Computers and Mathematics with Applications 59 (2010) 724–736
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Jordan type inequalities involving the Bessel and modified Bessel functions Ling Zhu Department of Mathematics, Zhejiang Gongshang University, Hangzhou, Zhejiang 310018, PR China
article
abstract
info
Article history: Received 26 March 2008 Received in revised form 25 August 2009 Accepted 26 August 2009
In this work, two general forms of Jordan’s inequalities for the Bessel and modified Bessel functions are established, and proved by using l’Hospital’s rule for monotonicity and some properties of the spherical Bessel and the modified spherical Bessel functions of the first kind. The applications of the results above give two new infinite series for the Bessel and modified Bessel functions. © 2009 Elsevier Ltd. All rights reserved.
Keywords: Lower and upper bounds Best constant General forms of Jordan type double inequalities Bessel, modified Bessel, spherical Bessel and modified spherical Bessel functions New infinite series for the Bessel and modified Bessel functions
1. Introduction The following result is known as Jordan inequality [1]: 2 sin x π ≤ < 1, 0 < x ≤ . π x 2 Feng [2], Mercer et al. [3], Qi et al. [4], Özban [5], Zhu [6–11], Debnath and Zhao [12], Zhao [13], Wu [14], Wu and Debnath [15–17], Wu and Srivastava [18], Sándor [19], and Li [20], Li and Li [21], and Baricz [22–25], Baricz and Neuman [26], Niu [27], Niu et al. [28], Qi and Niu [29] have done some refinement work on the inequality above. Recently, Niu [27] (or see [28]) established a new result about general form of Jordan’s inequalities as follows: Theorem 1. If 0 < x ≤ π /2 and n ∈ N, inequality 2
π
+
n X
αk (π 2 − 4x2 )k ≤
sin x x
k=0
<
2
π
+
n X
βk (π 2 − 4x2 )k
k=0
holds with equalities if and only if x = π /2, where the constants
αk =
i k+1 (−1)k X 2 k+i k a sin π , i −1 (4π )k k! i=1 π 2
βk = αk (1 ≤ k < n), with
aki
= (i + k − 1)
aki−−11
+
aki −1
βn =
1 − π2 −
n −1
P
αi π 2i
i =1
π 2n
(0 < i ≤ k) and ak0 = 1 in (1) are the best possible.
E-mail address:
[email protected]. 0898-1221/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2009.10.020
(1)
L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
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In fact, the author of this paper [11] establishes the following general form of Jordan’s inequality using l’Hospital’s rule for monotonicity and some properties of the spherical Bessel functions of the first kind jn (x): Theorem 2. Let 0 < x ≤ r ≤ π and N ≥ 0 be a natural number, then sin x
A2N (x) + α(r 2 − x2 )N +1 ≤
x
≤ A2N (x) + β(r 2 − x2 )N +1
with the equalities if and only if x = r, where A2N (x) = a0 = ak =
sin r
,
r 2k − 1 2kr 2
a1 = ak−1 −
sin r − r cos r 2r 3 1
4k(k − 1)r 2
Furthermore, α = aN +1 and β =
1−
k=0 ak r r 2(N +1)
ak (r 2 − x2 )k and
PN
k=0
,
ak−2 ,
PN
(2)
k = 1, 2, . . . .
2k
are the best constants in (2).
We note that Niu [27] obtains the following result: sin x x
=
∞ X
αk (π 2 − 4x2 )k =
k=0
∞ X
4k · αk
π 2
− x2
2
k=0
k (3)
while Zhu [11] establishes the new series expansion of (sin x)/x: sin x x
=
∞ X
ak ( r 2 − x 2 ) k .
(4)
k=0
Suppose that r = π /2 in (4). Then by the uniqueness principle of the power series representation of the same function, from (3) and (4) we have ak = 4k · αk . Thus if we let r = π /2 in Theorem 2, then the inequality (2) reduces to (1). At the same time, we find the interval (0, π] in Theorem 2 is larger than the one (0, π /2] in Theorem 1, and all corresponding coefficients are defined recursively in Theorems 1 and 2. Let Jp (x) ([30, p. 40] or see [31]) and Ip (x) ([30, p. 77] or see [31]) be the classical Bessel function and the modified Bessel function of the first kind of order p, respectively. Moreover, consider the normalized Bessel and modified Bessel function defined by Jp : R → (−∞, 1] and Ip : R → [1, +∞) respectively as
Jp (x) = 2p Γ (p + 1)x−p Jp (x)
(5)
Ip (x) = 2p Γ (p + 1)x−p Ip (x).
(6)
and
By (5) and (6) we have 1
1
Jp+ 1 (x) = 2p+ 2 Γ (p + 3/2)x−(p+ 2 ) Jp+ 1 (x) 2
(7)
2
and 1
1
Ip+ 1 (x) = 2p+ 2 Γ (p + 3/2)x−(p+ 2 ) Ip+ 1 (x). 2
(8)
2
Particularly for p = 0, p = 1 in (7) and (8) respectively, the functions Jp+ 1 (x) and Ip+ 1 (x) reduce to some elementary 2 2 functions, like [30, p.54]
J 1 (x) =
π
r
2x
2
I 1 (x) =
π
r
2x
2
J 1 (x) =
sin x
I 1 (x) =
sinh x
2
2
x
x
,
J 3 (x) =
3
2
,
I 3 (x) = 2
r
x 3 x
π 2x
r
J 3 (x) = 3
π 2x
sin x
2
I 3 (x) = −3 2
x3
−
cos x
sinh x x3
x2
−
,
cosh x x2
.
(9)
In this paper, we establish two general forms of Jordan type inequality for Jp+ 1 (x) and Ip+ 1 (x) (p ≥ 0 is a natural 2 2
number) by using l’Hospital’s rule for monotonicity and the properties of Bessel functions Jp (x), modified Bessel functions pπ Ip (x), spherical Bessel functions of the first kind jp (x) = 2x Jp+ 1 (x), and the modified spherical Bessel functions of the first kind ip (x) =
pπ
I
1 2x p+ 2
(x).
2
Theorem 3. Let 0 < x ≤ r ≤ π and N , p ≥ 0 be two natural numbers, then S2N (x) + α(r 2 − x2 )N +1 ≤ Jp+ 1 (x) ≤ S2N (x) + β(r 2 − x2 )N +1 2
(10)
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L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
holds with the equalities if and only if x = r, where S2N (x) =
N X
J0 bn ( r − x ) , 2
b0 = Jp+ 1 (r ),
2 n
2
n=0
bn =
2(p + n − 12 ) 2nr 2
1
b n −1 −
Furthermore, α = bN +1 and β =
4n(n − 1)r 2 1−
PN
n=0 bn r r 2(N +1)
b n −2 ,
p+ 12
b1 = −
(r )
,
2r
n = 2, 3, . . . .
(11)
2n
are the best constants in (10).
Theorem 4. Let 0 < x ≤ r < +∞ and N , p ≥ 0 be two natural numbers, we have (i) when N is even, the double inequality T2N (x) + α(r 2 − x2 )N +1 ≤ Ip+ 1 (x) ≤ T2N (x) + β(r 2 − x2 )N +1
(12)
2
holds with the equalities if and only if x = r, where T2N (x) =
N X
I0 dn (r − x ) , 2
d0 = Ip+ 1 (r ), 2
2 n
n=0
dn =
2(p + n − 12 ) 2nr 2
dn−1 +
1 4n(n − 1)r 2
dn−2 ,
d1 = −
p+ 21
2r
(r )
,
n = 2, 3, . . . .
(13)
(ii) when N is odd, the double inequalityP(12) is reversed. 2n 1− N n=0 dn r Furthermore, α = dN +1 and β = are the best constants in (12). r 2(N +1) When letting p = 0 in Theorems 3 and 4, we can obtain Theorem 2 for sin x/x and the following Theorem 5 for sinh x/x: Theorem 5. Let 0 < x ≤ r < +∞ and N ≥ 0 be a natural number, we have (i) when N is even, the double inequality C2N (x) + α(r 2 − x2 )N +1 ≤
sinh x x
≤ C2N (x) + β(r 2 − x2 )N +1
holds with the equalities if and only if x = r, where C2N (x) = c0 =
sinh r
cn+1 =
r
,
c1 = 1
4n(n + 1)r 2
sinh r − r cosh r
cn−1 +
2r 3 2n + 1 2(n + 1)r 2
PN
n =0
(14)
cn (r 2 − x2 )n and
,
cn ,
n = 1, 2, . . . .
(15)
(ii) when N is odd, the double inequalityP(14) is reversed. 2n 1− N n=0 cn r are the best constants in (14). Furthermore, α = cN +1 and β = r 2(N +1) During the course of the revision of this paper we found that new researches of Jordan type inequalities on the Bessel and modified Bessel functions have been developing by leaps and bounds from then on. In what follows we compare our Theorems 3 and 4 with the results of Baricz and Wu [25], and of Niu [27]. The generalized Bessel function of the first kind vp is defined by Baricz [43] as a particular solution of the generalized Bessel differential equation x2 y00 (x) + bxy0 (x) + [cx2 − p2 + (1 − b)p]y(x) = 0, where b, p, c ∈ R, and vp has the infinite series representation
vp (x) =
∞ X n=0
x 2n+p (−1)n c n , n!Γ (p + n + (b + 1)/2) 2
x ∈ R.
At the same time, the generalized and normalized Bessel function of the first kind is defined [32] as follows up (x) =
X (−c /4)n n ≥0
n!(κ)n
xn ,
x∈R
and satisfies the following differential equation xy00 (x) + κ y0 (x) + (c /4)y(x) = 0, where κ := p + (b + 1)/2 6= 0, −1, −2, . . ., (a)n = Γ (a + n)/Γ (a), a 6= 0, −1, −2, . . . is the well-known Pochhammer symbol defined in terms of Euler’s Γ -function.
L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
727
Furthermore, the function
λp (x) := up (x2 ) =
∞ X (−c /4)n n =0
n!(κ)n
x2n
(16)
has been considered in [32]. Noting that when c = b = 1 the function λp (x) reduces to the function Jp (x), and when c = −1, b = 1 then λp (x) becomes Ip (x). Several inequalities and representation formulas for generalized Bessel functions have been published already in the papers of Baricz [33–35,22,24], Baricz and Neuman [26]. All of these results have been collected in the doctoral thesis of Baricz [23]. In [22] and [24], Baricz first gives us some interesting results on Jordan type inequality for the function λp (x). Motivated by [22, Theorem 14] in his thesis Niu [27] obtains the following results. Theorem N. If n be a natural number, κ ≥ 1/2, and c ∈ [0, 1], then for all 0 < x ≤ r ≤ π /2, we have n X
σi (r 2 − x2 )i ≤ λp (x) ≤
i =0
n X
νi (r 2 − x2 )i ,
(17)
i=0
where the constants
σi =
c i λ (r ) i+p , 4 i!(κ)i
0 ≤ i ≤ n, n−1
1−
νi = σi (0 ≤ i < n − 1),
P
σl r 2l
l=0
νn =
r 2n
in (17) are the best possible. Moreover, if κ ≥ 0, c ≤ 0, 0 < x ≤ r < ∞ and n is odd, then (17) holds true, while if m is even, then (17) is reversed. In recent paper [25], Baricz and Wu obtain the following better results. Theorem BW. Let κ > 0, and jp,1 be the first positive zero of the Bessel function of the first kind Jp (x). Then for all c ∈ [0, 1] and 0 < x ≤ r ≤ jκ,1 the following sharp Jordan type inequalities hold Gp,m (x) + ςp (r )(r 2 − x2 )m+1 ≤ λp (x) ≤ Gp,m (x) + τp (r )(r 2 − x2 )m+1 ,
(18)
where Gp,m (x) =
m X
dp,i (r )(r 2 − x2 )i ,
i=0
m is a natural number and the coefficients dp,i (r ) are defined explicitly by dp,i (r ) =
c i λ (r ) i+p 4 i!(κ)i
for all 0 ≤ i ≤ m + 1, i ∈ N or recursively by dp,0 (r ) = λp (r ), dp,i+1 (r ) =
dp,1 (r ) =
c 4κ
λp+1 (r ),
κ +i−1 c dp,i (r ) − dp,i−1 (r ) (i + 1)r 2 4i(i + 1)r 2
for all 1 ≤ i ≤ n, i ∈ N+ . Moreover, if c ≤ 0, 0 < x ≤ r and m is even, then the Jordan type inequality (18) holds true, while if c ≤ 0, 0 < x ≤ r and m is odd, then the Jordan type inequality (18) is reversed. In each of cases the constants 1−
ςp (r ) = dp,m+1 (r ),
τp (r ) =
m P
dp,k (r )r 2k
k=0
r 2(m+1)
are the best possible in (18). Now, we compare the results of the Theorems 3 and 4 with of the Theorem BW. Since when choosing b = 2 and c = ±1 in (16) the λp (x) reduces to the functions Jp+ 1 (x) and Ip+ 1 (x), we have that Theorems 3 and 4 in this paper are two 2
2
particular cases of Theorem BW. In addition, in [25] the proof of the Jordan type inequalities is much simplified by using some monotonicity results from Baricz [32] while the one in this paper by using the similar technique as Zhu [9–11]. At the same time, these conditions in the Jordan type inequality have been relaxed in [25].
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L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
Next, let look at Theorem BW and Theorem N. In the case when c ∈ [0, 1], it is assumed that κ ≥ 1/2 and r ≤ π /2 in Theorem N. These conditions have been relaxed to κ > 0 and r ≤ jκ,1 in Theorem BW since the interval (0, jκ,1 ] is larger than the interval (0, π /2]. 2. Lemmas Lemma 1 ([36–38]). Let f , g : [a, b] → R be two continuous functions which are differentiable on (a, b) with g 0 6= 0 on (a, b). f (x)−f (b) f (x)−f (a) If f 0 /g 0 is increasing (or decreasing) on (a, b), we have that the functions g (x)−g (b) and g (x)−g (a) are also increasing (or decreasing) on (a, b). Lemma 2 ([30,31]). Let jn (x) be the spherical Bessel functions (sBfs) of the first kind, jn (x) = jn (x) = (−x)n
n
1 d
sin x
x dx
x
,
pπ
J
1 2x n+ 2
n = 0, 1, 2, . . . .
(19)
Lemma 3 ([10, Lemma 6]). Let jn (x) be the spherical Bessel functions (sBfs) of the first kind, jn (x) = n +2
x
(x), then
pπ
jn+1 (x), then
1 d
n
x dx
hn (x) = xj0 (x) = sin x,
J
1 2x n+ 2
(x), hn+1 (x) =
n = 0, 1, 2, . . . .
(20)
Lemma 4 ([39, p.279]). Let Jν (x) be the Bessel function. Then
1 d
n
x dx
[x−ν Jν (x)] = (−1)n x−ν−n Jν+n (x).
(21)
Lemma 5 ([39, p.279]). Let Jν (x) be the Bessel function. Then xJν−1 (x) + xJν+1 (x) = 2ν Jν (x).
(22)
In the last of this section, we show some corresponding results about the properties of the modified spherical Bessel functions (msBfs) of the first kind in (x). Lemma 6 ([30,31]). Let in (x) be the modified spherical Bessel functions of the first kind, in (x) = n
x
1 d n sinh x x dx x
pπ
, n = 0, 1, 2, . . . .
Lemma 7 ([40,41]). Let in (x) be the modified spherical Bessel functions of the first kind, in (x) = in+1 (x) = in−1 (x) −
2n + 1 x
I
1 2x n+ 2
pπ
I
1 2x n+ 2
in (x).
(x), then in (x) =
(x), then (23)
From the definition of in (x) in Lemma 6, we have Lemma 8. Let in (x) be the modified spherical Bessel functions of the first kind, in (x) =
pπ
I
1 2x n+ 2
xi0n (x) = nin (x) + xin+1 (x).
(x), then (24)
Lemma 9. Let in (x) be the modified spherical Bessel functions of the first kind, in (x) =
pπ
I
1 2x n+ 2
(x), kn+1 (x) = xn+2 in+1 (x), then
k0n+1 (x) = xkn (x) or
1 d x dx
kn+1 (x) = kn (x),
n = 0, 1, 2, . . . .
(25)
Proof. We know that k0n+1 (x) = xn+1 [(n + 2)in+1 (x) + xi0n+1 (x)], and xkn (x) = x(xn+1 in (x)) = xn+1 [xin (x)], so (25) holds when we prove
(n + 2)in+1 (x) + xi0n+1 (x) = xin (x)
(26)
L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
729
or
(n + 1)in (x) + xi0n (x) = xin−1 (x).
(27)
In fact, (27) comes from (23) and (24). Lemma 10. Let in (x) be the modified spherical Bessel functions of the first kind, in (x) =
pπ
then
n
1 d x dx
kn (x) = xi0 (x) = sinh x,
(28)
d x x [x2 i1 (x)] = 1x dx x2 cosh − sinh x x2 d N (ii) Supposing (28) holds for n = N or 1x dx kN (x) = sinh x. 1 d k x dx 1
( x) =
1 d x dx
1 d
(x), kn+1 (x) = xn+2 in+1 (x),
n = 0, 1, 2, . . . .
Proof. We prove (28) by using mathematical induction. h (i) When n = 1,
I
1 2x n+ 2
i
= sinh x.
By (25) we have
1 d
N +1
x dx
kN +1 (x) =
N
x dx
1 d
x dx
kN +1 (x) =
1 d
N
x dx
kN (x) = sinh x.
That is, (28) holds for n = N + 1. Lemma 11 ([39, p.280]). Let Jν (x) be the modified Bessel function. Then
n
1 d x dx
[x−ν Iν (x)] = x−ν−n Iν+n (x).
(29)
Lemma 12 ([39, p.280]). Let Jν (x) be the modified Bessel function. Then xIν−1 (x) − xIν+1 (x) = 2ν Iν (x).
(30)
3. A concise proof of Theorem 3 In this section, we shall prove Theorem 3 by using l’Hospital’s rule for monotonicity. J
(x)−S2N (x) p+ 1 2 (r 2 −x2 )N +1
f1 (x) g1 (x)
f1 (r ) = gf11 ((xx)− , where f1 (x) = Jp+ 1 (x) − S2N (x), and g1 (x) = (r 2 − x2 )N +1 ; )−g1 (r ) 2 Let limx→r − f1 (x) = 0, we have a0 = Jp+ 1 (r ) and
Let F (x) =
=
2
f10 (x) g10 (x)
d J 1 dx p+ 2
= 1
−2
(x) −
N
P
nbn (r 2 − x2 )n−1 (−2x)
n =1
(N + 1)(r 2 − x2 )N (−2x) N P 1 d J nbn (r 2 − x2 )n−1 1 ( x) − p + x dx 2
n=1
=
(N + 1)(r 2 − x2 )N f 2 ( x) f2 (x) − f2 (r ) = = , g2 (x) g2 (x) − g2 (r ) PN d where f2 (x) = −12 1x dx Jp+ 1 (x) − n=1 nbn (r 2 − x2 )n−1 , and g2 (x) = (N + 1)(r 2 − x2 )N ; 2
J0
Let limx→r − f2 (x) = 0, we have b1 = − 1
f2 (x)
(−2)2
0
g20 (x)
=
1 d 2 x dx
Jp+ 1 (x) − 2
N P
p+ 1 2
2r
(r )
and
n(n − 1)bn (r 2 − x2 )n−2
n =2
(N + 1)N (r 2 − x2 )N −1 f3 (x) − f3 (r ) f 3 ( x) = = , g3 (x) g3 (x) − g3 (r )
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L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
d 2
where f3 (x) = (−12)2
1 x dx
Jp+ 1 (x) −
PN
n =2
2
n(n − 1)bn (r 2 − x2 )n−2 , and g3 (x) = (N + 1)N (r 2 − x2 )N −1 ;
······; 1 d N x dx
1
fN (x) gN0 (x)
(−2)N
0
Jp+ 1 (x) − 2
=
N P
n(n − 1) · · · 2bn
n =N
(N + 1)N · · · 2(r 2 − x2 ) fN +1 (x) − fN +1 (r ) = = , gN +1 (x) gN +1 (x) − gN +1 (r ) PN d N Jp+ 1 (x) − n=N n(n − 1) · · · 2bn , and gN +1 (x) = (N + 1)N · · · 2(r 2 − x2 ). where fN +1 (x) = (−12)N 1x dx fN +1 (x)
2
1
Let limx→r − fN +1 (x) = 0, then bN = bN =
= = = = =
N
1 d x dx
(−2)N
J
p+ 1 2
(x)|x=r
. From (7), (21), and (22), we have
N!
N h
i 1 d −(p+ 12 ) + 1 x J 1 (x) |x=r p+ 2 (−2)N N ! 2 x dx 1 3 1 1 (−1)N x−(p+ 2 +N ) Jp+ 1 +N (x)|x=r 2p+ 2 Γ p + 2 (−2)N N ! 2 1 3 1 1 2p+ 2 Γ p + r −(p+ 2 +N ) Jp+ 1 +N (r ) 2 2N N ! 2 " # 2(p − 12 + N ) 1 3 p+ 12 −(p+ 12 +N ) Γ p+ r 2 Jp− 1 +N (r ) − Jp− 3 +N 2 2 2N N ! 2 r h i 1 − p+ 12 +N 3 3 r 2(p − 1/2 + N )(N − 1)!2N −1 r p− 2 +N bN −1 − (N − 2)!2N −2 r p− 2 +N bN −2 N 2 N! 2(p − 12 + N ) 1 b N −1 − b N −2 . 2 2Nr 4N (N − 1)r 2 1
1
2p+ 2 Γ
p+
1
At the same time, by (7), (21), and (20) we have fN0 +1 (x) gN +1 (x) 0
1
=
(−2)N +1
1 d N +1 x dx
Jp+ 1 (x) 2
(N + 1)! 1
1
1 d
N +1 h
i −(p+ 21 +N ) + 1 x J ( x ) 1 p+ 2 +N (−2)N +1 (N + 1)! 2 x dx 1 3 1 3 = 2p+ 2 Γ p + (−1)N +1 x−(p+ 2 +N ) Jp+ 3 +N (x) 2 (−2)N +1 (N + 1)! 2 3 J p+ 12 2 Γ (p + 2 ) p+ 23 +N (x) = N +1 2 (N + 1)! xp+ 32 +N =
2
p+ 21
Γ
p+
1
=
2p+ 2 Γ (p + 23 ) jp+N +1 (x) p π p+ 3 +N 2N +1 (N + 1)! x 2 2x
= = = =
2p+1 Γ (p + 23 ) 2 N +1
jp+N +1 (x) √ (N + 1)! π xp+N +1
2p+1 Γ (p + 23 ) 2 N +1
√ (N + 1)! π
2p+1 Γ (p + 23 )
√
xp+N +2 jp+N +1 (x) x2p+2N +3 hp+N +1 (x)
2N +1 (N + 1)! π x2p+2N +3 2p+1 Γ (p + 23 )
2p+1 Γ (p + 23 ) u1 (x) − u1 (0) u1 (x) = , √ √ 2N +1 (N + 1)! π q1 (x) 2N +1 (N + 1)! π q1 (x) − q1 (0)
where u1 (x) = hp+N +1 (x), and q1 (x) = x2p+2N +3 ; Then u01 (x) q1 (x) 0
=
d h dx p+N +1
(2p + 2N + 3)
( x)
x2p+2N +2
=
1 d x dx
hp+N +1 (x)
(2p + 2N + 3)x2p+2N +1
L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
u2 (x)
=
q2 (x)
where u2 (x) =
1 d x dx
u2 (x) − u2 (0)
=
q2 (x) − q2 (0)
731
,
hp+N +1 (x), and q2 (x) = (2p + 2N + 3)x2p+2N +1 ; 1 d 2 x dx
hp+N +1 (x)
u3 ( x ) u3 (x) − u3 (0) = = , (2p + 2N + 3)(2p + 2N + 1)x2p+2N −1 q3 (x) q3 (x) − q3 (0) d 2 where u3 (x) = 1x dx hp+N +1 (x), and q3 (x) = (2p + 2N + 3)(2p + 2N + 1)x2N −1 ; u02 (x) q2 (x) 0
=
······; u0p+N +1 (x)
1 d p+N +1 x dx
hp+N +1 (x)
=
qp+N +1 (x) 0
(2p + 2N + 3)!!x
sin x
=
(2p + 2N + 3)!!x
,
which is decreasing on (0, r ), where r ≤ π . Therefore, we obtain that F (x) is decreasing on (0, r ) by using Lemma 1 repeatedly. Furthermore, N P
1−
bn r 2n
n =0
lim F (x) =
lim F (x) = lim
Jp+ 1 (x) − S2N (x) 2
(
r2
x→ r −
x→ r −
,
r 2(N +1)
x→0+
−
)
x 2 N +1 1 x dx
(−2)N +1
gN0 +1 (x)
x→r −
d N +1
1
= lim
fN0 +1 (x)
= lim
Jp+ 1 (x) 2
(N + 1)!
x→ r −
= bN +1. Thus
1−
PN
n=0 bn r r 2(N +1)
2n
and bN +1 are the best constants in (10).
4. A concise proof of Theorem 4 Let F (x) =
I
(x)−T2N (x) p+ 1 2 (r 2 −x2 )N +1
f1 (x) g1 (x)
=
f1 (x)−f1 (r ) , g1 (x)−g1 (r )
=
where f1 (x) = Ip+ 1 (x) − T2N (x), and g1 (x) = (r 2 − x2 )N +1 ;. . . ; 2
By the same method in Section 3, we have
I0 d0 = Ip+ 1 (r ),
d1 = −
2
d N
1
dN =
1 x dx
(−2)N
p+ 21
(r )
2r
Ip+ 1 (x)|x=r 2
N!
,
.
From (8) and (29), and (30), we have dN =
= = = = =
1
1
(−2)N N ! 1
(−2)N N ! 1
2p+ 2 Γ p+ 12
2
Γ
1
(−2)N N ! 1
(−2)N N ! 1
2p+ 2 Γ p+ 12
2
Γ 1
p+
p+
p+
p+
1
+1 2 3
x
3
2
i
Ip+ 1 +N (x) |x=r 2
2 3
1
x−(p+ 2 ) Ip+ 1 (x) |x=r 2
x dx
−(p+ 12 +N )
2
N h
1 d
1
r −(p+ 2 +N ) Ip+ 1 +N (r ) 2
"
r
−(p+ 12 +N )
Ip− 3 +N (r ) − 2
h
3
2(p −
1 2
r
+ N)
# Ip− 1 +N 2
3
r −(p+ 2 +N ) (N − 2)!(−2)N −2 r p− 2 +N dN −2 − 2(p − 1/2 + N )(N − 1)!(−2)N −1 r p− 2 +N dN −1
(−2) ! 2(p − 21 + N ) NN
2Nr 2
d N −1 +
1 4N (N − 1)r 2
dN −2 .
i
732
L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736 1 d N x dx
At the same time, fN +1 (x) = (−12)N by Lemmas 11 and 6, we have fN0 +1 (x) gN0 +1 (x)
1 d N +1 x dx
1
(−2)N +1
=
Ip+ 1 (x) −
PN
2
n=N
n(n − 1) · · · 2bn , and gN +1 (x) = (N + 1)N · · · 2(r 2 − x2 ). Then
Ip+ 1 (x) 2
(N + 1)!
N +1 h i 1 d 1 + 1 x−(p+ 2 +N ) Ip+ 1 +N (x) N + 1 2 (−2) (N + 1)! 2 x dx 1 3 1 3 = 2p+ 2 Γ p + x−(p+ 2 +N ) Ip+ 3 +N (x) 2 (−2)N +1 (N + 1)! 2 1 2p+ 2 Γ (p + 23 ) Ip+ 23 +N (x) = (−2)N +1 (N + 1)! xp+ 32 +N 1
=
1
2p+ 2 Γ
1
2p+ 2 Γ (p + 23 )
=
p+
1
ip+N +1 (x)
(−2)N +1 (N + 1)!
3
pπ 2x
xp+ 2 +N
2p+1 Γ (p + 23 )
ip+N +1 (x) √ (−2)N +1 (N + 1)! π xp+N +1
=
2p+1 Γ (p + 23 )
=
√ (−2)N +1 (N + 1)! π p+1
2
=
(−2)
xp+N +2 ip+N +1 (x) x2p+2N +3
Γ (p + ) kp+N +1 (x) √ (N + 1)! π x2p+2N +3 3 2
N +1
2p+1 Γ (p + 23 )
=
(−2)
N +1
u1 (x) √ (N + 1)! π q1 (x)
2p+1 Γ (p + 23 )
=
(−2)
N +1
u1 (x) − u1 (0) , √ (N + 1)! π q1 (x) − q1 (0)
(31)
where u1 (x) = kp+N +1 (x), and q1 (x) = x2p+2N +3 ;. . . ; Then by Lemma 10 we have u0p+N +1 (x) q0p+N +1 (x)
=
d p+N +1 [ 1x dx ] kp+N +1 (x) sinh x = , (2p + 2N + 3)!!x (2p + 2N + 3)!!x
which is increasing on (0, r ). By (31), and using Lemma 1 repeatedly, we obtain (1) When N is even, the
fN0 +1 (x) 0 gN +1 (x)
is decreasing on [0, r ](r < +∞). So F (x) is decreasing too on [0, r ];
(2) When N is odd, we find that the
fN0 +1 (x) 0 gN +1 (x)
is increasing on [0, r ](r < +∞). So F (x) is also increasing on [0, r ].
Furthermore, 1− lim F (x) =
x→0+
N P
,
r 2(N +1)
lim F (x) = lim
x→r −
dn r 2n
n =0
x→ r −
Ip+ 1 (x) − T2N (x) 2
(
r2
−
1
= lim
(−2)N +1
)
x 2 N +1
d N +1
1 x dx
= lim
x→r −
fN0 +1 (x) gN0 +1 (x)
Ip+ 1 (x) 2
(N + 1)!
x→ r −
= d N +1 . 1−
PN
2n n=0 bn r r 2(N +1)
Thus and bN +1 are the best constants in (12). The proof of Theorem 4 is complete. 5. New infinite series for Jp+ 1 (x) and Ip+ 1 (x) 2
2
In this section, we prove two new infinite series of Jp+ 1 (x) and Ip+ 1 (x), and some property of the constants in these 2 2
series.
L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
733
Theorem 6. Let 0 < x ≤ r ≤ π and N , p ≥ 0 be two natural numbers, then
Jp+ 1 (x) = S2N (x) + R2N +2 ,
(32)
2
where, S2N (x) =
PN
n =0
bn (r 2 − x2 )n and
J0 b0 = Jp+ 1 (r ), 2 bn =
p+ 12
b1 = −
2(p + n − 12 ) 2nr 2
bn−1 −
(r )
2r
,
1 4n(n − 1)r 2
b n −2 ,
n = 2, 3, . . . .
The remainder term is R2N +2 =
2p+1 Γ (p + 23 )
2N +1
√ (N + 1)!(2p + 2N + 3)!! π
sin η
η
(r 2 − x2 )N +1 ,
0 < η < r ≤ π.
(33)
Proof. In the following, we shall give the remainder term in (32) by using Cauchy mean value theorem. J
Let
(x)−S2N (x) p+ 1 2 (r 2 −x2 )N +1
=
f1 (x) g1 (x)
=
f1 (x)−f1 (r ) , g1 (x)−g1 (r )
value theorem repeatedly we obtain
Jp+ 1 (x) − S2N (x) 2
(r 2 − x2 )N +1
= =
f10 (ξ1 ) g1 (ξ1 ) 0
f100 (ξ2 ) g1 (ξ2 ) 00
where f1 (x) = Jp+ 1 (x) − S2N (x), and g1 (x) = (r 2 − x2 )N +1 . By using Cauchy mean 2
f10 (ξ1 ) − f10 (r )
=
g10 (ξ1 ) − g10 (r ) f100 (ξ2 ) − f100 (r )
=
g100 (ξ2 ) − g100 (r )
······ = = =
2p+1 Γ (p + 23 )
√
jp+N +1 (ξ )
2N +1 (N + 1)! π ξ p+N +1
,
0<ξ
2p+1 Γ (p + 23 ) 2N +1
hp+N +1 (ξ ) √ (N + 1)! π ξ 2p+2N +3
2p+1 Γ (p + 23 )
2p+1 Γ (p + 32 ) u1 (ξ ) − u1 (0) u1 (ξ ) = , √ √ 2N +1 (N + 1)! π q1 (ξ ) 2N +1 (N + 1)! π q1 (ξ ) − q1 (0)
where u1 (ξ ) = hp+N +1 (ξ ), and q1 (ξ ) = ξ 2p+2N +3 . Then
Jp+ 1 (x) − S2N (x) 2
(
r2
−
)
x 2 N +1
=
2p+1 Γ (p + 23 ) 2N +1
2p+1 Γ (p + 32 ) u01 (η1 ) − u01 (0) u0 (η1 ) = N +1 √ 01 √ (N + 1)! π q1 (η1 ) 2 (N + 1)! π q01 (η1 ) − q01 (0)
2p+1 Γ (p + 32 ) u001 (η2 ) − u001 (0) u001 (η2 ) = √ √ 2N +1 (N + 1)! π q001 (η2 ) 2N +1 (N + 1)! π q001 (η2 ) − q001 (0) ······ 1 d p+N +1 hp+N +1 (x)|x=η 2p+1 Γ (p + 23 ) x dx = N +1 √ (2p + 2N + 3)!!η 2 (N + 1)! π
=
=
2p+1 Γ (p + 23 )
2p+1 Γ (p + 23 )
sin η
,
0 < η < r ≤ π.
(r 2 − x2 )N +1 ,
0 < η < r ≤ π.
√
2N +1 (N + 1)!(2p + 2N + 3)!! π
η
So R2N +2 =
2p+1 Γ (p + 23 )
2N +1
√ (N + 1)!(2p + 2N + 3)!! π
sin η
η
Since limN →+∞ R2N +2 = 0 from (33), we can obtain the result about a new expansion for Jp+ 1 (x) as follows 2
Theorem 7. Let 0 < |x| ≤ r ≤ π , then
Jp+ 1 (x) = 2
∞ X n=0
bn ( r 2 − x2 ) n ,
(34)
734
L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
where
J0 b0 = Jp+ 1 (r ), 2 bn =
p+ 21
b1 = −
2(p + n − 12 ) 2nr 2
b n −1 −
(r )
2r
,
1 4n(n − 1)r 2
b n −2 ,
n = 2, 3, . . . .
At the same time, the constants in series (34) have the following property ∞ X n =0
bn r 2n = lim Jp+ 1 (x) = 1. 2 x→0+
When letting p = 0 in Theorems 6 and 7, we obtain Theorem 8 ([10, Theorem 7]). Let 0 < x ≤ r ≤ π and N ≥ 0 be a natural number, then sin x x
= P2N (x) + R2N +2 ,
where, P2N (x) = a0 =
sin r r
PN
an (r 2 − x2 )n and
,
a1 =
n =0
2n + 1
an+1 =
2(n + 1)r 2
sin r − r cos r
an −
2r 3 1
,
4n(n + 1)r 2
an − 1 ,
n = 1, 2, . . . .
The remainder term is jN +1 (ξ ) (r 2 − x2 )N +1 , (N + 1)!(2ξ )N +1
R2N +2 =
0<ξ
or R2N +2 =
1
sin η
2N +1 (N + 1)!(2N + 3)!!
η
(r 2 − x2 )N +1 ,
0 < η < r ≤ π.
Theorem 9 ([10, Theorem 8]). Let 0 < |x| ≤ r ≤ π , then sin x x
∞ X
=
an (r 2 − x2 )n ,
n=0
where a0 =
sin r r
an+1 =
,
a1 =
2n + 1 2(n + 1)r 2
sin r − r cos r
an −
2r 3 1
,
4n(n + 1)r 2
an − 1 ,
n = 1, 2, . . . .
At the same time, the constants in series have the following property ∞ X
an r 2n = lim
sin x
x→0+
n =0
x
= 1.
Remark 2. In Li [19], the coefficients ak (k = 0, 1, 2, . . .) are expressed as a0 =
2
π
,
ak =
∞ X n=k
π 2n (−1)n n! , (2n + 1)!(n − k)! 2
k = 1, 2, . . . .
In this paper, we first show a practical recurrence formula of ak (k = 0, 1, 2, . . .). By the same method, we can obtain two corresponding results for Ip+ 1 (x). 2
Theorem 10. Let 0 < x ≤ r and N , p ≥ 0 be two natural numbers, then
Ip+ 1 (x) = T2N (x) + R2N +2 , 2
L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
where, T2N (x) =
n=0
d0 = Ip+1/2 (r ), dn =
735
dn (r 2 − x2 )n and
PN
d1 = −
2(p + n − 12 ) 2nr 2
Ip0 +1/2 (r ) 2r
,
1
d n −1 +
4n(n − 1)r 2
dn−2 ,
n = 2, 3, . . . .
The remainder term is R2N +2 =
2p+1 Γ (p + 23 )
√ (−2)N +1 (N + 1)!(2p + 2N + 3)!! π
sin η
η
(r 2 − x2 )N +1 ,
0 < η < r.
Theorem 11. Let 0 < |x| ≤ r, then
Ip+ 1 (x) = 2
∞ X
dn (r 2 − x2 )n ,
n=0
where
I0 d0 = Ip+ 1 (r ), 2 dn =
d1 = −
2(p + n − 12 ) 2nr 2
d n −1 +
p+ 21
(r )
2r
,
1 4n(n − 1)r 2
dn−2 ,
n = 2, 3, . . . .
At the same time, the constants in series have the following property ∞ X n =0
dn r 2n = lim Ip+ 1 (x) = 1. 2 x→0+
Letting p = 0 in Theorems 10 and 11 we have Theorem 12. Let 0 < x ≤ r and N ≥ 0 be a natural number, then sinh x x
= C2N (x) + R2N +2 ,
where C2N (x) = c0 =
PN
n =0
sinh r
cn+1 =
r
cn (r 2 − x2 )n and
,
c1 = 1
4n(n + 1)r 2
sinh r − r cosh r
cn−1 +
2r 3 2n + 1 2(n + 1)r 2
,
cn ,
n = 1, 2, . . . .
(35)
The remainder term is R2N +2 =
iN +1 (ξ ) ( r 2 − x 2 ) N +1 , (N + 1)!(2ξ )N +1
0<ξ
or R2N +2 =
1
sinh η
2N +1 (N + 1)!(2N + 3)!!
η
( r 2 − x 2 ) N +1 ,
0 < η < r.
At the same time sinh x x
=
∞ X
cn (r 2 − x2 )n
(36)
n =0
holds, and the constants cn in series (36) have the following property ∞ X n =0
cn r 2n = lim
x→0+
sinh x x
= 1.
Remark 3. In Copson [42, p. 110], the expressions of ck (k = 0, 1, 2, . . .) in (36) are not showed, we first give the expressions of ck (k = 0, 1, 2, . . .) in a practical recurrence formula in the new power series expansion of (sinh x)/x.
736
L. Zhu / Computers and Mathematics with Applications 59 (2010) 724–736
Remark 4. We note that the Taylor type formula with Lagrange type remainder term of order m about factor (r 2 − x2 ) and the new power series expansion of λp (x) are showed in the second part of Theorem 3 in [25], in which letting b = 2 and c = ±1 one can obtain Theorems 6, 7, 10 and 11 in this paper.
Acknowledgements The author is grateful to the referees for their helpful and constructive comments on this paper. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43]
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