GAMMA FUNCTION. n! = 1 ( log 0 s)n ds for n > 0 For the time being, we will refer to these as the product and integral definitions.
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1 GAMMA FUNCTION Abstract. I this paper we eplore the history ad properties of the Gamma fuctio i a aalytic umber theoretical cotet. We aalyze the behavior of the Gamma fuctio at its critical poits ad poits of discotiuity, ad discuss the covergece of the itegral.. Itroductio.. History ad Motivatio. I the early 6th cetury, Leohard Euler ad others attempted to epad the domai of the factorial to the real umbers. This caot be doe with elemetary fuctios, however, with the otios of its ad itegrals from the calculus, there were a few epressios developed. I 738, Euler published the followig forms of the geeralized factorial:! +/ +/! log s ds for > For the time beig, we will refer to these as the product ad itegral defiitios. I the 9th cetury, Carl Friedrich Gauss rewrote Euler s product defiitio to eted the domai to the comple plae, rather tha simply the real umbers. This epressio uses a itig process o a series of itermediate fuctios to represet the factorial. r!r Γ r r + Γ Γ r r This was also the time period that the otatio Γ was applied to the cocept. It was amed i 8 by Adrie-Marie Legedre, who also trasformed the itegral defiitio i a very simple way to eted its domai:! log s ds Let t log s. s e t
2 dt s ds et ds Γ t e t dt, which is the more commo itegral defiitio that we see today. A few years later, Karl Weierstrass rewored Gauss epressio i the followig maer: log r e Γ r + / + / /r e log r e γ e +/2+/3+...+/r e e /2 Γ r e γ + / + /2... e /r + /r Γ r Γ r e γ + r e /r, where γ r log r... is Euler s gamma costat. 2 3 r This formulatio is defied i terms of zeroes rather tha poles. This actually ispired Weierstrass to prove the Weierstrass Factorizatio Theorem, which says that ay etire fuctio ca be writte as a product of its zeroes over C. I 922, Bohr ad Mollerup cofroted the issue of whether these epressios of the gamma fuctio were equivalet, ad, relatedly, whether the Gamma fuctio as defied is uique. Their results are summarized i the followig theorem: Theorem. Bohr-Mollerup Theorem. If Γ Γ 2 o the itegers ad Γ Γ 2 is logarithmically cove for Re >, the Γ Γ 2 R..2. Log coveity. The previous theorem guaratees uiqueess for the Gamma fuctio if it is logarithmically cove. The followig theorem ad its proof ca be foud i Emil Arti s paper, The Gamma Fuctio. Defiitio.2. f is said to be logarithmically cove if the fuctio log f is cove. Theorem.3. Let f be a twice differetiable fuctio. The f is cove if ad oly if f for all i the domai. 2 r
3 Now we will show that the Gamma fuctio satisfies the above criterio for log coveity. Propositio.4. Γ is logarithmically cove for all >. Proof: d d log Γ Γ Γ + γ r + r r d 2 d log Γ i 2 It is clear that this epressio is strictly positive, so the Gamma fuctio is log cove. This establishes the uiqueess of the Gamma fuctio as the etesio of the factorial..3. Properties. The Gamma fuctio satisfies the followig fuctioal equatios: Γz + zγz Γ Γ! for N Γ zγz π siπz We epress the derivatives of the Gamma fuctio i terms of aother special fuctio, the Polygamma fuctio, Ψ. i Ψ m z m+ t m e zt dt e t More importat i our cotet is the followig form: Ψ m z d dz m+ log Γz. The digamma is the first polygamma fuctio, i the case m. Ψ z Γ z Γz Γ z Ψ zγz This leads us to use the followig epressio for the derivatives of gamma: Γ z t z e t log t dt. 3
4 The Gamma fuctio is aalytic for all values z C ecept for the o-positive itegers. At these poits the fuctio has simple poles. The residue for the poit at,, 2,... is!. 2. Notatio Here is some otatio that we will use later i this paper: For, s, ad real umbers, we defie: Ψ : Γ Γ E + : + E : Hs, : ζs : s s 3. Properties Here are some properties that we will use throughout this paper: Propositio 3.. For all real with >, we have: log 4
5 Proof. Because >, we have <. Therefore, we have the followig: Now itegratig both sides gives us: d log log log This proves our propositio. 2 2 d d Propositio 3.2. For all real the fuctio E + + is icreasig o the iterval [,. Proof. We have: E + + log E + log E + E + + log + + log
6 By propositio 3., we have: E + E > Our proof is doe because we have proved that E + > for all. Usig this propositio, we ca establish the boudaries for E + o [, : 2 E + < E + e Propositio 3.3. For all real the fuctio E is decreasig o the iterval,. Proof. We have: E log E log E E log log log < Sice E < for all >, our proof is doe. 2 Note that this propositio ca be used to establish the followig iequality: E > E e 6
7 Propositio 3.4. For every positive iteger, the fuctio g : log is icreasig. Proof. We have: g + g + log > This proves our propositio. 4. The behavior of the gamma fuctio ear its poits of discotiuity Now let us aalyze how the gamma fuctio behaves ear its poits of discotiuity. First let s prove the followig properties: 4. Γ γ 4.2 Γ γ Proof. We have: Γ Γ Γ + Γ + Γ + Γ Γ γ 7
8 Similarly: Γ + Γ Γ + Γ Γ Γ Γ γ These its tell us that the gamma fuctio behaves almost eactly lie y whe gets close to. Combiig these its gives us: [ Γ + Γ Γ + ] Γ 2γ Now let s see how the gamma fuctio behaves at other poits of discotiuity. Usig the recursive formula for gamma fuctio gives us: Γ + Γ + Γ + Γ Note that from 4. we have: [ Γ + Γ + + ] + γ + I this it, it does ot matter if goes to zero from the left or right side. Therefore, we have: + + Γ +, ad so: Similarly: + + Γ! + Γ! 8
9 These two it properties tell us that the gamma fuctio behaves almost eactly as the fuctio: y c, where c is a costat depedig + o, as gets close to the poit either from the left or right had side. For ay positive iteger, the it properties of the gamma fuctio show us that the graph is almost the same aroud. The differece i sig leads to the questio of whether or ot they cacel. We already have the aswer for the poit with the it we established earlier. They cacel each other, ad so we are left with the costat 2γ. Our tas ow is to fid the followig it LM or to prove that it does ot eist: We have: LM Γ + + Γ Γ Γ + Γ + Γ + Γ + + Γ 2 2! 2! 2! 2 Γ + + Γ c Γ + Γ c T where c are epressios ivolvig, ad T Γ + Γ. Note that whe is eve ad greater tha, we have: c Γ + Γ c 2γ So we are left with all the terms with odd, ad the term with, which we will tae ito accout later. Let s loo at the terms we have 9
10 left odd: T Γ Γ [ Γ + Γ + 2 ] Γ Γ T γ + γ So T will go to zero if 2. That leaves us with oly two terms left to cosider. For the term T, we have: c! Ad so: For the term with we have: c T 2! c T! Γ + Γ Ad so: c T 2γ! Now we have eough iformatio to calculate our origial it: Fially: LM! 2! 2! 2 c T c T + c T 2γ! + 2! Γ + + Γ 2H, 2γ! 5. Critical poits of the Gamma fuctio o the left We will aalyze the critical poits of the gamma fuctio o each oe-uit-wide iterval:,, 2,, 3, 2,.. ad so forth. For the th iterval, +, let: +, where is the coordiate of the critical poit of the fuctio o this iterval Γ
11 . Note that because, +, we have,. We will eplai later how we oly have oe critical poit per iterval. Now: Hece: Γ Γ Γ Γ + Γ Γ Γ + Γ Γ Γ Γ Γ + Γ Γ Ψ + Ψ Ψ + Ψ Let be the -coordiate of the critical poit i the th iterval, ad i the above equatio let. That meas +, as defied earlier. Therefore, we have: Ψ because Ψ Ψ Γ. I other words, Γ is a root of the equatio: 5. Ψ We will prove that 5. has oly oe root o,, which meas the value is uique, as is. Let s loo at our recursive formula for Ψ: Ψ + Ψ Ψ + + Ψ Ψ + Ψ ] [ Ψ Ψ Ψ +
12 with Ψ γ, a costat. Therefore: Ψ + Now let s tae the first derivative of Ψ usig the followig formula for the digamma fuctio: Ψ γ + + Ψ + > 2 This meas that Ψ is a icreasig fuctio, so whe goes from to, Ψ goes from to Ψ γ. O the other had: [ ] d d <, 2 which meas that the right side of 5. is decreasig with respect to. Whe, our right side RS becomes: RS H, Moreover: RS + + RS costat 2 Now let s summarize what we have foud. As goes from to, our left side icreases from to γ, ad our right side decreases from H, to. This fact guaratees us eactly oe solutio to our equatio. Assume that we have foud the root to our equatio. We will ow try to see whether + is smaller or greater tha. Note that i our equatio, if we start with ad eep icreasig the value of searchig for our root, we will have our left side is smaller tha our right side. That meas whe we come to a value of where we have our left side greater tha our right side, we ow that we have 2
13 passed our root, meaig our root is smaller tha the value of we are at right ow. Our + is the root of the followig equatio: Ψ + Note that whe, we have: RS + < Ψ LS So at the value, we have our left side greater tha our right side. This tells us that our root + is smaller tha. Now we have a strictly decreasig sequece { }. Now let s go bac to our equatio 5. ad multiply both sides by. We have: Ψ + l l l l [ l l 3 + ] l
14 Now chagig the otatio gives us: [ Ψ l Hl, ] with: l H, [ l Hl, ] l2 log + H, log log H, log Ψ log A A H, log + [ l Hl, ] l2 [ l Hl, ] l2 [ l Hl, ] Accordig to propositio 3.4, we have H, log log is a decreasig fuctio, so: H, log > H, log γ > ad also: H, log < H, log. Also otice that: Hs, < ζs s s Therefore: A H, log + A < + l ζl l2 < ζ2 + ζ2 ζ2 l l ζ2 l2 l 4 l2 [ l Hl, ] l2
15 The above iequality is true because < ζ2 π2, ad ζ2 > ζl 6 for all l 3. Remember that we are usig the equatio correspodig to, ad <. Therefore, we have: A < ζ2 < ζ2 because the fuctio f is icreasig o,. We have proved that there is eactly oe critical poit i each iterval. That meas ca be cosidered a costat, which meas A is bouded. Net we will show that Ψs is also bouded o,. From our recursive formula for Ψ, we have: Ψ + Ψ + Ψ + Ψ + Ψ + Ψ + Ψ The above argumet is valid because Ψ γ, a costat. Now the fuctio Ψ is cotiuous o,, ad Ψ ad its value at is γ. Therefore, Ψ is bouded o,. Usig these results gives us: log Ψ + A is also bouded. Now if goes to ifiity, the i order for log to be bouded, we must have: We showed earlier that H, log >. That meas A >. Now we have two boudaries for A: < A < ζ2 However,. This meas that: A Therefore: log Ψ + log + log 5 A
16 Fially, we have established a relatioship betwee ad, through a it property. Now we will try to do the same thig with the value of the gamma fuctio at those critical poits. We have: Γ Γ 2Γ 2. Γ Γ Let d be the value of the gamma fuctio at the critical poit i the th iterval. Note that with the defiitio of, the is the -coordiate for the critical poit o the th iterval. Therefore: d Γ Γ Let s loo at the followig epressio: B! We have: with: B! E e + ε e + ε C ε E e > 6
17 Now: C log C e + ε [ e + ε e [ + ε ] e e [ e E + Sice < 2 E ε ε e e e log E+ + ] ε e ε <... <, we have E e ] + e ε e ε > E ε e 2 e >... >, ad ε > ε 2 >... > ε. Usig the fact that E + > 2 for all >, we have: ε log C > e log 2 + Note that: ε log 2 e ε log 2 + e + ε log 2 H, + H, e ε log 2 H, log + log + e + H, log + log ε log 2 H, log + ε log 2 e e + H, log + log D ε log H, log γ 7 log
18 Usig these its gives us: D γ + + γ + Sice E + < e for all >, we have: log C < ε e + ε e + ε e + ε e H, + H, ε e H, log + log + + H, log + log ε e H, log + ε e log + H, log + log F Note that whe goes to ifiity, ε also goes to zero. Therefore, we have: F γ + + γ + Now we have D < log C < F ad D F. Therefore: Ad so: log C C e That meas:! 8 B e
19 Applyig this result gives us: Γ d Γ +! d log Γ + log!! d log e! d log e Ad ow we have established a relatioship betwee the value of the gamma fuctio at its critical poit i the th iterval ad the value of. 6. The blutess of the gamma fuctio o the egative side I this sectio, we will try to eplai why the graph of the gamma fuctio seems to flatte out whe we move to the left from oe iterval to the et. I order to do that, we cosider solvig the followig equatio i each iterval: 6. Γ α for some real ad positive α. Let deote the solutio to 6. o the correspodig iterval, +. We are just iterested i fidig the root that is o the right side of the critical poit of that iterval. I other words, if we are solvig the equatio 6. for the th iterval, the we oly loo for the root that qualifies: >, where is the critical poit of the gamma fuctio i the th iterval. We have show earlier that: Γ Γ + Γ Usig this properties times gives us: Γ Γ + Γ. Therefore: Γ Γ + Γ Now assume that our sequece { } is bouded by a positive costat s less tha. I the other words: s <, for all positive itegers. We will prove that whe is large eough, we caot fid a solutio 9
20 for equatio 6. i the th iterval. Of course this cotradicts with the fact that the domai of Γ is from egative ifiity to positive ifiity o ay itervals o the egative side. This cotradictio would cofirm that whe is large eough, the graph of the gamma fuctio will loo as flatteed as possible. I order to prove that we will ot have a solutio o the right side of the critical poit to 6. whe is large eough, it suffices to show that: Γ s < α, whe is sufficietly large. I the equatio we derived earlier, substitutig by s gives us: Γ Γ s + Γs s s s We will aalyze the umerator N ad the deomiator D separately. Note that sice s is a positive costat less tha, we have Γ s, Γs, ad s are costats also. We have: Therefore: N Γ s + Γs s Γ s + Γs s + Γs + s 2 < Γ s + Γs s + Γs 2 Γ s + Γs + ΓsH, s Now for our deomiator, we have: D! D N log H, log s s > s > s + s 2 s! 2 2
21 Combiig these results gives us: or: However:! log N log D! < s N! < D log s. This tells us that: N D I other words: Γ s. Note that what we have proved is pretty powerful. It tells us that o matter how close s is to, we always have Γ s as close to zero as possible whe is large eough. Equivaletly:. Our et goal is to aalyze how fast goes to. We will do that by aalyzig the sequece {y } defied as: y. Sice, we have: y. We have: Γ + Γ Γ α. Chagig the variable gives us: Γ y + Γ y Γ y + Γ y + Γ y y + Γ y y + Γ y y + + Γ y Γ y y y + + y + y + y + y + α α α A + B + C α, with A, B, ad C are the first, secod, ad third term respectively. We will tae the it whe goes to ifiity for both sides. Notice that: Γ y Γ γ, 2
22 ad: Γ y Γ. The first thig we ca easily otice is that: A. Dealig with B ad C requires us to prove somethig first. Recall that whe we have the sequece { } that qualifies: < <, ad log, we have:! e Whe we loo bac at the proof, we ca see that if we had log c istead of, we would have:! e c Now let s loo at the followig it:! +! 2 2! siπ π 2 From the it we just derived, we have: +! e c, Note that the secod to last equal sig should be justified more formally, or we ca use the similar proof that we did for the other case. Assume we have log c. Let: G +! 22
23 We have: B +! + + E + e ε H + with: ε e E + > Now: H log H e ε [ e ε e [ ε ] e e [ e E ε e ε ε e e log E e ε + ] e ] e ε e ε ε e e Sice < 2 <... <, we have E + E +, ad ε > ε 2 >... > ε. We have ε e E + < E + 2 <... <. Therefore, < e E + e 2 <. Hece, e ε > e > 2. 23
24 Therefore, E e ε < E 2 4. We have: log H > > ε e log 4 + ε log 4 + e ε log 4 e + ε log 4 H, + H, e ε log 4 H, log + log + e + H, log + log ε log 4 H, log ε log 4 e e + H, log + log D log Note that: ε log c H, log γ Usig these its gives us: D γ + c + γ + c c 24
25 Sice E > e for all >, we have: ε log H < e + ε e + ε e + ε e H, + H, ε e H, log + log + + H, log + log ε e H, log ε e log + H, log + log F Note that whe goes to ifiity, ε also goes to zero. Therefore, we have: F γ c + γ + c c Now we have D < log H < F ad D F c. Therefore: Ad so: That meas: log H c H ec +! G e c Now let s go bac to aalyze our terms B ad C. Notice that: ad: y + >! 2 < < y + 25
26 Therefore: We have: log y y log y y + A + B + C α A C + B C + α C + y + y + log y + Usig this it gives us: log log y! e ad: + y e! From our equatio: A+B+C α, multiplyig both sides by y ad taig the it whe goes to ifiity gives us: y 2! y αy 2 y + y + α y + y + y +! α y2! α 26
27 Now we will use this it property to prove that for sufficietly large, the sequece { } is strictly icreasig. Usig the recursive formula for the first derivative of the gamma fuctio times gives us: Γ Γ + α After dividig both sides by Γ, we have: Ψ + α Γ Moreover, Γ Γ. Therefore, we have: Ψ + α Γ or: 6.2 f where: 6.3 f Ψ + α Γ Note that i the equatio 6.2 our left had side is icreasig o the iterval, ad our right had side is decreasig o the iterval,. Now let s loo at equivalet equatio we eed to solve to fid a solutio, o the et iterval, to our origial equatio Γ α. That is: α f + Γ Lettig maes our left side become: f + Ψ α Γ + + O the other had, our right side would be: α Γ α + Γ For fidig +, if we start at the critical poit of this iterval, +, we will have our left side is zero ad our right side is positive. This meas that at ay poit, if we have the value of the left side still smaller 27
28 tha the right side, we ow that our solutio, +, is greater tha that poit. I other words, if we have: α Γ + + α + Γ the we ca coclude that +. The iequality gives us: We have: + α y + y y + α Γ Γ + Γ + 2 Γ y + y + y y + Γ y + y + y y + y + y 2! y +! y y + y + α Therefore, the it of the right side of our iequality whe goes to ifiity is zero. This meas that with is large eough, our iequality is qualified, sice α is positive. I other words, whe is sufficietly large, our sequece { } is strictly icreasig. 7. Properties ivolvig the gamma fuctio Propositio 7.. e log 2 d π2 6 + γ2 Proof. From the defiitio of the gamma fuctio, we ca come up with the formula for the secod derivative of the gamma fuctio, ad 28
29 that is: Γ t e t log 2 tdt Therefore: Γ e t log 2 tdt e log 2 d To calculate Γ, we first eed to calculate Ψ. We have the followig formula for Ψ: Ψ + γ + Ψ + Ψ r r r r + 2 Remember that: Ψ Γ Γ. Therefore: Therefore, we have: rr + π2 ζ2 r2 6 Ψ Γ Γ Γ 2 Γ 2 Ψ Γ γ 2 Γ π2 6 + γ2 e log 2 d Γ π2 6 + γ2 This proves our propositio. Propositio Γ γ 29
30 Proof. We have: Γ Γ Γ Γ γ Γ Γ Γ Our propositio is proved. I this comig sectio, I hope you will forgive me the mildest of abuses of otatio. Whe itegratig up to or over the discotious poits of Γ the proper otatio of the uderlyig itig process may be omitted for the sae of saity ad simplicity. 8. Area Uder the Curve Whe cosiderig b Γd several questios arise. a For what values of a, b does this itegral coverge? For what values of a, b, if ay, does this itegral diverge? As a does this itegral coverge or diverge? I order to better uderstad the aswers to these questios, it becomes ecessary to first eamie the behavior of Γear its poits of discotiuity. Propositio 8.. For all atural umbers,, ad for real, Γ! as. Formally, we claim Γ! <. Proof. We shall prove this by iductio. We first show that Γ <. We recall the recurrece relatio, Γ Γ + 3
31 Γ Γ + Γ + Γ Γ γ We assume that Γ! <. To complete our iductio we must show Γ + + +! <. Recallig the recurrece relatio of Γ, we may write Γ + Γ. + Γ + + Γ +! + + +! Γ +! + Γ +! Γ + + Γ + Sice by our iductive hypothesis Γ! <, we must ow show Γ Γ <. + 3
32 Γ Γ + Γ + + Γ +! + Γ! Γ!! + Now sice we have Γ! <, ad sice each term idividually diverges to ±, it immediately follows that. So we may coclude that Γ! Γ!! + +! Ad so, sice this it is fiite as well, we have Γ + + +! <, completig our iductio. I poit of fact, our method of solutio ot oly implies that this it is fiite, but provides a method of calculatig it. If we deote the value of this it by C Γ + +, we +! may establish a recurrece relatio, C γ C C +! Armed with this proof, we are ready to aswer some of our origial questios about the Gamma fuctio. 32
33 Propositio 8.2. b Γd diverges if b is either zero or a egative b δ iteger, ad for ay arbitrary δ,. Proof. Before we begi, we shall mae a defiitio which shall be used throughout the rest of this paper. We defie, E Γ! We are ow well-equipped to tacle the proof of our propositio. δ b b! b b δ Γd δ δ Γ bd b b! + E b d δ clearly diverges, so all we eed to demostrate is the covergace of E δ bd ad we will have show that Γ bd δ diverges. Certaily for all, E b is fiite, ad by Propositio 8. E b is fiite. Now, sice E b is cotiuous, we may coclude that E b is bouded o the iterval δ,. Sice we are itegratig a fuctio over a fiite iterval upo which it is bouded we may coclude that, δ b b δ E b d < Γd ± It is trivial to show that a igh-idetical proof could be used to show that a+δ Γd diverges as well if a is either or a egative iteger a ad δ,. Propositio 8.3. Although +δ Γd is ot well-defied for all egative itegers, ad δ,, the Cauchy pricipal value defied δ by δ Γ + Γ d is well-defied ad fiite. Before we begi the proof it is importat to stress that we may oly tae a pricipal value of this itegral. Γd ± ad +δ δ Γd, so the value of this itegral is ot well-defied uless we cosider its pricipal value. 33
34 Proof. δ Γ + Γ d δ δ δ δ! +! + E + E d + E + E d E d As before, for all, ad for all, E is fiite, ad by Propositio 8. E is fiite. Now, sice E is cotiuous, we may coclude that E is bouded o the iterval δ, δ. Sice we are itegratig a fuctio over a fiite iterval upo which it is bouded we may coclude that, δ δ δ E d < Γ + Γ d < Propositio 8.4. b Γd is fiite if either of a, b are or a egative a iteger, provided we tae the value of the itegral over ay poit of discotiuity to be the Cauchy pricipal value thereof. Proof. This is a immediate cosequece of propositio 8.3. deote b a We may write b a Γd Γd a+ a+ + b a+ Γd Let By propositio 8.3 we ow that the pricipal value of each of these itegrals is fiite. So, we have a fiite sum of fiite terms, which therefore must be fiite. Propositio 8.5. b Γd diverges if at least oe of a, b is either or a a egative iteger. Proof. If oly oe of a, b is either or a egative iteger, the this proof immediately follows from propositio 8.4. Without loss of geerality assume that b is either or a egative iteger. Further let us assume that b a > sice if it is ot, the this is true by propositio
35 b a Γd b δ a Γd + b b δ Γd We ow that b δ Γd is fiite by propositio 8.4 ad we ow a that b Γd diverges by propositio 8.2. So, b Γd diverges as b δ a well. Now let us cosider the case where both a, b are oe of either or a egative iteger. We eed to show that we caot tae a pricipal value i this case. Let b a. b a Γd a+ a a+ 2 a Γd Γd + a+ 2 a+ 2 Γd + a+ a+ 2 Γd We ow a+ 2 Γd to be fiite by propositio 8.4. So we must a+ 2 ow show that we caot tae a pricipal value for a+ 2 Γd + a a+ Γd. a+ 2 a+ 2 a Γd + a+ a+ 2 Γd Γ ad + Γ ad Γ a + d Γ a + d Γ a + Γ a + d a a! + a+ a +! + E a + E a+ d Now, sice 2 E a d ad 2 E a+ d are both fiite, we must have that b Γd coverges if ad oly if 2 a + a+ a d does. a! a+! It ca be clearly see that this oly happes whe, so we are doe. Propositio 8.6. b Γd does ot coverge for ay value of b. Proof. We will prove this by cotradictio. We assume that there eists a real umber c such that 35
36 b Γd c This implies that ε > a real umber N such that > N b Γd c < ε. However, all we have to do is choose a iteger > N ad b Γd diverges, therefore this caot be true. 36
In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008
I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces
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