2
u
+8 2
2
u+ +4
+8
;
2
+ 18
u
+2
;
2
A.2. Qn (u). Q0 (u) Q1 (u)
= =
1; 1!z;
Q2 (u)
=
2!z 2
z+ ;
Q3 (u)
=
3!z 3
6 z2 +
=
4
Q4 (u)
4!z
3
z +2
; z
2
z+
2
+5
:
12
GHISLAIN R. FRANSSENS
References [1] Abramowitz, M. and Stegun, I.A., Handbook of Mathematical Functions (9-th ed.), Dover, New York, 1970. [2] Atkinson, M.D., How to compute the series expansions of sec x and tan x, Am. Math. Monthly 93 (1986) 387–389. [3] Comtet, L., Advanced Combinatorics: The Art of Finite and In…nite Expansions (rev. enl. ed.), Reidel, DordrechtHolland, 1974. [4] Franssens, G.R., On a number pyramid related to Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles. Journal of Integer Sequences 9 (2006) Article 06.4.1. Available online at: http://www.cs.uwaterloo.ca/journals/JIS/. [5] Ho¤man, M.E., Derivative Polynomials for Tangent and Secant, Am. Math. Monthly 102 (1995) 23–30. [6] MacMahon, P.A., The divisors of numbers, Proc. London Math. Soc. 19 (1920) 305–340. [7] Sloane, N.J.A., The On-Line Version of the Encyclopaedia of Integer Sequences. Available online at: www.research.att.com/~njas/sequences/eisonline.html. Belgian Institute for Space Aeronomy, Ringlaan 3, B-1180 Brussels, Belgium,
[email protected], http://www.aeronomie.be/
