π d i (b i z) (n 1)π )... sin(θ + )


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1 SOME TRIGONOMETRIC IDENTITIES RELATED TO EXACT COVERS Joh Beebee Uiversity of Alaska, Achorage Jauary 18, 1990 Sherma K Stei proves that if si π = k si π b where i the b i are itegers, the are positive itegers, k is a costat, the { : b i } is a exact cover It is show here that if 0 b i < the k = 2 1, that the coverse is also true, ad a aalogous formula is cojectured for ifiite exact covers May well kow ad lesser kow trigoometric ad fuctioal idetities ca be derived from this result ad kow families of exact covers A procedure is give for costructig exact covers by iductio Prelimiaries Let d : b be the arithmetic progressio AP {x : x = b +αd, α Z} A fiite collectio of disjoit AP s { : b i : 1 i } is called a exact cover if each iteger belogs to precisely oe AP The umbers are the moduli ad the umbers b i are the offsets Sice : b i = : b i + γ it is usual to assume that the offsets have bee stadardied, 0 b i < A ecessary ad sufficiet coditio for a fiite collectio of AP s to be a exact cover is that 1/ = 1 ad, d j b i b j As a cosequece there is oly oe AP i a exact cover with b i = 0 mod If we let = the collectio of AP s is called a ifiite exact cover For a ifiite exact cover it is possible for either 1/ < 1, i which case the cover is called usaturated, or 1/ = 1, i which case it is saturated I [7] Corollary 65 Stei proves that if 1 si π = k π b i where the b i are itegers, the are positive itegers, ad k is a costat the { : b i } is a exact cover Trigoometric idetities like 1, for example 2 si θ = 2 1 si θ siθ + π siθ + 2π 1π siθ + appear frequetly i aalysis books See Hobso [3], p119 May other idetities ca be derived from 1 or 2 by takig derivatives of both sides, dividig both sides 1991 Mathematics Subject Classificatio 11B25,33A10,68T15,1104,3304 Key words ad phrases exact coverig systems,fuctioal idetities, trigoometric idetities 1 Typeset by AMSTEX
2 2 JOHN BEEBEE by si π/ ad takig lim 0, choosig special values for, etc See Hase [2], particularly formulas 24,29, ad 91 I show here that if { : b i : 1 i } is a exact cover with stadardied offsets the 1 is true with k = 2 1, ad derive a aalogous formula that I cojecture is true for the case = Costructio of Exact Covers I first show how to costruct exact covers by iductio o the umber of AP s i the cover There is a exact parallel betwee the method of costructio for exact covers give here ad the trigoometric proof of a idetity like 1 This parallel could be used to write a computer program to create a proof of ay such idetity Theorem 1 below says that to verify a idetity like 1 it is oly ecessary to show that { : b i : 1 i } is a exact cover The reductio of C mod q : k Let C be a exact cover, ad let q : k be a AP Every umber i q : k belogs to exactly oe AP i C, so for each iteger α there is a uique idex i ad iteger γ such that 3 k + αq = b i + γ, where we must have, q b i k For each i with, q b i k there is a uique particular solutio, α = a i to 3 with 4 0 a i < /, q ad a i q/, q = b i k/, q + γ i /, q ad α satisfies 3 if ad oly if α = a i mod /, q; ie α /, q : a i But every iteger α satisfies 3 for exactly oe i with, q b i k This proves that { /, q : a i :, q b i k} is a exact cover, called the reductio of C mod q : k See Simpso [5] Now let q be prime, let 0 k < q, ad let B k be the reductio of C mod q : k To idicate that the offset a i exists for a particular combiatio of i ad k I will write a i = a i k The 5 B k = { : a i k :, q = 1} { /q : a i k :, q = q ad q b i k} = A k A k is a exact cover If, q = q for some i, B k must have fewer AP s tha C, because the sum of the reciprocals of its moduli must be 1 Example 1 Let C = {6 : 0, 6 : 2, 10 : 1, 10 : 5, 10 : 7, 10 : 9, 15 : 13, 30 : 3, 30 : 4, 30 : 10, 30 : 16, 30 : 22, 30 : 23} = { : b i : 1 i 13} Let q = 2 The B 0 ad B 1 are calculated from 5 ad the offsets a i 0 ad a i 1 are calculated from 4 B 0 = A 0 A 0 = {15 : 14} {3 : 0, 3 : 1, 15 : 2, 15 : 5, 15 : 8, 15 : 11} = {d 7 : a 7 0} { /2 : a 1 0, d 2 /2 : a 2 0, d 9 /2 : a 9 0, 0 /2 : a 10 0, 1 /2 : a 11 0, 2 /2 : a 12 0} B 1 = A 1 A 1 = {15 : 6} {5 : 0, 5 : 2, 5 : 3, 5 : 4, 15 : 1, 15 : 11} = {d 7 : a 7 1} {d 3 /2 : a 3 1, d 4 /2 : a 4 1, d 5 /2 : a 5 1, d 6 /2 : a 6 1, d 8 /2 : a 8 1, 3 /2 : a 13 1}
3 SOME TRIGONOMETRIC IDENTITIES RELATED TO EXACT COVERS 3 Lemma 1 If, q = 1 the a i k exists for all k so : a i k A k for all k ad q 1 k=0 q : k + a i kq = : b i Proof If, q = 1 there is a solutio a i k ad γ i k to 4 for every k By 4, as k varies over its rage, γ i k mod q takes o each iteger value i the same rage Thus q 1 k=0 q : k + a i kq = q 1 k=0 q : b i + γ i k = : b i Example 2 I example 1 d 7, q = 15, 2 = 1 so 15 : 14 = d 7 : a 7 0 A 0 a5 : 6 = d 7 : a 7 1 A 1 ad d 7 2 : 0 + a d 7 2 : 1 + a = 30 : : 13 = 15 : 13 = d 7 : b 7 Lemma 2 If q there is exactly oe value of k, 0 k < q, for which a i k exists, so /q : a i k A k for this value of k ad : k + a i kq = : b i Proof Clearly q b i k for some value of k, so a i k exists for this value of k If a i r exists, 0 r < q, the q b i r ad q b i k, so q k r This implies k = r, so k is uique If q, : b i must be a subset of some AP q : k, 0 k < q By 3, k + a i kq = b i + γ Thus : k + a i kq = : b i Example 3 I example 1, 2 for i 7 Thus a i 0 exists for i = 1, 2, 9, 10, 11, 12 ad a i 1 exists for i = 3, 4, 5, 6, 8, 13 For istace, if k = 1 a = 8 the d 8 : 1 + a = 30 : = 30 : 3 = d 8 : b 8 The geeraliatio of the partitio {7} {1, 2, 9, 10, 11, 12} {3, 4, 5, 6, 8, 13} to arbitrary primes q ad exact covers C has appeared previously i the literature about Zám s cojecture See Simpso [6] ad Zám [8] The followig lemma describes a class of exact covers called the atural irreducible covers by Korec [4], who also describes costructios 1 ad 2 below Lemma 3 If q is a prime, the Z q = {q : k : 0 k < q} is a exact cover Also, Z 1 = {1 : 0} is a exact cover Costructio 1 If C 1 = { : b i : 1 i } af C 2 = {e j : c j : 1 j m} are exact covers, the C 3 = { : b i : i I} {e j d I : b I + c j d I : 1 j m} is a exact cover with +m1 AP s Korec writes C 3 = SplitC 1, b I, C 2 I call this costructio coverig d I : b I with C 2 Example 4 Let Z 2 = {2 : 0, 2 : 1} Cover 2 : 0 with B 0 from example 1, gettig the exact cover {30 : 28, 6 : 0, 6 : 2, 30 : 4, 30 : 10, 30 : 16, 30 : 22, 2 : 1} Cover 2 : 1 i the latter with B 1, gettig C = {30 : 28, 6 : 0, 6 : 2, 30 : 4, 30 : 10, 30 : 16, 30 : 22, 30 : 13, 10 : 1, 10 : 5, 10 : 7, 10 : 9, 30 : 3, 30 : 23} Costructio 2 Let C 1 = { : b i : 1 i } be a exact cover Suppose the uio of some subcollectio D = {j : b ij : 1 j m} is the AP d : b The C 3 = C 1 D {d : b} is a exact cover with m + 1 AP s I call this costructio the substitutio of d:b for D Note that d j ad d b ij b, so C 2 = {j /d : b ij b/d : 1 j m} is a exact cover, the reductio of C 1 mod d : b If we cover d : b with C 2, we just get C 1 back agai, so costructios 1 ad 2 are iverses to each other
4 4 JOHN BEEBEE Example 5 Let D = {30 : 28, 30 : 13} From example 2 30 : : 13 = 15 : 13, so substitute 15 : 13 for D i C from example 4 gettig C = {15 : 13, 6 : 0, 6 : 2, 30 : 4, 30 : 10, 30 : 16, 30 : 22, 10 : 1, 10 : 5, 10 : 7, 10 : 9, 30 : 3, 30 : 23} Except for the order of the AP s C is the same as C from example 1 The ituitive meaig of the followig lemma ats proof is that ay exact cover C ca be disassembled ad reassemble the same way I will choose q so that the covers B k always have fewer AP s tha C Thus it gives a theoretical method to costruct a exact cover by iductio Here it is used to prove Theorem 1 below, which shows that there is a exact parallel betwee the proof of idetities like 1 ad the costructio of exact covers Lemma 4 Suppose we have a list of all exact covers with M or fewer AP s Ay exact cover C with M +1 AP s, except Z M+1 if M +1 is prime ca be costructed by first usig costructio 1 to cover the AP s q : k Z q with q exact covers from the list, where q is a prime that is a proper divisor of some modulus of C This gives a exact cover C If q does ot divide every modulus of C, it will the be ecessary to apply costructio 2 to some subcollectios of C to get C Proof Let C = { : b i : 1 i M + 1} be a exact cover, where C is ot the atural irreducible cover Z M+1 describe Lemma 3 The there is a prime q which is a proper divisor of at least oe modulus of C Let B k be the reductio of C mod q : k, 0 k < q By the remark followig 5 the B k s have fewer AP s tha C ad hece are i the list of exact covers with M or fewer AP s The proof of the lemma will be completed by showig how to recostruct C from the B k s ad Z q The procedure is illustrate examples 4 ad 5 above ad example 6 below For each k cover the AP q : k Z q with B k From 5 ad costructio 1 we obtai the exact cover C = q 1 { q : k + a i kq :, q = 1} k=0 q 1 { : k + a i kq :, q = q ad q b i k} k=0 By lemma 1, q 1 k=0 q : k + a i kq = : b i for each i with, q = 1 Usig costructio 2, substitute : b i for D = { q : k + a i kq : 0 k < q}, gettig the ew exact cover C = { : b i :, q = 1} q 1 { : k + a i kq :, q = q ad q b i k} k=0 By lemma 2 if, q = q the : k + a i kq = : b i, so C = { : b i :, q = 1} { : b i :, q = q} = C
5 SOME TRIGONOMETRIC IDENTITIES RELATED TO EXACT COVERS 5 I the precedig proof all of the AP s except perhaps those belogig to C have stadardied offsets Thus, whe costructios 1 ad 2 are used to build a exact cover with stadardied offsets we ca assume that we are applyig costructios 1 ad 2 to AP s with stadardied offsets Example 6 Let C be as i example 1, but let q = 5 The B 0 = A 0 A 0 = {6 : 0, 6 : 4} {2 : 1, 6 : 2} B 1 = A 1 A 1 = {6 : 1, 6 : 5} {2 : 0, 6 : 3} B 2 = A 2 A 2 = {6 : 2, 6 : 0} {2 : 1, 6 : 4} B 3 = A 3 A 3 = {6 : 3, 6 : 1} {3 : 2, 6 : 0, 6 : 4} B 4 = A 4 A 4 = {6 : 4, 6 : 2} {2 : 1, 6 : 0} Cover the AP 5 : k Z 5 with B k, 0 k < 5 gettig C = {30 : 0, 30 : 20} {30 : 6, 30 : 26} {30 : 12, 30 : 2} {30 : 18, 30 : 8} {30 : 24, 30 : 14} {10 : 5, 30 : 10} {10 : 1, 30 : 16} {10 : 7, 30 : 22} {15 : 13, 30 : 3, 30 : 23} {10 : 9, 30 : 4} Now 30 : 0 30 : 6 30 : : : 24 = 6 : 0 ad 30 : : : 2 30 : 8 30 : 14 = 6 : 2 Makig these substitutios i C we get C = {6 : 0, 6 : 2} {10 : 5, 30 : 10, 10 : 1, 30 : 16, 10 : 7, Trigoometric Idetities 30 : 22, 15 : 13, 30 : 3, 30 : 23, 10 : 9, 30 : 4} = C Theorem 1 The set of AP s C = { : b i : 1 i }, with stadardied offsets, is a exact cover if ad oly if 6 si π = 2 1 si π b i If the offsets are ot stadardied the write b i = b i + η i, where 0 b i < The product o the right i 6 must the be multiplied by 1 ηi Proof If 6 holds the C is a exact cover because the eros of the fuctio o the left must be the same as the eros of the fuctio o the right ad have the same multiplicity The first proof of the coverse uses the iductive costructio of exact covers describe lemma 4 A differet proof is give after Corollary 1 below Substitute π = θ i 2, gettig the idetity 1 si π = 2 1 k=0 π k
6 6 JOHN BEEBEE Thus Theorem 1 is true for the covers Z q describe lemma 3 To complete the proof it is ecessary to show that if covers C 1 ad C 2 satisfy 6, the the cover C 3 obtaied from costructio 1 or 2 satisfies 6 If C = { : b i : 1 i } let C = 2 1 si π b i Cosider costructio 1 ad suppose 7 The si π = 2 1 si π d I b I = 2 m 1 si π b i m = 2 m 1 si π c j e j m si π c j b I e j d I m = 2 m 1 si π b I + c j d I e j d I Substitutig the latter i 7, si π = 2 m+ 2 i I si π b i m si π b I + c j d I = C 3 e j d I Now cosider costructio 2 Sice m j : b ij = d : b we ca show that d j ad d b ij b ad {j /d : b ij b/d : 1 j m} is a exact cover, the reductio of C 1 mod d : b Thus si π = 2 m 1 m = 2 m 1 m si π bij b/d j /d si π j b ij b d Substitutig b d for i the last equatio, si π m b = 2m 1 d si π j b ij Substitutig for m si π j b ij i 7, si π = 2 m si π d b si i i j π b i = C 3
7 SOME TRIGONOMETRIC IDENTITIES RELATED TO EXACT COVERS 7 If the offsets are ot stadardied, let b i = b i + η i, where 0 b i < The si π b i = 1 ηi si π b i, so si π = 1 ηi 2 1 si π b i Corollary 1 Suppose : b 1 is the uique AP i the exact cover { : b i : 1 i } with b 1 = 0 ad the offsets are stadardied The ad hece 2 = 1 8 si π = si π si πb i si π b i si πbi Proof Divide both sides of 6 by si π b 1 ad take lim 0 o both sides I ow give a proof of Corollary 1 ad hece Theorem 1 that is quite differet from the oe just give This proof should make the cojecture below more plausible Proof of Corollary 1 For each real umber ρ > 0 defie where the otatio f ρ = π ρ ρ k= k k = π ρ 1, k meas that the product is to be evaluated for itegers k, ρ k ρ Now let C = { : b i : 1 i } be a exact cover with b 1 = 0 The 9 f ρ = π ρ k= ρ 1 k ρ b i k= ρ b i 1 From Hobso [3] p 350 we kow that lim ρ f ρ = si π b i + k
8 8 JOHN BEEBEE But if b i 0 the ρ bi bi f ρ = π Hece Thus 10 Also bi f ρ/di = π b i 1 ρ b i = π b i 1 ρ b i bi ρ = f ρ 1 1 b i k kdi b i + k k 1 b i 1 + k b i + k ρ/d bi i = f ρ/di 1 / bi si π = lim d f bi ρ/d i ρ i = si πb i = si πb i 11 si π = π lim lim ρ/ ρ / lim ρ ρ ρ/d 1 / ρ b i k= ρ b i 1 1 b i + k b i + k b i + k 1 k k b i Takig the limit of both sides of 9 ad substitutig from 10 a1 we get si π = si π si π b i si πbi If we could somehow let as well, we could prove the cojecture below I will ow show that if b 1 = 0, the 2 = si π b i Let m = lcm{d 1 1, d 2,, d } d i Sice C is a exact cover, x 2m 1 = x 2m/di e 2πibi/di
9 SOME TRIGONOMETRIC IDENTITIES RELATED TO EXACT COVERS 9 See the proof of Theorem 51 i Stei [7] Let x = e πi/m i this equatio The e 2πi 1 = e 2πi/di e 2πibi/di Divide both sides by e 2πi/d1 e 2πib1/d1 = e 2πi/d1 1, ad take lim 0 of both sides, gettig = 1 e 2πibi/di = e πibi/di 2i si πb i / By Theorem 1 of Fraekel [1] with =1, if 0 b i <, the Thus b i / = = 2 1 e πi 1/2 e πi 1/2 b i / = 1 2 si πb i = 2 1 si πb i Corollary 2 If { : b i : 1 i } is a exact cover with stadardied offsets the si π 2 2b i 1 = 1, 2 cot π = 1 cot π b i, 3 csc 2 π = 1 d 2 i csc 2 π b i Proof of 1 Let = 1 i 6 2 Proof of 2 Take the logarithmic derivative of both sides of 6 Proof of 3 Take the derivative of 2 of corollary 2 If C = { : b i : 1 i } is a exact cover, ad q > 0 is relatively prime to all ad k is ay iteger, the C = { : k + qb i mod : 1 i } is a exact cover More geerally, if s a positive iteger ad b is ay iteger, the C, the reductio of C mod d : b is a exact cover Corollary 3 If si π = C, the si π = C = C Cojecture If { : b i : 1 i < } is a ifiite exact cover saturated or usaturated with b 1 = 0 mod ad stadardied offsets the si π = si π si π b i si πbi
10 10 JOHN BEEBEE Refereces 2 Eldo R Hase, A table of series ad products, Pretice Hall, E W Hobso, A treatise o plae ad advaced trigoometry, Dover, Iva Korec, Irreducible disjoit coverig systems, Acta Arithmetica XLN 1984, R J Simpso, Regular coverigs of the itegers by arithmetic progressios, Acta Arithmetica XLV 1985, RJ Simpso, Disjoit Coverig Systems of Cogrueces, Amer Math Mothly , Sherma K Stei, Uios of arithmetic sequeces, Mathematishe Aale , Š Zám, A survey of coverig systems of cogrueces, Acta Math Uiv Comeia , 5979
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