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In this paper, we use the analytic method of Odlyzko and Richmond to study the log-concavity of power series. If f ( z ) = (cid:2) n a n z n is an inﬁnite series with a n ≥ 1 and a 0 + · · · + a n = O ( n + 1) for all n , we prove that a super-polynomially long initial segment of f k ( z ) is log-concave. Furthermore, if there exists constants C > 1 and α < 1 such that a 0 + · · · + a n = C ( n + 1) − R n where 0 ≤ R n ≤ O (( n + 1) α ), we show that an exponentially long initial segment of f k ( z ) is log-concave. This resolves a conjecture proposed by Letong Hong and the author, which implies another conjecture of Heim and Neuhauser that the Nekrasov-Okounkov polynomials Q n ( z ) are unimodal for suﬃciently large n .


Introduction
A polynomial or infinite series p(z) is said to be unimodal / log-concave if the sequence of its coefficients is unimodal / log-concave.More explicitly, let a k denote the coefficient of z k in p(z).We say p(z) is unimodal if there exists an index m such that a 0 ≤ a 1 ≤ • • • ≤ a m and a m ≥ a m+1 ≥ • • • .We say p(z) is log-concave if a k ≥ 0 for all k ≥ 0, and a 2 k ≥ a k−1 a k+1 for all k ≥ 1.
Many methods have been developed to show the unimodality and log-concavity of various combinatorial polynomials.We refer to the excellent surveys of Brenti [3] and Stanley [10,11] for a collection of these methods, which encompass algebra, combinatorics, and geometry.For example, breakthrough works by Adiprasito, Braden, Huh, Katz, Matherne, Proudfoot, Wang, and many others [1,2,7] use methods in algebraic geometry to settle the Mason and Heron-Rota-Welsh conjecture on the log-concavity of the chromatic polynomial of graphs and the characteristic polynomial of matroids.
This paper settles the unimodality of the Nekrasov-Okounkov polynomials, an important family of polynomials in combinatorics and representation theory.The Nekrasov-Okounkov polynomials are defined by where λ runs over all Young tableaux of size n, and H(λ) denotes the multiset of hook lengths associated with λ.They are central in the groundbreaking Nekrasov-Okounkov identity [8] ∞ n=0 All data generated or analysed during this study are included in this published article. 1 In [4], Heim and Neuhauser posed the conjecture that Q n (z) are unimodal polynomials.In [6], the author and Letong Hong showed that for sufficiently large n, the unimodality of Q n (z) is implied by the following conjecture.
Conjecture 1.1.Let f (x) be the infinity series defined by where σ −1 (n) = d|n d −1 .Let a n,k be the coefficient of z n in f k (z).There exists a constant C > 1 such that for all k ≥ 2 and n ≤ C k , we have In other words, an exponentially long initial segment of f k (z) is log-concave as k goes to infinity.See [5] for more comments on this conjecture.
This conjecture is closely related to a theorem of Odlyzko and Richmond in 1985.In [9], they showed the following remarkable result: if p(z) is a degree d polynomial with non-negative coefficients such that the coefficients a 0 , a 1 , a d−1 , a d are all strictly positive, then there exists an n 0 such that p n (z) is log-concave for all n > n 0 .However, Odlyzko and Richmond pointed out that this result does not trivially generalize to infinite series.
We call an infinite series f (z) 1-lower bounded if for any n ≥ 0, the coefficient of a n z n be a 1-lower bounded infinite series.Suppose there exist constants c, C > 0 such that for any r ∈ (0, 1) and i ∈ {0, 1, 2, 3}, we have Let a n,k denote the coefficient of z n in the infinite series f k (z).Then there exists a constant A > 1 depending on c and C only such that To rephrase the conditions in the main theorem, we have a n z n be a 1-lower bounded infinite series.Suppose there exists a constant C > 0 such that for any n, we have Let a n,k denote the coefficient of z n in the infinite series f k (z).Then there exists a constant A > 1 depending only on C such that We use an ad-hoc argument to strengthen Theorem 1.2 for infinite series that satisfy a stronger condition.
Theorem 1.4.Let a n z n be a 1-lower bounded infinite series.Suppose there exists constants C > 1, D > 0 and α ∈ [0, 1) such that for all n, we have Let a n,k be the coefficient of z n in the infinite series f k (z).Then for sufficiently large k and any Here A is the constant As a corollary, we establish Conjecture 1.1.
Corollary 1.5.Let f, a be as defined in Conjecture 1.1.Then for any fixed η < η 0 , for all sufficiently large k and n ≤ η k , we have Here η 0 is the constant Combining with Theorem 1.3 of [6], we have shown that Theorem 1.6.The Nekrasov-Okounkov polynomial Q n (z) is unimodal for all sufficiently large n.

Acknowledgement
The author thanks Prof. Ken Ono and Letong Hong for their long support and helpful advice throughout this project.The author thanks the anonymous referees for their detailed suggestions.

Proof of Theorem 1.2 and Corollary 1.3
In this section, we show Theorem 1.2.We use essentially the same method as Odlyzko-Richmond [9], with some modifications to accommodate the different behavior of our infinite series f (z).We also note that our notation of k, n is reversed from the convention in [9].
Throughout the proof, for every positive integer i, let c i denote a small positive constant that depends on c, C only, and let C i denote a large positive constant that depends on c, C only.We denote c 0 = c and C 0 = C.To ensure acyclic constant choice, any c i , C i is determined by c j and C j for j < i.We also assume k is sufficiently large relative to all the constants.If n ≤ k 1/4 , the proof in [9] can be applied verbatim to prove Theorem 1.2.We now prove Theorem 1.2 when n ≥ k 1/4 .
The function f (z) is holomorphic for |z| < 1.For any r ∈ (0, 1), we have So it suffices to show that for any n ∈ [k 1/4 , A k 1/3 ], there exists an r ∈ (0, 1) such that for α ∈ [n − 1, n + 1], the real-valued function is positive and log-concave.We will show the stronger conclusion that F is concave, that is The crucial idea is to study the argument of f (re iθ ).We first note that f is nonzero in a neighborhood of z = 1 − r.
So we have In particular, Furthermore, ψ is odd.By Taylor's formula, there exists some ξ ∈ (0, θ) such that We compute that By Lemma 2.1, we have f (re iξ ) > c(1 − r)/2, and by assumption, We conclude that Substituting in, we get the desired estimate.
As A(0) = 0 and lim r→1 A(r) = ∞, there exists an r 0 ∈ (0, 1) such that A(r 0 ) = n k .We have r 0 = Θ(min(1, n/k)).As n ≥ k 1/4 , for k sufficiently large we have We take for a c 3 < c 2 to be determined later, and split the integral We now estimate the two summands.By Lemma 2.2, for any θ with |θ| < θ 0 we have In particular, if we take the c 3 in the definition of θ 0 to be min(c 2 , π/(8C 2 )), then Thus we conclude that We apply Lemma 2.1 to obtain We can extract the constant f k (r 0 ) and apply a change of variable So we obtain the estimate

By assumption we have
So we conclude that To estimate the second summand, which is the integral over |θ| ∈ (θ 0 , π), we need a lemma about the upper bound of f away from the positive real axis.
Lemma 2.3.For any r ∈ (0, 1) and θ ∈ [−π, π), we have Proof.As we assumed that a n ≥ 1 for every n, we have In particular, we get Thus we have (1 − r) 3 where the second inequality follows from as desired.
By the lemma, for every θ with |θ| > θ 0 = c 3 (1 − r 0 )(kr 0 ) −1/3 , we have Thus we conclude that |θ|∈(θ 0 ,π) Substituting the value of θ 0 , we get Finally, we combine the estimates (2) and (3) to conclude that If C 10 k < n, then for sufficiently large k, we have So there exists a c 11 > 0 such that if C 10 k < n ≤ e c 11 k 1/3 , then If k 1/4 ≤ n ≤ C 10 k, then for sufficiently large k, we have In both cases we have Thus, we have shown Theorem 1.2.
Corollary 1.3 is an easy corollary of Theorem 1.2.
Proof of Corollary 1.3.For each non-negative integer i and r ∈ (0, 1), we note that On one hand, we have On the other hand, by Abel summation, we have Thus for any i ≥ 0 and r ∈ (0, 1) we have The corollary follows by applying Theorem 1.2.

Proof of Theorem 1.4 and Corollary 1.5
In this section, we assume that f (z) = n a n z n satisfies the condition of Theorem 1.4:For all n, we have a n ≥ 1 and 0 We observe that f (z) also satisfies the condition in Corollary 1.3, so there exists a B > 1 such that a 2 n,k ≥ a n−1,k a n+1,k for all n ≤ B k 1/3 .We now use a different method to show that a 2 n,k ≥ a n−1,k a n+1,k for all B k 1/3 ≤ n ≤ A k /k 2 and k sufficiently large.

The inequality a
The key observation is that the second order difference a n+1,k − 2a n,k + a n−1,k can be bounded.
We introduce a notation: for any n ≥ 0, define a n = 1 and a n = a n − 1.Then we have x i + a (1)   x i For a tuple Then we have We prove a series of lemmas that gives the desired control over the second-order difference.
Lemma 3.1.There exists a constant C 1 > 0 such that for any n and k ≥ 2, we have where Proof.We take C 1 = D/ min(C − 1, 1), and argue by induction on k.For k = 2, the statement is clear as Now suppose the lemma holds for k ′ = k − 1.To prove the lemma for k, we observe x 1 a x 1 )(a ).
Using (4), we have where We first continue estimating the main term.By the induction hypothesis, we have (C − 1) By the induction hypothesis, the subtracted term is positive.Again by the induction hypothesis, we estimate that To estimate error term S n,k , we first note that a for any x 2 .By (4), we conclude that S n,k is non-negative.On the other hand, by (4) and the induction hypothesis we have We estimate that Combining all the estimates, we conclude that where Lemma 3.2.For any n and k ≥ 2, if a tuple I ∈ {0, 1} k has k 0 zeros and k 1 ones with k 0 ≥ 1, then where Proof.By definition, permuting the entries of I does not change the value of a I n , so without loss of generality we can assume n,I = 0, and the lemma is obvious.Now assume k 1 ≥ 1.We have By Lemma 3.1 we have the bound Thus S n,I ≥ 0. We also have the upper bound So we have the desired inequality We arrive at the crucial second-order difference estimates.Lemma 3.3.There exists a constant C 2 > 0 such that for any n ≥ −1 and k ≥ 3, if the tuple I ∈ {0, 1} k has k 0 zeros and k 1 ones with k 0 ≥ 3, then where Proof.If I ′ is the tuple obtained by removing two zeros from I, then Thus we find that = a I ′ n+1 .Applying Lemma 3.2, we obtain Finally, we note that n,I ) where 0 ≤ S The error term S where R (2) Proof.Throughout the proof, we use R i to denote the various error term that contribute to R We split the sum into two parts.Let S 1 be the set of I ∈ {0, 1} k with at least three ones, and let S 2 be the set of I ∈ {0, 1} k with at most 2 ones.Then Let k 1 (I) denote the number of ones in I.By Lemma 3.3, the second-order difference of the first term is where n,I ≤ The residue R 1 of ( 5) is bounded by The main term of (5) satisfies Note that Thus we conclude that where R 1 , R 2 are controlled by ( 6) and ( 7) respectively.
It remains to estimate By (4), we have a n ≤ C + Dn α ≤ (C + D)n α .Thus We can appeal to Lemma 3.2 to obtain Thus we obtain We conclude that Thus the second order difference R 3 of I∈S 2 a I n is bounded by for some constant E 1 .
We have thus finished the second order difference estimate where the errors R i satisfy ( 6), ( 7) and ( 8) respectively.We now note that the absolute value of each R i is at most a constant times Thus we obtain the desired estimate.
Using the identity We conclude an analogous estimate on the first-order difference.where R n,k satisfies R n,k ≤ E k 2 (n + k) 1−α + (n + k) 2+α A −(2+α)k for some constant E.
Again using the identity We conclude an analogous estimate on the zeroth-order difference.
Corollary 3.6.For any n ≥ 1 and k ≥ 3, we have where R n,k ) 2 where for each i ∈ {0, 1, 2} we have for constants E 0 , E 1 , E 2 .If k 5/(1−α) ≤ n ≤ A k k 2 then for sufficiently large k, we have for each i ∈ {0, 1, 2}.Therefore we get So {a n,k } is log-concave for k 5/(1−α) ≤ n ≤ η k k 2 .As we have shown that {a n,k } is log-concave for n ≤ B k 1/3 , where B > 1 is a constant, the two intervals glue together to obtain Theorem 1.4.

Finally, we conclude by Lemma 3 . 4 ,
Corollary 3.5 and Corollary 3.6 that for any n ≥ −1, we have