A series expansion of a logarithmic expression and a decreasing property of the ratio of two logarithmic expressions containing cosine

Yan-Fang Li; Feng Qi

doi:10.1515/math-2023-0159

Article Open Access

A series expansion of a logarithmic expression and a decreasing property of the ratio of two logarithmic expressions containing cosine

Yan-Fang Li and Feng Qi

Published/Copyright: December 8, 2023

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$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

In this study, by virtue of a derivative formula for the ratio of two differentiable functions and with aid of a monotonicity rule, the authors expand a logarithmic expression involving the cosine function into the Maclaurin power series in terms of specific determinants and prove a decreasing property of the ratio of two logarithmic expressions containing the cosine function.

Keywords: Maclaurin power series expansion; decreasing property; logarithmic expression; cosine function; derivative formula; ratio of two differentiable functions; monotonicity rule

MSC 2010: 41A58; 26A09; 33B10

1 Motivations

In [1, pp. 42 and 55], we find the Maclaurin power series expansions

(1) cos x = ∑ k = 0 ∞ ( − 1 ) k x 2 k ( 2 k ) ! = 1 − x 2 2 + x 4 24 − x 6 720 + x 8 40,320 − ⋯ , x ∈ R

and

ln cos x = − ∑ k = 1 ∞ 2 2 k − 1 ( 2 2 k − 1 ) k ( 2 k ) ! ∣ B 2 k ∣ x 2 k = − x 2 2 − x 4 12 − x 6 45 − 17 x 8 2,520 − ⋯ , x 2 < π 2 4 ,

where B 2 k denotes the Bernoulli numbers, which can be generated by:

z e z − 1 = ∑ k = 0 ∞ B k z k k ! = 1 − z 2 + ∑ k = 1 ∞ B 2 k z 2 k ( 2 k ) ! = 1 − z 2 + z 2 12 − z 4 720 + z 6 30,240 − z 8 1,209,600 + ⋯ , ∣ z ∣ < 2 π .

For more detailed information about B 2 k , please refer to the monograph [2] and articles [3,4].

Motivated by the recently published articles [5–7], we consider the following two problems in this study.

What is the Maclaurin power series expansion of the even function
(2) F ( x ) = ln 2 ( 1 − cos x ) x 2 , 0 < ∣ x ∣ < 2 π 0 , x = 0
around x = 0 ?
Is the even function
(3) R ( x ) = ln 2 ( 1 − cos x ) x 2 ln cos x , 0 < ∣ x ∣ < π 2 1 6 , x = 0 0 , x = ± π 2
decreasing on the close interval [ 0 , π 2 ] ?

2 Lemmas

For smoothly solving the aforementioned two problems, we need the following lemmas.

Lemma 1

Let u ( x ) and v ( x ) ≠ 0 be two n-time differentiable functions on an interval I for a given integer n ≥ 0 . Then, the nth derivative of the ratio u ( x ) v ( x ) is

(4) d n d x n u ( x ) v ( x ) = ( − 1 ) n ∣ W ( n + 1 ) × ( n + 1 ) ( x ) ∣ v n + 1 ( x ) , n ≥ 0 ,

where the matrix

W ( n + 1 ) × ( n + 1 ) ( x ) = U ( n + 1 ) × 1 ( x ) V ( n + 1 ) × n ( x ) ( n + 1 ) × ( n + 1 ) ,

the matrix U ( n + 1 ) × 1 ( x ) is an ( n + 1 ) × 1 matrix whose elements satisfy u k , 1 ( x ) = u ( k − 1 ) ( x ) for 1 ≤ k ≤ n + 1 , the matrix V ( n + 1 ) × n ( x ) is an ( n + 1 ) × n matrix whose elements are v ℓ , j ( x ) = ℓ − 1 j − 1 v ( ℓ − j ) ( x ) for 1 ≤ ℓ ≤ n + 1 and 1 ≤ j ≤ n , and the notation ∣ W ( n + 1 ) × ( n + 1 ) ( x ) ∣ denotes the determinant of the ( n + 1 ) × ( n + 1 ) matrix W ( n + 1 ) × ( n + 1 ) ( x ) .

Formula (4) is a reformulation of [8, p. 40, Exercise 5] (see also the papers [4,9,10] and those papers collected at the site [11].

Lemma 2

(Monotonicity rule for the ratio of two functions [12, Theorem 1.25]) For a , b ∈ R with a < b , let λ ( x ) and μ ( x ) be continuous on [ a , b ] , differentiable on ( a , b ) , and μ ′ ( x ) ≠ 0 on ( a , b ) . If the ratio λ ′ ( x ) μ ′ ( x ) is increasing on ( a , b ) , then both λ ( x ) − λ ( a ) μ ( x ) − μ ( a ) and λ ( x ) − λ ( b ) μ ( x ) − μ ( b ) are increasing in x ∈ ( a , b ) .

3 Maclaurin power series expansion

In this section, we give a solution to the first problem posed in the first section of this article.

Theorem 1

Let the real numbers

ω k = ( − 1 ) k ( k + 1 ) ( 2 k + 1 ) , k ≥ 0

and the determinants

E 2 n = − A 2 n − 1 , 1 B 2 n − 1 , 2 n − 1 ω n C 1 , 2 n + 1 , n ≥ 1 ,

where the matrices A 2 n − 1 , 1 , B 2 n − 1 , 2 n − 1 , and C 1 , 2 n − 1 for n ≥ 1 are defined by:

A 2 n − 1 , 1 = 0 ω 1 0 ω 2 ⋮ 0 ω n − 1 0 = a i , j 1 ≤ i ≤ 2 n − 1 j = 1 , a i , 1 = 0 , 1 ≤ i = 2 k − 1 ≤ 2 n − 1 ; ω k , 2 ≤ i = 2 k ≤ 2 n − 2 , B 2 n − 1 , 2 n − 1 = 0 0 ω 0 0 0 ⋯ 0 0 1 1 ω 0 0 ⋯ 0 2 0 ω 1 0 2 2 ω 0 ⋯ 0 0 3 1 ω 1 0 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ 2 n − 4 0 ω n − 2 0 2 n − 4 2 ω n − 3 ⋯ 0 0 2 n − 3 1 ω n − 2 0 ⋯ 0 2 n − 2 0 ω n − 1 0 2 n − 2 2 ω n − 1 ⋯ 2 n − 2 2 n − 2 ω 0 = b i , j 1 ≤ i , j ≤ 2 n − 1 , b i , j = 0 , 1 ≤ i < j ≤ 2 n − 1 ; i j ω k , 0 ≤ i − j = 2 k ≤ 2 n − 2 ; 0 , 1 ≤ i − j = 2 k − 1 ≤ 2 n − 3 , C 1 , 2 n − 1 = 0 2 n − 1 1 ω n − 1 0 2 n − 1 3 ω n − 2 ⋯ 0 2 n − 1 2 n − 3 ω 1 0 = c i , j i = 1 1 ≤ j ≤ 2 n − 1 , c 1 , j = 0 , 1 ≤ j = 2 k − 1 ≤ 2 n − 1 ; 2 n − 1 2 k − 1 ω n − k , 2 ≤ j = 2 k ≤ 2 n − 2 .

Then, the function F ( x ) defined by (2) can be expanded into the Maclaurin power series expansion:

(5) F ( x ) = − ∑ n = 1 ∞ E 2 n ( 2 n ) ! x 2 n = − x 2 12 − x 4 1,440 − x 6 90,720 − x 8 4,838,400 − ⋯

for ∣ x ∣ < 2 π .

Proof

On the interval ( 0 , π ) , directly differentiating yields

F ′ ( x ) = x sin x + 2 cos x − 2 x ( 1 − cos x ) = x sin x + 2 cos x − 2 x 3 1 − cos x x 2 ≜ u ( x ) v ( x ) ,

where

u ( x ) = x sin x + 2 cos x − 2 x 3 , x ≠ 0 0 , x = 0 and v ( x ) = 1 − cos x x 2 , x ≠ 0 1 2 , x = 0

have the series expansions

u ( x ) = 1 2 ∑ k = 0 ∞ ( − 1 ) k + 1 ( k + 2 ) ( 2 k + 3 ) x 2 k + 1 ( 2 k + 1 ) ! = 1 2 ∑ k = 0 ∞ ω k + 1 x 2 k + 1 ( 2 k + 1 ) !

and

v ( x ) = 1 2 ∑ k = 0 ∞ ( − 1 ) k ( k + 1 ) ( 2 k + 1 ) x 2 k ( 2 k ) ! = 1 2 ∑ k = 0 ∞ ω k x 2 k ( 2 k ) ! .

These two series expansions imply

u ( n ) ( 0 ) = 0 , n = 2 k ω k + 1 2 , n = 2 k + 1 and v ( n ) ( 0 ) = ω k 2 , n = 2 k 0 , n = 2 k + 1

for k , n ≥ 0 . Accordingly, making use of Formula (4) results in

F ( 2 n + 2 ) ( 0 ) = lim x → 0 u ( x ) v ( x ) ( 2 n + 1 ) = ( − 1 ) 2 n + 1 v 2 n + 2 ( 0 ) u ( 0 ) v ( 0 ) 0 ⋯ 0 u ′ ( 0 ) v ′ ( 0 ) 1 1 v ( 0 ) ⋯ 0 u ″ ( 0 ) v ″ ( 0 ) 2 1 v ′ ( 0 ) ⋯ 0 u ( 3 ) ( 0 ) v ( 3 ) ( 0 ) 3 1 v ″ ( 0 ) ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ u ( 2 n − 2 ) ( 0 ) v ( 2 n − 2 ) ( 0 ) 2 n − 2 1 v ( 2 n − 3 ) ( 0 ) ⋯ 0 u ( 2 n − 1 ) ( 0 ) v ( 2 n − 1 ) ( 0 ) 2 n − 1 1 v ( 2 n − 2 ) ( 0 ) ⋯ 0 u ( 2 n ) ( 0 ) v ( 2 n ) ( 0 ) 2 n 1 v ( 2 n − 1 ) ( 0 ) ⋯ 2 n 2 n v ( 0 ) u ( 2 n + 1 ) ( 0 ) v ( 2 n + 1 ) ( 0 ) 2 n + 1 1 v ( 2 n ) ( 0 ) ⋯ 2 n + 1 2 n v ′ ( 0 )

= − 1 ω 0 2 n + 2 0 ω 0 0 0 ⋯ 0 ω 1 0 1 1 ω 0 0 ⋯ 0 0 ω 1 0 2 2 ω 0 ⋯ 0 ω 2 0 3 1 ω 1 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ 0 ω n − 1 0 2 n − 2 2 ω n − 2 ⋯ 0 ω n 0 2 n − 1 1 ω n − 1 0 ⋯ 0 0 ω n 0 2 n 2 ω n ⋯ 2 n 2 n ω 0 ω n + 1 0 2 n + 1 1 ω n 0 ⋯ 0 = − 0 1 0 0 ⋯ 0 ω 1 0 1 1 0 ⋯ 0 0 ω 1 0 2 2 ⋯ 0 ω 2 0 3 1 ω 1 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ 0 ω n − 1 0 2 n − 2 2 ω n − 2 ⋯ 0 ω n 0 2 n − 1 1 ω n − 1 0 ⋯ 0 0 ω n 0 2 n 2 ω n ⋯ 2 n 2 n ω n + 1 0 2 n + 1 1 ω n 0 ⋯ 0 = − A 2 n + 1 , 1 B 2 n + 1 , 2 n + 1 ω n + 1 C 1 , 2 n + 1 = − E 2 n + 2

for n ≥ 0 . Consequently, we obtain

F ( x ) = ∑ n = 0 ∞ F ( n ) ( 0 ) n ! x n = ∑ n = 1 ∞ F ( 2 n ) ( 0 ) ( 2 n ) ! x 2 n = − ∑ n = 1 ∞ E 2 n ( 2 n ) ! x 2 n .

The required proof is thus complete.□

Remark 1

For n = 3 , the determinant E 6 is

E 6 = 0 0 0 ω 0 0 0 0 0 ω 1 0 1 1 ω 0 0 0 0 0 2 0 ω 1 0 2 2 ω 0 0 0 ω 2 0 3 1 ω 1 0 3 3 ω 0 0 0 4 0 ω 2 0 4 2 ω 2 0 4 4 ω 0 ω 3 0 5 1 ω 2 0 5 3 ω 1 0 = 0 1 0 0 0 0 − 1 6 0 1 0 0 0 0 − 1 6 0 1 0 0 1 15 0 − 1 2 0 1 0 0 1 15 0 − 1 0 1 − 1 28 0 1 3 0 − 5 3 0 = 1 126

and − E 6 6 ! = − 1 90,720 . This coincides with the coefficient of the term x 6 in the Maclaurin power series expansion (5).

Remark 2

On 11 September 2023, Gradimir V. Milovanović (Serbian Academy of Sciences and Arts) pointed out that

(6) E 2 n = ∣ B 2 n ∣ n , n ≥ 1 .

This intrinsic observation can be derived from the interesting, but not-easily-guessed, relation:

( − z i ) F ′ ( − z i ) = 2 z e z − 1 − 1 + z 2 , ∣ z ∣ < 2 π ,

where i = − 1 is the imaginary unit in the theory of complex numbers. Then, the series expansion (5) can be reformulated as:

(7) F ( x ) = − ∑ n = 1 ∞ ∣ B 2 n ∣ n x 2 n ( 2 n ) ! , ∣ x ∣ < 2 π .

We can also regard Relation (6) as a determinantal expression of the Bernoulli numbers B 2 n for n ≥ 1 . For known results of determinantal expressions of the Bernoulli numbers B 2 n , please refer to the literature [7,13–15], for example.

Remark 3

From the series expansion (7), we construct a positive and even function:

H m ( x ) = − m + 1 ∣ B 2 m + 2 ∣ ( 2 m + 2 ) ! x 2 m + 2 F ( x ) + ∑ n = 1 m ∣ B 2 n ∣ n x 2 n ( 2 n ) ! , 0 < ∣ x ∣ < 2 π 1 , x = 0

for m ≥ 1 . We now propose the following two problems.

Discuss the logarithmic convexity or logarithmic concavity of the even and positive function H m ( x ) on ( 0 , 2 π ) .
Expand the function H m ( x ) into a Maclaurin power series at x = 0 .

We believe that, making use of the series expansion

H m ( x ) = ∑ n = 0 ∞ m + 1 n + m + 1 ( 2 m + 2 ) ! ( 2 n + 2 m + 2 ) ! ∣ B 2 n + 2 m + 2 ∣ ∣ B 2 m + 2 ∣ x 2 n , ∣ x ∣ < 2 π ,

employing the derivative Formula (4), and with the help of a monotonicity rule in the articles [5,6] for the quotient of two power series, these two problems can be possibly solved.

4 Decreasing property

In this section, we solve the second problem posed in the first section of this article.

Theorem 2

The function R ( x ) defined by (3) decreasingly maps [ 0 , π 2 ] onto [ 0 , 1 6 ] .

First proof

Straightforward computation yields

F ′ ( x ) ( ln cos x ) ′ = cot x ( x sin x + 2 cos x − 2 ) x ( cos x − 1 )

and

F ′ ( x ) ( ln cos x ) ′ ′ = csc 2 x x 2 ( 1 − cos x ) 2 Y ( x ) ,

where

(8) Y ( x ) = x 2 sin x − x 2 sin x cos 2 x + 4 sin x cos 2 x − 2 x cos 2 x + 4 x cos x − 4 sin x cos x + 2 cos x sin 3 x − 2 x .

It is well known that

(9) sin x = ∑ k = 0 ∞ ( − 1 ) k x 2 k + 1 ( 2 k + 1 ) ! , x ∈ R .

In [1, p. 43], we find

(10) cos 2 x = 1 − ∑ k = 1 ∞ ( − 1 ) k + 1 2 2 k − 1 ( 2 k ) ! x 2 k , ∣ x ∣ < ∞

and

(11) cos 3 x = 1 4 ∑ k = 0 ∞ ( − 1 ) k 3 2 k + 3 ( 2 k ) ! x 2 k , ∣ x ∣ < ∞ .

Differentiating results in

(12) sin x cos x = 1 2 ∑ k = 0 ∞ ( − 1 ) k 2 2 k + 1 ( 2 k + 1 ) ! x 2 k + 1 , ∣ x ∣ < ∞

and

(13) sin x cos 2 x = 1 12 ∑ k = 0 ∞ ( − 1 ) k 3 2 k + 2 + 3 ( 2 k + 1 ) ! x 2 k + 1 , ∣ x ∣ < ∞ .

Theorem 2.1 in the study [16] states that

(14) sin x x ℓ = 1 + ∑ j = 1 ∞ ( − 1 ) j T ( ℓ + 2 j , ℓ ) ℓ + 2 j ℓ ( 2 x ) 2 j ( 2 j ) !

for ℓ ≥ 0 and x ∈ C , where

(15) T ( n , ℓ ) = 1 ℓ ! ∑ m = 0 ℓ ( − 1 ) m ℓ m ℓ 2 − m n .

Taking n = 4 + 2 j and ℓ = 4 in (16) gives

T ( 4 + 2 j , 4 ) = 1 4 ! ∑ m = 0 4 ( − 1 ) m 4 m ( 2 − m ) 4 + 2 j = 4 j + 1 − 1 3 .

Setting ℓ = 4 in (14) leads to

sin x x 4 = 1 + ∑ j = 1 ∞ ( − 1 ) j T ( 4 + 2 j , 4 ) 4 + 2 j 4 ( 2 x ) 2 j ( 2 j ) ! = ∑ j = 0 ∞ ( − 1 ) j 2 2 j + 3 ( 4 j + 1 − 1 ) ( 2 j + 4 ) ! x 2 j ,

which can be rearranged as:

sin 4 x = ∑ j = 0 ∞ ( − 1 ) j 2 2 j + 3 ( 2 2 j + 2 − 1 ) ( 2 j + 4 ) ! x 2 j + 4 , ∣ x ∣ < ∞ .

Differentiating gives

(16) cos x sin 3 x = ∑ k = 0 ∞ ( − 1 ) k 2 2 k + 1 ( 2 2 k + 2 − 1 ) ( 2 k + 3 ) ! x 2 k + 3 , ∣ x ∣ < ∞ .

Making use of the Maclaurin power series expansions (1), (9), (10), (11), (12), (13), and (16), we can expand the function Y ( x ) as:

Y ( x ) = x 9 2 ∑ k = 0 ∞ ( − 1 ) k + 1 2 4 k + 17 − 2 k 2 + 17 k + 54 3 2 k + 7 + ( k + 6 ) 4 k + 5 + 6 k 2 + 35 k + 34 ( 2 k + 9 ) ! x 2 k ≜ x 9 2 ∑ k = 0 ∞ ( − 1 ) k + 1 Ψ k x 2 k = − x 9 2 ∑ k = 0 ∞ Ψ 2 k + 1 Ψ 2 k Ψ 2 k + 1 − x 2 x 4 k .

By induction, we can verify that

(17) 2 4 k + 17 − 2 k 2 + 17 k + 54 3 2 k + 7 > 0 , k ≥ 0 .

Hence, the sequence

Ψ k = 2 4 k + 17 − 2 k 2 + 17 k + 54 3 2 k + 7 + ( k + 6 ) 4 k + 5 + 6 k 2 + 35 k + 34 ( 2 k + 9 ) ! > 0 , k ≥ 0 .

The inequality

Ψ 2 k Ψ 2 k + 1 > 3 > π 4 2 , k ≥ 0

is equivalent to

(18) Ψ 2 k > 3 Ψ 2 k + 1 , k ≥ 0 ,

i.e.,

( 8 k 2 + 42 k + 31 ) 2 8 k + 18 − ( 128 k 4 + 1216 k 3 + 4384 k 2 + 7142 k + 3969 ) 3 4 k + 7 + ( 8 k 3 + 66 k 2 + 175 k + 144 ) 2 4 k + 12 + 384 k 4 + 3136 k 3 + 8992 k 2 + 10274 k + 3515 > 0

for k ≥ 0 . By induction, we can verify that the sequence

(19) ( 8 k 2 + 42 k + 31 ) 2 8 k + 18 − ( 128 k 4 + 1216 k 3 + 4384 k 2 + 7142 k + 3969 ) 3 4 k + 7 = 16 ( 8 k 2 + 42 k + 31 ) 3 4 k + 7 4 4 k + 7 3 4 k + 7 − 128 k 4 + 1216 k 3 + 4384 k 2 + 7142 k + 3969 16 ( 8 k 2 + 42 k + 31 ) > 0

for k ≥ 1 . This means that Inequality (18) is valid for k ≥ 1 . Moreover, it is easy to see that

Ψ 0 = 19 360 = 0.0527 ⋯ > 3 Ψ 1 = 29 560 = 0.0517 ⋯ .

Hence, Inequality (18) is valid for k ≥ 0 .

Combining the aforementioned results, we conclude that Y ( x ) < 0 on [ 0 , π 2 ] . This means that the derivative ratio F ′ ( x ) ( ln cos x ) ′ is decreasing on ( 0 , π 2 ] . Applying Lemma 2 leads to the decreasing property of the function R ( x ) on ( 0 , π 2 ) . The first proof of Theorem 2 is complete.

Second proof

We start out this proof from considering the function Y ( x ) defined by (8). Straightforward differentiating and expanding give

Y ′ ( x ) = [ ( 6 x 2 + 4 ) cos x + ( 3 x 2 − 4 ) cos ( 2 x ) + 4 cos ( 3 x ) − 4 x sin x + 2 x sin ( 2 x ) + 3 x 2 − 4 ] sin 2 x 2 = 1 2 sin 2 x 2 ∑ k = 3 ∞ ( − 1 ) k 8 × 3 2 k − ( 6 k 2 + k + 8 ) 2 2 k − 8 ( k − 1 ) ( 6 k + 1 ) ( 2 k ) ! x 2 k ≜ 1 2 sin 2 x 2 ∑ k = 3 ∞ ( − 1 ) k W k x 2 k = 1 2 sin 2 x 2 ∑ k = 1 ∞ [ ( − 1 ) 2 k + 1 W 2 k + 1 x 4 k + 2 + ( − 1 ) 2 k + 2 W 2 k + 2 x 4 k + 4 ] = 1 2 sin 2 x 2 ∑ k = 1 ∞ ( W 2 k + 2 x 2 − W 2 k + 1 ) x 4 k + 2 = 1 2 sin 2 x 2 ∑ k = 1 ∞ W 2 k + 2 x 2 − W 2 k + 1 W 2 k + 2 x 4 k + 2

for x ∈ [ 0 , π 2 ] .

By induction, we obtain

(20) 7 × 3 2 k − ( 6 k 2 + k + 8 ) 2 2 k = 7 × 2 2 k 3 2 2 k − 6 k 2 + k + 8 7 > 0

and

(21) 3 2 k − 8 ( k − 1 ) ( 6 k + 1 ) > 0

for k ≥ 3 . This means that W k > 0 for k ≥ 3 .

In order to prove the inequality x 2 − W 2 k + 1 W 2 k + 2 < 0 for k ≥ 1 and x ∈ [ 0 , π 2 ] , it is sufficient to show

(22) W 2 k + 1 W 2 k + 2 > 3 > π 2 2 = 2.467 … , k ≥ 1 .

The inequality W 2 k + 1 > 3 W 2 k + 2 for k ≥ 1 is equivalent to

9 ( 16 k 2 + 28 k − 15 ) 3 4 k − ( 96 k 4 + 272 k 3 + 242 k 2 + 33 k − 57 ) 2 4 k + 1 − ( 384 k 4 + 896 k 3 + 608 k 2 + 54 k − 39 ) > 0 , k ≥ 1 .

By induction, we can verify that

(23) 8 ( 16 k 2 + 28 k − 15 ) 3 4 k − ( 96 k 4 + 272 k 3 + 242 k 2 + 33 k − 57 ) 2 4 k + 1 = ( 16 k 2 + 28 k − 15 ) 2 4 k + 3 3 2 4 k − 96 k 4 + 272 k 3 + 242 k 2 + 33 k − 57 4 ( 16 k 2 + 28 k − 15 ) > 0

and

(24) ( 16 k 2 + 28 k − 15 ) 3 4 k − ( 384 k 4 + 896 k 3 + 608 k 2 + 54 k − 39 ) = ( 16 k 2 + 28 k − 15 ) 3 4 k − 384 k 4 + 896 k 3 + 608 k 2 + 54 k − 39 16 k 2 + 28 k − 15 > 0

for k ≥ 1 . Consequently, Inequality (22) is valid for k ≥ 1 . This implies that the derivative Y ′ ( x ) is negative on ( 0 , π 2 ] and then that the function Y ( x ) is decreasing on [ 0 , π 2 ] . Due to Y ( 0 ) = 0 , the function Y ( x ) is negative on ( 0 , π 2 ] . Then, the derivative ratio F ′ ( x ) ( ln cos x ) ′ is decreasing on ( 0 , π 2 ] . Furthermore, using Lemma 2 leads to the decreasing property of the function R ( x ) on ( 0 , π 2 ) . The second proof of Theorem 2 is complete.

Remark 4

For t ≥ 0 , let

G 1 ( t ) = 2 4 t + 17 3 2 t + 7 − ( 2 t 2 + 17 t + 54 ) > 0 , t ≥ 0 .

It is clear that

G 1 ( 3 ) ( t ) = 1,048,576 2,187 4 3 2 t ln 4 3 3 > 0 , t ≥ 0 .

This means that the second derivative G 1 ″ ( t ) is increasing in t ≥ 0 . From

G 1 ″ ( 0 ) = 524,288 2,187 4 3 2 t ln 4 3 2 − 4 = 15.840 … ,

it follows that G 1 ″ ( t ) > 15 for t ≥ 0 . This means that the first derivative G 1 ′ ( t ) is increasing in t ≥ 0 . Since

G 1 ′ ( 0 ) = 262,144 2,187 ln 4 3 − 17 = 17.482 … ,

we deduce that the first derivative G 1 ′ ( t ) > 17 for t ≥ 0 . Hence, the function G 1 ( t ) is increasing in t ≥ 0 . From the fact that

G 1 ( 0 ) = 2 17 3 7 − 54 = 5.932 … ,

we see that G 1 ( t ) > 5 for t ≥ 0 . Consequently, Inequality (17) is alternatively proved.

Similarly, we can also prove Inequalities (20) and (21) alternatively.

Remark 5

For t ≥ 1 , let

G 2 ( t ) = ( 8 t 2 + 42 t + 31 ) 2 8 t + 18 3 4 t + 7 − ( 128 t 4 + 1,216 t 3 + 4,384 t 2 + 7,142 t + 3,969 ) .

By calculus, we arrive at

G 2 ( 1 ) = 262,144 27 4 3 4 − 16,839 = 13846.351 … , G 2 ′ ( 1 ) = 15,204,352 2,187 4 3 4 + 1,048,576 27 4 3 4 ln 4 3 − 20070 = 37212.729 … , G 2 ″ ( 1 ) = 64 2,187 65,536 4 3 4 + 5,308,416 4 3 4 ln 4 3 2 + 1,900,544 4 3 4 ln 4 3 − 601,425 = 79662.224 … ,

G 2 ( 3 ) ( 1 ) = 128 729 3,538,944 4 3 4 ln 4 3 3 + 1,900,544 4 3 4 ln 4 3 2 + 131,072 4 3 4 ln 4 3 − 59,049 , = 144599.308 … , G 2 ( 4 ) ( 1 ) = 1,024 2,187 5,308,416 4 3 4 ln 4 3 4 + 3,801,088 4 3 4 ln 4 3 3 + 393,216 4 3 4 ln 4 3 2 − 6,561 = 232812.497 … ,

and

G 2 ( 5 ) ( t ) = 2 18 3 7 ∑ k = 0 5 5 k ( 8 t 2 + 42 t + 31 ) ( k ) 4 3 4 t ( 5 − k ) > 0 , t ≥ 1 .

Discussing as done in Remark 4, we conclude that the function G 2 ( t ) is positive in t ≥ 1 . Inequality (19) is thus alternatively proved.

Similarly, we can also prove Inequalities (23) and (24) alternatively.

5 Conclusion

Let f ( x ) be an even, positive, and analytic function on ( − r , r ) such that f ( 0 ) = 1 and f ( 2 m ) ( 0 ) ≠ 0 for m ≥ 1 . Then,

f ( x ) = ∑ k = 0 ∞ f ( 2 k ) ( 0 ) x 2 k ( 2 k ) ! , x ∈ ( − r , r ) .

What are the Maclaurin power series expansions of the logarithmic expressions ln f ( x ) and ln 2 [ f ( x ) − 1 ] f ″ ( 0 ) x 2 ? What is the monotonicity of the even function ln 2 [ f ( x ) − 1 ] f ″ ( 0 ) x 2 ln f ( x ) on the interval ( 0 , r ) ? Generally, what about the properties for the logarithms of the normalized remainders

F n ( x ) = 0 , x = 0 ln ( 2 n ) ! f ( 2 n ) ( 0 ) 1 x 2 n f ( x ) − ∑ k = 0 n − 1 f ( 2 k ) ( 0 ) x 2 k ( 2 k ) ! , x ≠ 0

for n ≥ 0 and their ratios R m , n ( x ) = F n ( x ) F m ( x ) for n > m ≥ 0 ?

In [5], the special case f ( x ) = tan x x was discussed.

In [6], the special case f ( x ) = sin x x was investigated.

In this article, we studied the special case f ( x ) = cos x and R 0 , 1 ( x ) .

In subsequent articles, we will investigate more general cases.

# Dedicated to Professor Dr. Sever Silvestru Dragomir at Victoria University in Australia.

Acknowledgement

The authors are grateful to anonymous referees for their careful corrections, valuable comments, helpful suggestions, and intrinsic observations on the original version of this article.

Funding information: The authors state no funding involved.
Author contributions: All authors contributed equally to the manuscript and read and approved the final manuscript.
Conflict of interest: The authors state no conflicts of interest.
Data availability statement: Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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Received: 2023-05-16

Revised: 2023-11-08

Accepted: 2023-11-19

Published Online: 2023-12-08

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2023-0159

Keywords for this article

Maclaurin power series expansion; decreasing property; logarithmic expression; cosine function; derivative formula; ratio of two differentiable functions; monotonicity rule

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