# The Estermann phenomenon

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The Estermann phenomenon
```(September 14, 2015)
The Estermann phenomenon
Paul Garrett [email protected]
http://www.math.umn.edu/egarrett/
[This document is
http://www.math.umn.edu/˜garrett/m/mfms/notes 2015-16/02 estermann phenom.pdf]
The Estermann phenomenon is that not every natural Dirichlet series has a meromorphic continuation. One
need not look far:
Claim: (Estermann) Let d(n) be the number of positive divisors of n. The Dirichlet series
X d(n)3
ns
n
= ζ(s)4
Y
1 + 4p−s + p−2s
p
has a natural boundary along Re (s) = 0, in contrast to meromorphically continuable
X d(n)
n
ns
= ζ(s)2
and
X d(n)2
n
ns
=
ζ(s)4
ζ(2s)
Proof: The cases with meromorphic continuations are treated along the way to examination of the example
lacking meromorphic continuation. By the multiplicativity d(mn) = d(m)d(n) for coprime m, n,
X d(n)
n
Recall
1 + 2x + 3x2 + . . . =
ns
=
Y
1+
p
2
3
+ 2 + ...
s
p
p
d d 1
1
1 + x + x2 + x3 + . . . =
=
dx
dx 1 − x
(1 − x)2
Thus,
X d(n)
n
ns
=
Y
p
1
= ζ(s)2
(1 − p−s )2
Continuing,
X d(n)2
n
ns
=
1 + 2 2 x + 3 2 x2 + . . . =
For
1+
p
and
=
Y
22
32
+ 2 + ...
s
p
p
d d x
1 + x + x2 + x3 + . . .
dx dx
d
x
1
2x
1+x
1 − x2
=
+
=
=
2
2
3
3
dx (1 − x)
(1 − x)
(1 − x)
(1 − x)
(1 − x)4
X d(n)3
n
ns
=
Y
1+
p
23
33
+
+
.
.
.
ps
p2
similarly
1 + 2 3 x + 3 3 x2 + . . . =
=
1+x 1+x
1
3(1 + x)
d x·
=
+x·
+x·
dx
(1 − x)3
(1 − x)3
(1 − x)3
(1 − x)4
(1 − x2 ) + x(1 − x) + 3x(1 + x)
1 − x2 + x − x2 + 3x + 3x2
1 + 4x + x2
=
=
4
4
(1 − x)
(1 − x)
(1 − x)4
1
Paul Garrett: The Estermann phenomenon (September 14, 2015)
The numerator is not a cyclotomic polynomial, so is not a finite product-and-ratio of polynomials 1 − x` , so
there is no obvious analogous expression in terms of ζ(s), ζ(2s), ζ(3s), etc.
The polynomial 1 + 4x + x2 can be written as an arbitrarily large
of binomials 1 − x` , with
Pproduct-and-ratio
`+1
3
s
a leftover polynomial factor of the form 1 + cx
+ . . .. Thus, n d(n) /n can be written as an arbitrarily
1
large product-and-ratio of factors ζ(`s) together with a leftover Euler product convergent in Re (s) > `+1
.
To illustrate this, the first step would be to get rid of the 4x term by multiplying by (1 − x)4 :
(1 − x)4 · (1 + 4x + x2 ) = (1 − 4x + 6x2 − 4x3 + x4 )(1 + 4x + x2 ) = 1 − 9x2 + 16x3 − 9x4 + x6
Thus,
Y
Y
(1 + 4p−s + p−2s ) = ζ(s)4 · (1 − 9p−2s + 16p−3s − 9p−4s + p−4s )
p
p
Next, to get rid of the −9x2 term, multiply by (1 + x2 )9 = (1 − x4 )9 /(1 − x2 )9 , giving
Y
ζ(4s)9 Y
(1 + 4p−s + p−2s ) = ζ(s)4 ·
(1 + 16p−3s + . . .)
·
9
ζ(2s)
p
p
Since 1 + 4x + x2 is not a cyclotomic polynomial, this process does not terminate. Inductively, there is an
infinite increasing sequence of integers `j and non-zero integers ej such that
1 + 4x + x2 = (1 − x)e1 (1 − x2 )e2 (1 − x3 )e3 . . . (1 − x`j )e`j · 1 + x`j +1 Pj (x)
with (non-zero) polynomials Pj (x). Certainly
Dj (s) =
Y
1 + p−s(`j +1) Pj (p−s )
p
is absolutely convergent and non-vanishing for Re (s) >
1
`j +1 .
Thus, for every j, there is an expression
Y
Y
(1 + 4p−s + p−2s ) = Dj (s) ·
ζ(`i · s)ei
p
(for Re (s) >
1
`j +1 )
1≤i≤j
On one hand, this gives a meromorphic continuation to Re (s) > `j 1+1 . On the other hand, since the exponents
ei are non-zero, the infinitely-many zeros of ζ(s) in the critical strip make the zeros of ζ(` · s) bunch up just
to the right of Re (s) = 0 as ` → ∞.
///
[0.0.1] Remarks: Continuing
in this vein, [Kurokawa 1985a,b] showed that
P
for k ≥ 3, where f (z) =
an e2πinz is a modular form,
P akn
ns
has a natural boundary
Bibliography
[Backlund 1914] R. Backlund, Sur les zéros de la fonction ζ(s) de Riemann, C.R. 158 (1914), 1979-81.
[Backlund 1918] R. Backlund, Über die Nullstellen der Riemannschen Zetafunktion, Acta Math. 41 (1918),
345-75.
[Estermann 1928] T. Estermann, On certain functions represented by Dirichlet series, Proc. London Math.
Soc. 27 (1928), 435-448.
[Kurokawa 1985a,b] N. Kurokawa On the meromorphy of Euler products, I, II, Proc. London Math. Soc.
53 (1985) 1-49, 209-236.
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