padic Egyptian Fractions


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1 padic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Setup 3 4 pgreedy Algorithm 5 5 pegyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction is sum of distinct positive unit frctions In mny cses given frction hs severl representtions, for exmple = or more simply, = The former is otined from the greedy lgorithm while the ltter comes from performing the greedy 11dic lgorithm tht we will introduce in Section 4 1
2 2 Trditionl Egyptin Frctions nd Greedy Algorithm Proposition 1 (Clssicl Division Algorithm For ll positive integers, Z there exist unique positive integers q nd r such tht = q r with r strictly less thn Proof This proof is similr to the stndrd proof of the originl clssicl division lgorithm Let nd e positive integers To show existence, consider the set S = {n 0 : n Z} Clerly there is t lest one nonnegtive entry since lrge multiple of must exceed Among these entries there is lest clled r tht s gurnteed y the WellOrdering Principle Then y definition for some integer, denoted q, we hve r = q or = q r s desired As with the originl clssicl lgorithm we require 0 r < The first inequlity is given since it s n element of S To prove tht the second holds, suppose for contrdiction tht r Then r = q implies r = (q 1 0 which shows tht r > r S which contrdicts the minimlity of r Like the clssicl division lgorithm we otin uniqueness Suppose tht = q r = q r Sutrcting gives (q q = r r so r r Which is impossile since r r < unless r r = 0 nd so r = r s needed Then, q = q nd q = q s well Definition 1 (The Greedy Algorithm The Greedy Algorithm pplied to strts y using the modified division lgorithm to find the unique q 1 nd r 1 such tht = q 1 r 1 Next, we pply the modified division lgorithm to q 1 nd r 1 Continuing we hve, = q 1 r 1 q 1 = r 1 q 2 r 2 q 1 q 2 q 3 q n 2 = r n 2 q n 1 r n 1 q 1 q 2 q 3 q n 2 q n 1 = r n 1 q n 0 with r n = 0 2
3 The lgorithm eventully termintes since the modified division lgorithm requires ech reminder to e less thn the previous nd therefore yields decresing sequence of positive integers Notice tht rerrnging the terms from the lgorithm yields, = 1 + r 1 q 1 q 1 r 1 = 1 + r 2 q 1 q 2 q 1 q 2 r n 2 = 1 q 1 q 2 q 3 q n 2 r n 1 r n 1 + q n 1 q 1 q 2 q 3 q n 2 q n 1 ( q 1 q 2 q 3 q n 2 q n 1 = 1 q n which shows tht the sum of the reciprocls of the q i is equl to ie = n To e n Egyptin frction expnsion ech term must e distinct Indeed, q i > q i 1 since q 1 = nd q 2 = q 1 r 1 = r 1 /q 1 >, the lst inequlity follows from r 1 q 1 < 1 q i Exmple 1 Performing the Clssicl Greedy Algorithm on following expnsion, 3 Setup 5 11 = gives the Definition 2 (Order The order of some integer n with respect to prime p, denoted ord p (n, is the power of p tht exctly divides n We define ( ord p = ord p ( ord p ( 3
4 Definition 3 (Unit prt Let r = Q nd p prime The unit prt of r, denoted r is given y r := rp ordp(r Note tht ord p (â = 0 nd â nd p re reltively prime Throughout we will write ord p ( = α nd ord p ( = β Proposition 2 If, Q then ord p ( = ord p ( + ord p ( Proof Using units we write = âp α nd = p β Then ord p ( = ord p ( â p α+β = α + β Lemm 1 If, Q nd p prime then ord p ( + min{ord p (, ord p (} We hve equlity when ord p ( ord p ( Proof Tking, Q we ssume without loss of generlity tht β α Using units we write = âp α nd = p β ( ord p ( + = ord p âp α + p β ( = ord p p (âp β α β + = β + ord p (âp α β + Proposition 2 Should α = β, we hve ord p ( + = β + ord p (â + min {α, β} In the cse where α β we hve tht α β is positive Then the order of âp α β + is zero nd ord p ( + = min{α, β} 4
5 4 pgreedy Algorithm Theorem 2 (pdivision Algorithm Let, p Z with p prime nd Z[ 1] = p {mpk m, k Z} Then, there exist unique r Z nd q Z[ 1] such p tht = q r with 0 r < p nd ord p (r > ord p ( Proof Existence Let, Z nd p prime Since â nd p re reltively prime ny power of p hs n inverse mod â Tke 0 r < â such tht r p β α 1 mod â For some integer m we cn write, rp β+α+1 ( = âm rp α+1 + p β = âmp β with q = mp β α nd r = rp α+1 p β = âp α mp β α rp α+1 = q r Uniqueness Given, Z nd p prime suppose tht there exist corresponding q 1, r 1, q 2, r 2 such tht = q 1 r 1 = q 2 r 2 Then, (q 1 q 2 = r 1 r 2 (1 By (1 we hve tht r 1 r 2 mod â nd r 1 r 2 mod p α+1 since y ssumption ord p (r i > Therefore, r 1 r 2 mod p Thus, r 1 = r 2 or only one is etween 0 nd p; so r 1 = r 2 nd q 1 = q 2 nd we hve uniqueness s desired Note, it s possile tht r r The vlue of ord p (r cn vry depending on the choice of prime Consider the cse where = 23 nd = 60 If we tke p = 5 then = , = Then, r (12( (mod 23 nd ord 5 (r = 0 However, if p = 3 then = , = Then, nd the order of r is 1 r (20( (mod 23 5
6 Definition 4 A jump occurs when fter performing the pdivision Algorithm we hve ord p (r 1 Lemm 3 If p > â then r = r, ie there is no jump Proof Simply note tht r is tken mod â; hence r < â < p Definition 5 (pgreedy Algorithm Let,, p Z with p prime Theorem 2 gives unique q 1 Z[ 1 p ] nd r 1 Z such tht = q 1 r 1 Repeting on q 1 nd r 1 gives, = q 1 r 1 q 1 = r 1 q 2 r 2 q 1 2 q n 2 = r n 2 q n 1 r n 1 q 1 q 2 q n 1 = r n 1 q n 0 until r n = 0 Note y??( we gin get = 1 q i Exmple 2 Consider the frction 12 Performing the Greedy 3Adic Algorithm yields, 5 ( 8 5 = r 1 = 3 3 ( ( = nd 12 5 = In the clssicl cse typiclly the q s re ny integer greter thn one; these come from expnding frction whose size is less thn one We get q = 1 only if the frction eing expnded is greter thn one (The clssicl greedy lgorithm lwys tkes the next lrgest unit frction, including 1 The padic q s from the ove exmple look very different ut re in fct typicl in tht their order is decresing nd we will see lter tht they re otined from expnding frctions whose order is nonnegtive Frctions with negtive order led to the typicl padic cse where q = p s shown 6
7 elow Even though the padic cse dels with order, it should e noted tht s elements of the padic field Q p, frctions with negtive order re considered greter thn one, nd q = p leds to the term 1/p ppering in the expnsion which is nlogous to 1/1 ppering in clssicl expnsion since under the padic norm 1/p hs size p Exmple 3 Consider the frction 85 Adic Algorithm yields, nd = Performing the Greedy = 85(7 469 r 1 = = 469( r 2 = 49 ( = = Theorem 4 The pgreedy Algorithm termintes in finite numer of steps Proof Note tht r k is tken mod r k 1 with representtives in {0, 1, 2,, r k 1 1} nd therefore, r k r k < r k 1 Since r k is nonnegtive, this is decresing sequence of nturl numers tht must end with zero Eventully r n nd r n re identiclly zero nd the lgorithm termintes Corollry 5 The pgreedy Algorithm termintes in t most â steps This follows from the fct tht the sequence of the unit prts of the reminders r 1, r 2,, 0 is defined so tht r i is tken mod r i 1 with r 0 := â Definition 6 (Pure pegyptin Frction For fixed prime p, if = n with l i strictly incresing nonnegtive nd m i positive reltively prime to p then the sum is pegyptin frction expnsion for Lemm 6 ( (Clssifiction Let,, p Z with p prime Suppose tht l = ord p 1 Let q denote the quotient from the pdic division lgorithm nd q the quotient from the clssicl division lgorithm on p Then, q = pq 7 p l i m i
8 Proof Since l 1 we cn write = â p l p = â p l 1 The pdic division lgorithm gives r nd q such tht, p l = âq rp r p l 1 (mod â q = p l + rp â Applying the modified clssicl division lgorithm yields q nd r such tht, p l 1 = âq r Finlly, r = âq l 1 p r p l 1 r (mod â r = r q = p l 1 + r â (smllest nonnegtive representtives lwys used q p = p l + rp â = p l + rp â = q ( Corollry 7 If ord p 1 nd if no jumps occur thn ech term of the padic Greedy Algorithm for is equl to the corresponding term in the clssicl Greedy Algorithm on p divided y p For exmple, ( consider the frction = 5 = 51 If we tke p = then ord p = 2 Applying the 11Adic Greedy Algorithm to 5 yields 121 r 1 = 4, r 2 = 3, r 3 = 0 ll of order zero Using the q i we cn write, = Recll exmple 1 gve p = 5 11 = Note, the restriction on jumps is sufficient ut not necessry, for n exmple use 22 with p =
9 Lemm 8 Suppose tht ord p ( 0, then the quotient from the division lgorithm is not p Proof If ord p ( > 0 then ordp (q = β α 0 Suppose ord p ( = 0 For ske of contrdiction, suppose tht q = p Since q = mp β α nd the order is zero we hve tht q = m = p nd rp + = âm = âp however this is impossile since ord p (rp + = 0 1 = ord p (âp ( Theorem 9 Let,, p Z with p prime Suppose tht ord p 1 nd p 1, then dividing y using the padic Division Algorithm yields q = p nd = 1 p + r p Proof The frction p 1 nd therefore its Clssicl Greedy Egyptin Frction Expnsion egins with 1 By ssumption ord ( p 1 so y Lemm 6 the quotient from using the padic Divison Algorithm to divide y is p Theorem 10 If, p Z with p prime nd Z[ 1 p ] nd 0 ord p ( then pplying the padic Division lgorithm to divide y yields q nd r such tht ord p (q ord p ( ord p ( < ord p (r Proof The padic Division lgorithm sys tht r = rp ordp(+1 which llows us to write ord p ( < ord p ( + 1 ord p (r By hypothesis we lredy hve ord p ( ord p ( Therefore, it remins to show tht ord p (q 0 Drwing on the pdivision lgorithm gin we hve, ord p (q = ord p ( + r ord p ( = ord p ( ord p ( since ord p ( < ord p (r 0 Hence, ord p (q = ord p ( + ord p (q ord p ( 9
10 Theorem 11 Let,, p Z with p prime Then performing the padic Greedy Algorithm yields quotients such tht q i = p for i K where K is the lest K such tht one of the following is true: p K p ord p ( p K p p 1 (2 0 (3 Proof First note tht K lwys exists since the frction in condition 2 is decresing For 0 i < K we hve tht the negtions of 2 nd 3 re stisfied Should K = 0 then either p 1 in which cse the quotient from the Greedy Expnsion ( isn t 1 nd therefore q 1 from the padic Greedy Algorithm isn t p or ord p 0 nd y Lemm 8 q1 p Suppose tht K 1 Recll tht t ech step of the padic Greedy Algorithm we perform the padic Division Algorithm on r K q 1 q K = K p p K = p Note tht K is the first integer such tht p K fils to stisfy the hypotheses p of theorem 9 Susequent reminders re either strictly less thn p K or p their orders re greter y Lemm 10 5 pegyptin Trditionl In oth the clssicl nd the pdic cse, repeted ppliction of modified division lgorithm yielded the following sum, = n The clssicl q i re lwys integers so the sum is n Egyptin frction expnsion The pdic cse yields q i tht force powers of p to pper in the numertor of ech term Even though this sum isn t immeditely n Egyptin frction expnsion, it ecomes one once you divide out y the highest power of p present mong the terms Of course it s no longer n Egyptin 1 q i 10
11 frction expnsion for ut it is vlid expnsion for nother rtionl, nmely p ordp(qn It s possile in some cses to reverse this process y strting with frction nd for ech prime tht mkes up, remove it nd find the corresponding pdic Egyptin frction expnsion for wht remins A frction works if the desired power k of p in the denomintor is greter thn or equl to the order of the lst term in the expnsion so tht dividing out doesn t leve ny powers of p ehind The order of 1/q n sets the lowest power of p tht is llowed in the denomintor Exmple 4 Suppose tht we wnt n Egyptin frction expnsion for 189 = From the primes in the denomintor we see tht we cn either perform the 3Adic lgorithm on or the 7Adic lgorithm on In the 3Adic cse 7 27 we will e successful if the finl term in the expnsion hs no more thn 3 powers of p = 3 in the numertor, while the 7Adic cse would require tht the finl term hve order t most one The 7Adic cse yields, Dividing y 7 we find tht 27 = = which won t work for our purposes since there re still terms of positive order in the expnsion The 3Adic cse yields, which when divided y 3 3 yields, 7 = =
12 which is n Egyptin frction expnsion tht s significntly more lnced thn the one otined from the clssicl greedy lgorithm: 189 = The kinds of frctions tht cn e expnded in this wy depend on the order of q n Corollry 12 Inspecting susequent steps of the pgreedy lgorithm we see tht, ord p (q k = ord p (q 1 q 2 q k 1 ord p (r k 1 Proof Since q k r k 1 = q 1 q k 1 + r k nd ord p (q 1 q k 1 < ord p (r k y Lemm 1 ord p (q k = ord p ( q1 q k 1 r k 1 = ord p (q 1 q 2 q k 1 ord p (r k 1 Theorem Let,, p Z with p prime nd ord p ( = β, ord p ( = α Then, n 1 ord p (q n = (β n 1 ord p (r i α 1 2 i with q n nd r i defined s per the pgreedy lgorithm in Definition 5 Proof We proceed y induction on n Bse Cse: n = 1 By Theorem 2 the order of q 1 is β α = (β α 1 Inductive Cse: Suppose for some n 1 tht ord p (q n = n 1 (β α 1 By corollry 12 we lredy hve tht Then, n 1 ord p (r i 2 i ord p (q n+1 = ord p (q 1 q n ord p (r n ord p (q n+1 = ord p (q 1 q n + ord p (r n 1 ord p (r n 1 ord p (r n = ord p (q 1 q n 1 + ord p (q n + ord p (r n 1 ord p (r n 1 ord p (r n 12
13 = 2 ord p (q n + ord p (r n 1 ord p (r n Corollry 12 = 2 ord p (q n + ord p (r n 1 ord p (r n ord p (r n 1 1 Since r = rp α+1 n 1 = (β n ord p (r i α 1 ord 2 i p (r n 1 n 1 = (β n α 1 = n (β α 1 n Thus the order of q n is given y, ord p (r i 2 i ord p (r i 2 i ord p (q n = n 1 (β α 1 2n ord p (r n 2 n n 1 ord p (r i 2 i Corollry 14 If ord p ( = 0 then 6 Conclusion n 1 ord p (q n = 1 2 n 1 2 n+i 1 ord p (r i The clssicl lgorithm cn hndle irrtionl numers since since q = / where = 1 nd irrtionl Although padic numers hve n ssocited norm, there re mny rtionls tht stisfy x p = 1 eg ll rtionls of order 0 so it s difficult to pick the correct one There is lso no cler wy to define wht it mens for one to mod out y â if it s irrtionl The clssicl greedy lgorithm for egyptin frctions lwys tkes the next lrgest unit frction Therefore, if term is deleted the expnsion for the remining terms will simply e the originl expnsion with the one term deleted The sme ppers to e true for the padic
14 lgorithm, however if one defines next lrgest unit frction s the next 1/q (in the clssicl cse this corresponds to the next lrgest y solute vlue thn ll you know is tht some 1/q is less thn /, however there s no wy to test this unless one ctully crries out the lgorithm Of course, one cnnot expect perfect nlogy to lwys exist 2 n 1 ( The unit prt of q n cn e roughly pproximted y 2 n 1 â Assuming tht ord p ( = 0 nd tht p > â llows us to determine the order of q n exctly nd we hve tht the 1/q terms grow s p 2n 1 1 ( 2 n 1 2 â n 1 hence if /â p/2 we otin prticulrily lnced terms The padic lgorithm yields n expnsion with t most â terms It remins to e seen if these exmples lwys exist It would lso py to investigte the verge lengths of expnsions nd compre it to the verge length of sequences if r i is ssumed to e rndom integer mod r i 1 14
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