Equations and Inequalities Involving v p (n!)

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1 Equatios ad Iequalities Ivolvig v (!) Mehdi Hassai Deartmet of Mathematics Istitute for Advaced Studies i Basic Scieces Zaja, Ira mhassai@iasbs.ac.ir Abstract I this aer we study v (!), the greatest ower of rime i factorizatio of!. We fid some lower ad uer bouds for v (!), ad we show that v (!) = 1 + O(l ). By usig above metioed bouds, we study the equatio v (!) = v for a fixed ositive iteger v. Also, we study the triagle iequality about v (!), ad show that the iequality v(!) > q vq(!) holds for rimes < q ad sufficietly large values of Mathematics Subject Classificatio: 05A10, 11A41, 26D15, 26D20. Keywords: factorial fuctio, rime umber, iequality. 1 Itroductio As we kow, for every N,! = Let v (!) be the highest ower of rime i factorizatio of! to rime umbers. It is well-kow that (see [3] or [5]) [ ] [ l l ] [ ] v (!) = =, (1)

2 i which [x] is the largest iteger less tha or equal to x. A elemetary roblem about! is fidig the umber of zeros at the ed of it, i which clearly its aswer is v 5 (!). The iverse of this roblem is very ice; for examle fidig values of i which! termiates i 37 zeros [3], ad geerally fidig values of such that v (!) = v. We show that if v (!) = v has a solutio the it has exactly solutios. For doig these, we eed some roerties of [x], such as [x] + [y] [x + y] (x, y R), (2) ad [ x = ] [ ] [x] (x R, N). (3) 2 Estimatig v (!) Theorem 1 For every N ad rime, such that, we have: 1 l l < v (!) 1 1. (4) Proof: Accordig to the relatio (1), we have v (!) = m [ ] i which m = [ l l ], ad sice x 1 < [x] x, we obtai cosiderig m 1 = 1 1 m 1 m 1 m < v (!), we yield that 1 (1 1 ) m < v (!) m ad combiig this iequality with l 1 < m l l l m 1, 1 (1 1 ), m comletes the roof. Corollary 1 For every N ad rime, such that, we have: Proof: By usig (4), we have ad this yields the result. v (!) = 0 < + O(l ). 1 v 1 (!) l 2 < 1 l,

3 Note that the above corollary asserts that! eds aroximately i zeros [1]. 4 Corollary 2 For every N ad rime, such that, ad for all a (0, + ) we have: 1 1 ( ) l a + l a 1 < v (!). (5) Proof: Cosider the fuctio f(x) = l x. Sice, f (x) = 1 x 2, l x is a cocave fuctio ad so, for every a (0, + ) we have l x l a + 1 (x a), a combiig this with the left had side of (4) comletes the roof. 3 Study of the Equatio v (!) = v Suose v N is give. We are iterested to fid the values of such that i factorizatio of!, the highest ower of, is equal to v. First, we fid some lower ad uer bouds for these s. Lemma 1 Suose v N ad is a rime ad v (!) = v, the we have 1 + ( 1)v < v l(1+( 1)v) l (1+( 1)v) l l. (6) Proof: For Provig the left had side of (6), use right had side of (4) with assumtio v (!) = v, ad for rovig the right had side of (6), use (5) with a = 1+( 1)v. The lemma 1 suggest a iterval for the solutio of v (!) = v. I the ext lemma we show that it is sufficiet oe check oly multiles of i above iterval. Lemma 2 Suose m N ad is a rime, the we have v ((m + )!) v ((m)!) 1. (7) Proof: By usig (1) ad (2) we have v ((m + )!) = [ ] m + [ ] m + [ ] = 1 + v ((m)!), ad this comletes the roof. 3

4 I the ext lemma, we show that if v (!) = v has a solutio, the it has exactly solutios. I fact, the ext lemma asserts that if v ((m)!) = v holds, the for all 0 r 1, v ((m + r)!) = v also holds. Lemma 3 Suose m N ad is a rime, the we have v ((m + 1)!) v (m!), (8) ad v ((m + 1)!) = v ((m)!). (9) Proof: For rovig (8), use (1) ad (2) as follows [ ] m + 1 [ ] m [ ] 1 v ((m + 1)!) = + = [ ] m = v (m!). For rovig (9), it is eough to show that for all k N, [ m+ 1 ] = [ m this by iductio o k; for k = 1, clearly [ m+ 1 ] = [ m [ ] m + 1 = +1 This comletes the roof. So, we have roved that [ m+ 1 ] = [ ] m+ 1 = [ m ] ad we do ]. Now, by usig (3) we have ] = [ m ] [ m = Theorem 2 Suose v N ad is a rime. For solvig the equatio v (!) = v, it is sufficiet to check the values = m, i which m N ad [ ] [ 1 + ( 1)v v + m + ] l(1+( 1)v) 1 1 l l. (10) 1 (1+( 1)v) l Also, if = m is a solutio of v (!) = v, the it has exactly solutios = m+r, i which 0 r 1. Note ad Problem 1 As we see, there is o guaratee for existig a solutio for v (!) = v. I fact we eed to show that {v (!) N} = N; however, comutatioal observatios suggest that = 1+( 1)v usually is a solutio, such that x is the earest iteger to x, but we ca t rove it. Note ad Problem 2 Other roblems ca lead us to other equatios ivolvig v (!); for examle, suose, v N give, fid the value of rime such that v (!) = v. Or, suose ad q are rimes ad f : N 2 N is a rime value fuctio, for which s we have v (!) + v q (!) = v f(,q) (!)? Ad may other roblems! +1 ]. 4

5 4 Triagle Iequality Cocerig v (!) I this sectio we are goig to comare v ((m + )!) ad v (m!) + v (!). Theorem 3 For every m, N ad rime, such that mi{m, }, we have v ((m + )!) v (m!) + v (!), (11) ad v ((m + )!) v (m!) v (!) = O(l(m)). (12) Proof: By usig (1) ad (2), we have [ ] m + [ ] m v ((m + )!) = + Also, by usig (4) ad (11) we obtai [ ] = v (m!) + v (!). 0 v ((m + )!) v (m!) v (!) < l(m) l this comletes the roof. 3 + l(m) l 2, More geerally, if 1, 2,, t N ad is a rime, i which mi{ 1, 2,, t }, by usig a extesio of (2), we obtai v (( k )!) v ( k!), ad by usig this iequality ad (4), we yield that 0 v (( k )!) ad cosequetly we have v (( v ( k!) < k l( 1 2 t ) l k )!) v ( k!) = O(l( 1 2 t )). 2k 1 + l( 1 2 t ), l Note ad Problem 3 Suose f : N t N is a fuctio ad is a rime. For which 1, 2,, t N, we have v ((f( 1, 2,, t )!) f(v ( 1!), v ( 2!),, v ( t!))? Also, we ca cosider the above questio i other view oits. 5

6 5 The Iequality v (!) > q v q(!) Suose ad q are rimes ad < q. Sice v (!) v q (!), comarig v(!) ad q vq(!) become a ice roblem. I [2], by usig elemetary roerties about [x], it is cosidered the iequality v(!) > q vq(!) i some secial cases, beside it is show that 2 v 2(!) > 3 v 3(!) holds for all 4. I this sectio we study v(!) > q vq(!) i more geeral case ad also rerove 2 v 2(!) > 3 v 3(!). Lemma 4 Suose ad q are rimes ad < q, the Proof: Cosider the fuctio q 1 > q 1. f(x) = x 1 x 1 (x 2). A simle calculatio yields that for x 2 we have f (x) = x x 2 x 1 (x l x x + 1) (x 1) 2 < 0, so, f is strictly decreasig ad f() > f(q). This comletes the roof. Theorem 4 Suose ad q are rimes ad < q, the for sufficietly large s we have v(!) > q vq(!). (13) Proof: Sice < q, the lemma 4 yields that q 1 q 1 that for > N we have > 1 ad so, there exits N N such Thus, ad therefor, So, we obtai ( q 1 q 1 ) (q 1) q 1 ( 1)(q 1). (q 1) q( 1) ( 1)(q 1) (q 1) q, l l q q 1. q 1 q 1 q 1 q 1, ad cosiderig this iequality with (4), comletes the roof. 6

7 Corollary 3 For = 2 ad 4 we have Proof: It is easy to see that for 30 we have 2 v 2(!) > 3 v 3(!). (14) ( 4 3 ) , ad by theorem 4, we yield (14) for 30. For = 2 ad 4 < < 30 check it by a comuter. A Comutatioal Note. I the theorem 4, the relatio (13) holds for > N (see its roof). We ca check (13) for N at most by checkig the followig umber of cases: R(N) := # {(, q, ), q P, = 3, 4,, N, ad < q N}, i which P is the set of all rimes. If, π(x) = The umber of rimes x, the we have R(N) = N # {(, q), q P, ad < q } = 1 2 =3 But, clearly π() < ad this yields that R(N) < N 3 6. Of course, we have other bouds for π() sharer that such as [4] π() < ( l l ) l 2 ( ), ad by usig this boud we ca fid sharer bouds for R(N). N π()(π() 1). Ackowledgemet. I deem my duty to thak A. Abedi-Zade ad Y. Rudghar- Amoli for their commets o the Note ad Problem 1. =3 Refereces [1] Adrew Adler ad Joh E. Coury, The Theory of Numbers, Bartlett Publishers,

8 [2] Io Bălăceoiu, Remarkable iequalities, Proceedigs of the First Iteratioal Coferece o Smaradache Tye Notios i Number Theory (Craiova, 1997), , Am. Res. Press, Luto, AZ, [3] David M. Burto, Elemetary Number Theory (Secod Editio), Uiversal Book Stall, [4] P. Dusart, Iégalités exlicites our ψ(x), θ(x), π(x) et les ombres remiers, C. R. Math. Acad. Sci. Soc. R. Ca. 21 (1999), o. 2, [5] Melvy B. Nathaso, Elemetary Methods i Number Theory, Sriger,