Searching for the maximal valence of harmonic polynomials: a new example
SSEARCHING FOR THE MAXIMAL VALENCE OF HARMONICPOLYNOMIALS: A NEW EXAMPLE
SEUNG-YEOP LEE, ANDRES SAEZ
Abstract.
We find a new lower bound for the maximal number of zeros to harmonic poly-nomials, p ( z ) + q ( z ), when deg p = n and deg q = n − Introduction and result
Given two polynomials p ( z ) and q ( z ) of degrees n and m respectively, the maximal numberof roots (i.e. maximal valence) of the harmonic polynomial , p ( z ) + q ( z ), is not known [11] exceptfor a few cases (e.g. when m = n − m = 1 [9, 10]). See also [2, 5, 6, 7, 8, 12].Recently there have been several results [1, 3, 4] on the lower bounds of the maximal valence,see Table 1. Table 1.
The known maximal valence of p + q (deg p, deg q ) ( n, m ) ( n, n −
1) ( n, n −
3) ( n, ≥ m + m + n n ≥ n − n + O (1) 3 n − p, deg q ) =( n, n − n let us define two polynomials, p ( z ) = S ( z ) + T ( z ) and q ( z ) = S ( z ) − T ( z ), where(1) S ( z ) = iz n , T ( z ) = i (cid:0) z + 1 (cid:1) n − (cid:0) z − ( n − (cid:1) . It follows that deg p = n and deg q = n −
2. Since the maximal valence for ( n, m ) = (3 ,
1) isknown (see the above table), we only consider n ≥ Theorem.
Given n ≥ , let the polynomials p and q be given as above. Let k max ( n ) be definedby k max ( n ) = max ≤ k ≤ n/ (cid:26) k : ( n −
2) cot (cid:18) k − n − π (cid:19) − n cot (cid:18) πkn (cid:19) > (cid:27) . Then the total number of zeros, counting the multiplicity, of p ( z ) + q ( z ) is given by n − n + 2 + 4 k max ( n ) . The asymptotic behavior of k max ( n ) as n → ∞ is given by k max ( n ) = (cid:18) − X π (cid:19) n + O (1) ≈ . n + O (1) where X ≈ . is the unique solution to the equation X = cos X . a r X i v : . [ m a t h . C V ] D ec SEUNG-YEOP LEE, ANDRES SAEZ
Table 2.
The number of zeros, n − n + 2 + 4 k max ( n ), of p + q for small n ’s n Number of zeros4 105 176 267 378 549 6910 8611 10512 12613 14914 17415 20116 23417 26518 29819 333 n Number of zeros20 37021 40922 45023 49724 54225 58926 63827 68928 74229 79730 85431 91732 97833 104134 110635 1173
Remark 1.
For general n and m , there exists a conjecture by Wilmshurst [13] on the largestvalence of the harmonic polynomials. Though the conjecture has been disproved [4, 3] for anumber of cases, it has not been checked for many other cases including the case considered inthis paper. Our theorem says that the maximal valence is greater at least by4 k max ( n ) − ≈ . n + O (1) as n → ∞ than the conjectured value of n − n + 4. Our theorem also improves upon the more recentconjecture by the authors that suggests n − n/ O (1) for the asymptotic maximal valenceas n grows to the infinity. In fact, the currect project is motivated by the latter conjecture. Remark 2.
The specific harmonic polynomials that we consider in this paper are not new. Thesame polynomials appeared in [4]. However, instead of obtaining a lower bound on the number ofroots, here we obtain the exact number of roots for the given polynomials. (If one naively appliesthe method in [4] one would only get n − n + 2 as a lower bound.) It is curious whether thesimilar analysis (i.e. exact counting) can be done for the case of m = n − Remark 3.
We note that our valence, n − n + 2 + 4 k max ( n ), is not be the maximal valence. Infact, for some n ’s, we could find harmonic polynomials with higher valence. The example shownin Figure 1 is generated by defining p ( z ) = S ( z )+ T ( z ) and q ( z ) = S ( z ) − T ( z ), where S ( z ) = iz and T ( z ) = i ( z + e i ) ( z − e i ). The number of zeros is 128, two more than 126. We conjecturethat that the maximal valence is either n − n +2+4 k max ( n ) or n − n +4+4 k max ( n ) dependingon n . Remark 4.
We conjecture that, given the degrees, n = deg p and m = deg q , there exists nopolynomial formula in n and m that gives the maximal valence of the harmonic polynomial p + q for all (except possibly for a finite number of cases) n and m . If there exists such a polynomialformula, say P ( n, m ), then for m = n −
2, we must have P ( n, n −
2) = n + An + B for some EARCHING FOR THE MAXIMAL VALENCE OF HARMONIC POLYNOMIALS: A NEW EXAMPLE 3 - - - - Figure 1.
The zero sets of Im T (black) and Re S (red) as defined in Remark 3. (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) Figure 2.
The zero sets of Re S (red) and Im T (black) for n = 6 and 12 (from left)constants A and B . The constants A and B must be intergers since P ( n, n −
2) must be aninteger for each n . Our theorem indicates that the only possibilities are either A = 0 or A = − P (3 ,
1) = 7 and P (4 , ∈ { , , } , one must have A = 0 and B = −
2. And this gives P ( n, n −
2) = n − Acknowledgement.
The authors would like to thank Prof. Dmitry Khavinson and Prof.Catherine B´en´eteau for the motivating discussions in the early stage of the project. The firstauthor was supported by Simons Collaboration Grants for Mathematicians.
SEUNG-YEOP LEE, ANDRES SAEZ Proof
We assume n ≥ p ( z )+ q ( z ) = 2 Re S ( z )+2 i Im T ( z ). Therefore, the zeros of p ( z )+ q ( z )are exactly the intersection points of { z : Re S ( z ) = 0 } and { z : Im T ( z ) = 0 } . In Figure 2 theformer set is depicted in the red lines and the latter set in the black curves. The zero set of Re S is given explicitly by the union of 2 n rays, i.e., { z : Re S ( z ) = 0 } = n (cid:91) k = − n +1 { re iπk/n : 0 ≤ r < ∞} . Therefore, to find the zeros of p ( z ) + q ( z ), it is enough to find the number of intersections oneach ray, i.e. N k := { r ∈ (0 , ∞ ) : Im T (cid:0) re iπk/n (cid:1) = 0 } , for each k = − n + 1 , · · · , n .From the expression of T in (1), we obtain N = 1 and N n = n − T ( re − iπk/n ) = Im T ( re iπk/n ) and, therefore, N k = N − k . As a result weonly need to find N k for k = 1 , · · · , n − Lemma 1.
For ≤ k ≤ n − , N k is given by the number of zeros of A : (0 , πk/n ) → R where A ( θ ) = A ( θ ; k ) = tan[( n − θ ] + n − θ − n cot πkn . Proof.
Let us evaluate arg T (cid:0) re iπk/n (cid:1) = ( n − θ ( r ) + φ ( r ) mod 2 π, where we define the two (angular) variables, θ ( r ) = θ ( r ; k ) = arg (cid:0) re iπk/n + 1 (cid:1) ,φ ( r ) = φ ( r ; k ) = arg (cid:0) re iπk/n − ( n − (cid:1) + π . (2)This gives tan arg T (cid:0) re iπk/n (cid:1) = tan (cid:0) ( n − θ ( r ) (cid:1) + tan φ ( r )1 − tan (cid:0) ( n − θ ( r ) (cid:1) tan φ ( r ) . (3)Using the identitiestan θ ( r ) = tan arg (cid:0) re iπk/n + 1 (cid:1) = r sin (cid:0) πkn (cid:1) r cos (cid:0) πkn (cid:1) + 1 , tan φ ( r ) = tan (cid:16) arg (cid:0) re iπk/n − ( n − (cid:1) + π (cid:17) = − r cos (cid:0) πkn (cid:1) − ( n − r sin (cid:0) πkn (cid:1) , one can relate φ and θ by tan φ ( r ) = n − θ ( r ) − n cot πkn . Using this relation, the numerator oftan arg T ( re iπk/n ) in (3) is written astan (cid:0) ( n − θ ( r ) (cid:1) + tan φ ( r ) = tan (cid:0) ( n − θ ( r ) (cid:1) + n − θ ( r ) − n cot πkn , which is exactly A ( θ ( r )) as defined in the lemma. Since we have Im T ( z ) = 0 (note T ( z ) (cid:54) = 0away from the real axis) if and only if tan arg T ( z ) = 0, N k is given by the number of zeros of For ( n, m ) = (4 ,
2) the maximal valence achieved so far is 12. The maximal valence for even n needs to beeven due to the argument principle. See, for example, [1]. EARCHING FOR THE MAXIMAL VALENCE OF HARMONIC POLYNOMIALS: A NEW EXAMPLE 5 Π Π Π Π Π Figure 3.
Plots of A ( θ ) for n = 12 and k = 1 (left) and k = 5 (right) A ( θ ( r )) over r ∈ (0 , ∞ ). It is simple to check that the denominator in (3) does not vanish whenthe numerator vanishes.Lastly, from (2), one can see that θ ( r ) (the angle from − r ) increases from zero to πk/n monotonically as r moves from zero to ∞ . (cid:4) In the rest of the proof, we will use elementary argument (e.g. the mean value theorem andthe intermediate value theorem) to count the zeros of A ( θ ). See Figure 3 for some plots of A .One notices that A has simple poles (of negative residue) where tan[( n − θ ] has poles, i.e., θ = 1 / n − π, / n − π, · · · , k poles − / n − π, where k poles is the largest integer such that k poles − / n − < kn . We have(4) k poles = k poles ( k ) = (cid:40) k when k < n/ ,k − k ≥ n/ . We also get the boundary behavior of A : lim θ → A ( θ ) = + ∞ and(5) A (cid:16) πkn (cid:17) = − tan (cid:16) πkn (cid:17) − cot (cid:16) πkn (cid:17) = − πk/n ) = (cid:40) ≤ k < n/ ,> k ≥ n/ . (When k = n/
2, lim θ → πk/n A ( θ ) = + ∞ .) To find the critical points of A , we evaluate A (cid:48) ( θ ) = ( n − (cid:18) [( n − θ ] − θ (cid:19) , that becomes zero whencos[( n − θ ] = ± sin θ ⇐⇒ ( n − θ = π ± θ + πj (for j ∈ Z ) , or, equivalently, when θ = j − / n π where j = 1 , · · · , k,j − / n − π where j = 1 , · · · , k crit . SEUNG-YEOP LEE, ANDRES SAEZ
We note that k crit is the largest integer such that k crit − / n − < kn or, equivalently, k crit = k crit ( k ) = k when k < n/ k − n/ ≤ k < n/ k − k ≥ n/ A (cid:18) j − / n − π (cid:19) = ( n −
2) cot (cid:18) j − / n − π (cid:19) − n cot (cid:16) πkn (cid:17) ,A (cid:18) j − / n π (cid:19) = n cot (cid:18) j − / n π (cid:19) − n cot (cid:16) πkn (cid:17) > . (6)The last inequality is from the monotonicity of cot( x ) over 0 < x < π . Lemma 2.
For ≤ k ≤ n − we have A (cid:18) j − / n − π (cid:19) > for ≤ j ≤ min { k crit , k − } . Proof.
Since A (cid:0) j − / n − π (cid:1) is monotonic in j , it is enough to prove that A (cid:18) k (cid:48) − / n − π (cid:19) = ( n −
2) cot (cid:18) k (cid:48) − / n − π (cid:19) − n cot (cid:16) πkn (cid:17) > , for k (cid:48) = min { k crit , k − } . Using the following identity,(7) ( n −
2) cot θ − n cot θ = ( n −
1) sin( θ − θ ) − sin( θ + θ )sin θ sin θ , the above inequality in question becomes F ( k ) := ( n −
1) sin (cid:18) n − k n − n π (cid:19) − sin (cid:18) k − / n − π + πkn (cid:19) > k < n/ ,F ( k ) := ( n −
1) sin (cid:18) n − k n − n π (cid:19) − sin (cid:18) k − / n − π + πkn (cid:19) > k ≥ n/ . We have F ( k ) > k > C n := n (2 n − n − F ( k ) contribute positively (we defined C n such that the second term of F ( k ) vanishes when k = C n ). For k ≤ C n , since the first term in F ( k ) decreases monotonicallyin k and is given by (using sin x > xπ for 0 < x < π/ n −
1) sin π n − > k = C n ,F ( k ) > k ≤ C n .Similarly, F ( k ) > k ≥ n/ F ( k ) contribute positively. (cid:4) From (6) and Lemma 2, the only possible critical point θ such that A ( θ ) ≤ θ = k − / n − π, k < n . Note that this is bigger than k − / n − π which is the location of the rightmost pole of A . Since allthe other critical values are positive, it follows that, between any successive poles of A (exceptbetween θ = 0 and θ = π n − ), there is exactly one root of A , by applying the intermediate EARCHING FOR THE MAXIMAL VALENCE OF HARMONIC POLYNOMIALS: A NEW EXAMPLE 7 value theorem and the mean value theorem. Therefore, there are total k poles − A thatare located to the left of the rightmost pole. By applying the intermediate value theorem with(5), for all k ≥ n/
2, there is at least one root of A to the right of the rightmost pole of A .Combining with (4) there are at least k − A for all ≤ k ≤ n −
1. The total numberof zeros of p + q is therefore at least(8) n + 2 n − (cid:88) k =1 ( k −
1) = n − n + 2 . For k < n/
2, again applying the intermediate value and the mean value theorem between therightmost pole and kπ/n , there are no zero when k ≥ n/ k < n/ A (cid:16) k − / n − π (cid:17) < , two zeros when k < n/ A (cid:16) k − / n − π (cid:17) > . Note that the second inequality is exactly the condition to define k max ( n ) in the main theorem.Using the identity (7), the inequality, A (cid:0) k − / n − π (cid:1) >
0, holds if and only if( n −
1) sin (cid:18) n − k n − n π (cid:19) − sin (cid:18) k − / n − π + πkn (cid:19) > . For k < n/
4, the left hand side is monotonically decreasing in k and, therefore, the inequality issatisfied exactly for k ≤ k max ( n ). Therefore we get 4 k max ( n ) more roots of p + q than (8). Thisproves our theorem except the statement about the asymptotic behavior.In terms of the new parameter, γ = k/n , the left hand side of the above inequality becomes( n −
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