Powers of Euler's q-Series

0.16 MB·English

by  Steven Finch

Powers of Euler’s q-Series
Steven Finch
7
0
January 19, 2007
0
2
Abstract. What are the asymptotic moments of coefficients obtained
n
Ja awshaenneewxpmanudltinipglica∞mtiv=e1(r1ep−reqsmen)ℓtaitniosnerfioers?coAeffifecwienetxsawmhpelnesℓa=re1g0ivaennd, ℓas=w1e4ll.
8 Q
1 Starting with Euler’s result:

]
T ∞ ∞
(1 qm) = ( 1)kqk(3k+1)/2
N
− −
h. mY=1 kX=−∞
t and Jacobi’s cubic analog:
a
m
∞ ∞
[ (1 qm)3 = ( 1)k(2k+1)qk(k+1)/2,

− −
2
m=1 k=0
v Y X
1 it is natural to wonder about the corresponding square and ℓth powers for ℓ 4. We
5 ≥
will consider thecoefficients a oftheq-series expansion (ontheright-handside)
2 { n}∞n=0
1 for fixed ℓ. Estimates of the magnitude of a finite subsequence a of coefficients
n n N
0 { } ≤
include
7
0 a , a2, max a
/ | n| n n N | n|
h n N n N ≤
X≤ X≤
t
a and we will analyze the asymptotics (when possible) of these as N . Much of
m → ∞
the material presented here is perhaps not new. Our contribution (as in [1, 2]) is only
:
v to collect results in one place.
i
X
1. First Power
r
a Let the coefficients a satisfy [3, 4, 5]
{ n}∞n=0
∞ ∞
a qn = (1 qm).
n
−
n=0 m=1
X Y
Euler’s pentagonal-number theorem gives that
( 1)k if 24n+1 = (6k +1)2 for some < k < ,
a = − −∞ ∞
n 0 otherwise
(cid:26)
0 Copyright c 2007 by Steven R. Finch. All rights reserved.
(cid:13)
1
Powers of Euler’s q-Series 2
and hence, by elementary considerations,
1
a (24N)1/2
n
| | ∼ 3
n N
X≤
as N . Also a = f(24n+1), where f is multiplicative and
n
→ ∞
1 if p 1,11mod12 and r is even,
≡
f(pr) = ( 1)r/2 if p 5,7mod12 and r is even,
 − ≡
0 otherwise

for p prime and r ≥ 0. Of course, n N an oscillates between −1,0,1 and never
converges. ≤
P
2. Second Power
Let the coefficients a satisfy [6, 7, 8]
{ n}∞n=0
∞ ∞
a qn = (1 qm)2.
n
−
n=0 m=1
X Y
No simple formula for a is known, but a = f(12n+ 1), where f is multiplicative
n n
and
1 if p 7,11mod12 and r is even,
≡
( 1)r/2 if p 5mod12 and r is even,
 − ≡
f(pr) = r +1 if p 1mod12 and ( 3)(p 1)/4 1modp,
 −
 ( 1)r(r +1) if p ≡ 1mod12 and (−3)(p 1)/4 ≡ 1modp,
 −
− ≡ − ≡ −
0 otherwise.



We examine 
∞ f(n) ∞ 1 ∞ r +1
| | = 1+ 1+
ns p2rs · prs
! !
n=1 p 5,7,11 r=1 p 1 r=1
X m≡Yod12 X mYo≡d12 X
1 1
1 1 − 1 1 −
= 1 1 1+ 1
− ps +1 − ps · ps 1 − ps
p 5,7,11(cid:18) (cid:19)(cid:18) (cid:19) p 1 (cid:18) − (cid:19)(cid:18) (cid:19)
m≡Yod12 mYo≡d12
1/2 1/2 1 1/2
1 1 1 − 1 −
= 1 1 1+ 1
− 2s − 3s ps − ps
(cid:18) (cid:19) (cid:18) (cid:19) p 5,7,11(cid:18) (cid:19) (cid:18) (cid:19)
m≡Yod12
3/2
1 −
1 ζ(s)1/2
· − ps ·
p 1 (cid:18) (cid:19)
mYo≡d12
Powers of Euler’s q-Series 3
and deduce, via the Selberg-Delange method [9, 10, 11], that
N
a C
| n| ∼ (lnN)1/2
n N
X≤
as N , where [12]
→ ∞
1 1/2 3/2
12 1 − 1 − 1 −
C = 1+ 1 1
√3π p − p · − p
p 5,7,11(cid:18) (cid:19) (cid:18) (cid:19) p 1 (cid:18) (cid:19)
m≡Yod12 mYo≡d12
= 3.3215840614847482046694103....
Also,
∞ f(n)2 ∞ 1 ∞ (r +1)2
= 1+ 1+
ns p2rs · prs
! !
n=1 p 5,7,11 r=1 p 1 r=1
X m≡Yod12 X mYo≡d12 X
1 2
1 1 − 2 1 −
= 1 1 1+ 1
− ps +1 − ps · ps 1 − ps
p 5,7,11(cid:18) (cid:19)(cid:18) (cid:19) p 1 (cid:18) − (cid:19)(cid:18) (cid:19)
m≡Yod12 mYo≡d12
1 2
1 1 1 − 1 1 −
= 1 1 1+ 1+ 1 ζ(s)
− 2s − 3s ps · ps − ps ·
(cid:18) (cid:19)(cid:18) (cid:19)p 5,7,11(cid:18) (cid:19) p 1 (cid:18) (cid:19)(cid:18) (cid:19)
m≡Yod12 mYo≡d12
and hence
a2 DN
n ∼
n N
X≤
where [12]
1 2
1 − 1 1 −
D = 4 1+ 1+ 1
p · p − p
p 5,7,11(cid:18) (cid:19) p 1 (cid:18) (cid:19)(cid:18) (cid:19)
m≡Yod12 mYo≡d12
= 2.6339157938496334172500926....
Also,
ln(N)
ln max a ln(2)
n
n N | | ∼ ln(ln(N))
(cid:18) ≤ (cid:19)
but a more precise asymptotic statement is evidently open [11]. Finally,
a O N exp c(lnN)3/5(lnlnN) 1/5
n −
∼ −
n N
X≤ (cid:0) (cid:0) (cid:1)(cid:1)
and a stronger conjectured bound a = O(Nd) with d < 1 is equivalent to a
n N n
weak form of the generalized Riemann≤hypothesis [13].
P
Powers of Euler’s q-Series 4
2.1. Variation One. Let the coefficients b satisfy [14, 15]
{ n}∞n=0
∞ ∞
b qn = (1 qm)(1 q2m).
n
− −
n=0 m=1
X Y
It can be shown that b = g(8n+1), where g is multiplicative and
n
1 if p = 2 and r = 1,
1 if p 5,7mod8 and r is even,
 ≡
( 1)r/2 if p 3mod8 and r is even,
g(pr) =  − ≡
 r+1 if p 1mod8 and ( 4)(p 1)/8 1modp,
 −
 ( 1)r(r+1) if p ≡ 1mod8 and (−4)(p 1)/8 ≡ 1modp,
−
− ≡ − ≡ −
0 otherwise.





For reasons ofspace, we mention only the mean-square result:
b2 EN
n ∼
n N
X≤
where [12]
1 2
1 − 1 1 −
E = 4 1+ 1+ 1
p · p − p
p 3,5,7(cid:18) (cid:19) p 1 (cid:18) (cid:19)(cid:18) (cid:19)
m≡Yod8 mYo≡d8
= 1.7627471740390860504652186....
2.2. Variation Two. Let the coefficients c satisfy [14, 16]
{ n}∞n=0
∞ ∞
c qn = (1 qm)(1 q3m).
n
− −
n=0 m=1
X Y
It can be shown that c = h(6n+1), where h is multiplicative and
n
1 if p 5mod6 and r is even,
≡
r +1 if p 1mod6 and 2(p 1)/3 1modp,
−
 ≡ ≡
h(pr) = 1 if p 1mod6, 2(p 1)/3 1modp and r 0mod3,
 −
 1 if p ≡ 1mod6, 2(p 1)/3 6≡ 1modp and r ≡ 1mod3,
 −
− ≡ 6≡ ≡
0 otherwise.



Observe that h(p) assumes two distinct values over the set p 1mod6, which
| | ≡
is more complicated than f(p) over p 1mod12 and g(p) over p 1mod8.
| | ≡ | | ≡
Consequently, the asymptotics for c2 are more subtle than those for a2
n N n n N n
and b2. ≤ ≤
n N n P P
≤
P
Powers of Euler’s q-Series 5
3. Third Power
Let the coefficients a satisfy [17, 18]
{ n}∞n=0
∞ ∞
a qn = (1 qm)3.
n
−
n=0 m=1
X Y
Jacobi’s triple-product identity gives that
( 1)k(2k +1) if 8n+1 = (2k +1)2 for some 0 k < ,
a = − ≤ ∞
n 0 otherwise
(cid:26)
and hence, by elementary considerations,
1
a 2N, a2 (8N)3/2, max a (8N)1/2
| n| ∼ n ∼ 6 n N | n| ∼
n N n N ≤
X≤ X≤
as N . Also a = f(8n+1), where f is multiplicative and
n
→ ∞
pr/2 if p 1mod4 and r is even,
≡
f(pr) = ( 1)r/2pr/2 if p 3mod4 and r is even,
 − ≡
0 otherwise.

Of course, 
a = ( 1)k(k +1) if (2k +1)2 8N +1 < (2k +3)2
n
− ≤
n N
X≤
and thus a diverges because
n N n
≤
P 1 1
limsup a = 1, liminf a = 1.
n n
N √2N N √2N −
→∞ nX≤N →∞ nX≤N
4. Fourth Power
Let the coefficients a satisfy [6, 19]
{ n}∞n=0
∞ ∞
a qn = (1 qm)4.
n
−
n=0 m=1
X Y
No simple formula for a is known, but a = f(6n+1), where f is multiplicative and
n n
( 1)r/2pr/2 if p 5mod6 and r is even,
− ≡
(x +√3iy )r+1 (x √3iy )r+1
f(pr) =  δ p p − p − p if p 1mod6,
p,r
 2√3iy ≡
 p

0 otherwise



Powers of Euler’s q-Series 6
where i is the imaginary unit and (x ,y ) is the unique pair of positive integers for
p p
which p = x2 + 3y2. Also, δ = 1 when r is odd and x 1mod3; otherwise
p p p,r − p 6≡
δ = 1.
p,r
It turns out that ∞n=1f(n)n−s is the L-series for the elliptic curve 36A1:
P y2 = x3 +1
and hence a theorem of Rankin [20] implies that
a2 CN2.
n ∼
n N
X≤
We examine
∞ f(n)2 ∞ pr
= 1+
ns+1 p2r(s+1)
!
n=1 p 5 r=1
X mYo≡d6 X
r+1 r+1 2
x +√3iy x √3iy
1 ∞ p p − p − p
1
·  − 12y2 h(cid:0) (cid:1) pr(s+(cid:0)1) (cid:1) i 
p 1 p r=1
mYo≡d6 X 
 1 
1 −
= 1
− p2s+1
p 5 (cid:18) (cid:19)
mYo≡d6
p2(s+1) ps+1 +x2 +3y2
p p
· 2
ps+1 x2 +3y2 x2 +(cid:0) 3y2 2ps+1 (cid:1)x2 3y2 +p2(s+1)
mYpo≡d16 − p p p p − p − p
h i
(cid:2) (cid:0)1 (cid:1)(cid:3) (cid:0) (cid:1) (cid:0) (cid:1)
1 −
= 1
− p2s+1
p 5 (cid:18) (cid:19)
mYo≡d6
p2(s+1)(ps+1 +p)
· [ps+1 p] p2 2ps+1 x2 3y2 +p2(s+1)
p 1 − − p − p
mYo≡d6
(cid:2) (cid:0) (cid:1) (cid:3)
1
1 1 1 1 −
= 1 1 1 1
− 2s − 3s − ps − p2s+1
(cid:18) (cid:19)(cid:18) (cid:19) p 5 (cid:18) (cid:19)(cid:18) (cid:19)
mYo≡d6
1 2ps 1 x2 3y2 1
1+ 1+ − p − p − ζ(s)
· ps p2s 2ps 1 x2 3y2 +1 ·
p 1 (cid:18) (cid:19) − (cid:0)− p − (cid:1) p !
mYo≡d6
(cid:0) (cid:1)
Powers of Euler’s q-Series 7
and thus it seems that
C = 6 1 1 1 1 −1 1+ 1 1+ 2 x2p −3yp2 −1 .
p 5 (cid:18) − p(cid:19)(cid:18) − p3(cid:19) · p 1 (cid:18) p(cid:19) p2 −(cid:0)2 x2p −3(cid:1)yp2 +1!
mYo≡d6 mYo≡d6
(cid:0) (cid:1)
The language of modular forms is now unavoidable: let
∞
η(t) = eπit/12 1 e2πkit , Im(t) > 0
−
k=1
Y(cid:0) (cid:1)
denote the Dedekind eta function. The unique cusp form of weight 3, level 12 and
Nebentypus character ( 3/ ) is [21]
− ·
∞ ∞
η(2t)3η(6t)3 = q (1 q2m)3 (1 q6m)3, q = e2πit
− −
m=1 m=1
Y Y
and possesses the expansion
1
L (s) = 1+ 1 −1 1 1 −1 1 2 x2p −3yp2 + 1 − ;
xy 3s 1 − p2(s 1) · − ps p2(s 1)
(cid:18) − (cid:19) p 5 (cid:18) − (cid:19) p 1 (cid:0) (cid:1) − !
mYo≡d6 mYo≡d6
hence
L (2) = 3 1 1 −1 1+ 2 x2p −3yp2 −1
xy 4 − p2 · p2 2 x2 3y2 +1
p 5 (cid:18) (cid:19) p 1 −(cid:0) p − (cid:1)p !
mYo≡d6 mYo≡d6
(cid:0) (cid:1)
= 0.7372929961855962401764261...;
hence [12]
1
1 1 1 − 1
C = 8L (2) 1 1 1 1+
xy − p − p2 − p3 · p
p 5 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) p 1 (cid:18) (cid:19)
mYo≡d6 mYo≡d6
= 4.6417183001293981350615666....
Observe, however, that in the vicinity of s = 1, the series
2ps 1 x2 3y2 1
− p − p −
p2s 2ps 1 x2 3y2 +1
p − (cid:0)− p − (cid:1) p
X
is not absolutely convergent since either x2(cid:0)or 3y2 is(cid:1)often of size p. The requirement
p p
that ( f(n)2n s 1)ζ(s) 1 is analytic consequently might not hold. Empirically, we
− − −
compute
P 1
a2 4.877...
N2 n ≈
n N
X≤
which raises further doubt about the correctness of our value 4.641... for C.
Powers of Euler’s q-Series 8
4.1. Variation One. Let the coefficients b satisfy [22, 23]
{ n}∞n=0
∞ ∞
b qn = (1 qm)2(1 q2m)2.
n
− −
n=0 m=1
X Y
It can be shown that b = g(4n+1), where g is multiplicative and
n
( 1)r/2pr/2 if p 3mod4 and r is even,
− ≡
(u +iv )r+1 (u iv )r+1
g(pr) =  ε p p − p − p if p 1mod4,
p,r
 2ivp ≡

0 otherwise

where (up,vp) is the unique pair of positive integers for which p = u2p +vp2, up is odd
and v is even. Also, ε = ( 1)(up+vp 1)/2 when r is odd and ε = 1 when r is even.
p p,r − p,r
−
It turns out that ∞n=1g(n)n−s is the L-series for the elliptic curve 32A2:
P y2 = x3 x
−
and hence Rankin’s theorem implies that
b2 DN2
n ∼
n N
X≤
where seemingly
D = 4 1 1 1 1 −1 1+ 1 1+ 2 u2p −vp2 −1 .
− p − p3 · p p2 2 u2 v2 +1
p 3 (cid:18) (cid:19)(cid:18) (cid:19) p 1 (cid:18) (cid:19) −(cid:0) p −(cid:1)p !
mYo≡d4 mYo≡d4
(cid:0) (cid:1)
The unique cusp form of weight 3, level 16 and Nebentypus character ( 4/ ) is [21]
− ·
η(4t)6 = q ∞ 1 q4m 6
−
m=1
Y (cid:0) (cid:1)
and possesses the expansion
1
L (s) = 1 1 −1 1 2 u2p −vp2 + 1 − ;
uv
− p2(s 1) · − ps p2(s 1)
p 3 (cid:18) − (cid:19) p 1 (cid:0) (cid:1) − !
mYo≡d4 mYo≡d4
hence
L (2) = 1 1 −1 1+ 2 u2p −vp2 −1
uv − p2 · p2 2 u2 v2 +1
p 3 (cid:18) (cid:19) p 1 −(cid:0) p −(cid:1)p !
mYo≡d4 mYo≡d4
(cid:0) (cid:1)
= 0.8593982272525466034362619...;
Powers of Euler’s q-Series 9
hence [12]
1
1 1 1 − 1
D = 4L (2) 1 1 1 1+
uv − p − p2 − p3 · p
p 3 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) p 1 (cid:18) (cid:19)
mYo≡d4 mYo≡d4
= 1.9533514553987911733090376....
Again, analyticity requirements might not hold and we empirically compute
1
b2 2.188...
N2 n ≈
n N
X≤
which raises doubt about our value 1.953... for D.
4.2. Variation Two. Let the coefficients c satisfy [24, 25]
{ n}∞n=0
∞ ∞
c qn = (1 qm)2(1 q3m)2.
n
− −
n=0 m=1
X Y
It can be shown that c = h(3n+1), where h is multiplicative and
n
( 1)r/2pr/2 if p 2mod3 and r is even,
− ≡
r+1 r+1
 zp +3√3iwp zp 3√3iwp
−
h(pr) =  2 − 2
 ! !
 δ˜ if p 1mod3,
 p,r 3√3iw ≡
p
0 otherwise





where (z,w ) is the unique pair of positive integers for which 4p = z2 +27w2. Also,
p p p p
˜ ˜
δ = 1 when r is odd and x 1mod3; otherwise δ = 1.
p,r p p,r
− ≡
It turns out that ∞n=1h(n)n−s is the L-series for the elliptic curve 27A3:
P y2 +y = x3
and hence Rankin’s theorem implies that
c2 EN2
n ∼
n N
X≤
where seemingly
E = 3 1 1 1 1 −1 1+ 1 1+ (zp2 −27wp2)/2−1 .
− p − p3 · p p2 (z2 27w2)/2+1
p 2 (cid:18) (cid:19)(cid:18) (cid:19) p 1 (cid:18) (cid:19)(cid:18) − p − p (cid:19)
mYo≡d3 mYo≡d3
Powers of Euler’s q-Series 10
The vector space of cusp forms of weight 3, level 27 and Nebentypus character ( 3/ )
− ·
is three-dimensional. A certain basis element is given by [21, 26]
η(9t)η(3t)2 η(3t)3 +9η(27t)3
and possesses the expansion (cid:0) (cid:1)
L (s) = 1 1 −1 1 zp2 −27wp2 + 1 −1;
zw
− p2(s 1) · − 2ps p2(s 1)
p 2 (cid:18) − (cid:19) p 1 (cid:18) − (cid:19)
mYo≡d3 mYo≡d3
hence
L (2) = 1 1 −1 1+ (zp2 −27wp2)/2−1
zw − p2 · p2 (z2 27w2)/2+1
p 2 (cid:18) (cid:19) p 1 (cid:18) − p − p (cid:19)
mYo≡d3 mYo≡d3
= 1.0403374913367121372113004...;
hence [12]
1
1 1 1 − 1
E = 3L (2) 1 1 1 1+
zw
− p − p2 − p3 · p
p 2 (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) p 1 (cid:18) (cid:19)
mYo≡d3 mYo≡d3
= 1.0526097875093498936749762....
Again, analyticity requirements might not hold and we empirically compute
1
c2 1.290...
N2 n ≈
n N
X≤
which raises doubt about our value 1.052... for E.
5. Sixth Power
Let the coefficients a satisfy [6, 27]
{ n}∞n=0
∞ ∞
a qn = (1 qm)6.
n
−
n=0 m=1
X Y
No simple formula for a is known, but a = f(4n+1), where f is multiplicative and
n n
if p 3mod4
pr ≡
and r is even,

f(pr) = (v2 u2 +2iu v )r+1 (v2 u2 2iu v )r+1
 ( 1)r p − p p p − p − p − p p if p 1mod4,

 − 4iupvp ≡
0 otherwise




