A NOTE ON THE BECKER-STARK TYPE INEQUALITIES Текст научной статьи по специальности «Математика»

58

Probl. Anal. Issues Anal. Vol. 11 (29), No 1, 2022, pp. 58-66

DOI: 10.15393/j3.art.2022.10770

UDC 517.1, 517.521

R. M. Dhaigüde, Y. J. Bagul A NOTE ON THE BECKER-STARK TYPE INEQUALITIES

Abstract. This note is devoted to establishing the sharp bounds for the function x/ tg x, thus refining the well-known Becker-Stark's inequality.

Keywords: Becker-Stark inequality, tangent function, monotonici-ty of functions, Bernoulli numbers

2020 Mathematical Subject Classification: 26A48, 26D05, 33B10

1. Introduction. The inequality

At2 -v2 ■x2

. I -' iA_' /1 iA_' , / \ / ^ \

i - — < — < ^ - x e (0,^/2) (1)

n2 tgx 8 2

is known in the literature as the Becker-Stark inequality. Here tg denotes the trigonometric tangent function. It was proved in [6]. Z.-H. Yang et. al. in [15] prove that

1 - ^ < tgs<1 - iT;- e (0^/2)- (2)

while Chen and Cheung [7] show that

Ax2\ x (

< < i -/ tg x V n

l 4x2 \ x i 4x2\^2/i2

with the best possible constants 1 and ^2/12. The lower bound in all the three inequalities listed above is one and the same. However, the upper bound in (3) is sharper than those in (1) and (2). The upper bounds in (2)-(3) are not sharp as x ^ n/2-. Researchers obtained different

© Petrozavodsk State University, 2022

generalizations and refinements of inequality (1). The details can be seen in [3-10], [13], [15-18] and the references therein.

In view of obtaining refinement of the lower bounds in inequalities (1)-(3) and the sharp upper bound for x/ tgx as x ^ k/2-, we propose the following theorem:

Theorem 1. For x G (0,^/2), we have

/ \ (jL_iï T2 X / 4X*\ 4ln(x2/8) T2

1--- ^ -2 ^x < -< 1--- e -2 x . (4)

V n2 J tg x \ n2 J v y

The bounds in (4) are polynomial-exponential in nature. Since (t4 — 3) x<2 > 1, the lower bound in (4) is sharper than the one in (1) - (3). With the help of any plotting software, it can be observed that the upper bound in (4) is sharper than the corresponding upper bound in (3) for x G (£i,^/2), where £1 ~ 0.8496. The constant — 3) (or a similar) one is interesting, and as a complement to (4), we present simple polynomial bounds for x/ tan x as the following double inequality:

Proposition 1. For x G (0,^/2), we have

1 -

4x2\r ( 4 1 \ 2i x ( 4x2\r ( 4 1 \

4^) + aX< tgï< X^) 1-Ci42- 2)

tg X V 'K2

1- (4--) x

As 1 + (^42 — |) x2 > 1, we conclude that the lower bound in (5) is sharper than the one in (1)-(3). And again, with the help of any plotting software, one can observe that the upper bound in (5) is sharper than the corresponding upper bound in (3) for x G (£2,^/2), where £2 ~ 0.9721.

2. Preliminaries and lemmas. We recall the formula for the simple geometric series:

1 1+ X + X2 + x3 + ..., |x| < 1 (6)

1 — X

and

T ^ 2 f22fc-i - i)

^=1+£ ^-m-1 ^^ ■ w<'- <7>

where B2k are the even indexed Bernoulli numbers. The expansion (7) can be found in [11, 1.411]. We also need the following lemmas for proving our main results. The Lemma 1 is known as l'Hôpital's rule of monotonicity. We refer to [2] for more details.

Lemma 1. Let f1(x) and f2(x) be two real-valued functions that are continuous on [a,b] and differentiable on (a,b), where —<x < a <b < and g'(x) = 0, for all x E (a, b). Let

AW= ffl " fft E (a, t),

f2(X) - J2(0)

B(x) = fi(x) - fi(b) xE (ab) B(X) h(x) - h(b),X E (a,b)-

Then we have

(i) A(x) and B(x) are increasing on (a,b) if f1(x)/f2(x) is increasing on (a, b).

(ii) A(x) and B(x) are decreasing on (a,b) if f1(x)/f2(x) is decreasing on (a, b).

The strictness of the monotonicity of A(x) and B (x) depends on the strictness of monotonicity of f[(x)/f2(x).

Lemma 2. For all integers k ^ 1, we have

^ < 2(2fc)! 1 -

(2n)2k 1 - 2^-2fc' w

where ß = 2 + (ln(1 - 6/n2))/ln2 w 0.6491.

Lemma 2 appears in [1] and Lemma 3 is proved by L. Zhu et. al. in [17].

Lemma 3. For < n/2, we have

I 2 a 2\tg X 2 I V"^ 2k

(n2 — 4x2)-= n2 + > akx ;

rp ' J

X

k=1

22k+2(22k+2 - 1W2, , 4 ■ 22k(22k - 1), ,

where ak = —— —m— 1 < 0,

k = 1, 2, 3,...

3. Proofs of the main results. In this section, we prove our main results.

Proof of Theorem 1. Consider

tg xn2 — 4 x2 n f2(x)

where f1(x) = ln (-

x

-) and f2(x) = x2 with /i(0+) = 0 and

-.tgXK2 — 4x2 >

/2(0) = 0. Differentiation of the numerator and denominator with respect to x gives

IM

f2 (x)

tg x — x sec2 x + 8x

x tgx

n2

- 4x2

2x

1 tgx — x sec2 x

2 x2 tg x 1 sin 2 x 2 x

+

+

n2 — 4x2

4

4 x2 sin 2x n 2

(1 — X)

1

4x2

1

n 2 x

4

+ -T

1

4x2 sin 2x n 2

— ^2

Using (6) and (7), we write

/: 1 1 fl+ ^22^-'—1) ^

2( x) 4 x2 4 x2

fc=i

(2 )!

A v (~\2k

n2 ^U /

k=0

4

^ -2fc J^ o2fc-^o2fc-1

x2k-

22

n2 ^ n2k k=0

2 22

k=1

(22fc 1 1)IR I 2fc-2

— |^D2fc |x

(2 )!

^ o2fc+2 £ -

n

2k+2"

k=0

<x

£ 2

k=0

2k+1

k=0

2 (22fc+1 — 1)

where Ck = 22fc+1

Ln2fc+2 (2k + 2)! 2 (22fc+1 — 1)

(2 k + 2)!

| B2k+2 |

„2fc

x

:= E

Ck x

„2k

k=0

n2k+2 (2k + 2)! From (8), we have

| B2k+2 |

| B2k+2 | <

2(2 fc + 2)!

1

2(2 k + 2)!

1

22fc+2n2fc+2 1 — 2ß-2k-2 n2k+2 (22k+2 — 2^) where ß = 2 + (ln(1 — 6/n2))/ln2 w 0.6491.

4

1

Clearly, - 1 < 22k+1 for k ^ 0, i.e., 2^ - 1 < 22k+2 - 22k+1 or 22k+i _ i < 22k+2 — 2?. From this, we write:

1

<

1

22k+2 _ 2ß 22k+l — 1 '

Then

lB2k+2 1 <

2(2k + 2)!

1

n2k+2 (2?k+2 - 2ß )

<

2(2k + 2)!

2k+2 (22k+l — 1) '

■n

which leads to ck > 0 for k ^ 0. Hence, f[(x)/f2(x) is strictly increasing on (0,^/2). By Lemma 1, f (x) is also strictly increasing on (0,^/2). So, f (0+) < f (x) < f (-k/2-). The limits f (0+) = 4/^2 - 1/3 and f (^/2-) = 4ln(^2/8)/^2 prove the assertion. □

Proof of Proposition 1. Let

2

= (^{J2- 4x2) - ^* e (M/2).

By Lemma 3, we get

g(x) = (

K2 + J2 T=1 akx

2k

-1 X2

- ^ akx k=1

2k

'K2X2 + Y1 ak X2k+2 k=1

where ak < 0 for k =1, 2, 3,... Then

k2x2 + £ akx2k+2 1 k=1

g(x)

—■n

— X

- E akX2k k=1

Y, akX2k-2 k=1

This implies that g(x) is strictly increasing on (0,^/2). So, we have 0(0+) < g(x) < g(n/2-). With the limits #(0+) = 4/^2 - 1/3 and g(n/2—) = 1/2 — 4/'K2, we end the proof. □

A graphical illustration of the lower and upper bounds of x/ tg x appeared in (1)-(5) is given in Figure 1 and Figure 2.

1

0

0 _ ID O

- x/tg(x) V\

- x/tg(x) C\J -- 1 - x2/3 •• (1 - 4x2/rc2)*2"2

-- 1 - 4x2/ n2 \ ° v-

(1 - 4x2/ rc2)exp((4/ n2 - l/3)x2) • - • (1 - 4x2/ji2)exp(4log(ji2/8)/j[2x2) v

(1 - 4x2/*2)(l + (4/*2 - l/3)x2) O O (1 - 4x2/n2)(1 -(41 n2 -1 /2)x2) i

Figure 1: Graphs of the lower bounds of (1) - (5), x e (0,n/2).

Figure 2: Graphs of the upper bounds of (2)-(5), x e (0,n/2).

x

x

4. Extended inequalities via monotonically stratified functions. We extend inequalities (4) and (5) to a wider range of parameters using the technique of the minimax approximant given in [12]. For this, consider the family of continuous functions

x ( 4x2\ 2

0„(x) =--(1--— )epx secx

sinx V n2 /

on (0,n/2) for every p e R+. Clearly, the family of functions 0p(x) is decreasingly stratified for p e R+. If

4 1 , 4ln(n2/8)

A = — — - = 0.071951... and B =-= 0.085117...,

n2 3 n2

then we have 0A(0+) = 0b(0+) = (n/2—) = 0 and 0A(n/2—) e R+. Moreover, the functions 0p(x) are continuous with respect top e (A, B) for every x e (0,n/2) and 0p(n/2—) is continuous with respect to p e (A,B). So, using Theorem 1' and Theorem 2' of [12], we have the following statements:

Statement 1. If p e (0, A], where A = ^ — 3 = 0.071951..., then

4>P(x) > Mx) = — - (1 - 4x22)e(0^ secx > 0 sinx V k2 )

for x G (0, k/2).

Statement 2. If p e [B, ro), where B = 4ln(^/8) = 0.085117..., then

, . . X ( 4x2 \ 4 1n(x2/S) x2

<pp(x) ^ <pB(x) =--1--tt e secx < 0

sin x V /

for X e (0,n/2).

Similarly, if we consider

x 4x2 \

■iL(x) =--v(1--— ) (1 + px2) sec x, p e R+

sin X -K2 J

and

4 1 14

C =—: - - = 0.071951..., D =---- = 0.094715...,

n2 3 2 n2

then it is again obvious that the family of functions ^p(x) is decreasingly stratified and satisfies the assumptions of Theorem 1' of [12]. So, using Theorem 2' of [12], we state the following:

Statement 3. If p e (0, C] where C = 4 - | = 0.071951..., then

^p(x) > ^c(x) = — - (1 - %)[1 + (- ^X2 sinx V n2 / L \n2 3/

for X e (0,n/2).

Statement 4. If p e [D, ro), where D = 2 - 4 = 0.094715 ..., then

sec x > 0

^P(X) ^ ^D (X) = — - (1 - [1 + f1 - -2) X2

sin x V n2 / - V2 n2J

14

sec x

for X e (0,n/2).

Acknowledgement. The authors would like to thank the referees for their valuable comments and suggestions.

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Received August 17, 2021. In revised form,, January 09, 2022. Accepted January 24, 2022. Published online February 3, 2022.

Ramkrishna M. Dhaigude Department of Mathematics

Government Vidarbha Institute of Science and Humanities Amravati(M. S.)-444604, India E-mail: rmdhaigude@gmail.com

Yogesh J. Bagul Department of Mathematics K. K. M. College, Manwath Dist: Parbhani(M. S.)-431505, India E-mail: yjbagul@gmail.com