Functions with derivatives given by polynomials in the ... - Springer Link

Analysis Mathematica, 33(2007), 17–36 DOI: 10.1007/s10474-007-0102-5

Functions with derivatives given by polynomials in the function itself or a related function GHISLAIN R. FRANSSENS Belgian Institute for Space Aeronomy, Ringlaan 3, 1180 Brussels, Belgium, e-mail: [email protected] Received June 27, 2006.

A b s t r a c t . We construct the set of holomorphic functions S1 = {f : Ωf ⊆ 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 = {g : Ωg ⊆ C → C} 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.

1. Introduction We consider the set of holomorphic functions S1 = {f : Ωf ⊆ C → C} such that the n-th order derivative D zn 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 = {g : Ωg ⊆ C → C} such that the n-th order derivative D zn g(z) is given by the product of g with a polynomial in f ∈ S1 of degree n. The main topic of this paper is to investigate the two polynomial sequences involved. 0133–3852/$ 20.00 c 2007 Akad´emiai Kiad´o, Budapest

18

Ghislain R. Franssens

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 (non-trivial!) 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 Hoffman 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 efficiently compute the coefficients in the Maclaurin series of sec(x) and tan(x) was addressed by Atkinson in [2]. In Section 2, we explicitly construct the sets S 1 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 different 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 definitions as introduced in [4].

2. Derivative polynomials 2.1. The function set S1 Consider holomorphic functions f : Ωf ⊆ C → C, such that z 7→ u = f (z), for which (1)

Dzn f (z) = (−1)n Pn+1 (f (z)),

∀ n ∈ N,

with Pn+1 (u) a polynomial in u with constant coefficients and of degree n + 1. For any f satisfying (1), it is necessary and sufficient that f satisfies the first order differential equation (2)

Dz f (z) = −P2 (f (z)).

Derivatives given by polynomials in the function itself

19

Without lack of generality, we may restrict ourselves to normed polynomials P2 (u) (having a leading coefficient equal to 1), as we can always absorb this coefficient by a multiplicative rescaling of z. With P2 (u) = (u − ρ1 )(u − ρ2 ),

ρ1 , ρ2 ∈ C,

the general solution of (2) is (3) 8 >
:ρ +

ρ1 −ρ2 (z−z0 )) 2 ρ1 −ρ2 (z−z0 ))−(f (z0 )−ρ1 ) exp(− 2 (z−z0 ))

(z−z0 ))−ρ2 (f (z0 )−ρ1 ) exp(−

if ρ1 6= ρ2 , if ρ1 = ρ2 , ρ.

1+(f (z0 )−ρ)(z−z0 )

We will need below also the compositional inverse function f −1 : Ωf −1 ⊂ C → C

u 7→ z = f −1 (u),

such that

given by (4)

f

−1

(u) =

(

u0 −ρ1 1 1 (ln u−ρ f −1 (u0 ) − ρ1 −ρ u−ρ2 − ln u0 −ρ2 ) 2 1 f −1 (u0 ) + ( u−ρ − u01−ρ )

if ρ1 6= ρ2 , if ρ1 = ρ2 , ρ.

In (3) and (4) is z0 ∈ C arbitrary, but such that u0 , f (z0 ) 6= ρi ,

∀ i ∈ {1, 2}.

The following equality holds: (5)

Du f −1 (u) = −

1 . P2 (u)

2.2. The function set S2 Consider holomorphic functions g : Ωg ⊆ C → C

such that

z 7→ v = g(z),

which satisfies the first order differential equation (6)

Dz g(z) = −f (z)g(z),

and where f is given by (3). The n-th derivative of g then satisfies the equality (7)

Dzn g(z) = (−1)n Qn (f (z))g(z),

∀ n ∈ N,

with Qn (u) a polynomial in u of degree n. The general solution of (6) is, with c ∈ C\{0},

20

Ghislain R. Franssens

(8) =c

g(z) = 8 > < > :

ρ +ρ (ρ1 −ρ2 ) exp(− 1 2 2 (z−z0 )) ρ −ρ ρ1 −ρ2 (f (z0 )−ρ2 ) exp(+ 2 (z−z0 ))−(f (z0 )−ρ1 ) exp(− 1 2 2 exp(−ρ(z−z0 ))

(z−z0 ))

if ρ1 6= ρ2 , if ρ1 = ρ2 , ρ.

1+(f (z0 )−ρ)(z−z0 )

2.3. Generating functions for Pn+1 (u) and Qn (u)

The two differentiation rules (1) and (7) generate polynomials P n+1 (u) and Qn (u), which will be the main focus of this paper. T h e o r e m 1. The polynomial sequences P n+1 (u) and Qn (u) have as generating functions ∞ X

F (t, u) ,

(9)

Pn+1 (u)

n=0

G(t, u) ,

(10)

∞ X

Qn (u)

n=0

tn = f (f −1 (u) − t), n!

g(f −1 (u) − t) tn = , n! g(f −1 (u))

valid, for any fixed u, in a sufficiently small neighbourhood of t = 0. P r o o f . (i) Using (1), the generating function F (t, u) becomes F (t, u) =

∞ n X t

n! n=0

Pn+1 (u) =

∞ X (−t)n

n=0

n!

Dzn f (z) = f (z − t),

−1

and substituting z = f (u) yields (9). (ii) Using (7), the generating function G(t, u) becomes G(t, u) =

∞ n X t

n=0

n!

Qn (u) =

∞ X (−t)n Dzn g(z)

n=0

n!

g(z)

=

g(z − t) , g(z)

and substituting z = f −1 (u) yields (10).

u t

In [5, Theorem 2.1], an alternative proof of (9)–(10) is given, based on the properties (35)–(36) below. Substituting (3) and (4) in (9) yields explicitly the following: (11) 8 F (t, u) =

> < > :

ρ1 −ρ2 2 ρ1 −ρ2 (u−ρ1 ) exp(+ 2

ρ2 (u−ρ1 ) exp(+

u−ρ(u−ρ)t 1−(u−ρ)t

ρ1 −ρ2 2 ρ1 −ρ2 t)−(u−ρ2 ) exp(− 2 t)

t)−ρ1 (u−ρ2 ) exp(−

t)

if ρ1 6= ρ2 ,

if ρ1 = ρ2 , ρ.

Derivatives given by polynomials in the function itself

21

Also, substituting (8) and (4) in (10) gives (12) G(t, u) =

8 > < > :

(ρ2 −ρ1 ) exp( ρ −ρ (u−ρ1 ) exp(+ 1 2 2 ρt

ρ1 +ρ2 2

t)

t)−(u−ρ2 ) exp(−

ρ1 −ρ2 2

t)

e 1−(u−ρ)t

if ρ1 6= ρ2 , if ρ1 = ρ2 , ρ.

From (11)–(12) it follows that the power series (9) and (10) converge absolutely and uniformly for (13)

|t|
0 P2 (u)An−1 (u − ρ1 , u − ρ2 ),

∀ n ∈ N,

where the bivariate polynomials An−1 (u − ρ1 , u − ρ2 ) are given by (20). P r o o f . Combining (31) with (27), we get Du F (t, u) = (B(u − ρ1 , u − ρ2 , t))2 . Substitution of the Maclaurin series (9) and expression (18) gives ∞ X tn tn An (u − ρ1 , u − ρ2 ) , Du Pn+1 (u) = n! n=0 n! n=0 ∞ X

or, since t is arbitrary,

Du Pn+1 (u) = An (u − ρ1 , u − ρ2 ),

∀ n ∈ N.

Using the recurrence relation (41), we get Pn+2 (u) = P2 (u)An (u − ρ1 , u − ρ2 ). Substituting n → n − 1 and since P1 (u) = u, we obtain (51).

u t

26

Ghislain R. Franssens

For u = 0, it follows from (51) that Pn+1 (0) = δn>0 (−1)n+1 ρ1 ρ2 An−1 (ρ1 , ρ2 ).

(52)

In particular, for −ρ1 = ρ = ρ2 , due to (22), (52) reduces to “ ”B n+1 n+1 (53) Pn+1 (0) = −δn>0 2n+1 2n+1 − 1 ρ . n+1 Let ρ1 6= ρ2 and ρ1 ρ2 6= 0. If we choose z0 such that u0 = f (z0 ) = 0, combining (1) and (52) gives lim Dzn f (z) = −δn>0 ρ1 ρ2 An−1 (ρ1 , ρ2 ),

(54)

z→z0

which yields the Taylor series of f (z) about z 0 , ρ1 −ρ2 2 exp(+ ρ1 −ρ f (z) 2 (z − z0 )) − exp(− 2 (z − z0 )) = = 2 2 ρ1 ρ2 (z − z0 )) − ρ1 exp(− ρ1 −ρ (z − z0 )) ρ2 exp(+ ρ1 −ρ 2 2

(55)

=−

∞ X

An−1 (ρ1 , ρ2 )

n=1

(z − z0 )n . n!

T h e o r e m 4. We have! n X n −n (56) Qn (u) = 2 (ρ1 + ρ2 )n−k Bk (u − ρ1 , u − ρ2 ), k k=0

∀ n ∈ N,

where the bivariate polynomials Bk (u − ρ1 , u − ρ2 ) are given by (19).

P r o o f . Substitution of the Maclaurin series (10) in the left-hand side 2 t) and (17) in the right-hand of (16) and the Maclaurin series for exp( ρ1 +ρ 2 side of (16), gives ∞ X

Qn (u)

n=0

=

∞ ∞ “ X tk X ρ 1 + ρ 2 ”l t l tn = 2−k Bk (u − ρ1 , u − ρ2 ) = n! k=0 k! l=0 2 l!

n ∞ “X X

n=0

!

“ ρ + ρ ”n−k ” tn n −k 1 2 2 Bk (u − ρ1 , u − ρ2 ) . k 2 n!

k=0

Since t is arbitrary, we have Qn (u) = 2

−n

n X

!

n (ρ1 + ρ2 )n−k Bk (u − ρ1 , u − ρ2 ), k

k=0

∀ n ∈ N.

In particular, for ρ1 = ρ = −ρ2 , (56) reduces to (57)

Qn (u) = 2−n Bn (u − ρ, u + ρ) . For u = 0 it follows from (56) that

(58)

Qn (0) = 2

−n

n X

(−1)

k=0

k

!

n (ρ1 + ρ2 )n−k Bk (ρ1 , ρ2 ). k

u t

Derivatives given by polynomials in the function itself

27

In particular, for −ρ1 = ρ = ρ2 , due to (21), (58) reduces to Qn (0) = En ρn .

(59)

Let ρ1 6= ρ2 . If we choose z0 such that u0 = f (z0 ) = 0

and

g(z0 ) 6= 0,

combining (7) and (58) gives lim D n g(z) z→z0 z

(60)

“

= 2

−n

n X

!

” n (−(ρ1 + ρ2 ))n−k Bk (ρ1 , ρ2 ) g(z0 ), k

k=0

which yields the Taylor series of g(z) about z 0 , 2 (ρ2 − ρ1 ) exp(− ρ1 +ρ g(z) 2 (z − z0 )) = = ρ1 −ρ2 2 g(z0 ) ρ2 exp(+ ρ1 −ρ 2 (z − z0 )) − ρ1 exp(− 2 (z − z0 ))

(61)

=

∞ “ X

n=0

2

−n

n X

k=0

!

” (z − z )n n 0 (−(ρ1 + ρ2 ))n−k Bk (ρ1 , ρ2 ) . k n!

4.3.2. Second form T h e o r e m 5. The following equality holds: (62)

Pn+1 (u) =

n+1 X

pn+1 uk , k

∀n ∈ N

k=0

where (63)

pn+1 k

= δn=0 δk=1 + δn>0 (−1)

n+1−k

n X l=1

An,l Vk

!

n + 1 − l, l , ρ1 , ρ 2

An,l are the Eulerian numbers (see [7, Sloane’s A008292]) and V k n+1−l,l ρ1 ,ρ2 are the coefficients of the normed polynomial with roots ρ 1 and ρ2 having respective multiplicities n + 1 − l and l. `

P r o o f . From (51) and (20) it follows that Pn+1 (u) = uδn=0 + δn>0 P2 (u)

n−1 X

An,k+1 (u − ρ1 )n−1−k (u − ρ2 )k =

k=0

= uδn=0 + δn>0

n−1 X

An,k+1 (u − ρ1 )n+1−(k+1) (u − ρ2 )k+1 =

k=0

= uδn=0 + δn>0

n X l=1

An,l (u − ρ1 )n+1−l (u − ρ2 )l .

´

28

Ghislain R. Franssens

Using (u − ρ1 )

n+1−l

l

(u − ρ2 ) =

n+1 X

Vk

k=0

we get Pn+1 (u) = uδn=0 + δn>0

n X

An,l (−1)

n+1−k

l=1

= uδn=0 + δn>0

n+1 X“

(−1)

n+1−k

Vk

n X

An,l Vk

l=1

By (62) and (63) with Vn+1

n+1 X k=0

k=0

n + 1 − l, l ρ1 , ρ 2

!

n + 1 − l, l k u , ρ1 , ρ 2

!

=1

and

n X

!

n + 1 − l, l k u = ρ1 , ρ 2 !

n + 1 − l, l ” k u . ρ1 , ρ 2

u t

An,l = n!

l=1

(see [4, eq. (3.26)]), we find that Pn+1 (u) = n! un+1 + O(un ).

(64)

u t

T h e o r e m 6. The following equality holds: (65)

Qn (u) =

n X

qkn uk ,

∀ n ∈ N,

k=0

where qkn

(66)

=2

−n

n X

m=k

m X m − l, l n , (ρ1 + ρ2 )n−m (−1)m−k Bm+1,l+1 Vk ρ1 , ρ 2 m l=0

!

!

Bm,l are the MacMahon numbers (see [7, Sloane’s A060187]) and V k m−l,l ρ1 ,ρ2 are the coefficients of the normed polynomial with roots ρ 1 and ρ2 having respective multiplicities m − l and l. `

P r o o f . Substituting (19) in (56) gives Qn (u) = 2−n

n X

m=0

m X n (ρ1 + ρ2 )n−m Bm+1,l+1 (u − ρ1 )m−l (u − ρ2 )l . m l=0

!

Expanding the second sum in powers of u, we get Qn (u) = =2

−n

n X

m=0

m m X X n m − l, l uk . (ρ1 + ρ2 )n−m Bm+1,l+1 (−1)m−k Vk m ρ , ρ 1 2 l=0 k=0

!

!

´

Derivatives given by polynomials in the function itself

29

Exchanging the summation order yields Qn (u) = =

n X

k=0

“

2

−n

n X

m=k

m X n m − l, l ” k (ρ1 + ρ2 )n−m (−1)m−k Bm+1,l+1 Vk u . m ρ1 , ρ 2 l=0 u t

!

!

By (65) and (66) and the fact that n X

Bn+1,l+1 = 2n n!

l=0

(see [4, eq. (3.21)]), we find that Qn (u) = n!un + O(un−1 ).

(67)

In particular, for −ρ1 = ρ = ρ2 , (66) reduces to qkn

(68)

= (−1)

n−k −n

2

n X

Bn+1,l+1 Vk

l=0

!

n − l, l . −ρ, ρ

5. Special subsets of S1 ∪ S2 Put f (z0 ) , u0 and z0 = 0. In this section, we consider a number of important subsets of functions having polynomial derivative expressions. 5.1. The case ρ1 = ρ2 = ρ Without loss of generality, we may assume ρ = 0. Let u 0 6= 0. The functions of this subset are of the form 1 , (69) f (z) = 1/u0 + z 1

; 1/u0 + z and the generating functions for the polynomials are u (71) F (t, u) = , 1 − ut (70)

g(z) = c

(72)

G(t, u) =

1 . 1 − ut

30

Ghislain R. Franssens

The polynomials are easily found to be (73)

Pn+1 (u) = n! un+1 ,

(74)

Qn (u) = n! un ,

and the differentiation formulas (1) and (7) are trivial. 5.2. The case ρ1 = 0 and ρ2 6= 0 Without loss of generality, we may assume ρ 2 = 1. Let u0 6= 1

and

a , −(1 − 1/u0 ) 6= 0.

The functions of this subset are of the form 1 (75) f (z) = , 1 + ae−z e−z ; 1 + ae−z and the generating functions for the polynomials are 1 (77) F (t, u) = , 1 − (1 − 1/u)et (76)

g(z) = c

(78)

G(t, u) =

1 et . u 1 − (1 − 1/u)et

In this case the polynomials are the following: (79)

Pn+1 (u) = δn=0 u + δn>0 u(u − 1)An−1 (u, u − 1),

(80)

Qn (u) = 2

−n

n X

k=0

!

n Bk (u, u − 1). k

The differentiation formulas (1) and (7) become (81) Dzn (82)

“

” 1 1 = δn=0 + δn>0 (−1)n −z 1 + ae 1 + ae−z

Dzn Since

“ −ae−z ”

1 + ae−z

n −n

= (−1) 2

n X

k=0

Pn−1

! Pk

n k

An,l+1 (−ae−z )l+1 , (1 + ae−z )n+1

l=0

−z l+1 ) l=0 Bk+1,l+1 (−ae . −z k+1 (1 + ae )

−ae−z 1 = − 1, −z 1 + ae 1 + ae−z

Derivatives given by polynomials in the function itself

31

the right-hand sides of the expressions (81) and (82) must be equal for n > 0. This implies the following relation between Eulerian and MacMahon polynomials: (83)

n

2 An−1 (1, y) =

n X

k=0

!

n (1 − y)n−k Bk (1, y), k

∀ n ∈ Z+ , ∀ y ∈ C.

For y = 1, (83) is an identity due to [4, eqs. (3.21) and (3.26)], while for y = −1 we find, due to [4, eqs. (3.22) and (3.27)], a well-known identity between Euler and Bernoulli numbers. For y = 0, (83) reduces to a well-known binomial identity. Substitution of the expression [4, eq. (4.16)] for A n−1 (u, u − 1) in (79) and using the recursion relation for the Stirling numbers S(n, k) of the second kind [1, p. 824, 24.1.4, II, A], gives (84)

Pn+1 (u) =

n+1 X

(−1)n+1−q (q − 1)! S(n + 1, q)uq .

q=1

This standard form for Pn+1 (u) then yields the following equivalent expression for (81): (85)

Dzn

“

n ” “ ”k+1 X 1 1 k . = (−1) k! S(n + 1, k + 1) 1 + ae−z 1 + ae−z k=0

Functions like (75)–(76), when regarded as functions 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 ρ1 = −1 and ρ2 = 1 Let u0 6= ±1. The functions of this subset are of the form sinh(z) + u0 cosh(z) (86) f (z) = , cosh(z) + u0 sinh(z) u0 ; cosh(z) + u0 sinh(z) and the generating functions for the polynomials are − sinh(t) + u cosh(t) (88) F (t, u) = , cosh(t) − u sinh(t) (87)

g(z) = c

(89)

G(t, u) =

1 . cosh(t) − u sinh(t)

32

Ghislain R. Franssens

5.3.1. Differentiation formulas First form. The polynomials Pn+1 (f (z)) and Qn (f (z)) become (90)

Pn+1 (f (z)) = δn=0 f (z) + + δn>0 (u20 − 1)

(91)

Pn−1 k=0

Qn (f (z)) = 2

−n

An,k+1 (u0 + 1)n−1−k (u0 − 1)k e+(n−1−2k)z , (cosh(z) + u0 sinh(z))n+1

Pn

+ 1)n−k (u0 − 1)k e+(n−2k)z . (cosh(z) + u0 sinh(z))n

k=0 Bn+1,k+1 (u0

(i) Taking the limit u0 → 0 in the expressions (86)–(87) and (90)–(91) gives for the differentiation formulas (1) and (7), (i.1) for n even, Dzn tanh(z) =

(92) = δn=0 tanh(z) − δn>0 (93)

Dzn sech(z) = 2−n

Pn−1

k=0 (−1)

Pn

k=0 (−1)

k

k

An,k+1 sinh((n − 1 − 2k)z) , coshn+1 (z)

Bn+1,k+1 cosh((n − 2k)z) ; coshn+1 (z)

(i.2) for n odd, (94)

(95)

Dzn tanh(z) Dzn sech(z)

=

Pn−1

= −2

k=0 (−1)

−n

k

An,k+1 cosh((n − 1 − 2k)z) , coshn+1 (z)

Pn

k=0 (−1)

k

Bn+1,k+1 sinh((n − 2k)z) . coshn+1 (z)

(ii) Taking the limit u0 → ∞ in the expressions (86)–(87) and (90)–(91) gives for the differentiation formulas (1) and (7), Dzn coth(z) = δn=0 coth(z) +

(96)

+ δn>0 (−1) (97)

Dzn csch(z)

n

Pn−1 k=0

n −n

= (−1) 2

An,k+1 cosh((n − 1 − 2k)z) , sinhn+1 (z) Pn

k=0

Bn+1,k+1 cosh((n − 2k)z) . sinhn+1 (z)

Formulas (92)–(97) express the n-th derivative of the hyperbolic functions tanh, sech, coth and csch as a simple rational function of the functions sinh and cosh, involving the Eulerian and MacMahon number triangles.

Derivatives given by polynomials in the function itself

33

Second form. From (1), (7) and (62), (65) it follows that Dzn tanh(z)

(98)

= (−1)

n

n+1 X

pn+1 tanhk (z), k

k=0

Dzn sech(z) = (−1)n (sech(z))

(99)

n X

qkn tanhk (z);

k=0

and (100)

Dzn coth(z) = (−1)n

n+1 X

pn+1 cothk (z), k

k=0

(101)

Dzn csch(z) = (−1)n (csch(z))

n X

qkn cothk (z).

k=0

pn+1 k

qkn

The constants and satisfy the recursion relations (49)–(50), with in this case σ = 0 and π = −1. Explicit expressions for p n+1 are given k by (63), with ρ1 = −1 and ρ2 = 1, and for qkn by (68), with ρ = 1. 5.4. The case ρ1 = −i and ρ2 = i Let u0 6= ±i. The functions of this subset are of the form sin(z) − u0 cos(z) , (102) f (z) = − cos(z) + u0 sin(z) u0 (103) g(z) = c ; cos(z) + u0 sin(z) and the generating functions for the polynomials are sin(t) + u cos(t) , (104) F (t, u) = cos(t) − u sin(t) (105)

G(t, u) =

1 . cos(t) − u sin(t)

5.4.1. Differentiation formulas First form. The polynomials Pn+1 (f (z)) and Qn (f (z)) become (106)

Pn+1 (f (z)) = δn=0 f (z) + + δn>0 (u20 + 1)

(107)

Pn−1

Qn (f (z)) = 2

k=0

−n

An,k+1 (u0 + i)n−1−k (u0 − i)k e+i(n−1−2k)z , (cos(z) + u0 sin(z))n+1

Pn

k=0

Bn+1,k+1 (u0 + i)n−k (u0 − i)k e+i(n−2k)z . (cos(z) + u0 sin(z))n

34

Ghislain R. Franssens

(i) Taking the limit u0 → 0 in the expressions (102)–(103) and (106)– (107) gives for the differentiation formulas (1) and (7), (i.1) for n even, Dzn tan(z) = δn=0 tan(z) −

(108) − δn>0 (−1) (109)

Dzn sec(z)

Pn−1

k=0 (−1)

n/2

=2

−n

(−1)

n/2

k

An,k+1 sin((n − 1 − 2k)z) , cosn+1 (z)

Pn

k=0 (−1)

k

Bn+1,k+1 cos((n − 2k)z) ; cosn+1 (z)

(i.2) for n odd, (110) (111)

Dzn tan(z)

= (−1)

(n−1)/2

Pn−1

k=0 (−1)

Dzn sec(z) = 2−n (−1)(n−1)/2

Pn

k

An,k+1 cos((n − 1 − 2k)z) , cosn+1 (z)

k=0 (−1)

k

Bn+1,k+1 sin((n − 2k)z) . cosn+1 (z)

(ii) Taking the limit u0 → ∞ in the expressions (102)–(103) and (106)– (107) gives for the differentiation formulas (1) and (7), (112)

Dzn cot(z) = δn=0 cot(z) + + δn>0 (−1)

n

Pn−1 k=0

An,k+1 cos((n − 1 − 2k)z) , sinn+1 (z)

Pn

Bn+1,k+1 cos((n − 2k)z) . sinn+1 (z) Formulas (108)–(113) express the n-th derivative of each of the trigonometric functions tan, sec, cot and csc as a rational function of the functions sin and cos, with the numerator in the form of a Fourier series with amplitudes given by the Eulerian and MacMahon number triangles. (113)

Dzn csc(z) = (−1)n 2−n

k=0

Second form. From (1), (7) and (62), (65) it follows that (114)

Dzn tan(z) =

n+1 X

(−1)n+1−k pn+1 tank (z), k

k=0

(115)

Dzn sec(z)

= (sec(z))

n X

(−1)n−k qkn tank (z);

k=0

and (116)

Dzn cot(z) = (−1)n

n+1 X k=0

pn+1 cotk (z), k

Derivatives given by polynomials in the function itself

Dzn csc(z) = (−1)n (csc(z))

(117)

n X

35

qkn cotk (z).

k=0

pn+1 k

qkn

The constants and satisfy the recursion relations (49)–(50), in are given this case with σ = 0 and π = +1. Explicit expressions for p n+1 k n by (63), with ρ1 = −i and ρ2 = i, and for qk by (68), with ρ = i. Appendix: Some examples of Pn+1 (u) and Qn (u) Define σ , ρ1 + ρ2 and π , ρ1 ρ2 . Case Pn+1 (u): P1 (u) = u, P2 (u) = 1! u2 − σu + π, P3 (u) = 2! u3 − 3σu2 + (σ 2 + 2π)u − σπ, P4 (u) = 3! u4 − 12σu3 + (7σ 2 + 8π)u2 − σ(σ 2 + 8π)u + (σ 2 + 2π)π, P5 (u) = 4! u5 − 60σu4 + 10(5σ 2 + 4π)u3 − 15σ(σ 2 + 4π)u2 + + (σ 4 + 2π(11σ 2 + 8π))u − σ(σ 2 + 8π)π. Case Qn (u): Q0 (u) = 1, Q1 (u) = 1! u, Q2 (u) = 2! u2 − σu + π, Q3 (u) = 3! u3 − 6σu2 + (σ 2 + 5π)u − σπ, Q4 (u) = 4! u4 − 36σu3 + 14(σ 2 + 2π)u2 − σ(σ 2 + 18π)u + (σ 2 + 5π)π.

References [1] M. Abramowitz and I. A. Stegun, Handbook of mathematical functions (9th ed.), Dover (New York, 1970). [2] M. D. Atkinson, How to compute the series expansions of sec x and tan x, Amer. Math. Monthly, 93(1986), 387–389. [3] L. Comtet, Advanced combinatorics: the art of finite and infinite expansions (rev. and enl. ed.), Reidel (Dordrecht, Holland, 1974). [4] G. R. Franssens, On a number pyramid related to binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, J. Integer Seq., 9(2006), Article 06.4.1; http://www.cs.uwaterloo.ca/journals/JIS. [5] M. E. Hoffman, Derivative polynomials for tangent and secant, Amer. Math. Monthly, 102(1995), 23–30.

36

Ghislain R. Franssens: Derivatives given by polynomials in the function itself

[6] P. A. MacMahon, The divisors of numbers, Proc. London Math. Soc., 19(1920), 305–340. [7] N. J. A. Sloane, The on-line version of the encyclopaedia of integer sequences, available online at: www.research.att.com/∼njas/sequences/eisonline.html.

O funkcih, q~ proizvodna — mnogoqlen ot samo$ i funkcii, libo e$ i rodstvenno$ i ISLEN R. FRANSSENS

V rabote soderits konstrukci mnoestva S1 = {f : Ωf ⊆ C → C} takih golomorfnyh funkci$ i, n-a proizvodna kado$ i iz kotoryh est~ mnogoqlen stepeni n + 1 ot to$ i funkcii. Postroeno take i mnoestvo S2 = {g : Ωg ⊆ C → C} takih golomorfnyh funkci$ i, n-a proizvodna kado$ i iz kotoryh ravna proizvedeni to$ i funkcii na mnogoqlen stepeni n ot lementa mnoestva S 1 . Obedinenie mnoestv S1 ∪ S2 soderit vse giperboliqeskie i trigonometriqeskie funkcii. Izuqats svo$ istva sootvetstvuwih mnogoqlenov, dl kotoryh poluqeny vnye vyraeni. V qastnosti, poluqeny vnye i legantnye predstavleni dl proizvodnyh n-ogo pordka ot giperboliqeskih funkci$ i tanh, sech, coth i csch, i trigonometriqeskih tan, sec, cot i csc.