A Lower Bound on the Sinc Function and Its Application

Abstract

A lower bound on the sinc function is given. Application for the sequence {𝑏_𝑛}^∞_𝑛=1 which related to Carleman inequality is given as well.

1. Introduction

The sinc function is defined to be (𝑥)=(𝑥)sincsin𝑥𝑥≠0,1𝑥=0. (1)

This function plays a key role in many areas of mathematics and its applications [1–6].

The following result that provides a lower bound for the sinc is well known as Jordan inequality [7]: (𝑥)⩾2sinc𝜋𝜋,𝑥∈0,2, (2) where equality holds if and only if 𝑥=𝜋/2.

This inequality has been further refined by many authors in the past few years [8–35].

In [36], it was proposed that (𝑥)⩾𝜋sinc²−𝑥²𝜋²+𝑥²,𝑥≠0. (3)

We noticed that the lower bound in (3) is the fractional function. Similar result has been reported as follows [1]: sinc(𝑥)⩾5353+9𝑥²1,0⩽𝑥⩽3. (4)

To the best of the authors’ knowledge, few results have been obtained on fractional lower bound for the sinc function. It is the first aim of the present paper to establish the following fractional lower bound for the sinc function.

Theorem 1. For any 𝑥∈[0,𝜋], one has 𝑠𝑖𝑛𝑐(𝑥)⩾16𝜋⁴3𝜋²+𝑥²²−1. (5)

In [37], Yang proved that for any positive integer 𝑚, the following Carleman type inequality holds: ^∞_𝑛=1𝑎₁𝑎₂⋯𝑎_𝑛^1/𝑛<𝑒^∞_𝑛=11−^𝑚_𝑘=1𝑏_𝑘(𝑛+1)^𝑘𝑎_𝑛, (6) whenever 𝑎_𝑛⩾0, 𝑛=1,2,3,…, with 0<∑^∞_𝑛=1𝑎_𝑛<∞, where 𝑏₀𝑏=1,_𝑛=1𝑛1−𝑛+1^𝑛−2_𝑘=0𝑏_{𝑛−1−𝑘},().𝑘+1𝑛=1,2,… (7)

From a mathematical point of view, the sequence {𝑏_𝑛}^∞_𝑛=1 has very interesting properties. Yang [38] and Gyllenberg and Ping [39] have proved that, for any positive integer 𝑛, 𝑏_𝑛𝑏>0,_𝑛<1𝑛().𝑛+1 (8)

In [40], the authors proved that lim_𝑛→∞𝑏_𝑛+1𝑏_𝑛=1,𝑒𝑏_𝑛=¹₀𝑥^𝑛−2ℎ(𝑥)𝑑𝑥,𝑛⩾2, (9)(10) where ℎ(𝑥)=𝑥^𝑥+1(1−𝑥)^1−𝑥sinc(𝜋𝑥). (11)

As an application of Theorem 1, it is the second aim of the present paper to give a better upper bound on the sequence {𝑏_𝑛}^∞_𝑛=1.

Theorem 2. For any positive integer 𝑛⩾2, one has 𝑒𝑏_𝑛<1𝑛()−𝑛+12−4/𝜋𝑛().𝑛+1)(𝑛+2 (12)

2. The Proof of Theorem 1

The proof is not based on (3). We first prove the following result.

Lemma 3. For any 𝑥∈(𝜋−1/3,𝜋], one has 𝑠𝑖𝑛𝑐(𝑥)⩾16𝜋⁴3𝜋²+𝑥²²−1. (13)

Proof. Set 𝑥=𝜋−𝑡, 0⩽𝑡<1/3. Then inequality (13) is equivalent to 𝜋−𝑡+sin𝑡⩾16𝜋⁴(𝜋−𝑡)3𝜋²+(𝜋−𝑡)²². (14) To prove (14) by (4), it is enough to prove that (𝜋−𝑡)+53𝑡53+9𝑡²⩾16𝜋⁴(𝜋−𝑡)3𝜋²+(𝜋−𝑡)²²; (15) namely, (𝜋−𝑡)53+9𝑡²+53𝑡53+9𝑡²⩾16𝜋⁴(𝜋−𝑡)3𝜋²+(𝜋−𝑡)²². (16) Next we prove (16). Let 𝑔(𝑡)=(𝜋−𝑡)53+9𝑡²3𝜋²+(𝜋−𝑡)²²+53𝑡3𝜋²+(𝜋−𝑡)²²−16𝜋⁴(𝜋−𝑡)53+9𝑡². (17) We need only to prove that 𝑔(𝑡)⩾0. Elementary calculations reveal that 𝑔(𝑡)=𝑡²−9𝑡⁵+45𝜋𝑡⁴−144𝜋²𝑡³+53𝜋+252𝜋³𝑡²−4𝜋²36𝜋²+53𝑡+636𝜋³. (18) Noting that, for 0⩽𝑡<1/3, we have −9𝑡⁵1>−,27−144𝜋²𝑡³>−16𝜋²3,−4𝜋²36𝜋²4+53𝑡>−3𝜋²36𝜋²+53. (19) Thus, from (19) and (18), we get 𝑔(𝑡)⩾𝑡²636𝜋³−1−2716𝜋²3−4𝜋²36𝜋²+533⩾0. (20) This completes the proof. Now we prove Theorem 1.

Proof. By using the power series expansions of sin(𝑥) and 16𝜋⁴/(3𝜋²+𝑥²)², we find that 1+sinc(𝑥)−16𝜋⁴3𝜋²+𝑥²²=29+^∞_𝑛=1(−1)^𝑛−1𝑢_𝑛𝑥²𝜋^2𝑛, (21) where 𝑢_𝑛=16(𝑛+1)3^𝑛+2−𝜋^2𝑛()!.2𝑛+1 (22) Set 𝑥²/𝜋²=𝑡, 0⩽𝑡⩽1. Consider the function 𝑓(𝑡) defined by 𝑓(𝑡)=29+^∞_𝑛=1(−1)^𝑛−1𝑢_𝑛𝑡^𝑛. (23) From (21), we get 𝑓(0)=2/9 and 𝑓(1)=0. Lemma 3 implies 𝑓(𝑡)⩾0,𝑡₀<𝑡⩽1, (24) where 𝑡₀=11−3𝜋²≈0.79⋯. (25) Elementary calculations reveal that for 𝑛⩾4, 16(2𝑛+1)3³>3𝜋^2𝑛()!.2𝑛+1 (26) Hence, for 𝑛⩾4, we have 𝑢_𝑛𝑢>0,_𝑛−𝑢_𝑛+1=16(2𝑛+1)3^𝑛+3−𝜋^2𝑛()!+𝜋2𝑛+1(2𝑛+3^2𝑛+2)!>0. (27) Therefore, 𝑓(𝑡)⩾29+⁶_𝑛=1(−1)^𝑛−1𝑢_𝑛𝑡^𝑛. (28) If we set 𝑔(𝑡)=29+⁶_𝑛=1(−1)^𝑛−1𝑢_𝑛𝑡^𝑛, (29) then we have 𝑔(0)=29>0,𝑔(1)<𝑓(1)=0. (30) The intermediate value theorem implies that there must be at least one root 𝑐 with (0,1) such that 𝑔(𝑐)=0. Using Maple, we find that on the open interval (0,1) the equation 𝑔(𝑡)=0 has a unique real root 𝑡₁≈0.89⋯.
Hence, from (28) we get 𝑓(𝑡)⩾0,𝑡∈0,𝑡₁. (31) By (21), (24), and (31), Theorem 1 follows.

3. The Proof of Theorem 2

First, we need an auxiliary result.

Lemma 4. For any 𝑥∈[0,1/2], one has 𝑠𝑖𝑛𝑐²(𝜋𝑥)⩾1−2𝑥+𝑥²1−𝑥+𝑥². (32)

Proof. By letting 𝑥=1/2−𝑡/2𝜋, 0⩽𝑡⩽𝜋, the requested inequality can be equivalently written as 𝑡cos𝑡⩾²2+8𝜋⁴𝑡²+3𝜋²−5𝜋²+22, (33) so it suffices to show that the function 𝐺(𝑡)𝑡=cos𝑡−²2−8𝜋⁴𝑡²+3𝜋²+5𝜋²+22 (34) is negative on 0⩽𝑡⩽𝜋. Theorem 1 implies 𝐺^′(𝑡)<0. (35) Hence, 𝐺(𝑡)⩾𝐺(𝜋)=0. (36) The required inequality follows. Now we prove Theorem 2.

Proof. Let 𝐻(𝑥)=𝑥^𝑥(1−𝑥)^−𝑥sinc(𝜋𝑥),0<𝑥<11𝑥=0,1. (37) We first consider the case 0⩽𝑥⩽1/2.
Taking the natural log gives ln𝐻(𝑥)=(𝑥−1)ln𝑥−𝑥ln(1−𝑥)+lnsin𝜋𝑥−ln𝜋. (38) Taking the second derivative of both sides of (38), we have 𝐻𝐻^′′−𝐻^′2𝐻²=𝑥²−𝑥+1𝑥²(1−𝑥)²−𝜋²csc²(𝜋𝑥). (39) By Lemma 4, it follows that 𝐻𝐻^′′−𝐻^′2𝐻²>0. (40) Thus, 𝐻^′′>0, (41) and therefore for 0⩽𝑥⩽1/2, we have 𝐻(𝑥)⩽()𝐻(0)11−2𝑥+2𝑥𝐻2=42−𝜋(−𝑥)+1. (42) For the case 1/2<𝑥⩽1, since 𝐻(1/2)=2/𝜋, 𝐻(1)=1, and 𝐻 is concave up, it follows that 𝐻(𝑥)()𝐻1⩽21−𝑥2+()𝐻(1)=42𝑥−12−𝜋(𝑥−1)+1. (43) Using (10) from (42) and (43), we have 𝑒𝑏_𝑛=¹₀𝑥^𝑛−2ℎ(𝑥)𝑑𝑥=¹₀𝐻(𝑥)𝑥^𝑛−1−𝑥^𝑛⩽𝑑𝑥_01/241−2−𝜋𝑥𝑥^𝑛−1−𝑥^𝑛+𝑑𝑥_11/241+2−𝜋(𝑥−1)𝑥^𝑛−1−𝑥^𝑛=1𝑑𝑥𝑛()−4𝑛+12−𝜋2^𝑛+2−𝑛−4𝑛(𝑛+1)(𝑛+2)2^𝑛+1<1𝑛()−𝑛+12−4/𝜋𝑛().𝑛+1)(𝑛+2 (44) This proves Theorem 2.

Conflict of Interests

There is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are very grateful to the anonymous referees and the editor for their insightful comments and suggestions. The work of the second author was supported by a Grant of the Romanian National Authority for Scientific Research, CNCS-UEFISCDI project no. PN-II-ID-PCE-2011-3-0087.