The zero distribution of polynomials with a three-term recurrence Khang Tran Truman State University October 20, 2014 1 / 24 An objective We study the zero distribution of a sequence of polynomials Hm (z) generated by ∞ 1 ∑ Hm (z)t m = 1 + B(z)t + A(z)t n m=0 where A(z) and B(z) are polynomials with complex coefficients. For example, when B(z) = z 3 + i, A(z) = z 2 + 1, and n = 5 the zeros of H100 (z) are plotted below. 1.0 0.5 -1.0 -0.5 0.5 1.0 -0.5 -1.0 2 / 24 Main theorem Theorem Let Hm (z) be a sequence of polynomials whose generating function is ∞ 1 ∑ Hm (z)t m = 1 + B(z)t + A(z)t n m=0 where A(z) and B(z) are polynomials in z with complex coefficients. There is a constant C = C (n) such that for all m > C , the zeros of Hm (z) which satisfy A(z) 6= 0 lie on a fixed curve C given by ℑ B n (z) =0 A(z) and 0 ≤ (−1)n ℜ B n (z) nn ≤ A(z) (n − 1)n−1 and are dense there as m → ∞. 3 / 24 An example Consider the sequence of polynomials, Hm (z), generated by ∞ 1 ∑ Hm (z)t m = 1 + (z 2 − 2z + 3)t + z 2 t 2 . m=0 The zeros of H50 (z) are plotted below. Figure: The zeros of H50 (z) 4 / 24 An example We recall that ∞ 1 ∑ Hm (z)t m = 1 + (z 2 − 2z + 3)t + z 2 t 2 . m=0 Let z = x + iy . We note that B 2 (z) A(z) B 2 (z) ℜ A(z) ℑ = = 2y (x 2 + y 2 − 3)P (x 2 + y 2 )2 P2 − Q2 (x 2 + y 2 )2 where P = 3x − 2x 2 + x 3 − 2y 2 + xy 2 Q = y (x 2 + y 2 − 3). 5 / 24 Some ideas We will show that the number of zeros of Hm (z) on the curve C is at least the degree of Hm (z). Hence this number equals the degree and all the zeros of Hm (z) lie on this curve. Let z be a zero of Hm (z) and ti = ti (z), 0 ≤ i < n, be the zeros of D(t, z) = 1 + B(z)t + A(z)t n . We will consider the quotients of zeros ti /tj since these quotients seem to lie on a fixed curve independent of A(z), B(z). 6 / 24 Distribution of quotients of zeros Figure: Quotients of zeros when n = 3 7 / 24 Distribution of quotients of zeros Figure: Quotients of zeros when n = 4 8 / 24 Distribution of quotients of zeros 1.5 1.0 0.5 -1.5 -1.0 0.5 -0.5 1.0 -0.5 -1.0 -1.5 Figure: Quotients of zeros when n = 5, m = 50 9 / 24 Distribution of quotients of zeros 1.0 0.5 -1.0 0.5 -0.5 1.0 -0.5 -1.0 Figure: Quotients of zeros when n = 6, m = 30 10 / 24 The q-discriminant Definition (Ismail) The q-discriminant, Discx (P; q), of a polynomial Pn (x) of degree n and leading coefficient p is defined as p 2n−2 q n(n−1)/2 ∏ (q −1/2 xi − q 1/2 xj )(q 1/2 xi − q −1/2 xj ) 1≤i<j≤n where x1 , . . . , xn are the zeros of Pn (x). 11 / 24 The q-disriminant of the denominator Proposition The q-discriminant of D(t) = 1 + Bt + At n is n n−1 n n (1 − q n−1 )n−1 n−2 B q n−1 A(1 − q ) Disct (D(t); q) = ±A + (−1) . (1 − q)n−1 (1 − q)n Corollary If q is a quotient of zeros ti /tj of D(t, z) then (−1)n B n (z) (1 − q n )n = . A(z) (1 − q)q n−1 (1 − q n−1 )n−1 12 / 24 Our problem We want to show the number of zeros of Hm (z) on C is at least the degree of Hm (z) which is ( mb if nb > a pa + rb if nb ≤ a where m = pn + r , 0 ≤ r < n, and a, b are the degrees of A(z) and B(z) respectively. In fact, we will count the number of quotients of zeros q = ti /tj of D(t, z) on the arc e 2iθ , 0 ≤ θ ≤ π/n. This relates to the number of zeros of Hm (z) on C by (−1)n B n (z)(1 − q)q n−1 (1 − q n−1 )n−1 = (1 − q n )n A(z). It is not difficult to show the special point θ = π/n gives rb zeros z of Hm (z) on C . It remains to show the number of quotients q on the arc e 2iθ , 0 ≤ θ < π/n is p = bm/nc. 13 / 24 Distribution of quotients of zeros 1.5 1.0 0.5 -1.5 -1.0 0.5 -0.5 1.0 -0.5 -1.0 -1.5 Figure: Quotients of zeros when n = 5, m = 50 14 / 24 Distribution of quotients of zeros 1.0 0.5 -1.0 0.5 -0.5 1.0 -0.5 -1.0 Figure: Quotients of zeros when n = 6, m = 30 15 / 24 A lemma Lemma For a fixed z, let t0 , . . . , tn−1 be the zeros in t of D(t, z) = 1 + B(z)t + A(z)t n and qi = ti /t0 , 0 ≤ i < n. If z is a point on C such that q1 = e 2iθ , θ ∈ R then z is a zero of Hm (z) if and only if n−1 1 h(θ ) := ∑ m+1 0 =0 Pθ (ζk ) k=0 ζk where ζ0 , . . . , ζn−1 are the zeros of Pθ (ζ ) = ζ n − sin nθ sin(n − 1)θ ζ+ . sin θ sin θ 16 / 24 A lemma Lemma If 0 ≤ θ < π/n then, besides the two zeros e ±iθ , the remaining n − 2 zeros of the polynomial Pθ (ζ ) = ζ n − sin nθ sin(n − 1)θ ζ+ sin θ sin θ lie outside the closed unit disk, i.e., |ζ | > 1. 17 / 24 Consequences of the two lemmas Recall that we want to count the number of zeros θ on [0, π/n) of n−1 1 h(θ ) := ∑ m+1 0 =0 Pθ (ζk ) k=0 ζk where ζ0 , . . . , ζn−1 are the zeros of Pθ (ζ ) = ζ n − sin(n − 1)θ sin nθ ζ+ . sin θ sin θ Since ζ0,1 = e ±iθ and |ζk | > 1 if 2 ≤ k < n, in general the function h(θ ) is dominated by the two summands when ζk = e ±iθ . 18 / 24 Consequences of the two lemmas The sum of these two terms is 2n cos(m + n)θ − 2 cos(m + 1)θ sin nθ / sin θ . |Pθ0 (e iθ )|2 The graph of this sum when m = 40, n = 4 is 4 2 0.2 0.4 0.6 0.8 -2 -4 Figure: The sum of two main terms 19 / 24 A technical case When θ is close to π/n, say θ = π/n − l/m, we have 2 l lπ(cos π/n − ek ) + On ζk = ek 1 + m sin π/n m2 where ek = e (2k−1)iπ/n , 0 ≤ k ≤ n − 1, and the first two terms in n−1 1 h(θ ) = ∑ m+1 0 Pθ (ζk ) k=0 ζk sum of the two terms when ζk = e ±iθ will not dominate sum when m is big. In this case the sign of h(θ ) is the same as that of 1 n−1 −m−n lnek 1 n−1 −m−n ∞ (lnek )j ∑ ek e = n ∑ ek ∑ j! n k=0 j=0 k=0 20 / 24 A sketch of the proof Let m = nq + r . With the fact that ( 0 a ∑ ek = n(−1)a/n k if n - a if n|a the double summation becomes ∞ (ln)jn+r ∑ (−1)j+p+1 (jn + r )! . j=0 This is an alternating series whose sign is the sign of the term when j = h where l = h + r /n. 21 / 24 Future development If the generating function is ∞ ∑ m=0 Hm (z)t m = ∑ Ck (z)t k 1 + B(z)t + A(z)t n then the zeros of Hm (z) will approach the same curve C and a finite set of points when m → ∞ (a joint work with Robert Boyer). An exprimental work suggests that if Ck (z)/B k (z)C0 (z) is real whenever z ∈ C then the zeros of Hm (z) lie exactly on C (except a finite set of points). It is a necessary and sufficient condition when n = 2. The author believes the same argument will work when the generating function is 1/(1 + B(z)t l + A(z)t n ). When the generating function is 1/ ∑ Ak (z)t k , can we find a condition on Ak (z) so that the zeros of Hm (z) lie on an explicit curve? 22 / 24 Future development For example when the generating function is 1/(1 + C (z)t + B(z)t 2 + A(z)t 3 ), an experimental work suggests that if B(z)C (z)/A(z) ≡ K ∈ R, K ≤ 0 or K ≥ 9, then the zeros lie on the curve given by 3 3 B (z) C 3 (z) B (z) C 3 (z) ℑ + = 0, ℜ − ≥ 0, A2 (z) A(z) A2 (z) A(z) and ℜ B 3 (z) C 3 (z) + A2 (z) A(z) ≥ K 2 + 18K − 27 . 4 If A(z), B(z), and C (z) are any polynomials with complex coefficients then an experimental work suggests that the zeros approach a portion of a curve given defined by an explicit equation involving the imaginary and the real parts of B 3 (z)/A2 (z), C 3 (z)/A(z), B(z)C (z)/A(z). Can we find this explicit portion? (i.e. can we find the inequality conditions?) 23 / 24 Thank you Thank you. 24 / 24

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