# A Lower Bound on the Sinc Function and Its Application

Authors: Yue Hu; Cristinel Mortici
Journal: The Scientific World Journal
 (2014)

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## Abstract

A lower bound on the sinc function is given. Application for the sequencebnn=1∞ which related to Carleman inequality is given as well.


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## Body

## 1. Introduction

The sinc function is defined to be(1)sinc(x)={sin(x)xx≠0,1x=0.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]:
(2)sinc(x)⩾2π,x∈[0,π2],
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
(3)sinc(x)⩾π2-x2π2+x2,x≠0.We noticed that the lower bound in (3) is the fractional function. Similar result has been reported as follows [1]:
(4)sinc(x)⩾5353+9x2,0⩽x⩽13.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 anyx∈[0,π], one has
(5)sinc(x)⩾16π4(3π2+x2)2-1.In [37], Yang proved that for any positive integer m, the following Carleman type inequality holds:
(6)∑n=1∞‍(a1a2⋯an)1/n<e∑n=1∞‍(1-∑k=1m‍bk(n+1)k)an,
whenever an⩾0, n=1,2,3,…, with 0<∑n=1∞an<∞, where
(7)b0=1,bn=1n(1n+1-∑k=0n-2‍bn-1-kk+1),(n=1,2,…).From a mathematical point of view, the sequence{bn}n=1∞ has very interesting properties. Yang [38] and Gyllenberg and Ping [39] have proved that, for any positive integer n,
(8)bn>0,bn<1n(n+1).In [40], the authors proved that
(9)lim⁡n→∞bn+1bn=1,(10)ebn=∫01‍xn-2h(x)dx,n⩾2,
where
(11)h(x)=xx+1(1-x)1-xsinc(πx).As an application of Theorem1, it is the second aim of the present paper to give a better upper bound on the sequence {bn}n=1∞.Theorem 2.
For any positive integern⩾2, one has
(12)ebn<1n(n+1)-2-4/πn(n+1)(n+2).


## 2. The Proof of Theorem1

The proof is not based on (3). We first prove the following result.Lemma 3.
For anyx∈(π-1/3,π], one has
(13)sinc(x)⩾16π4(3π2+x2)2-1.Proof.
Setx=π-t, 0⩽t<1/3. Then inequality (13) is equivalent to
(14)π-t+sint⩾16π4(π-t)(3π2+(π-t)2)2.
To prove (14) by (4), it is enough to prove that
(15)(π-t)+53t53+9t2⩾16π4(π-t)(3π2+(π-t)2)2;
namely,
(16)(π-t)(53+9t2)+53t(53+9t2)⩾16π4(π-t)(3π2+(π-t)2)2.
Next we prove (16). Let
(17)g(t)=(π-t)(53+9t2)(3π2+(π-t)2)2+53t(3π2+(π-t)2)2-16π4(π-t)(53+9t2).
We need only to prove that g(t)⩾0. Elementary calculations reveal that
(18)g(t)=t2(-9t5+45πt4-144π2t3+(53π+252π3)t2-4π2(36π2+53)t+636π3).
Noting that, for 0⩽t<1/3, we have
(19)-9t5>-127,-144π2t3>-16π23,-4π2(36π2+53)t>-43π2(36π2+53).
Thus, from (19) and (18), we get
(20)g(t)⩾t2(636π3-127-16π23-4π2(36π2+53)3)⩾0.
This completes the proof. Now we prove Theorem 1.Proof.
By using the power series expansions ofsin(x) and 16π4/(3π2+x2)2, we find that
(21)1+sinc(x)-16π4(3π2+x2)2=29+∑n=1∞‍(-1)n-1un(x2π2)n,
where
(22)un=16(n+1)3n+2-π2n(2n+1)!.
Set x2/π2=t, 0⩽t⩽1. Consider the function f(t) defined by
(23)f(t)=29+∑n=1∞‍(-1)n-1untn.
From (21), we get f(0)=2/9 and f(1)=0. Lemma 3 implies
(24)f(t)⩾0,t0<t⩽1,
where
(25)t0=(1-13π)2≈0.79⋯.
Elementary calculations reveal that for n⩾4,
(26)16(2n+1)33>(3π2)n(2n+1)!.
Hence, for n⩾4, we have
(27)un>0,un-un+1=16(2n+1)3n+3-π2n(2n+1)!+π2n+2(2n+3)!>0.
Therefore,
(28)f(t)⩾29+∑n=16‍(-1)n-1untn.
If we set
(29)g(t)=29+∑n=16‍(-1)n-1untn,
then we have
(30)g(0)=29>0,g(1)<f(1)=0.
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
(31)f(t)⩾0,t∈[0,t1].
By (21), (24), and (31), Theorem 1 follows.


## 3. The Proof of Theorem2

First, we need an auxiliary result.Lemma 4.
For anyx∈[0,1/2], one has
(32)sinc2(πx)⩾1-2x+x21-x+x2.Proof.
By lettingx=1/2-t/2π, 0⩽t⩽π, the requested inequality can be equivalently written as
(33)cos⁡t⩾t22+8π4t2+3π2-5π2+22,
so it suffices to show that the function
(34)G(t)=cos⁡t-t22-8π4t2+3π2+5π2+22
is negative on 0⩽t⩽π. Theorem 1 implies
(35)G′(t)<0.
Hence,
(36)G(t)⩾G(π)=0.
The required inequality follows. Now we prove Theorem 2.Proof.
Let(37)H(x)={xx(1-x)-xsinc(πx),0<x<11x=0,1.
We first consider the case 0⩽x⩽1/2.
Taking the natural log gives(38)ln⁡H(x)=(x-1)ln⁡x-xln⁡(1-x)+ln⁡sinπx-ln⁡π.
Taking the second derivative of both sides of (38), we have
(39)HH′′-H′2H2=x2-x+1x2(1-x)2-π2csc2(πx).
By Lemma 4, it follows that
(40)HH′′-H′2H2>0.
Thus,
(41)H′′>0,
and therefore for 0⩽x⩽1/2, we have
(42)H(x)⩽(1-2x)H(0)+2xH(12)=(2-4π)(-x)+1.
For the case 1/2<x⩽1, since H(1/2)=2/π, H(1)=1, and H is concave up, it follows that
(43)H(x)⩽2(1-x)H(12)+(2x-1)H(1)=(2-4π)(x-1)+1.
Using (10) from (42) and (43), we have
(44)ebn=∫01‍xn-2h(x)dx=∫01‍H(x)(xn-1-xn)dx⩽∫01/2‍[1-(2-4π)x](xn-1-xn)dx+∫1/21‍[1+(2-4π)(x-1)](xn-1-xn)dx=1n(n+1)-(2-4π)2n+2-n-4n(n+1)(n+2)2n+1<1n(n+1)-2-4/πn(n+1)(n+2).
This proves Theorem 2.



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