Bounds for the Positive nth-root of Positive Integers

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1 Pure Mathematical Scieces, Vol. 6, 07, o., HIKARI Ltd, Bouds for the Positive th-root of Positive Itegers Rachid Marsli Mathematics ad Statistics Departmet Kig Fahd Uiversity of Petroleum ad Mierals Dhahra, 36, Kigdom of Saudi Arabia Copyright c 07 Rachid Marsli. This article is distributed uder the Creative Commos Attributio Licese, which permits urestricted use, distributio, ad reproductio i ay medium, provided the origial work is properly cited. Abstract I this paper, the author derives, a upper ad a lower bouds for the positive th-root of positive itegers. I the secod sectio, the mai result is applied to geerate some iequalities ivolvig Gamma futio. Mathematics Subject Classificatio: 33B5 Keywords: Gamma; product; root Itroductio The motivatio behid this result is Fermat last theorem itself. We discovered k the iequalities ( + j ) < k k < ( + ) while tryig to fid j bouds A ad B for x + y, satisfyig the coditio A = B, where x, y are itegers, ad. is the greatest iteger fuctio. Ufortuately, as it will be see i Corollary, the absolute errors give by the obtaied bouds diverge at ifiity. Therefore they caot satisfy the above coditio. However, because of the geeral aspect of this iequalities, i the sese that they apply to all positive itegers greater tha, ad because they reflect a coectio betwee aalysis ad umber theory, from the viewpoit that the bouds are product series ad k takes iteger values; we foud it worthy to explore

2 48 Rachid Marsli them further ad look for their applicatios. This leaded as to some results ivolvig gamma fuctio, product series as well as the its stated i the last theorem of this paper. Mai result From the idetity b a = (b a) b k a k, we have k=0 x + x =. () ( x + ) k ( x) k k=0 Usig the fact that x < x + for every positive iteger ad every positive real umber x, we have ( x ) < k=0 ( x + ) k ( x ) k < k=0 ( x + ), k=0 that is ( x) < ( k x + ) k ( x) k < ( x + ). () k=0 From () ad (), it follows that ( x + ) < x + x < which implies (, (3) x) x + x + (x + ) < x + < x ( + x ). (4) By actig o the left iequality above, we get x ( + (x + ) ) < x + < x ( + x ). (5) Now, i (5), replace x, cosecutively, by,, 3,..., k to get ( + ) < < ( + )

3 Bouds for the positive th-root of positive itegers 49 ( + 3 ) < 3 < ( + ) 3( + 4 ) < 4 < 3( + 3 )... k ( + k ) < k < k ( + (k ) ). Usig iductio, we obtai our mai result which we state as a theorem. Theorem.. For every positive itegers k ad both, k ( + j ) < k < k ( + j ). (6) Example ( + 5j ) < < 6 ( + 5j ).59. I what follows, we deote k (+ j ), respectively, by A,k, B,k. Remarks. a k (+ j ), k ad ad C,k. Uless stated otherwise, ad k are itegers. Clearly, A,k, B,k ad C,k all coverge to as.. The divergece of the products of the form ( + ), where a is aj + b a positive real umber ad b is a real umber, ca be deduced from (6). Cosequetly, ( aj + b ) = 0. (7) 3. The followig idetities are used thereafter, i some proofs ad calculatios. A,k = k ( j ), (8)

4 50 Rachid Marsli ad C,k = k ( j + ), (9) C,k A,k = k ( )(k + ) k ( (j) ). (0) 4. The right iequality i (6) ca be also show based o these two facts: k j + k =, () j ad + j < ( + j ). () As show i the ext corollary, Theorem. ca be used to determie some k bouds for the products ( + k j ) ad ( j ). Corollary. k + k k < k ( + j ) < k ( (j) ) k, (3) ad k < k ( j ) < k k + k ( (j) ) k. (4) The right iequalities i (3) ad (4) ca be relaxed by omittig the term k ( (j) ). Theorem.3. ad C,k A,k = si(π/) π( ) >, (C,k A,k ) =.

5 Bouds for the positive th-root of positive itegers 5 Proof. We use the well kow sie product formula that ca be foud i [3] ; ( x j ) = si(πx) πx. (5) Hece C,k A,k = [ k ( )(k + ) = si(π/) π( ). k ( (j) )] The quotiet C,k A,k icreases i terms of k, because C,k+ /A,k+ = k + k + k + C,k /A,k k + k + k >. Therefore sup( C,k ) = si(π/) k A,k π( ), (6) ad sice C, = + A, >, it must be si(π/) π( ) >. (7) Also, [C,k A,k ] = [ = = k ( )(k + ) si(π/) ( ) π/ k = ; by usig (7). [ k ( + k j ) ( + j )] k ( (j) ) ] k ( j ) ( j ) ; by usig (5),

6 5 Rachid Marsli Remark.4. By (7), we have < si(π/) π/ <, (8) from which, it follows that for every iteger m, we have < m ( (j) ) <. (9) Now, we prove some facts related to the absolute ad relative errors give by A,k ad C,k o B,k. The followig iequality kow as Beroulli s iequlity will be of some use for us: For all positive itegers j ad, with, we have j < ( j ). (0) Theorem.5. Both B,k A,k ad C,k B,k icrease i terms of k, with ad < B,k A,k < < C,k B,k < π( ) si(π ), () π( ) si(π ). () Proof. Iequality () esures that To show that B,k A,k C,k A,k is a icreasig sequece of k. icreases i terms of k, we observe that k = k j, (3) ad we use Beroulli s iequality (0) which implies > (, (4) ) j j that is j > j = + j. (5)

7 Bouds for the positive th-root of positive itegers 53 Therefore Sice B, A, > ad C, B, B,k+ /A,k+ B,k /A,k >. >, we coclude that B,k A,k > ad C,k B,k >. Combiig the above iequalities with formula (6), we have < B,k A,k C,k B,k = C,k A,k < si(π/) π ( ). This completes the proof. Corollary. ad (B,k A,k ) =, (C,k B,k ) =. Proof. From Theorem.5, we have that 0 < [B,k A,k ] < si(π/) π ( ). Therefore, (B,k A,k ) = ( B,k A,k ) (A,k ) = ( B,k A,k ). =. The secod assertio of the corollary is proved similarily. Example.6. The followig short example, gives a little sight o the relative umerical behavior of the above series as well as the absoute ad relative errors i approximatig B,k from the left by A,k ad from the right by C,k. For = 0 ad k = 3000, we have A,k B,k C,k B,k A,k B,k A,k B.k C,k B,k C,k B,k B,k, 469, 4930,

8 54 Rachid Marsli 3 Coectio with Gamma fuctio. I this sectio, we use the mai theorem to prove some iequalities ivolvig the real Gamma fuctio defied by Γ(x) = t x e t dt, x R {0,,,...}. The followig two formulas, which ca be foud i [3], are amog the wellkow properties of Γ(x). Γ(x + ) = xγ(x), for x R {0,,,...} (6) ad Γ(x) Γ( x) = π si(π x), for x / Z. (7) We remaied the reader that, as before ad uless stated otherwise, ad k are itegers. Theorem 3.. ( )k k Γ(k) Γ( ) Γ(k ) < k < Γ(k + ) Γ(k) Γ( + ). (8) Proof. Usig (6), we have from which Γ(k + ) = ([k ] + )([k ] + )... ([] + )Γ( + ), k ( + j ) = k (j + ) (k )! = Γ(k + ) Γ(k) Γ( + ), (9) ad k ( + j ) = k! = k (j ) ( )k k Γ(k) Γ( ) Γ(k ). (30) The we use the mai theorem to obtai the result.

9 Bouds for the positive th-root of positive itegers 55 I what follows, we use Algebra operatios uder the radical i the mai theorem, to geerate more iequalities ivolvig Gamma fuctio ad closed form products. Theorem 3.. Let s ad t be two differet itegers. The ( )(t)(st)! ( + )(t )(s )!(t )! < Γ(st + + ) Γ( + ) Γ(s + )Γ(t + ) < Proof. Theorem. implies that (st)! ( )(s )!(t )!. (3) s ( + j ) t ( + st j ) < ( + j ), (3) ad st ( + s j ) < ( + t j ) ( + j ). (33) A little work o (3) ad (33), usig (8), (9) ad (0), leads to the followig iequalities s [ + ] (j) t [ ] st [ ] + < + j j s < st So that we have s [ ] + < (j) s s [ ] t (j) ( )(t) t [ ] st [ ] + < + < ( + )(t ) j j The we use (9) to complete the proof. [ ] +. (34) j t [ ] +. (35) j Additio of itegers uder the radical i the mai theorem, may lead to some iequalities such as the followig. Theorem 3.3. Let s ad t be two itegers, with s t. The + s ( + (s+t) j ) < ( + ( + ) ) < j t ( + j ). (36)

10 56 Rachid Marsli Proof. I the mai theorem, replace k cosecutively, by s, t ad (s + t), to get s ad ( + j ) + That implies (s+t) t ( + s j ) < s + t < ( + t j ) + ( + j ), ( + j ) < s + t < (s+t) ( + j ). [ + ] [ + (s + t) ] [ s ( + t j ) + ( + (s+t) )] < ( + j j ). The observe that to deduce (s+t) (s+t) Sice s t, we have s ( + j ) < + ( + j ) < + (s+t) ( + j ), [ s ( + t j ) + ( + ] j ). ( + s j ) [ ( + t j ) + ( + t )] ( + j j ), ad the theorem follows. Corollary 3. Γ ( (s) + + ) S! Γ ( s + + ) (S)! Proof. I (36), replace t by s to obtai, + < ( + j ) (s) s ( + j ) < =. (37) ( + ) The use formula (9), ad take the it as.. (38)

11 Bouds for the positive th-root of positive itegers 57 Next, we use the mai theorem to geerate some iequalities ivolvig positive th-root of ratioal umbers. Theorem 3.4. Let p ad q be positive itegers with p > q. The p ( + p j ) ( + (j) ) < j=q p q ( + j ). (39) < p Proof. If q, the usig the left iequality i the mai theorem, we have p < p ( q j ) ( + q, j ) i.e. j=q p ( + p j ) ( + (j) ) < q p q. The above iequality holds for the case where q =. The right iequlity i (39), is obtaied by usig idetities () ad (). Corollary 4. Let α, p ad q be positive itegers with p > q. The p [ q < ( + ] p α j ) < β q, (40) ad q p < with β = si(π/) π( ) [ ( ] α j ). < β q p, (4) p Proof. Sice q =, we replace p by ad q by α q i (39), we let α q α, the we apply formula (5). To prove the iequalities i (4), we take the reciprocals of the terms i (40), the we use the fact that α ( (j) ) = α α q α ( (j) ) ( (j) ) =. (4)

12 58 Rachid Marsli Example 3.5. Here is a short example that shows the umerical aspect of the product i formula 40, whe α takes values, 0, 00, 000. For = 5, p = 7 ad q =, we have , ad α α ( + ) 5j α Some few extra examples give us the impressio that the product i (40) decreases, i the geeral case, toward subject of the followig theorem. p q as α teds to ifiity. This is the Theorem 3.6. Let α, p ad q be positive itegers with p > q. The [ ( + ] p j ) = q. (43) ad α [ ( ] α j ) Proof. Sice ( + j ) Now, α j = m = q p. (44) < exp( ) it follows that j m i=0 ( + j ) < exp( j ). x i δx, with x i = q+ i α, m = α(p q) ad δx = α. Therefore, by the defiitio of Reima itegral, we have m p δx = α x i x dx = l(p q ), so that α i= q ( + j ) exp(l( p q )) = p q. The the poof of (43) is completed by usig Corollary 4. The proof of (44) is doe by takig the reciprocals of the terms i (43) ad usig (4). Ackowledgemets. The author would like to express sicere appreciatio to Dr. Abdul-Aziz Al-Assaf, Dr. Frak Hall ad Dr. Garbor Korvi.

13 Bouds for the positive th-root of positive itegers 59 Refereces [] E.T. Whitaker, G.N. Watso, A Course of Moder Aalysis, Cambridge, 90. [] M. Abramowitz, I.A. Stegu, Hadbook of Mathematical Fuctios, Dover, New York, 97. [3] W. Freede, M. Guttig, Special Fuctios of Mathematical (Geo- )Physics, Spriger, Received: Jauary 7, 07; Published: February 4, 07