COEFFICIENTS IN POWERS OF THE LOG SERIES
Donald M. Davis
Department of Mathematics, Lehigh University, Bethlehem, PA, 18015, USA [email protected]
Received: 5/7/10, Accepted: 12/30/10, Published: 2/4/11
Abstract
We determine thep-exponent in many of the coefficients of�(x)t, where�(x) is the power series for log(1+x)/xandtis any integer. In our proof, we introduce a variant of multinomial coefficients. We also characterize the power series x/log(1 +x) by certain zero coefficients in its powers.
1. Main Divisibility Theorem
The divisibility by primes of the coefficients in the integer powers�(x)tof the power series for log(1 +x)/x, given by
�(x) :=
∞ �
i=0
(−1)i x i
i+ 1,
has been applied in several ways in algebraic topology. See, for example, [1] and [4]. Our main divisibility result, Theorem 1, says that, in an appropriate range, this divisibility is the same as that of the coefficients of (1±xp−1
p )
t. Herepis any prime andtis any integer, positive or negative. We denote byνp(−) the exponent ofpin an integer and by [xn]f(x) the coe
fficient ofxn in a power seriesf(x).
Theorem 1. If tis any integer and 1≤m≤pνp(t), then
νp�[x(p−1)m]�(x)t�=νp(t)−νp(m)−m.
Thus, for example, if ν3(t) = 2, then, for m = 1, . . . ,9, the exponent of 3 in [x2m]�(x)tis, respectively, 1, 0,−2,−2,−3,−5,−5,−6, and−9, which is the same as in (1±x32)
t. In Section 3, we will discuss what we can say about ν
p([xn]�(x)t) whennis not divisible by (p−1) andn <(p−1)pνp(t)
.
complex projective spaces described in [4]. The coefficients studied here can be directly related to Stirling numbers and generalized Bernoulli numbers ([3, Chapter 6]), but it seems that our divisibility results are new in any of these contexts.
Proving Theorem 1 led the author to discover an interesting modification of multinomial coefficients.
Definition 2. For an orderedr-tuple of nonnegative integers (i1, . . . , ir), not all 0,
we define
c(i1, . . . , ir) := ( �
ijj)(�ij−1)! i1!· · ·ir!
.
Note thatc(i1, . . . , ir) equals (�ijj)/�ij times a multinomial coefficient.
Sur-prisingly, these numbers satisfy the same recursive formula as multinomial coeffi -cients.
Definition 3. For positive integersk≤r, letEk denote the orderedr-tuple whose only nonzero entry is a 1 in positionk.
Proposition 4. If I = (i1, . . . , ir) is an ordered r-tuple of nonnegative integers with �ij >1, then
c(I) = � ik>0
c(I−Ek). (1)
If we think of a multinomial coefficient� �ij
i1,···,ir �
:= (i1+· · ·+ir)!/((i1)!· · ·(ir)!)
as being determined by the unordered r-tuple (i1, . . . , ir) of nonnegative integers,
then it satisfies the recursive formula analogous to that of (1). For a multino-mial coefficient, entries which are 0 can be omitted, but that is not the case for c(i1, . . . , ir).
Proof of Proposition 4. The right hand side of (1) equals
�
k ik (
�
ij−2)! (i1)!· · ·(ir)!
�
j
ijj−k
= (
�
ij−2)! (i1)!· · ·(ir)!
���
ik� ��ijj�−�ikk�
= (
�
ij−2)! (i1)!· · ·(ir)!
��
ijj� ��ij−1�,
which equals the left hand side of (1).
Corollary 5. If�ij>0, thenc(i1, . . . , ir)is a positive integer.
Corollary 6. For any ordered r-tuple(i1, . . . , ir)of nonnegative integers and any primep,
νp��ij�≤νp��ijj�+νp
� �ij
i1,· · ·, ir �
. (2)
Proof. Multiply numerator and denominator of the definition ofc(i1, . . . , ir) by�ij
and apply Corollary 5.
The proof of Theorem 1 utilizes Corollary 6 and also the following lemma.
Lemma 7. If tis any integer and �ij≤pνp(t), then
νp
�
t
t−�ij, i1, . . . , ir �
=νp(t) +νp
� �
ij i1, . . . , ir
�
−νp��ij�. (3)
Proof. For any integer t, the multinomial coefficient on the left hand side of (3) equals t(t−1)· · ·(t + 1−�ij)/�ij!, and so the left hand side of (3) equals νp(t(t−1)· · ·(t+ 1−�ij))−�νp(ij!). Sinceνp(t−s) =νp(s) provided 0< s < pνp(t)
, this becomes νp(t) +νp((�ij −1)!)−�νp(ij!), and this equals the right hand side of (3).
Proof of Theorem 1. By the multinomial theorem, [x(p−1)m]�(x)t= (
−1)(p−1)m� I
TI,
where
TI =
�
t
t−�ij, i1, . . . , ir �
1
�
(j+ 1)ij, (4)
with the sum taken over all I = (i1, . . . , ir) satisfying �ijj = (p−1)m. Using
Lemma 7, we have
νp(TI) =νp(t) +νp
� �
ij i1, . . . , ir
�
−νp��ij�−�ijνp(j+ 1).
IfI=mEp−1, thenνp(TI) =νp(t)+0−νp(m)−m. The theorem will follow once we show that all otherIwith�ijj= (p−1)msatisfyνp(TI)>νp(t)−νp(m)−m. Such I must haveij >0 for somej �=p−1. This is relevant because 1
p−1j ≥νp(j+ 1)
with equality if and only ifj =p−1. ForI such as we are considering, we have
νp(TI)−(νp(t)−νp(m)−m) (5)
= νp
� �ij
i1, . . . , ir �
−νp(�ij)−�ijνp(j+ 1) +νp(�ijj) + 1 p−1
�
ijj
≥ �ij( 1
p−1j−νp(j+ 1)) > 0.
2. Zero Coefficients
While studying coefficients related to Theorem 1, we noticed the following result about occurrences of coefficients of powers of the reciprocal log series which equal 0.
Theorem 8. Ifm is odd andm >1, then[xm]� x
log(1+x) �m
= 0, while if mis even andm >0, then[xm+1]� x
log(1+x) �m
= 0.
Moreover, this property characterizes the reciprocal log series.
Corollary 9. A power seriesf(x) = 1 +� i≥1
cixi withc
1�= 0 has[xm](f(x)m) = 0 for all odd m > 1 and [xm+1](f(x)m) = 0 for all even m > 0 if and only if f(x) = 2c1x
log(1+2c1x).
Proof. By Theorem 8, the reciprocal log series satisfies the stated property. Now assume thatf satisfies this property and letnbe a positive integer and�= 0 or 1. Since
[x2n+1]f(x)2n+�
= (2n+�)(2n+�−1)c1c2n+ (2n+�)c2n+1+P,
where P is a polynomial inc1, . . . , c2n−1, we see that c2n andc2n+1 can be
deter-mined from theci withi <2n.
Our proof of Theorem 8 is an extension of arguments of [1] and [2]. It benefited from ideas of Francis Clarke. The theorem can be derived from results in [3, Chapter 6], but we have not seen it explicitly stated anywhere.
Proof of Theorem 8. Letm >1 and
�
x log(1 +x)
�m
=�
i≥0
aixi.
Lettingx=ey−1, we obtain
�
ey−1 y
�m
=�
i≥0
ai(ey
−1)i. (6)
Let j be a positive integer, and multiply both sides of (6) by ymey/(ey−1)j+1,
obtaining
(ey
−1)m−j−1ey = ym� i≥0
ai(ey
−1)i−j−1ey (7)
= ym
aj e
y
ey−1+
�
i�=j ai
i−j d dy(e
y−1)i−j
Since the derivative of a Laurent series has no y−1-term, we conclude that the
coefficient ofym−1on the right-hand side of (7) is aj[y−1](1 + 1
y y
ey−1) =aj.
The Bernoulli numbersBn are defined by eyy−1 = �Bn
n!y
n. Since y ey−1+
1 2y is
an even function of y, we have the well-known result thatBn = 0 if n is odd and n >1.
Let
j=
�
m modd
m+ 1 meven.
For thisj, the left-hand side of (7) equals
�
1 +�Bi
i! y
i−1 modd
−dyd(ey−1)−1=−�(i−1)Bi
i! y
i−2 meven,
and comparison with the coefficient ofym−1in (7) implies �
am= Bm
m! = 0 modd
am+1=−mB(m+1)!m+1 = 0 meven,
yielding the theorem.
3. Other Coefficients
In this section, a sequel to Theorem 1, we describe what can be easily said about νp([x(p−1)m+∆]�(x)t) when 0<
∆< p−1 andm < pνp(t). This is not relevant in
the motivating case,p= 2. Our first result says that these exponents are at least as large as those of [x(p−1)m]�(x)t. Heretcontinues to denote any integer, positive or negative.
Proposition 10. If0<∆< p−1andm < pνp(t), then
νp�[x(p−1)m+∆]�(x)t�≥νp(t)−νp(m)−m.
Proof. We consider termsTI as in (4) with�ijj= (p−1)m+∆. Similarly to (5), we obtain
νp(TI)−(νp(t)−νp(m)−m) = νp
� �
ij i1, . . . , ir
�
−νp��ij�−�ijνp(j+ 1)
+νp(m) +m. (8)
ForI= (i1, . . . , ir), let
�
νp(I) := νp
� �
ij i1, . . . , ir
�
−νp��ij�
= νp
�
1 ij
� �
ij−1 i1, . . . , ij−1, . . . , ir
��
,
for anyj. Thus
�
νp(I)≥ −min j
νp(ij). (9)
Ignoring the termνp(m), the expression (8) is
≥ν�p(I) +�ij( 1
p−1j−νp(j+ 1))−p∆−1. (10)
Note that �
ij( 1
p−1j−νp(j+ 1))−p∆−1 =m− �
ijνp(j+ 1)
is an integer and is greater than−1, and hence is≥0. Thus we are done if�νp(I)≥0. Now suppose ν�p(I) = −e with e >0. By (9), all ij are divisible by pe. Thus (p−1)�ij(p−11 j−νp(j+ 1)) is a positive integer and divisible bype. Hence it is
≥pe. Therefore, (10) is an integer which is strictly greater than −e+ pe
p−1−1 = −e+
e−1 �
k=1
pk+ 1
p−1. Since it is an integer, we can replace the 1
p−1 by 1, and obtain
the nonnegative expression e−1 �
k=0
(pk
−1). We obtain the desired conclusion, that, for
eachI, (10), and hence (8), is≥0.
Finally, we address the question of when does equality occur in Proposition 10. We give a three-part result, but by the third it becomes clear that obtaining additional results is probably more trouble than it is worth.
Proposition 11. In Proposition 10,
(a) the inequality is strict (�=) ifm≡0 (p);
(b) equality holds if∆= 1 andm�≡0,1 (p);
(c) if∆= 2andm�≡0,2 (p), then equality holds if and only if 3m�≡5 (p). Proof. We begin as in the proof of Proposition 10, and note that, using (2), the value of (8) is greater than or equal to
νp(m)− ∆ p−1+
�
ij( 1
p−1j−νp(j+ 1))−νp((p−1)m+∆). (11)
In (b) and (c), we exclude consideration of the case wherem ≡∆ (p) because thenνp((p−1)m+∆)>0 causes complications.
(b) If∆= 1 andm�≡0,1 (p), then forI=E1+mEp−1, (8) equals
νp(m+ 1)−νp(m+ 1)−m+νp(m) +m= 0,
while for otherI, (11) is
0−p−11 +�ij(p−11 j−νp(j+ 1))>0.
(c) Assume∆= 2 and m�≡0,2 (p). Then
T2E1+mEp−1+TE2+mEp−1 =
t(t−1)· · ·(t−m−1) 2!m!
1 4pm+
t(t−1)· · ·(t−m) m!
1 3pm = (−1)m t
pm �1
8(−m−1 +A) + 1
3(1 +B) �
= (−1)m t
24pm(−3m+ 5 + (3A+ 8B)). (12)
Here Aand B are rational numbers which are divisible byp. This is true because νp(t)>νp(i) for alli≤m. Sincep >3, (12) hasp-exponent greater than or equal to νp(t)−m, with equality if and only if 3m−5 �≡0 (p). Using (11), the other termsTI satisfy
νp(TI)−(νp(t)−m)≥�ij( 1
p−1j−νp(j+ 1))−p−12 >0.
References
[1] M.F. Atiyah and J.A. Todd,On complex Stiefel manifolds, Proc. Camb. Phil. Soc.56(1960), 342-353.
[2] D.M. Davis,Divisibility of generalized exponential and logarithmic coefficients, Proc. Amer. Math. Soc.109(1990), 553-558.
[3] R.L. Graham, D.E. Knuth, and O. Patashnik,Concrete mathematics: a foundation for com-puter science, Addison-Wesley (1989).
[4] F. Sigrist and U. Suter, On immersions of CPn in R2n
−2α(n)
