On some properties of digamma and polygamma functions

Transcription

1 J. Math. Aal. Appl O some properties of digamma ad polygamma fuctios Necdet Batir Departmet of Mathematics, Faculty of Arts ad Scieces, Yuzucu Yil Uiversity, 65080, Va, Turkey Received 4 December 2005 Available olie 22 Jue 2006 Submitted by H.M. Srivastava Abstract I this ote we preset some ew ad structural iequalities for digamma, polygamma ad iverse polygamma fuctios. We also exted, geeralize ad refie some kow iequalities for these importat fuctios Elsevier Ic. All rights reserved. Keywords: Polygamma fuctios; Digamma fuctio; Iverse polygamma fuctios; Complete mootoicity; Gautschi iequality; Riema zeta-fuctio ad gamma fuctio. Itroductio For x>0the Euler s gamma fuctio Ɣ ad psi fuctio or digamma fuctio ψ are defied as Ɣx = 0 u x e u du ad ψx= Ɣ x Ɣx. The psi fuctio has the followig itegral ad series represetatios: ψx= γ + 0 e t e xt e t dt = γ x + x x + = x > 0.. address: ecdet_batir@hotmail.com X/$ see frot matter 2006 Elsevier Ic. All rights reserved. doi:0.06/j.jmaa

2 N. Batir / J. Math. Aal. Appl For basic properties of these fuctios see []. Polygamma fuctios ψ are defied to be derivatives of psi fuctio, that is, ψ x = ψ x, =, 2, 3,... The followig itegral ad series represetatios are valid for x>0ad =, 2, 3,...: t ψ x = e t e xt dt =! 0 k=0..2 x + k + Polygamma fuctios arise aturally i the study of beta distributios probability models for radom variables restricted to [0, ]. They play a cetral role i the theory of special fuctios ad have may applicatios i mathematical physics ad statistics. They are helpful tools to approximate classical fuctios ad costats. They are also coected to may special fuctios such as Riema-zeta fuctio ad Clause s fuctio etc. Aother brach of mathematics i which these fuctios are used is iequality theory. Horst Alzer has used polygamma fuctios to prove may basic iequalities o the classical gamma fuctio extesively, see [5 9]. Over the past fiftee years may authors have ivestigated these fuctios ad obtaied may remarkable iequalities, see [2 4,0,3 5,7,9]. It is the aim of this ote to cotiue the study of these fuctios ad to establish some ew iequalities ad to refie ad geeralize some kow iequalities ivolvig polygamma fuctios. Before we prove our mai theorems we eed to preset several lemmas which play key roles i the proofs of the mai results. Lemma.. Let 2 be a iteger. The for all positive real umbers x we have < [ψ x] 2 ψ xψ + x < +..3 This result is kow ad has bee proved by Alzer ad Wells, see [3, Corollary 2.3]. Lemma.2. Let be a iteger. The for all positive real umbers x we have ψ + x <! / ψ x +/..4 Proof. From the left iequality of.3 we have for all, ψ xψ +2 x + [ ψ + x ] 2 < 0. Divide both sides of this iequality by ψ xψ + x to obtai ψ +2x ψ + x + ψ +x > 0, ψ x sice ψ xψ + x < 0. This ca be rewritte as d dx log [ψ x] + [ψ + x] = d dx log [ψ x] + [ψ + x] < 0. Thus the mappig x [ψ x] + [ψ + x]

3 454 N. Batir / J. Math. Aal. Appl is strictly icreasig for x>0. From the asymptotic expressio [, p. 260] [ ψ m x m m! x m + m! ] 2x m+ + 2k + m! B 2k 2k!x 2k+m k= as x, m =, 2, 3,..., we get lim x xm ψ m x = m m!, m=, 2, 3,..., ad hece we obtai [ψ x] + lim x [ψ + x] = lim x [x ψ x] +! [x + = ψ + x] +. This implies that [ψ x] + [ψ + x] <!, + x >0 ad =, 2, 3,... Now a direct computatio proves Lemma.2. Remark.3. We ote that for = i.4 we get ψ x 2 + ψ x > 0..5 Thus, Lemma.2 provides a geeralizatio of.5. We also ote that this iequality has bee used by the author of the preset paper [2,3] ad Alzer [4, Theorem 4,8] to prove may iterestig iequalities for the gamma ad polygamma fuctios. Lemma.4. Let s ad t be positive real umbers, with t>s. The for ay positive iteger we have [ [ [x + tx + s] ] /+ lim A x x + t x + s x] = s + t 2, where A =[t s] /+. Proof. We ca write [ [ [x + tx + s] ] /+ lim A x x + t x + s x] [ Ax + s /+ x [x + s/x + t] /+ ] = lim x [x + s/x + t] /+. Settig y = x + s/x + t ad the multiplyig by y /+ both of the umerator ad deomiator, this becomes lim x [ [ [x + tx + s] A x + t x + s ] + x ] = + At s + y + ty s y / y + lim. y y If we apply l Hospital s rule to the last limit ad let y =, we arrive at the desired limit.

4 N. Batir / J. Math. Aal. Appl Lemma.5. Let x be a positive real umber ad θ be defied by θ x = ψ! x The the followig facts are valid for θ : a θ u > 0 for all u>0. b 0 θ u < /2 for all positive itegers ad all u [0,. c θ u is strictly decreasig for u>0 ad =, 2, 3,... d θ u < 0 for all u>0. x, =, 2, 3,....6 Proof..6 is equivalet to! x = ψ x + θ x..7 Usig the followig differece equatio ψ x + ψ x =! x,.8 see [, p. 260, 6.4.6],.7 becomes for all positive itegers, ψ x + ψ x = ψ x + θ x. This implies by mea value theorem for differetiatio that 0 <θ x < for all x>0. Settig i.7 [!] / x = [ ψ u] /, we fid that [!] / ψ [ ψ u] / + θ [!] / [ ψ u] / = ψ u. Sice the mappig u ψ u is bijective this leads to [!] / [!] / θ [ ψ u] / = u [..9 ψ u] / Differetiatig both sides of.9 we get θ! / [ ψ u] +/ [ ψ u] / =! /..0 ψ + u By Lemma.2 we fid that for all u>0, θ! / [ ψ u] / > 0. But sice the mappig u ψ u from 0, to 0, is bijective we coclude θ u > 0 for all u>0. This proves a. Sice θ is bouded ad strictly icreasig the limit of θ x as x teds to ifiity exists. Now we shall prove b. Replace x by x + i.7 to obtai ψ x + + θ x + =! x +.

5 456 N. Batir / J. Math. Aal. Appl Hece, by.8 we have! x + θ x ψ x + θ x + =! x +, so that we get! + θ x + = x. ψ x + θ x +!/x + Sice lim x θ x + = lim x θ x this becomes by.7 lim x θ x + = lim x Simplifyig this idetity we fid that [ /+ lim x θ x = lim x! x!! x+ x + + /x + x ]. + x. It is easy to prove that this limit teds to /2. This proves b by the help of a. Now we shall prove c. Differetiatig.7, we get! x + = + θ x ψ + x + θ x.. Now replace by + i.7 to get ψ + x + θ+ x =! x +..2 From. ad.2 we obtai θ x = ψ +x + θ + x. ψ + x + θ x Sice θ x > 0forx>0 by a ad x ψ + x is strictly decreasig this yields θ x =, 2, 3,...is strictly decreasig o 0,, provig c. Differetiate both sides of.0 to obtai θ! / [ ψ u] / 2 [ [ψ u] / = [!] 2/ ψ + u Usig Lemma., we coclude for u>0 that θ! / [ ψ u] / < 0. ] 2 ψ u ψ + u [ + [ ψ+ u ] ] 2 ψ uψ +2 u. Proceedig as above we obtai θ u < 0 for all u>0, which proves d. This completes the proof of Lemma.5. A fuctio f is said to be strictly completely mootoic o a iterval I if f has derivatives of all orders o I which alterate i sig, that is f x > 0 for all x I ad = 0,, 2, 3,...

6 N. Batir / J. Math. Aal. Appl Let I be a iterval, s,t I ad f : I R be cotiuous ad strictly mootoic. The by mea value theorem for itegratio there exists a uique ξ [s,t] for which t fudu= fξ. t s s ξ is called itegral f -mea of s ad t, ad is deoted by I f = I f s, t = f t fudu..3 t s I f has the followig property: s Lemma.6. Let I be a iterval ad f : I R be a icreasig fuctio such that f is completely mootoic o I. The x I f x + s,x + t x is icreasig ad cocave o I. This result is kow ad due to Elezovic et al. [6, Corollary ]. 2. Mai results We are i a positio to state ad prove our mai results. I 2000 N. Elezovic et al. [5] discovered a upper boud for ψ i terms of the ψ-fuctio. They proved the iequality ψ x < exp ψx x > 0 2. holds. Alzer [2, Theorem 4.8] used the iequality i.5 to show that there exists bouds for higher derivatives similar to 2.. More precisely, he proved! exp ψx + < ψ x <! exp ψx 2.2 holds for all x>0ad =, 2, 3,... Our first theorem i this sectio refies the left iequality of 2.2. We proved more ad we showed that eve a refied form of 2.2 is a special case of a more geeral iequality for polygamma fuctios. Ideed it is a simple cosequece of the iequality 0 <θ k x θ x < /2 fork = 0, where θ 0 ad θ are as defied by 2.4 ad.6, respectively. Usig this iequality for k, we ca obtai bouds for ψ x i terms of ψ k x as stated i the secod theorem. Theorem 2.. Let be a positive iteger ad x be a positive real umber. The the double iequality! exp ψx + /2 < ψ x <! exp ψx 2.3 holds. Proof. We defie for x>0, θ 0 x = ψ log x x. 2.4 First we shall show that θ x < θ 0 x for x>0, where θ x is as defied by.6 with =. 2.4 is equivalet to ψ x + θ 0 x = log x. 2.5

7 458 N. Batir / J. Math. Aal. Appl Differetiatio of 2.5 yields x = + θ 0 x ψ x + θ 0 x. 2.6 From.7 with = we fid that x = ψ x + θ x. 2.7 From 2.6 ad 2.7 we get θ 0 x = ψ x + θ x ψ x + θ 0 x. I [3] it has bee proved that θ 0 x > 0forx>0. From this ad the fact that ψ is strictly icreasig o 0, we get θ x < θ 0 x for all x>0. This proves by the help of Lemma.5c that θ x < θ 0 x x > 0, =, 2, 3,... Replacig the values of θ x ad θ 0 x give i.6 ad 2.4 here we get ψ! x <ψ log x. 2.8 Replacig x by e ψx here we fid that ψ!e ψx <x. 2.9 If is a eve iteger, the mappig x ψ x is icreasig ad hece we get from 2.9 that ψ x >!e ψx. Similarly if is a odd iteger we get from 2.9 that ψ x <!e ψx. From the last two iequalities we prove the right iequality of 2.3. Now we will prove the left iequality of 2.3. I [3, Theorem 2.] the author of this paper proved that 0 <θ 0 x < /2for all x>0. By Lemma.5b we have 0 <θ x < /2 forx>0 ad =, 2, 3,..., hece we have to have for all x>0, θ 0 x θ x < /2 =, 2, 3, Usig.6 ad 2.4 this ca be writte as ψ log x < ψ! x + 2. Replacig x by e ψx here we get for x>0, x 2 <ψ!e ψx. 2. For odd s 2. becomes for x>/2, ψ x /2>! exp ψx.

8 N. Batir / J. Math. Aal. Appl Replacig x by x + /2 here we get for x>0, ψ x >! exp ψx + / Similarly, for eve s we fid from 2. that ψ x /2<! exp ψx, x >/2. Replacig x by x + /2 here we get for x>0, ψ x <! exp ψx + / alog with 2.3 proves the left iequality of 2.3. Theorem 2.2. Let x be a positive real umber ad be a positive iteger. The the followig double iequality holds for all k =, 2, 3,..., : ψk x + /2 /k! k < ψ x ψ k x /k <! k! k. 2.4 k! Proof. Usig mootoicity of the sequece θ x =, 2, 3,..., where θ is as defied i.6, we get for x>0ad >k, ψ! x <ψ k k k! x k. 2.5 Puttig k k! /k ψ k x for x i 2.5 we obtai for odd s ψ x <!ψ kx k k! x > 0, 2.6 ad for eve s ψ x >!ψ kx k k! x > ad 2.7 together prove the right iequality of 2.4. By Lemma.5b we have 0 < θ x < /2 for all x>0 ad =, 2, 3,... This fact allows us to write θ k x θ x < /2. Usig the values of θ x ad θ k x give i.6 we write ψ! x >ψ k k k! x k Puttig k k! ψ k x /k for x i 2.8 we get for x>/2ad odd s ψ k x /k ψ x /2>! k 2.9 k!

9 460 N. Batir / J. Math. Aal. Appl ad for eve s ψ k x /k ψ x /2<! k k! Replacig x by x + /2 i 2.9 ad 2.20 we reach the left iequality of 2.4. Remark 2.3. a Settig k = ad = 2 i 2.4, we get the iequalities ψ x + ψ x 2 > 0 ad ψ x + ψ x + /2 2 < 0. Hece 2.4 gives a coverse ad aother geeralizatio of.5. Thus, it will be iterestig to ivestigate the best possible costats α ad β such that the iequalities ψ x + ψ x + α 2 > 0 ad ψ x + ψ x + β 2 < 0 hold for all x>0. b Settig x = i 2.4 we get the followig bouds for ζ+ i terms of ζk +, [ k 2 k+ ζk + 2 k+] /k [ ] /k, <ζ+ < kζk + for k =, 2, 3,...,, where ζ is the Riema-zeta fuctio defied by for Re s> ζs= k= k s. Usig mootoicities of θ ad θ fuctio. we get the followig ice iequalities for polygamma Theorem 2.4. Let be a positive iteger ad x be a positive real umber. The the followig iequalities for the polygamma fuctios hold: a α< ψ x + / ψ x / <β, 2.2 b where the costats α =!ζ+ / ad β =! / are best possible. ψ + x < ψ x / ψ x ψ x + c ψ + x + > ψ x + Proof. Let where ψ x + ψ x / gx = θ hx + θ hx, 2.24 hx =! / [ ψ x] /

10 N. Batir / J. Math. Aal. Appl Differetiatig both sides of 2.24 ad the usig.0 we get after a little simplificatio g x = h x + θ hx + h xθ θ hx = hx + θ hx θ hxθ. hx + Sice the mappig x ψ x is strictly decreasig, h is strictly icreasig o 0,. But sice θ is strictly decreasig by Lemma.5d, we coclude from the last equality that g x < 0 for all x>0ad therefore g is strictly decreasig, which implies g <gx<g0 for x>0. It is clear from 2.24 ad 2.25 that g = 0, ad by.9 ad from the series represetatio.2 that [!] / g0 = θ [ ψ ] / = [ζ + ] / which proves a. Now applyig mea value theorem to θ o the iterval [hx, hx + ], we fid that θ hx + θ hx = hx + hx θ h x + δx, where 0 <δx< for all x>0ad h is as defied i Substitutig the value of θ give i.6 here we get hx + hx = θ h x + δx Sice θ is strictly decreasig ad h is strictly icreasig o 0,, we obtai from 2.26 that hx + hx <θ hx ad hx + hx >θ hx +. Replacig the values of h ad θ give i 2.25 ad.0 respectively ad the simplifyig the lastig expressio these become ψ + x ψ x /.ψ x ψ x + ad ψ + x + > ψ x + So b ad c are proved. ψ x + ψ x /. There exists a extesive ad rich literature for polygamma fuctios, but the iverse polygamma fuctios have ot bee ivestigated ad almost othig is kow about them. I the followig theorem, which is a ice applicatio of Theorem 2.4a, we provide explicit bouds for them.

11 462 N. Batir / J. Math. Aal. Appl Theorem 2.5. Let be a positive iteger ad x be a positive real umber. The we have! + [ x x / + α ] + < x [ ψ <! + x x / + β ] + where the costats α =! / ad β =!ζ+ / are best possible. Proof. Settig x = ψ t i 2.2 ad the usig.8 we get [ ] /! /!ζ+ < t ψ t + t / <! /. Simplifyig these iequalities ad the settig t = x fiishes the proof. The followig theorem gives ew bouds for digamma fuctio i terms of ψ -fuctio. Theorem 2.6. Let x be a positive real umber. The the followig double iequality for digamma fuctio holds: γ + xψ x/2<ψx+ < γ + xψ x Here γ is Euler s costat. Proof. By. we have the followig series represetatios for x>0: ψx + = γ + ωk ωk + x, 2.28 k= where ωt = /t. By the mea value theorem for differetiatio we have a μ = μk = μk, x such that 0 <μk<xad x ωk ωk + x = k + μk Employig 2.29 i 2.28 we fid that ψx + = γ + x k + μk k= From 2.29 we obtai μk = kk + x k. It is ot difficult to show that k μk is strictly icreasig o, ad lim μk = x k 2. Hece, from 2.30 we get γ + x k= k + μ 2 <ψx+ < γ + x k + μ 2. By the facts μ = x +, μ = x/2, ad.2 with = we fiish the proof of Theorem 2.6. k=

12 N. Batir / J. Math. Aal. Appl The secod Gautschi Kershaw iequality for the Ɣ-fuctio states that ψx + logɣx + logɣx + s s < <ψ x + s +, s 2 where 0 <s<, see [5,8]. I the followig theorem we establish a exteded form of these iequalities for polygamma fuctios. Theorem 2.7. Let x ad y be positive real umbers ad be a positive iteger. The we have x + y ψ + < ψ x ψ y 2 x y < ψ + S + x, y, 2.3 where for p R, a p b p S p a, b = pa b /p is Stolarsky s mea of a ad b, see for details [20]. Proof. From the series represetatios.2 we ca write + ψ x ψ y =! σk+ x σk+ y, 2.32 k=0 where σu = /u +. By the mea value theorem for differetiatio we have η = ηk = ηk,x,y such that η is betwee x ad y for which + x y σk+ x σk+ y = k + ηk +2 Hece, 2.32 ca be rewritte as +2 ψ x ψ y = +! From 2.33 we get ηk = σ x y k+y k+x σ u du k. k= k + ηk +2 Sice k σ k is completely mootoic o 0,, ηk is strictly icreasig by Lemma.6. From 2.33 we get + y xk + x + k + y + +2 ηk = k. k + y + k + x + From this idetity ad Lemma.4 we fid that + y xxy + η0 = y + x

13 464 N. Batir / J. Math. Aal. Appl ad lim ηk = x + y. k 2 Now we fid from 2.34 that +! k + η +2 < ψ x ψ y k=0 <+! k + η0 +2. k=0 The proof follows from 2.35, 2.36 ad series represetatio I [3, Theorem 2.b] the author of the preset paper discovered the followig bouds for the digamma fuctio: If a = log 2 ad b = 0, the the followig iequalities hold: a log e /x <ψx<b log e /x, x > Thus, it is atural to ask whether the umbers log 2 ad 0 ca be replaced by better costats. I the followig theorem we prove that the right-had side of 2.38 is sharp, but the costat log 2 o the left-had side ca be improved. Theorem 2.8. The psi fuctio satisfies the followig double iequality for x>0: α log e /x <ψx<β log e /x, 2.38 where α = γ γ is Euler s costat ad β = 0 are best possible costats. Proof. Replacig x by e ψx i 2.4, we get θ 0 e ψx = x e ψx. Now we defie for x>0, gx = θ 0 e ψx+ θ 0 e ψx Differetiatio yields g x = v x v x +, 2.40 where vx = e ψx. By differetiatio twice we fid that v x = [ ψ x 2 + ψ x ] e ψx. By.5 we have ψ x 2 + ψ x > 0, which proves that v is strictly icreasig, ad hece g is strictly decreasig by I [3, Theorem 2.b] the author of this paper proved that lim x θ 0 x = /2. Employig 2.39 we obtai from mootoicity of g that g0 = e γ <gx<g = 0 x > Sice gx = e ψx e /x, iequality 2.4 allows us to coclude that 2.38 holds with the best possible costats α = γ ad β = 0.

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