International Mathematical Forum, 5, 2010, no. 27, 1303 - 1322
On Symmetric Polynomials and Number Theory
Rafael Jakimczuk
Divisi´on Matem´atica, Universidad Nacional de Luj´an Buenos Aires, Argentina [email protected]
Abstract
It is well known that the symmetric function xk1xk2 ...xkh 1 2 h
in n indeterminates x1,x2,...,xn may be expressed as a polynomial in terms of the more simple symmetric functions
xk xk ... xk 1 + 2 + + n
In this article we obtain some results on these polynomials. Finally, we apply these results to some problems in number theory on sums of products of positive integers. For example, let A be the set of the first n primes, that is A = {p1,p2,...,pn}. Let An be the sum of all possible products of k different primes in A. We prove
k k k 2k k (p1 + p2 + ...+ pn) n pn n log n An ∼ ∼ ∼ k! k!2k k!2k Mathematics Subject Classification: 11B99, 11N45
Keywords: Symmetric polynomials, sums of products of positive integers, formulas, asymptotic formulas
1 Symmetric Polynomials
A polynomial in the indeterminates x1,x2,...,xn which is unchanged by any permutation of the indeterminates is called a symmetric polynomial. The following symmetric polynomials are very important in the theory of these polynomials. 1304 R. Jakimczuk
Definition 1.1 Given the indeterminates x1,x2,...,xn, let us consider the set of all possible monomials such that they have L (L ≥ 1) different indetermi- nates, t (t ≥ 1) indeterminates, and furthermore L1 different indeterminates have exponent K1, L2 different indeterminates have exponent K2 , ..., Lm different indeterminates have exponent Km (K1 >K2 >...> Km). The sum of all these monomials with coefficient 1 is a polynomial in the indeterminates x1,x2,...,xn which we shall denote
(L1,K1 : L2,K2 : ...: Lm,Km) (1)
Note that L1 + L2 + ...+ Lm = L (2)
L1K1 + L2K2 + ...+ LmKm = t (3) Furthermore, clearly the number of terms (or monomials) in this polynomial is n(n − 1) ...(n − (L − 1)) (n ≥ L) (4) L1!L2! ...Lm!
Example 1.2 If the indeterminates are x1,x2,x3,x4 we have
, x2 x2 x2 x2 (1 2) = 1 + 2 + 3 + 4
(2, 1) = x1x2 + x1x3 + x1x4 + x2x3 + x2x4 + x3x4
(4, 1) = x1x2x3x4
, , , x3x3x2x x3x3x2x x3x3x2x x3x3x2x x3x3x2x (2 3:1 2:1 1) = 1 2 3 4 + 1 2 4 3 + 1 3 2 4 + 1 3 4 2 + 1 4 2 3 x3x3x2x x3x3x2x x3x3x2x x3x3x2x x3x3x2x + 1 4 3 2 + 2 3 1 4 + 2 3 4 1 + 2 4 1 3 + 2 4 3 1 x3x3x2x x3x3x2x + 3 4 1 2 + 3 4 2 1
Consider the polynomials of the form, H ,r xr xr ... xr r =(1 )= 1 + 2 + + n
It is well known that the polynomial (1) may be expressed as a polynomial of rational coefficients in terms of the Hr where r takes certain values. That is
P (Hi ,Hi ,...,Hi ) L ,K L ,K ... L ,K 1 2 k ( 1 1 : 2 2 : : m m)= A (5) P H ,H ,...,H H ,H ,...,H Where ( i1 i2 ik ) is a polynomial of integer coefficients in i1 i2 ik and A is a positive integer. The following combinatorial general formulas (L ≥ 2) will be our key lemma. Symmetric polynomials and number theory 1305
Lemma 1.3 If L1 ≥ 2,
L1(L1,K1 : L2,K2 : ...: Lm,Km) H L − ,K L ,K ... L ,K = K1 ( 1 1 1 : 2 2 : : m m)
− (1,K1 + K1 : L1 − 2,K1 : L2,K2 : ...: Lm,Km)
− (1,K1 + K2 : L1 − 1,K1 : L2 − 1,K2 : ...: Lm,Km) − ...
− (1,K1 + Km : L1 − 1,K1 : L2,K2 : ...: Lm − 1,Km) (6)
If L1 =1,
(1,K1 : L2,K2 : ...: Lm,Km) H L ,K ... L ,K = K1 ( 2 2 : : m m)
− (1,K1 + K2 : L2 − 1,K2 : ...: Lm,Km) − ...
− (1,K1 + Km : L2,K2 : ...: Lm − 1,Km) (7)
Proof. These formulas can be proved without difficulty from definition 1.1. See for example [6] (page 82, exercise 5), where a similar formula more complicated is established. Our formulas are more simple since we assume K1 >K2 >...> Km (see definition 1.1).
Remark 1. Note that the polynomial in the left side of (6) or (7) has L different indeterminates in its monomials (see definition 1.1). On the other hand, the polynomials in the right side of (6) or (7) have L − 1 different indeterminates in their monomials. This fact is important in the proof by mathematical induction on L of a set of properties of these polynomials (see below).
Remark 2. Note that the polynomial in the left side of (6) or (7) has t indeterminates in its monomials (see definition 1.1). Note that (see (6)) the polynomial (L1 −1,K1 : L2,K2 : ...: Lm,Km) has t−K1 indeterminates in its monomials and the m polynomials with sign minus have t indeterminates in their monomials. Analogously (see (7)) the polynomial (L2,K2 : ...: Lm,Km) has t − K1 indeterminates in its monomials and the m − 1 polynomials with sign minus have t indeterminates in their monomials.
If we apply (6) or (7) repeatedly, clearly we obtain the polynomial (5). Since in each application L decreases in one and if L = 2 formulas (7) and (6) give
,K ,K H ,K − ,K K H H − H (1 1 :1 2)= K1 (1 2) (1 1 + 2)= K1 K2 K1+K2 (8)
,K H ,K − ,K K H2 − H 2(2 1)= K1 (1 1) (1 1 + 1)= K1 K1+K1 (9) 1306 R. Jakimczuk
Example 1.4 a) Given the indeterminates x1,x2,...,xn consider the poly- nomial (2, 2:1, 1) where L =3. Formula (6) gives
2(2, 2:1, 1) = H2(1, 2:1, 1) − (1, 4:1, 1) − (1, 3:1, 2:0, 1)
= H2(1, 2:1, 1) − (1, 4:1, 1) − (1, 3:1, 2) (10)
Formula (8) gives (1, 2:1, 1) = H2H1 − H3
(1, 4:1, 1) = H4H1 − H5
(1, 3:1, 2) = H3H2 − H5 Substituting these formulas into (10) we obtain the polynomial (5) that corre- spond to (2, 2:1, 1)
P (H ,H ,H ,H ,H ) H2H − 2H H − H H +2H (2, 2:1, 1) = 1 2 3 4 5 = 2 1 3 2 4 1 5 (11) 2 2 b) Given the indeterminates x1,x2,...,xn consider the polynomial
(2, 3:1, 2:1, 1) where L =4. Formula (6) gives
2(2, 3:1, 2:1, 1) = H3(1, 3:1, 2:1, 1) − (1, 6:1, 2:1, 1) − (1, 5:1, 3:1, 1) − (1, 4:1, 3:1, 2) (12)
Formula (7) gives
(1, 3:1, 2:1, 1) = H3(1, 2:1, 1) − (1, 5:1, 1) − (1, 4:1, 2)
(1, 6:1, 2:1, 1) = H6(1, 2:1, 1) − (1, 8:1, 1) − (1, 7:1, 2)
(1, 5:1, 3:1, 1) = H5(1, 3:1, 1) − (1, 8:1, 1) − (1, 6:1, 3)
(1, 4:1, 3:1, 2) = H4(1, 3:1, 2) − (1, 7:1, 2) − (1, 6:1, 3) (13) Formula (8) gives (1, 2:1, 1) = H2H1 − H3
(1, 5:1, 1) = H5H1 − H6
(1, 4:1, 2) = H4H2 − H6
(1, 8:1, 1) = H8H1 − H9
(1, 7:1, 2) = H7H2 − H9 Symmetric polynomials and number theory 1307
(1, 3:1, 1) = H3H1 − H4
(1, 6:1, 3) = H6H3 − H9
(1, 3:1, 2) = H3H2 − H5 (14) Substituting (14) into (13) we obtain
, , , H H H − H2 − H H − H H H (1 3:1 2:1 1) = 3 2 1 3 5 1 4 2 +2 6
(1, 6:1, 2:1, 1) = H6H2H1 − H6H3 − H8H1 − H7H2 +2H9
(1, 5:1, 3:1, 1) = H5H3H1 − H5H4 − H8H1 − H6H3 +2H9
(1, 4:1, 3:1, 2) = H4H3H2 − H5H4 − H7H2 − H6H3 +2H9 (15) Finally, substituting (15) into (12) we obtain the polynomial (5) that corre- spond to (2, 3:1, 2:1, 1) , , , H2H H − H3 − H H H − H H H − H H H (2 3:1 2:1 1) = 3 2 1 3 2 5 3 1 2 4 3 2 6 2 1
+5H6H3 +2H8H1 +2H7H2 +2H5H4 − 6H9) /2 (16) c) We have (formula (6))
2(2, 1) = H1(1, 1) − (1, 2) = H1H1 − H2 (17)
We have (formulas (6) and (7))
3(3, 1) = H1(2, 1) − (1, 2:1, 1)
(1, 2:1, 1) = H2H1 − H3 Therefore
3(3, 1) = H1(2, 1) − H2H1 + H3 (18) We have (formulas (6) and (7))
4(4, 1) = H1(3, 1) − (1, 2:2, 1)
(1, 2:2, 1) = H2(2, 1) − (1, 3:1, 1)
(1, 3:1, 1) = H3H1 − H4 Therefore
4(4, 1) = H1(3, 1) − H2(2, 1) + H3H1 − H4 (19) In this way we obtain the general recursive formula
k k+1 k(k, 1) = H1(k − 1, 1) − H2(k − 2, 1) + ...+(−1) Hk−1H1 +(−1) Hk (20) 1308 R. Jakimczuk where k ≥ 2. (17) gives H2 − H (2, 1) = 1 2 (21) 2! (18) and (21) give H3 − 3H H +2H (3, 1) = 1 2 1 3 (22) 3! (19), (21) and (22) give H4 − 6H H2 +8H H +3H2 − 6H (4, 1) = 1 2 1 3 1 2 4 (23) 4! (20) with k =5, (23), (22) and (21) give H5 − 10H H3 +20H H2 +15H 2H − 30H H − 20H H +24H (5, 1) = 1 2 1 3 1 2 1 4 1 3 2 5 5! (24) . . From (6), (7), (8) and (9) can be proved by mathematical induction on L the following properties. These properties are our main results in this section.
Property 1. In (5) we have A = L1!L2! ...Lm!.
For example: In (11) we have 2=2! 1!. In (16) we have 2=2!1!1!. In (21) we have 2!. In (22) we have 3!. In (23) we have 4!. In (24) we have 5!.
Proof. The property is true if L = 2 since (see (8) and (9)) H H − H K1 K2 K1+K2 (1,K :1,K )=HK HK − HK K = (25) 1 2 1 2 1+ 2 1!1! 2 2 H − HK K H − HK K (2,K )= K1 1+ 1 = K1 1+ 1 (26) 1 2 2! Suppose that the property is true for L − 1 ≥ 2, we shall prove that the property is true for L. From the inductive hypothesis we can write (see (6), (5) and remarks 1 and 2)
P Hi ,Hi ,...,Hi 0( 1 2 k0 ) (L1 − 1,K1 : L2,K2 : ...: Lm,Km)= (27) (L1 − 1)!L2! ...Lm!
P Hi ,Hi ,...,Hi 1( 1 2 k1 ) (1,K1 +K1 : L1 −2,K1 : L2,K2 : ...: Lm,Km)= (28) 1!(L1 − 2)!L2! ...Lm!
(1,K1 + K2 : L1 − 1,K1 : L2 − 1,K2 : ...: Lm,Km) P H ,H ,...,H 2( i1 i2 ik ) = 2 (29) 1!(L1 − 1)!(L2 − 1)! ...Lm! Symmetric polynomials and number theory 1309
. .
(1,K1 + Km : L1 − 1,K1 : L2,K2 : ...: Lm − 1,Km) Pm(Hi ,Hi ,...,Hi ) = 1 2 km (30) 1!(L1 − 1)!L2! ...(Lm − 1)!
Consequently (6) becomes (see (5))
HK P Hi ,Hi ,...,Hi 1 0( 1 2 k0 ) (L1,K1 : L2,K2 : ...: Lm,Km)= L1!L2! ...Lm! (L − 1)P (Hi ,Hi ,...,Hi ) L P (Hi ,Hi ,...,Hi ) − 1 1 1 2 k1 − 2 2 1 2 k2 L1!L2! ...Lm! L1!L2! ...Lm! L P H ,H ,...,H m m( i1 i2 ikm ) 1 − ...− HK P Hi ,Hi ,...,Hi = 1 0( 1 2 k0 ) L1!L2! ...Lm! L1!L2! ...Lm! − L − P Hi ,Hi ,...,Hi − L P Hi ,Hi ,...,Hi − ... ( 1 1) 1( 1 2 k1 ) 2 2( 1 2 k2 ) P (Hi ,Hi ,...,Hi ) − L P H ,H ,...,H 1 2 k m m( i1 i2 ikm ) = (31) L1!L2! ...Lm! as we desired. Analogously, from the inductive hypothesis we can write (see (7), (5) and remarks 1 and 2)
P Hi ,Hi ,...,Hi 0( 1 2 k0 ) (L2,K2 : ...: Lm,Km)= L2! ...Lm!
P Hi ,Hi ,...,Hi 2( 1 2 k2 ) (1,K1 + K2 : L2 − 1,K2 : ...: Lm,Km)= 1!(L2 − 1)! ...Lm! . . P H ,H ,...,H m( i1 i2 ikm ) (1,K1 + Km : L2,K2 : ...: Lm − 1,Km)= 1!L2! ...(Lm − 1)! Consequently (7) becomes (see (5))
HK P Hi ,Hi ,...,Hi 1 0( 1 2 k0 ) (1,K1 : L2,K2 : ...: Lm,Km)= 1!L2! ...Lm! L P H ,H ,...,H ( i1 i2 ik ) LmPm(Hi ,Hi ,...,Hi ) − 2 2 2 − ...− 1 2 km 1!L2! ...Lm! 1!L2! ...Lm! 1 HK P Hi ,Hi ,...,Hi − L P Hi ,Hi ,...,Hi = 1 0( 1 2 k0 ) 2 2( 1 2 k2 ) 1!L2! ...Lm! P (Hi ,Hi ,...,Hi ) − ...− L P H ,H ,...,H 1 2 k m m( i1 i2 ikm ) = (32) 1!L2! ...Lm! 1310 R. Jakimczuk as we desired. The property is thus proved.
Property 2. P H ,H ,...,H The polynomial ( i1 i2 ik ) in (5) has an unique mono- L H HL1 HL2 ...HLm mial with factors i. This monomial is K1 K2 Km and its coefficient is 1. The number of factors in the others monomials is less than L.
H2H For example: In (11) this unique monomial is 2 1. In (16) this unique H2H H H5 monomial is 3 2 1. In (24) this unique monomial is 1 .
Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 ≥ 2, we shall prove that the property is also true for L. P Hi ,Hi ,...,Hi From the inductive hypothesis the polynomial (see (27)) 0( 1 2 k0 ) L − H HL1−1HL2 ...HLm has an unique monomial with 1 factors i, namely K1 K2 Km . This monomial has coefficient 1 and the number of factors Hi in the rest of the monomials is less than L − 1. Consequently the polynomial (see (31)) HK P Hi ,Hi ,...,Hi L Hi 1 0( 1 2 k0 ) will have an unique monomial with factors , HL1 HL2 ...HLm namely K1 K2 Km . This monomial will have coefficient 1 and the num- ber of factors Hi in the rest of the monomials will be less than L. The rest of the polynomials in (31) (inductive hypothesis) have in their monomials a number of factors Hi less than or equal to L − 1. As we desired. The proof using (32) is the same. The property is proved.
Property 3. P H ,H ,...,H The sum of the coefficients in the polynomial ( i1 i2 ik ) is zero.
For example in (23) we have 1 − 6+8+3− 6=0.
Proof. The property is true if L = 2 (see (25) and (26)). The proof is now an immediate consequence of the inductive hypotheis and (31) (or (32)). The property is proved.
Property 4. If L is even: P H ,H ,...,H The monomials, in the polynomial ( i1 i2 ik ), that have an odd number of factors Hi have negative coefficient. P H ,H ,...,H The monomials, in the polynomial ( i1 i2 ik ), that have an even number of factors Hi have positive coefficient. On the other hand, if L is odd: P H ,H ,...,H The monomials, in the polynomial ( i1 i2 ik ), that have an odd number of factors Hi have positive coefficient. P H ,H ,...,H The monomials, in the polynomial ( i1 i2 ik ), that have an even Symmetric polynomials and number theory 1311
number of factors Hi have negative coefficient.
For example see (11), (16), (21), (22), (23) and (24).
Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 even. Then, an immediate consequence of the inductive hypothesis and (31) (or (32)) is that the property is also true for L odd. Suppose that the property is true for L − 1 odd. Then, an immediate consequence of the inductive hypothesis and (31) (or (32)) is that the property is also true for L even. The property is proved.
Property 5. The sum of the absolute values of the coefficients in the poly- P H ,H ,...,H nomial ( i1 i2 ik ) is L!.
For example: In (16) where L = 4 we have 1+1+2+2+1+5+2+2+2+6 = 4!. In (22) where L = 3 we have 1 + 3 + 2 = 3!.
Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 ≥ 2. Then, an immediate consequence of the property 4 and the inductive hy- pothesis is that the sum of the absolute values of the coefficients in the poly- P H ,H ,...,H nomial ( i1 i2 ik ) will be (see (31) and (2))
(L − 1)! + (L1 − 1)(L − 1)! + L2(L − 1)! + ...+ Lm(L − 1)!
=(1+L1 − 1+L2 + ...+ Lm)(L − 1)! = L! as desired. The proof using (32) is the same. The property is proved.
Property 6. P H ,H ,...,H In each monomial of the polynomial ( i1 i2 ik ) the sum of the subscripts i in the factors Hi is t (see (3)). Therefore each monomial can be associated to a partition of t.
For example in (16), where t = 9 , the sum of subscripts in each monomial H2H H is 9. The monomial 3 2 1 is associated to the partition 3 + 3 + 2 + 1 = 9. H3 The monomial 3 is associated to the partition 3 + 3 + 3 = 9. The monomial H5H3H1 is associated to the partition 5 + 3 + 1 = 9. The monomial H7H2 is associated to the partition 7 + 2 = 9. The monomial H9 is associated to the partition 9 = 9.
Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 ≥ 2. 1312 R. Jakimczuk
Then, an immediate consequence of the inductive hypothesis, remark 2 and (31) (or (32)) is that the property is also true for L. The property is proved.
Property 7. P H ,H ,...,H Consider the polynomial (5), that is ( i1 i2 ik ), and consider the set of L numbers ⎧ ⎫ L L L ⎨⎪ 1 2 m ⎬⎪ K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 2 2 ⎭⎪ whose sum is t (see (3)). Now, consider a partition of this set. If we sum the numbers in each subset of the partition we obtain a partition of t. All monomials in the polynomial (5) are obtained from all these possible partitions of t.
Example 1.5 Consider the polynomial (2, 2:1, 1) in example 1.4 a). The set of L =3elements is {2, 2, 1} and t =5. All possibles partitions of this set are:
{2, 2, 1}
{2, 2}∪{1} {2, 1}∪{2} {2}∪{2}∪{1} Consequently we obtain the following partitions of 5.
5=5
4+1=5 3+2=5 2+2+1=5 P H ,H ,...,H All possible monomials in the polynomial ( i1 i2 ik ) will be
H5
H4H1
H3H2 H2H 2 1 See (11). Symmetric polynomials and number theory 1313
In the case of the polynomials (k, 1) of example 1.4 c) where L = t = k the set is ⎧ ⎫ k ⎨⎪ ⎬⎪ 1,...,1, ⎩⎪ ⎭⎪
Consequently all possible partitions of this set give us all possible partitions of t. See (21), (22), (23) and (24).
Proof. The property is true if L = 2 (see (25) and (26)). Suppose that the property is true for L − 1 ≥ 2. Then, an immediate consequence of the inductive hypothesis , (27), (28), (29), ... (30) and (31) is that the property is also true for L. P Hi ,Hi ,...,Hi Note that to the polynomial (27), that is 0( 1 2 k0 ), correspond the set of L − 1 numbers ⎧ ⎫ L − L L ⎨⎪ 1 1 2 m ⎬⎪ K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 2 2 ⎭⎪
whose sum is t − K1 P Hi ,Hi ,...,Hi Note that to the polynomial (28), that is 1( 1 2 k1 ) , correspond the set of L − 1 numbers ⎧ ⎫ L − L L ⎨⎪ 1 2 2 m ⎬⎪ K + K , K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 1 1 2 2 ⎭⎪ whose sum is t P Hi ,Hi ,...,Hi Note that to the polynomial (29), that is 2( 1 2 k2 ) , correspond the set of L − 1 numbers ⎧ ⎫ L − L − L ⎨⎪ 1 1 2 1 m ⎬⎪ K + K , K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 2 1 1 2 2 ⎭⎪ whose sum is t . . P H ,H ,...,H Note that to the polynomial (30), that is m( i1 i2 ikm ) , correspond the set of L − 1 numbers ⎧ ⎫ L − L L − ⎨⎪ 1 1 2 m 1 ⎬⎪ K + Km, K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 1 2 2 ⎭⎪ whose sum is t 1314 R. Jakimczuk
P H ,H ,...,H Finally, note that to the polynomial ( i1 i2 ik) (see (5) and (31)) correspond the set of L numbers ⎧ ⎫ L L L ⎨⎪ 1 2 m ⎬⎪ K ,...,K , K ,...,K ,...,Km,...,Km ⎩⎪ 1 1 2 2 ⎭⎪ whose sum is t (see the enunciate of property 7). The proof using (32) is the same. The property is proved.
Property 8. The polynomial (5) is the zero polynomial if the number of indeterminates xi in the Hi is less than L.
For example, consider the polynomial (11) where L = 3. Consequently:
(x2)2x − 2x3x2 − x4x +2x5 1 1 1 1 1 1 1 =0 2 (x2 + x2)2(x + x ) − 2(x3 + x3)(x2 + x2) − (x4 + x4)(x + x )+2(x5 + x5) 1 2 1 2 1 2 1 2 1 2 1 2 1 2 =0 2
Proof. This property also can be proved without difficulty from (6) or (7). Note that if the number of indeterminates xi is L − 1 then the right side in (6) or (7) is the zero polynomial.
In the following section we shall need these properties.
2 Applications to the Number Theory
Now, we shall denote the polynomial
(L1,K1 : L2,K2 : ...: Lm,Km) in the form (L1,K1 : L2,K2 : ...: Lm,Km)(x1,...,xn)
Consider a set of different positive integers {a1,a2,...,an}. If we put xi = ai then (see definition 1.1)
(L1,K1 : L2,K2 : ...: Lm,Km)(a1,...,an) is the sum of all possible products which have t factors ai , L different factors ai and furthermore L1 different factors ai have exponent K1, L2 different factors ai have exponent K2 , ..., Lm different factors ai have exponent Km. Symmetric polynomials and number theory 1315
Given a strictly increasing sequence an of positive integers we are interested in this section in the asymptotic study ( or exact study, if is possible) of the sequence, (L1,K1 : L2,K2 : ...: Lm,Km)(a1,...,an) Our objective is to obtain in short proofs general formulas applicable to very general sequences an. The properties obtained in the former section will be very useful in the proofs of our main theorems in this section.
We shall need the following well known lemma (see [4], page 332). Lemma 2.1 ∞ a ∞ b Let i=1 i and i=1 i be two series of positive terms such ai ∞ i→∞ bi that lim bi =1. Then if i=1 is divergent, the following limit holds