A Series Expansion of a Logarithmic Expression and a Decreasing Property of the Ratio of Two Logarithmic Expressions Containing Sine

: In the paper, by virtue of a derivative formula for the ratio of two differentiable functions and with the help of a monotonicity rule, the authors expand a logarithmic expression involving the sine 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 sine function. These results are interesting purely in pure mathematics


1.
What is the Maclaurin power series expansion of the even function around x = 0? 2.
Is the even function , |x| ∈ (0, π) 3 10 , decreasing on the close interval [0, π]? It is clear that the logarithmic expression Q(x) defined in (3) is more complicated than the logarithmic expression ln sin x x . Therefore, it is interesting purely in pure mathematics to expand Q(x) into a Maclaurin power series expansion around the origin x = 0 and to compare these two logarithmic expressions by considering the decreasing property of their quotient R(x).
In this paper, after preparing the general formula for higher derivatives of the quotient of two differentiable functions and reciting two monotonicity rules in Section 2, we will give an answer to the first problem by presenting a Maclaurin power series expansion of the even function Q(x) around x = 0 in Section 3, as well as provide an answer to the second problem by verifying the decreasing property of the quotient R(x) in Section 4.

Lemmas
For smoothly solving the above 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( where the matrix for 1 ≤ ≤ 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).
Lemma 3 (Monotonicity rule for the ratio of two functions [7] (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

Maclaurin Power Series Expansion
In this section, we solve the first problem posed in the first section of this paper.
where the matrices A 2n−1,1 , B 2n−1,2n−1 , and C 1,2n−1 for n ≥ 1 are defined by Then the function Q(x) defined by (3) can be expanded into the Maclaurin power series expansion x 4 + 1 756,000 x 6 + 89 3,104,640,000 Proof. The first derivative of Q(x) is for n ≥ 1 and, due to the evenness of Q(x) on R, Q (2n+1) (0) = 0 for n ≥ 0. Consequently, we obtain The required proof is thus complete.

Remark 1.
For n = 3, the determinant E 6 is

Decreasing Property
In this section, we solve the second problem posed in the first section of this paper. Theorem 2. The function R(x) defined by (4) decreasingly maps [0, π] onto 0, 3 10 .
In [1] (p. 43), we find Differentiating results in and sin x cos 2 x = 1 12 Theorem 2.1 in the paper [8] reads that for ≥ 0 and z ∈ C, where Taking n = 4 + 2j and = 4 in (15) gives which can be rearranged as Differentiating gives Making use of the power series expansions (8)-(13) and (16), we can expand the function h(x) into We now consider the functions Then the function h(x) can be expressed as By induction, we can verify that for k ≥ 0. Therefore, we acquire that Θ k > 0 for k ≥ 0. The partial sum The inequality In other words, this inequality can be formulated as From (7), it follows that the derivative ratio, is decreasing on (0, π). Using Lemma 3, we derive that the function R(x) defined by (4) is decreasing on [0, π]. The proof of Theorem 2 is complete.
Second proof. We just state the outline and sketch of this proof because the method and approach are similar to the first proof in this paper. Differentiating gives By induction, we can verify that Therefore, the sequence W k is positive for k ≥ 0. The partial sum For k ≥ 3, the inequality W 2k W 2k+1 > 10 > π 2 is valid. Hence, the derivative h (x) is positive on [0, π]. Then the function h(x) is increasing on [0, π]. Since h(0) = 0, the function h(x) is positive on (0, π]. From the derivative in (7), it follows that the derivative ratio is decreasing on (0, π). Employing Lemma 3, we deduce that the function R(x) defined by (4) is decreasing on [0, π]. The second proof of Theorem 2 is complete.