Partitions into Values of a Polynomial

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1 Partitions into Values of a Polynomial Ayla Gafni University of Rochester Connections for Women: Analytic Number Theory Mathematical Sciences Research Institute February 2, 2017

2 Partitions A partition of a number n is a non-increasing sequence of positive integers whose sum is equal to n. The individual summands of a partition are called parts. We denote the number of partitions of n by p(n).

3 Partitions A partition of a number n is a non-increasing sequence of positive integers whose sum is equal to n. The individual summands of a partition are called parts. We denote the number of partitions of n by p(n). Example: p(4) = 5: 4, 3 + 1, 2 + 2, ,

4 Partitions A partition of a number n is a non-increasing sequence of positive integers whose sum is equal to n. The individual summands of a partition are called parts. We denote the number of partitions of n by p(n). Example: p(4) = 5: 4, 3 + 1, 2 + 2, , We can study other partition functions by restricting the set of allowable parts: Partitions into odd numbers, squares, primes, etc.

5 Partitions into powers Fix k 2. Let p k (n) denote the number of partitions whose parts are all perfect k-th powers.

6 Partitions into powers Fix k 2. Let p k (n) denote the number of partitions whose parts are all perfect k-th powers. Example: p 2 (10) = 4: 9 + 1, , ,

7 Partitions into powers Fix k 2. Let p k (n) denote the number of partitions whose parts are all perfect k-th powers. Example: p 2 (10) = 4: 9 + 1, , , Example: p 3 (35) = 7: , , , , , , 1+ +1

8 Partitions into powers Fix k 2. Let p k (n) denote the number of partitions whose parts are all perfect k-th powers. Example: p 2 (10) = 4: 9 + 1, , , Example: p 3 (35) = 7: , , , , , , There are similarities to Waring s problem, but we are not restricting the number of parts.

9 Hardy & Ramanujan Theorem (Hardy - Ramanujan, 1918) ( p(n) 1 ) 2n 4n 3 exp π 3 [Notation: f(x) g(x) means that lim x f(x) g(x) = 1]

10 Hardy & Ramanujan Theorem (Hardy - Ramanujan, 1918) ( p(n) 1 ) 2n 4n 3 exp π 3 Claim (Hardy and Ramanujan, 1918) ( ( 1 log p k (n) (k + 1) k Γ ) ( ζ )) k/(k+1) n 1/(k+1). k k [Notation: f(x) g(x) means that lim x f(x) g(x) = 1]

11 E. Maitland Wright In 1934, E. Maitland Wright published a paper in Acta Mathematica, entitiled Asymptotic partition formulae III: Partitions into k-th powers. The main result of this paper is an asymptotic formula for the p k (n).

12 E. Maitland Wright In 1934, E. Maitland Wright published a paper in Acta Mathematica, entitiled Asymptotic partition formulae III: Partitions into k-th powers. The main result of this paper is an asymptotic formula for the p k (n). In 2014, R.C. Vaughan revisited the problem, providing a much simpler proof for p 2 (n) with a decent error term.

13 Partitions into k-th powers Theorem (G., 2016) p k (n) = ( ) exp k+1 ζ( k+1 k 2 k )Γ( 1 k )X 1 k 1 2 (2π) k+2 2 X 3 2 Y 1 2 ( π J c h h=1 Y h 2 ) + O( 1 Y J ) Here X and Y satisfy X C 1 n k/(k+1) and Y C 2 n 1/(k+1).

14 Partitions into k-th powers Theorem (G., 2016) p k (n) = ( ) exp k+1 ζ( k+1 k 2 k )Γ( 1 k )X 1 k 1 2 (2π) k+2 2 X 3 2 Y 1 2 ( π J c h h=1 Y h 2 ) + O( 1 Y J ) Here X and Y satisfy X C 1 n k/(k+1) and Y C 2 n 1/(k+1). Consequence ( ( 1 log p k (n) (k + 1) k Γ ) ( ζ )) k/(k+1) n 1/(k+1). k k [Notation: f(x) g(x) means that lim x f(x) g(x) = 1]

15 Partitions into Polynomials Fix a polynomial f(x) = a(x α 1 )(x α 2 ) (x α k ), and let S f = {f(m) : m N}.

16 Partitions into Polynomials Fix a polynomial f(x) = a(x α 1 )(x α 2 ) (x α k ), and let S f = {f(m) : m N}. Define p f (n) to be the number of partitions of n whose parts are all members of S f.

17 Partitions into Polynomials Fix a polynomial f(x) = a(x α 1 )(x α 2 ) (x α k ), and let S f = {f(m) : m N}. Define p f (n) to be the number of partitions of n whose parts are all members of S f. Suppose that f is integer-valued, GCD(S f ) = 1, R(α j ) < 1 for each j.

18 Partitions into Polynomials Fix a polynomial f(x) = a(x α 1 )(x α 2 ) (x α k ), and let S f = {f(m) : m N}. Define p f (n) to be the number of partitions of n whose parts are all members of S f. Suppose that f is integer-valued, GCD(S f ) = 1, R(α j ) < 1 for each j. Goal: Find an asymptotic formula for p f (n).

19 Example Let f(x) = 1 2 x(x + 1). Then S f = {1, 3, 6, 10,...}. Thus p f (n) counts the number of partitions into triangle numbers. p f (10) = 7: 10, , , , , ,

20 Special Cases (Vaughan, 2014) f(x) = x 2. Partitions into squares ( ( )) 1 3 2/3 log p 2 (n) 3 πζ n 4 1/3 2

21 Special Cases (Vaughan, 2014) f(x) = x 2. Partitions into squares ( ( )) 1 3 2/3 log p 2 (n) 3 πζ n 4 1/3 2 (G., 2016) f(x) = x k. Partitions into k-th powers ( ( 1 log p k (n) (k + 1) k Γ ) ( ζ )) k k+1 1 n k+1. k k

22 Special Cases (Vaughan, 2014) f(x) = x 2. Partitions into squares ( ( )) 1 3 2/3 log p 2 (n) 3 πζ n 4 1/3 2 (G., 2016) f(x) = x k. Partitions into k-th powers ( ( 1 log p k (n) (k + 1) k Γ ) ( ζ )) k k+1 1 n k+1. k k (Berndt, Malik, Zaharescu, 2017) f(x) = (bx + a) k, (a, b) = 1. Partitions into powers of an arithmetic progression ( ( 1 log p k;a,b (n) (k + 1) bk Γ ) ( ζ )) k k+1 1 n k+1. k k

23 General Polynomials Let f(x) = a(x α 1 ) (x α k ).

24 General Polynomials Let f(x) = a(x α 1 ) (x α k ). Claim The methods of Vaughan, G., and Berndt-Malik-Zaharescu can be extended to prove that ( ( 1 log p f (n) (k + 1) k Γ ) ( ζ )) k/(k+1) (n k k a ) 1/(k+1)

25 General Polynomials Let f(x) = a(x α 1 ) (x α k ). Claim The methods of Vaughan, G., and Berndt-Malik-Zaharescu can be extended to prove that ( ( 1 log p f (n) (k + 1) k Γ ) ( ζ )) k/(k+1) (n k k a This implies that the number of partitions of n into triangle numbers [f(x) = 1 2x(x + 1)] grows roughly like the number of partitions of 2n into squares [f(x) = x 2 ]. ) 1/(k+1)

26 Generating Functions The partition function p(n) has generating function Ψ(z) = p(n)z n = n=0 (1 z n ) 1. n=1

27 Generating Functions The partition function p(n) has generating function Ψ(z) = p(n)z n = n=0 (1 z n ) 1. n=1 Restricting parts to values of f(n), we get a generating function for p f (n): Ψ f (z) = p f (n)z n = n=0 (1 z f(n) ) 1. n=1

28 Generating Functions The partition function p(n) has generating function Ψ(z) = p(n)z n = n=0 (1 z n ) 1. n=1 Restricting parts to values of f(n), we get a generating function for p f (n): Ψ f (z) = p f (n)z n = n=0 (1 z f(n) ) 1. n=1 We can use Cauchy s Theorem to extract the coefficient of z n : p f (n) = 1 0 ρ n Ψ f (ρe 2πiΘ )e 2πinΘ dθ.

29 Generating Functions We have Ψ f (z) = (1 z f(n) ) 1. n=1

30 Generating Functions We have Define Ψ f (z) = Φ f (z) = (1 z f(n) ) 1. n=1 j=1 n=1 1 j zjf(n).

31 Generating Functions We have Define Then Ψ f (z) = Φ f (z) = (1 z f(n) ) 1. n=1 j=1 n=1 1 j zjf(n). Ψ f (z) = exp(φ f (z))

32 Generating Functions We have Define Then Ψ f (z) = Φ f (z) = (1 z f(n) ) 1. n=1 j=1 n=1 1 j zjf(n). Ψ f (z) = exp(φ f (z)) and our problem is now p f (n) = 1 0 ρ n exp(φ f (ρe(θ)) 2πinΘ) dθ.

33 Hardy-Littlewood Circle Method p f (n) = 1 0 ρ n exp(φ f (ρe(θ)) 2πinΘ) dθ. The critical step is to consider objects of the form x n e(θf(x)). If Θ is close to a rational number with small denominator, then we can obtain good estimates for the integrand. These regions make up the Major Arcs, and contribute to the main term. The gaps between the major arcs are called Minor Arcs. [Notation: e(α) = e 2πiα ]

34 Hardy-Littlewood Circle Method p f (n) = 1 0 ρ n exp(φ f (ρe(θ)) 2πinΘ) dθ. The critical step is to consider objects of the form x n e(θf(x)). If Θ is close to a rational number with small denominator, then we can obtain good estimates for the integrand. These regions make up the Major Arcs, and contribute to the main term. The gaps between the major arcs are called Minor Arcs. We will split the integral into three parts: 1 = M(1,0) M\M(1,0) m [Notation: e(α) = e 2πiα ]

35 Main Term Heuristics Let f(x) = a(x α 1 ) (x α k ). We need to estimate Φ f (z) = j=1 n=1 1 j zjf(n).

36 Main Term Heuristics Let f(x) = a(x α 1 ) (x α k ). We need to estimate Φ f (z) = j=1 n=1 1 j zjf(n). The main contribution to p f (n) comes from computing the residues of Γ(s)ζ(1 + s)m s n=1 f(n) s.

37 Main Term Heuristics Let f(x) = a(x α 1 ) (x α k ). We need to estimate Φ f (z) = j=1 n=1 1 j zjf(n). The main contribution to p f (n) comes from computing the residues of Γ(s)ζ(1 + s)m s n=1 f(n) s. The dominant singularity is a simple pole at s = 1/k.

38 Main Term Heuristics Let f(x) = a(x α 1 ) (x α k ). We need to estimate Φ f (z) = j=1 n=1 1 j zjf(n). The main contribution to p f (n) comes from computing the residues of Γ(s)ζ(1 + s)m s n=1 f(n) s. The dominant singularity is a simple pole at s = 1/k. The residue of n=1 f(n) s at s = 1/k is 1 ka 1/k.

39 Special Cases Again Recall: f(x) = a(x α 1 ) (x α k ), residue of 1 n=1 f(n) s at s = 1/k is ka 1/k.

40 Special Cases Again Recall: f(x) = a(x α 1 ) (x α k ), residue of 1 n=1 f(n) s at s = 1/k is ka 1/k. Powers: f(x) = 1x k ( ( 1 log p f (n) (k+1) 1k Γ ) ( ζ )) k/(k+1) n 1/(k+1). k k

41 Special Cases Again Recall: f(x) = a(x α 1 ) (x α k ), residue of 1 n=1 f(n) s at s = 1/k is ka 1/k. Powers: f(x) = 1x k ( ( 1 log p f (n) (k+1) 1k Γ ) ( ζ )) k/(k+1) n 1/(k+1). k k Powers in an arithmetic progression: f(x) = (bx + a) k = b k (x + a b )k ( ( 1 log p f (n) (k+1) bk Γ ) ( ζ )) k/(k+1) n 1/(k+1). k k

42 General Polynomials Let f(x) = a(x α 1 ) (x α k ). Claim The methods of Vaughan, G., and Berndt-Malik-Zaharescu can be extended to prove that ( ( 1 log p f (n) (k + 1) k Γ ) ( ζ )) k/(k+1) (n k k a ) 1/(k+1)

43 Last Remarks The heuristics above predict that the main term of the log p f (n) will depend only on the degree and leading coefficient of the polynomial f.

44 Last Remarks The heuristics above predict that the main term of the log p f (n) will depend only on the degree and leading coefficient of the polynomial f. The full asymptotic formula will depend on all the roots of f.

45 Last Remarks The heuristics above predict that the main term of the log p f (n) will depend only on the degree and leading coefficient of the polynomial f. The full asymptotic formula will depend on all the roots of f. We need to study the analytic continuation and the residues of Z f (s) = f(n) s. n=1

46 Last Remarks The heuristics above predict that the main term of the log p f (n) will depend only on the degree and leading coefficient of the polynomial f. The full asymptotic formula will depend on all the roots of f. We need to study the analytic continuation and the residues of Z f (s) = f(n) s. n=1 The error term will be calculated using delicate Hardy-Littlewood method arguments.

47 Thank You!

GOLDBACH S PROBLEMS ALEX RICE

GOLDBACH S PROBLEMS ALEX RICE Abstract. These are notes from the UGA Analysis and Arithmetic Combinatorics Learning Seminar from Fall 9, organized by John Doyle, eil Lyall, and Alex Rice. In these notes,