# On the Study of Trigonometric Polynomials Using Strum Sequence

Authors: Saiful R. Mondal; Kottakkaran Sooppy Nisar; Thabet Abdeljawad
Journal: Journal of Mathematics
 (2020)

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## Abstract

This article constructs trigonometric polynomials of the sine and cosine whose sums are nonnegative. As an application, those nonnegative trigonometric sums are used to study the geometric properties of complex polynomials in the unit disk. The Strum sequences are used to prove the main outcome.


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## Body

## 1. Introduction

The trigonometric polynomials with nonnegative sums and its enormous application in various branches of mathematics and science are well known. There are several theories in the literature to find necessary (or/and sufficient) conditions by which(1)Cθ=∑n=0mancosnθ,Sθ=∑n=1mansinnθ,θ∈0,π,are nonnegative or positive. This is a century old problem and many mathematicians [1–10] have contributed numerous results. The geometric function theory (univalent functions) has a very close association with the positive trigonometric sums. Both areas have taken and given to each other as evident in the work [11–22]. Here, we intend to present few more results of this interplay.An analytic functionf is univalent in the unit disk D=z∈ℂ:z<1 if for z1≠z2 implies fz1≠fz2. Let S be the class of all univalent functions with normalization f0=0=f′0−1 and have the series form(2)fz=z+∑n=2∞anzn.Denote byA, the class of all analytic functions of form (2). A domain D∈ℂ is said to be starlike with respect to a point z0∈D if the line segment joining any arbitrary chosen points in D with z0 lies completely in D. The domain D will be convex if it is star-like with respect to all of its points. Analytically, these classes are characterized as(3)f∈S∗⟺Rezf′zfz>0,f∈C⟺Re1+zf″zf′z>0,where S∗ and C, respectively, denote the class of star-like and convex functions.The aim of this article is to construct the star-like polynomials. It is known that partial sum of a univalent function inD need not be univalent. For example, the Koebe function kz=z1−z−2 is an extremal function of many classes in the univalent functions theory defined on D. However, it can be observed that S2k,z=z+2z2, the 2nd partial sum of the Taylor series for kz is univalent only on the disk z:z<1/4. Thus, the geometric properties of an analytic function do not inherit to its partial sum. These motivate us to construct the polynomial which are star-like on D.The following result related to the starlikeness off is required in sequel.Lemma 1.
(see [23]). If f′ is typically real in D, that is, ImzImf′z>0 for z∈D and satisfies Ref′z>0, z∈D, then f is star-like in D.It is clear from Lemma1 that the positivity of cosine and sine sums has a significant role to study the star-like polynomials. During this investigation, we observed that, along with other conditions, the monotonicity of an is pivotal for the positivity or nonnegativity for the sine and cosine sums. This makes us think about what will be the case if we chose ak randomly (especially nonmonotone)? How we can prove the positivity of those trigonometric sums with arbitrary coefficients? Can those positive sums have an application in the study of geometric function theory?Here, we will provide some affirmative answers for the above questions. The work is motivated from the recent work by Kwong [24], where several trigonometric polynomials with positive sums are given.


## 2. Statements of Main Results and Their Consequences

Now, we will state our main results. The first result will give the positivity of cosine and sine sum with nonmonotonic coefficients.Theorem 1.
Forθ∈0,π, the cosine sum(4)C1θ=1+cosθ2+cos2θ6+cos3θ4+cos4θ9+cos5θ8+cos6θ7,and the sine sum(5)S1θ=sinθ2+sin2θ6+sin3θ4+sin4θ9+sin5θ8+sin6θ7,are positive.The basic identitycos2π−θ=cosθ implies that C1θ>0 for all θ∈0,2π.In the sense of complex function theory, the above two positive trigonometric sums can be described as follows. On the boundary of the unit circlez=1, that is, for z=eiθ,0≤θ≤2π, Refeiθ>0; while Imfeiθ>0 when θ∈0,π. Now, by reflection principal, it follows that Imfeiθ<0 when θ∈π,2π. Here,(6)fz╔1+z2+z26+z34+z49+z58+z67.Denotegz:=∫0zftdt. Then, clearly Reg′z=Refreiθ and Img′z=Imfreiθ, 0≤r<1. Since Refeiθ>0, the minimum principal for harmonic functions yield Refeiθ>0; similarly, Img′z>0 when Imz>0.The above fact together with Lemma1 gives the following result.Theorem 2.
The polynomial(7)gz=z+z24+z318+z416+z545+z648+z749,is star-like with respect to the origin in D.Theorem1 can also be validated from Figures 1(a) and 1(b), while starlikeness of g can be seen in Figure 2.Now, by the same argument as in Theorem1, several other examples can be constructed by suitable positive cosine and sine sums. Now, we state our second theorem.Figure 1
Image of (a)C1θ and (b) S1θ.
(a)(b)Figure 2
Image ofD under the mapping g.Theorem 3.
The polynomial(8)g1z=z+8z231+12400z3+5332−45z4,maps the unit disk D to a star-like domain.Our third result can be stated as follows.Theorem 4.
The polynomial(9)hnz=z+∑k=1n2z2k+132k+13k+1+∑k=0n2z2k+232k+22k+1,maps the unit disk to a star-like domain for n=1,2.Before we give analytical Proof of Theorem 4 in Section3, the starlikeness of h1 and h2 can be seen in Figures 3 and 4, respectively.Figure 3
Image ofD under the mapping h1.Figure 4
Image ofD under the mapping h2.


## 3. Proof of the Main Results

Proof of Theorem 1.
To prove identity (1), we need to recall the classical identity:(10)cosα+cosβ=2cosα+β2cosα−β2,1±cosnα≥0,for allα∈0,2π,n=0,1,2,….
This implies(11)cosθ2+cos3θ4+cos5θ8=184cosθ+2cos3θ+cos5θ=183cosθ+cosθ+cos3θ+cos3θ+cos5θ=cosθ83+2cos2θ+2cos4θ.
Now, we can rewrite (1) as(12)1+cosθ2+cos2θ6+cos3θ4+cos4θ9+cos5θ8+cos6θ7=181+cosθ3+2cos2θ+2cos4θ+58−112cos2θ−536cos4θ+17cos6θ=181+cosθ1+cos2θ2+3cos22θ+131504+1121−cos2θ+5361−cos4θ+171+cos6θ>0,hence proved.
Next, we will prove the positivity of the sine sum in (5). From the well-known trigonometric identities, one can obtain the following identities:(13)S1θ=sinθ504441+424cosθ+378cos2θ+256cos3θ+126cos4θ+144cos5θ.
It is known thatsinθ is positive for 0<θ<π, and hence S1θ>0 if(14)441+424cosθ+378cos2θ+256cos3θ+126cos4θ+144cos5θ=189+376cosθ−252cos2θ−1856cos3θ+1008cos4θ+2304cos5θ>0.
Denotey:=cosθ. Then, clearly y∈−1,1 and (14) is equivalent to(15)P0y=189+376y−252y2−1856y3+1008y4+2304y5>0.
SinceP00=189>0, inequality (15) holds if P0 have no zero in −1,1. We will prove these facts by constructing Strum sequences to determine the number of zeroes of a polynomial in an open interval.
The Strum sequence for the polynomialP0 is given as follows:(16)P1y=P0′y=376−504y−5568y2+4032y3+11520y4,P2y=−remP0y,P1y=−912150−15481y50+1344y225+20324y325,P3y=−remP1y,P2y=−2865119800025816561−85877316000y25816561+36060640800y225816561,P4y=−remP2y,P3y=−1578653330785493685111287932421064450−171856954494801497753y33863797263193350,P5y=−remP3y,P4y=995719081958136788394316950594748001144025837067465055541881465258369.
Here, byremPy,Qy, we mean the remainder of the long division of P by the polynomial Q. Now, by Strum theorem the number of roots of P0 in −1,1 is equal to σ−1−σ1, where σa is the number of change of sign by the Strum sequence at the point a. From Table 1, it follows that σ−1−σ1=2−2=0. Thus, P0 does not have any zero on −1,1.Table 1
Change the signs of Strum sequence forP0.
y=−1y=1P0y++P1y−+P2y+−P3y+−P4y++P5y++Proof of Theorem 2.
Clearly,(17)g1′z=1+1631z+1400z2+20332−45z3.
Now, forz=eiθ, it follows that(18)Reg1′eiθ=1+1631cosθ+1400cos2θ+20332−45cos3θ>1−1631−1400−20332−45≈0.0412016,and for 0<θ<π,(19)Img1′eiθ=1631sinθ+1400sin2θ+20332−45sin3θ=sinθ1631+1200cosθ+20332−454cos2θ−1>sinθ1631−1200−20332−45≈0.0709597sinθ>0.
The result follows from Lemma1.Proof of Theorem.
3. Forz=eiθ, it follows from (3) that(20)Rehn′eiθ=1+23∑k=1ncos2kθ3k+1+23∑k=0ncos2k+1θ2k+1,for θ∈0,2π.
Clearly, forn=1, identity (20) gives(21)Reh1′eiθ=1+23cosθ+16cos2θ+29cos3θ=56+13cos2θ+89cos3θ=518+cos2θ31+cosθ+591+cos3θ>0.
Similarly, forn=1, we have(22)Imh1′eiθ=23sinθ+16sin2θ+29sin3θ=sinθ94+3cosθ+8cos2θ.
Ifθ∈0,π, then (22) implies that Imh2′eiθ>0, while if θ∈π,2π, then Imh2′eiθ<0. Thus, we can conclude that ImzImh2′z>0. Finally, the conclusion follows from Lemma 1.
Next, we will consider the case forn=2. In this case,(23)Reh2′eiθ=1+23cosθ+16cos2θ+29cos3θ+221cos4θ+215cos5θ=56+13cos2θ+89cos3θ+221cos4θ+215cos5θ=31630+cos2θ31+cosθ+591+cos3θ+2211+cos4θ+2151+cos5θ>0.
Similarly,(24)Imh1′eiθ=23sinθ+16sin2θ+29sin3θ+221sin4θ+215sin5θ=sinθ315322+165cosθ+224cos2θ+60cos3θ+84cos4θ=sinθ315182−15cosθ−224cos2θ+240cos3θ+672cos4θ.
It is well known thatsinθ is positive for θ∈0,π. Thus, it is enough to show that(25)182−15cosθ−224cos2θ+240cos3θ+672cos4θ>0,which is equivalent to prove(26)Qy=182−15y−224y2+240y3+672y4>0.
SinceQ0=182>0, we need to prove that Qy does not have any zero in −1,1. We apply the Strum theorem for this purpose. A differentiation of (26) gives(27)Q1y=Q′y=−15−448y+720y2+2688y3.
Next to form the Strum sequence, we need to find the remainder ofQy and Q1y as(28)Q2y=−remQy,Q1y=−40843224+5y4+1793y214,and continuing, similar to the Strum theorem, we have(29)Q3y=−remQ1y,Q2y=−284283300292259−108408708803214849y,Q4y=−remQ2y,Q3y=1444199218806969984783946058169165696.
The change of sign of the Strum sequenceQ,Q1,Q2,Q3,Q4, respectively, at y=−1 and y=1 can be seen in Table 2. Since σ−1−σ1=2−2=0, the polynomial Q does not have any zero on −1,1.
Finally, we haveImh2′eiθ>0 for all θ∈0,π, and by Lemma 1, it can be concluded that h2D is star-like with respect to the origin.Table 2
Change the signs of Strum sequence forQ.
y=−1y=1Qy++Q1y−+Q2y−−Q3y+−Q4y++


## 4. Concluding Remark

From the above three examples, it can be concluded that the trigonometric sine and cosine sum can also be positive with some nonmonotone coefficients.In Theorem4, though we prove that hn is star-like for n=1,2, but the graphical experiment Figure 5 shows that hn is star-like for all n∈ℕ. However, Rehn′eiθ is negative for some range of θ when n is larger (greater than 28, see Figure 6). Thus, our method is not applicable on hn for larger n.Figure 5
Image ofD under the mapping h28.Figure 6
Image ofReh28′eiθ, 3.1≤θ≤3.17.The results in this article indicate that several polynomials can be constructed which maps unit disk to a star-like domain by using positivity of cosine and sine sum.



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