The monotonicity of ratios involving arc tangent function with applications

Lemma 1

[23] For –∞ ≤ a < b ≤ ∞, let f and g be differentiable functions on (a, b) with f(a⁺) = g(a⁺) = 0 or f(b^–) = g(b^–) = 0. Assume that g^′(x) ≠ 0 for each x in (a, b). If f^′/g^′ is increasing (decreasing) on (a, b) then so is f/g.

Lemma 2

[24] Let a_n and b_n (n = 0, 1, 2, ...) be real numbers and let the power series A(t) = ∑n=1∞ a_ntⁿ and B(t) = ∑n=1∞ b_ntⁿ be convergent for |t| < r. If b_n > 0 for n = 0, 1, 2, ..., and a_n/b_n is strictly increasing (or decreasing) for n = 0, 1, 2, ..., then the function t ↦ A(t)/B(t) is strictly increasing (or decreasing) on (0, r).

Lemma 3

[25] The following expansions

hold for |x| < π, where B_n is the Bernoulli numbers.

Lemma 4

Let

Then φ₂(t) > 0 for t ∈ (0, π /2), and φ₁(t)/φ₂(t) is strictly decreasing from (0, π/2) onto (4/π², 15/32).

Proof

We write φ₂(t) as

Differentiating and expanding in power series by (2.3) and (2.4) yield

where s = 2t ∈ (0, π). Then φ₄(2t) = φ₄(s) > φ₄(0) = 0, which implies φ₂(t) > 0 for t ∈ (0, π/2).

Similarly, φ₁(t) can be written as

where s = 2t. Then φ₁(t)/φ₂(t)

where

Thus, to prove the desired monotonicity, it suffices to prove that φ₃(s)/φ₄(s) is strictly increasing on (0, π). Expanding in power series by (2.4), (2.1) and (2.2) leads to

Taking into account (2.7) and (2.8) we get

By Lemmas 1 and 2, it is enough to show that the sequence {u_n/v_n}_n≥2 is increasing. A direct verification yields

Using the binomial theorem we arrive at

which gives the increasing property of the sequence {u_n/v_n}_n≥2.

An easy calculation gives

which completes the proof.□

Remark 1

Using the methods from [26, 27, 28, 29], one can directly prove the Lemma 4.

Theorem 1

1. If a ≥ 5/8, then the ratio

   R1x=1x1−a+23ax2+a2arctan⁡x

   is strictly increasing from (0, ∞) onto (1, πa/6 ). Therefore, the double inequality

   x1−a+2ax2/3+a2<arctan⁡x<aπ26x1−a+2ax2/3+a2 (3.1)

   holds for x > 0.

2. If 3π2+2−22 /(3π²) = a₁ < a < 5/8, then there is an x₀ > 0 such that R₁ is strictly decreasing on (0, x₀) and increasing on (x₀, ∞). So the double inequality

   R1x0×x1−a+2ax2/3+a2≤arctan⁡x<max1,aπ2/6×x1−a+2ax2/3+a2 (3.2)

   holds for x > 0. In particular, for a = 6/π², we have

   c21x1−6/π2+4x2/π2+36/π4≤arctan⁡x<x1−6/π2+4x2/π2+36/π4, (3.3)

   where c₂₁ = R₁(x₀) = 0.9976914... is the best possible.

3. If 0 < a ≤ a₁ = 3π2+2−22 /(3π²) = 0.598..., then R₁(x) is strictly decreasing from (0, ∞) onto ( πa/6 , 1). Then the double inequality

   aπ2/6×x1−a+2ax2/3+a2arctan⁡x<x1−a+2ax2/3+a2 (3.4)

   holds for x > 0.

Proof

Making a change of variable x = tan t for t ∈ (0, π/2) yields

where

Differentiation yields

where φ₁(t) and φ₂(t) are defined by (2.5) and (2.6), respectively. As shown in Lemma 4, φ₂(t) > 0 for t ∈ (0, π/2), φ₁(t)/φ₂(t) is strictly decreasing from (0, π/2) onto (4/π², 15/32). Then the relation (3.5) can be written as

Therefore, (p^′(t)/q^′(t))^′ > (<) 0 if and only if

while a ∈ (16/(3π²), 5/8), there is a t₁ ∈ (0, π/2) such that (p^′(t)/q^′(t))^′ < 0 for t ∈ (0, t₁) and (p^′(t)/q^′(t))^′ > 0 for t ∈ (t₁, π/2).

On the other hand, we easily get

Now, we distinguish three cases to determine the monotonicity of p/q.

1. a ≥ 5/8. By the relation (1.22), we have Hp,q′ = (p^′/q^′)^′q > 0, and so H_p,q(t) > H_p,q(0⁺) = 0 for t ∈ (0, π/2). The relation (1.21) in combination with q^′(t) > 0 and H_p,q(t) > 0 leads to (p(t)/q(t))^′ > 0 for t ∈ (0, π/2). Hence, for t ∈ (0, π/2),

   1=limt→0ptqt<ptqt<limt→π/2ptqt=aπ26,

   which is equivalent to (3.1). This proves the first assertion of this theorem.

2. 0 < a ≤ 16/(3π²). Likewise, we deduce that (p(t)/q(t))^′ < 0 for t ∈ (0, π/2). So the double inequality (3.4) holds for x > 0.

3. 16/(3π²) < a < 5/8. As shown previously, by the relation (1.22) it is seen that Hp,q′ < 0 for t ∈ (0, t₁) and Hp,q′ > 0 for t ∈ (t₁, π/2). Since H_p,q(0⁺) = 0 and

   Hp,qπ2−=a+263πa−1>0if a1<a<58,≤0if 163 π 2<a≤a1,

   where a₁ = 3π2+2−22 /(3π²) = 0.598..., we find that

4. 16/(3π²) < a ≤ a₁. We have H_p,q(t) < 0 for t ∈ (0, π/2), so is (p(t)/q(t))^′, which implies the double inequality (3.4) holds for x > 0.

   Combining Case 2 and Subcase 3.1 gives the third assertion of this theorem.

5. a₁ < a < 5/8. There is a t₀ ∈ (t₁, π/2) such that H_p,q(t) < 0 for t ∈ (0, t₀) and H_p,q(t) > 0 for t ∈ (t₀, π/2), and so is (p(t)/q(t))^′. That is, the ratio p/q is decreasing on (0, t₀) and increasing on (t₀, π/2). This leads to

   pt0qt0<ptqt<limt→0ptqt=1 for t∈0,t0,pt0qt0<ptqt<limt→π/2ptqt=aπ26 for t∈t0,π2,

   that is,

   pt0qt0≤ptqt<max1,aπ26 for x∈0,∞,

   which is equivalent to the double inequality (4.2), where x₀ = tan t₀. In particular, for a = 6/π², solving the equation H_p,q(t) = 0 gives t₀ = 1.2900104..., then x₀ = tan t₀ = 3.467 341..., so c₂₁ = R₁(x₀) = 0.9976914..., which proves the second assertion of this theorem.

   Thus we completes the proof.□

Corollary 1

The following inequalities

hold for all x > 0. All the bounds are sharp.

Remark 2

It is easy to verify that

that is, a ↦ A_(2,0)(x; a) and a ↦ a A_(2,0)(x; a) are decreasing and increasing on (0, ∞), respectively. Then taking a = 2/3, ∞ in inequalities (3.1) gives

for x > 0; taking a = 1/2, 7/12 in (3.4) yields

for x > 0.

Proposition 1

The double inequality

holds for x > 0 with the best constants

where a₁ = 3π2+2−22 /(3π²) and R₁(x₀) is given in Theorem 1.

Corollary 2

Let a, b > 0. The double inequality

holds for x > 0 if and only if a ≥ 5/8 = 0.625 and 0 < b ≤ 6/π² = 0.607....

Remark 3

Clearly, the lower and upper bounds given in (3.11) for a = 5/8 and b = 6/π², that are, A_(3,0)(x) and A_(2,1)(x), are the sharpest lower and upper bounds of arctan x of the form B_a,b,c(x).

Remark 4

The first inequality of (3.11) for a = 5/8 was first presented in [7], while the second one of (3.11) for b = 6/π² appeared in [9]. It is easy to check that

for all x > 0, the upper bound in (3.11) for b = 6/π² is better than one in (1.4).

Corollary 3

Let a, b > 0. The double inequality

holds for x > 0 if and only if 0 < a ≤ a₁ = 3π2+2−22 /(3π²) and b ≥ 6/π².

Proof

We only prove that the left hand side inequality of (3.12) holds for x > 0 if and only if 0 < a ≤ a₁. The sufficiency follows from Proposition 1 and the increasing property of a ↦ a A_(2,0)(x; a) on (0, ∞). Suppose that the left hand side inequality of (3.12) holds for all x > 0. If a > a₁, then a ≥ 5/8 or a₁ < a < 5/8. If a ≥ 5/8, then by Theorem 1, the second inequality of (3.1) holds for all x > 0, which yields a contradiction with the assumption. If a₁ < a < 5/8, then from Subcase 3.2 we see that

which is equivalent to

where x₀ = tan t₀, which also yields a contradiction with the assumption, and the necessity follows. This completes the proof.□

Remark 5

Corollary 3 offers a new family of lower bounds in the form of λB_a,b,c(x) (λ > 0 with λ ≠ 1) for arctan x. And, a sharp lower bound is

Theorem 2

1. If a ≥ a₀ = 2 3π2+2−22 /π⁴ = 0.363..., then the function

   R2x=1x4π2+4π2x2+aarctan⁡x

   is strictly decreasing from (0, ∞) onto (1, 4/π² + a ). Therefore, the double inequality

   x4/π2+4x2/π2+a<arctan⁡x<4/π2+ax4/π2+4x2/π2+a (4.1)

   holds for x > 0.

2. If 32/π⁴ < a < a₀ = 2 3π2+2−22 /π⁴, then there is an x₀ > 0 such that R₂ is strictly increasing on (0, x₀) and decreasing on (x₀, ∞). So the inequalities

   min1,4/π2+a×x4/π2+4x2/π2+a<arctan⁡x≤R2x0×x4/π2+4x2/π2+a (4.2)

   hold for x > 0, where x₀ is the unique solution of the equation R2′ (x) = 0 on (0, ∞). In particular, for a = (1 – 4/π²)², we have

   x4/π2+4x2/π2+1−4/π22<arctan⁡x≤c12x4/π2+4x2/π2+1−4/π22, (4.3)

   where c₁₂ = R₂(x₀) = 1.0026766... is the best possible.

3. If 0 < a ≤ 32/π⁴ = 0.328..., then R₂(x) is strictly increasing from (0, ∞) onto (4/π² + a , 1). Therefore, the double inequality

   4/π2+ax4/π2+4x2/π2+a<arctan⁡x<x4/π2+4x2/π2+a (4.4)

   holds for x > 0.

Proof

Making a change of variable x = tan t for t ∈ (0, π/2) yields

where

Differentiation yields