An improved Vietoris sine inequality

Available online at www.sciencedirect.com

ScienceDirect Journal of Approximation Theory 189 (2015) 29–42 www.elsevier.com/locate/jat

Full length article

An improved Vietoris sine inequality Man Kam Kwong Department of Applied Mathematics, The Hong Kong Polytechnic University, Hunghom, Hong Kong Received 16 June 2013; received in revised form 15 February 2014; accepted 26 August 2014 Available online 28 September 2014 Communicated by Karlheinz Groechenig

Abstract We prove that if {ak } is a sequence of positive numbers, such that

√ (2 j − 1) j + 1 2j − 1 a2 j+1 ≤ a2 j ≤ a2 j−1 √ 2j 2j j for all j = 1, 2, . . ., then for all n = 1, 2, . . ., x ∈ [0, π], n 

ak sin(kx) ≥ 0.

k=1

  3 , √1 , √ 5 , . . . = {1, 0.5, 0.707, 0.530, 0.577, 0.481, . . .} where An example is {ak } = 1, 21 , √1 , √ 3 6 √ 3 2 4 2 √ ak = 1/ (k + 1)/2 for odd k, and (k − 1)/(k k/2), for even k. This improves the well-known Vietoris sine inequality, by relaxing the requirement that {an } has to be a nonincreasing sequence. The proof is based on a Luk´acs-type inequality and a result on positive trigonometric sums with “convex” coefficients (both established recently by the authors), the classical Sturm Theorem on the number of real roots of a polynomial, and a well-known comparison principle. The symbolic manipulation software MAPLE is used amply for various computations. c 2014 Elsevier Inc. All rights reserved. ⃝

MSC: 26D05; 42A05 Keywords: Trigonometric sums; Positivity; Inequalities

E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jat.2014.08.005 c 2014 Elsevier Inc. All rights reserved. 0021-9045/⃝

30

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

1. Introduction We use bold capital “sums” of  N letters to denote finite or infinite  real numbers or functions. For example, A =  k=1 ak (often abbreviated as ak ) or 2 = ∞ k=1 θk (x). Partial sums are written as A(n) = nk=1 ak . Subscripted letters, such as 21 and 22 , refer to different sums. A is called a P-sum (PS for short) if all of its finite partial sums, A(n) are nonnegative. A sum 8 = φk (x) of real functions defined on an interval I is said to be a PS in I if it is a PS for each fixed x ∈ I . Let N0 be a positive integer. A is called a PS for n > N0 if A(n) ≥ 0 for all n > N0 . Our usage of the term “sum” means more than just the “total” obtained by adding up all the terms (or the “sum” of the series if the infinite sum converges). A “sum” is endowed with a partial sum structure. PS is thus not an intrinsic property of the “total” of the sum, but depends on how the sum is decomposed into its terms. The following is a classical result.  Theorem A (Vietoris [20]). The sum ak sin(kx) is a PS if ak ↘ 0

and

a2 j ≤

2j − 1 a2 j−1 2j

for j = 1, 2, . . . , .

(1.1)

Vietoris’ work had been ignored for many years before it was noticed by Askey (see [5, 7], in which a simplified proof is given). Positive trigonometric polynomials have important applications in various areas, including geometric function theory (Gluchoff and Hartmann [15], Ruscheweyh and Salinas [18]), and signal processing (Dumitrescu [11]). See also Dimitrov and Merlo [10], and Fern´andez-Dur´an [13]. Extensions to orthogonal polynomials play a role in the proof of the Bieberbach conjecture; see [6]. See also [14].  There is an analogous cosine inequality (namely that a1 + ak cos(kx) is also a PS), but we are only concerned with the sine sum in this paper. Many improvements on the cosine inequality are known. For a good historical survey of the subject, see Koumandos [16]. See also Brown [9]. There was little progress in the case of the sine inequality, until 1995 when Belov [8] gave a necessary and sufficient condition for PS, under the monotonicity requirement.  Theorem B (Belov [8]). Assume ak ↘ 0. Then ak sin(kx) is a PS iff n  (−1)k−1 kak ≥ 0

(1.2)

k=1

for all n ≥ 2. (For the cosine analog, (1.2) is a sufficient but not necessary condition.) Alzer et al. [1] refine the Vietoris sine inequality by obtaining a fourth-degree polynomial lower bound. For further progress in the subject area, please refer to [16]. For some recent, related work by Alzer and the author, see [2,3]. All existing examples of Vietoris-type PS sine and cosine polynomials have nonincreasing coefficients. Vietoris’ Theorem is a corollary of Belov’s Theorem. Indeed, the latter leaves no more room for further improvement, unless the monotonicity assumption on ak is lifted. In this less restrictive situation, (1.2) is no longer sufficient for PS; in fact, no useful sufficient conditions are known except some that are trivially implied by the Vietoris–Belov result. The main purpose of this paper is to extend the Vietoris sine inequality to include PS with coefficients that are not monotonically decreasing. A significant contribution of the current work is a new approach in establishing the desired inequality. The new approach appears to have more room for improvement and can be further exploited to obtain better inequalities.

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

31



 3 5 Let {γk } = 1, 21 , √1 , √ , √1 , √ , . . . = {1, 0.5, 0.707, 0.530, 0.577, 0.481, . . .} be the 3 6 3 2 4 2 sequence of positive numbers defined by  1   k is odd √ (k + 1)/2 γk = (1.3) k−1   k is even.  √ k k/2 Our main result is  Theorem 1. (i) 8 = γk sin(kx) is a PS in [0, π]. (ii) A sum 9 = ak sin(kx) is a PS in [0, π] if, for j = 1, 2, . . . , √ (2 j − 1) j + 1 2j − 1 a2 j−1 . a2 j+1 ≤ a2 j ≤ √ 2j 2j j

(1.4)

The second conclusion follows from the first by applying the Comparison Principle (Lemma 2 of Section 3) to Φ and Ψ . Theorem 1 includes the classical Vietoris result because the first inequality in condition (1.4) is automatically satisfied when ak are nonincreasing. Our proof of Theorem 1 does not depend on knowing an explicit closed form of the particular sum in question, as does the classical proof of the Vietoris result (see [5]). On the other hand, Theorem 1 is independent of Belov’s Theorem, the proof of which strongly depends on the monotonicity of the coefficients. Belov’s Theorem cannot be applied to show that 8 is a PS. Yet there are sums that satisfy the Belov condition but not (1.4). It is an interesting question to ask if a result can be established to subsume both Belov’s result and Theorem 1. By applying the reflection x → (π − x) to the sum, we see that Theorem 1 is equivalent to asserting that  2= (−1)k+1 γk sin(kx) (1.5) is a PS in [0, π]. In fact, in the proof of Theorem 1, we mostly focus on 2 instead of 8. Define ψ j (x) = sin((2 j − 1)x) −

2j − 1 sin(2 j x). 2j

(1.6)

Then 2(n) has the representation   ψ2 (x) ψn˜ (x) sin(nx) 2(n) = ψ1 (x) + √ + · · · + √ + √ , n˜ + 1 2 n˜

(1.7)

where n˜ = ⌊n/2⌋ denote the largest integer less than or equal to n/2, and the notation [·] means that the term is added only if n is an odd integer. 2. Sturm’s Theorem and trigonometric polynomials Given a specific sine polynomial, there is an algorithm to determine in a finite number of steps, whether the sum is positive or not in a given interval. (In particular, we are mostly concerned with the interval (0, π ) in this paper.) The idea is to expand each sin(kx) into a product of sin(x) and a polynomial in y = cos(x). The sine polynomial is thus positive in (0, π ) if and only if the corresponding algebraic polynomial in y is nonnegative in (−1, 1), or, if and only if the polynomial has no real roots in (−1, 1) and is positive at some point in the interval.

32

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

A well-known result of Sturm (see, for example, [19, Chapter IX, Section 68]) furnishes a procedure to find the number of real roots of a given polynomial equation p(y) = cn y n + cn−1 y n−1 + · · · + c1 y + c0 = 0

(2.1)

in a given interval of real numbers [α, β]. We can assume, without loss of generality, that p(y) does not vanish at the endpoints, α and β. If it does, then it has a factor of the form (y − α) or (y − β) and we can study the quotient of p(y) divided by the factor instead. One constructs, starting with p(y) and its derivative p ′ (y), a sequence of polynomials using a form of Euclid’s algorithm. By counting the number of sign changes of the values of these polynomials at the two endpoints of the interval, the exact number of real roots of p(y) in [α, β] can be determined. This procedure of using Sturm’s Theorem to prove the positivity of a specific sine polynomial is rigorous. All the computation involved can be carried out by pen and paper if desired. However, in practice, the computations involved in Sturm’s algorithm are quite timeconsuming and error-prone. Fortunately, most symbolic manipulation software provide commands to automate the procedure. In our research, we have used the software MAPLE 13, which provides the following useful commands: “expand” to expand sine polynomials, “sturmseq” to compute the Sturm sequence of the polynomial under study, and “sturm” to count the number of roots in a given interval. We have written a simple MAPLE procedure to automate the Sturm procedure to affirm the positivity of 8(n), for n = 3, . . . , 30. The proof of Lemma 3 in the next section is also established using the Sturm procedure. The details of computer programs are omitted in this article. We refer the readers to [17] for a more indepth explanation of the Sturm procedure and the above-mentioned computer program. See also the related article [4]. 3. Preliminary lemmas We collect some preliminary lemmas in this section before giving the proof of the main result in the next section. The first lemma is well-known, especially to those who are familiar with the Vietoris inequalities. We include it for completeness and ease of reference.  Lemma ak be a given PS and {βk } be a nonincreasing sequence of positive numbers.  1. Let Then βk ak is also a PS. Proof. This is a simple consequence of the summation by parts formula. Alternatively, it can be proved using induction, noting that n  k=1

βk a k =

n−1 n   (βk − βn )ak + βn ak .  k=1

k=1

An immediate corollary of Lemma 1 is Lemma 2 (Comparison Principle). Given a PS another sequence of coefficients {bk }, such that



ak φk (x), ak ̸= 0 in some interval I , and

bk ↘ 0. ak ∞ Then k=1 bk φk (x) is also a PS in I . Note that condition (3.1) requires that bk and ak have the same sign for all k.

(3.1)

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

Lemma 3. The functions  4   1 1 θ4 (x) := √ − √ ψ j (x) k 5 j=1   15  1 1 ψ j (x) θ15 (x) := √ − 4 k j=1

33

(3.2) (3.3)

are increasing in the intervals I3 = [9π/64, π/2], and I4 = [π/2, 3π/4], respectively. Consequently, θ4 (x) ≥ θ4 (9π/64) > 0.27, x ∈ I3 . θ15 (x) ≥ θ15 (π/2) > 0.48, x ∈ I4 .

(3.4) (3.5)

Proof. The notations I3 and I4 are explained in the next section. Numerical evidence clearly supports the conclusions. Both θ4 (x) and θ(15) are finite sine polynomials and the Sturm procedure can be used to give a rigorous proof of the desired result.  The next lemma is due to Fej´er [12]. Lemma 4. Let b1 , . . . , bn be nonnegative real numbers satisfying 2bk ≤ bk−1 + bk+1

for k = 2, . . . , n − 1.

(i) If 2bn ≤ bn−1 , then, for x ∈ [0, π], n  bk sin(kx) ≥ 0. k=1

(ii) If bn ≤ bn−1 , then, for x ∈ [0, π], n−1 

bk sin(kx) +

k=1

1 bn sin(nx) ≥ 0. 2

Weaker versions of the following two lemmas are obtained in [3]. The proofs are almost the same. The first lemma is a Luk´acs-type inequality. Lemma 5. For any 0 < a ≤ 2n + 6, n = 1, 2, . . . , x ∈ [0, π/2], n  (n − k + a) sin(kx) ≥ 0.

(3.6)

k=1

Proof. We rewrite the left hand side of (3.6) as     n n   (n − k + 1) sin(kx) + (a − 1) sin(kx) . k=1

(3.7)

k=1

Suppose we are able to show that (3.6) is true for the particular choice a = 2n+6, or equivalently,     n n   h(x) = (n − k + 1) sin(kx) + (2n + 5) sin(kx) ≥ 0. (3.8) k=1

k=1

Consider this as a sum of two terms. The first term is the classical Luk´acs sum and is known to be positive. This two-term sum is, therefore, a PS. Since a − 1 ≤ 2n + 5, we can use Lemma 2 to conclude that (3.7) is also a PS and (3.6) is proved. Hence, it remains to prove (3.8).

34

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

The simple Sturm procedure can be used to verify (3.8) for small n. In the rest of the proof, we assume that n ≥ 13. In [0, π/n], sin(kx) ≥ 0 for all k ≤ n. Hence, all terms in h(x) are nonnegative, and (3.8) is trivially true. We only have to worry about x ∈ [π/n, π/2]. We assume that the standard technique of summing the trigonometric polynomials    sin(kx), cos(kx), k sin(kx),   is known. One method is to consider them as the real and imaginary parts of ekx and kekx . Using these facts, we obtain h(x) =

(3n + 6) sin(x) + (2n + 5) sin(nx) − (2n + 6) sin((n + 1)x)

4 sin2 (x/2) Thus, it suffices to prove that, for x ∈ [π/n, π/2], k(x) := sin(x) +

.

2n + 5 2n + 6 sin(nx) − sin((n + 1)x) ≥ 0. 3n + 6 3n + 6

For x ∈ [0, π/2], sin(nx) − sin((n + 1)x) = 2 sin

x  2

 cos

(2n + 1)x 2

 ≥ −2 sin

x  2

.

Hence,  sin((n + 1)x) 2n + 5  sin(nx) − sin(n + 1)x − 3n + 6 3n + 6 x  2(2n + 5) 1 ≥ sin(x) − sin − 3n + 6 2 3n + 6 := q(x).

k(x) = sin(x) +

(3.9)

It is easy to verify that, for x ∈ [0, π/2],   x  x  (2n + 5) − 2 cos q ′′ (x) = sin 2  2(3n + 6)  2  √ (2n + 5) x ≤ − 2 sin 2(3n + 6) 2 ≤ 0. Hence, q(x) is concave and in [π/n, π/2], it attains its minimum at one of the two endpoints. At the right endpoint (since n ≥ 13),  π  (3 − 2√2)n − 5(√2 − 1) = ≥ 0. q 2 3n + 6 At the left endpoint, we make use of the inequalities y − y 3 /6 ≤ sin(y) ≤ y to get π  π 1  π 3 2n + 5  π  1 q ≥ − − − n n 6 n 3n + 6 n 3n + 6 2(π − 1)n 3 + 2π n 2 − π 3 n − 2π 3 = 6(n + 2)n 3 > 0. Hence, in [π/n, π/2], q(x) ≥ 0, and by (3.9), we also get k(x) ≥ 0.



By comparison, in the classical Luk´acs result, the condition on a is much stronger, namely, 0 < a ≤ 1, but the inequality (3.6) holds on the wider interval [0, π].

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

35

Lemma 6. Let {ak : k = 1, . . . , n} be a sequence of positive numbers satisfying 2ak ≤ ak−1 + ak+1

for k = 2, . . . , n − 1

(3.10)

and 2n + 6 an−1 . 2n + 7 Then, for x ∈ [0, π/2],

(3.11)

an ≤

n 

ak sin(kx) ≥ 0.

(3.12)

k=1

Proof. We set λ = an−1 − an . Condition (3.11) implies that λ > 0. If Lemma 6 is true under the assumption that λ = 1, then it is true for all positive λ. Indeed, we only have to consider the sequence ak∗ = ak /λ (k = 1, . . . , n). Hence, we may assume that an−1 = an + 1. Substituting this into (3.11) yields an ≤ 2n + 6.

(3.13)

Let x ∈ (0, π/2]. We obtain n 

ak sin(kx) =

k=1

n 

(n − k + an ) sin(kx) +

k=1

n−2 

bk sin(kx),

(3.14)

k=1

where bk = ak − (n − k + an ) for k = 1, . . . , n − 2. In view of (3.13) we conclude from Lemma 4 that the first sum on the right hand side of (3.14) is positive. Since bk−1 + bk+1 − 2bk = ak−1 + ak+1 − 2ak ≥ 0 for k = 2, . . . , n − 3, and bn−3 − 2bn−2 = an−3 + an−1 − 2an−2 ≥ 0, it follows from Lemma 4 that also the second sum on the right hand side of (3.14) is nonnegative. Thus, (3.12) holds.  S1 =

 2k − 1 √ sin(kx) k k

  3 is not a PS in [0, π], because the second partial sum S1 (2) = sin(x) + √ sin(2x) is negative 2 2 in a left neighborhood of π . It is also not a PS in [0, π/2] because the partial sum S1 (4) is negative in a neighborhood of π/2. However, we can use Lemma 6 to show that Lemma 7. S1 (n) is a PS in [0, π/2] for n > 4. Proof. Lemma 6 cannot be applied directly to the sum S1 because the condition (3.10) is satisfied only for k > 2, but not for k = 2. Furthermore, (3.11) is satisfied for n > 2, but not for n = 2. For n = 5, 6, 7, 8 and 9, the conclusion of the lemma can be verified using the Sturm procedure. For the remainder of the proof, we assume that n ≥ 10.

36

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

Let bk denote the coefficients of S1 . We construct a new sum S2 = bk for k ≥ 9, and



ak sin(kx), where ak =

ak = (b9 − b10 )(10 − k) + b10 for k = 1, . . . , 8. In other words, we are replacing the first eight coefficients with new numbers so that the points (k, ak ), k = 1, . . . , 10 are collinear. It is easy to verify that conditions (3.10), and (for n ≥ 10) (3.11) are satisfied by ak . Thus, Lemma 6 applies to give S2 (n) ≥ 0, for n ≥ 10. Let s(x) = S1 (n) − S2 (n) = S1 (8) − S2 (8) 8  = (ak − bk ) sin(kx).

(3.15)

k=1

Again, the Sturm procedure can be used to show that s(x) ≥ 0 in [0, π/2]. Thus S1 (n) = s(x) + S2 (n) ≥ 0, for n ≥ 10, in [0, π/2], as desired.  4. Proof of the main result In this section, we assume n ≥ 31. For a fixed n, we decompose [0, π] into five subintervals: 5 

Ii = [0, π/n] ∪ [π/n, 9π/64] ∪ [9π/64, π/2] ∪ [π/2, 3π/4] ∪ [3π/4, π].

i=1

We are going to show that 2 is a PS in each of them. 4.1. Interval I1 = [0, π/n] For any j ≤ n, ˜ ψ ′j (x) = (2 j − 1)(cos((2 j − 1)x) − cos(2 j x)) = 2(2 j − 1) sin(x/2) sin((4 j − 1)x/2) ≥0

(4.1) √ in [0, π/2 j] ⊃ I1 . Thus ψ j (x) ≥ 0 in I1 . When n is odd, the extra term sin(nx)/ n˜ + 1 is also nonnegative in I1 . Since every term on the right hand side of (1.7) is nonnegative, so is 2(n). 4.2. Interval I2 = [π/n, 9π/64] Let us first consider some simpler sums. The partial sums of 21 = sin(x) − sin(2x) + sin(3x) − sin(4x) + · · · have the explicit form 21 (n) =

sin(x/2) − (−1)n sin((2n + 1)x/2) , 2 cos(x/2)

from which we deduce 21 (n) ≥

sin(x/2) − 1 . 2 cos(x/2)

Note that the expression on the right hand side is an increasing function of x ∈ [0, π].

(4.2)

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

The Fej´er–Jackson–Grownwall sum by 2x, we see that



37

sin(kx)/k is a well-known PS in [0, π]. Replacing x

sin(2x) sin(4x) sin(6x) sin(8x) sin(2nx) ˜ + + + + ··· + 2 4 6 8 2n˜  x n˜ cos(2ks) ds =

22 (n) ˜ =

0 k=1 x sin((2n˜

 =

0

+ 1)s) x ds − 2 sin(s) 2

is a PS in [0, π/2] (but not in [π/2, π]). Define 23 (n) = 21 (n) + 22 (n) ˜ = ψ1 (x) + ψ2 (x) + · · · + ψ⌊n/2⌋ (x) + [sin(nx)] .

(4.3)

We claim that 23 is a PS in I2 . Let us first consider the subinterval J1 = [π/n, π/8]. From (4.2), we have 1 21 (n) ≥ − . 2 Hence, we are done if we can show that, for x ∈ J1 , 1 . 2 In view of (4.3), this is implied by  x sin(ns) f (x) := ds ≥ 1 + x, 0 sin(s) 22 (n) ˜ ≥

(4.4)

(4.5)

(4.6)

for x ∈ [π/(n + 1), π/8], which is in turn implied by the stronger estimate f (x) ≥ 1 +

π ≈ 1.392699. 8

(4.7)

Note that we only need to define f (x) for x ∈ [0, π/2]. Since f ′ (x) = sin(nx)/ sin(x), the critical points of f (x) in [0, π/2] are π/n, 2π/n, 3π/n, . . . . The even-order points x1 = 2π/n, x2 = 4π/n, . . . are local minima and the odd-order ones are local maxima. A lower bound for f (x) in [π/(n + 1), ∞) is, therefore, f (x) ≥ inf { f (π/n + 1), f (x1 ), f (x2 ), . . .} . At the left endpoint    π/(n+1) π sin(ns) f = ds n+1 sin(s) 0  π/(n+1) sin(ns) > ds s 0   nπ = Si n+1

(4.8)

38

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

> Si



31π 32



> 1.85.

(4.9)

Using integration by parts, we see that  xk+1 sin(ns) f (xk+1 ) − f (xk ) = ds sin(s) xk  xk+1 (1 − cos(ns)) cos(s) ds =− n sin2 (s) xk < 0.

(4.10)

Hence, f (x1 ) < f (x2 ) < f (x3 ) < · · ·. Let us estimate f (x1 ). In [π/n, 2π/n], sin(ns) ≤ 0 and s/ sin(s) ≤ (π/8)/ sin(π/8). Hence,     2π/n   2π/n  sin(ns) sin(ns) s ds = ds sin(s) s sin(s) π/n π/n    2π/n   π/8 sin(ns) ≥ ds sin(π/8) π/n s   π/8 =− (Si(π ) − Si(2π )). sin(π/8) It follows that π/n

2π/n



  sin(ns) s ds s sin(s) π/n 0   π/8 > Si(π ) − (Si(π ) − Si(2π )) sin(π/8)

f (x1 ) =



sin(ns) ds + sin(s)



> 1.40679.

(4.11)

The desired assertion (4.7) and, hence, the claim now follow from (4.8)–(4.11). For the remaining subinterval J2 = I2 \ J1 = [π/8, 9π/64], (4.4) can be improved to 21 (n) ≥

sin(π/16) − 1 ≥ −0.42. 2 cos(π/16)

The inequality (4.5) we need to establish the claim is now relaxed by replacing the right hand side with 0.42. The analogous inequality (4.6) becomes f (x) ≥ 0.84 + x, which is implied by f (x) ≥ 0.84 +

9π ≈ 1.28179, 64

which is verified by (4.10) and (4.11). This completes the proof of the claim. Now that we know 23 is a PS in I2 , an application of Lemma 2, comparing 23 with 2, shows that the latter is also a PS in I2 .

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

39

4.3. Interval I3 = [9π/64, π/2] In I3 , 23 is no longer a PS and we need a different comparison PS. In I3 , (4.2) gives the lower bound 21 (n) ≥

sin(π/8) − 1 > −0.334090. 2 cos(π/8)

(4.12)

Since 22 is a PS, it follows from (4.3) that 23 (n) ≥ 21 (n) > −0.334090.

(4.13)

We construct a new sum 1 24 (n) = θ4 (x) + √ 23 (n), 5 where θ4 (x) is defined by (3.2). From (4.13) and (3.4), we have 24 (n) ≥ θ4 (π/4) −

0.334090 > 0, √ 5

implying that 24 , after being rewritten in the form   ψ⌊n/2⌋ (x) ψ2 (x) ψ3 (x) sin(nx) 24 (n) = ψ1 (x) + √ + √ · · · + + , √ √ 3 5 5 5

(4.14)

is a PS in I3 for n > 8. Then we compare 2 (in the form of (1.7)) with 24 (in the form of (4.14)) to conclude that 2 is a PS in I3 . 4.4. Interval I4 = [π/2, 3π/4] In I4 , (4.2) gives 21 (n) ≥

sin(π/4) − 1 > −0.207107, 2 cos(π/4)

which is a better estimate than (4.12). On the other hand, 22 is no longer a PS in I4 . Hence, we need to find a lower bound of 22 (n) before we can deduce a lower bound for 23 (n). By definition, 22 (n) is an odd function with respect to the point x = π/2. If M is an upper bound of 22 (n) in [π/4, π/2], then −M will serve as a lower bound of 22 (n) in I4 . We again make use of the alternative representation of 22 ; see (4.3), and (4.6). For x ∈ [π/4, π/2], x π ≤ f (x) − . (4.15) 2 8 Arguing as in Subsection 4.2, we see that the points x¯1 = π/n, x¯2 = 3π/n, . . . are local maxima of f (x) and that f (x¯1 ) > f (x¯2 ) > · · ·. Hence, f (x¯1 ) = f (π/n) is the absolute maximum of f (x) in [0, ∞). In particular, it is an upper bound of f (x) in [π/4, π/2]. We estimate f (π/n) as follows.  π   π/n  sin(ns)   s  f = ds n s sin(s) 0 π ≤ Si(π ) n sin(π/n) 22 (n) = f (x) −

40

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

π Si(π ) 31 sin(π/31) ≤ 1.86. ≤

(4.16)

It follows from (4.15) and (4.16) that, in [π/4, π/2], π ≤ 1.47. (4.17) 8 Hence a lower bound of 22 (n) in I4 is −1.47, yielding the following lower bound of 23 (n) in I4 22 (n) ≤ 1.86 −

23 (n) ≥ −1.47 − 0.207107 > −1.68. Next, we construct the sum 1 23 (n). 4 The estimates (4.17) and (3.5) give 26 (n) = θ15 (x) +

1.68 > 0. 4 It follows that 26 , after being rewritten in the appropriate way, is a PS in I4 for n > 30. Using the comparing principle as in the previous subsection, we conclude that 2 is a PS in I4 . 26 (n) ≥ 0.48 −

4.5. Interval I5 = [3π/4, π] Lemma 8. If 2 is a PS in [0, π/4], then it is also a PS in I5 . Proof. Note that  2k − 1 8=2+ √ sin(ky), k k where y = 2x, y ∈ [0, π/2]. By Lemma 7, the last sum is a PS for y ∈ [0, π/2] (or for x in [0, π/4]). By assumption, Θ is a PS in [0, π/4]. Hence, 8 being the sum of two PS in [0, π/4] is also a PS in [0, π/4]. The reflection relation between 8 and 2 then implies that 2 is a PS in I5 .  Upon the verification of Lemma 8, the proof of Theorem 1 is now completed. 5. Remarks and examples Example 1. Let 0 < α < β be constants. Then   ψ1 (x) ψ2 (x) ψn˜ (x) sin(nx) +√ + ··· +  +  √ β −α 2β − α nβ ˜ −α (n˜ + 1)β − α is a PS. Some concrete examples are   sin(nx) ψ2 (x) ψ3 (x) ψn˜ (x) ψ1 (x) + √ + √ · · · + √ + √ n 2n˜ − 1 3 5 and   ψ2 (x) ψ3 (x) ψn˜ (x) sin(nx) + √ ψ1 (x) + √ + √ · · · + √ . (3n − 1)/2 3n˜ − 2 7 4

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

41

Remark 2. One of our main goals in this paper is to obtain a first example of Vietoris-type sine PS with non-monotone coefficients. There is no attempt to push for the most general result. For instance, numerics suggests that the sum   ψ1 (x) ψ2 (x) ψn˜ (x) sin(nx) √ + √ + ··· + √ + √ n˜ + 1 3 2 n˜ is a PS. If this is true, then the comparison principle yields Theorem 1 as a corollary. Numerics also indicates that   sin(nx) ψ2 (x) ψn˜ (x) ψ1 (x) + + ··· + + 3α (2n˜ − 1)α nα is a PS for α > 1/8, a result that also implies Theorem 1. Remark 3. Theorem 1 relaxes the first condition, in (1.1), of the Vietoris result. It is natural to ask whether the second condition in (1.1) can also be relaxed by replacing some of the factors ρ j = 2 2j−1 j with larger constants. It is easy to see that the answer is no. Suppose we replace one of them with a larger ratio ρ¯m > ρm . Let 9 be a sum with coefficients ak such that a2 j = ρ j a2 j−1 for j = 1, . . . , m − 1 and a2m = ρ¯m a2m−1 . Then the partial sum 9(2m) takes negative values for x in a left neighborhood of π because the derivative of 9(2m) at x = π is negative. Remark 4. a0 +

n 

ak cos(kx)

k=1

is a PS if the condition (1.1) is satisfied. A very natural question is whether our Theorem 1 has a cosine counterpart, namely, whether γk cos(kx) is a PS, if γ0 = 1 and γk is given by (1.3) for k = 1, 2, . . . . The answer is also no. For x = π , the cosine series becomes γ2 − γ3 + γ4 − γ5 + · · · and every partial sum with an even number of terms is negative, because γ2 < γ3 , γ4 < γ5 , etc. Acknowledgments The author is thankful to Horst Alzer for many inspiring discussions on the subject of inequalities, especially trigonometric inequalities. He is also thankful to the referees for many constructive suggestions that greatly improved the presentation of the paper. The research of the author is supported by the Hong Kong Government GRF Grant PolyU 5003/12P and the Hong Kong Polytechnic University Grants G-YK49 and G-U751. References [1] H. Alzer, S. Koumandos, M. Lamprecht, A refinement of Vietoris’ inequality for sine polynomials, Math. Nachr. 283 (2010) 1549–1557. [2] H. Alzer, Man Kam Kwong, On two trigonometric inequalities of Tur´an, Preprint. [3] H. Alzer, Man Kam Kwong, Rogosinski-Szeg¨o type inequalities for trigonometric sums, J. Approx. Theory (2014) http://dx.doi.org/10.1016/j.jat.2014.04.007. in press. [4] H. Alzer, Kwong Man Kam, Sturm theorem and a refinement of Vietoris inequality for cosine polynomials, 2014. arXiv:1406.069 [math CA].

42

M.K. Kwong / Journal of Approximation Theory 189 (2015) 29–42

[5] R. Askey, Orthogonal Polynomials and Special Functions, in: Reg. Conf. Ser. Appl. Math., vol. 21, SIAM, Philadelphia, PA, 1975. [6] R. Askey, G. Gasper, Inequalities for polynomials, in: A. Baernstein II, D. Drasin, P. Duren, A. Marden (Eds.), The Bieberbach Conjecture, in: Math. Surveys and Monographs, vol. 21, Amer. Math. Soc., Providence, RI, 1986, pp. 7–32. [7] R. Askey, J. Steinig, Some positive trigonometric sums, Trans. Amer. Math. Soc. 187 (1974) 295–307. [8] A.S. Belov, Examples of trigonometric series with nonnegative partial sums, Math. USSR Sb. 186 (1995) 21–46. (in Russian) 186, 485–510 (1995) (English translation). [9] G. Brown, Positivity and boundedness of trigonometric sums, Anal. Theory Appl. 23 (2007) 380–388. [10] D.K. Dimitrov, C.A. Merlo, Nonnegative trigonometric polynomials, Constr. Approx. 18 (2002) 117–143. [11] B.A. Dumitrescu, Positive Trigonometric Polynomials and Signal Processing Applications, Springer, Berlin, 2007. [12] L. Fej´er, Einige S¨atze, die sich auf das Vorzeichen einer ganzen rationalen Funktion beziehen, Monatsh. Math. Phys. 35 (1928) 305–344. [13] J.J. Fern´andez-Dur´an, Circular distributions based on nonnegative trigonometric sums, Biometrics 60 (2004) 499–503. [14] G. Gasper, Nonnegative sums of cosine, ultraspherical and Jacobi polynomials, J. Math. Anal. Appl. 26 (1969) 60–68. [15] A. Gluchoff, F. Hartmann, Univalent polynomials and non-negative trigonometrics sums, Amer. Math. Monthly 195 (1998) 508–522. [16] S. Koumandos, Inequalities for trigonometric sums, in: Nonlinear Analysis, in: Springer Optimization and its Applications, vol. 68, 2012, pp. 387–416. [17] M.K. Kwong, Nonnegative trigonometric polynomials, Sturm’s theorem, and symbolic computation, 2014. arXiv:1402.6778 [math CA]. [18] S. Ruscheweyh, L. Salinas, Stable functions and Vietoris’ theorem, J. Math. Anal. Appl. 291 (2004) 596–604. [19] B.L. van der Waerden, Algebra I, Springer Verlag, Berlin, 1971. ¨ [20] L. Vietoris, Uber das Vorzeichen gewisser Trigonometrische Summen, S.-B. Osterreich. Akad. Wiss. 167 (1958) 125–135. Teil II: Anz. Osterreich. Akad. Wiss., (1959), 192–193.