Introduction f x xG f x - Brandeis Universitypeople.brandeis.edu/~gessel/homepage/papers/Lagrange.pdf · Abstract. We give a survey of the Lagrange inversion formula, including di

LAGRANGE INVERSION

IRA M GESSELlowast

Abstract We give a survey of the Lagrange inversion formula including different versionsand proofs with applications to combinatorial and formal power series identities

1 Introduction

The Lagrange inversion formula is one of the fundamental formulas of combinatorics Inits simplest form it gives a formula for the power series coefficients of the solution f(x) ofthe function equation f(x) = xG(f(x)) in terms of coefficients of powers of G Functionalequations of this form often arise in combinatorics and our interest is in these applicationsrather than in other areas of mathematics

There are many generalizations of Lagrange inversion multivariable forms [28] q-analogues[22 23 25 71] noncommutative versions [6 7 23 56] and others [29 43 45] In this paper wediscuss only ordinary one-variable Lagrange inversion but in greater detail than elsewherein the literature

In section 2 we give a thorough discussion of some of the many different forms of Lagrangeinversion prove that they are equivalent to each other and work through some simpleexamples involving Catalan and ballot numbers We address a number of subtle issues thatare overlooked in most accounts of Lagrange inversion (and which some readers may wantto skip) In sections 3 we describe applications of Lagrange inversion to identities involvingbinomial coefficients Catalan numbers and their generalizations In section 4 we giveseveral proofs of Lagrange inversion some of which are combinatorial

A number of exercises giving additional results are includedAn excellent introduction to Lagrange inversion can be found in Chapter 5 of Stanleyrsquos

Enumerative Combinatorics Volume 2 Other expository accounts of Lagrange inversioncan be found in Hofbauer [35] Bergeron Labelle and Leroux [5 Chapter 3] Sokal [68] andMerlini Sprugnoli and Verri [51]

11 Formal power series Although Lagrange inversion is often presented as a theorem ofanalysis (see eg Whittaker and Watson [76 pp 132ndash133]) we will work only with formalpower series and formal Laurent series A good account of formal power series can be foundin Niven [55] we sketch here some of the basic facts Given a coefficient ring C whichfor us will always be an integral domain containing the rational numbers the ring C[[x]] offormal power series in the variable x with coefficients in C is the set of all ldquoformal sumsrdquosuminfin

n=0 cnxn where cn isin C with termwise addition and multiplication defined as one would

expect using distributivitysuminfin

n=0 anxn middotsuminfin

n=0 bnxn =

suminfinn=0 cnx

n where cn =sumn

i=0 aibnminusiDifferentiation of formal power series is also defined termwise A series

suminfinn=0 cnx

n has amultiplicative inverse if and only if c0 is invertible in C We may also consider the ring of

Date June 6 2016

formal Laurent series C((x)) whose elements are formal sumssuminfin

n=n0cnx

n for some integern0 ie formal sums

suminfinn=minusinfin cnx

n in which only finitely many negative powers of x havenonzero coefficients Henceforth will omit the word ldquoformalrdquo and speak of power series andLaurent series

We can iterate the power series and Laurent series ring constructions obtaining for ex-ample the ring C((x))[[y]] of power series in y whose coefficients are Laurent series in xIn any (possibly iterated) power series or Laurent series ring we will say that a set fα ofseries is summable if for any monomial m in the variables the coefficient of m is nonzero inonly finitely many fα In this case the sum

sumα f is well-defined and we will say that

sumα fα

is summable If we writesum

α fα as an iterated sum then the order of summation is irrele-vant If f(x) =

sumn cnx

n is a Laurent series in C((x)) and u isin C where C may be a powerseries or Laurent series ring then we say that that the substitution of u for x is admissibleif f(u) =

sumn cnu

n is summable and similarly for multivariable substitutions Admissiblesubstitutions are homomorphisms If u is a power series or Laurent series g(x) then f(g(x))if summable is called the composition of f and g If f(x) = c1x + c2x

2 + middot middot middot where c1 isinvertible in C then there is a unique power series g(x) = cminus11 x+ middot middot middot such that f(g(x)) = xthis implies that g(f(x)) = x We call g(x) the compositional inverse of f(x) and writeg(x) = f(x)〈minus1〉 For simplicity we will always assume that if f(x) = c1x+ c2x

2 + middot middot middot wherec1 6= 0 then c1 is invertible (Since C is an integral domain we can always adjoin cminus11 to Cif necessary)

The iterated power series rings C[[x]][[y]] and C[[y]][[x]] are essentially the same in thatboth consist of all sums

summnge0 cmnx

myn We may therefore write C[[x y]] for either of

these rings However the iterated Laurent series C((x))[[y]] C((y))[[x]] and C[[y]]((x))are all different in the first we have (x minus y)minus1 =

suminfinn=0 y

nxn+1 in the second we have(xminus y)minus1 = minus

suminfinn=0 x

nyn+1 and in the third xminus y is not invertibleIt is sometimes convenient to work with power series in infinitely many variables for

example we may consider the power seriessuminfin

n=0 rntn where the rn are independent inde-

terminates Although we donrsquot give a formal definition of these series they behave in ourapplications exactly as expected

We use the notation [xn] f(x) to denote the coefficient of xn in the Laurent series f(x)An important fact about the coefficient operator that we will use often without commentis that [xn]xkf(x) = [xnminusk] f(x)

The binomial coefficient(ak

)is defined to be a(aminus1) middot middot middot (aminusk+ 1)k if k is a nonnegative

integer and 0 otherwise Thus the binomial theorem (1 + x)a =suminfin

)xk holds for all a

2 The Lagrange inversion formula

21 Forms of Lagrange inversion We will give several proofs of the Lagrange inversionformula in section 4 Here we state several different forms of Lagrange inversion and showthat they are equivalent

Theorem 211 Let R(t) be a power series not involving x Then there is a unique powerseries f = f(x) such that f(x) = xR(f(x)) and for any Laurent series φ(t) and ψ(t) notinvolving x and any integer n we have

[xn]φ(f) =1

n[tnminus1]φprime(t)R(t)n where n 6= 0 (211)

[xn]φ(f) = [tn](1minus tRprime(t)R(t)

)φ(t)R(t)n (212)

φ(f) =sumn

xn [tn](1minus xRprime(t))φ(t)R(t)n (213)

[xn]ψ(f)

1minus xRprime(f)= [tn]ψ(t)R(t)n (214)

[xn]ψ(f)

1minus fRprime(f)R(f)= [tn]ψ(t)R(t)n (215)

We show here these formulas are equivalent in the sense that any one of them is easilyderivable from any other proofs of these formulas are given in section 4 It is clear that(214) and (215) are equivalent since x = fR(f) Taking ψ(t) = (1 minus tRprime(t)R(t))φ(t)shows that (212) and (215) are equivalent

To derive (213) from (214) we rewrite (214) as

1minus xRprime(f)=sumn

xn [tn]ψ(t)R(t)n (216)

Until now we have assumed that φ(t) and ψ(t) do not involve x We leave it to the readerto see that in (216) this assumption can be removed Then (213) follows from (216) bysetting ψ(t) = (1minus xRprime(t))φ(t) and similarly (216) follows from (213)

Although we allow R to be an arbitrary power series in Theorem 211 if R has constantterm 0 then f(x) = 0 so we may assume now that R has a nonzero constant term andthus f and R(f) are nonzero Then the equation f(x) = xR(f(x)) may be rewritten asfR(f) = x So if we set g(t) = tR(t) then we have g(f) = x and thus g = f 〈minus1〉 Itis sometimes convenient to rewrite the formulas of Theorem 211 using g rather than RSince 1 minus tRprime(t)R(t) = tgprime(t)g(t) formula (212) takes on a slightly simpler form (whichwill be useful later on) when expressed in terms of g rather than R

[xn]φ(f) = [tnminus1]φ(t)gprime(t)

)n= [tminus1]

φ(t)gprime(t)

g(t)n+1 (217)

For future use we note also the corresponding form for (215)

[xnminus1]ψ(f)

fgprime(f)= [tn]ψ(t)

)n= [t0]

g(t)n (218)

To show that (211) and (212) are equivalent using (217) in place of (212) we showthat

[tminus1]φprime(t)

g(t)n= n [tminus1]

φ(t)gprime(t)

g(t)n+1(219)

Butφprime(t)

g(t)nminus nφ(t)gprime(t)

g(t)n+1=d

and the coefficient of tminus1 in the derivative of any Laurent series is 0 so (219) follows Thisshows that (211) and (212) are equivalent if n 6= 0 If φ is a power series then thecoefficient of x0 in φ(f) is simply the constant term in φ but if φ is a more general Laurentseries then the constant term in φ(f) is not so obvious and cannot be determined by (211)In equation (228) we will give a formula for the constant term in φ(f) for all φ

The case φ(t) = tk of (211) with R(t) = tg(t) may be written

[xn] fk =k

n[tnminusk]

n[tminusk] g(t)minusn (2110)

In other words if g = f 〈minus1〉 and for all integers k fk =sum

n ankxn and gk =

sumn bnkx

n then

ank =k

nbminuskminusn (2111)

for n 6= 0 Equation (2111) is known as the Schur-Jabotinsky theorem (See Schur [66equation (10)] and Jabotinsky [37 Theorem II])

22 Polynomials We now give a slightly more general form of Lagrange inversion basedon the fact that if two polynomials agree at infinitely many values than they are identicallyequal This will imply that to prove our Lagrange formulas for all n it is sufficient to provethem in the case in which n is a positive integer (Some proofs require this restriction)

By linearity the formulas of Section 21 are implied by the special cases in which φ(t)and ψ(t) are of the form tk for some integer k and these special cases (especially k = 0 andk = 1) are particularly important These special cases of (211) (214) and (215) areespecially useful and may be written

[xn] fk =k

n[tnminusk]R(t)n where n 6= 0 (221)

[xn]fk

1minus xRprime(f)= [tnminusk]R(t)n (222)

[xn]fk

1minus fRprime(f)R(f)= [tnminusk]R(t)n (223)

In these formulas let us assume that R(t) has constant term 1 (It is not hard to modifyour approach to take care of the more general situation in which the constant term of R(t)is invertible) Then the coefficient of x in f(x) is 1 so f(x)x has constant term 1 If we setn = m+ k in (221) (222) and (223) then the results may be written

[xm](fx)k =k

m+ k[tm]R(t)m+k where m+ k 6= 0 (224)

[xm](fx)k

1minus xRprime(f)= [tm]R(t)m+k (225)

[xm](fx)k

1minus fRprime(f)R(f)= [tm]R(t)m+k (226)

It is easy to see that in each of these equations for fixed m both sides are polynomials in kThus if these equalities hold whenever k is a positive integer then they hold as identities ofpolynomials in k Moreover although (224) is invalid for k = minusm if m gt 0 we may takethe limit as k rarr minusm with lrsquoHopitalrsquos rule to obtain

[xm](fx)minusm = [x0] fminusm = minusm [tm] logR (227)

(Note that (227) does not hold for m = 0) By linearity (227) yields a supplement to(211) that takes care of the case n = 0

[x0]φ(f) = [t0]φ(t) + [tminus1]φprime(t) logR (228)

We can also differentiate (224) with respect to k and then set k = 0 to obtain

[xm] log(fx) =1

m[tm]R(t)m for m 6= 0 (229)

Returning to (221)ndash(223) we see that if they hold when n and k are positive integersthen they also hold when n and k are arbitrary integers (Note that if n lt k then everythingis zero)

23 A simple example Catalan numbers The Catalan numbers Cn may defined bythe equation

c(x) = 1 + xc(x)2 (231)

for their generating function c(x) =suminfin

n=0Cnxn The quadratic equation (231) has two

solutions(1plusmnradic

1minus 4x)(2x) but only the minus sign gives a power series so

c(x) =1minusradic

1minus 4x

Unfortunately (231) is not of the form f(x) = xR(f(x)) so we cannot apply directly anyof the versions of Lagrange inversion that we have seen so far

One way to apply Lagrange inversion is to set f(x) = c(x)minus 1 so that f = x(1 + f)2 Wemay then apply Theorem 211 to the case R(t) = (1 + t)2 The equation f = x(1 + f)2 hasthe solution

f(x) = c(x)minus 1 = xc(x)2 =1minusradic

1minus 4x

2xminus 1

Then (211) with φ(t) = (1 + t)k gives for n gt 0

[xn] c(x)k = [xn](1 + f)k =1

n[tnminus1] k(1 + t)kminus1(1 + t)2n

n[tnminus1]

(2n+ k minus 1

nminus 1

Thus since the constant term in c(x)k is 1 we have

c(x)k = 1 +infinsumn=1

(2n+ k minus 1

nminus 1

The sum may also be written

c(x)k =infinsumn=0

(2n+ k

infinsumn=0

(2n+ k minus 1

)xn (232)

These formulas are valid for all k except where n = minusk2 in the first sum in (232) orn = minusk in the second sum in (232) These coefficients are called ballot numbers and fork = 1 (232) gives the usual formula for the Catalan numbers Cn = 1

Equation (212) with φ(t) = (1 + t)k gives a formula for the ballot number as a differenceof two binomial coefficients

[xn] c(x)k = [tn]1minus t1 + t

(1 + t)k(1 + t)2n

= [tn](1minus t)(1 + t)2n+kminus1

(2n+ k minus 1

)minus(

2n+ k minus 1

nminus 1

and equation (213) with φ(t) = (1 + t)k gives another such formula

[xn] c(x)k =sumn

xn [tn]((1 + t)2n+k minus 2x(1 + t)2n+k+1

)=sumn

)minus 2

(2n+ k minus 1

nminus 1

Finally (229) gives

[xm] log(fx) = [xm] 2 log c(x) =1

m[tm](1 + t)2m =

log c(x) =infinsumm=1

Since R(t) = (1 + t)2 we have Rprime(t) = 2(1 + t) so 1 minus xRprime(f) = 1 minus 2xc(x) =radic

1minus 4xThus (214) with ψ(t) = (1 + t)k gives

[xn]c(x)kradic1minus 4x

= [tn](1 + t)k(1 + t)2n =

(2n+ k

c(x)kradic1minus 4x

=infinsumn=0

(2n+ k

)xn (233)

Equating coefficients of xn in c(x)kc(x)l = c(x)k+l and using (232) gives the convolutionidentity sum

(2i+ k

)middot l

2j + l

(2j + l

2n+ k + l

(2n+ k + l

Similarly using (232) and (233) we getsumi+j=n

(2i+ k

)(2j + l

(2n+ k + l

These convolution identities are special cases of identities discussed in section 33

Exercise 231 Derive these formulas for c(x) in other ways by applying Lagrange inversionto the equations f = x(1minus f) and f = x(1 + f 2)

24 A generalization There is another way to apply Lagrange inversion to the equationc(x) = 1 + xc(x)2 that while very simple has far-reaching consequences Consider theequation

F = z(1 + xF 2) (241)

where we think of F as a power series in z with coefficients that are polynomials in x Wemay apply (221) to (241) to get

[zn]F k =k

n[tnminusk](1 + xt2)n

The right side is 0 unless nminus k is even and for n = 2m+ k we have

[z2m+k]F k =k

2m+ k[t2m](1 + xt2)2m+k =

(2m+ k

Thus we have (if k is an integer but not a negative even integer)

F k =infinsumm=0

(2m+ k

)xmz2m+k (242)

Now let c be the result of setting z = 1 in F (an admissible substitution) so by (242) wehave

ck =infinsumm=0

(2m+ k

and by (241) we have c = 1 + xc2 Moreover as we have seen before c = 1 + xc2 has aunique power series solution

The same idea works much more generally but we must take care that the substitution isadmissible For example we can solve f = x(1 + f) by Lagrange inversion but we cannotset x = 1 in the solution

The case in which the coefficients of R(t) are indeterminates is easy to deal with

Theorem 241 Suppose that R(t) =suminfin

n=0 rntn where the rn are indeterminates Then

there is a unique power series f satisfying f = R(f) If φ(t) is a power series then

φ(f) = φ(0) +infinsumn=1

n[tnminus1]φprime(t)R(t)n (243)

and for any Laurent series φ(t) and ψ(t) we have

φ(f) = [t0]φ(t) + [tminus1]φprime(t) log(Rr0) +sumn6=0

φ(f) =sumn

[tn](1minus tRprime(t)R(t)

)φ(t)R(t)n =

[tn](1minusRprime(t))φ(t)R(t)n (245)

1minusRprime(f)=

1minus fRprime(f)R(f)=sumn

[tn]ψ(t)R(t)n (246)

In (245) and (246) the sum is over all integers n

Proof These formulas follow from equations (211) to (215) on making the admissiblesubstitution x = 1 where for (244) we have included the correction term given by (228)modified to take into account that the constant term of R(t) is r0 rather than 1 Uniquenessfollows by equating coefficients of the monomials ri00 r

i11 middot middot middot on both sides of f = R(f) which

gives a recurrence that determines them uniquely

We would like to relax the requirement in Theorem 241 that the rn be indeterminatesTo do this we can take any of the formulas of Theorem 241 and apply any admissiblesubstitution for the rn For example the following result while not the most general possibleis sometimes useful

n=0 rntn where the coefficients lie in some power

series ring C[[u1 u2 ]] and that each rn with n gt 0 is divisible by some ui Then there isa unique power series f satisfying f = R(f) and formulas (243) to (246) hold

We note that more generally any admissible substitution will yield a solution of f = R(f)to which these formulas hold but uniqueness is not guaranteed For example the equationf = x+ yf 2 has the unique power series solution

f =1minusradic

1minus 4xy

2y=infinsumn=0

)xn+1yn

The admissible substitution y = 1 gives f = 12(1minusradic

1minus 4x) as a power series solution of theequation f = x + f 2 However the equation f = x + f 2 has another power series solutionf = 1

radic1minus 4x)

Exercise 243 The equation f = x+f 2 has two power series solutions f = 12(1plusmnradic

1minus 4x)However according to Theorem 211 the equivalent equation f = x(1 minus f) has only onepower series solution Explain the discrepancy

25 Explicit formulas for the coefficients It is sometimes useful to have an explicitformula for the coefficients of fk where f = xR(f) With R(t) =

suminfinn=0 rnt

n if we expandR(t)n by the multinomial theorem then (221) gives

fk =sum

n0+n1+middotmiddotmiddot=nn1+2n2+3n3+middotmiddotmiddot=nminusk

k(nminus 1)

n0n1 rn00 r

n11 middot middot middotxn (251)

We might also want to express the coefficients of fk in terms of the coefficients of g = f 〈minus1〉Suppose that g(x) = xminus g2x2 minus g3x3 minus middot middot middot where the minus signs and the assumption thatthe coefficient of x in g is 1 make our formula simpler with no real loss of generality Then(221) gives

[xm]fk =k

m[tmminusk]

1minus g2tminus g3t2 minus middot middot middot

Expanding by the binomial theorem and simplifying gives

fk =sum

n2+n3+middotmiddotmiddot=nminusmn2+2n3+middotmiddotmiddot=mminusk

k(nminus 1)

mn2n3 middot middot middotgn22 g

n33 middot middot middotxm

where the sum is over all m n and n2 n3 satisfying the two subscripted equalities Ifwe replace m with n0 then we may write the formula as

fk =sum

n0+n2+n3+middotmiddotmiddot=n2n2+3n3+middotmiddotmiddot=nminusk

k(nminus 1)

n0n2n3 middot middot middotxn0gn2

2 gn33 middot middot middot (252)

and we see that the coefficients here are exactly the same as the coefficients in (251) (withn1 = 0)

Exercise 251 Explain the connection between (251) and (252) without using Lagrangeinversion

26 Derivative formulas Lagrange inversion especially in its analytic formulations isoften stated in terms of derivatives We give here several derivative forms of Lagrangeinversion

Theorem 261 Let G(t) =suminfin

n=0 gntn where the gi are indeterminates Then there is a

unique power series f satisfying

f = x+G(f)

and for any power series φ(t) and ψ(t) we have

φ(f) =infinsumm=0

(φ(x)

(1minusGprime(x)

)G(x)m

) (261)

φ(f) = φ(x) +infinsumm=1

dmminus1

dxmminus1

(φprime(x)

) (262)

andψ(f)

1minusGprime(f)=

infinsumm=0

(ψ(x)

) (263)

Proof We first prove (263) By (246) we have

1minusGprime(f)=infinsumn=0

[tn]ψ(t)(x+G(t))n =infinsumm=0

infinsumn=0

)xnminusmψ(t)G(t)m

So to prove (263) it suffices to prove that

(ψ(x)

)=infinsumn=0

)xnminusmψ(t)G(t)m

We show more generally that for any power series α(t) we have

dxmα(x)

m=infinsumn=0

)xnminusmα(t) (264)

But by linearity it is sufficient to prove (264) for the case α(t) = tj where both sides areequal to

)xjminusm

Next (261) follows from (263) by taking ψ(t) = (1minusGprime(t))φ(t)

Finally we derive (262) from (261) Writing D for ddx φ for φ(x) and G for G(x) wehave

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

As before applying an admissible substitution allows more general coefficients to be usedAs an application of these formulas let us take G(x) = zH(x) and consider the formula

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

Applying (262) and (263) to the left side and (263) to the right and then equatingcoefficients of zn gives the convolution identity

nsumm=0

)dmminus1

dxmminus1(φprime(x)H(x)m)

dnminusm

dxnminusm(ψ(x)H(x)nminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

and similarly expanding φ(f)ψ(f) in two ways using (262) givesnsum

)dmminus1

dnminusmminus1

dxnminusmminus1(ψprime(x)H(x)nminusm

)=dnminus1

dxnminus1((φ(x)ψ(x))primeH(x)n

(266)Here dmminus1φprime(x)dx for m = 0 is to be interpreted as φ(x) Formulas (265) and (266)were found by Cauchy [10] a formula equivalent to (265) had been found earlier by Pfaff[58] A detailed historical discussion of these identities and generalizations has been givenby Johnson [40] see also Chu [12 14] and Abel [2] Our approach to these identities hasalso been given by Huang and Ma [36]

Exercise 262 With the notation of Theorem 261 show that for any positive integer k

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

In this section we describe some applications of Lagrange inversion

31 A rational function expansion It is surprising that Lagrange inversion can giveinteresting results when the solution to the equation to be solved is rational

We consider the equationf = 1 + a+ abf

with solution

f =1 + a

1minus ab

We apply (246) with R(t) = 1 + a + abt and ψ(t) = tr(1 + bt)s Here we have 1 + bf =(1 + b)(1minus ab) and 1minusRprime(f) = 1minus ab Then

(1 + a)r(1 + b)s

(1minus ab)r+s+1=sumn

[tn]tr(1 + bt)s(1 + a+ abt)n

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

For another approach to this identity see Gessel and Stanton [26]

32 The tree function In applying Lagrange inversion the nicest examples are those inwhich the series R(t) has the property that there is a simple formula for the coefficients ofR(t)n and these simple formulas usually come from the exponential function or the binomialtheorem In this section we discuss the simplest case in which R(t) = et Later in section35 we discuss a more complicated example involving the exponential function

Let T (x) be the power series satisfying

T (x) = xeT (x)

Equivalently T (x) = (xeminusx)〈minus1〉 and thus T (xeminusx) = x Then by the properties of exponen-tial generating functions (see eg Stanley [70 Chapter 5]) T (x) is the exponential gener-ating function for rooted trees and eT (x) = T (x)x is the exponential generating function forforests of rooted trees We shall call T (x) the tree function and we shall call F (x) = eT (x)

the forest function The tree function is closely related to the Lambert W function [16 17]which may be defined by W (x) = minusT (minusx) Although the Lambert W function is betterknown we will state our results in terms of the tree and forest functions

Applying (221) with R(t) = et gives

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

for all positive integers k and

F (x)k = ekT (x) =infinsumn=0

k(n+ k)nminus1xn

n(323)

for all k Equation (322) implies that there are knnminuskminus1(nk

)forests of n rooted trees on

n vertices and equation (323) implies that there are k(n + k)nminus1 forests with vertex set1 2 n+ k in which the roots are 1 2 k

An interesting special case of (323) is k = minus1 which may be rearranged to

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

Equation (324) may be interpreted in terms of prime parking functions [70 Exercise 549fp 95 Solution p 141]

Applying (226) gives

F (x)k

1minus T (x)=infinsumn=0

(n+ k)nxn

n (325)

A Lacasse [46 p 90] conjectured an identity that may be written as

U(x)3 minus U(x)2 =infinsumn=0

nn+1xn

n (326)

U(x) =infinsumn=0

Proofs of Lacassersquos conjecture were given by Chen et al [11] Prodinger [59] Sun [73] andYounsi [80] We will prove (326) by showing that both sides are equal to T (x)(1minusT (x))3

To do this we first note that the right side of (326) is

infinsumn=0

(nminus 1)nxn

eminusT (x)

1minus T (x)=eminusT (x)T (x)T prime(x)

(1minus T (x))2 (327)

Differentiating (321) with respect to x gives

T prime(x) =infinsumn=0

(n+ 1)nxn

which by (325) is equal to eT (x)(1 minus T (x)) Thus (327) is equal to T (x)(1 minus T (x))3But by (325) U(x) = 1(1 minus T (x)) from which it follows easily that U(x)3 minus U(x)2 =T (x)(1minus T (x))3

The series T (x) and U(x) were studied by Zvonkine [79] who showed that DkU(x) andDkU(x)2 where D = xddx are polynomials in U(x)

The seriessuminfin

n=0(n + k)n+mxnn where m is an arbitrary integer can be expressed interms of T (x) (The case k = 0 has been studied by Smiley [67]) We first deal with thecase in which m is negative

Theorem 321 Let l be a positive integer Then for some polynomial pl(u) of degree lminus 1with coefficients that are rational functions of k we have

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

The first three polynomials pl(u) are

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

Before proving Theorem 321 let us check (328) for l = 1 and l = 2 The case l = 1 isequivalent to (323) For l = 2 we have

ekT (x) = F (x)k = k

infinsumn=0

(n+ k)nminus1xn

ekT (x)T (x) = F (x)k middot xF (x) = xF (x)k+1

= x(k + 1)infinsumn=0

(n+ k + 1)nminus1xn

= (k + 1)infinsumn=0

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

((n+ k)nminus1 minus n(n+ k)nminus2

=infinsumn=0

(n+ k)nminus2xn

The general case can be proved in a similar way (See Exercise 324) However it isinstructive to take a different approach using finite differences (cf Gould [31]) that we wewill use again in section 33

Let s be a function defined on the nonnegative integers The shift operator E takes s tothe function Es defined by (Es)(n) = s(n + 1) We denote by I the identity operator thattakes s to itself and by ∆ the difference operator E minus I so (∆s)(n) = s(n + 1) minus s(n) Itis easily verified that if s is a polynomial of degree d gt 0 with leading coefficient L then∆s is a polynomial of degree d minus 1 with leading coefficient dL and if s is a constant then∆s = 0 The kth difference of s is the function ∆ks Thus if s is a polynomial of degree dwith leading coefficient L then ∆ks = 0 for k gt d and ∆ds is the constant dL

Since the operators E and I commute we can expand ∆k = (E minus I)k by the binomialtheorem to obtain

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

We may summarize the result of this discussion in the following lemma

Lemma 322 Let s be a polynomial of degree d with leading coefficient L Then

ksumi=0

(minus1)kminusi(k

)s(n+ i)

is 0 if k gt d and is the constant dL for k = d

Proof of Theorem 321 If we set x = ueminusu in Theorem 321 and use the fact that T (ueminusu) =u we see that Theorem (321) is equivalent to the formula

eminuskuinfinsumn=0

(n+ k)nminusl(ueminusu)n

n= pl(u)

We have

eminuskuinfinsumn=0

(n+ k)n+m(ueminusu)n

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

If m = minusl is a negative integer and j ge l then (n+ k)j+m = (n+ k)jminusl is a polynomial in nof degree less than j so the inner sum in (329) is 0 Thus (328) follows with

pl(u) =lminus1sumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)minus(lminusj)

We cannot set k = 0 in (328) since the n = 0 term on the left is kminusl However it is nothard to evaluate

suminfinn=1 n

nminuslxnn

Theorem 323 Let ql(u) be the result of setting k = 1 in pl(u) Then

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

Proof We haveinfinsumn=1

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

By (328) with k = 1 this is

xeT (x)ql(T (x)) = T (x)ql(T (x))

Exercise 324 Prove Theorem 321 by finding a formula for the coefficient tj(n) of xnnin ekT (x)T (x)j and showing that (n + k)nminusl can be expressed as a linear combination withcoefficients that are rational functions of k of t0(n) tlminus1(n)

Next we considersuminfin

n=0(n+ k)n+mxnn where m is a nonnegative integer (We evaluatedthe case k = 0m = 1 in our discussion of Lacassersquos conjecture)

Theorem 325 Let m be a nonnegative integer Then there exists a polynomial rm(u k)with integer coefficients of degree m in u and degree m in k such that

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

The first three polynomials rm(u k) are

r0(u k) = 1

r1(u k) = k + (1minus k)u

r2(u k) = k2 + (1 + 3k minus 2k2)u+ (2minus 3k + k2)u2

Proof sketch We give here a sketch of a proof that tells us something interesting about thepolynomials rm(u k) for a more direct approach see Exercise 326

As in the proof of Theorem 321 we set x = ueminusu and consider the sum on the left side of(329)

Following Carlitz [9] we define the weighted Stirling numbers of the second kind R(n j k)by

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

(For k = 0 R(n j k) reduces to the ordinary Stirling number of the second kind S(n j))Equation (3210) implies that the R(n j k) is a polynomial in k with integer coefficientsThen (329) is equal to

suminfinj=0R(j + m j k)uj It is not hard to show that for fixed m

R(j +m j k) is a polynomial in j of degree 2mThus

infinsumj=0

R(j +m j k)uj =rm(u k)

(1minus u)2m+1

for some polynomial rm(u k) of degree at most 2mWe omit the proof that rm(u k) actually has degree m in u

For k = 1 the coefficients of rm(u 1) are positive integers sometimes called second-orderEulerian numbers see for example [32 p 270] and [24]

Exercise 326 Give an inductive proof of Theorem 325 using the fact that if W (m k) =suminfinn=0(n+ k)n+mxnn then dW (m k)dx = W (m+ 1 k + 1)

We can get convolution identities by applying (323) and (325) to

F (x)k+l

1minus T (x)= F (x)k

F (x)l

1minus T (x)

F (x)k+l = F (x)kF (x)l

The first identity yields

(n+ k + l)n =nsumi=0

)k(i+ k)iminus1(nminus i+ l)nminusi

and the second yields

(k + l)(n+ k + l)nminus1 =nsumi=0

)k(i+ k)iminus1l(nminus i+ l)nminusiminus1

Note that these are identities of polynomials in k and l If we set k = x and l = yminusn in thefirst formula we get the nicer looking

(x+ y)n =nsumi=0

)x(x+ i)iminus1(y minus i)nminusi (3211)

Replacing x with xz and y with yz in (3211) and multiplying through by zn gives thehomogeneous form

(x+ y)n =nsumi=0

)x(x+ iz)iminus1(y minus iz)nminusi (3212)

which was proved by N H Abel in 1826 [1] Note that for z = 0 (3212) reduces to thebinomial theorem Riordan [64 pp 18ndash27] gives a comprehensive account of Abelrsquos identityand its generalizations though he does not use Lagrange inversion

Exercise 327 (Chu [13]) Prove Abelrsquos identity (3211) using finite differences (Startby expanding (y minus i)nminusi = [(x+ y)minus (x+ i)]nminusi by the binomial theorem)

33 Fuss-Catalan numbers The Fuss-Catalan (or Fuszlig-Catalan) numbers of order p (alsocalled generalized Catalan numbers) are the numbers 1

(pn+1n

(pminus1)n+1

) which re-

duce to Catalan numbers for p = 2 They were first studied by N Fuss in 1791 [21] As weshall see they are the coefficients of the power series cp(x) satisfying the functional equation

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

1minus xcp(x)pminus1

as was shown using Lagrange inversion by Liouville [49]An account of these generating functions can be found Graham Knuth and Patashnik

[32 pp 200ndash204]It follows easily from (331) that

cp(x)minus 1 = xcp(x)p =

(1 + x)p

)〈minus1〉 (332)

xcp(x)pminus1 =(x(1minus x)pminus1

)〈minus1〉

xcp(xpminus1) = (xminus xp)〈minus1〉

cprimep(x) =cp(x)p

1minus pxcp(x)pminus1

Lagrange inversion gives

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

for all k With R(t) = 1 + xtp we have Rprime(t) = pxtpminus1 so

1minusRprime(cp(x)) = 1minus pxcp(x)pminus1 = 1minus p(cp(x)minus 1)cp(x) = 1minus p+ pcp(x)minus1

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

1minus pxcp(x)pminus1=

cp(x)k+1

1minus (pminus 1)(cp(x)minus 1) (334)

Equivalentlyinfinsumn=0

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

)(x(1minus x)pminus1

(1minus px)(1minus x)k

The convolution identities obtained from (333) and (334) known as Rothe-Hagen iden-tities [65 33 30] aresum

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

Exercise 331 Show that cminusp(x) = 1cp+1(minusx)

Exercise 332 Prove that cp+q(x) = cp(xcp+q(x)q

)(a) combinatorially (b) algebraically

(c) using Lagrange inversion

Exercise 333 Prove that (xcp(x

a)b)〈minus1〉

= xcabminusp+1(minusxa)b

(a) algebraically (b) using Lagrange inversion In particular as noted by Dennis Stanton iff(x) = xc(x)3 then f(x)〈minus1〉 = minusf(minusx)

Exercise 334 (Mansour and Sun [50 Example 56] Sun [72]) Show that

1minus xc3

(1minus x)3

)= c2(x)

Exercise 335 Prove (333) and (334) by finite differences

Exercise 336 (Chu [13]) Prove the Hagen-Rothe identities by finite differences

Next we prove Jensenrsquos formula [39]nsuml=0

(j + pl

)(r minus plnminus l

nsumi=0

(j + r minus inminus i

)pi (335)

By (334) we haveinfinsuml=0

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

(1minus pxcp(x)pminus1)2

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixicp(x)j+k+(pminus1)i

=infinsumi=0

pixiinfinsumm=0

(pm+ j + k + (pminus 1)i

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

Equating coefficients of xn on both sides givesnsuml=0

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

Setting k = r minus pn gives (335)We also have analogues of Theorems 321 and 325 for Fuss-Catalan numbers

Theorem 337 Let i and j be nonnegative integers with i lt j Theninfinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

is a polynomial uij(x) in x of degree j minus iminus 1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)xn

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

n ((pminus 1)n+ j)(minus1)mminusn

((pminus 1)n+ i+m

mminus n

For m ge j minus i the coefficient of xm may be rearranged to(mminus (j minus i)

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

The sum is the mth difference of a polynomial of degree less than m and is therefore 0 Form = j minus iminus 1 the coefficient of xjminusiminus1 reduces to

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

which is nonzero by the well-known identity

msumn=0

(minus1)n(m

)minus1

so the degree of the polynomial is not less than j minus iminus 1

The first few values of these polynomials are

uii+1(x) =1

uii+2(x) =1

(i+ 1)(i+ 2)minus pminus 1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 1)(i+ 2)(i+ 3)minus (pminus 1)(p+ 2i+ 4)

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

As a simple example of Theorem 337 the number of 2-stack-sortable permutations of1 2 n is

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

(see [57 Sequence A000139]) so by Theorem 337 with p = 3 i = 0 and j = 2suminfinn=0 anx

n(1 + x)3n+1 is a polynomial of degree 1 which is easily computed to be 2 minus xThen by (332) we find that

infinsumn=0

anxn = 3c3(x)minus c3(x)2

which can be checked directly from (333)There is a result similar to Theorem 337 for i ge j which we state without proof

Theorem 338 Let i and j be nonnegative integers with i ge j Then

(1minus (pminus 1)x)2(iminusj)+1

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

is a polynomial in x of degree at most iminus j

Exercise 339 Prove Theorem 338

34 Narayana and Fuss-Narayana numbers The Narayana numbers may be definedby N(n i) = 1

)(niminus1

)for n ge 1 They have many combinatorial interpretations in terms

of Dyck paths ordered trees binary trees and noncrossing partitionsIt is not hard to see from the formula for Narayana numbers that N(n i) = N(n n+1minusi)

A generating function for the Narayana that exhibits this symmetry is given by the solutionto the equation

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

n[tnminusk](1 + xt)n(1 + yt)n

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

N(n i+ 1)xiynminusiminus1

Equation (341) can be solved explicitly to give

f =1minus xminus y minus

radic(1minus xminus y)2 minus 4xy

Exercise 341 Prove that

(1 + xf)r(1 + yf)sradic(1minus xminus y)2 minus 4xy

=sumij

(r + i+ j

)(s+ i+ j

(1 + xf)r(1 + yf)s =sumij

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

The first formula is equivalent to a well-known generating function for Jacobi polynomialssee Carlitz [8]

We may generalize (341) to

f = (1 + x1f)r1(1 + x2f)r2 middot middot middot (1 + xmf)rm (342)

for which Lagrange inversion gives

fk =infinsumn=k

sumi1+middotmiddotmiddot+im=nminusk

)middot middot middot(rmn

)xi11 middot middot middotximm (343)

These numbers reduce to Catalan numbers for k = m = 1 r1 = 2 and to Narayana numbersfor k = 1m = 2 r1 = r2 = 1 For k = 1m = 2 r1 = 1 they are sometimes called Fuss-Narayana numbers see Armstrong [3] Cigler [15] Edelman [19] Eu and Fu [20] and Wang[75] The numbers for k = 1 ri = 1 for all i have been called generalized Fuss-Narayananumbers by Lenczewski and Sa lapata [48] they have also been studied by Edelman [19]Stanley [69] and Xu [78] The case k = 1m = 3 r2 = minusr1 r3 = 1 of these numbers wasconsidered by Krattenthaler [44 equation (31)] We note that if r1 = middot middot middot = rm then fis a symmetric function of x1 xm and this symmetric function arises in the study ofalgebraic aspects of parking functions [69]

If we set ri = minussi in (342) and replace xi with minusxi and f with g then (342) becomes

(1minus x1g)s1(1minus x2g)s2 middot middot middot (1minus xmg)sm (344)

and with the formula(minusai

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

)middot middot middot(smn+ im minus 1

These numbers reduce to Catalan numbers for k = m = s1 = 1 For s1 = middot middot middot = sm = 1 theyhave been considered by Aval [4] and (for k = 1) by Stanley [69]

Of special interest are the cases of (342) that reduce to a quadratic equation since inthese cases there are simple explicit formulas for f If we take m = 1 r1 = r2 = 1 andr3 = minus1 then with a change of variable names and one sign we have

f =(1 + xf)(1 + yf)

1minus zf

with the solution

radic(1minus xminus y)2 minus 4xy minus 4z

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

Another case of (342) that reduces to a quadratic is f =radic

(1 + xf)(1 + yf) we leave thedetails to the reader

35 Raneyrsquos equation G Raney [62] considered the equation

f =infinsumm=1

AmeBmf (351)

in which f is a power series in the indeterminates Am and Bm He used Pruferrsquos corre-spondence to give a combinatorial derivation of a formula for the coefficients in this powerseries

We can use Lagrange inversion to give a formula for the coefficients of f

Theorem 351 Let f be the power series in Am and Bm satisfying (351) and let k bea positive integer Let i1 i2 and j1 j2 be nonnegative integers only finitely many ofwhich are nonzero If i1+i2+middot middot middot = k+j1+j2+middot middot middot then the coefficient of Ai11 A

i22 middot middot middotB

j22 middot middot middot

in fk is

k(i1 + i2 + middot middot middot minus 1)

i1 i2 middot middot middotij11j1

ij22j2middot middot middot

and if i1 + i2 + middot middot middot 6= k + j1 + j2 + middot middot middot then the coefficient is zero

Proof Applying equation (243) to (351) gives

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

sumi1+i2+middotmiddotmiddot=n

i1 i2 middot middot middotAi11 A

i22 middot middot middot ei1B1tei2B2t middot middot middot

=infinsumn=k

n[tnminusk]

i22 middot middot middot

(i1B1t)j1

(i2B2t)j2

j2middot middot middot

=infinsumn=k

j1+j2+middotmiddotmiddot=nminusk

k(nminus 1)

ij11j1

and the formula follows

A combinatorial derivation of Raneyrsquos formula has also been given by D Knuth [42 Section2344]

4 Proofs

In this section we give three several proofs of the Lagrange inversion formula

41 Residues The simplest proof of Lagrange inversion is due to Jacobi [38] We definethe residue res f(x) of a Laurent series f(x) =

sumn fnx

n to be fminus1Jacobi proved the following change of variables formula for residues

Theorem 411 Let f be a Laurent series and let g(x) =suminfin

n=1 gnxn be a power series with

g1 6= 0 Thenres f(x) = res f(g(x))gprime(x)

Proof By linearity it is sufficient to prove the formula when f(x) = xk for some integer kIf k 6= minus1 then resxk = 0 and

res g(x)kgprime(x) = resd

dxg(x)k+1(k + 1) = 0

since the residue of a derivative is 0

If k = minus1 then resxk = 1 and

res g(x)kgprime(x) = res gprime(x)g(x) = resg1 + 2g2x+ middot middot middotg1x+ g2x2 + middot middot middot

= res1

xmiddot g1 + 2g2x+ middot middot middotg1 + g2x+ middot middot middot

Jacobirsquos paper [38] contains a multivariable generalization of Theorem 411 see also Gessel[28] and Xin [77]

Now let f(x) and g(x) be compositional inverses Then for any Laurent series φ

[xn]φ(f) = resφ(f)

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

This is equation (217) which we have already seen is equivalent to the other forms ofLagrange inversion

Exercise 412 Let f be a Laurent series and let g(x) =suminfin

n=m gnxn be a Laurent series

with gm 6= 0 Show that if f(g(x)) is well-defined as a Laurent series then

m res f = res f(g(x))gprime(x)

Exercise 413 (Hirzebruch [34] see also Kneezel [41]) Use the change of variables formula(Theorem 411) to show that the unique power series f(x) satisfying

for all n ge 1 is f(x) = x(1minus eminusx)

42 Induction In this proof and the next we consider the equation f = xR(f) whereR(t) is a power series If R(t) has no constant term then f = 0 and the formulas are trivialSo we may assume that R(t) has a nonzero constant term so f exists and is unique since itis the compositional inverse of xR(x)

We now give an inductive proof of (212) for any power series φ(t)

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

(As noted in section 2 this implies that (421) holds more generally when φ(t) is a Laurentseries)

We first take care of the case in which φ(t) = 1 The case φ(t) = 1 n = 0 is trivial Ifφ(t) = 1 and n gt 0 we have(

1minus tRprime(t)

)R(t)n = R(t)n minus tRprime(t)R(t)nminus1

= R(t)n minus t

dtR(t)n (422)

Now for any power series u(t)

(u(t)minus t

nuprime(t)

With(422) this proves (421) for φ(t) = 1 n gt 0

Now we prove the formula (421) by induction on n It is clear that (421)holds for n = 0Now let us suppose that for some nonnegative integer m (421) holds for all φ when n = mWe now want to show that (421) holds for all φ when n = m+ 1 By linearity and the caseφ(t) = 1 it is enough to prove (421) for n = m + 1 and φ(t) = tk where k ge 1 In thiscase we have

[xm+1] fk = [xm+1] fkminus1 middot xR(f)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

43 Factorization Another proof is based on a version of the ldquofactor theoremrdquo if f =xR(f) then tminus f divides tminus xR(t) This proof is taken from Gessel [27] but it is similar toLagrangersquos original proof [47]

We will prove (221) which gives a formula for fk where f = f(x) satisfies f = xR(f)First we recall Taylorrsquos theorem for power series if P (t) is a power series in t and α is an

element of the coefficient ring then

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

as long as this sum is summable (The case P (t) = tm of (431) is just the binomial theoremand the general case follows by linearity)

Now let us apply (431) with P (t) = t minus xR(t) and α = f where f = xR(f) so thatP (f) = 0 Then we have

tminus xR(t) = 0 + (tminus f)(1minus xRprime(f)

)+ (tminus f)2S(x t)

= (tminus f)Q(x t) (432)

where S(x t) is a power series and Q(x t) is a power series with constant term 1Equation (432) is an identity in the ring C[[x t]] which is naturally embedded in the

ring C((t))[[x]] of power series in x with coefficients that are Laurent series in t In thisring series like

suminfinn=0(xt)

n are allowed even though they have infinitely many negativepowers of t since the coefficient of any power of x is a Laurent series in t We now do somecomputations in C((t))[[x]]

By (432) we have 1minus xR(t)t = (1minus ft)Q(x t) Since xR(t)t and ft are divisible byx and Q(x t) is a power series in x and t with constant term 1 we may take logarithms toobtain

minus log(1minus xR(t)t) = minus log(1minus ft)minus logQ(x t) (433)

Note that logQ(x t) is a power series in x and t and so has no negative powers of t Now weequate coefficients of xntminusk on both sides of (433) where n and k are both positive integersOn the left we have [tminusk](R(t)t)nn = [tnminusk]R(t)nn and on the right we have [xn] fkkThus

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

Exercise 431 (Gessel [27]) Derive (222) similarly

44 Combinatorial proofs There are several different combinatorial proofs of Lagrangeinversion They all interpret the solution f of f = xR(f) or f = R(f) as counting certaintrees Here f may be interpreted as either an ordinary or exponential generating functionand thus different types of trees may be involved

In ordinary generating function proofs f will count unlabeled ordered trees (also calledplane trees) which are rooted trees in which the children of each vertex are linearly ordered(See Figure 1) More generally fk will count k-tuples of ordered trees which we also call

Figure 1 An ordered tree

forests of ordered treesIf f = xR(f) where R(t) =

suminfini=0 rit

i then the coefficient of xn in f is the sum of theweights of the ordered trees with a total of n vertices where the weight of a tree is theproduct of the weights of its vertices and the weight of a vertex with i children is ri (Forexample the tree of Figure 1 has weight r30r

22) So to give a combinatorial proof of formula

(221) we show that the sum of the weights of all k-tuples of ordered trees with a total ofn vertices is (kn) [tnminusk]R(t)n or equivalently by (251) the number of k-tuples of orderedtrees with ni vertices with i children for each i is equal to

n0 n1 n2

)(441)

ni and nminus k =sumk

ini (442)

and is zero otherwise W T Tutte [74] gave the case k = 1 of (441) which he derived (ina roundabout way) from Lagrange inversion

We can also work with exponential generating functions One way to do this is considerthe equation f = x

suminfini=0 sif

ii where x is the exponential variable and the sn are weightsThen by the properties of exponential generating functions (see eg [70 Chapter 5]) fcounts labeled rooted trees where a vertex with i children is weighted si More preciselythe coefficient of xnn in fkk is the sum of the weights of all forests of k rooted trees withvertex set [n] = 1 2 n Then Lagrange inversion in the form (251) is equivalent toassertion that the number of forests of k rooted trees with vertex set [n] in which ni verticeshave i children is equal to

(nminus 1)

(k minus 1) 0n01n12n2 middot middot middot

)(443)

with the same conditions on n0 n1 as before (See Stanley [70 p 30 Corollary 535])

Another exponential generating function approach is to consider the equation

f =infinsumi=0

where we work with exponential generating functions in the variables s0 s1 Then bythe properties of multivariable exponential generating functions the coefficient of

in fkk is the number of forests of k rooted trees in which for each i ni vertices have ichildren and these vertices are labeled 1 2 ni (Since vertices with the same label havedifferent numbers of children there are no nontrivial label-preserving automorphisms of suchtrees) For example such a forest with k = 2 n0 = 5 n2 = 3 and ni = 0 for i isin 0 2 isshow in Figure 2 Then Lagrange inversion in the form (251) (with x = 1) is equivalent to

Figure 2 A forest with n0 = 5 n2 = 3

the assertion that the number of such forests is

(nminus 1)

(k minus 1) 0n01n12n2middot middot middot (444)

where n = n0 + n1 + middot middot middot and n1 + 2n2 + 3n3 + middot middot middot = nminus k

45 Raneyrsquos proof The earliest combinatorial proof of Lagrange inversion is that ofRaney [61] We sketch here a proof that is based on Raneyrsquos though the details are dif-ferent (See also Stanley [70 pp 31ndash35 and 39ndash40]) We define the suffix code c(T ) for anordered tree T to be a sequence of nonnegative integers defined recursively If the root r ofT has j children and the trees rooted at the children of r are T1 Tj then c(T ) is theconcatenation c(T1) middot middot middot c(Tj)j More generally the suffix code for a k-tuple (T1 Tk) ofordered trees is the concatenation c(T1) middot middot middot c(Tk) We defined the reduced code of a tree ork-tuple of trees to be the sequence obtained from the suffix code by subtracting 1 from eachentry

For example the suffix code of the k-tuple of trees in Figure 3 is 0 0 2 0 1 and the reducedcode is 1 1 1 1 0 where 1 denotes minus1

The following lemma can be proved by induction

Lemma 451

(i) A forest of ordered trees is uniquely determined by its reduced code

Figure 3 A forest of ordered trees

(ii) A sequence a1a2 middot middot middot an of integers greater than or equal to minus1 is the reduced code ofan ordered k-forest if and only if a1 + middot middot middot+ an = minusk and a1 + middot middot middot+ ai is negative fori = 1 n

We also need a lemma due to Raney that generalizes the ldquocycle lemmardquo of Dvoretzkyand Motzkin [18] It can be proved by induction or in other ways (See eg Stanley [70pp 32ndash33])

Lemma 452 Let a1 middot middot middot an be a sequence of integers greater than or equal to minus1 withsum minusk lt 0 Then there exactly k integers i with 1 le i le n such that the sequenceai middot middot middot ana1 middot middot middot aiminus1 has all partial sums negative

We can now prove Lagrange inversion We want to prove that the sum of the weights ofall k-forests with n vertices is kn times

[tnminusk]R(t)n = [tminusk]

where R(t) =suminfin

n=0 sntn Let us define the weight of a sequence a1 middot middot middot an of integers to be

the product sa1+1 middot middot middot san+1 Then by Lemma 451 the sum of the weights of all k-forestswith n vertices is the sum of the weights of all sequences of integers of length n with entriesgreater than or equal to minus1 with sum minusk and with all partial sums negative It is clearthat [tminusk](R(t)t)n is the sum of the weights of all sequences of integers of length n withentries greater than or equal to minus1 and with sum minusk But by Lemma 452 a proportionkn of these sequences have all partial sums negative

46 Proofs by labeled trees We can derive Lagrange inversion by counting labeled treeswith the following result which seems to have first been proved by Moon [52]

Theorem 461 Let n be a positive integer and let d1 d2 dn be positive integers Thenthe number of (unrooted) trees with vertex set [m] in which vertex i has degree di is themultinomial coefficient (

mminus 2

d1 minus 1 dm minus 1

)(461)

ifsumm

i=1 di = 2(mminus 1) and is 0 otherwise

We will prove Theorem 461 a little later we first look at some of its consequences Acorollary of Theorem 461 allows us to count forests of rooted trees

Corollary 462 Let e1 e2 en be nonnegative integers with e1 + e2 + middot middot middot = n minus k andlet k be a positive integer Then the number of forests of k rooted trees with vertex set [n]in which vertex i has ei children is (

nminus 1

k minus 1 e1 en

Proof Let T be a tree on [n+ 1] in which vertex n+ 1 has degree k and vertex i has degreeei+1 for 1 le i le n (so that

sumni=1 ei = nminusk) Removing vertex n+1 from T and rooting the

resulting component trees at the neighbors of n + 1 in T gives a forest F of k rooted treesin which vertex i has ei children and this operation gives a bijection from trees on [n + 1]in which vertex n + 1 has degree k and vertex i has degree ei + 1 for 1 le i le n to the setof rooted forests of k trees on [n] in which vertex i has ei children The result then followsfrom Lemma (461)

It follows from Corollary 462 that the number of forests of k rooted trees in which foreach i ni vertices have i children and these vertices are labeled 1 2 ni is given by (444)which as we saw earlier is equivalent to Lagrange inversion

Now to count k-forests on [n] in which ni vertices have i children we first assign thenumber of children to each element of [n] which can be done in

)ways For each

assignment the number of trees is

(nminus 1)

(k minus 1) 0n01n1 middot middot middot

Multiplying these factors gives (443)We now present sketches of three proofs of Theorem 461 They all depend on the fact

that a tree with at least two vertices must have at least one leaf (vertex with no children)The first proof uses Pruferrsquos correspondence [60] Given a tree T1 with vertex set [n] let

a1 be the least leaf of T1 and let b1 be the unique neighbor of a1 in T1 Let T2 be the resultof removing a1 and its incident edge from T1 Let a2 be the least leaf of T2 and let b2 beits neighbor in T2 Continue in this way to define b3 middot middot middot bnminus2 Then the Prufer code ofT1 is the sequence (b1 b2 middot middot middot bnminus2) It can be shown that the map that takes a tree to itsPrufer code is a bijection from the set of trees with vertex set [n] for n ge 2 to the set ofsequences b1b2 bnminus2 of elements of [n] with the property that a vertex of degree d appearsdminus1 times in the Prufer code Then as noted by Moon [52] Theorem 461 is an immediateconsequence since the multinomial coefficient counts Prufer codes of trees in which vertex ihas degree di

The second proof is by induction and is due to Moon [53] (see also [54 p 13]) For m ge 2let T (m d1 dm) be the number of trees on [m] in which vertex i has degree di and letU(m d1 dm) be the multinomial coefficient (461) which is 0 if any di is less than 1 orifsum

i di 6= 2(nminus 1)Clearly T (m d1 dm) is equal to U(m d1 dm) for m = 2 Now suppose that n gt 2

and that T (m d1 dm) is equal to U(m d1 dm) for m = n minus 1 and all choices ofd1 dm

We next observe that if dn = 1 then

T (n d1 dn) =nminus1sumi=1

T (nminus 1 d1 di minus 1 dnminus1) (462)

since every tree in which n is a leaf is obtained by joining vertex n with an edge to somevertex of a tree on [nminus1] increasing by 1 the degree of this edge and leaving the other degreesunchanged Then by the inductive hypothesis and a well-known recurrence for multinomialcoefficients T (n d1 dn) is equal to U(n d1 dn) We have assumed that dn = 1 but

since T (n d1 dn) is symmetric in d1 dn and every tree has at least one leaf the resultholds without this assumption

Renyi [63] (see also Moon [54 p 13]) gave an elegant variation of this proof He showed byinduction that for m ge 2 the number of trees on [m] with degree sequence (d1 dm) ietrees in which vertex i has degree di is the coefficient of xd1minus11 middot middot middotxdmminus1m in (x1+ middot middot middot+xm)mminus2The result clearly holds for m = 2 so suppose that n gt 2 and that the result holds whenm = n minus 1 To show that it holds for m = n we note that every tree on [n] has at leastone leaf so by symmetry we may assume without loss of generality that dn = 1 Thus weneed only show that the number of trees on [n] with degree sequence (d1 dnminus1 1) is the

coefficient of xd1minus11 middot middot middot xdnminus1minus1nminus1 in

(x1 + middot middot middot+ xnminus1)nminus2 = (x1 + middot middot middot+ xnminus1) middot (x1 + middot middot middot+ xnminus1)

nminus3

But every tree on [n] in which vertex n is a leaf is obtained by joining vertex n by an edgeto some vertex of a tree on [n minus 1] increasing by 1 the degree of this edge and leaving theother degrees unchanged Thus by the induction hypothesis the contribution from trees inwhich vertex n is joined to vertex i is xi(x1 + middot middot middot+ xnminus1)

nminus3 and the result follows

Acknowledgment I would like to thank Sateesh Mane and an anonymous referee for helpfulcomments

References

[1] N H Abel Beweis eines Ausdruckes von welchem die Binomial-Formel ein einzelner Fall ist J ReineAngew Math 1 (1826) 159ndash160

[2] Ulrich Abel A generalization of the Leibniz rule Amer Math Monthly 120 (2013) no 10 924ndash928[3] Drew Armstrong Generalized noncrossing partitions and combinatorics of Coxeter groups Mem Amer

Math Soc 202 (2009) no 949 x+159[4] Jean-Christophe Aval Multivariate Fuss-Catalan numbers Discrete Math 308 (2008) no 20 4660ndash

4669[5] F Bergeron G Labelle and P Leroux Combinatorial Species and Tree-like Structures Encyclopedia

of Mathematics and its Applications vol 67 Cambridge University Press Cambridge 1998 Translatedfrom the 1994 French original by Margaret Readdy With a foreword by Gian-Carlo Rota

[6] Christian Brouder Alessandra Frabetti and Christian Krattenthaler Non-commutative Hopf algebraof formal diffeomorphisms Adv Math 200 (2006) no 2 479ndash524

[7] Jean-Paul Bultel Combinatorial properties of the noncommutative Faa di Bruno algebra J AlgebraicCombin 38 (2013) no 2 243ndash273

[8] L Carlitz The generating function for the Jacobi polynomial Rend Sem Mat Univ Padova 38 (1967)86ndash88

[9] Weighted Stirling numbers of the first and second kind I Fibonacci Quart 18 (1980) no 2147ndash162

[10] Augustin-Louis Cauchy Exercises de mathematiques vol 1 De Bure freres 1826 Application du calculdes residus a la sommation de plusieurs suites Oeuvres completes drsquoAugustin Cauchy Serie 2 tome 6pp 62ndash73

[11] William Y C Chen Janet F F Peng and Harold R L Yang Decomposition of triply rooted treesElectron J Combin 20 (2013) no 2 Paper 10 10 pp

[12] Wenchang Chu Leibniz inverse series relations and Pfaff-Cauchy derivative identities Rend Mat Appl(7) 29 (2009) no 2 209ndash221

[13] Elementary proofs for convolution identities of Abel and Hagen-Rothe Electron J Combin 17(2010) no 1 Note 24 5 pp

[14] Derivative inverse series relations and Lagrange expansion formula Int J Number Theory 9(2013) no 4 1001ndash1013

[15] J Cigler Some remarks on Catalan families European J Combin 8 (1987) no 3 261ndash267[16] R M Corless G H Gonnet D E G Hare D J Jeffrey and D E Knuth On the Lambert W function

Adv Comput Math 5 (1996) no 4 329ndash359[17] Robert M Corless David J Jeffrey and Donald E Knuth A sequence of series for the Lambert W

function Proceedings of the 1997 International Symposium on Symbolic and Algebraic Computation(Kihei HI) ACM New York 1997 pp 197ndash204

[18] A Dvoretzky and Th Motzkin A problem of arrangements Duke Math J 14 (1947) 305ndash313[19] Paul H Edelman Chain enumeration and noncrossing partitions Discrete Math 31 (1980) no 2

171ndash180[20] Sen-Peng Eu and Tung-Shan Fu Lattice paths and generalized cluster complexes J Combin Theory

Ser A 115 (2008) no 7 1183ndash1210[21] Nicolao Fuss Solutio quaestionis quot modis polygonum n laterum in polygona m laterum per diago-

nales resolvi queat Nova Acta Academiae Sci Petropolitanae 9 (1791) 243ndash251[22] Adriano M Garsia A q-analogue of the Lagrange inversion formula Houston J Math 7 (1981) no 2

205ndash237[23] Ira Gessel A noncommutative generalization and q-analog of the Lagrange inversion formula Trans

Amer Math Soc 257 (1980) no 2 455ndash482[24] Ira Gessel and Richard P Stanley Stirling polynomials J Combinatorial Theory Ser A 24 (1978)

no 1 24ndash33[25] Ira Gessel and Dennis Stanton Applications of q-Lagrange inversion to basic hypergeometric series

Trans Amer Math Soc 277 (1983) no 1 173ndash201[26] Short proofs of Saalschutzrsquos and Dixonrsquos theorems J Combin Theory Ser A 38 (1985) no 1

87ndash90[27] Ira M Gessel A factorization for formal Laurent series and lattice path enumeration J Combin Theory

Ser A 28 (1980) no 3 321ndash337[28] A combinatorial proof of the multivariable Lagrange inversion formula J Combin Theory Ser

A 45 (1987) no 2 178ndash195[29] Ira M Gessel and Gilbert Labelle Lagrange inversion for species J Combin Theory Ser A 72 (1995)

no 1 95ndash117[30] H W Gould Final analysis of Vandermondersquos convolution Amer Math Monthly 64 (1957) 409ndash415[31] Eulerrsquos formula for nth differences of powers Amer Math Monthly 85 (1978) no 6 450ndash467[32] Ronald L Graham Donald E Knuth and Oren Patashnik Concrete Mathematics A Foundation for

Computer Science second ed Addison-Wesley Publishing Company Reading MA 1994[33] S J Hagen Johann G Synopsis der Hoeheren Mathematik Arithmetische und Algebraische Analyse

vol 1 Felix L Dames Berlin 1891[34] F Hirzebruch The signature theorem reminiscences and recreation Prospects in mathematics (Proc

Sympos Princeton Univ Princeton NJ 1970) Princeton Univ Press Princeton NJ 1971 pp 3ndash31 Ann of Math Studies No 70

[35] Josef Hofbauer Lagrange-inversion Sem Lothar Combin 6 (1982) Art B06a httpwwwemisdejournalsSLCopaperss06hofbauerhtml

[36] Jianfeng Huang and Xinrong Ma Two elementary applications of the Lagrange expansion formula JMath Res Appl 35 (2015) no 3 263ndash270

[37] Eri Jabotinsky Representation of functions by matrices Application to Faber polynomials Proc AmerMath Soc 4 (1953) 546ndash553

[38] C G J Jacobi De resolutione aequationum per series infinitas Journal fur die reine und ange-wandte Mathematik 6 (1830) 257ndash286 Gesammelte Werke vol 6 pp 26ndash61 G Reimer Berlin (1891)reprinted by Chelsea Publishing Company New York (1969)

[39] J L W V Jensen Sur une identite drsquoAbel et sur drsquoutres formules analogues Acta Math 26 (1902)307ndash318

[40] Warren P Johnson The PfaffCauchy derivative identities and Hurwitz type extensions Ramanujan J13 (2007) no 1ndash3 167ndash201

[41] Dan Kneezel Hirzebruchrsquos motivation of the Todd class MathOverflow httpmathoverflownetq60478 (version 2011-04-03)

[42] Donald E Knuth The Art of Computer Programming Vol 1 Fundamental Algorithms third edAddison-Wesley Reading MA 1997

[43] Ch Krattenthaler Operator methods and Lagrange inversion a unified approach to Lagrange formulasTrans Amer Math Soc 305 (1988) no 2 431ndash465

[44] Christian Krattenthaler The F -triangle of the generalised cluster complex Topics in Discrete Mathe-matics Algorithms Combin vol 26 Springer Berlin 2006 pp 93ndash126

[45] Gilbert Labelle Some new computational methods in the theory of species Combinatoire enumerative(Montreal Que 1985Quebec Que 1985) Lecture Notes in Math vol 1234 Springer Berlin 1986pp 192ndash209

[46] Alexandre Lacasse Bornes PAC-Bayes et algorithmes drsquoapprentissage PhD thesis Universite LavalQuebec 2010

[47] Joseph Louis Lagrange Nouvelle methode pour resoudre les equations litterales par le moyen des seriesMemoires de lrsquoAcademie Royale des Sciences et Belles-Lettres de Berlin 24 (1770) 251ndash326 Oeuvrescomplete tome 3 Paris 1867 5ndash73

[48] Romuald Lenczewski and Rafa l Sa lapata Multivariate Fuss-Narayana polynomials and their applicationto random matrices Electron J Combin 20 (2013) no 2 Paper 41 14 pp

[49] J Liouville Remarques sur un memoire de N Fuss J Math Pures Appl 8 (1843) 391ndash394[50] Toufik Mansour and Yidong Sun Bell polynomials and k-generalized Dyck paths Discrete Appl Math

156 (2008) no 12 2279ndash2292[51] D Merlini R Sprugnoli and M C Verri Lagrange inversion when and how Acta Appl Math 94

(2006) no 3 233ndash249 (2007)[52] J W Moon The second moment of the complexity of a graph Mathematika 11 (1964) 95ndash98[53] Enumerating labelled trees Graph Theory and Theoretical Physics Academic Press London

1967 pp 261ndash272[54] Counting Labelled Trees Canadian Mathematical Monographs No 1 Canadian Mathematical

Congress Montreal Que 1970[55] Ivan Niven Formal power series Amer Math Monthly 76 (1969) 871ndash889[56] Jean-Christophe Novelli and Jean-Yves Thibon Noncommutative symmetric functions and Lagrange

inversion Adv in Appl Math 40 (2008) no 1 8ndash35[57] The On-Line Encyclopedia of Integer Sequences published electronically at httpoeisorg 2016[58] J F Pfaff Allgemeine Summation einer Reihe worinn hohere Differenziale vorkommen Archiv der

reinen und angewandten Mathematik 1 (1795) 337ndash47[59] Helmut Prodinger An identity conjectured by Lacasse via the tree function Electron J Combin 20

(2013) no 3 Paper 7 3 pp[60] Heinz Prufer Neuer Beweis eines Satzes uber Permutationen Arch Math Phys 27 (1918) 742ndash744[61] George N Raney Functional composition patterns and power series reversion Trans Amer Math Soc

94 (1960) 441ndash451[62] A formal solution of

suminfini=1 Aie

BiX = X Canad J Math 16 (1964) 755ndash762[63] Alfred Renyi On the enumeration of trees Combinatorial Structures and their Applications (Proc

Calgary Internat Conf Calgary Alta 1969) Gordon and Breach New York 1970 pp 355ndash360[64] John Riordan Combinatorial Identities John Wiley amp Sons Inc New York-London-Sydney 1968[65] Henrico Augusto Rothe Formulae De Serierum Reversione Demonstratio Universalis Signis Localibus

Combinatorio-Analyticorum Vicariis Exhibita Sommer Leipzig 1793[66] Issai Schur On Faber polynomials Amer J Math 67 (1945) 33ndash41[67] Leonard M Smiley Completion of a rational function sequence of Carlitz 2000 arXivmath

0006106[mathCO][68] Alan D Sokal A ridiculously simple and explicit implicit function theorem Sem Lothar Combin 61A

(200911) Art B61Ad 21 pp[69] Richard P Stanley Parking functions and noncrossing partitions Electron J Combin 4 (1997) no 2

Research Paper 20 4 pp The Wilf Festschrift (Philadelphia PA 1996)[70] Enumerative Combinatorics Vol 2 Cambridge Studies in Advanced Mathematics vol 62

Cambridge University Press Cambridge 1999 With a foreword by Gian-Carlo Rota and appendix 1by Sergey Fomin

[71] Dennis Stanton Recent results for the q-Lagrange inversion formula Ramanujan revisited (Urbana-Champaign Ill 1987) Academic Press Boston MA 1988 pp 525ndash536

[72] Yidong Sun A simple bijection between binary trees and colored ternary trees Electron J Combin 17(2010) no 1 Note 20 5

[73] A simple proof of an identity of Lacasse Electron J Combin 20 (2013) no 2 Paper 11 3pp

[74] William T Tutte The number of planted plane trees with a given partition Amer Math Monthly 71(1964) no 3 272ndash277

[75] Chao-Jen Wang Applications of the Goulden-Jackson Cluster Method to Counting Dyck paths by Oc-currences of Subwords PhD thesis Brandeis University 2011

[76] E T Whittaker and G N Watson A Course of Modern Analysis Third Edition Cambridge UniversityPress 1920

[77] Guoce Xin A residue theorem for Malcev-Neumann series Adv in Appl Math 35 (2005) no 3271ndash293

[78] Dapeng Xu Generalizations of two-stack-sortable permutations 2002 arXivmath0209313[mathCO][79] Malik Younsi An algebra of power series arising in the intersection theory of moduli spaces of curves

and in the enumeration of ramified coverings of the sphere 2004 arXivmath0403092[mathAG][80] Proof of a combinatorial conjecture coming from the PAC-Bayesian machine learning theory

2012 arXiv12090824[mathCO]

Department of Mathematics Brandeis University Waltham MA 02453-2700E-mail address gesselbrandeisedu

1 Introduction

11 Formal power series


21 Forms of Lagrange inversion

22 Polynomials

23 A simple example Catalan numbers

24 A generalization

25 Explicit formulas for the coefficients

26 Derivative formulas

3 Applications

31 A rational function expansion

32 The tree function

33 Fuss-Catalan numbers

34 Narayana and Fuss-Narayana numbers

35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization

44 Combinatorial proofs

45 Raneys proof

46 Proofs by labeled trees

References

formal Laurent series C((x)) whose elements are formal sumssuminfin

n=n0cnx

n for some integern0 ie formal sums

suminfinn=minusinfin cnx

n in which only finitely many negative powers of x havenonzero coefficients Henceforth will omit the word ldquoformalrdquo and speak of power series andLaurent series

We can iterate the power series and Laurent series ring constructions obtaining for ex-ample the ring C((x))[[y]] of power series in y whose coefficients are Laurent series in xIn any (possibly iterated) power series or Laurent series ring we will say that a set fα ofseries is summable if for any monomial m in the variables the coefficient of m is nonzero inonly finitely many fα In this case the sum

sumα f is well-defined and we will say that

sumα fα

is summable If we writesum

α fα as an iterated sum then the order of summation is irrele-vant If f(x) =

sumn cnx

n is a Laurent series in C((x)) and u isin C where C may be a powerseries or Laurent series ring then we say that that the substitution of u for x is admissibleif f(u) =

sumn cnu

n is summable and similarly for multivariable substitutions Admissiblesubstitutions are homomorphisms If u is a power series or Laurent series g(x) then f(g(x))if summable is called the composition of f and g If f(x) = c1x + c2x

2 + middot middot middot where c1 isinvertible in C then there is a unique power series g(x) = cminus11 x+ middot middot middot such that f(g(x)) = xthis implies that g(f(x)) = x We call g(x) the compositional inverse of f(x) and writeg(x) = f(x)〈minus1〉 For simplicity we will always assume that if f(x) = c1x+ c2x

2 + middot middot middot wherec1 6= 0 then c1 is invertible (Since C is an integral domain we can always adjoin cminus11 to Cif necessary)

The iterated power series rings C[[x]][[y]] and C[[y]][[x]] are essentially the same in thatboth consist of all sums

summnge0 cmnx

myn We may therefore write C[[x y]] for either of

these rings However the iterated Laurent series C((x))[[y]] C((y))[[x]] and C[[y]]((x))are all different in the first we have (x minus y)minus1 =

suminfinn=0 y

nxn+1 in the second we have(xminus y)minus1 = minus

suminfinn=0 x

nyn+1 and in the third xminus y is not invertibleIt is sometimes convenient to work with power series in infinitely many variables for

example we may consider the power seriessuminfin

n=0 rntn where the rn are independent inde-

terminates Although we donrsquot give a formal definition of these series they behave in ourapplications exactly as expected

We use the notation [xn] f(x) to denote the coefficient of xn in the Laurent series f(x)An important fact about the coefficient operator that we will use often without commentis that [xn]xkf(x) = [xnminusk] f(x)

The binomial coefficient(ak

)is defined to be a(aminus1) middot middot middot (aminusk+ 1)k if k is a nonnegative

integer and 0 otherwise Thus the binomial theorem (1 + x)a =suminfin

)xk holds for all a

21 Forms of Lagrange inversion We will give several proofs of the Lagrange inversionformula in section 4 Here we state several different forms of Lagrange inversion and showthat they are equivalent

Theorem 211 Let R(t) be a power series not involving x Then there is a unique powerseries f = f(x) such that f(x) = xR(f(x)) and for any Laurent series φ(t) and ψ(t) notinvolving x and any integer n we have

[xn]φ(f) =1

n[tnminus1]φprime(t)R(t)n where n 6= 0 (211)

)φ(t)R(t)n (212)

φ(f) =sumn

[xn]ψ(f)

)n= [tminus1]

φ(t)gprime(t)

g(t)n+1 (217)

[xnminus1]ψ(f)

)n= [t0]

g(t)n (218)

[tminus1]φprime(t)

g(t)n= n [tminus1]

φ(t)gprime(t)

g(t)n+1(219)

Butφprime(t)

g(t)n+1=d

[xn] fk =k

n[tnminusk]

n ankxn and gk =

sumn bnkx

n then

ank =k

[xn] fk =k

[xn]fk

[xm](fx)k =k

[xm](fx)k

[xm] log(fx) =1

m[tm]R(t)m for m 6= 0 (229)

c(x) = 1 + xc(x)2 (231)

c(x) =1minusradic

1minus 4x

2xminus 1

[xn] c(x)k = [xn](1 + f)k =1

n[tnminus1]

(2n+ k minus 1

nminus 1

(2n+ k minus 1

nminus 1

c(x)k =infinsumn=0

(2n+ k

infinsumn=0

(2n+ k minus 1

)xn (232)

(1 + t)k(1 + t)2n

(2n+ k minus 1

)minus(

2n+ k minus 1

nminus 1

[xn] c(x)k =sumn

)=sumn

)minus 2

(2n+ k minus 1

nminus 1

Finally (229) gives

m[tm](1 + t)2m =

= [tn](1 + t)k(1 + t)2n =

(2n+ k

c(x)kradic1minus 4x

=infinsumn=0

(2n+ k

)xn (233)

(2i+ k

)middot l

2j + l

(2j + l

2n+ k + l

(2n+ k + l

(2i+ k

)(2j + l

(2n+ k + l

F = z(1 + xF 2) (241)

[zn]F k =k

[z2m+k]F k =k

2m+ k[t2m](1 + xt2)2m+k =

(2m+ k

F k =infinsumm=0

(2m+ k

)xmz2m+k (242)

ck =infinsumm=0

(2m+ k

φ(f) =sumn

)φ(t)R(t)n =

1minusRprime(f)=

[tn]ψ(t)R(t)n (246)

f =1minusradic

1minus 4xy

2y=infinsumn=0

)xn+1yn

radic1minus 4x)

suminfinn=0 rnt

fk =sum

k(nminus 1)

n0n1 rn00 r

[xm]fk =k

m[tmminusk]

fk =sum

k(nminus 1)

fk =sum

k(nminus 1)

f = x+G(f)

φ(f) =infinsumm=0

(φ(x)

(1minusGprime(x)

)G(x)m

) (261)

dmminus1

dxmminus1

(φprime(x)

) (262)

andψ(f)

1minusGprime(f)=

infinsumm=0

(ψ(x)

) (263)

infinsumn=0

)xnminusmψ(t)G(t)m

(ψ(x)

)=infinsumn=0

)xnminusmψ(t)G(t)m

dxmα(x)

m=infinsumn=0

)xjminusm

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

nsumm=0

)dmminus1

dnminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

)dmminus1

dnminusmminus1

)=dnminus1

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

with solution

f =1 + a

1minus ab

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

)φ(t)R(t)n (212)

φ(f) =sumn

[xn]ψ(f)

)n= [tminus1]

φ(t)gprime(t)

g(t)n+1 (217)

[xnminus1]ψ(f)

)n= [t0]

g(t)n (218)

[tminus1]φprime(t)

g(t)n= n [tminus1]

φ(t)gprime(t)

g(t)n+1(219)

Butφprime(t)

g(t)n+1=d

[xn] fk =k

n[tnminusk]

n ankxn and gk =

sumn bnkx

n then

ank =k

[xn] fk =k

[xn]fk

[xm](fx)k =k

[xm](fx)k

[xm] log(fx) =1

m[tm]R(t)m for m 6= 0 (229)

c(x) = 1 + xc(x)2 (231)

c(x) =1minusradic

1minus 4x

2xminus 1

[xn] c(x)k = [xn](1 + f)k =1

n[tnminus1]

(2n+ k minus 1

nminus 1

(2n+ k minus 1

nminus 1

c(x)k =infinsumn=0

(2n+ k

infinsumn=0

(2n+ k minus 1

)xn (232)

(1 + t)k(1 + t)2n

(2n+ k minus 1

)minus(

2n+ k minus 1

nminus 1

[xn] c(x)k =sumn

)=sumn

)minus 2

(2n+ k minus 1

nminus 1

Finally (229) gives

m[tm](1 + t)2m =

= [tn](1 + t)k(1 + t)2n =

(2n+ k

c(x)kradic1minus 4x

=infinsumn=0

(2n+ k

)xn (233)

(2i+ k

)middot l

2j + l

(2j + l

2n+ k + l

(2n+ k + l

(2i+ k

)(2j + l

(2n+ k + l

F = z(1 + xF 2) (241)

[zn]F k =k

[z2m+k]F k =k

2m+ k[t2m](1 + xt2)2m+k =

(2m+ k

F k =infinsumm=0

(2m+ k

)xmz2m+k (242)

ck =infinsumm=0

(2m+ k

φ(f) =sumn

)φ(t)R(t)n =

1minusRprime(f)=

[tn]ψ(t)R(t)n (246)

f =1minusradic

1minus 4xy

2y=infinsumn=0

)xn+1yn

radic1minus 4x)

suminfinn=0 rnt

fk =sum

k(nminus 1)

n0n1 rn00 r

[xm]fk =k

m[tmminusk]

fk =sum

k(nminus 1)

fk =sum

k(nminus 1)

f = x+G(f)

φ(f) =infinsumm=0

(φ(x)

(1minusGprime(x)

)G(x)m

) (261)

dmminus1

dxmminus1

(φprime(x)

) (262)

andψ(f)

1minusGprime(f)=

infinsumm=0

(ψ(x)

) (263)

infinsumn=0

)xnminusmψ(t)G(t)m

(ψ(x)

)=infinsumn=0

)xnminusmψ(t)G(t)m

dxmα(x)

m=infinsumn=0

)xjminusm

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

nsumm=0

)dmminus1

dnminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

)dmminus1

dnminusmminus1

)=dnminus1

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

with solution

f =1 + a

1minus ab

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

[xn] fk =k

n[tnminusk]

n ankxn and gk =

sumn bnkx

n then

ank =k

[xn] fk =k

[xn]fk

[xm](fx)k =k

[xm](fx)k

[xm] log(fx) =1

m[tm]R(t)m for m 6= 0 (229)

c(x) = 1 + xc(x)2 (231)

c(x) =1minusradic

1minus 4x

2xminus 1

[xn] c(x)k = [xn](1 + f)k =1

n[tnminus1]

(2n+ k minus 1

nminus 1

(2n+ k minus 1

nminus 1

c(x)k =infinsumn=0

(2n+ k

infinsumn=0

(2n+ k minus 1

)xn (232)

(1 + t)k(1 + t)2n

(2n+ k minus 1

)minus(

2n+ k minus 1

nminus 1

[xn] c(x)k =sumn

)=sumn

)minus 2

(2n+ k minus 1

nminus 1

Finally (229) gives

m[tm](1 + t)2m =

= [tn](1 + t)k(1 + t)2n =

(2n+ k

c(x)kradic1minus 4x

=infinsumn=0

(2n+ k

)xn (233)

(2i+ k

)middot l

2j + l

(2j + l

2n+ k + l

(2n+ k + l

(2i+ k

)(2j + l

(2n+ k + l

F = z(1 + xF 2) (241)

[zn]F k =k

[z2m+k]F k =k

2m+ k[t2m](1 + xt2)2m+k =

(2m+ k

F k =infinsumm=0

(2m+ k

)xmz2m+k (242)

ck =infinsumm=0

(2m+ k

φ(f) =sumn

)φ(t)R(t)n =

1minusRprime(f)=

[tn]ψ(t)R(t)n (246)

f =1minusradic

1minus 4xy

2y=infinsumn=0

)xn+1yn

radic1minus 4x)

suminfinn=0 rnt

fk =sum

k(nminus 1)

n0n1 rn00 r

[xm]fk =k

m[tmminusk]

fk =sum

k(nminus 1)

fk =sum

k(nminus 1)

f = x+G(f)

φ(f) =infinsumm=0

(φ(x)

(1minusGprime(x)

)G(x)m

) (261)

dmminus1

dxmminus1

(φprime(x)

) (262)

andψ(f)

1minusGprime(f)=

infinsumm=0

(ψ(x)

) (263)

infinsumn=0

)xnminusmψ(t)G(t)m

(ψ(x)

)=infinsumn=0

)xnminusmψ(t)G(t)m

dxmα(x)

m=infinsumn=0

)xjminusm

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

nsumm=0

)dmminus1

dnminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

)dmminus1

dnminusmminus1

)=dnminus1

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

with solution

f =1 + a

1minus ab

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

[xm] log(fx) =1

m[tm]R(t)m for m 6= 0 (229)

c(x) = 1 + xc(x)2 (231)

c(x) =1minusradic

1minus 4x

2xminus 1

[xn] c(x)k = [xn](1 + f)k =1

n[tnminus1]

(2n+ k minus 1

nminus 1

(2n+ k minus 1

nminus 1

c(x)k =infinsumn=0

(2n+ k

infinsumn=0

(2n+ k minus 1

)xn (232)

(1 + t)k(1 + t)2n

(2n+ k minus 1

)minus(

2n+ k minus 1

nminus 1

[xn] c(x)k =sumn

)=sumn

)minus 2

(2n+ k minus 1

nminus 1

Finally (229) gives

m[tm](1 + t)2m =

= [tn](1 + t)k(1 + t)2n =

(2n+ k

c(x)kradic1minus 4x

=infinsumn=0

(2n+ k

)xn (233)

(2i+ k

)middot l

2j + l

(2j + l

2n+ k + l

(2n+ k + l

(2i+ k

)(2j + l

(2n+ k + l

F = z(1 + xF 2) (241)

[zn]F k =k

[z2m+k]F k =k

2m+ k[t2m](1 + xt2)2m+k =

(2m+ k

F k =infinsumm=0

(2m+ k

)xmz2m+k (242)

ck =infinsumm=0

(2m+ k

φ(f) =sumn

)φ(t)R(t)n =

1minusRprime(f)=

[tn]ψ(t)R(t)n (246)

f =1minusradic

1minus 4xy

2y=infinsumn=0

)xn+1yn

radic1minus 4x)

suminfinn=0 rnt

fk =sum

k(nminus 1)

n0n1 rn00 r

[xm]fk =k

m[tmminusk]

fk =sum

k(nminus 1)

fk =sum

k(nminus 1)

f = x+G(f)

φ(f) =infinsumm=0

(φ(x)

(1minusGprime(x)

)G(x)m

) (261)

dmminus1

dxmminus1

(φprime(x)

) (262)

andψ(f)

1minusGprime(f)=

infinsumm=0

(ψ(x)

) (263)

infinsumn=0

)xnminusmψ(t)G(t)m

(ψ(x)

)=infinsumn=0

)xnminusmψ(t)G(t)m

dxmα(x)

m=infinsumn=0

)xjminusm

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

nsumm=0

)dmminus1

dnminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

)dmminus1

dnminusmminus1

)=dnminus1

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

with solution

f =1 + a

1minus ab

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

(1 + t)k(1 + t)2n

(2n+ k minus 1

)minus(

2n+ k minus 1

nminus 1

[xn] c(x)k =sumn

)=sumn

)minus 2

(2n+ k minus 1

nminus 1

Finally (229) gives

m[tm](1 + t)2m =

= [tn](1 + t)k(1 + t)2n =

(2n+ k

c(x)kradic1minus 4x

=infinsumn=0

(2n+ k

)xn (233)

(2i+ k

)middot l

2j + l

(2j + l

2n+ k + l

(2n+ k + l

(2i+ k

)(2j + l

(2n+ k + l

F = z(1 + xF 2) (241)

[zn]F k =k

[z2m+k]F k =k

2m+ k[t2m](1 + xt2)2m+k =

(2m+ k

F k =infinsumm=0

(2m+ k

)xmz2m+k (242)

ck =infinsumm=0

(2m+ k

φ(f) =sumn

)φ(t)R(t)n =

1minusRprime(f)=

[tn]ψ(t)R(t)n (246)

f =1minusradic

1minus 4xy

2y=infinsumn=0

)xn+1yn

radic1minus 4x)

suminfinn=0 rnt

fk =sum

k(nminus 1)

n0n1 rn00 r

[xm]fk =k

m[tmminusk]

fk =sum

k(nminus 1)

fk =sum

k(nminus 1)

f = x+G(f)

φ(f) =infinsumm=0

(φ(x)

(1minusGprime(x)

)G(x)m

) (261)

dmminus1

dxmminus1

(φprime(x)

) (262)

andψ(f)

1minusGprime(f)=

infinsumm=0

(ψ(x)

) (263)

infinsumn=0

)xnminusmψ(t)G(t)m

(ψ(x)

)=infinsumn=0

)xnminusmψ(t)G(t)m

dxmα(x)

m=infinsumn=0

)xjminusm

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

nsumm=0

)dmminus1

dnminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

)dmminus1

dnminusmminus1

)=dnminus1

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

with solution

f =1 + a

1minus ab

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

F = z(1 + xF 2) (241)

[zn]F k =k

[z2m+k]F k =k

2m+ k[t2m](1 + xt2)2m+k =

(2m+ k

F k =infinsumm=0

(2m+ k

)xmz2m+k (242)

ck =infinsumm=0

(2m+ k

φ(f) =sumn

)φ(t)R(t)n =

1minusRprime(f)=

[tn]ψ(t)R(t)n (246)

f =1minusradic

1minus 4xy

2y=infinsumn=0

)xn+1yn

radic1minus 4x)

suminfinn=0 rnt

fk =sum

k(nminus 1)

n0n1 rn00 r

[xm]fk =k

m[tmminusk]

fk =sum

k(nminus 1)

fk =sum

k(nminus 1)

f = x+G(f)

φ(f) =infinsumm=0

(φ(x)

(1minusGprime(x)

)G(x)m

) (261)

dmminus1

dxmminus1

(φprime(x)

) (262)

andψ(f)

1minusGprime(f)=

infinsumm=0

(ψ(x)

) (263)

infinsumn=0

)xnminusmψ(t)G(t)m

(ψ(x)

)=infinsumn=0

)xnminusmψ(t)G(t)m

dxmα(x)

m=infinsumn=0

)xjminusm

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

nsumm=0

)dmminus1

dnminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

)dmminus1

dnminusmminus1

)=dnminus1

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

with solution

f =1 + a

1minus ab

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

f =1minusradic

1minus 4xy

2y=infinsumn=0

)xn+1yn

radic1minus 4x)

suminfinn=0 rnt

fk =sum

k(nminus 1)

n0n1 rn00 r

[xm]fk =k

m[tmminusk]

fk =sum

k(nminus 1)

fk =sum

k(nminus 1)

f = x+G(f)

φ(f) =infinsumm=0

(φ(x)

(1minusGprime(x)

)G(x)m

) (261)

dmminus1

dxmminus1

(φprime(x)

) (262)

andψ(f)

1minusGprime(f)=

infinsumm=0

(ψ(x)

) (263)

infinsumn=0

)xnminusmψ(t)G(t)m

(ψ(x)

)=infinsumn=0

)xnminusmψ(t)G(t)m

dxmα(x)

m=infinsumn=0

)xjminusm

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

nsumm=0

)dmminus1

dnminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

)dmminus1

dnminusmminus1

)=dnminus1

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

with solution

f =1 + a

1minus ab

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

fk =sum

k(nminus 1)

f = x+G(f)

φ(f) =infinsumm=0

(φ(x)

(1minusGprime(x)

)G(x)m

) (261)

dmminus1

dxmminus1

(φprime(x)

) (262)

andψ(f)

1minusGprime(f)=

infinsumm=0

(ψ(x)

) (263)

infinsumn=0

)xnminusmψ(t)G(t)m

(ψ(x)

)=infinsumn=0

)xnminusmψ(t)G(t)m

dxmα(x)

m=infinsumn=0

)xjminusm

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

nsumm=0

)dmminus1

dnminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

)dmminus1

dnminusmminus1

)=dnminus1

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

with solution

f =1 + a

1minus ab

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

)= Dmminus1

(D(φGm)

)= Dmminus1

(φprimeGm

m+ φGprime

Gmminus1

(mminus 1)

Dmminus1(φprimeGm

)minusDmminus1

(φGprime

Gmminus1

(mminus 1)

φ+infinsumm=1

Dmminus1(φprimeGm

infinsumm=0

)minusinfinsumm=0

(φGprime

φ(f) middot ψ(f)

1minus zH prime(f)=

φ(f)ψ(f)

1minus zH prime(f)

nsumm=0

)dmminus1

dnminusm

dxn(φ(x)ψ(x)H(x)n

)(265)

)dmminus1

dnminusmminus1

)=dnminus1

G(f)k =infinsumm=0

(m+ k)m

dxmG(x)m+k

3 Applications

with solution

f =1 + a

1minus ab

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

(1 + a)r(1 + b)s

=sumni

)aitr(1 + bt)s+i

=sumnij

)ai(s+ i

)bjtr+j

=sumij

(r + j

)(s+ i

T (x) = xeT (x)

T (x) =infinsumn=1

nnminus1xn

n(321)

and more generally

T (x)k

k=infinsumn=k

knnminuskminus1(n

n(322)

k(n+ k)nminus1xn

n(323)

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

F (x) =

(1minus

infinsumn=1

(nminus 1)nminus1xn

)minus1 (324)

F (x)k

(n+ k)nxn

n (325)

nn+1xn

n (326)

U(x) =infinsumn=0

infinsumn=0

(nminus 1)nxn

eminusT (x)

(1minus T (x))2 (327)

(n+ 1)nxn

The seriessuminfin

infinsumn=0

(n+ k)nminuslxn

n= ekT (x)pl(T (x)) (328)

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

p1(u) =1

p2(u) =1

k2minus u

k(k + 1)

p3(u) =1

k3minus (2k + 1)

k2(k + 1)2u+

k(k + 1)(k + 2)

infinsumn=0

(n+ k)nminus1xn

(n+ k + 1)nminus1xn

n(n+ k)nminus2xn

ekT (x)(

k2minus T (x)

k(k + 1)

infinsumn=0

=infinsumn=0

(n+ k)nminus2xn

(∆ks)(n) =ksumi=0

(minus1)kminusi(k

)s(n+ i)

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

ksumi=0

(minus1)kminusi(k

)s(n+ i)

eminuskuinfinsumn=0

n= pl(u)

We have

eminuskuinfinsumn=0

n=infinsumn=0

(n+ k)n+mun

neminus(n+k)u

=infinsumj=0

jsumn=0

(minus1)jminusn(j

)(n+ k)j+m

jsumn=0

(minus1)jminusn(j

suminfinn=1 n

nminuslxnn

infinsumn=1

nnminuslxn

n= T (x)ql(T (x))

nnminuslxn

infinsumn=0

(n+ 1)nminuslxn

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

infinsumn=0

(n+ k)n+mxn

n= ekT (x)

rm(T (x) k)

(1minus T (x))2m+1

r0(u k) = 1

R(n j k) =1

jsumi=0

(minus1)jminusi(j

)(k + i)n

so thatinfinsumn=0

R(n j k)xn

n= ekx

(ex minus 1)j

j (3210)

infinsumj=0

(1minus u)2m+1

F (x)k+l

F (x)l

1minus T (x)

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

(x+ y)n =nsumi=0

(pn+1n

(pminus1)n+1

) which re-

cp(x) = 1 + xcp(x)p (331)

or equivalently

cp(x) =1

(1 + x)p

)〈minus1〉 (332)

)〈minus1〉

cprimep(x) =cp(x)p

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

cp(x)k =infinsumn=0

(pn+ k

)xn (333)

and thus by (246)

infinsumn=0

(pn+ k

cp(x)k

cp(x)k+1

(pn+ k

(1 + x)p

(1 + x)k+1

1minus (pminus 1)x

andinfinsumn=0

(pn+ k

(pi+ k

)middot l

pj + l

(pj + l

pn+ k + l

(pn+ k + l

and sumi+j=n

(pi+ k

)(pj + l

(pn+ k + l

a)b)〈minus1〉

1minus xc3

(1minus x)3

)= c2(x)

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

(j + pl

nsumi=0

)pi (335)

(pl + j

)xlinfinsumm=0

(pm+ k

cp(x)j+k

=cp(x)j+k

infinsumi=0

pixicp(x)(pminus1)i

=infinsumi=0

pixiinfinsumm=0

=infinsumn=0

xnnsumi=0

(pn+ j + k minus i

nminus i

(pl + j

)(p(nminus l) + k

nminus l

nsumi=0

(pn+ j + k minus i

nminus i

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

=infinsumn=0

(pn+ i)

infinsuml=0

(minus1)l(pn+ i+ l

=infinsumm=0

xmmsumn=0

(pn+ i)

((pminus 1)n+ i+m

mminus n

msumn=0

(minus1)mminusn(m

)((pminus 1)n+ i+m

mminus (j minus i)

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

msumn=0

(minus1)mminusn(m

(pminus 1)n+ j

msumn=0

(minus1)n(m

)minus1

uii+1(x) =1

uii+2(x) =1

(i+ 2)(p+ i+ 1)x

uii+3(x) =1

(i+ 2)(i+ 3)(p+ i+ 1)(p+ i+ 2)x

+(pminus 1)2

(i+ 3)(p+ i+ 2)(2p+ i+ 1)x2

an = 2(3n)

(n+ 1) (2n+ 1)= 4

n (2n+ 2)

infinsumn=0

(pn+ i)

n ((pminus 1)n+ j)

(1 + x)pn+i+1

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

)(niminus1

f = (1 + xf)(1 + yf) (341)

fk =infinsumn=0

=infinsumn=0

n[tnminusk]

)xiyjti+j

=infinsum

i+ j + k

(i+ j + k

)(i+ j + k

In particular

f =infinsum

i+ j + 1

(i+ j + 1

)(i+ j + 1

=infinsumn=1

nminus1sumi=0

=sumij

(r + i+ j

)(s+ i+ j

[(r + i+ j minus 1

)(s+ i+ j

)minus(r + i+ j

)(s+ i+ j minus 1

j minus 1

)]xiyj

=sumij

rs+ ri+ sj

(r + i+ j)(s+ i+ j)

(r + i+ j

)(s+ i+ j

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

fk =infinsumn=k

)= (minus1)i

(a+iminus1i

) (343) becomes

gk =infinsumn=k

(s1n+ i1 minus 1

f =(1 + xf)(1 + yf)

1minus zf

with the solution

2(xy + z)

and (343) gives

fk =infinsumn=k

sumi1+i2+i3=nminusk

)(n+ i3 minus 1

)xi1yi2zi3

f =infinsumm=1

AmeBmf (351)

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

in fk is

fk =infinsumn=k

n[tnminusk]

AmeBmt

)n=infinsumn=k

n[tnminusk]

=infinsumn=k

n[tnminusk]

(i1B1t)j1

(i2B2t)j2

=infinsumn=k

k(nminus 1)

ij11j1

4 Proofs

sumn fnx

dxg(x)k+1(k + 1) = 0

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

= res1

xn+1= res

φ(f(g))gprime

gn+1= res

φ(x)gprime

[xn]φ(f) = [tn]

(1minus tRprime(t)

)φ(t)R(t)n (421)

1minus tRprime(t)

= R(t)n minus t

dtR(t)n (422)

(u(t)minus t

nuprime(t)

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

= [xm] fkminus1R(f)

= [tm]

(1minus tRprime(t)

)tkminus1R(t)R(t)m

= [tm+1]

(1minus tRprime(t)

)tkR(t)m+1

P (t) =infinsumn=0

(tminus α)n

nP (n)(α) (431)

suminfinn=0(xt)

[xn] fk =k

n[tnminusk]R(t)n

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

which is (221)

suminfini=0 rit

n0 n1 n2

)(441)

ini (442)

suminfini=0 sif

(nminus 1)

)(443)

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

f =infinsumi=0

(nminus 1)

Lemma 451

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

mminus 2

)(461)

ifsumm

nminus 1

k minus 1 e1 en

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

)ways For each

(nminus 1)

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

nminus3

References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

suminfini=1 Aie

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

1 Introduction




22 Polynomials


24 A generalization



3 Applications





35 Raneys equation

4 Proofs

41 Residues

42 Induction

43 Factorization


45 Raneys proof


References

Documents

Introduction f x xG f x - Brandeis Universitypeople.brandeis.edu/~gessel/homepage/papers/Lagrange.pdf · Abstract. We give a survey of the Lagrange inversion formula, including di