The number of unimodular roots of some reciprocal polynomials
Dragan Stankov



Number theory









[label1]Partially supported by Serbian Ministry of Education and Science, Project 174032.


















,
dstankov@rgf.bg.ac.rs




Katedra Matematike RGF-a,
University of Belgrade,
Belgrade, Đušina 7,
Serbia















We introduce a sequence P_2n of monic reciprocal polynomials with integer coefficients having the central coefficients fixed. We prove that the ratio between number of
nonunimodular roots of P_2n and its degree d has a limit when d tends to infinity. We present an algorithm for calculation the limit and a numerical method for its approximation. If P_2n is the sum of a fixed number of monomials we determine the central coefficients such that the ratio has the minimal limit. We generalise the limit of the ratio for multivariate polynomials. Some examples suggest a theorem for polynomials in two variables which is analogous to Boyd's limit formula for Mahler measure.


0.5



























§ INTRODUCTION

If P(x)=a_dx^d+a_d-1x^d-1+⋯ +a_1x+a_0 (a_d  0) has zeros α_1,α_2,…,α_d then the Mahler measure of P(x) is 
    M(P(x)) = |a_d|∏_j=1^dmax(1, |α_j|).

Let I(P) denote the the number of complex
zeros of P(x) which are <1 in modulus, counted with multiplicities.
Let U(P) denote the number of zeros of P(x) which are =1 in modulus,
(again, counting with multiplicities). Such zeros 
are called unimodular.
Let E(P) denote the number of complex
zeros of P(x) which are >1 in modulus,, counted with multiplicities.
Then it is obviously that I(P)+U(P)+E(P)=d.
Pisot number can be defined as a real algebraic integer greater than 1 having the minimal polynomial P(x) of degree d such that I(P)=d-1.
Salem number 
is a real algebraic integer > 1 having the minimal polynomial P(x) of degree d such that U(P)=d-2, I(P)=1.

We say that a polynomial of degree d is reciprocal if
P(x) = x^dP(1/x).
If moduli of coefficients are small then a reciprocal polynomial has many unimodular roots. A Littlewood polynomial is a polynomial all of whose coefficients are 1 or -1. Mukunda <cit.>
showed that every self-reciprocal Littlewood polynomial of odd
degree at least 3 has at least 3 zeros on the unit circle. Drungilas <cit.> proved that every
self-reciprocal Littlewood polynomial of odd degree n ≥ 7 has at least 5 zeros on the unit
circle and every self-reciprocal Littlewood polynomial of even degree n ≥ 14 has at least
4 unimodular zeros. In <cit.> two types of very special Littlewood polynomials are considered: Littlewood polynomials with one sign change in the sequence of coefficients and Littlewood
polynomials with one negative coefficient. The numbers U(P) and I(P) of such Littlewood
polynomials P are investigated. In <cit.> Borwein, Erdélyi, Ferguson and Lockhart showed that there exists a cosine polynomials ∑_m=1^Ncos(n_mθ)
with the n_m integral and all different so that the number of its real zeros in [0,2π) is
O(N^9/10(log N)^1/5) (here the frequencies n_m = n_m(N) may vary with N). However, there
are reasons to believe that a cosine polynomial ∑_m=1^Ncos(n_mθ) always has many zeros in
the period.

Clearly,
if α_j, is a root of a reciprocal P(x) then 1/α_j is also a root of P(x) so that I(P)=E(P).
Let n, k, a_0,a_1,…,a_k, be integers such that n>k≥ 0, and let P_2n(x) be a monic, reciprocal polynomial with integer coefficients

    P_2n=x^n(x^n+a_0+1/x^n+∑
    _j=1^k a_j(x^j+1/x^j)).

Let C(P)=I(P)+E(P)/2n be the ratio between number of nonunimodular zeros of P and its degree. Actually, it is the probability that a randomly chosen zero is not unimodular, and C(P)=E(P)/n.



§ THE MAIN THEOREM



If k>0 is an integer then for all fixed integers a_j, j=1, …,k there is a limit C(P_2n) when n tends to infinity.


The theorem will be proved if we show that 1-C(P_2n)  has a limit when n tends to ∞. Since 1-C(P_2n)=U(P_2n)/2n we have to count the unimodular roots of P_2n(x).
If we use the substitution x=e^it in the equation P_2n(x)=0 we get

    e^int(2cos nt +a_0+∑_j=1^k 2 a_jcos jt)=0

Since e^int 0 it follows that the equation is equivalent to

    cos nt =-a_0/2-∑_j=1^k a_jcos jt.

From the substitution x=e^it it follows that x is unimodular if and only if t is real so that we have to count the real roots of (<ref>) (t∈ [0,2π)).
If Γ_1 is the graph of f_1(t)=cos nt 
and Γ_2 is the graph of f_2(t)=-a_0/2-∑_j=1^k a_jcos jt, the function on the right side of equation (<ref>), then U(P) is equal to the number of intersection points of these two graphs.
These intersection points are obviously settled between lines y=-1 and y=1.
Graph Γ_2 of the continuous function f_2 is fixed i.e. does not depend on n, therefore we can introduce a partition of [0,2π] using points 0=t_0<t_1<…<t_p=2π such that |f_2(t_j)|=1, 0<j<p. Let us consider subintervals I_j=[t_j-1,t_j] such that if t∈ I_j then |f_2(t)|<1, j∈ J={j_1,j_2,…,j_r}⊆{1,2,…,p}.

A part of the graph of f_1(t)=cos nt such that (k-1)π/n ≤ t ≤ kπ/n, k∈ℤ is k-th branch of cos nt. The interval [(k-1)π/n,kπ/n] is the domain of the k-th branch.

Each branch of cos nt obviously has exactly one intersection point with the t-axis. We are going to prove that if n is large enough then each branch of cos nt also has
exactly one intersection point with Γ_2.
We need the next lemma which will be proved in the next subsection.


For all B_1, B_2>0 and ε such that 1> ε>0, there is n_0∈ℕ such that if n≥ n_0 then

(1) if |cos(nt)|<1-ε then n|sin(nt)|>B_1,

(2) if |cos(nt)|>1-ε then n^2|cos(nt)|>B_2.







We will also need the following claims.

  
  * There is a bound B_1 of the modulus of the first derivative of f_2(t). Indeed
  
  |f'_2(t)|= |∑_j=1^k ja_jsin jt|≤∑_j=1^k j|a_j|=:B_1.
  
  * There is a bound B_2 of the modulus of the second derivative of f_2(t). Indeed
  
  |f”_2(t)|= |∑_j=1^k j^2a_jcos jt|≤∑_j=1^k j^2|a_j|=:B_2.

  
  * The first derivative of f_2(t) has a finite number of roots on [0,2π] so that
there is ε_j>0 such that 1-ε_j is greater than the value at each local maximum and -1+ε_j is less than the value at each local minimum of f_2(t) on (t_j-1,t_j).

  
  * If the domain of a branch of cos nt is the subset of the interior of I_j then cos nt-f_2(t) has values of the opposite sign at the end points of the domain so that the branch has at least one intersection point with Γ_2.


Since f_2(t) is continuous at t_j-1 and t_j it follows that
there are δ_1j>0, δ_2j>0 such that if t∈(t_j-1,t_j-1+δ_1j) or t∈(t_j-δ_2j,t_j) then 1-|f_2(t)|<ε_j. If we bring to mind (iii) it follows that f_2(t) is monotonic on (t_j-1,t_j-1+δ_1j) and on (t_j-δ_2j,t_j). Therefore we can choose δ_1j>0, δ_2j>0 such that |f_2(t_j-1+δ_1j)|=1-ε_j, |f_2(t_j-δ_2j)|=1-ε_j





Using Lemma <ref> (1) there is n_j such that if n≥ n_j and |f_1(t)|<1-ε_j then |f_1'(t)|>B_1. It follows that on E_j:=[t_j-1+δ_1j,t_j-δ_2j] a branch of cos nt and Γ_2 can not have more than one intersection point: if they have two intersection points M_1, M_2 then using the mean value theorem for the continuous function f_1 the slope S of the line M_1M_2 is greater than B_1 in modulus. Using the mean value theorem again for the continuous function f_2 it follows that there is a point t such that f'_2(t)=S so that |f'_2(t)|=|S|>B_1 which is the contradiction with (i).

It remains to be proved that if the domain of a branch is the subset of D_1j=(t_j-1,t_j-1+δ_1j] or of D_2j=[t_j-δ_2j,t_j) then the branch of cos nt and Γ_2 can not have more than one intersection point. Let 1>f_2(t)>1-ε_j and let the branch has an adjacent branch such that the union of its domains is [(k-1)π/n,(k+1)π/n]⊂ D_1j and k is even. Then using Lemma <ref> (2) it follows that if cos(nt)>1-ε then f”_1(t)-f”_2(t)=-n^2cos nt-f”_2(t)<-B_2-f”_2(t) is negative. Therefore f_1(t)-f_2(t) is a concave function so that its graph can have at most two intersection points with the line y=0. If such an adjacent branch does not exist which means that t_j-1∈ [kπ/n,(k+1)π/n], k is even, then we can prove the concavity of f_1(t)-f_2(t) in the same manner. We conclude that if t_j-1, the start point of I_j, is in the domain of a branch of cos nt then the branch can have 0, 1, or 2 intersection points with Γ_2 (see fig. 1).



If -1<f_2(t)<-1+ε_j after showing the convexity of f_1(t)-f_2(t) on D_1 the claim follows in the similar manner.
Analogously we prove the claim if the domain of a branch is the subset of D_2 as well as the claim for the end point of I_j: if t_j is in the domain of a branch of cos nt then the branch can have 0, 1, or 2 intersection points with Γ_2.


We conclude that if n is large enough then each branch of cos nt, such that the start and the end point of I_j are not elements of its domain, has exactly one intersection point with Γ_2.  
Thus the number U_j of intersection points of Γ_1 and Γ_2 differs to the number V_j of intersection points of Γ_1 and the t-axis, t∈ I_j, by 0,1 or 2, because in the beginning and at the end of I_j branches are not complete (see fig. 1).
If we take the sum U_j and V_j over all r subintervals then it is clear that 
U(P_2n) differs to the number V(P_2n)=∑_j∈ J V_j 
by a number ≤ 2r. Since 2r does not depend on n it follows that

    (lim_n→∞(1-C(P_2n))=)lim_n→∞U(P_2n)/2n=lim_n→∞V(P_2n)/2n(=lim_n→∞∑_j∈ J V_j/2n).

Since the intersection points of the graphs of y=cos nt and the t-axis are obviously uniformly distributed on I_j we conclude

    lim_n→∞∑_j∈ J V_j/2n=∑_j∈ J |I_j|/2π.







 §.§ Proof of Lemma <ref>



Using the symmetry and the periodicity of cos nt it is enough to prove the claim for the first branch of cos nt, t∈ [0,π/n].
For an arbitrarily chosen ε>0 and n∈ N we determine τ such that |cos(nτ)|=1-ε.
It follows that τ=arccos(1-ε)/n or τ=arccos(-1+ε)/n so that

(1) if t∈ (τ,π/n-τ) then nsin nt>nsin nτ=n sin(arccos(1-ε)) →∞ when
n →∞. Therefore the claim follows immediately if we chose 
    n_1=⌈B_1/sin(arccos(1-ε))⌉ .


(2) if t∈ (0,τ)⋃(π/n-τ,π/n) then n^2|cos nt|>n^2|cos nτ|=n^2 cos(arccos(1-ε))=n^2 (1-ε) →∞ when
n →∞. Therefore the claim follows immediately if we chose 
    n_2=⌈√(B_2/1-ε)⌉ .


It remains to take n_0=max(n_1,n_2).





 §.§ Algorithm for determination lim_n→∞C(P_2n)


In the proof of Theorem 1 we actually declared steps of an algorithm for determination lim_n→∞C(P_2n):


  * determine all real roots t_j of the equations f_2(t)=1 and f_2(t)=-1 ,


  * arrange them as an increasing sequence 0=t_0<t_1<…<t_p=2π,


  * determine I_j=[t_j-1,t_j] such that if t_j-1<t<t_j then |f_2(t)|<1, j∈ J={j_1,j_2,…,j_r}⊆{1,2,…,p},


  * calculate lim_n→∞C(P_2n)=1-∑_j∈ J (t_j-t_j-1)/(2π).


If f_2(t) is defined:


    f_2(t) = 
                   1,     |f_2(t)|≥ 1
    
                   0,    otherwise

then

    lim_n→∞C(P_2n)=1/2π∫_0^2πf_2(t)dt.




§ APPROXIMATING LIM_N→∞C(P_2N)


The equation f_2(t)=± 1 i.e. -a_0/2-∑_j=1^k a_jcos jt=± 1 is algebraic in cos t so that t_j can be expressed by arccosine of an algebraic real number α∈ [-1,1] thus only solutions of this kind should be taken into account.

We can approximate numerically the integral in (<ref>) i.e. lim_n→∞C(P_2n). Suppose the interval [0,2π] is divided into p equal subintervals of length Δ t=2π/p so that we introduce a partition of [0,2π] 0=t_0<t_1<…<t_p=2π such that t_j-t_j-1=Δ t. Then we chose numbers ξ_j∈ [t_j,t_j-1]   and count all ξ_j such that |f_2(ξ_j)|>1, j=1,2,…,p. If there are s such ξ_j then lim_n→∞C(P_2n) is approximately equal to s/p.


    lim_n→∞C(P_2n)≈1/p∑_j=1^p f_2(j2π/p)

where we chosed ξ_j=2jπ/p.



 §.§ Small limit points of C(P_2n)






In the case of trinomials i.e. if  k=0, |a_0|≤2 then all roots of P_2n(x)=x^2n+a_0x^n+1 obviously are unimodular. If |a_0|>2 then P_2n does not have any unimodular root so that C(P) tends either to zero or to one as n approaches infinity.

In the case of quadrinomials i.e.
if  k=1, a_0=0, a_1=± 1 then P_2n(x)=x^2n± x^n+k± x^n-k+1=(x^n-k± 1)( x^n+k± 1) so that obviously all roots are unimodular.
If |a_1|>1 then 
    C(x^2n+a_1 x^n+k+a_1 x^n-k+1)=2arccos(1/a_1)/π
 so that it has the minimum value 2/3 when a_1=2 and C(P) tends to one as a_1 approaches infinity.

If we exclude trinomials and quadrinomials then it is clear that the limit points of C(P_2n) are always greater than zero.
A natural question that arises here is what is the smallest value, greater than 0, of the limit points of C(P_2n)?



 §.§ Hexanomials with smallest limit points of C(P_2n)


Between all pentanomials x^2n+a_kx^n+k+a_0x^n +a_kx^n-k+1 an exhaustive search such that k=1,2,…,10, a_k=±1,± 2,…± 10 suggests that C(x^2n+x^n+1+x^n +x^n-1+1) has the minimal limit point. It is equal to 
    1/πarccos(1/2)=1/3.


We have submitted an exhaustive search between all hexanomials x^2n+a_j_2x^n+j_2+a_j_1x^n+j_1+a_j_1x^n-j_1 +a_j_2x^n-j_2+1 such that j_1=1,2,…,10,  j_2=j_1+1,j_1+2,…,10, a_j_1=±1,± 2,…± 10,a_j_2=±1,± 2,…± 10 suggests that C(x^2n+x^n+3+x^n+1 +x^n-1+x^n-3+1) has the minimal limit point.

Using the algorithm we solve the equation
cos3t+cos t=±1. Since cos 3t=4cos^3 t-3cos t if we substitute cos t=x we get algebraic equations 4x^3-2x=±1 each of them with unique real solution ±α where

    α=√(√(57)/288+23/864)+1/6/√(√(57)/72+1/8)≈ 0.885,   C(P_2n)→2/πarccos(α) ≈ 0.308799876.






Each intersection point of graph of f_1=cos 60 t and f_2=-cos t-cos 2t (see fig. 2) corresponds to an unimodular root of the reciprocal polynomial x^120+x^63+x^61+x^59+x^57+1. Nonunimodular roots have arguments in [-θ,θ] or in [π-θ,π+θ] where θ=arccos(α)≈ 0.485 ≈ 27.8^∘ (see fig. 3). Since there are 41 intersection point on [0,π] and f_1, f_2 are both even it follows that there are 120-2·41=38 nonunimodular roots so that C(P_120)=38/120≈ 0.317 is close to the limit of
C(P_2n)→2/πarccos(α)=2/πθ≈ 0.308799876. 






 §.§ Heptanomials with smallest limit points of C(P_2n)


Between all heptanomials x^2n+a_j_2x^n+j_2+a_j_1x^n+j_1+a_0x^n+a_j_1x^n-j_1 +a_j_2x^n-j_2+1 an exhaustive search such that j_1=1,2,…,10,  j_2=j_1+1,j_1+2,…,10, a_0=±1,± 2,…± 10, a_j_1=±1,± 2,…± 10,a_j_2=±1,± 2,…± 10 suggests that C(x^2n+x^n+4+x^n+2+x^n +x^n-2+x^n-4+1) has the minimal limit point.

Using the algorithm we solve the equation
cos4t+cos 2t+1/2=±1. If we develop cos 4t and cos2 t and substitute cos t=x we get biquadratic equation 8x^4-6x^2+1/2=1 with four real solutions ±√(3/8±√(13)/8) and 8x^4-6x^2+1/2=-1 without any real solution. Using symmetry we can show that

    C(P_2n)→2/πarccos(√(3/8+√(13)/8)) ≈ 0.2741871146.




 §.§ Octanomials with smallest limit points of C(P_2n)


Between all octanomials
x^2n+a_j_3x^n+j_3+a_j_2x^n+j_2+a_j_1x^n+j_1+a_j_1x^n-j_1 +a_j_2x^n-j_2+a_j_3x^n-j_3+1
an exhaustive search such that j_1=1,2,…,10,  j_2=j_1+1,j_1+2,…,10, j_3=j_2+1,j_2+2,…,10,
a_j_1=±1,± 2,…± 10,a_j_3=±1,± 2,…± 10, a_j_3=±1,± 2,…± 10, suggests that C(x^2n+x^n+5+x^n+3+x^n+1+x^n-1 +x^n-3+x^n-5+1) has the minimal limit point.

Using the algorithm we solve the equation
cos5t+cos 3t+cos t=±1. If we develop cos 5t and cos 3t and substitute cos t=x we get pentic equation 16x^5-16x^3+3x=± 1 with two real solutions ± 0.92757157104393247625. Using symmetry we can show that

    C(P_2n)→2/πarccos(0.92757157104393247625) ≈ 0.24378469902904315.




 §.§ Nonanomials with smallest limit points of C(P_2n)


Between all nonanomials x^2n+a_j_3x^n+j_3+a_j_2x^n+j_2+a_j_1x^n+j_1+a_0x^n+a_j_1x^n-j_1 +a_j_2x^n-j_2+a_j_3x^n-j_3+1 an exhaustive search such that j_1=1,2,…,10,  j_i=j_i-1+1,j_i-1+2,…,10, i=2,3; a_j=±1,± 2,…± 10, j=0,j_1,j_2,j_3 suggests that C(x^2n+x^n+6+x^n+4+x^n+2+x^n +x^n-2+x^n-4+x^n-6+1) has the minimal limit point.

Using the algorithm we solve the equation
cos6t+cos4t+cos 2t+1/2=±1. If we develop cos6t, cos 4t and cos2 t and substitute cos t=x we get bicubic equation 32x^6-40x^4+12x^2-1/2=1 with two real solutions

    α_1,2=±√(60√(√(29)/384+61/3456)+144√(61√(29)/663552+3035/5971968)+7)/12√(√(29)/384+61/3456)

and 32x^6-40x^4+12x^2-1/2=-1 without any real solution. Using symmetry we can show that

    C(P_2n)→2/πarccos(|α_1|) ≈ 0.21854988117598984.




 §.§ Decanomials with smallest limit points of C(P_2n)


Between all decanomials x^2n+a_j_4x^n+j_4+a_j_3x^n+j_3+a_j_2x^n+j_2+a_j_1x^n+j_1+a_j_1x^n-j_1 +a_j_2x^n-j_2+a_j_3x^n-j_3+a_j_4x^n-j_4+1 an exhaustive search such that j_1=1,2,…,10,  j_i=j_i-1+1,j_i-1+2,…,10, i=2,3,4; a_j=±1,± 2,…± 10, j=j_1,j_2,j_3,j_4 suggests that C(x^2n+x^n+7+x^n+5+x^n+3+x^n+1+x^n-1 +x^n-3+x^n-5+x^n-7+1) has the minimal limit point.

Using the algorithm we solve the equation
cos7t+cos5t+cos 3t+cos t=±1. If we develop cos7t, cos 5t and cos3 t and substitute cos t=x we get two equations 64x^7-96x^5+40x^3-4x=±1 with two real solutions
± 0.9521755884525. Using symmetry we can show that

    C(P_2n)→2/πarccos(0.95217558845251615756) ≈ 0.19768155115418617.


We remark that this is the smallest limit point of C(P_2n), n→∞ we know.



 §.§ Polynomials with smallest limit points of C(P_2n)


Our calculations suggest that the next conjecture seems to be true:

If P_2n is a sum of 2k+3 monomials, i.e. a_0 0, then the sequence C(x^2n+x^n+2k+⋯+x^n+4+x^n+2+x^n +x^n-2+x^n-4+⋯+x^n-2k+1) tends to the smallest limit, greater than zero, of C(P_2n), n→∞.

Let k≥ 1 be an integer. If P_2n is a sum of 2k+4 monomials, i.e. a_0=0, then the sequence C(x^2n+x^n+2k+1+⋯+x^n+5+x^n+3+x^n+1+x^n-1 +x^n-3+x^n-5+⋯+x^n-2k-1+1) tends to the smallest limit, greater than zero, of C(P_2n), n→∞.


But in the case of dodecanomials we found that C(x^2n+x^n+9+x^n+7+2x^n+5+2x^n+3+2x^n+1+2x^n-1 +2x^n-3+2x^n-5+x^n-7+x^n-9+1) tends to 2arccos(0.943468)/π=0.215085 which is smaller than 0.226163=2(arccos(0.966357)+arccos(0.877575)-arccos(0.919147))/π the limit of C(x^2n+x^n+9+x^n+7+x^n+5+x^n+3+x^n+1+x^n-1 +x^n-3+x^n-5+x^n-7+x^n-9+1).
Nevertheless the conjecture seems to be true for many k.


C(x^2n+x^n+9+x^n+7+2x^n+5+2x^n+3+2x^n+1+2x^n-1 +2x^n-3+2x^n-5+x^n-7+x^n-9+1) → 2arccos(0.943468)/π =0.215085< 0.226163= 2(arccos(0.966357)+arccos(0.877575)-arccos(0.919147))/π the limit of C(x^2n+x^n+9+x^n+7+x^n+5+x^n+3+x^n+1+x^n-1 +x^n-3+x^n-5+x^n-7+x^n-9+1).


It is natural to ask: do exist

    lim_k→∞lim_n→∞C(x^n(x^n+1/x^n+∑
    _j=1^k (x^2j-1+1/x^2j-1))),


    lim_k→∞lim_n→∞C(x^n(x^n+1+1/x^n+∑
    _j=1^k (x^2j+1/x^2j)))?


It is an easy exercise to prove that

    cos t+cos3t+⋯+cos(2m-1)t=sin2mt/2sin t  ,


    1/2+cos 2t+cos4t+⋯+cos2mt=sin(2m+1)t/2sin t  .


These formulae enable us to calculate f_2 and C(P_2n) much faster. Our experiments with k≈ half of million, n≈ one hundred million suggest that these limits exist and that they are both equal to 0.20885.



§ EXTENSION OF MAHLER MEASURE


The definition of the Mahler measure could be extended to polynomials in several
variables. We recall Jensen's formula which states that
1/2π∫_0^2πlog |P(e^ iθ)|θ̣= log |a_0| + ∑_j=1^d logmax (|α_j|, 1)
Thus

    M(P) = exp{1/2π∫_0^2πlog |P(e^ iθ)|θ̣},

so M(P) is just the geometric mean of |P(z)| on the torus T.

Hence a natural
candidate for M(F) is

    M(F) = exp{1/(2π)^r∫_0^2πθ̣_1⋯∫_0^2πlog |F(e^ iθ_1,…,e^ iθ_r)|θ̣_r }.


The smallest known Mahler measures in two variables are

    M((x + 1)y^2 + (x^2 + x + 1)y + x(x + 1)) = 1.25542…
 and

    M(y^2 + (x^2 + x + 1)y + x^2) = 1.28573…


Boyd proved (1981)<cit.> the next

As m→∞, M(P (x, x^m))→ M(P(x,y)).



Let Q(x_1,x_2,…,x_r)=∑_j=1^ka_jx_1^e_j1x_2^e_j2⋯ x_r^e_jr, a_j∈ℝ,
e_ji∈ℤ, j=1,2,…,k, i=1,2,…,r and let W(x_1,x_2,…,x_r)=

    =(x_1^nx_2^n⋯ x_r^n)(x_1^nx_2^n⋯ x_r^n+x_1^-nx_2^-n⋯ x_r^-n+Q(x_1,x_2,…,x_r)+Q(x_1^-1,x_2^-1,…,x_r^-1)).

where n∈ℕ is greater than max(|e_ji|) so that W is a multivariate polynomial. Let g_2(x_1,x_2,…,x_r):=-1/2(Q(x_1,x_2,…,x_r)+Q(x_1^-1,x_2^-1,…,x_r^-1)) and

    g_2(x_1,x_2,…,x_r) := 
                   1,     |g_2(x_1,x_2,…,x_r)|≥ 1
    
                   0,    otherwise,

then we can define

    LC(W):=1/(2π)^r∫_0^2π∫_0^2π⋯∫_0^2πg_2(exp(it_1),exp(it_2),…,exp(it_r))ṭ_1ṭ_2⋯ṭ_r.

If r=1, Q(x_1)=a_0/2+∑_j=1^k a_jx_1^j then W(x_1)=P_2n(x_1), g_2(exp(it))=f_2(t) and g_2(exp(it))=f_2(t) so that, recalling (<ref>),
we conclude that LC(P_2n)=lim_n→∞C(P_2n).



If Q(x,y)=x+y+1 we can prove that the Boyd's property for LC is valid: LC(W (x, x^m))→ LC(W(x,y)) as m→∞. Indeed
LC(x^2n+x^n(x+ x^m+2+x^-1+x^-m))+1)=lim_n→∞C(x^2n+x^n(x+ x^m+2+x^-1+x^-m))+1)=1/2π∫_0^2πf_2m(t), where f_2m(t)=-1-cos(mt)-cos(t).
Since cos(t)=-cos(π-t) it follows that cos(2(m_1+1)t)=-cos(π-2(m_1+1)t)=-cos((2m_1+1)(π-t)). Therefore if m is odd then for each interval I=[a,b]⊆ [0,π] such that |f_2m(t)|>1, a<t<b, there is the interval I'=[π-b,π-a] of the equal length such that |f_2m(t)|≤ 1, t∈ I'. We conclude that 1/π∫_0^πf_2m(t)=0.5 for m odd so that LC(W (x, x^m))→ 0.5 as m→∞.

On the other hand

    LC(W)     =1/(2π)^2∫_0^2π∫_0^2π-1/2(exp(it_1)+exp(it_2)+2+exp(-it_1)+exp(-it_2))ṭ_1ṭ_2 
        =1/(2π)^2∫_0^2π∫_0^2π(-1-cos t_1-cos t_2)ṭ_1ṭ_2.



Since if t_1,t_2∈ [0,π] then |1+cos t_1+cos t_2|≥ 1 is equivalent with t_2≤π-t_1 and using the symmetry of the set {(t_1,t_2)∈ [0,2π]× [0,2π]:-1-cos t_1-cos t_2≥ 1} it follows that

    LC(W)=4/(2π)^2∫_0^π(∫_0^π-t_1ṭ_2)ṭ_1=1/π^2∫_0^π(π-t_1)ṭ_1=1/2.

This example as well as numerical approximations of LC of many other polynomials in two variables using the formula (<ref>) and the definition (<ref>) suggest us that the Boyd's limit formula in Theorem <ref> is also valid for LC i.e. we propose the following

As m→∞, LC(W (x, x^m))→ LC(W(x,y)).



To prove Theorem <ref> we use two lemmas which Everest and Ward proved in <cit.>.
For the sake of completeness we cite proofs of the lemmas in the Appendix. Denote as usual the (multiplicative) circle group by K=𝕊^1, and the torus by K^2=𝕊^1×𝕊^1. For an integrable function f:K→ℂ, write

    ∫_0^1f(e^2π i θ)dθ=∫ f(x)dμ_K= ∫ fdμ_K


and for an integrable function g:K^2→ℂ, write

    ∫_0^1∫_0^1g(e^2π i θ_1,e^2π i θ_2)dθ_1dθ_2=∫ g(x_1,x_2)dμ_K^2= ∫ gdμ_K^2.

We will use the Lebesgue measure μ_K on the circle to evaluate the measure of disjoint unions of intervals (whose measure is simply the sum of the lengths).


Let ϕ:K^2→ℝ be any continuous function. Then

    lim_N→∞∫ϕ(x,x^N)dμ_K=∫ϕ dμ_K^2





Let ϕ:K^2→ℝ be any Riemann-integrable function and δ>0 be given. There are finite trigonometric series P(x_1,x_2)=∑_||n||<Ma_nx_1^n_1x_2^n_2 and Q(x_1,x_2)=∑_||n||<Mb_nx_1^n_1x_2^n_2 with the property that

    P(x_1,x_2)≤ϕ(x_1,x_2)≤ Q(x_1,x_2)

for all (x_1,x_2)∈ K^2 and

    ∫(Q-P)dμ_K^2<δ,
 
where ||n||=max{|n_1|,|n_2|}, M>0.

The finite sums P, Q used to bound ϕ are called trigonometric polynomials since in the additive group notation the monomial x_1^n_1x_2^n_2  corresponds to e^2π i(n_1θ_1+n_2θ_2) under the correspondence x_1=e^2π i θ_1, x_2=e^2π i θ_2.

Proof of Theorem <ref>.
If r=2 in (<ref>) then function g_2(x_1,x_2)
is not continuous but is Riemann-integrable (since it is bounded and the set of discontinuities
of g_2 has measure 0). By Lemma <ref> there are finite trigonometric series P_2(x_1,x_2)=∑_||n||<Ma_nx_1^n_1x_2^n_2 and Q_2(x_1,x_2)=∑_||n||<Mb_nx_1^n_1x_2^n_2 with the property that

    P_2(x_1,x_2)≤g_2(x_1,x_2)≤ Q_2(x_1,x_2)

for all (x_1,x_2)∈ K^2 and

    ∫(Q_2-P_2)dμ_K^2<δ.

It follows that

    ∫ P_2(x,x^m)dμ_K≤∫g_2(x,x^m)dμ_K≤∫ Q_2(x,x^m)dμ_K.

Function P_2, Q_2 are continuous so by Lemma <ref>

    ∫ P_2(x,x^m)dμ_K→∫ P_2(x_1,x_2)dμ_K^2,


    ∫ Q_2(x,x^m)dμ_K→∫ Q_2(x_1,x_2)dμ_K^2.

Since ∫ Q_2dμ_K^2-∫ P_2dμ_K^2<δ and δ>0 was arbitrary, (<ref>), (<ref>) and (<ref>) then show that


    ∫g_2(x,x^m)dμ_K→∫g_2(x_1,x_2)dμ_K^2.

Recalling (<ref>), (<ref>) and using the substitutions t=2πθ, t_1=2πθ_1, t_2=2πθ_2 it follows that

    ∫ g_2(x,x^m)dμ_K=∫_0^1 g_2(e^2π iθ,e^2mπ iθ)dθ=1/2π∫_0^2π g_2(e^it,e^m it)dt=LC(W(x,x^m))

and

    ∫g_2(x_1,x_2)dμ_K^2=∫_0^1∫_0^1 g_2(e^2π iθ_1,e^2π iθ_2)dθ_1 dθ_2=1/4π^2∫_0^2π∫_0^2π g_2(e^it_1,e^it_2)dt_1dt_2=LC(W(x_1,x_2)),

hence the claim follows.





Acknowledgements: I am grateful to Jonas Jankauskas for his careful reading and for sending me a sketch of a proof of Theorem <ref>.




§ APPENDIX



 §.§ Proof of Lemma <ref>


By the Stone-Weierstrass approximation theorem (see <cit.> Appendix B) for any ϵ>0 there is an M and there are coefficients {a_n}_||n||<M for which

    |ϕ(x_1,x_2)-∑_||n||<Ma_n x_1^n_1x_2^n_2| <ϵ

for all x_1,x_2 ∈ K, where ||n||=max{|n_1|,|n_2|}. Since this estimate is uniform, it is enough to show that the required convergence happens for the function

    G(x_1^n_1x_2^n_2)=∑_||n||<Ma_n x_1^n_1x_2^n_2.

Now

    ∫ G d μ_K^2= ∑_||n||<M∫ a_n x_1^n_1x_2^n_2 d μ_K^2=a_0.

On the other hand

    ∫ G(x,x^N) d μ_K= ∑_||n||<M∫ a_n x_1^n_1+Nn_2 d μ_K=∑_n:n_1+Nn_2=0 a_n.

For fixed M, when m is large n_1+Nn_2=0 with ||n||<M if and only if n_1=n_2=0, so

    lim_m→∞∫ G(x,x^N) d μ_K=a_0=∫ G d μ_K^2  .





 §.§ Proof of Lemma <ref>


By the Stone-Weierstrass approximation theorem (see <cit.> Appendix B), it is enough to find continuous functions P and Q with the stated properties. Since ϕ is Riemann -integrable it is bounded above by R say, and there is a finite collection of rectangles {A_1,A_2,…,A_n} with the property that

    ∑_i=1^narea A_i ·sup_(x_1,x_2)∈ A_i{ϕ(x_1,x_2)}-∫ϕ d μ_K^2<δ/4.

The word "rectangle" means a set of the form

    { (e^2π i θ_1,e^2π i θ_2):θ_j∈[a_j,b_j]]⊂𝕋.}

Now define


    Q(x_1,x_2) = sup_(x_1,x_2)∈ A_i{ϕ(x_1,x_2)}   if (x_1,x_2)∈ A_i∖⋃_j i A_j,
    
                   R    otherwise.


Then Q≥ϕ and their integrals are within δ/4. Now approximate Q from above by a continuous function Q which is equal to Q except very close to the boundary of each rectangle A_i, and fills in continuously to reach the value R on the boundary. This can be done by keeping ∫ Qdμ_K^2 -∫Qdμ_K^2<δ/4, which gives a continuous Q with Q≥ϕ and ∫ Qdμ_K^2 -∫ϕ dμ_K^2<δ/2.

Repeating the argument from below (or simply repeating the argument for -ϕ) gives P.



















































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