# Uniqueness of the Sum of Points of the Period-Five Cycle of Quadratic Polynomials

Authors: Pekka Kosunen
Journal: Journal of Complex Analysis
 (2017)

---

## Abstract

It is well known that the sum of points of the period-five cycle of the quadratic polynomialfc(x)=x2+c is generally not one-valued. In this paper we will show that the sum of cycle points of the curves of period five is at most three-valued on a new coordinate plane and that this result is essentially the best possible. The method of our proof relies on a implementing Gröbner-bases and especially extension theory from the theory of polynomial algebra.


---

## Body

## 1. Introduction

The dynamics of quadratic polynomials is commonly studied by using the family of mapsfc(x)=x2+c, where c∈C and xi+1=fc(xi)=xi2+c. In the article [1] we presented the corresponding iterating system on a new coordinate plane using the change of variables(1)u=x+y=x0+x1v=x+y2+y-x2=x0+x12+x1-x02to the (x,y)-plane model (see [2]). In this new (u,v)-plane model, equations of periodic curves are of remarkably lower degree than in earlier models. Now the dynamics of the (u,v)-plane is determined by the iteration of the function(2)Gu,v=Ru,v,Qu,v=-u+v+uvu,u2-u+v-u2v-uv+uv2+v2u,which is a two-dimensional quadratic polynomial map defined in the complex 2-space C2. The new iteration system is defined recursively as follows:(3)R0u,v,Q0u,v=u,v=u0,v0,R1u,v,Q1u,v=Ru,v,Qu,v=u1,v1,Rn+1u,v,Qn+1u,v=GRnu,v,Qnu,v=un+1,vn+1,where(4)Rn+1u,v=Qnu,v-1+Qnu,vRnu,vQn+1u,v=Rn+1u,v1+Qnu,v-Rnu,v,and n∈N∪{0}. Now (u,v) is fixed Gn, so Gn(u,v)=(u,v), if and only if (Rn(u,v),Qn(u,v))=(u,v). The set of such points is the union of all orbits, whose period divides n, and the set of periodic points of period n are the points with exact period dividing n.In complex dynamics, the sum of period cycle points has been a commonly used parameter in many connections (see, e.g., [2–6]). In the article [5] Giarrusso and Fisher used it for the parameterization of the period 3 hyperbolic components of the Mandelbrot set. Later, in the article [2], Erkama studied the case of the period 3-4 hyperbolic components of the Mandelbrot set on the (x,y)-plane and completely solved both cases.Moreover, Erkama [2] has shown that the sum of periodic orbit points(5)Sn=x0+x1+x2+⋯+xn-2+xn-1is unique when n=3 or n=4. Conversely, the sum of cyclic points of periods three and four determines these orbits uniquely. In the period-five case this situation changes and the sum of the cycle points is no longer unique. We can see this property in the articles [3, 6], in which Brown and Morton have formed the so calledtrace formulas in the cases of periods five and six using c and the sum of period cycle points as parameters. In this paper we will show that, by implementing the change of variables (1), we obtain a new coordinate plane where the sum of period-five cycle points is at most three-valued and show that no better result is obtainable in this coordinate plane. This is done by applying methods of polynomial algebra (without the classical trace formula), as our proof relies on the use of the elimination theory and especially the extension theorem [7]. The extension theorem tells us the best possible result (which the trace formula does not necessarily do) due to the use of Gröbner-basis. In the next section we present the most central tools and constructions related to these theorems.


## 2. A Brief Introduction to the Elimination Theorem

We start with theHilbert basis theorem: EveryidealI⊂C[x1,…,xn] has a finite generating set. That is, I=〈g1,…,gt〉 for some g1,…,gt∈I. Hence 〈g1,…,gt〉 is the ideal generated by the elements g1,…,gt; in other words g1,…,gt is the basis of the ideal. The so calledGröbner-basis has proved to be especially useful in many connections [7], for example, in kinematic analysis of mechanisms (see [8, 9]). In order to introduce this basis we need the following constructions.Letf∈C[x1,…,xn] be the polynomial given by(6)f=∑αaαxα,where aα∈C, α=(α1,…,αn), and xα=x1α1·x2α2⋯xnαn is a monomial. Then themultidegree of f is(7)mdegf=max⁡α∣aα≠0,theleading coefficient of f is(8)LCf=amdegf,theleading monomial of f is(9)LMf=xmdegf,and theleading term of f is(10)LTf=LCfLMf.To calculate a Gröbner-basis of an ideal we need to order terms of polynomials by using amonomial ordering. A Gröbner-basis can be calculated by using any monomial ordering, but differences in the number of operations can be very significant. An effective tool to calculate the Gröbner-basis is the softwareSingular, which has been especially designed for operating with polynomial equations. Next we will define a monomial ordering of nonlinear polynomials.Relation< is thelinear ordering in the set S, if x<y, x=y, or y<x for all x,y∈S. A monomial ordering in the set Nn is a relation ≺ if(1)
≺ is linear ordering,(2)
implicationxα≺xβ⇒xα+γ≺xβ+γ holds for all α,β,γ∈Nn,(3)
xα>1.To compute elimination ideals we needproduct orderings. Let ≻A be an ordering for the variable x, and let ≻B be ordering for the variable y in the ring C[x1,…,xn,y1,…,ym]. Now we can define the product ordering as follows:(11)xαyβ≻xγyδifxα≻Axγorxα=xγ,yβ≻Byδ.There are several monomial orders but we need only thelexicographic order≺lex in the elimination theory. Let α,β∈Nn. Then we say that xα≺lexxβ if xα1=xβ1,…,xαk-1=xβk-1,xαk⪰xβk and xαk+1≺xβk+1. One of the most important tools in the elimination theory is theGröbner-basis of an ideal: Fix a monomial order. A finite subset(12)GI=g1,…,gt⊂Iof an ideal I is said to be a Gröbner-basis (or standard basis) if(13)LTg1,…,LTgt=LTI.Based on the Hilbert basis theorem we know that every ideal I⊂C[x1,…,xn] has a Gröbner-basis GI={g1,…,gs} so that(14)GI=I.It is essential to construct also an affine variety corresponding to the ideal. Letf1,…,fs be polynomials in the ring C[x1,…,xn]. Then we set(15)Vf1,…,fs=a1,…,an∈Cn:fia1,…,an=0∀1≤i≤s,and we call V(f1,…,fs) as theaffine variety defined by f1,…,fs. Now if I=〈f1,…,fs〉, V(I)=V(f1,…,fs) and naturally we obtain the variety of the ideal as the variety of its Gröbner-basis: V(I)=V(〈GI〉).When we consider ideals and their algebraic varieties we are sometimes just interested about polynomialsf∈C[x1,…,xn], which belong to the original ideal f∈I but contain only certain variables of the ring variables of C[x1,…,xn]. For this purpose we needelimination ideals. Let I=〈f1,…,fs〉⊂C[x1,…,xn]. The k:th elimination ideal Ik is the ideal of C[xk+1,…,xn] defined by(16)Ik=I∩Cxk+1,…,xn.Next we give an importantelimination theorem which we use in our proof.Theorem 1 (the elimination theorem).
LetI⊂C[x1,…,xn] be an ideal and let G be a Gröbner-basis of I with respect to lexicographic order, where x1≻x2≻⋯≻xn. Then, for every 0≤k≤n, the set(17)Gk=G∩Cxk+1,…,xnis a Gröbner-basis of the k:th elimination ideal Ik.The elimination theorem is closely related to theextension theorem, which tells us the correspondence between varieties of the original ideal and the elimination ideal. In other words, if we apply this theorem to a system of equations we see whether the partial solution V(Ik) of the system of equations is also a solution of the whole system V(I).Theorem 2 (the extension theorem).
LetI=〈f1,…,fs〉⊂C[x1,…,xn] and let I1 be the first elimination ideal of I. For each 1≤i≤s write fi in the form(18)fi=gix2,…,xnx1Ni+termsinwhichdeg⁡x1<Ni,where Ni≥0 and gi⊂C[x2,…,xn], gi≠0. Suppose that we have a partial solution (a2,…,an)∈V(I1). If (a2,…,an)∉V(g1,…,gs), then there exists a1∈C such that (a1,a2,…,an)∈V(I).


## 3. On Properties of Points Sums of Periods 3–5 Cycles

In this section we first prove the uniqueness properties of points sums of cycles of period three and four by using methods from polynomial algebra in a new way. After this we concentrate on the period-five case and show that the sum of period-five cycle points is at most three-valued. The next result shows the relation between the sums of cycle points of the(x,y)-plane [2] and the (u,v)-plane [1].Theorem 3.
Letx0,x1,x2,…,xn-1 be the period-n orbit points. If(19)Sn=x0+x1+x2+⋯+xn-2+xn-1,then by transformation of (1) and (3)(20)Sn=12Sn1=12Sn2,where(21)Sn1=u0+u1+u2+⋯+un-2+un-1,Sn2=v0+v1+v2+⋯+vn-2+vn-1.Proof.
By writing out both components we obtain(22)Sn1=u0+u1+u2+⋯+un-2+un-1=x0+x1+x1+x2+x2+x3+⋯+xn-2+xn-1+xn-1+xn=x0+x1+x1+x2+x2+x3+⋯+xn-2+xn-1+xn-1+x0=2x0+x1+x2+x3+⋯+xn-2+xn-1and similarly(23)Sn2=v0+v1+v2+⋯+vn-2+vn-1=x0+x2+x1+x3+x2+x4+⋯+xn-2+xn+xn-1+xn+1=2x0+x1+x2+⋯xn-2+xn-1.


### 3.1. The Uniqueness of Cycle Points Sums of Periods Three and Four Orbits

The sums of points of the periods three and four cycles is obtained in [2] as(24)S3=x0+x1+x2,S4=x0+x1+x2+x3.According to Theorem3 and by using the formula (3) we obtain on the (u,v)-plane(25)S3u,v=12u0+u1+u2=u2-u+v+2uv2u,S4u,v=12u0+u1+u2+u3=--v+u-u2+u2v-uv2-v2u.Based on article [1], the equations of periodic orbits of period three and four are P3(u,v)=0 and P4(u,v)=0, where(26)P3u,v=uv+1+v,P4u,v=-u2v+u2v2-u+uv+uv2-v2-v3-uv3=u2-v2+v+uv3-v2-v+1+v3+v2.Now we form polynomials B3(u,v,S3)=0 and B4(u,v,S4)=0 based on formulas (25) as(27)B3u,v,S3=2uS3-u2-u+v+2uv=-u2+2S3+1-2vu-v,B4u,v,S4=uS4-v+u-u2+u2v-uv2-v2=-1+vu2+S4-v2+1u-v-v2.Based on the previous equations we can form the pair of equations(28)P3u,v=0B3u,v,S3=0,P4u,v=0B4u,v,S4=0and obtain the ideals(29)I3=P3u,v,B3u,v,S3=uv+1+v,-u2+2S3+1-2vu-v,I4=P4u,v,B4u,v,S4=u2-v2+v+uv3-v2-v+1+v3+v2,-1+vu2+S4-v2+1u-v-v2.We eliminate from these ideals the variable u and obtain the Gröbner-basis of the eliminated ideals I3u and I4u to calculate the Gröbner-basis of the ideals I3 and I4 using the Singular program ([10]). Gröbner-bases of the ideals I3 and I4, by using the ordering ≺lex, where S3≺lexv≺lexu and S4≺lexv≺lexu, are(30)G3=g31,g32,where(31)g31=v3-2v2S3-2vS3-3v-1,g32=u+v2-2vS3-2S3-2,and(32)G4=g41,g42,g43,g44,where(33)g41=v4-v3S4+v3-v2S4-v2-v,g42=uv2-uvS4-u,g43=u2S4-uv3+uv2S4-uv2+uvS4+uv-uS42+u-v3+v2S4-2v2+vS4-v,g44=u2v-u2-uv2+uS4+u-v2-v.Thus g31 and g41 depend only on the variables v and S5. Based on the elimination Theorem 1 the set(34)G3u=G3∩Cv,S3=g31is the Gröbner-basis of the elimination ideal I3u and so V(I3u)=V(g31). At the same way the set(35)G4u=G4∩Cv,S4=g41is the Gröbner-basis of the elimination ideal I4u and so V(I4u)=V(g41). In the case g31=0 it follows that(36)S3=-1-3v+v32vv+1.If g41=0 we have(37)S4=-v4-v3+v2+v-v3-v2=v2-1v.As we can see, in both cases the sum of the points of cycles of the given period is unique. In other words, the orbit sums S3 and S4 uniquely determine the orbit. If we eliminate in the first case the variable v instead of the variable u, we obtain the Gröbner-basis(38)G3v=u3-2u2S3-2uS3-3u-1,which gives the same result as (36). However, the same procedure in the period four case produces the Gröbner-basis(39)G4v=u5S4-2u4S42+u3S43-u3S42-2u3S4-4u3+u2S43+2u2S42+4u2S4+uS4=u3+u2S43+-u3+2u2-2u4S42+u5-2u3+u+4u2S4-4u3and this is of higher degree than (37).


### 3.2. On the Uniqueness of the Cycle Points Sum of Period-Five Orbits

Next we prove that, in the case of period-five cycles, the sum of period-five points is at most three-valued. We use in this proof the Gröbner-basis of an ideal, like before in periods three and four cases, which produce for us the Gröbner-basis of the elimination ideal. Because this method relies on bases, the following result is optimal.Theorem 4.
The sum of period-five cycle points is at most three-valued.Proof.
By article [1], the equation for period-five orbit on the (u,v)-plane is of the form P5(u,v)=0, where(40)P5u,v=u7-v4+2v3-v2+u63v5-8v4+5v3+v2-v+u53v6+14v5-12v4-5v3+7v2-v+u4v7-12v6+18v5+6v4-16v3+3v2+2v+u34v7-16v6+19v4-5v3-4v2+2v+1+u26v7-6v6-12v5+6v4+4v3-2v2+u4v7+3v6-4v5-2v4+v3+v7+2v6+v5.According to the Theorem 3, the sum(41)S5=x0+x1+x2+x3+x4of the period-five points satisfies (42)Sn=12Sn1=12Sn2,and based on the formula (3) we obtain(43)S5u,v=12u0+u1+u2+u3+u4=-3u2+4u2v+3uv-4u3v32u2+2u2v4+4uv4+u3-2u4v-2u3v2u2+2u2v2-2uv2+2v3+2u4v22u2+6u3v2-8u2v3-2uv3+2v42u2on the (u,v)-plane. We form from this the polynomial(44)B5u,v,S5=u42v2-2v+u3-4v3+6v2-2v+1+u2-2S5+2v4-8v3+2v2+4v-3+u4v4-2v3-2v2+3v+2v4+2v3.Now we can form the pair of equations(45)P5u,v=0B5u,v,S5=0,and the two polynomials P5(u,v) and B5(u,v,S5) form an ideal(46)I5=P5u,v,B5u,v,S5=a7u7+a6u6+a5u5+a4u4+a3u3+a2u2+a1u+a0,b4u4+b3u3+b2u2+b1u+b0,where(47)a0=v7+2v6+v5a1=4v7+3v6-4v5-2v4+v3a2=6v7-6v6-12v5+6v4+4v3-2v2a3=4v7-16v6+19v4-5v3-4v2+2v+1a4=v7-12v6+18v5+6v4-16v3+3v2+2va5=3v6+14v5-12v4-5v3+7v2-va6=3v5-8v4+5v3+v2-va7=-v4+2v3-v2b0=2v3b1=4v4-2v3-2v2+3vb2=-2S5+2v4-8v3+2v2+4v-3b3=-4v3+6v2-2v+1b4=2v2-2v.We eliminate from this the variable u by forming the Gröbner-basis G5u of the elimination ideal I5u in order to calculate the Gröbner-basis G5 of the ideal I5 using Singular program. We obtain the Gröbner-basis of the ideal I as(48)G5=g51,g52,g53,g54,g55,g56using ordering ≺lex, where S5≺lexv≺lexu. Here g51, g52, g53, g54, and g55 depend on the variables u, v, and S5, and g56 depends only on the variables v and S5. By the elimination theorem the set(49)G5u=G∩Cv,S5=g56is the Gröbner-basis of the elimination ideal I5u and so V(I5u)=V(g56). Now the Gröbner-basis of the elimination ideal I5u is of the form(50)G5u=v6v+12c0v15+c1v14+c2v13+c3v12+c4v11+c5v10+c6v9+c7v8+c8v7+c9v6+c10v5+c11v4+c12v3+c13v2+c14v+c15,where(51)c0=27c1=-162S5c2=252S52-432S5-684c3=280S53+2592S52+4128S5+556c4=-1264S54-5760S53-8712S52+236S5+4002c5=1440S55+5888S54+6864S53-8440S52-19596S5-4336c6=-704S56-2816S55+320S54-8380+19584S53+37536S52+11528S5c7=128S57+512S56-3328S55-18112S54-30144S53+1120S52+39192S5+14868c8=1664S56+7488S55+7824S54-21520S53-64076S52-38238S5+4003c9=-256S57-1152S56+1952S55+19360S54+44040S53+22980S52-29970S5-19924c10=-1216S56-6336S55-11216S54+5848S53+46108S52+43516S5+5736c11=128S57+640S56-160S55-8208S54-25384S53-25368S52+3504S5+10380c12=256S56+1664S55-16730S5+4432S54+2056S53-11160S52-4909c13=96S55+1104S54+4240S53+6396S52+2070S5-1934c14=216S53+1068S52+1974S5+1347c15=-27.
By (50) G5u is formed as a product of three terms. We denote the last of these terms in (50) by C(v,S5). Now we obtain the variety V(Iu) of the elimination ideal as the union of three varieties corresponding to the factors of G5u as follows:(52)VI5u=Vv6∪Vv+12∪VCv,S5=0,S5,-1,S5∪VCv,S5.Note that G5u is of degree 23 with respect to the variable v and of degree 7 with respect to the variable S5. We denote, according to the extension theorem,(53)fi=giv,S5uNi+terms  such  thatdeg⁡u<Ni,i=1,2,where(54)g1=a7=-v4+2v3-v2,g2=b4=2v2-2v.The corresponding varieties are(55)Vg1=0,S5,1,S5=Vg2,so(56)Vg1,g2=Vg1∩Vg2=0,S5,1,S5.In other words for all v≠0 and v≠1 we have (v,S5)∉V(g1,g2) and in that case by the extension theorem then there exists u∈C so that (u,v,S5)∈V(I5), so all partial solutions V(I5u)=((v,S5)∣v≠0,v≠1) extend as solutions of the original system (45). Since the term C(v,S5) is of degree 15 with respect to the variable v, it follows by the fundamental theorem of algebra that the equation C(v,S5)=0 has at most 15 different roots. For example, for the value S5=0 we obtain the Gröbner-basis of the elimination polynomial (57)G5u=27v15-684v13+556v12+4002v11-4336v10-8380v9+14868v8+4003v7-19924v6+5736v5+10380v4-4909v3-1934v2+1347v-27,for which the variety V(I5u) includes 15 different values. From these five are real and the rest ten are complex numbers. According to the extension theorem, for every pair of points (v1,0),…,(v15,0) we find the corresponding value of the variable u so that (u1,v1,0),…,(u15,v15,0)∈V(I5). Consequently the sum of period-five cycle points attains the same value at most three times.We obtain also the same result if we eliminate the variablev from the pair of equations (45) using the ordering ≺lex, where S5≺lexu≺lexv.


## 3.1. The Uniqueness of Cycle Points Sums of Periods Three and Four Orbits

The sums of points of the periods three and four cycles is obtained in [2] as(24)S3=x0+x1+x2,S4=x0+x1+x2+x3.According to Theorem3 and by using the formula (3) we obtain on the (u,v)-plane(25)S3u,v=12u0+u1+u2=u2-u+v+2uv2u,S4u,v=12u0+u1+u2+u3=--v+u-u2+u2v-uv2-v2u.Based on article [1], the equations of periodic orbits of period three and four are P3(u,v)=0 and P4(u,v)=0, where(26)P3u,v=uv+1+v,P4u,v=-u2v+u2v2-u+uv+uv2-v2-v3-uv3=u2-v2+v+uv3-v2-v+1+v3+v2.Now we form polynomials B3(u,v,S3)=0 and B4(u,v,S4)=0 based on formulas (25) as(27)B3u,v,S3=2uS3-u2-u+v+2uv=-u2+2S3+1-2vu-v,B4u,v,S4=uS4-v+u-u2+u2v-uv2-v2=-1+vu2+S4-v2+1u-v-v2.Based on the previous equations we can form the pair of equations(28)P3u,v=0B3u,v,S3=0,P4u,v=0B4u,v,S4=0and obtain the ideals(29)I3=P3u,v,B3u,v,S3=uv+1+v,-u2+2S3+1-2vu-v,I4=P4u,v,B4u,v,S4=u2-v2+v+uv3-v2-v+1+v3+v2,-1+vu2+S4-v2+1u-v-v2.We eliminate from these ideals the variable u and obtain the Gröbner-basis of the eliminated ideals I3u and I4u to calculate the Gröbner-basis of the ideals I3 and I4 using the Singular program ([10]). Gröbner-bases of the ideals I3 and I4, by using the ordering ≺lex, where S3≺lexv≺lexu and S4≺lexv≺lexu, are(30)G3=g31,g32,where(31)g31=v3-2v2S3-2vS3-3v-1,g32=u+v2-2vS3-2S3-2,and(32)G4=g41,g42,g43,g44,where(33)g41=v4-v3S4+v3-v2S4-v2-v,g42=uv2-uvS4-u,g43=u2S4-uv3+uv2S4-uv2+uvS4+uv-uS42+u-v3+v2S4-2v2+vS4-v,g44=u2v-u2-uv2+uS4+u-v2-v.Thus g31 and g41 depend only on the variables v and S5. Based on the elimination Theorem 1 the set(34)G3u=G3∩Cv,S3=g31is the Gröbner-basis of the elimination ideal I3u and so V(I3u)=V(g31). At the same way the set(35)G4u=G4∩Cv,S4=g41is the Gröbner-basis of the elimination ideal I4u and so V(I4u)=V(g41). In the case g31=0 it follows that(36)S3=-1-3v+v32vv+1.If g41=0 we have(37)S4=-v4-v3+v2+v-v3-v2=v2-1v.As we can see, in both cases the sum of the points of cycles of the given period is unique. In other words, the orbit sums S3 and S4 uniquely determine the orbit. If we eliminate in the first case the variable v instead of the variable u, we obtain the Gröbner-basis(38)G3v=u3-2u2S3-2uS3-3u-1,which gives the same result as (36). However, the same procedure in the period four case produces the Gröbner-basis(39)G4v=u5S4-2u4S42+u3S43-u3S42-2u3S4-4u3+u2S43+2u2S42+4u2S4+uS4=u3+u2S43+-u3+2u2-2u4S42+u5-2u3+u+4u2S4-4u3and this is of higher degree than (37).


## 3.2. On the Uniqueness of the Cycle Points Sum of Period-Five Orbits

Next we prove that, in the case of period-five cycles, the sum of period-five points is at most three-valued. We use in this proof the Gröbner-basis of an ideal, like before in periods three and four cases, which produce for us the Gröbner-basis of the elimination ideal. Because this method relies on bases, the following result is optimal.Theorem 4.
The sum of period-five cycle points is at most three-valued.Proof.
By article [1], the equation for period-five orbit on the (u,v)-plane is of the form P5(u,v)=0, where(40)P5u,v=u7-v4+2v3-v2+u63v5-8v4+5v3+v2-v+u53v6+14v5-12v4-5v3+7v2-v+u4v7-12v6+18v5+6v4-16v3+3v2+2v+u34v7-16v6+19v4-5v3-4v2+2v+1+u26v7-6v6-12v5+6v4+4v3-2v2+u4v7+3v6-4v5-2v4+v3+v7+2v6+v5.According to the Theorem 3, the sum(41)S5=x0+x1+x2+x3+x4of the period-five points satisfies (42)Sn=12Sn1=12Sn2,and based on the formula (3) we obtain(43)S5u,v=12u0+u1+u2+u3+u4=-3u2+4u2v+3uv-4u3v32u2+2u2v4+4uv4+u3-2u4v-2u3v2u2+2u2v2-2uv2+2v3+2u4v22u2+6u3v2-8u2v3-2uv3+2v42u2on the (u,v)-plane. We form from this the polynomial(44)B5u,v,S5=u42v2-2v+u3-4v3+6v2-2v+1+u2-2S5+2v4-8v3+2v2+4v-3+u4v4-2v3-2v2+3v+2v4+2v3.Now we can form the pair of equations(45)P5u,v=0B5u,v,S5=0,and the two polynomials P5(u,v) and B5(u,v,S5) form an ideal(46)I5=P5u,v,B5u,v,S5=a7u7+a6u6+a5u5+a4u4+a3u3+a2u2+a1u+a0,b4u4+b3u3+b2u2+b1u+b0,where(47)a0=v7+2v6+v5a1=4v7+3v6-4v5-2v4+v3a2=6v7-6v6-12v5+6v4+4v3-2v2a3=4v7-16v6+19v4-5v3-4v2+2v+1a4=v7-12v6+18v5+6v4-16v3+3v2+2va5=3v6+14v5-12v4-5v3+7v2-va6=3v5-8v4+5v3+v2-va7=-v4+2v3-v2b0=2v3b1=4v4-2v3-2v2+3vb2=-2S5+2v4-8v3+2v2+4v-3b3=-4v3+6v2-2v+1b4=2v2-2v.We eliminate from this the variable u by forming the Gröbner-basis G5u of the elimination ideal I5u in order to calculate the Gröbner-basis G5 of the ideal I5 using Singular program. We obtain the Gröbner-basis of the ideal I as(48)G5=g51,g52,g53,g54,g55,g56using ordering ≺lex, where S5≺lexv≺lexu. Here g51, g52, g53, g54, and g55 depend on the variables u, v, and S5, and g56 depends only on the variables v and S5. By the elimination theorem the set(49)G5u=G∩Cv,S5=g56is the Gröbner-basis of the elimination ideal I5u and so V(I5u)=V(g56). Now the Gröbner-basis of the elimination ideal I5u is of the form(50)G5u=v6v+12c0v15+c1v14+c2v13+c3v12+c4v11+c5v10+c6v9+c7v8+c8v7+c9v6+c10v5+c11v4+c12v3+c13v2+c14v+c15,where(51)c0=27c1=-162S5c2=252S52-432S5-684c3=280S53+2592S52+4128S5+556c4=-1264S54-5760S53-8712S52+236S5+4002c5=1440S55+5888S54+6864S53-8440S52-19596S5-4336c6=-704S56-2816S55+320S54-8380+19584S53+37536S52+11528S5c7=128S57+512S56-3328S55-18112S54-30144S53+1120S52+39192S5+14868c8=1664S56+7488S55+7824S54-21520S53-64076S52-38238S5+4003c9=-256S57-1152S56+1952S55+19360S54+44040S53+22980S52-29970S5-19924c10=-1216S56-6336S55-11216S54+5848S53+46108S52+43516S5+5736c11=128S57+640S56-160S55-8208S54-25384S53-25368S52+3504S5+10380c12=256S56+1664S55-16730S5+4432S54+2056S53-11160S52-4909c13=96S55+1104S54+4240S53+6396S52+2070S5-1934c14=216S53+1068S52+1974S5+1347c15=-27.
By (50) G5u is formed as a product of three terms. We denote the last of these terms in (50) by C(v,S5). Now we obtain the variety V(Iu) of the elimination ideal as the union of three varieties corresponding to the factors of G5u as follows:(52)VI5u=Vv6∪Vv+12∪VCv,S5=0,S5,-1,S5∪VCv,S5.Note that G5u is of degree 23 with respect to the variable v and of degree 7 with respect to the variable S5. We denote, according to the extension theorem,(53)fi=giv,S5uNi+terms  such  thatdeg⁡u<Ni,i=1,2,where(54)g1=a7=-v4+2v3-v2,g2=b4=2v2-2v.The corresponding varieties are(55)Vg1=0,S5,1,S5=Vg2,so(56)Vg1,g2=Vg1∩Vg2=0,S5,1,S5.In other words for all v≠0 and v≠1 we have (v,S5)∉V(g1,g2) and in that case by the extension theorem then there exists u∈C so that (u,v,S5)∈V(I5), so all partial solutions V(I5u)=((v,S5)∣v≠0,v≠1) extend as solutions of the original system (45). Since the term C(v,S5) is of degree 15 with respect to the variable v, it follows by the fundamental theorem of algebra that the equation C(v,S5)=0 has at most 15 different roots. For example, for the value S5=0 we obtain the Gröbner-basis of the elimination polynomial (57)G5u=27v15-684v13+556v12+4002v11-4336v10-8380v9+14868v8+4003v7-19924v6+5736v5+10380v4-4909v3-1934v2+1347v-27,for which the variety V(I5u) includes 15 different values. From these five are real and the rest ten are complex numbers. According to the extension theorem, for every pair of points (v1,0),…,(v15,0) we find the corresponding value of the variable u so that (u1,v1,0),…,(u15,v15,0)∈V(I5). Consequently the sum of period-five cycle points attains the same value at most three times.We obtain also the same result if we eliminate the variablev from the pair of equations (45) using the ordering ≺lex, where S5≺lexu≺lexv.



---

*Source: 8474868-2017-11-23.xml*