DERIVATIVES OF TANGENT FUNCTION AND TANGENT NUMBERS
arXiv:1202.1205v5 [math.CA] 3 Jun 2015
FENG QI
Abstract. In the paper, by induction, the Faà di Bruno formula, and some techniques in the theory
of complex functions, the author finds explicit formulas for higher order derivatives of the tangent and
cotangent functions as well as powers of the sine and cosine functions, obtains explicit formulas for
two Bell polynomials of the second kind for successive derivatives of sine and cosine functions, presents
curious identities for the sine function, discovers explicit formulas and recurrence relations for the tangent
numbers, the Bernoulli numbers, the Genocchi numbers, special values of the Euler polynomials at zero,
and special values of the Riemann zeta function at even numbers, and comments on five different forms of
higher order derivatives for the tangent function and on derivative polynomials of the tangent, cotangent,
secant, cosecant, hyperbolic tangent, and hyperbolic cotangent functions.
1. Main results
It is well known that the tangent function tan x can be expanded into the Maclaurin series
∞
∞
X
(−1)k−1 22k 22k − 1 B2k 2k−1 X
x2k−1
π
x
=
T2k−1
, |x| < ,
tan x =
(2k)!
(2k − 1)!
2
k=1
k=1
see [1, p. 75, 4.3.67] and [5, p. 259], where T2k−1 are called the tangent numbers or zag numbers and Bn
for n ≥ 0 are the Bernoulli numbers which may be defined by the power series expansion
∞
∞
X
xn
x2k
x X
x
B
B
=
=
1
−
+
, |x| < 2π.
n
2k
ex − 1 n=0
n!
2
(2k)!
k=1
The tangent numbers T2k−1 may also be defined combinatorially as the numbers of alternating permutations on 2k − 1 = 1, 3, 5, 7, . . . symbols (where permutations that are the reverses of one another
counted as equivalent). The first few tangent numbers T2k−1 for k = 1, 2, . . . , 5 are 1, 2, 16, 272, 7936.
It is clear that
2k−1 2k
2 −1
k−1 2
B2k = lim tan(2k−1) x.
(1.1)
T2k−1 = (−1)
x→0
k
Consequently, one way to compute the tangent numbers T2k−1 and the Bernoulli numbers B2k is to
find explicit formulas for tan(2k−1) x. To the best of the author’s ability, explicit formulas for the n-th
derivatives (tan x)(n) of the tangent function tan x can not be searched out anywhere.
It is well known that the Riemann zeta function ζ(s) may be defined by
∞
X
1
ζ(s) =
, ℜ(s) > 0,
ks
k=1
that the Genocchi numbers Gk are given by the generating function
∞
X
2z
zk
=
G
, |z| < π,
k
ez + 1
k!
k=1
and that the Euler polynomials Ek (x) are defined by
∞
X
2exz
zk
=
Ek (x) ,
z
e +1
k!
|z| < π.
k=0
2010 Mathematics Subject Classification. Primary 11B68, 11B83, 33B10; Secondary 11C08, 11M06, 26A24.
Key words and phrases. explicit formula; recurrence; derivative; derivative polynomial; tangent; cotangent; Riemann
zeta function; Bernoulli number; Genocchi number; Bell polynomial of the second kind; Euler polynomial; induction; Faà
di Bruno formula.
This paper was typeset using AMS-LATEX.
1
2
F. QI
The tangent numbers T2k−1 , the Bernoulli numbers B2k , the Genocchi numbers Gk , the Euler polynomials
Ek (x), and the Riemann zeta function ζ(s) have closely relations.
In combinatorics, the Bell polynomials of the second kind, or say, the partial Bell polynomials, denoted
by Bn,k (x1 , x2 , . . . , xn−k+1 ) for n ≥ k ≥ 0, are defined by
X
Bn,k (x1 , x2 , . . . , xn−k+1 ) =
1≤q≤n,ℓ
Pn q ∈{0}∪N
iℓq =n
Pq=1
n
q=1 ℓq =k
n!
Qn−k+1
q=1
ℓq !
n−k+1
Y q=1
xq ℓq
.
q!
See [5, p. 134, Theorem A]. In combinatorial analysis, the Faà di Bruno formula plays an important role
and may be described in terms of the Bell polynomials of the second kind Bn,k by
n
X
dn
f
◦
h(t)
=
f (k) (h(t))Bn,k h′ (t), h′′ (t), . . . , h(n−k+1) (t) .
d tn
(1.2)
k=1
See [5, p. 139, Theorem C].
In this paper, by induction, the Faà di Bruno formula, and some techniques in the theory of complex
functions, we will establish general and explicit formulas for the n-th derivatives of tan x, cot x, sink x,
and cosk x for k ∈ N, presents curious identities for the sine function, and obtain explicit formulas for
two Bell polynomials of the second kind
π
Bm,k − sin x, − cos x, sin x, cos x, . . . , − sin x + (m − k)
2
and
π
.
Bm,k cos x, − sin x, − cos x, sin x, . . . , − cos x + (m − k)
2
By applying these formulas, we will also derive explicit formulas and recurrence relations for the tangent
numbers T2n−1 , the Bernoulli numbers B2n , the Genocchi numbers G2n , special values E2n−1 (0) of the
Euler polynomials at 0, and special values ζ(2n) of the Riemann zeta function ζ(z) at even numbers 2n.
Finally, we will comment on five different forms of higher order derivatives for tan x and on derivative
polynomials of tan x, cot x, sec x, csc x, tanh x, and coth x.
Our main results may be stated as the following theorems.
Theorem 1.1. For n ∈ N, derivatives of the tangent and cotangent functions may be computed by
( 1
1 + (−1)n
1
1 − (−1)n π
1 + (−1)n
(n)
n
tan x =
1+
an, 1+(−1) sin
x+
2
cosn+1 x 2
2
2
2
2
1+(−1)n
1
)
2 n−1−
X 2
1 + (−1)n
1 − (−1)n π
+
an,2k+ 1+(−1)n sin 2k +
x+
(1.3)
2
2
2
2
k=1
and
(n)
cot
(−1)n
x=
sinn+1 x
( 1 + (−1)n
1
1 + (−1)n
1+
an, 1+(−1)n cos
x
2
2
2
2
1+(−1)n
1
)
2 n−1−
2
X
1 + (−1)n
k
x , (1.4)
+
(−1) an,2k+ 1+(−1)n cos 2k +
2
2
k=1
where
ap,q = (−1)
for p > q ≥ 0.
1
2
p
q− 1+(−1)
2
p−q−1
2
p−q
[1 − (−1)
]
X
ℓ=0
(−1)ℓ
p
p+1
p−q−1
−ℓ+1
2
ℓ
(1.5)
DERIVATIVES OF TANGENT AND APPLICATIONS
3
Theorem 1.2. Let k, m be nonnegative integers such that (k, m) 6= (0, 0).
For all k, m ≥ 0, the sine and cosine functions satisfy
k
(−1)k X
π
dm sink x
m
q k
=
(2q − k) cos (m − k) + (2q − k)x ,
(−1)
d xm
2k q=0
2
q
(1.6)
k dm cosk x
1 X k
π
m
(2q − k) cos m + (2q − k)x ,
= k
d xm
2 q=0 q
2
(1.7)
π
k
m
(2q − k) sin (m − k) + (2q − k)x = 0,
(−1)
2
q
q=0
(1.8)
k
X
q
and
k X
k
q=0
π
(2q − k)m sin m + (2q − k)x = 0.
q
2
(1.9)
For m ≥ k ≥ 1, the Bell polynomials of the second kind Bm,k satisfy
π
Bm,k − sin x, − cos x, sin x, cos x, . . . , − sin x + (m − k)
2
k
ℓ ℓ
k
X
X
ℓ
π
1
(−1) k
(−1)
m
k
(2q − ℓ) cos m + (2q − ℓ)x , (1.10)
cos x
=
k!
2ℓ
2
ℓ cosℓ x q=0 q
ℓ=0
and
π
Bm,k cos x, − sin x, − cos x, sin x, . . . , − cos x + (m − k)
2
ℓ
k
k
X 1 k
(−1)
π
1 X
q ℓ
m
=
(−1)
sink x
(2q
−
ℓ)
cos
(m
−
ℓ)
+
(2q
−
ℓ)x
. (1.11)
q
k!
2ℓ ℓ sinℓ x q=0
2
ℓ=0
For n ∈ N, derivatives of the tangent and cotangent functions may be computed by
n+1
k
ℓ X1X
π
dn tan x
1 X ℓ
(−1)ℓ k
n+1
(2q
−
ℓ)
sin
=
−
n
+
(2q
−
ℓ)x
d xn
k
2ℓ
2
ℓ cosℓ x q=0 q
(1.12)
ℓ
n+1
k
dn cot x X 1 X 1 k
π
1 X
n+1
q ℓ
(2q − ℓ)
sin (n − ℓ) + (2q − ℓ)x .
(−1)
=
q
d xn
k
2ℓ ℓ sinℓ x q=0
2
(1.13)
k=1
ℓ=0
and
k=1
ℓ=0
Theorem 1.3. For m ∈ N, the tangent numbers T2m−1 may be computed by
T2m−1 =
m−1
m−k−1
X
X
2m
k
2m−1
ℓ 2m
(m − k − ℓ)2m−1
(m − k)
+2
(−1)
(−1)
(−1)
ℓ
k
m−1
X
k=0
k
k=1
(1.14)
ℓ=0
and
m
T2m−1 = (−1)
2m
X
ℓ=1
1
(−1) ℓ
2
ℓ
1
1
−
ℓ
m+1
ℓ 2m + 1 X ℓ
(2q − ℓ)2m .
q
ℓ
q=0
(1.15)
and satisfy a recurrence
T2m+1
m X
2m
=
T2k−1 T2(m−k)+1 .
2k − 1
k=1
(1.16)
4
F. QI
Theorem 1.4. For m ∈ N, the Bernoulli numbers B2m may be computed by
"m−1
X
m
m−1
k 2m
B2m = (−1)
(m − k)2m−1
(−1)
k
22m−1 22m − 1 k=0
#
m−1
m−k−1
X
X
2m−1
k
ℓ 2m
(m − k − ℓ)
+2
(−1)
(−1)
ℓ
k=1
and
2m
B2m = −
22m−1
and the sequence B2m =
X
1
m
(−1)ℓ ℓ
2
22m − 1 ℓ=1
(1.17)
ℓ=0
1
1
−
ℓ
m+1
22m −1
2m B2m
ℓ 2m + 1 X ℓ
(2q − ℓ)2m
q
ℓ
q=0
satisfies a recurrence
m X
2m
B2(m+1) = −
B2k B2(m−k+1) .
2k − 1
(1.18)
(1.19)
k=1
For m ∈ N, special values ζ(2m) of the zeta function ζ(z) at z = 2m may be computed by
"m−1
X
π 2m m
k 2m
(m − k)2m−1
ζ(2m) =
(−1)
(2m)!(22m − 1)
k
k=0
#
m−1
m−k−1
X
X
k
2m−1
ℓ 2m
+2
(−1)
(m − k − ℓ)
(−1)
ℓ
k=1
(1.20)
ℓ=0
and
2m
ζ(2m) = (−1)m
X
π 2m m
1
(−1)ℓ ℓ
2m
(2m)!(2 − 1)
2
ℓ=1
1
1
−
ℓ
m+1
2m
ℓ 2m + 1 X ℓ
(2q − ℓ)2m
ℓ
q
q=0
−1)
and the sequence Z2m = (−1)m (2m)!(2
(2π)2m m ζ(2m) satisfies a recurrence
m X
2m
Z2(m+1) =
Z2k Z2(m−k+1) .
2k − 1
(1.21)
(1.22)
k=1
For m ∈ N, the Genocchi numbers G2m may be calculated by
"m−1
X
m
m
k 2m
G2m = (−1) 2(m−1)
(m − k)2m−1
(−1)
k
2
k=0
+2
m−1
X
(−1)k
k=1
and
G2m =
and the sequence G2m =
2m
2m X
22m−1
1
m G2m
ℓ=1
1
(−1) ℓ
2
ℓ
1
1
−
ℓ
m+1
m−k−1
X
ℓ=0
#
2m
(m − k − ℓ)2m−1
(−1)ℓ
ℓ
ℓ 2m + 1 X ℓ
(2q − ℓ)2m
q
ℓ
q=0
satisfies a recurrence
m 2m
1X
G2(m+1) =
G2k G2(m−k+1) .
4
2k − 1
(1.23)
(1.24)
(1.25)
k=1
For m ∈ N, special values E2m−1 (0) of the Euler polynomials Em (x) at x = 0 may be calculated by
"m−1
X
1
m
k 2m
E2m−1 (0) = (−1) 2m−1
(m − k)2m−1
(−1)
2
k
k=0
#
m−1
m−k−1
X
X
2m
+2
(1.26)
(−1)k
(m − k − ℓ)2m−1
(−1)ℓ
ℓ
k=1
ℓ=0
DERIVATIVES OF TANGENT AND APPLICATIONS
5
and
E2m−1 (0) =
1
22m−1
2m
X
ℓ=1
1
(−1) ℓ
2
ℓ
1
1
−
ℓ
m+1
ℓ 2m + 1 X ℓ
(2q − ℓ)2m
q
ℓ
q=0
and the sequence E2m−1 (0) satisfies a recurrence
m 1X
2m
E2m+1 (0) =
E2k−1 (0)E2(m−k)+1 (0).
2
2k − 1
(1.27)
(1.28)
k=1
2. Comparisons and highlights
Before proving our main results, we compare them with some known conclusions for showing their
highlights.
2.1. Applications of the Bell polynomials (1.10) and (1.11). By the Faá di Bruno formula (1.2)
and the Bell polynomials (1.10) and (1.11) in Theorem 1.2, it is easy to write
(n)
n
k
ℓ 1 X X (−1)ℓ k
π
1 X ℓ
1
n
(n)
(2q − ℓ) cos n + (2q − ℓ)x
=
(sec x) =
cos x
cos x
2ℓ
2
ℓ cosℓ x q=0 q
k=1 ℓ=0
and
(n)
(csc x)
=
1
sin x
(n)
ℓ
k
n
π
1 XX 1 k
1 X
n
q ℓ
(2q − ℓ) cos (n − ℓ) + (2q − ℓ)x .
(−1)
=
q
sin x
2ℓ ℓ sinℓ x q=0
2
k=1 ℓ=0
Consequently, we may find explicit formulas for the secant and cosecant numbers. Due to the limitation
of length, we do not write down them in details.
The formulas (1.10) and (1.11) answer a problem in [39, Section 5]. These formulas, together with the
Faà di Bruno formula (1.2), may be applied to calculate the n-th derivatives of functions of the forms
f (sin x) and f (cos x), such as (sin x)α , (cos x)α , (sec x)α , (csc x)α , e± sin x , e± cos x , ln(cos x), ln(sin x),
ln(sec x), ln(csc x), sin(sin x), cos(sin x), sin(cos x), and cos(cos x).
2.2. Five different forms of higher order derivatives for tan x. On [41, pp. 28–31, Chapter II],
see also its second edition [42], by virtue of the formula
n
k
k
n α
X
dn y
(−1)k X
k−α d (u ) d y
α k
=
u
(−1)
d xn
k! α=1
d xn d uk
α
k=1
in [41, p. 12, (83)], where y = φ(u) and u = f (x), the formulas
k
2n
X
X
d2n tan x
1
α k
n+1 2n
k+1
α2n ,
(−1)
= (−1)
2
sec
x sin[(k − 1)x]
d x2n
2k
α
α=1
k=1
k
2n+1
2n+1
X
X
1
d
tan x
n+1 2n+1
k+1
α k
α2n+1 ,
=
(−1)
2
sec
x
cos[(k
−
1)x]
(−1)
d x2n+1
2k
α
α=1
k=1
X
n
k
X 1
1−(−1)n
1
π 1 − (−1)n
dn tan x
k+1
n
α k
2 n+
2
=
(−1)
sec
x
sin
+
(k
−
1)x
2
αn
(−1)
d xn
2k
2
2
α
α=1
k=1
2n
k
2n
X 1 X
d tan x
k 2n
= (−1)n−1 22n sec2 x
α N2β+1 ,
(−1)α
2n
k
dx
2 α=1
α
k=1
2n+1
k
X 1 X
d2n+1 tan x
n−1 2n+1
2
α k
= (−1)
2
sec x
α2n+1 N2β ,
(−1)
d x2n+1
2k α=1
α
k=1
n
k
n
X
1 X
d tan x
⌊ n+2
⌋ n
2
α k
2
= (−1)
2 sec x
αn N2β+ 1+(−1)n
(−1)
2
d xn
2k α=1
α
k=1
(2.1)
(2.2)
(2.3)
(2.4)
(2.5)
(2.6)
6
F. QI
were obtained for n ∈ N, where
⌊ k−1
2 ⌋
X
β k−1
(−1)
N2β =
tan2β x,
2β
⌊ k−2
2 ⌋
N2β+1 =
β=0
X
β
(−1)
β=0
k−1
tan2β+1 x,
2β + 1
and ⌊x⌋, whose value is the biggest integer not more than x, stands for the floor function of x.
The formula (1.3) may be separately written as
n−1
X
1
a2n−1,2k cos(2kx)
2n
cos x
tan(2n−1) x =
(2.7)
k=0
and
tan(2n) x =
1
cos2n+1 x
for n ∈ N, where
n−1
X
a2n,2k+1 sin[(2k + 1)x]
(2.8)
k=0
a1,0 = 1,
a2n−1,0 = 2n
n−2
X
ℓ=0
for n > 1, and
ap,q = (−1)
1
2
p−
3+(−1)p
2
2n − 1
(n − ℓ − 1)2n−2
(−1)
ℓ
ℓ
p−q−1
2
2
X
ℓ=0
(−1)
p−q−1
−ℓ
2
p
p+1
p−q−1
−ℓ+1
2
ℓ
for 0 < q < p with p − q being a positive odd number. See the first version of the preprint [30]. It is
clear that, to some extent, the forms of the formulas (2.7) and (2.8) are apparently simpler than those
of the formulas (2.1) and (2.2). On the other hand, there are 2n and 2n + 1 terms in the formulas (2.1)
and (2.2) respectively, but only n terms in both of the formulas (2.7) and (2.8). This means that the
formula (1.3) in Theorem 1.1 is essentially simpler than (2.3).
Let u = u(x) and v = v(x) 6= 0 be differentiable functions. In [2, p. 40], the formula
u
v
0
...
0 u′
v′
v
...
0 n ′′
′
n u′′
v
2v
...
0 d
(−1)
u
= n+1 (2.9)
n
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
dx
v
v
. (n−1)
n−1 (n−2)
u
v (n−1)
...
v 1 v
(n)
n (n−1)
n
(n)
′
u
v
...
1 v
n−1 v
for the nth derivative of the ratio u(x)
v(x) was listed. For easy understanding and convenient availability, we
now reformulate the formula (2.9) as
(−1)n dn u
A(n+1)×1 B(n+1)×n (n+1)×(n+1) ,
=
(2.10)
n
n+1
dx
v
v
where | · |(n+1)×(n+1) denotes a determinant and the matrices
A(n+1)×1 = (aℓ,1 )0≤ℓ≤n
and
B(n+1)×n = (bℓ,m )0≤ℓ≤n,0≤m≤n−1
satisfy
ℓ (ℓ−m)
aℓ,1 = u (x) and bℓ,m =
v
(x)
m
under the conventions that v (0) (x) = v(x) and that pq = 0 and v (p−q) (x) ≡ 0 for p < q. See also [37,
sin x
Lemma 2.1]. Applying the formula (2.10) to tan x = cos
x acquires
π
π
ℓ
(ℓ)
aℓ,1 = sin x = sin x + ℓ
and bℓ,m =
(2.11)
cos x + (ℓ − m) .
2
2
m
(ℓ)
DERIVATIVES OF TANGENT AND APPLICATIONS
7
Therefore, we can find an alternative form, the fifth form, for higher order derivatives of tan x. The first
four forms for higher order derivatives of tan x are (1.3), (1.12), (2.3), and (2.6). These five different
forms come from the induction and different formulas for higher order derivatives of composite functions.
2.3. Derivative polynomials. Suppose f is a function whose derivative is a polynomial in f , that is,
f ′ (x) = P (f (x)) for some polynomial P . Then all the higher order derivatives of f are also polynomials
in f , so we have a sequence of polynomials Pn defined by f (n) (x) = Pn (f (x)) for n ≥ 0. As usual, we
call Pn (u) the derivative polynomials of f .
In terms of this terminology, some results in [15, 12, 43] may be restated as follows: when λ > 0 and
t 6= − lnαλ or when λ < 0 and t ∈ R, the derivative polynomials of the function λeαt1 −1 are
(−1)n αn
n+1
X
(m − 1)!S(n + 1, m)um
(2.12)
m=1
in the variable u for α 6= 0 and n ∈ N, where
S(q, m) =
m
m q
1 X
ℓ
(−1)m−ℓ
m!
ℓ
ℓ=1
for 1 ≤ m ≤ q are the Stirling numbers of the second kind which may be generated by
∞
X
(ex − 1)k
xn
=
S(n, k) ,
k!
n!
k ∈ N.
n=k
In [17, p. 25, (5)] and [17, Theorem 3.2], it was obtained that the derivative polynomials Pn and Qn
defined by
dn (sec x)
dn (tan x)
=
P
(tan
x)
and
= Qn (tan x) sec x
n
d xn
d xn
for n ≥ 0 are polynomials of degree n + 1 and n respectively and satisfy the recurrences
n X
n
Pn+1 (u) =
Pk (u)Pn−k (u) + δ0n ,
k
k=0
n X
n
Qn+1 (u) =
Pk (u)Qn−k (u),
(2.13)
k
k=0
n X
n
Pn+1 (u) = 1 + u2
Qk (u)Qn−k (u),
k
k=0
where
δij =
(
0, i 6= j,
1, i = j,
P0 (u) = u,
and P1 (u) = 1 + u2 .
In [18], more recurrences for the derivative polynomials Pn and Qn were obtained and they were applied
to combinatorics and number theory. We observe that the formulas
n (n) X
n
(n+1)
2
tan
x = tan x
=
tan(k) x tan(n−k) x,
(2.14)
k
k=0
n X
n
(n+1)
(n)
sec
x = (sec x tan x) =
sec(k) x tan(n−k) x,
k
k=0
n X
n
(n)
tan(n+1) x = sec2 x
=
sec(k) x sec(n−k) x
k
k=0
may be used to straightforwardly and simply recover the formulas in (2.13) for n ∈ N.
Let
d
, y = tan x, and z = sec x.
D=
dx
8
F. QI
For n ≥ 0, define
(Dy)0 (y) = y,
(Dy)(y) = D(y 2 ),
(Dy)0 (z) = z,
n+1
n
(Dy)(z) = D(yz),
(Dy)
(y) = (Dy)(Dy) (y) = D y(Dy) (y) ,
(Dy)n+1 (z) = (Dy)(Dy)n (z) = D y(Dy)n (z) ,
(yD)n+1 (y) = (yD)(yD)n (y) = yD (yD)n (y) ,
(yD)n+1 (z) = (yD)(yD)n (z) = yD (yD)n (z) .
n
(2.15)
In [21, 22], the above quantities were computed in terms of polynomials in variables y and z and connected
with the Eulerian numbers and polynomials and others. We observe the following contradiction: since
(Dy)2 (y) = (Dy)(Dy)(y) = (Dy)D y 2
and
it follows that
(Dy)2 (y) = D y(Dy)(y) = D yD y 2 = (Dy)D y 2 + yD2 y 2 ,
(Dy)D y 2 = (Dy)D y 2 + yD2 y 2 ,
that is,
yD2 y 2 = 0,
which means that y 2 = ax+b, where a, b ∈ C. Therefore, the authors of the papers [21, 22, 24] and related
ones using the definitions in (2.15) and their analogues should select a better manner and choose suitable
symbols to express their ideas. The first author of the papers [21, 22, 24] told the current author on May
24, 2015 that he explained in [23, Theorem 10] the above easily-confused definitions more explicitly.
In [3, 4], the formulas
m
m
d
d
coth x = Cm (coth x),
tanh x = Cm (tanh x),
dx
dx
m
m
d
d
sech x = (sech x)Sm (tanh x),
csch x = (csch x)Sm (coth x),
dx
dx
m
X
k!
1 + iz
m+1
m
m+1 m
= −i
(−2) (iz + 1)
S(m, k)(iz − 1)k ,
Pm (z) = i
2 (1 − iz)ωm −
2
2k
k=0
Qm (z) = im Sm (iz)
were established for m ≥ 1, where C0 (z) = 1,
m
X
z−1
k!
Cm (z) = (−2)m (z + 1)ωm
= (−2)m (z + 1)
S(m, k)(z − 1)k , m ≥ 1,
2
2k
k=0
#
"
m
m
m
X m
X
X m
z+1
j
k−j
k
(z + 1)j , m ≥ 0,
(−1) j!
S(k, j)2
=
Sm (z) =
2 ωk −
k
2
k
j=0
k=j
k=0
and
ωn (x) =
n
X
S(n, k)k!xk
k=0
are called the geometric polynomials. We notice that the formulas
n X
(n)
n
tanh(k) x tanh(n−k) x,
tanh(n+1) x = − tanh2 x
=−
k
k=0
n X
(n)
n
tanh(n+1) x = sech2 x
=
sech(k) x sech(n−k) x,
k
k=0
and
sech(n+1) x = −(sech x tanh x)(n) = −
n
X
k=0
sech(k) x tanh(n−k) x
DERIVATIVES OF TANGENT AND APPLICATIONS
9
for n ≥ 1 imply trivially the recurrences
n X
n
Cn+1 (u) = −
Ck (u)Cn−k (u),
k
k=0
n X
n
Cn+1 (u) = 1 + u2
Sk (u)Sn−k (u),
k
k=0
n X
n
Sn+1 (u) = −
Ck (u)Sn−k (u)
k
(2.16)
k=0
for n ∈ N.
Because
n X
(n)
n
cot
x = − cot x
=−
cot(k) x cot(n−k) x,
k
k=0
n X
n
(n)
cot(n+1) x = − csc2 x
=−
csc(k) x csc(n−k) x,
k
(n+1)
2
k=0
and
csc(n+1) x = −(cot x csc x)(n) = −
n X
n
k=0
for n ∈ N, if we define polynomials Pn and Qn by
k
cot(k) x csc(n−k) x
dn (csc x)
dn (cot x)
=
P
(cot
x)
and
= Qn (cot x) csc x,
n
d xn
d xn
then the polynomials Pn and Qn for n ∈ N meet the recurrences
n X
n
Pn+1 (u) = −
Pk (u)Pn−k (u),
k
k=0
n X
n
2
Pn+1 (u) = − 1 + u
Qk (u)Qn−k (u),
k
k=0
n X
n
Qn+1 (u) = −
Pk (u)Qn−k (u).
k
(2.17)
k=0
What are the relations among Pn , Pn , and Cn ? What are the relations among Qn , Qn , and Sn ?
Because they have similar recurrences in (2.13), (2.16), and (2.17), we conjecture that their differences
are just a minus.
Finally we just mention that the asymptotic distribution of zeros of some of the above derivative
polynomials were investigated in [7].
3. Proofs of main results
We are now in a position to prove Theorems 1.1 to 1.4.
Proof of Theorem 1.1. We prove this theorem by mathematical induction.
It is easy to obtain that
2 sin x
1
and (tan x)′′ = 2 tan x sec2 x =
cos2 x
cos3 x
This means a1,0 = 1 and a2,1 = 2. Therefore, the formula (1.3) is valid for n = 1, 2.
Assume that the formulas (1.3) and (1.5) are valid for some n > 1. By this inductive hypothesis and
a direct differentiation, we have
)′
(
n−1
X
(2n) ′
1
(2n+1)
a2n,2ℓ+1 sin[(2ℓ + 1)x]
tan
x = tan
x =
cos2n+1 x
(tan x)′ = sec2 x =
ℓ=0
10
F. QI
=
n−1
X
a2n,2ℓ+1
ℓ=0
=
=
1
cos2(n+1) x
1
cos2(n+1) x
sin[(2ℓ + 1)x]
cos2n+1 x
n−1
X
′
a2n,2ℓ+1 {(n + 1 + ℓ) cos(2ℓx) + (ℓ − n) cos[2(ℓ + 1)x]}
ℓ=0
(
(n + 1)a2n,1 +
n−1
X
[(n + 1 + ℓ)a2n,2ℓ+1
ℓ=1
)
− (n + 1 − ℓ)a2n,2ℓ−1 ] cos(2ℓx) − a2n,2n−1 cos(2nx)
and
(2n+2)
tan
(2n+1)
x = tan
′
x =
"
#′
n
X
1
a2n+1,2ℓ cos(2ℓx)
cos2n+2 x
ℓ=0
′
n
X
cos(2ℓx)
=
a2n+1,2ℓ
cos2n+2 x
ℓ=0
=
=
1
cos2n+3 x
1
cos2n+3 x
n
X
a2n+1,2ℓ {(n + 1 − ℓ) sin[(2ℓ + 1)x] − (n + 1 + ℓ) sin[(2ℓ − 1)x]}
ℓ=0
(
[2(n + 1)a2n+1,0 − (n + 2)a2n+1,2 ] sin x +
n−1
X
[(n + 1 − ℓ)a2n+1,2ℓ
ℓ=1
)
− (n + 2 + ℓ)a2n+1,2(ℓ+1) ] sin[(2ℓ + 1)x] + a2n+1,2n sin[(2n + 1)x] .
By straightforward computation, it is not difficult to see that
(n + 1)a2n,1 = a2n+1,0 ,
(n + 1 + ℓ)a2n,2ℓ+1 − (n + 1 − ℓ)a2n,2ℓ−1 = a2n+1,2ℓ ,
−a2n,2n−1 = a2n+1,2n ,
2(n + 1)a2n+1,0 − (n + 2)a2n+1,2 = a2n+2,1 ,
(n + 1 − ℓ)a2n+1,2ℓ − (n + 2 + ℓ)a2n+1,2(ℓ+1) = a2n+2,2ℓ+1 ,
a2n+1,2n = a2n+2,2n+1 ,
where 1 ≤ ℓ ≤ n − 1. From these recurrence relations, the formulas (1.3) and (1.5) may be derived
straightforwardly.
The formula (1.4) can be proved by induction as the proof of the formulas (1.3) and (1.5). However,
it is easy to be derived from the formulas (1.3) and (1.5) by considering cot x = − tan x + π2 and
cot(n) x = − tan(n) x + π2 for n ∈ N. Theorem 1.1 is thus proved.
Proof of Theorem 1.2. In [5, p. 133], it was listed that
1
k!
∞
X
tm
xm
m!
m=1
!k
=
∞
X
n=k
Bn,k (x1 , x2 , . . . , xn−k+1 )
tn
n!
(3.1)
DERIVATIVES OF TANGENT AND APPLICATIONS
11
π
for k ≥ 0. Taking xm = cos x + m 2 in (3.1) yields
∞
X
n
π
t
Bn,k − sin x, − cos x, sin x, cos x, . . . , − sin x + (n − k)
2
n!
n=k
" ∞
#
k
k
1
t
t
π tm
1 X
−2 sin
sin
=
cos x + m
+x
=
k! m=1
2 m!
k!
2
2
k
[cos(t + x) − cos x]k
(−1)k X
k
=
=
cosℓ (t + x) cosk−ℓ x.
(−1)ℓ
k!
k!
ℓ
(3.2)
ℓ=0
Since cos x =
eix +e−ix
,
2
we have
ℓ ℓ 1 ix
1 X ℓ (2q−ℓ)ix
1 X ℓ qix −(ℓ−q)ix
−ix ℓ
e
e
=
e
(e
+
e
)
=
2ℓ
2ℓ q=0 q
2ℓ q=0 q
(cos x)ℓ =
and
ℓ ℓ 1 X ℓ
1 X ℓ
dm (cos x)ℓ
m m (2q−ℓ)ix
(2q − ℓ) i e
= ℓ
(2q − ℓ)m e[πm/2+(2q−ℓ)x]i
= ℓ
d xm
2 q=0 q
2 q=0 q
ℓ π
π
1 X ℓ
m
cos m + (2q − ℓ)x + i sin m + (2q − ℓ)x
(2q − ℓ)
= ℓ
2 q=0 q
2
2
which implies the formula (1.7) and the equation (1.9).
Differentiating m ≥ k times with respect to t on the very ends of (3.2) and employing (1.7) result in
∞
X
n−m
π
t
Bn,k − sin x, − cos x, sin x, cos x, . . . , − sin x + (n − k)
2
(n − m)!
n=m
k
k dm cosℓ (t + x)
(−1)k X
cosk−ℓ x
(−1)ℓ
=
k!
d tm
ℓ
ℓ=0
X
k
ℓ k X
(−1)
ℓ
π
k 1
m
=
(2q
−
ℓ)
cos
m
+
(2q
−
ℓ)(t
+
x)
cosk−ℓ x.
(−1)ℓ
k!
2
ℓ 2ℓ q=0 q
ℓ=0
Further taking the limit t →0 and rearranging reveal (1.10).
Taking xm = sin x + m π2 in (3.1) yields
∞
X
n
π
t
Bn,k cos x, − sin x, − cos x, sin x, . . . , − cos x + (n − k)
2
n!
n=k
" ∞
#
k
k
1 X
t
t
π tm
1
=
2 sin
cos
=
+x
sin x + m
k! m=1
2 m!
k!
2
2
k
[sin(t + x) − sin x]k
(−1)k X
k
=
=
sinℓ (t + x) sink−ℓ x.
(−1)ℓ
k!
k!
ℓ
ℓ=0
Since sin x =
eix −e−ix
,
2i
we have
(sin x)ℓ =
ℓ
ℓ
1 X
(−1)ℓ X
qix −(ℓ−q)ix
ℓ−q ℓ
q ℓ
e
e
=
e(2q−ℓ)ix
(−1)
(−1)
(2i)ℓ q=0
(2i)ℓ q=0
q
q
(3.3)
12
F. QI
and
ℓ
(−1)ℓ X
dm (sin x)ℓ
q ℓ
=
(2q − ℓ)m im e(2q−ℓ)ix
(−1)
d xm
(2i)ℓ q=0
q
ℓ
(−1)ℓ X
q ℓ
=
(2q − ℓ)m im−ℓ e(2q−ℓ)ix
(−1)
2ℓ q=0
q
ℓ
(−1)ℓ X
q ℓ
(2q − ℓ)m e[(m−ℓ)π/2+(2q−ℓ)x]i
(−1)
=
2ℓ q=0
q
=
ℓ
(−1)ℓ X
π
π
m
q ℓ
cos
(m
−
ℓ)
+
(2q
−
ℓ)x
+
i
sin
(m
−
ℓ)
+
(2q
−
ℓ)x
(2q
−
ℓ)
(−1)
2ℓ q=0
2
2
q
which means the formula (1.6) and the equation (1.8).
Differentiating m ≥ k times with respect to t on the very ends of (3.3) and utilizing (1.6) lead to
n−m
∞
X
t
π
Bn,k cos x, − sin x, − cos x, sin x, . . . , − cos x + (n − k)
2
(n
− m)!
n=m
m ℓ
k
k d sin (t + x)
(−1)k X
sink−ℓ x
(−1)ℓ
=
k!
d xm
ℓ
ℓ=0
k ℓ
k X
π
k 1 X
(−1)
m
q ℓ
(2q − ℓ) cos (m − ℓ) + (2q − ℓ)(t + x) sink−ℓ x.
(−1)
=
k!
2
q
ℓ 2ℓ q=0
ℓ=0
Further letting t → 0 gives the formula (1.11).
Applying f (u) = ln u and u = h(t) = cos t in (1.2) and using the formula (1.10) give
n+1
X
dn tan t
dn+1 ln(cos t)
=
−
=
−
(ln u)(k) Bn+1,k cos′ t, cos′′ t, . . . , cos(n−k+2) t
d tn
d tn+1
k=1
n+1
X (−1)k−1 (k − 1)!
π
−
sin
t,
−
cos
t,
sin
t,
cos
t,
.
.
.
,
−
cos
t
+
(n
−
k)
B
=−
n+1,k
uk
2
k=1
n+1
k
ℓ X1X
1 X ℓ
π
(−1)ℓ k
n+1
=
(2q
−
ℓ)
cos
(n
+
1)
+
(2q
−
ℓ)t
.
ℓ cosℓ t q=0 q
k
2ℓ
2
k=1
ℓ=0
The formula (1.12) follows.
Applying f (u) = ln u and u = h(t) = sin t in (1.2) and using the formula (1.11) generate
n+1
dn+1 ln(sin t) X
dn cot t
=
=
(ln u)(k) Bn+1,k sin′ t, sin′′ t, . . . , sin(n−k+2) t
n
n+1
dt
dt
k=1
n+1
X (−1)k−1 (k − 1)!
π
=
Bn+1,k cos t, − sin t, − cos t, sin t, . . . , − sin t + (n − k)
uk
2
k=1
n+1
k
ℓ
X1X
1 X
1 k
π
q ℓ
n+1
=
(−1)
(2q − ℓ)
sin (n − ℓ) + (2q − ℓ)t .
k
2ℓ ℓ sinℓ t q=0
2
q
k=1
ℓ=0
The formula (1.13) follows. The proof of Theorem 1.2 is complete.
Proof of Theorem 1.3. Taking n = 2m − 1 and letting x → 0 in (1.3) yield
T2m−1 = lim tan(2m−1) x =
x→0
m−1
X
1
a2m−1,0 +
a2m−1,2k ,
2
k=1
DERIVATIVES OF TANGENT AND APPLICATIONS
where
a2m−1,0 = 2
m−1
X
(−1)ℓ
ℓ=0
and
k
a2m−1,2k = (−1) 2
m−k−1
X
ℓ=0
The formula (1.14) follows.
Taking x → 0 in (1.12) gives
13
2m
(m − ℓ)2m−1
ℓ
2m
(m − k − ℓ)2m−1 .
(−1)
ℓ
ℓ
ℓ 2m
k
X
d2m−1 tan x
π
1 X (−1)ℓ k X ℓ
2m
(2q
−
ℓ)
sin
=
−
(2m
−
1)
x→0
d x2m−1
k
2ℓ
2
ℓ q=0 q
lim
k=1
ℓ=0
= (−1)m
2m
X
k=1
= (−1)m
2m
X
ℓ k
1 k X ℓ
1X
(2q − ℓ)2m
(−1)ℓ ℓ
k
2 ℓ q=0 q
(−1)ℓ
ℓ=1
= (−1)m
2m
X
ℓ=1
= T2m−1 .
ℓ=0
2m
ℓ X
1 X ℓ
1 k
2m
(2q
−
ℓ)
2ℓ q=0 q
k ℓ
k=ℓ
ℓ 1 2m + 1 − ℓ 2m + 1 X ℓ
(−1)ℓ ℓ
(2q − ℓ)2m
2 (2m + 1)ℓ
ℓ
q
q=0
The formula (1.15) follows.
Taking x → 0 in (2.14) figures out
m m X
(2k−1) (2m−2k+1) X
2m
2m
(2m+1)
=
y
T2k−1 T2(m−k)+1 .
T2m+1 = lim y
=
lim y
x→0
2k − 1
2k − 1 x→0
k=1
k=1
The recurrence (1.16) follows. The proof of Theorem 1.3 is thus proved.
Proof of Theorem 1.4. From (1.1), it follows that
B2k =
(−1)k−1 k
T2k−1 .
22k−1 22k − 1
Combining this with the formulas (1.14) and (1.15) gains the formulas (1.17) and (1.18).
Substituting (1.1) into (1.16) and rearranging yield (1.19).
In [1, p. 807, 23.2.16], it is listed that for n ≥ 1
ζ(2n) =
(2π)2n
|B2n |.
2(2n)!
(3.4)
See also [40, p. 332]. Substituting the formulas (1.17) and (1.18) into (3.4) results in (1.20) and (1.21).
Substituting (3.4) into (1.19) and rearranging yield (1.22).
The Genocchi numbers satisfy G1 = 1, G2n+1 = 0, and
G2n = 2 1 − 22n B2n = 2nE2n−1 (0)
for n ∈ N. By similar arguments as above, we may obtain the formulas (1.23), (1.24), (1.26), and (1.27)
and the recurrence relations (1.25) and (1.28). The proof of Theorem 1.4 is complete.
4. Remarks
Finally, we give several remarks on related results.
Remark 4.1. Let y = tan x. The relation (2.14) may be rewritten as
y (n) = Y AY T
for n ≥ 2, where Y T stands for the transpose of the 1 × n matrix
Y = y, y ′ , y ′′ , . . . , y (n−2) , y (n−1)
14
F. QI
and A is a n × n square matrix whose elements ak,ℓ satisfy

 n − 1 , k + ℓ = n + 1,
ℓ−1
ak,ℓ =

0,
k + ℓ 6= n + 1.
Remark 4.2. By the formulas in (2.11), the tangent numbers T2m−1 for m ∈ N may be
0
1
0
0
...
1
0
1
0
.
..
0
−1
0
1
.
.
.
0
− 31
0
...
T2m−1 = − −1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
0
(−1)m−1
0
(−1)m−2 2m−2
...
2
(−1)m−1
0
(−1)m−1 2m−1
0
.
..
1
represented as
0
0
0
0 .
. .
1
0
Remark 4.3. The general and explicit formula (1.4) for derivatives of the cotangent function cot x in
Theorem 1.1 has been applied in [32] to establish limit formulas for ratios of polygamma functions at
their singularities.
Remark 4.4. The formulas (1.20), (1.21), and (1.22) show the irrationality of ζ(2n) for n ∈ N. However,
a problem about the irrationality of ζ(2n + 1) for n ≥ 2 is still keeping open, to the best of the author’s
knowledge. For more information, please refer to [19, 20] and closely related references therein.
Remark 4.5. In [14], the method and idea to prove the formulas (1.6) and (1.7) were employed to express
the Wallis ratio formula as a sum.
Remark 4.6. In [15], among other things, by establishing the derivative polynomials (2.12) for λ = α = 1,
that is,
(q)
m
q+1
X
1
1
q
= (−1)
(m − 1)!S(q + 1, m) t
et − 1
e −1
m=1
for q ∈ {0} ∪ N, an explicit formula
B2k = 1 +
2k−1
X
m=1
S(2k + 1, m + 1)S(2k, 2k − m)
−
2k
m
2k
2k X S(2k, m)S(2k + 1, 2k − m + 1)
,
2k
2k + 1 m=1
m−1
k∈N
for computing the Bernoulli numbers B2k in terms of the Stirling numbers S(q, m) of the second kind
was derived. For more information on closely related problems, please refer to [12, 43] and the references
cited therein.
Remark 4.7. In [29], based on establishment of the general and explicit formula
(n)
n+1
1
(−1)n X αn,q
=
, n ∈ N,
ln x
xn q=2 (ln x)q
where
αn,2 = (n − 1)!
(4.1)
and, for n + 1 ≥ q ≥ 3,
αn,q = (q − 1)!(n − 1)!
n−1
X
ℓ1 =1
ℓq−4 −1
ℓq−3 −1
ℓ1 −1
X
X
1
1 X
1
1
···
,
ℓ1
ℓ2
ℓq−3
ℓq−2
ℓ2 =1
ℓq−3 =1
(4.2)
ℓq−2 =1
among other things, it was discovered that the Stirling numbers of the first kind s(n, q) for 1 ≤ q ≤ n
may be computed by
s(n, q) = (−1)n+q (n − 1)!
n−1
X
ℓ1 =1
ℓq−3 −1
ℓq−2 −1
ℓ1 −1
X
X
1
1
1
1 X
···
,
ℓ1
ℓ2
ℓq−2
ℓq−1
ℓ2 =1
ℓq−2 =1
ℓq−1 =1
DERIVATIVES OF TANGENT AND APPLICATIONS
15
equivalently,
n−k
(−1)
n−1
X 1
s(n, k)
m−(k−1) s(m, k − 1)
(−1)
,
=
(n − 1)!
m
(m − 1)!
2 ≤ k ≤ n,
m=k−1
and that the Bernoulli numbers of the second kind bn for n ≥ 2 may be computed by
!
n
X
1
αn,k − nαn−1,k
n 1
bn = (−1)
,
+
n! n + 1
k!
k=2
where the Stirling numbers of the first kind s(n, k) and the Bernoulli numbers of the second kind bn for
n ≥ 0 may be generated respectively by
∞
X
[ln(1 + x)]m
s(k, m) k
=
x ,
m!
k!
|x| < 1
k=m
and
∞
X
x
=
bn xn ,
ln(1 + x) n=0
|x| < 1,
and αn,k are defined by (4.1) and (4.2).
Remark 4.8. In [31], three integral representations for the Stirling numbers of the first kind s(n, k) were
created, one of them is
Z ∞ Z 1
k dn−k
n
xu−1
−u
s(n, k) =
lim
t
dt e du
(4.3)
k x→0 d xn−k
1/e
0
for 1 ≤ k ≤ n. By virtue of (4.3), the sequence
(
n s(n
(−1)
+ k, k)
n+k
k
)
n≥0
for any given k ∈ N is proved to be logarithmically convex.
Remark 4.9. In [38], it was obtained that the Bernoulli numbers of the second kind bn may be represented
by
Z ∞
1
bn = (−1)n+1
d t, n ∈ N.
(4.4)
2 + π 2 }tn
{[ln(t
−
1)]
1
As a result of the integral representation (4.4), the sequence {(−1)n bn+1 }n≥0 of the Bernoulli numbers
of the second kind was proved to be completely monotonic. In [38], among other things, the sequence
{(−1)q q!bq+1 }q≥0 is proved to be logarithmically convex.
Remark 4.10. The general and explicit formulas for the n-th derivatives of the exponential function e±1/x
were established in [6, 44] and applied in [27, 29, 33, 34, 36] and related references therein.
Remark 4.11. Recently, some results on explicit formulas and inequalities for the Bell numbers, the Bell
polynomials Bn,k , the Bernoulli numbers B2n , the Cauchy numbers, the Euler polynomials En , and the
Stirling numbers were discovered or recovered in [8, 9, 10, 11, 12, 13, 15, 16, 25, 26, 28, 31, 35] and closely
related references therein.
Remark 4.12. This paper is a corrected, expanded, and revised version of the preprint [30].
Acknowledgments. The author thanks Professor Sergei M. Sitnik in Voronezh Institute of the Ministry
of Internal Affairs of Russia for his providing the formula (2.9) and the reference [2] on September 25,
2014.
The author appreciates anonymous referees for their careful corrections to and valuable comments on
the original version of this paper.
16
F. QI
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Institute of Mathematics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China;
College of Mathematics, Inner Mongolia University for Nationalities, Tongliao City, Inner Mongolia Autonomous Region, 028043, China; Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300387, China
E-mail address: [email protected], [email protected], [email protected]
URL: http://qifeng618.wordpress.com