Formal Power Series
Author(s): Ivan Niven
Source: The American Mathematical Monthly, Vol. 76, No. 8 (Oct., 1969), pp. 871-889
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2317940
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FORMAL POWER SERIES
IVAN NIVEN, University
ofOregon
1. Introduction.Our purpose is to develop a systematic theory of formal
power series. Such a theoryis known, or at least presumed, by many writers
on mathematics,who use it to avoid questions of convergencein infiniteseries.
What is done here is to formulatethe theoryon a properlogical basis and thus
to reveal the absence of the convergencequestion. Thus "hard" analysis can be
replaced by "soft" analysis in many applications.
John Riordan [4] has discussed these matters in a chapter on generating
functions,but his interestis in the applications to combinatorial problems. A
more abstract discussion is given by de Branges and Rovnyak [1]. Many examples of the use of formalpower series could be cited fromthe literature;we
mentiononly two, one by John Riordan [5] the other by David Zeitlin [6].
The scheme of the paper is as follows.The theoryof formalpower series is
developed in Sections 3, 4, 5, 6, 7, 11, and 12. Applications to number theory
and combinatorialanalysis are discussed in Sections 2, 8, 9, 10, and in the last
part of 11.
The paper is self-containedinsofar as it pertains to the theory of formal
power series. However, in the applications of this theory,especially in the application to partitionsin Section 9, we do not repeat here the fundamentalresults needed fromnumber theory.Thus Sections 9 and 10 may be difficultfor
a reader who is not too familiarwith the basic theoryof partitionsand the sum
of divisors function.This difficulty
can be removed by use of the specificreferences given in these sections; only a fewpages of fairlystraightforward
material
are needed as background.In Section 11 on the otherhand, the backgroundmaterial is set forthin detail because the source is not too readily available.
2. An example from algebra. To motivate the theory we begin with an
illustrationfromalgebra, to be found in Jacobson [2, p. 19]. Let qn denote the
number of ways of associating an n-producta1a2a3. . . an in a nonassociative
system. For example q3=2 because a,(a2a3) and (aja2)a3 are the only possibilities. Similarly q4=5 because of the cases aj(a2(a3a4)), aj((a2a3)a4), (aja2)(a3a4),
(a,(a2a3))a4, ((ala2)a3)a4. For n>2 it is easy to establish the recursive formula
n-1
(1)
qn =
j=1
qjqn-j,
q
by the followingargument.In imposinga systemof parentheseson aja2a3
..
. an
Professor
Nivenhas workedin numbertheorysincehis Chicagodissertation
on theWaring
problemunderL. E. Dickson.He was a fellowat Pennsylvania,
working
withH. Rademacher,
and
sincehas beenat Illinois,Purdue,and Oregon,interrupted
by leavestwiceto Berkeley.He was the
in 1960and has longbeenan activeparticipant
MAA Hedricklecturer
in theMAA. His sixbooks
includetheCarus MonographIrrationalNumbers
and (withH. S. Zuckerman)Introduction
to the
Editor.
Theory
ofNumbers.
871
[October
FORMAL POWER SERIES
872
to make it a well-definedn-product,we can begin by writing
(a1a2 ...
(2)
...
aj)(aj+laj+2
an).
Now the numberof ways of associating the product aja2
*ajis q; by definition, and likewise the second factor in (2) can be associated in qn-Jways.
Hence (2) can be associated in qjqn-jways, and formula(1) followsby considering the possible values forj. Now definethe power series
f(x)
(3)
-
qjxi.
j=l
Taking for granted (for the moment) the multiplicationof power series, we
see that forn 2 2 the coefficientof xn in {f(x) }2 iS
qlqn-1
+
+
q2qn-2
+
q3qn-3
* * * +
qn-lql.
But this is qn by (1), and so we see that {f(x)}2=f(x)-x
Solving this quadratic equation forf we get
f(x) = f = .{1 ? (1 -4x)
(4)
orf2-f+x=0.
1/2J.
The binomial theoremgives
(1- 4x)1/2
= 1
+
2
+ -(-4x)
2(21)
+
22)
2
~~2!2
..
(-4x)
2+
2f n + 1)
~
-
ni
.
(-4x)n +*X
The coefficientof xn here can be simplifiedby multiplyingnumerator and
denominatorby 2n to give
(1)(-1)(-3)(-5)
*
2ff-n!
(-2n + 3) _
(4)n
1*3*5 *
=_---.2n
(2n
(2n
-
3)
n!
-
2)!
(n )2n-1(n -1)!
(2n - 2)!
n!(n-1)!
In view of the minus sign here we see that (4) holds with the minus sign and
of Xn in (4) we get the simple formula
not the plus sign. Comparing coefficients
forqn,
(5)
qn
=
(2n
-
2)!
(-1
This analysis, however,leaves a numberof questions unanswered.Why can
on the two
we solve the quadratic to derive (4)? Why can we equate coefficients
1969]
FORMAL
POWER
873
SERIES
sides of (4) to obtain (5)? To avoid hard analysis in answeringsuch questions,
we now develop a theoryof formalpower series that involves no questions of
convergenceor divergence.At the end of Section 5 we shall returnto the question of the validity of the procedureleading to formula(5).
3. Formal power series. Define axto be an infinitesequence of complex numbers
Ca= [ao,al, a2,a3, ...
(6)
By P we denote the class of all such infinitesequences a, and these are the formal power series. There are three subsets of P that play a significantrole:
P,: those sequences a all of whose componentsaj are real numbers;
P1: those sequences a with aO = 1;
Po: those sequences a with ao = 0.
Although we have specified that the componentsaj in the elements of P are
complex nuinbers, the theory could be developed with the as in any integral
domain.
If I3CP, say ,3= tbo,bi, b2,b3,
], defineaddition by
a + f = [ao + bo,a, + bi, a2+ b2,*
Define multiplicationby
ai8 =
aobo,albo+ aob1,a2bo+ alb1+ aob2,,
n
E
ajbnj*
The definitionof equality is that a=j3 if and only if aj=bj forall j, i.e., j=O,
1, 2, 3,
It is not difficultto establish that the set P is a commutative ring witha
unit. The zero element and the unit element are
z = [0,0,
,0
O,
]
and u = [1, O. O.O.
the additive inverseof a is -a
Given any a = [ao, a,, a2, a3, *
[-ao, -a1,
-a2, -a3, . . . ]. The verificationof the associative propertyof multiplication
is not difficult,
and it is the only propertyof any depth in establishingthat P is
a commutativering.
Moreover,ao = z ifand only ifa= z orI =z. If a = z or ,B=z it is obvious that
af3=z. To establish the converse, suppose that ao4=z but aoz and j3z. Let
j be the least nonnegative integersuch that aj#O, and similarlylet k be the
least nonnegative integer such that bk#0. Then the component in the
(J+k+1)-th position in af3is
a,j
r-O
-=
ajbk.#0,
which contradictsaoj = z.
It followsthat if aoj = ay and a --z then j=y, and P is an integraldomain.
FORMALPOWER SERIES
874
[October
Given any a in P, there correspondsa multiplicativeinversea-' if there is an
elementa1 in P such that
= i =
Cl!-' = a-'
THEOREM
1. If a = [ao, a,,
a2,
[1, O,, O,
], ao
*
Proof. Denote a-' by [cO,cl, c2, ...
infinitesystemof equations
* *.
1 exists if and only
if ao#0.
]. We see that a a-'= u amounts to an
n
aoco
=
E
j=O
1, alco + aoc1 = 0, ...,
aCn
These equations can be solved successivelyforco,cl, c2,
=
0.
ifand only ifao0 0.
. Thenfor any
LEMMA2. Let 13CP1, so thatj3 is of theform [1, bl, b2,b3, *
. Also cl=n bi
c3,
positiveintegern we see thatO3nEPi, say 03n= [1, C1,C2, C
is an approand for each k ?2 we haveck= n bk+fn ,k (b1,b2,
bk1) wherefn,,k
in bi,b2, * * , bk-l
priatepolynomial
,
Proof. This result can be readily established by induction on n.
THEOREM3. Let a CPi, say a= [I, a1, a2, a3, . . . ], and let n be any positive
], such thatOn =a.
integer.Then thereis a unique 3E&P1,say ,B-[1, b1,b2,b3,
ln
Definea = .
Proof. Using Lemma 2 we can solve the equations
nb1= a1,nb2+ f2,n(b1)= a2, .
. .,
nbk+ fk,n(b1,
b2,
bk-1) = ak,
successively forb1,b2,b3,
THEOREM 4. For any positiveintegern and aCPi,
Define a -n = (a n)-1 and a0 = u.
we have (a-1)
=
(an)-1.
= u. (Anotherway
a-'
Proof.We see that a n(a-1)n =a * a *. . a a- *a(2-l'
of establishingTheorem 4 is to observe that P1 is a multiplicativegroup.)
n > 0. To any a (P1 therecorresponds
THEOREM 5. Let m and n be any integers,
3
=
amn/n.
a unique 13CP1 such thatatn=j3n, i.e.,
Proof. This is a corollaryof Theorem 3 with a in that theorem replaced
by atm.
4. A power series notation. Let X denote the particular element [0, 1, 0, 0,
]of P so that
0,
2=
AS
[OXOX1 OOX*l ..
*1]X
I =
[OJ0X 07 1, OXOX *.**.]
and in general Xn-1 is the sequence with zeros in all positions except the nth,
where 1 occurs. We now introducethe notation
00
(7)
, ajXi = ao + ajX +
i=O
a2X2 +
a3X'3 +**
19691
875
FORMAL POWER SERIES
]. What this amounts to is an agreement that a1 in (7)
fora= [ao, a,, a2,
stands for [aj, 0, O, O, . . . ] and that XO= [1, 0, 0, 0, . . . ]. Thus we are not
extendingthe integraldomain P to a vector space by introducingscalar multiplication; this could be done, but all we intend by (7) is an alternative, convenient notation forthe elementsof P. Thus z and u can now be writtensimply
as 0 and 1. The definitionsof addition, multiplication,and equality of elements
of P can be rewrittenas follows.With a as in (7) and
00
=
[bo b, b2,
. .
=E
j=o
bjXj,
then
00
a + ,B= E (aj + bj)X', a
j=O0jO
00
7
E akbf.k)
=E
k0
X0,
and a=,B ifand only ifaj = bj forall j = 0, 1, 2, 3,
For example, in the earlier notation we could write
L1, -1,
,0 ,0 ,
. . .
[1,0 ,0 ,0 ,0 , * * . J.
] =
]*[1, 1, 1, 1, 1,*
This can now be written as (1 -X)(1+X +X2+X3+ * * *) = 1, or (1 -X)-1
= 1 +X+X2+X3+
*-*. A general binomial theorem is established later, in
Theorems 11 and 17.
THEOREM 6. Let n be any positiveinteger,leta(=P, and 13Pr, so thata and ,B
are real sequences.If n is odd, aCn= gn implies a = 1. If n is even,an = 13nimplies
a=13 or a= -P.
Proof.We may presumea 0- and i#- O. For ifaz= 0, forexample, thenall = 0,
3n= 0 and so 13=0, a =,B. Let X denote the nth root of unity
w=
e2lifn =
cos(27r/n)+ i sin(27r/n).
ThenaO-fn3=o can be factoreda -n n = nlJ1(a-cif) = 0. If xi is notrealthen
a -cif3 0 because ai and ,Bare real sequences with ci O0 and j3#0. If n is odd,
coiis real only in the case j = n and hence
n#= X0, a
a -
- 8 = O, a
=
.
If n is even, xi is real in the two cases j = n and j= n/2, leading to the conclusion
that ca= 3 or ca= -1.
Consideran infinite
sequencecil,a12, 03,
co
(8)
ak
=
jiO
ajkX',
*
-of
elementsofP, say
k =1, 2, 3, *
A sequenceoil, a2, a13,
DEFINITION.
as in (8) is said to be a sequence
toany integerr ? 0 thereis an integerN = N(r)
admittingadditionif corresponding
such thatfor all n> N, ao=aln=
= arn = O
a2n = .*
876
[October
FORMAL POWER SERIES
If this condition is satisfiedwe also say that 2aj is an admissiblesum, and
we can write
00
D
ai
=
ZSTX
,
j=1
where foreach integerr ? 0 the coefficientST is the coefficientof Xr in the finite
i.e.,
+aN,
sum a1+a2+
Sr =
+ arv.
ar2 +
arl+
ofXr in every finitesum oz1+az2+ *
We note that s, is the coefficient
n_N.
+az with
be a sequenceofelementsofP admittingaddition.
LEMMA 7. Let aol,aZ2, a*3,
of theao'sin thesense thatgivenany j there
Let /31,32, /33,* * be a rearrangement
is also a sequenceadmitting
existsa unique k such thataj = /3k. Then /,32, /3*,
addition,and
al +
+ /2 + f
=/1
*
a3 +
!2 +
+
large the
Proof. Let r be any given nonnegative integer.For n sufficiently
ofXrin the finitesum
equals the coefficient
of'X in a1 +a2 +a3 + *
coefficient
...
+a,. Similarly for n sufficientlylarge the coefficientof Xr in
al +a2+
And clearly
...
+/3n.
...
equals the coefficientof X' in f1+/2+
+
31+/2+/3
. ..
...
+i3 have identical termsin Xr.
+a, and /1+/2+
al+a2+
Next we get a resultanalogous to Lemma 7 formultiplication.Consider an
of elements of P of the form
infinitesequence yi, 'Y2 'Y3..
00
(9)
Yk -
k = 1 2, 3,*.
cik Xi,
j=1
Note that the sums begin withj= 1. If this is a sequence admittingaddition,
then we say that the related sequence
(10)
I +
71X1
+
+
Y2,I
73,
is a sequence admittingmultiplication.Furthermore,we write
co
][
kl1
(I +
-Yk)
=
+
00
E
jl1
qj Xi,
of XT in any finiteproduct fl[k (1 +'k) with n suffiwhere q, is the coefficient
ciently large that Cjk=0 for 1 <j<? r if k> n. Then it is clear that we can state
a result analogous to Lemma 7 as follows:
so is any rearrangeLEMMA 7a. If (10) is a sequenceadmittingmultiplication,
ment1+&1, 1 + 82, 1 + 83, * * * of (10), and
1969]
877
FORMALPOWER SERIES
co
00
II (1 + a*) = II (1 + h).
k-1
k=l
5. Formal derivatives. Given any a in P, say a =
derivative D (a) and the scalar S(a) by
j O aj Xi, definethe
00
(11)
D(a)
=
S(a)
EjajXi',
j=1
= aO.
Define D2(a) =D(D(a)),
and in general for any positive integer n, the nth
derivative is Dn(a). Taking D(ax) =a for convenience, we can now write a
McLaurin series expansion.
THEOREM
8. a =
nZ0(1/n!) S(Dn(a))*Xn.
The proofof this is quite easy.
3EEP thenD (a +f3) = D (a) +D(Q3) and D (a 3) = aD (13)
=
and
nan-ID
(a) for any positiveintegern. Also if ca-I existsthen
+3D(a),
D(an)
=
and
D
-_c-2D
D(at1)
(a)
(a-n) = _na-- 1D (a).
THEOREM 9. If aSEP,
Proof.The formulaforD (a 13)can be establishedeasily by comparingcoefficients of Xn.By using inductionon n we get the formulaforD (an). Next if we
a a-1= 1 we get the formulaforD(ac-1). Finally, a-n = (a1-)n can
differentiate
be used to write
D(a-n) = D((a-1)n) = n(a-1)f-1D(c-1) =
na-n-D(a).
THEOREM 10. Let aEEP, so thatS(a) = 1. For any rational numberr, D(ao)
-
rar-1D (a).
Proof. By Theorem 5 there is a unique meaning fora,'. If r = m/n where m
and n are integerswe can write
D((a-r)n)
= n(acr)n-1D(a%)
D((ar)n)
= D(alm) = man-lD(a)
by Theorem 9. The result followsat once.
A simple version of the binomial theorem can be easily obtained from
Theorems 8 and 10, as follows:
THEOREM 11. For any rational numberr and any complexnumberk,
(1 + kX)r= 1 + r(kX)+ r(r- 1) (kX)2 +
2!
P
f F1)
(r- 2)..
+(rkn+
)=
Proof. First note that D(1 +kX)" ==r(l +kX)r-1D(1 +kX) = rk(1+kX)1'-, and
878
[October
FORMAL POWER SERIES
so by inductionon n,
Dn(1
+
kX)r
= r(r -
1)(r
(r - n + 1)kn(1 +
2)
-
kX)?1t.
is a unique element of P1 by Theorems 3 and 5, and so
Now (1 +kX)r-n
S(1 + k/) r-n = 1. It followsthat
S(Dn(I
+ kX)I) = r(r - 1)(r
-
* (r - n + 1)kn.
2)
Now use Theorem 8 with a replaced by (1 +kX)r, and the result follows.
The formof the binomial theoremjust established is sufficientin most applications,forexample, to justifythe argumentgiven in Section 2. To see this,
we replace equation (3) with this definitionof a,
00
a =
j=1
qXi)
wherethe qj have the same meaningas in Section 2. Then the analysis following
equation (3) leads to a2 = a -X. From this we can write 4a2 - 4a + 1 = 1- 4X, or
(1
2a)2
-
((1 -4X)
=
12)2
By Theorem 6 it followsthat 1- 2a = (1- 4X)1/2,and so by Theorem 11 we
conclude that
1
-
2q,X -
2q2X2 -
1+
2q3X3 -
- (-4)X) +
2
2
(3
2!1
(-4X<)2 + '(?-2
3!
2
(4X)
From the definitionof equality in Section 3 we can now equate the coefficients
of Xnto get equation (5).
We want to get a more general formof the binomial theorem,namely the
expansionof (1 +a)r wherea EP0, so that S(a) = 0. To do thiswe definea formal
logarithm.But firstwe establish one more resultabout derivatives.
is an admissiblesum of elementsof P in
THEOREM 12. If a1+a2+a3+
thesense of Section4, then
+ D(a2) + D(a3) +*a
D(al + ae2+ aS + *aa)=D(ai1)
.
Proof. For any nonnegativeintegerr the coefficientof XT in the infinitesum
ofXT in the finitesum a,+a2+
* * * +an
* * * equals the coefficient
al1+a2+as3+
provided n > N= N(r). Hence the coefficientsof 'r-l are equal in the equation
in Theorem 12. But this holds forall r, so the result follows.
6. Logarithmsand the binomial theorem.A formallogarithmis not defined
forany element of P, but only foraeCP1, so that S(a) =1. For any aEGPI, say
a = 1 +,B with j3EP0, define
L(a) = L(1+i3)
=
22 +
I3
- i4
+
1=
(1)
ji-
j++1i/j
19691
879
FORMAL POWER SERIES
notingthat thisis an admissiblesum as in Section 4. Thus L is a formallogarithmic functionfromP1 to Po.
THEOREM
13. D(L(a))
=O'1D(a).
Proof. With a = 1 +1 we use Theorem 12 to write
D (L
D (L(1 +
))
[-
3
4
= D(A) + D(- 1 2) + D(13#3)
+ D(=
D(fl) - #D(fl)+ 32D(Q)
=D(fl) [1 - ' + ,B2 - #3 +
434)
+
*
f3D(03) + * -
-
.
]
= D(8) *(1 + 1)-1 = D(a)* ccl
because D(a) =D(3)
THEOREM
14. If
by definition.
aCEfPP
and 'yEPi thenL-(cy) =L(a)+L('y).
Proof. We use Theorems 13 and 9 to observe that
D(L(a'y)) = (ay)-4D(ay) = (ay)-l{aD(y) + yD(a)J
= &c'D(a) + y-'D(y)
= D(L(a)) + D(L(y))
= D(LQ(a) + L(y)).
Now L(xy) and L(a) +L(7) are elementsin Po, and it is clear fromthe definition
of a derivative that if 0,GPo and 02CPo and D(01) = D(02), then 01=02
THEOREM
15. For any rationalnumberr, L(a,) = rL(a).
Proof.By definitionL(1) = 0. Then a *a' = 1 impliesL(a) +L(ac-) =L(x a-')
=L(1) =0 and so L(a-')=
-L((a). For any integern we have L(an) =nLQ(a) by
induction. If r = m/n where m and n are integerswe see that
mL(a) = L(am) =
L((ar)n)
= nL(ar).
THEOREM 16. L(a) = 0 if and onlyif a =1. Also if L(a) = L(O) thena =j.
Proof. If L(a) =0 then D(L(a))
and hence D (a) = 0 and a = 1.
=D(O) =0 and so ao'D(a)
=0. But a'-00
THEOREM 17. If r is rational,if 1 is an elementof Po so that S( 3) = 0, then
(12)
(1 + fB)r 1 +
r,8 + r(r
r(r-1)(r-2)
2!
1)
#2 +...
* * * (r-n
Proof. For conveniencewe write
+1)
n
880
FORMAL
(r
POWER
...
r(r-l)(r-2)
[October
SERIES
(r-n + 1)
Let y denote the rightside of equation (12) so that
D(y)
=
j( >i
D(j3)
>j
=
(1+3)D&) D(y3)
-
i-1 + D(13) E
D(3)zj (')'-r
+ D(f) E (ji-i1)
1
jl
j=-21-
= D(l) r + D(O)
j1+ _
+
(j)
(j
D(3) r+ D(3)
-rDQ3)
(r)
r(j
_
)
=ryD(,8).
Multiplyingby y-'(l +3)-l we get Ty-D('y) = r(1 +3)-'D()
= r(1 +3)-D(1
+O).
But D(L('y)) =y-'DQ(y) and D(L((1+j3)r)) =D(rL(1+3))=r(1+3)-1D(1+3),
and so D(L('y)) =D(L((1+f)-r)). Since L('y) and L(1+13)r are in Po it follows
that L(y) =L((1+13)r), and so y= (1+1)r by Theorem 16.
7. The exponential function.Let f3be an element of Po, so that S(
Then we define
+ - + -o+
E(f) = 1 +
2!
31
=
0.
F3 n
n=~on
so that E is a functionfromPo to Pi. Since E(J3),as defined,is an admissiblesum,
we can apply Theorem 12 to get
D(E(#)) = D()
THEOREM
{1
+:
+ -
3!
}
D)E
18. If EQ(3)=E('y) then 3 ='y.
Proof. We observe that D (E(3)) = D (E(Qy)), so that D (f) *E () = D (y) *E (y).
But E(3) #0 so that E(3) and E(y) can be cancelled giving D f3)= D (y), and
hence i ='y.
THEOREM 19. If 13CPo thenL(E(Q)) =f. If oaCPi thenE(L(a)) =a. Thus L
and E are inversefunctions,L beingone-to-one
fromPi ontoPO, and E one-to-one
fromPo ontoP1.
19691
881
FORMAL POWER SERIES
Proof. By Theorem 13 we see that
D(L(E(s3))) = {E(3)}-'.D(E(Q))
= D(f).
E(fl)}-1.E(0).D(0)
={
It followsthat L(E(3)) =f3. Next, given any a in Pi suppose that E(L(a)) =a,.
Then L(E(L(a))) =L(a,) and so L(a) =L(a,). Hence a = a1 by Theorem 16.
THEOREM
E(,y).
20. Given0CEPo, yEzPo,thenE(3+'y) =E(Q)
Proof. By Theorems 14 and 19 we see that
L(EQ(3)*E(y)) = L(EQ())
+ L(E(y)) = A + y.
Taking the exponential functionof each side, and using Theorem 19 again, we
get the result.
By Theorems 15 and 19 we see that arc=E(rL (a)) for any ac-Pi and any
rational r. This equation we take as the definitionof a r forany complex number
= ar+8 followat once forcomplex
r, so that such propertiesof exponentsas a"r a8W
we note that Theorem 10 can
this
s.
Also
of
definition
by use
numbersr and
to
number
thus
r;
be extended any complex
D(ar) = D(E(rL (a))) = E(rL (a))
D(rL(a))
= arra-clD((a)
=
rar-lD(a).
Also Theorem 15 extends to any complex r by use of Theorem 19. Finally,
Theorem 17 holds forcomplex r; in fact the proofof this result needs no alteration forthis generalizationin view of the extendedversionsof Theorems 10 and
15 just mentioned.
8. An application to recurrencefunctions.For any given a, b, xo, xl define
a sequencexo,X1,
relationxn+i=axn+bXn-ifor
by the recurrence
X2, X3, - * *
n = 1, 2, 3, * * *. The Fibonacci sequence is thespecial case witha = b = xo= xi = 1.
The problem is to determinexn explicitlyin terms of a, b, xo, xi. If we define
a=Xo+XiX+X21X2
+ X3X3 +
a
(13)
* * *
we see that
aXa - bX2a = xo +
-
we see that (13) can be writtenas
If ki and k2are the roots of k2-ak-b=O
a(l
(14)
k,X)(1
-
k2X) = xo +
-
axo)X.
-
(x1
(xi
-
axo)X.
CASE 1. Suppose that k1= k2.Then we see that
a = Txo+
(15)
(xi -
axo)X} (1
-
k1X)-2.
Now by Theorem 11 or Theorem 17 we have
(1-k1X)
-2
1
+
2
~~~~2
2kiX + 3kX
+ 14kl+5kX
4 4
+***,
and so equating coefficientsof XI in (15) we get
(16)
= xo(n + 1)kt + n(x1
x.A
Xn= nXlkl
-
(n
-
-
axo)k"1
n
1)xokl.
or
882
[October
FORMAL POWER SERIES
CASE
2. Suppose that k134k2.Multiplyingthe identity
ki-k2 =ki(I
k2X)-
kiX)
k2(1-
by (1 - kX)-l (1 - k2X)-l we get
(k1 - k)(1
-
k,)'l-
k1(l
k2)=
-
ki)1-
k2(1-
l
k2-
Multiplyingthis into (14) we have
(17)
(k1 - k2)a
xo + (xi
=
-
axo)X {k1(l
-
k2(1
ki)1-
Also we use kl(l-ki'X)1'=ki+k2lX+kl2+k4jX3 +
ing coefficientsof 'XI,in (17) we have
(k1-
X =
xo(k"+
-
kX-
-
V
.
Equat-
k`+ ') + (xi - axo) (kn-kn
or
(18)
x0(kn+~1 kn1
X=
+ (xi
-
axo)(kun&
-n)/k
k2).
The results (16) and (18) are well known; an alternative derivationis given in
[3, page 100]. An entirelydifferentway of treatingequation (13) is as follows.
We can write
a
(19)
(1
=
-
aX -
+ (xi
bX2)-l{Xo
-
axo)XI[
Now by Theorem 17 we have
(1
-
aX - b2-l
1 + (aX + WA)+ (aX + bX2)2+ (aX + bA2)3+
The coefficientof Xnhere is
an +
+
)an2b
(f
(fl
n2 a4b 2 +
)an-61
(f
+
.
of 'Xnin (19) gives therefore
Equating coefficients
=
(XO
xo
n
j=
-2
-j(I(n-1)
a?&2ibi+
(xi
/2]
axo)
/n
-
-1)an-12ibi.
Finally, let us returnto the method used for deriving (16) and (18). This
method can be used with recurrencerelations of higher order. Consider for
example any given real (or complex) numbersxopXl, X2,a, b, c and a recurrence
relation
n
Xn2~ axn+l + bXn+ CXn-1,
If we definea= xo+x1X+x2X'+x3V'+
(20)
a(l
-
aX - bX2- CX3)= xo + (xi
=
1, 2, 3,
we note that
-
axo)X+ (X2- ax,
-
bxo)>%2.
1969]
883
FORMAL POWER SERIES
has roots kl, k2,k3say, then (13) can be re-
If the equation k3-ak2-bk-c=O
writtenas
(21) a(1
k1X)(1-
-
= xo - (x
k3X)
k2X)(1 -
-
axo)X+
(X2-
ax1- bxo)X2.
There are now threecases depending on the nature of the roots ki, k2,k3: three
equal roots, two equal roots, or distinct roots. The case of equal roots follows
the pattern of equation (15),
CZ= [xo+ (x1 - axo)X+ (X2
ax1 - bxo)X2]. (1
-
kX)-3.
In the other two cases it is a matterof partial fractionexpansions, in the sense
that constants ql, q2, q3, q4, q6, q6 can be found so that
(1
- k,X)-2(1
(1-A)-'(l
- k2X)-
- k2X)-(1
ql(l
=
- k3X)-l
=
kiX)-1+
kiX)-2 + q3(1
q2(1-
- k2X),
q4(1-kX)-?+qb(l-k2XA)-I+q6(1-k3\)-li
in the case of two equal roots or the case of distinct roots, respectively.
For example if a=6, b=-11, c=6 then we find that ki=1, k2=2, k3=3,
q4=2, q5= -4, q6=9/2. Then (21) impliesthat
a = [xo+ (xi
Xn =
6xo)X+ (X2- 6x, + JJx0)X2]
*[(I - X)-1 - 4(1 - 2X)-1 + 9 (I
X,(2 - 4-2n + -&3-3n)+ (xi - 6xo)(2 - 4,2n-' + 9.3n-- )
+ (X2- 6x1 + 11xo)(I - 4.2 n-2 +
-
3-)-]
2.3n-2)
9. An applicationto partitions.The notation p(n) representsthe numberof
ways that a positive integern can be writtenas a sum of positive integers.Two
iftheydifferonly in the orderof theirsummands. As
partitionsare not different
usual, we definep(O) = 1.
for every positive integer j. Then
Let aj denote 1+Xi+X2i+X3i+ *
* is a sequence admitting multiplicationin the sense of (10) in
ali, a2, a03.,
Section 4. By the standard argument,forexample in [3, pp. 226, 227], we have
00
(22)
o1 -a2 -a3
.-
I
i=l_
But also we see that a,(1 -Xi)
=1
00
(23)
00
=-H ac
so that aj
p(k)xk.
k=O
(1-Xi)-', and
00
HaI = 11(1 -Xi)-'.
jl1
j=-
Next let qe(n) denote the numberof partitionsof any positive integern into
an even numberof distinctsummands,and similarlylet q0(n) be the numberof
partitionsof n into an odd number of distinct summands. It is customary to
take qe(O) =1 and q0(O) = 0. Then the coefficientof Xn in the expansion of the
admissible product
884
[October
FORMAL POWER SERIES
=
***
(1-X)(1-X2)(1-X)
Il (1i-xi)
i=1
is seen to be qe(n) -qO(n) by a simple combinatorialargument. It followsthat
II (1 00
(24)
j=1
00
X') =
E
n=O
{qe(n) - qO(n) xn.
By use of graphs of partitions it can be proved, cf. [3, pp. 224-226], that
qe(n)-qO(n) = (-1)i ifn is of the form(3j2+j)/2 or (3j2-j)/2 forsome nonnegative integerj, and qe(n) -q?(n) = 0 otherwise.It is easy to prove that the sets of
positive integers
{(3j2
}, {(3j2
+ j)/2; j = 1,2,3,*
- j)/2; j = 1, 2, 3,*
are distinct,and hence (24) can be writtenas
00
11 (1
(25)
xi) = 1
-
00
+
= 1-
E
(-l)i(X(3w2+i)/2 + X(3i-j)j2)
+ XI
X-X2
X12 - X15 +
+ X7
This with (22) and (23) implies that
{1 + E
(1X
+
(-l)i(X(3i2+i)/2
- X2 + X
E
_(3j2_i)12)}
p(k)x\k =
)Ep(k)Xk
+ X7_ X12 - X15 +
1
=1.
ofXn on the leftside of this equation is
For any positive integern, the coefficient
*
p(n)-p(n-1)-p(n-2)+p(n-5)+p(n-7)-p(n-12)-p(n-15)+
Thus we have proved the followingwell-knownresult of Euler [3, p. 235].
THEOREM 21. For any positiveintegersn,
p(n) =p(n
-
1) + p(n
-
2)
co
=
E
j=1
(_-)i+l{p(n
(3j2
-
p(n
-
5)
-
p(n
+ j)/2) + p(n
-
-
7) + * *
(3j2
- j)/2)1
withp(t) =0 if t<O, so thatthesum is finite.
It should be emphasized that the proofgiven here of Theorem 21 is not new.
The proofabove is simplythe usual one formulatedin termsof the "soft" analysis of formalpower series.
10. An applicationto the sum of divisors function.For any positive integer
n let o-(n) denote the sum of the positive divisors of n; forexample o(6) = 1+2
+3+6. We establisha knownrecurrencerelation [3, p. 236] fora(n), and again
the positive integersof the form(3k2-k)/2 and (3k2+k)/2 play a role, namely,
the positive integers1, 2, 5, 7, 12, 15, 22, 26,
FORMALPOWER SERIES
1969]
885
THEOREM 22. For any positive integer k,
a (k) -
a (k - 1)
cr(k- 2) + a (k - 5) + ar(k- 7)F(-1)J1k ifk = (3j2 + j)/2 or k = (3j2 - j)/2,
-
O otherwise.
Proof. Define f = I=
-D(L(/3))
=
(1 -Xi) so that L(3) = ,t
1
-
L(1-Xi),
k
=
, j(l
-
j=l
k
{jX0-l + jX2i-1 + j\3l
=
j=l
+ jX4j--+
..
.
oo
= ?f(n)n-1,
n=1
wheref (n) is seen to be the sum of all positive divisorsof n that do not exceed
k. Thus we have f(n) =o (n) if n < k, and so we can write
oo
k
2 c(n)Xn-1 +
-f?r1DQ3)=
(26)
n=1
?
f(n)Xn-1.
n=k+l
Now equation (24) can be writtenwith a finiteproduct
(27)
=
.
k
II (1-
j=1
X) =E
o
n=O
, q()
-
() I X
where q,(n) denotes the numberof partitionsof n into an even numberof distinct summands <k, and qk(n) denotes the number of partitionsof n into an
odd number of distinct summands ?<k. Define qk(O)=1 and q?(O) =0. If n < k
we note that qk(n) =qe(n) and qk(n) =q0(n), so (27) can be writtenas
k
(28)
E=
?
n=O
qe(n) -
coo
qo(n)} X + E
n=k+l1
{ qe(n) - qo(n)}X
We now equate the coefficientsof X-k-1in -D(O) and in the product
3(-3-'D(f3)). From (28) it is clear that the coefficientof Xk-1 in -D(j3) is
-k{qe(k)
- q?(k)}
= [-(-)Wk
if k
O otherwise.
=
From (28) and (26) the coefficientof X-'-1in f(-f-1D(,8))
o-(k){qe(O)
a
-
qO(O)} + oa(k - 1){qe(l)
=
s(k)o-ae(kh-e1)o-p(kr-2)
e(kd-5)
.(k-7)
and so the theoremis proved.
-
(3j2
+
j)/2
is
qO(1)} + oa(k - 2){qe(2) - qO(2)} +
+
+
886
[October
FORMAL POWER SERIES
11. Trigonometricfunctionsand differentialequations. We now returnto
the general theoryof formalpower series and make the definitions
{ E(ix)
sina
-
E(- ia) }/2i
00
=
E
{1(-1)ka2k+l /(2k
+ 1)!
k=0
{E(ia) + E(-ia))}/2
Cosa
=
E
{(_1)ka2k}/(2k)!,
k=O
where a is any element in Po. Thus sin a is in Po, but cos a is in P1, so we can
definesec a = (cos a)-1 and tan a = (sin a) (cos a)-1. However, we cannot now
definecosec a and cot a, but in the next section we extend the theory,to ennow apply, such
compass these two functions.All the rules of differentiation
as D (sin a) = (cos a)D (a).
equations with conThe standard theoryof homogeneouslinear differential
stant coefficientsis valid. For example, in the second order case, let a and b be
any complex numbers,and let r1and r2 be the roots of x2+ax+b = 0. Then a
solution forp in P of the equation D2(p) +aD(p) +bp =0 is
p
=
clE(r,X) + c2E(r2X)
with arbitraryconstants c1 and c2. It is easy to prove that this is the general
solutionifr15 r2. If r1= r2the generalsolutionis of coursep = c1E(r1X)+c2XE(r1X).
We now give a briefsketch of the use of a differentialequation to solve a
combinatorialproblem,as in Andre [7, p. 172]. Our approach differsfromthat
of Andre in that we treat the differentialequation in a purely formalsense,
which he did not. For n> 2 let bnbe the number of permutationsa,, a2, . . .
, n such that aj> aj-, ifj is even, and aj < aj-, ifj is odd. Call such
an of 1, 2, *
a permutation an E-permutation. Similarly, say that a1, a2, . . ., an is an
0-permutationof 1, 2, *. . , n ifaj > aj-, ifj is odd, and aj < aj-1 ifj is even. Note
n+l-a2,
that if a1, a2, * * *, an is an 0-permutation then n+1-ai,
n+1-an is an E-permutation,and conversely.Thus thereis a one-to-onecorrespondence between E-permutations and 0-permutations; there are bnof each
type. Define bo= 1 and b1= 1.
Next, consider the numberof 0-permutationswith a1= n. It is not difficult
to see that thereare bn-1 of these. Also, thereare no E-permutationswith a1= n.
Turning to permutationswitha2 = n, there are no 0-permutationsof this type.
However, the number of E-permutations with a2 = n is (n - I)bn-2, or what is
the same thing (n-1)bibn-2; the reason for this is that a1 can be any element
among 1, 2, * * * , n-I and the restcan be set up as a3, a4, . . ., an in bn2ways.
A similarargumentshows that thereare no E-permutationswitha3 = n, whereas
the number of 0-permutationswith a3 = n is (n)b2bn-3.Thus by consideringall
E-permutations and all 0-permutations with successively a, = n, then a2 =n,
relation
thena3=n,=
and finallyan= n, we are led to the recurrence
*
,*
2bn=
n-1
j=o
(
1)
1
b n_j_or 2ncn =
n~t-1
C;Cn-j-1,
j-o
1969]
POWER
FORMAL
SERIES
887
where cnis definedas bn/n!forall nonnegativeintegersn. Taking a to be the
formalpower series
co
a
=
ECnXn
n=O
we can readily verifythat the differentialequation 2D(a) = a2+1 holds. Now
it is easy to verifyfromthe definitionsof the formaltrigonometricfunctions
that sin2X+COS2X = 1, sec2X = 1 +tan 2X, D (tanX) = sec2 X, D (secX) = sec X tanX.
Thus the unique formalsolution of the differentialequation is a= tanX+secx.
(Andre gives the solution of the differentialequation as a =tan (X/2+7r/4)
which has no meaning in our formaldefinitionof the trigonometricfunctions.
The usual formulafortan(a +f) in termsof tan a and tan A is valid, but tan 7r/4
=1 cannot be established in the formal theory. In fact tan 7r/4is not even
definedbecause 7r/4is not an element of Po, although it is an element of P.)
Thus we have
= tan X + sec X.
bnXn/n!
CZ=
Now the power series for tan X has odd powers of X only, with coefficients
closely connected with the Bernoulli numbers [8, p. 268]. Similarly the power
series forsec X has even powers of X only, with coefficients
related to the Euler
numbers [8, p. 269]. Thus Andre was able to relate the combinatorialnumbers
bn to the Bernoulli numbers for odd n, and to the Euler numbers for even n.
(A differentapproach to this problem has been given recentlyby R. C. Entringer [9].)
From our point of view in this paper, the importantaspect of this is that
Andre's conclusions can be drawn with only a formaluse of calculus and differentialequations and without any convergence questions in the use of a2,
the square of a power series, in the differentialequation. The series expansions
fortan X and sec X come fromthose forsin X and cos X, and these are defined
in termsof the exponential functionsE(iX) and E(-iX). The formalstructure
carriesthe entireargument,with no need forthe classical infinitesimalcalculus.
Of course, such relations as sin2 X+cos2 X= 1 have meaning only in terms of
formalpower series in this context and not in termsof the geometryof rightangled triangles.
12. Extension to a field.Since the set of formalpowerseries P is a commutative integraldomain, it can be imbedded in a fieldP* in the classical manner by
use of pairs of elements, cf. [2, pp. 87-92]. This construction is very well
known in the extensionof the integersto the rational numbers.Thus P* is the
fieldof all pairs (a, O) with aCP, fE3P and A0. Addition and multiplication
are definedby
(all, 01) +
(a1i 1)
)
(a2, l2)
=
* (C2, I 2) =
(alfl2 +
a2f1, /012),
(a1la2, #13@2).
888
FORMAL
[October
POVWER SERIES
Two elements(a,, #I3)and (a2, 32) are said to be equal ifand onlyifai/2=a2013
If 3 = 1 we agree to write a for (ae, (a) (a, 1), so that P is a subset of P*.
Similarly we agree to write
(29)
ajXi'X,l- as
j=.
z
j=o
ajXir,
where r is a positive integer.We prove in Theorem 23 that every element of P
can be writtenin this way, so that P* can be thoughtof as the class of Laurent
power series expansions, with a finitenumber of negative exponents allowed.
To do this we firstdefinethe degree of a for any a in P, aO 0. If a =2ajXi
then the degree of ae,writtendeg(ae), is the subscriptof the firstnonzero coefficient in the sequence of coefficientsao, a,, a2, * a . If a, and a2 are nonzero
elementsof P it followsthat deg(ala2) = deg(al) +deg(a2). This definitionis extended toP*as follows:if (a,1)CeP*witha-oe#OtOhen deg(a, j) =deg(Q)-deg(3).
Degree is well-defined,because if (al, t1) = (a2, 32) thena1/2 =a2f1 and so we have
deg(al) + deg(l32) = deg(a2) + deg(Q,),
deg(al)
deg(Q,) = deg(at2)
-
deg(Q2).
-
Next for any (a, f3) in P* with a0O, let deg(a)=m, deg(3)=n so that
deg(a, O)=m -n. Then we see that j =Xn3,1where j1 has degree 0, so that f1
has an inverse. It follows that (a, )=(a,
'Xnll) = (c43B1 Xn). Now ac-' has
degree m, so it can be writtenin the form
cali
=
j=m
ajX
am 0 0.
Thus we have
00
(30)
(a,,6)
=
EajX~n
am.
0,
j=m
by virtue of (29).
THEOREM
unique.
23. The representation
(30) of any nonzeroelement(a, O) of P* ts
Proof. Suppose that (a, i3) can also be writtenas
00
(a,X :)
=
~E
CjX'n,
Ch ? 0.
j=h
By the invariance of degree under differentrepresentationswe see that m-n
=h-n and m =h. Also we have
(o,
= (n=
a
(fE C2jx Xkn)
1969]
FORMAL
POWER
SERIES
889
and so by the definitionof equality in P*,
00
=
.E ajXi+n
j=m
00
?
j-m
cjxj+n.
The theoremfollowsby the definitionof equality in P.
In the preceding section we saw that the trigonometricfunctionssin a,
cos a, tan a, and sec a could be definedforany elementa of P, but not cosec a
and cot a. If a$O0 we can definethe latter two functionsfromP to P*; thus
cosec a = (1, sin a),
cot a = (cos a, sin a).
A simple calculation shows that
cosec X = X-1+ (X/6)+ (7X3/360)+
Finally, we note that the theoryof formalpower series, developed here in
analogy to power series in a single variable, can be extended in a similar way
to the multiplevariable case.
byNSF GrantGP 6510.
Worksupported
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1. Louis de Brangesand JamesRovnyak,Square SummablePowerSeries,Holt, Rinehart
and Winston,New York,1966.
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to theTheoryof Numbers,2nd ed.,
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, Combinatorial
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Wiley,New York,1968.
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alternees,
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New York,
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interpretation
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