Enumerative & Asymptotic Combinatorics

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1 C50 Eumerative & Asymptotic Combiatorics Stirlig ad Lagrage Sprig 2003 This sectio of the otes cotais proofs of Stirlig s formula ad the Lagrage Iversio Formula. Stirlig s formula Theorem 1 (Stirlig s Formula)! ( ) 2π e Proof Cosider the graph of the fuctio y = logx betwee x = 1 ad x =, together with the piecewise liear fuctios show i Figure 1.. Figure 1: Stirlig s formula 1

2 Let f (x) = logx, let g(x) be the fuctio whose value is logm for m x < m + 1, ad let h(x) be the fuctio defied by the polygo with vertices (m,logm), for 1 m. Clearly g(x) dx = log2 + + log = log!. 1 The differece betwee the itegrals of g ad h is the sum of the areas of triagles with base 1 ad total height log; that is, 1 2 log. Some calculus 1 shows that the differece betwee the itegrals of f ad g teds to a fiite limit c as. Fially, a simple itegratio shows that We coclude that 1 f (x) dx = log + 1. so that log! = log log + (1 c) + o(1),! C+1/2 e. To idetify the costat C, we ca proceed as follows. Cosider the itegral Itegratio by parts shows that I = π/2 0 si xdx. I = 1 I 2, 1 Let F(x) = f (x) g(x). The covexity of logx shows that F(x) 0 for all x [m,m + 1]. For a upper boud we use the fact, a cosequece of Taylor s Theorem, that for x [m,m + 1]. The F(x) = logx logm log logx logm + x m m logm + 1 m ( )(x m) 1m ( m log ) 1 m 2m 2, where the last iequality comes from aother applicatio of Taylor s Theorem which yields log(1+x) x x 2 /2 for x [0,1]. Now (1/m 2 ) coverges, so the itegral is bouded. 2

3 ad hece O the other had, from which we get ad so I 2 = (2)!π (!) 2, I 2+1 = 22 (!) 2 (2 + 1)!. I 2+2 I 2+1 I 2, (2 + 1)π 4( + 1) 24 (!) 4 (2)!(2 + 1)! π 2, lim 2 4 (!) 4 (2)!(2 + 1)! = π 2. Puttig! C +1/2 /e i this result, we fid that so that C = 2π. C 2 ( e 4 lim ) (2+3/2) = π 2 2, Lagrage Iversio A formal power series over a field, with zero costat term ad o-zero term i x, has a iverse with respect to compositio. The associative, closure, ad idetity laws are obvious, ad the rule for fidig the iverse i characteristic zero is kow as Lagrage iversio. We work over R for coveiece. The theorem The basic fact ca be stated as follows. Propositio 2 Let f be a formal power series over R, with f (0) = 0 ad f (0) 0. The there is a uique formal power series g such that g( f (x)) = x; the coefficiet of y i g(y) is [ d 1 ( ) x ] / dx 1!. f (x) x=0 3

4 This ca be expressed i a more coveiet way for our purpose. Let φ(x) = x f (x). The the iverse fuctio g is give by the fuctioal equatio The Lagrage iversio has the form where g(y) = yφ(g(y)). b y g(y) = 1!, [ d 1 ] b = φ(x) dx 1. x=0 Example: Cayley s Theorem The expoetial geeratig fuctio for rooted trees satisfies the equatio T (x) = xexp(t (x)). With φ(x) = exp(x), we fid that the coefficiet of y /! i T (y) is [ d 1 ] exp(x) = 1. dx 1 Now there are ways to root a give tree; so the umber of trees is 2, provig Cayley s Theorem. Proof of the theorem The proof of Lagrage s iversio formula ivolves a cosiderable detour. The treatmet here follows the book by Goulde ad Jackso. Throughout this sectio, we assume that the coefficiets form a field of characteristic zero; for coveiece, we assume that the coefficiet rig is R. First, we exted the otio of formal power series to formal Lauret series, defied to be a series of the form f (x) = a x, m where m may be positive or egative. if the series is ot idetically zero, we may assume without loss of geerality that a m 0, i which case m is the valuatio of f, writte m = val( f ). 4 x=0

5 We defie additio, multiplicatio, compositio, differetiatio, etc., for formal Lauret series as for formal power series. I particular, f (g(x)) is defied for ay formal Lauret series f,g with val(g) > 0. (This is less trivial tha the aalogous result for formal power series. I particular, we eed to kow that g(x) m exists as a formal Lauret series for m > 0. It is eough to deal with the case m = 1, sice certaily g(x) m exists. If val(g) = r, the g(x) = x r g 1 (x), ad so g(x) 1 = x r g 1 (x) 1, ad we have see that g 1 (x) 1 exists as a formal power series, sice g 1 (0) is ivertible. We deote the derivative of the formal Lauret series f (x) by f (x). We also itroduce the followig otatio: [x ] f (x) deotes the coefficiet of x i the formal power series (or formal Lauret series) f (x). The case = 1 is especially importat, as we lear from complex aalysis. The value of [x 1 ] f (x) is called the residue of f (x), ad is also writte as Res f (x). Everythig below higes o the followig simple observatio, which is too trivial to eed a proof. Propositio 3 For ay formal Lauret series f (x), we have Res f (x) = 0. Now the followig result describes the residue of the compositio of two formal Lauret series. Theorem 4 (Residue Compositio Theorem) Let f (x), g(x) be formal Lauret series with val(g) = r > 0. The Res( f (g(x))g (x)) = r Res( f (x)). Proof It is eough to cosider the case where f (x) = x, sice Res is a liear fuctio. Suppose that 1, so that the right-had side is zero. The Res(g (x)g (x)) = 1 ( ) d + 1 Res dx g+1 (x) = 0. So cosider the case where = 1. Let g(x) = ax r h(x), where a 0 ad h(0) = 1. The g (x)g(x) 1 = d dx logg(x) = d (loga + r logx + logh(x)) dx = r x + d dx logh(x), so Resg (x)g(x) 1 = r = r Resx 1. 5

6 Note that we have cheated slightly i the first lie of this argumet: logg(x) may ot exist as a formal Lauret series; but it is the case that for if the equatio f (x) = g(x)h(x) holds for formal Lauret series, the f (x)/ f (x) = g (x)/g(x) + h (x)/h(x), ad i the precedig argumet it is the case that logh(x) exists (this is obtaied by substitutig y = h(x) 1 i log(1 + y)) ad its derivative is h (x)/h(x). Cosider this poit carefully; a error here would lead to the icorrect coclusio that Res(g (x)/g(x)) = 0. From the Residue Compositio Theorem, we ca prove a more geeral versio of Lagrage Iversio. Theorem 5 (Lagrage Iversio) Let φ be a formal power series with val(φ) = 1, The the equatio g(x) = xφ(g(x) has a uique formal power solutio g(x). Moreover, for ay Lauret series f, we have { 1 [x [x ] f (g(x)) = 1 ]( f (x)φ(x) ) if val( f ) ad 0, f (0) + Res( f (x)log(φ(0) 1 φ(x)) if = 0. Proof Let Φ(x) = x/φ(x), so that Φ(g(x)) = x ad val(φ(x)) = 1. The g is the iverse fuctio of Φ. We have [x ] f (g(x)) = Resx 1 f (g(x)) = ResΦ(y) 1 Φ (y) f (y), where we have made the swubstitutio x = Φ(y) (so that y = g(x)) ad used the Residue Compositio Theorem. For 0, we have [x ] f (g(x)) = 1 [y 1 ] f (y) ( Φ(y) ) = 1 [y 1 ] f (y)φ(y) = 1 [y 1 ] f (y)φ(y). Here, i the secod lie, we have used the fact that Res( f (x)g(x)) = Res( f (x)g (x)), 6

7 a cosequece of the fact that Res( f (x)g(x)) = 0; i the third lie we use the fact that Φ(x) = x/φ(x). For = 0, we have [x 0 ] f (g(x)) = [y 0 ] f (y) [y 1 ] f (y)φ (y)φ(y) 1 = f (0) + Res( f (y)log(φ(y)φ 1 (0)), usig the same priciple as before ad the fact that (log(φ(y)φ 1 (0))) = φ (y)φ(y) 1. Takig f (x) = x i this result gives the form of Lagrage Iversio quoted earlier. We proceed to a applicatio, also take from Goulde ad Jackso, of the Residue Compositio Theorem. Example: a biomial idetity that k=0 k=0 We use the Residue Compositio Theorem to prove )( ) ( ) j + k 2 j =. 2 2 ( k + 1 We begi with the sum of the odd terms i (1 + x) 2+1 : ( ) x 2k = 1 ( (1 + x) 2+1 (1 x) 2+1). 2k + 1 2x Call the right-had side of this equatio f (x). Now, if S is the sum that we wat to evaluate, the S = [y 2 ](1 + y) j ( ) (1 + y) k k=0 2k + 1 = Resy (2+1) (1 + y) j f ((1 + y) 1/2 ). Now we do the followig rather strage thig: make the substitutio y = z 2 (z 2 2). The val(y(z)) = 2, ad (1 + y) 1/2 = 1 z 2. So the Residue Compositio Theorem gives ( S = Res(z 2 1) 2 j 1 (z 2 2) ) z 4+2 z = Res(z 2 1) 2 j z (4+1) = [z 4 ](z 2 1) 2 j ( ) 2 j =, 2 as required. (I the secod lie we have used the fact that (z 2 2) (2+1) is a formal power series ad so its residue is zero.) 7