Peter Bala, Nov

Transcription

1 Fractional iteration of a series inversion operator Peter Bala, Nov We consider an operator on formal power series, closely related to the series reversion operator, and show how to dene comple iterates α of. The power series epansion of α (f()) is found. Eamples include the generalized binomial and generalized eponential series of Lambert. We begin by recalling some facts about formal power series [4]. Let g() be a formal power series with comple coecients of the form g() = + g g There is a unique formal power series h(), called the compositional inverse of g(), such that h(g()) = g(h()) =. The process of nding the coecients of the series h() from those of g() is called series reversion or series inversion. We denote h() by Rev(g()), where Rev is the series reversion operator. Let f() = 1 + f 1 + f f (1) be a formal power series with constant term 1. We can raise a power series of this type to an arbitrary comple power α. ndeed, if we write f in the form f() = 1 + S(), with S() = n 1 f n n, then we have, by virtue of the generalized binomial theorem, f α () = 1 + n 1 ( ) α S n (), n Collecting together the powers of we obtain f α () = n 0 p n (α) n, a well-dened formal power series, where p n (t) is a sequence of convolution polynomials [3]. 1

2 Denition 1. We dene a series inversion operator, acting on power series f() of the form (1), by (f()) = 1 ( ) Rev. (2) f() The Lagrange inversion formula gives the coecents in the power series epansion of (f()) as [ n ] (f()) = 1 n + 1 []n f() n+1. The power series (f()) has the same form as f() with constant term 1. Thia allows us to iterate the inversion operator. Eample 1. Let f() = 1/(1 ) = , the generating function for the all 1's sequence, A Then (f()) = 1 Rev ((1 )) = is the generating function for the Catalan numbers, A One (of the many) combinatorial interpretations of the Catalan numbers is as the number of full binary trees with n internal vertices. By a slight abuse of notation we display this relationship between the two sequences diagramatically as A A teration of the operator produces the following diagram of OES sequences. A A A A A A Combinatorially, these sequences enumerate k-ary trees for k = 2, 3, 4,... The opertor is invertible with inverse denoted 1. t follows from (2) that 1 (f()) = which we can write in the more suggestive form Rev(f()), 1 (f()) = ( ( )) 1 1 Rev f 1, (3) () 2

3 where the eponent 1 occurring on the right-hand side means reciprocation of a function. Denition 2. Let α C. Let f() be a power series of the form (1). We dene α, the fractional inversion operator of order α, by putting 0 (f()) = f() and setting α (f()) = ((f α ())) 1 α = ( 1 Rev ( f α () )) 1 α for α 0. (4) Clearly, this denition agrees with (3) when α = 1, and we also have 1 =. We have used the same misnomer here as in fractional calculus by referring to the operator α as a fractional inversion operator, even when α is not a rational number. n order for the denition of fractional inversion to be useful we would like the operator n to equal the n-fold iterate of when n is a positive integer, and equal the n-fold iterate of 1 when n is a negative integer. This is true and is an immediate consequence of the following result. Theorem 1. Let α, β be comple numbers. Then α β = α+β. Proof. The result is obviously true when either α = 0 or β = 0 and is straightforward to verify when α + β = 0. From now on we assume that α, β and α + β are all non-zero. Let f() = 1 + f 1 + f f be a formal power series and dene g() = 1 ( ) Rev f β. (5) () From the denition (4) of the fractional inversion operator we have β (f()) = ( ( )) 1 1 Rev β f β () = g 1 β (). 3

4 Hence f we put then (6) reads α ( β (f() ) = α ( g 1 β () ) = ( ( )) 1 1 Rev α. (6) g α β () h() = 1 ( ) Rev g α β () (7) α ( β (f() ) = h 1 α (). (8) Again, by the denition of the fractional inversion operator, we have f we put then (9) becomes α+β (f()) = ( ( 1 Rev f α+β () H() = 1 ( ) Rev f α+β () )) 1 α+β. (9) (10) α+β (f()) = H() 1 α+β. (11) Comparing (8) and (11), the proof that α β = α+β will be established if we can show h() 1 α = H() 1 α+β. (12) Now from (5), the power series g() is the inverse of /f β (), which leads to the functional equation ( ) g f β = f β () () or equivalently g 1 β ( ) f β () = f(). (13) Similarly, from (7) we nd that the power series h() satises the functional equation ( ) h 1 α = g 1 g α β () (14) β () 4

5 while from (10), the power series H() satises the functional equation ( ) H 1 α+β f α+β = f(). (15) () n (14), replace with f β () h 1 α and then use (13) to nd that h satises ( ) f α+β () = f(). (16) Thus from (15) and (16) we have H 1 α+β () = f (Rev ( )) f α+β () proving (12) and completing the proof of the Theorem. = h 1 α (), Corollary 1. Let α, β be comple numbers. The operators α and β commute α β = β α. Corollary 2. For a positive integer n, the operator n equals the n-fold iterate of and the operator n equals the n-fold iterate of 1. n =... (n factors) n = (n factors) Eercise 1. Let R denote the reciprocation operator: R : f() 1/f(). Show R α and α R are both idempotent operators. Eercise 2. Show Theorem 1 has the following generalization: Let f() = 1 + f 1 + f f be a formal power series. Let α, β, r be comple numbers. Then α ( ( β (f) ) r ) = ( rα+β (f) ) r. Corollary 2 reveals the surprising property of the inversion operator that its n-th iterate can be calculated from a single application of. The net result gives an alternative way of computing n (f()) that also requires only a single application of. 5

6 Theorem 2. Let f() = 1 + f 1 + f f be a formal power series. Let n be a positive integer. Then (i) (ii) n (f()) = [(f( n ))] 1 n n (f()) = [ 1 (f( n )) ] 1 n Proof. We prove (i), the proof of (ii) being eactly similar. The claim is that n (f()) is obtained from (f( n )) (a power series in n ) on replacing with 1 n. ( ) Let G() = n (f()) and H() = (f( n )). We show G() = H 1 n. By denition (4) of the fractional inversion operator we have ( ( )) 1 1 G() = Rev n. (17) f n () t follows that G n () is the series reversion of /f n (), and consequently G n () f n (G n ()) Thus G() satises the functional equation =. G() = f (G n ()). (18) Moreover, this functional equation determines G() in terms of f() because, starting from (18), we can reverse the above steps to nd G() is given by (17). t follows from H() = (f( n )) that H() is the series reversion of /f( n ), and hence giving Thus H() f ( n H n ()) = H() = f ( n H n ()). ( ) H 1 n = f ( ( )) H n 1 n. (19) Comparing (18) and (19) we see that G() and H( 1 n ) satisfy the same functional equation, which, as we noted above, has a unique solution in terms of f(). Consequently G() = H( 1 n ). 6

7 We turn our attention to the series epansion of t (f()) and its powers. We shall need the following version of the Lagrange-Bürmann formula for formal power series (see [1, Theorem 1.2.4] or [5]): f f() = 1 + f 1 + f f and H() = h 0 + h 1 + h h ( are formal power series and G() = Rev [ n ]H(G()) = 1 n f() ) then [ n 1 ] H ()f() n, for n, k > 0. Theorem 3. Let f() = 1 + f 1 + f f be a formal power series and let f() t = n 0 p n (t) n, where p n (t) is a family of convolution polynomials. Then for r C, (i) ( t (f())) r = n 0 r nt + r p n(nt + r) n (The denominator nt + r cancels with the corresponding factor in the numerator polynomial p n (nt + r), so no problem arises if nt + r is zero.) (ii) log ( t (f()) ) = n 1 1 nt p n(nt) n. Proof. (i) The result is clearly true if t = 0. Assume now t is nonzero. By denition ( ( )) 1 1 t (f()) = Rev t. f() t Let g() = ( t (f())) t so that g() = Rev ( ) f() t. Now apply the Lagrange-Bürmann formula with H() = k, k > 0 to obtain [ n ] (g()) k = 1 n [n 1 ]k k 1 f() nt 7

8 or equivalently Replace n with n + k to get [ n k ] (g()) k = k n [n k ]f() nt. [ n ] (g()) k = Thus we have the series epansion = k n + k [n ]f() (n+k)t k n + k p n((n + k)t). g() k = ( t (f())) kt = n 0 k n + k p n((n + k)t) n. (20) Equation (20) has been derived on the assumption that k is a positive integer but also holds for arbitrary comple k, since when k is a comple number the coecients of g() k are polynomials in k that equal the polynomials k n+k p n((n + k)t) for innitely many values of k and so must be identically equal polynomials. Let r C and set k = r/t in (20) to nd n particular, (ii) t follows from (21) that ( t (f())) r = n 0 t (f()) = n 0 r nt + r p n(nt + r) n. (21) 1 nt + 1 p n(nt + 1) n log ( t (f()) ) t (f()) r 1 = lim r 0 r = 1 nt p n(nt) n. n 1 One consequence of Theorem 3 (i) is that if p n (t) is a family of convolution polynomials then q n (r) := r nt+r p n(nt + r), regarded as polynomials in the variable r, is another family of convolution polynomials [2, pp ]. 8

9 Theorem 3 coupled with Eercise 2 allows us to write down the eect of the fractional inversion operator s applied to powers of t (f()): s (( t (f()) ) r) = ( rs+t (f()) ) r = n 0 r n(rs + t) + r p n(n(rs + t) + r) n. We conclude by looking at two classical families of series which may be dened using the fractional inversion operator. Eample 2. Take f() = 1 +. Let B t () denote the power series t (f()) = t (1 + ). Now f() t = ( ) t n, so in this case p n (t) = ( t n n) is a n 0 falling factorial polynomial. Theorem 3 gives the epansion B t () = ( ) 1 nt + 1 n (22) nt + 1 n n 0 with powers given by B t () r = ( ) r nt + r n. (23) nt + r n n 0 Also log (B t ()) = n 1 ( ) 1 nt n. nt n This latter series (with t replaced by 1 + t) is the eponential generating function for A The series B t () are called generalized binomial series and have a long history dating back to Lambert. See [2, Section 5.4 and Section 7.5] and A For an alternative approach to proving (23), avoiding the Lagrange-Bürmann formula and using only basic calculus, see 'The power series for the inverse function of y(1-y)^t' by N. D. Elkies, available online at harvard.edu/~elkies/misc/catalan.pdf. Eample 3. Takef() = e. Let E t () denote the power series t (f()) = t (e ). We have f() t = e t = t n n n!, so in this case p n(t) = t n n 0 is a monomial. Theorem 3 gives the epansion E t () = n 1 n (nt + 1) n!. (24) n 0 n the particular case t = 0 we have E 0 () = e. Graham et al. [2, Section 5.4] call E t () a generalized eponential series. See A

10 The powers of the generalized eponential series may also be written down E t () r = n 0 n 1 n r(nt + r) n!. (25) Furthermore, log (E t ()) = n 1 n 1 n (nt) n!. (26) n particular, log (E 1 ()) = n 1 n 1 n n n! (27) is Euler's tree function. See A REFERENCES [1]. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Dover Publications nc., Mineola, NY, Reprint of the 1983 original. [2] R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA. Second ed [3] D. E. Knuth, Convolution polynomials, Mathematica J. 2.1 (1992), no. 4, [4] Wikipedia, Formal Power Series [5] Wikipedia, Lagrange inversion theorem 10