INVESTIGATING ZETA FUNCTIONS
AMANDA KNECHT
Abstract. In this paper we investigate the Riemann zeta function, ζR (s) =
P
∞
−s
. The Riemann Hypothesis, which remains unproven after 143 years,
n=0 n
states that the nontrivial complex roots of ζ have real parts equal to one-half.
Our study primarily involves the function ξR used in the functional equation of
the Riemann Zeta Function. The roots of ξR correspond exactly to the roots of
ζR , and the function satisfies the equation ξR (s) = ξR (s−1). We study the slopes
of the level sets < (ξR (s)) = 0 and = (ξR (s)) = 0 in order to find information
about the intersection points of the level sets, because these intersection points
are the roots of ξR . We then extend our study of the Riemann zeta function to
the more general Epstein zeta function, Z (Y, s). The function Z (Y, s) depends
on a positive, symmetric n × n real matrix Y and s ∈ C. We define an Epstein ξ
function, ξE , as we did for the Riemann Zeta Function and investigate the level
sets < (ξE (Y, s)) = 0 and = (ξE (Y, s)) = 0 in order to discover how the matrix
Y affects the location of the function’s roots.
Contents
1. Introduction
2. The Riemann Zeta Function
3. Consequences of the Slope Conjecture
4. The Epstein Zeta Function
References
1
3
13
13
24
1. Introduction
For references pertaining to the early history of the Riemann zeta function, we
refer to Edwards [Ed] and Titchmarsh [Ti] and to Terras [Te] for background on
the Epstein zeta function. In 1859 Bernhard Riemann wrote an 8-page paper,
“ On the Number of Primes Less Than a Given Magnitude,” which has made a
great impact on modern mathematics . In this paper Riemann studies the Euler
product formula:
∞
X
Y
1
1
=
s
1
n
1
−
n=1
p prime
ps
which is absolutely convergent for all s ∈ C such that
formula to define a new function.
1We
1
< (s) > 1. He uses this
will use < (s) and = (s) to denote the real and imaginary parts, respectively, of a complex
number s.
1
2
AMANDA KNECHT
Definition 1. Riemann’s zeta function on the half plane < (s) > 1 is
ζ (s) =
∞
X
1
.
s
n
n=1
Riemann analytically continues the zeta function above so that it is absolutely
convergent over the entire complex plane except for a simple pole at s = 1. The
paper also introduces the functional equation:
s
1−s
− 1−s
− 2s
2
ζR (s) = π
Γ
ζR (1 − s) .
π Γ
2
2
R∞
Here Γ is Euler’s Gamma Function and can be written as Γ (s) = 0 e−t ts−1 dt. By
multiplying the functional equation by 21 s (s − 1), Riemann was able to eliminate
the poles of the function and create a new function, ξR , defined by:
s −s
1
ξR (s) = s (s − 1) Γ
π 2 ζR (s) .
2
2
Riemann goes on to show that none of the roots s of ξR (s) = 0 lie on the half plane
< (s) > 1. And through a simple investigation, one can see that ξR (s) = ξR (1 − s),
so 1 − s is a root if and only if s is a root. Thus, it can easily be shown that there
are no roots on the half plane < (s) < 0. Whence, all roots of ξR must lie on the
strip 0 ≤ < (s) ≤ 1. He then states, without proof, that for 0 < = (s) < T and
T
T
T
< (s) = 21 the number of roots of ξR (s) is approximately 2π
log 2π
− 2π
. And this
idea led to the famous Riemann Hypothesis, which states that all the roots of ξR
lie on the line < (s) = 21 . It is known that ζR has zeros at every even negative
integer, and these zeros are called trivial. The remaining zeros correspond exactly
to the zeros of ξR , and so we can restate the Riemann Hypothesis by saying that
the nontrivial roots of ζR have real part equal to 12 .
The images included in the appendix are used in our research in order to better
understand the functions we analize. The first image is a map of the complex plane
where each quadrant is assigned a color. When graphing a function, f (z), the original value z is denoted by its location on the graph, and the function value, f (z),
is denoted by color. Thus, the image tells us the quadrant each point is mapped
to by f. Where two colors meet is a level set <(f (z)) = 0 or =(f (z)) = 0, and
when all four colors meet Re(f (z)) = =(f (z)) = 0. One can see from image 2 that
the Riemann Hypothesis appears to be true. But this hypothesis pertains only to
the Riemann zeta function, which is a specific case of a Selberg Zeta function.
The Selberg Zeta function is defined on all manifolds and can be written as
X 1
,
λs
λ6=0
where the λ are the eigenvalues of the Laplacian counted with multiplicity. From
studying the geometry of flat tori, Epstein’s zeta function, ζE (Y, s), was created.
This more general ζE (Y, s) can be studied much like ζR (s).
In this paper, we will investigate ξR , so that we can better understand why the
INVESTIGATING ZETA FUNCTIONS
3
Riemann Hypothesis appears to be true. We will then extend our study into the
Epstein zeta function in order to understand why the Riemann Hypothesis does
not hold for all values of Y inputted into ζE (Y, s).
2. The Riemann Zeta Function
Our study of the Riemann Zeta function,
∞
X
1
ζR (s) =
,
s
n
n=1
rests primarily on the study of the equations:
Z ∞
X
u
1
1
1
2
u
2
+ 2w = + 4 w −
e 4 cosh(uw)
(1)
ξR
e−k πe du.
2
2
16
0
k≥1
(2)
ξC
1
+w
4
1
=
2 w2 −
Z
+4
1
16
∞
u
X
e 4 cosh(uw)
0
e−k
2 πeu
du.
k≥1
We derive equation (2) from the Circle Zeta function in which the λ represent the
eigenvalues of the Laplacian of the circle.
X 1
(3)
ζC (s) =
.
s
λ
λ6=0
R∞
Knowing that ΓR(s) = 0 e−u us−1 du, we can change
variables letting u = λt to
R ∞ −λt
1
s ∞ −λt s−1
−s
find Γ (s) = λ 0 e t dt. Then λ = Γ(s) 0 e ts−1 dt. Thus, we can also
write the circle zeta function as:
!
Z ∞
X
1
ts−1
(4)
ζC (s) =
e−λt dt.
Γ (s) 0
λ6=0
Substituting λ = n2 ), with multiplicity two, into equations (3) and (4
∞
X
1
= 2ζR (2s) .
ζC (s) = 2
2s
n
n=1
1
ζC (s) =
Γ (s)
Z
0
∞
ts−1
2
∞
X
!
−n2 t
e
dt.
n=1
Now we introduce a new function, K (t), based on Jacobi’s theta function [Te]:
√ X
∞
X
−k2 π 2
π
−n2 t
K (t) + 1 := 1 + 2
e
= √
e t .
t k∈Z
n=1
√
√
π
π
K (t) = √ − 1 + K (t) + 1 − √ .
t
t
4
AMANDA KNECHT
Plugging this equation into the circle zeta function:
Z ∞
1
ζC (s) =
ts−1 K(t)dt
Γ(s) 0
√
Z π
1
π
s−1
√ − 1 dt
=
t
Γ(s) 0
t
√ Z π
π
1
s−1
K (t) + 1 − √ dt
t
+
Γ (s) 0
t
!
Z ∞
∞
X
1
2
+
ts−1 2
e−n t dt.
Γ (s) π
n=1
Looking at the first two integrals above separately,
(1)
1
Γ (s)
0
√
π
1
s−1
√ − 1 dt =
t
Γ (s)
t
Z
π
π
Z
!
1√
ts− 2 π ts π
−
|0
s
s − 12
1
πs
1
−
=
f or <(s) > 1/2.
Γ (s) s − 12
s
(2)
1
Γ (s)
0
√ π
t
K (t) + 1 − √ dt =
t
!
Z π
X −k2 π2 √π
π2
1
√ dt
=
ts−1 2
e t
Let v =
Γ (s) 0
t
t
k≥1
!
√
Z ∞ 2 s−1
X
1
π
π −π 2
−k2 v
q
=
2
e
dv
2
Γ(s) π
v
π2 v
k≥1
v
!
1
Z
X
π (2s− 2 ) ∞ (−s− 21 )
2
2
e−k v dv.
=
v
Γ (s) π
k≥1
s−1
Thus, we write the circle zeta function as,
ζC (s) =
πs
1
1
−
Γ (s) s − 12
s
!
1
Z
X
π (2s− 2 ) ∞ (−s− 21 )
2
+
v
2
e−k v dv
Γ (s) π
k≥1
!
Z ∞
∞
X
1
2
+
v s−1 2
e−n v dv.
Γ (s) π
n=1
INVESTIGATING ZETA FUNCTIONS
5
From the functional equation of ζC we can write,
ξC (s) : = π −s Γ (s) ζC (s) .
!
Z ∞
X
1
1
1
1
2
= − − 1
+ π −( 2 −s)
v (( 2 −s)−1) 2
e−k v dv
s
−
s
π
2
k∈Z
!
Z ∞
∞
X
2
e−n v dv.
v s−1 2
+π −s
π
n=1
Now we let v = πa to find,
Z ∞
X
1
1
1
2
ξC (s) = − + 1
a−1 as + a 2 −s 2
+
e−k πa da.
s
−s
1
2
k≥1
And letting s =
ξC
1
4
+w
1
4
+ w,
=
R ∞ −3
P
2
+ 1 a 4 (aw + a−w ) 2 k≥1 e−k πa da
P
R ∞ −3
2
= 2 w21− 1 + 1 a 4 ew log a + e−w log t 2 k≥1 e−k πa da
(
)
16
R ∞ −3
P
2
= 2 w21− 1 + 1 a 4 cosh (w log a) k≥1 e−k πa da.
(
)
16
−1
1
+w
4
−
1
1
−w
4
Finally, we will let u = log a to find that ξC can be written as
Z ∞
X
u
1
1
2
u
4 cosh (uw)
ξC
e
e−k πe du.
+w =
+
4
1
2
4
2 w − 16
0
k≥1
It is easy to show that ξC 41 + w = ξC 14 − w , so the function reflects across the
line < (z) = 41 . This equation can be related back to the Riemann Zeta Function
as follows:
ξC (s) = π −s Γ (s) 2ζR (2s)
s −s
1
ξR (s) = s (s − 1) Γ
π 2 ζR (s)
2
2
s
2ξR (s) π 2
ζR (s) =
s (s − 1) Γ 2s
2ζR (2s) =
ξC (s) =
4ξR (2s) π s
2s (2s − 1) Γ (s)
4π −s Γ (s) ξR (2s) π s
2s (2s − 1) Γ (s)
ξC (s) =
2ξR (2s)
s (2s − 1)
Letting s = w + 14 :
ξC
1
w+
4
ξR 2w + 12
=
1
w2 − 16
6
AMANDA KNECHT
ξR
1
2w +
2
=
1
w −
16
2
ξC
1
w+
4
We know:
Z ∞
X
u
1
−k2 πeu
4 cosh (uw)
ξC
e
=
+
4
e
du.
1
2 w2 − 16
0
k≥1
u P
2
u
So letting h (u) = e 4 k≥1 e−k πe , we get:
ξR
1
+w
4
1
2w +
2
(5)
ξR
=
1
2w +
2
1
w2 −
16
"
1
2
2 w −
Z
1
16
+4
#
∞
cosh (uw) h (u) du
0
Z ∞
1
1
2
= +4 w −
cosh (uw) h (u) du.
2
16
0
This new function, ξR , reflects across the line <(s) = 1/2.
The level sets < ξR 21 + 2w
= 0 represent the values of w ∈ C around
which < ξR 21+ 2w changes from positive to negative. Whereas, the sets where
= ξR 21 + 2w = 0 represent the values of w ∈ C around which = ξR 12 + 2w
changes from positive to negative. These level sets occur under two situations.
(1) They are found on the critical line, the line < (s) = 21 , when < (w) = 0.
Substituting w = 2ıy, y ∈ R, into the functional equation(5 ), we find:
Z ∞
u
1
1
1
2
ξR
e 4 cosh (ı2uy) h (u) du.
+ 2ıy = + 4 −4y −
2
2
16
0
Because cosh (ı2uy) = cos (2uy),
Z ∞
u
1
1
1
2
ξR
+ 2ıy = + 4 −4y −
e 4 cos (2uy) h (u) du
2
2
16
0
1
which is always real. So = ξR 2 + 2ıy = 0 for all y ∈ R .
(2) The general equations satisfied when the ξ function has real or imaginary
parts equal to zero can be found by substituting w = x + ıy, x, y ∈ R and
x 6= 0, into equation (5):
Z ∞
1
1
1
2
ξR 2 (x + ıy) +
= + 4 (x + ıy) −
cosh (u (x + ıy)) h (u) du.
2
2
16
0
We know cosh (ux + ıuy) = cosh (ux) cos (uy) + ı sinh (ux) sin (uy) . So
1
ξR 2x + 2ıy +
=
2
Z ∞
1
2
2
4 x −y −
cosh (ux) cos (uy) + ı sinh (ux) sin (uy) h (u) du
16
0
Z ∞
1
cosh (ux) cos (uy) + ı sinh (ux) sin (uy) h (u) du.
+ + 8ıxy
2
0
INVESTIGATING ZETA FUNCTIONS
7
We can now find the real and imaginary parts of ξR .
R∞
1
1
< ξR
cosh (ux) cos (uy) h (u) du
+ 2x + 2ıy
= 4 x2 − y 2 − 16
0
2
R∞
+ 21 − 8xy 0 sinh (ux) sin (uy) h (u) du.
R∞
1
1
sinh(ux) sin(uy) h(u)du
= ξR
+ 2x + 2ıy
= 4 x2 − y 2 − 16
0
2
R∞
+8xy 0 cosh(ux) cos(uy) h(u)du.
These can be studied in order to find where < (ξR ) = = (ξR ) = 0.
The first quantities we want to study are the slopes of the equations < (ξR ) = 0
and = (ξR ) = 0. We know that the gradient of a function F (x, y) equals (Fx , Fy )
and is perpendicular to a level set, F (x, y) = C for some constant C. So to
find the slope of a level set of the function, we must find a vector perpendicular to the gradient of the set. The vector (−Fy , Fx ) satisfies this need because
x
(Fx , Fy ) · (−Fy , Fx ) = 0. The slope of this vector, −F
, is the slope of F (x, y) = C.
Fy
In order to find the slopes of < ξ 12 + 2x + 2ıy = 0 and = ξ 21 + 2x + 2ıy =
0, we only need to think of them as functions of x and y and apply what we know
from above. The equations for the partial derivatives are:
1
∂
+ 2x + 2ıy
=
< ξR
∂x
2
Z ∞
8x
cosh(ux) cos(uy)h (u) du
0
Z ∞
1
2
2
+4 x − y −
sinh(ux) cos(uy)uh(u)du
16
0
Z ∞
Z ∞
−8y
sinh(ux) sin(uy)h (u) du − 8xy
cosh(ux) sin(uy)h (u) du.
0
0
∂
1
< ξR
+ 2x + 2ıy
=
∂y
2
Z ∞
Z ∞
1
2
2
−8y
cosh(ux) cos(uy)h (u) du − 4 x − y −
cosh(ux) sin(uy)uh(u)du
16
0
0
Z ∞
Z ∞
−8x
sinh(ux) sin(uy)h (u) du − 8xy
sinh(ux) cos(uy)h (u) du.
0
0
∂
1
= ξR
+ 2x + 2ıy
=
∂x
2
Z ∞
Z ∞
1
2
2
8x
sinh(ux) sin(uy)h (u) du + 4 x − y −
cosh(ux) sin(uy)uh(u)du
16
0
0
Z ∞
Z ∞
+8y
cosh(ux) cos(uy)h (u) du + 8xy
sinh(ux) cos(uy)uh (u) du
0
0
8
AMANDA KNECHT
∂
1
= ξR
+ 2x + 2ıy
=
∂y
2
Z ∞
Z ∞
1
2
2
−8y
sinh(ux) sin(uy)h (u) du + 4 x − y −
sinh(ux) cos(uy)uh(u)du
16
0
0
Z ∞
Z ∞
cosh(ux) cos(uy)h (u) du − 8xy
cosh(ux) sin(uy)uh (u) du
+8x
0
0
With these equations, we can represent the slopes of the functions where the real
parts or the imaginary parts of ξR are zero by
∂
− ∂x
< ξR 21 + 2x + 2ıy
mRe (x + ıy) = ∂ 1
<
ξ
+
2x
+
2ıy
R
∂y
2
and
∂
− ∂x
= ξR
mIm (x + ıy) = ∂ = ξR
∂y
1
2
+ 2x + 2ıy
.
1
+
2x
+
2ıy
2
1
Implicitly we assume
that
the
domain
of
m
is
w
|
<
ξ
+
2w
=
0
and the
Re
2
1
domain of mIm is w | = ξ + 2w = 0 .
2
∂
1
1
+
2w
=
0
is
negative
if
<
ξ
+
2x
+
2ıy
and
The
slope
of
<
ξ
R
2 ∂x
2
1
∂
1
<
ξ
+
2x
+
2ıy
have
the
same
sign,
and
likewise
for
the
slope
of
=
ξ
+ 2w =
R 2
∂y
2
0. Image 3, which is a graph of ξE , helps us visualize these slopes.
From analyzing these and other graphs, we were able to develop the following
conjectures.
Conjecture 1. (Slope Conjecture) For = (w) > 0, mRe 12 + 2w and mIm 12 + 2w
are positive when < (w) < 0 and negative when < (w) > 0.
All analytic functionssatisfy the equality f (−w) = f (−w), and thus our function f (w) = ξR 21 + 2w satisfies f (−w) = f (w). Thus, by symmetry it suffices
to prove just one of the cases for < (w). From this conjecture, we can deduce that
as < (w) approaches zero the slopes must
also approach zero or become undefined.
If we wish to prove that mRe 21 + 2w approaches zero as Re (w) approaches zero,
we must show that as <
(w) approaches zero,
1
∂
<
ξ
+
2x
+
2ıy
approaches zero faster than the partial derivative with
R 2
∂x
respect to y approaches zero. Otherwise, the slope becomes undefined.
Conjecture 2. The partial derivative with respect to x of < ξ 21 + 2x + 2ıy = 0
approaches zero faster than the partial derivative with respect to y as x approaches
zero.
Another way to look at the slopes of these level sets is to notice that two vectors
are perpendicular when one vector’s slope is the negative
reciprocal of the
other
1
1
vector’s. So for all w ∈ C satisfying < ξR 2w + 2 = = ξR 2w + 2 = 0,
the equation mRe (w)mIm (w) = −1 must also be satisfied. We know from before
that this occurs when <(w) = 0. But the Slope Conjecture implies that this never
occurs outside of the critical line because the slopes have the same sign. So for all
INVESTIGATING ZETA FUNCTIONS
9
points w ∈ C − 0 satisfying either < ξ 12 + 2w = 0 or Im ξ 12 + 2w = 0,
mRe (w)mIm (w) is nonnegative. But in order to prove that the level sets are never
perpendicular outside of the critical strip, we only need to show a weaker case,
the product of the slopes,mRe (w)mIm (w), for a fixed w is always greater than −1.
Let
Z
∞
A =
cosh(ux) cos(uy)h(u)du,
0
Z
∞
sinh(ux) cos(uy)uh(u)du,
B =
0
Z
∞
C =
sinh(ux) sin(uy)h(u)du,
0
Z
∞
cosh(ux) sin(uy)h(u)du.
D =
0
Plugging these into the equations for the real and imaginary parts of ξR and the
partial derivatives we find
1
1
< ξR
+ 2x + 2ıy
=
4 x2 − y 2 − 16
A + 12 − 8xyC
2
1
1
= ξR
+ 2x + 2ıy
=
4 x2 − y 2 − 16
C + 8xyA
2
1
∂
1
+ 2x + 2ıy
= 8xA + 4 x2 − y 2 − 16
< ξR
B − 8yC − 8xyD
∂x
2
∂
1
1
< ξR
+ 2x + 2ıy
= −8xC − 4 x2 − y 2 − 16
D − 8yA − 8xyB
∂y
2
∂
1
1
= ξR
+ 2x + 2ıy
= 8xC + 4 x2 − y 2 − 16
D + 8yA + 8xyB
∂x
2
∂
1
1
= ξR
+ 2x + 2ıy
= 8xA + 4 x2 − y 2 − 16
B − 8yC − 8xyD.
∂y
2
(x2 −y2 − 161 )A
1
Now for x, y ∈ R such that < ξR 2x + 2ıy + 12 = 0, C =
+ 16xy
.
2xy
For x, y ∈ C such that = ξR 2x + 2ıy + 21 = 0, C = x2−2xyA
. Whence, we
( −y2 − 161 )
can rewrite the slopes of the level sets we have been studying in terms of A, B,
and D.
10
AMANDA KNECHT
Conjecture 3. (Weaker statement that implies the Slope Conjecture)
Define A, B and D as before and write
1
−8x(A−Dy)−4(x2 −y 2 − 16
)[B−A/x]+ 2x1
F (x, y) =
1
1
2
2
−8y(A+By)−4x(x −y − 16 )(A/y+D)− 2y
2
G(x, y) =
1
−16x2 yA−8y (x2 −y 2 − 16
)(A+Bx)−4D(x2 −y2 − 161 )
2
1
−16xy 2 A+8x(x2 −y 2 − 16
)(A−Dy)+4B (x2 −y2 − 161 )
.
Then F (x, y) · G(x, y) > −1 if x 6= 0 and w = x + ıy is either in the domain of
mRe 21 + 2w or in the domain of mIm 12 + 2w .
Now, instead of only thinking about the derivatives
withd respect to x 1and
d
1
y, we can also analyze the vectors dw < ξR 2w + 2 and dw = ξR 2w + 2 .
Some simple observations make our study of these vectors easier. First, we know
(z)
(z)
that < (f (z)) = f (z)+f
and =(f (z)) = f (z)+f
for any function f (z). Now
2
2
when f (z) is analytic,
∂
< (f
∂z
∂
∂z
f (z) = 0. Thus we find that for analytic functions
∂
d
(z) and ∂z
= (f (z)) = −ı
f (z) . We know from before that,
2 dz
Z ∞
1
1
1
ξR 2w +
cosh(uw) h (u) du.
= + 4 w2 −
2
2
16
0
(z)) =
1 ∂
f
2 ∂z
Thus,
Z ∞
Z ∞
∂
1
1
2
ξR 2w +
= 8w
cosh(uw) h(u)du+4 w −
sinh(uw) uh(u)du.
∂w
2
16
0
0
Applying the formulae above to ξR , we find:
Z ∞
Z ∞
1
∂
1
2
cosh(uw) h(u)du+2 w −
sinh(uw) uh(u)du.
< ξR 2w +
= 4w
∂w
2
16
0
0
Z ∞
Z ∞
∂
1
1
2
= ξR 2w +
sinh uw uh (u) du.
= −4ıw
cosh uw h (u) du−2ı w −
∂w
2
16
0
0
Now in order to understand these partial derivatives, we returned to analyzing
the graphs of ξR . This
time we concentrated on the how the lines < ξR 2w + 21 =
1
0 and = ξR 2w +
= 0 change as the point w changes. We know that
2
∂
∂z
=
1
2
∂
∂x
∂
− ı ∂y
. Thus both
∂
∂x
and
∂
∂y
∂
∂z
∂
if ∂z
are negative if and only if
is in
∂
∂
the second quadrant. Likewise, both ∂x
and ∂y
are positive if and only
is in
the fourth quadrant. We can use this information and the analysis of the graphs
to make the following conjecture.
Conjecture 4. (Equivalent to the Slope Conjecture) For < (w) > 0 and
= (w) > 0:
∂
• If < ξR 2w + 21 = 0, then the vector ∂w
< ξR 2w + 12 is in the second quadrant when = ξR 2w + 21 > 0 and in the fourth quadrant when
= ξR 2w + 12 < 0.
1
∂
• If = ξR 2w + 2 = 0, then the vector ∂w
= ξR 2w + 12 is in the second quadrant when < ξR 2w + 21 < 0 and in the fourth quadrant when
< ξR 2w + 12 > 0.
INVESTIGATING ZETA FUNCTIONS
11
Now that we have discovered where these level sets intersect, we will turn our
attention to the roots created at the intersection of the level sets <(f (z)) = 0 and
=(f (z)) = 0 of an analytic function f (z). Images 4 and 5 let us visualize roots of
an analytic function. These images suggest the following lemmas.
Lemma 1. If the gradient of an analytic function, f (z), is not equal to zero
at z0 , then the lines < (f (z)) = < (f (z0 )) and = (f (z)) = = (f (z0 )) intersect
perpendicularly.
Proof: We know that a function f (x, y) = u (x, y) + ıv (x, y) is analytic if it
satisfies the Cauchy Riemann equations:
ux = vy
uy = −vx .
We see that the gradients of u (x, y) and v (x, y) are (ux , uy ) and (vx , vy ), respectively. Because we are assuming our function f (x, y) is analytic, we use the Cauchy
Riemann equations to notice that the gradient of v (x, y) is also (−uy , ux ) . And we
can represent the vectors tangent to the level sets by u⊥ (z0 ) = (−uy (z0 ) , ux (z0 ))
and v⊥ (z0 ) = (ux (z0 ) , uy (z0 )) . Now we see that when the gradient is non-zero,
they form perpendicular angles because (−uy (z0 ) , ux (z0 )) · (ux (z0 ) , uy (z0 )) = 0.
Lemma 2. Roots of analytic functions are simple if and only if the gradients of
the real and imaginary parts of the function are not zero vectors.
Proof: Let f (z) be an analytic function with a root at z0 .
First suppose z0 is a simple root. Then we can write f (z) = (z − z0 ) g (z) where
g (z0 ) is not equal to zero. Letting z = x + ıy, z0 = x0 + ıy0 , and g (z) =
g1 (z) + ıg2 (z) we find:
f (z) = [(x − x0 ) g1 (z) − (y − y0 ) g2 (z)] + ı[(x − x0 ) g2 (z) + (y − y0 ) g1 (z)].
Taking partial derivatives of the real and imaginary parts gives us:
∇< = (x − x0 ) g1x (z) − (y − y0 ) g2x (z) + g1 (z) , (x − x0 ) g1y (z) − (y − y0 ) g2y (z) − g2 (z) .
∇= = (x − x0 ) g2x (z) + (y − y0 ) g1x (z) + g2 (z) , (x − x0 ) g2y (z) + (y − y0 ) g1y (z) + g1 (z) .
So , ∇Re (z0 ) = (g1 (z0 ) , −g2 (z0 )) and ∇Im (z0 ) = (g2 (z0 ) , g1 (z0 )) . And g (z0 ) 6=
0 implies that g1 (z0 ) and g2 (z0 ) are never simultaneously zero. Therefore the
gradients are not zero vectors.
Now suppose the gradients are not zero vectors. We can write f (z) = (z − z0 ) g (z)
as before, but this time (z − z0 ) might divide g (z). Just as before, ∇Re (z0 ) =
(g1 (z0 ) , −g2 (z0 )) and ∇Im (z0 ) = (g2 (z0 ) , g1 (z0 )) . Because g (z0 ) = g1 (z0 ) +
ıg2 (z0 ) and the gradients are not zero vectors, g (z0 ) is not equal to zero. Thus,
(z − z0 ) does not divide g (z), so z0 is a simple root.
Lemma 3. The mth -order roots of an analytic function occur only when the level
sets, < (f (z)) = 0 and = (f (z)) = 0, of the function intersect to create 4m angles
π
measuring 2m
radians each.
12
AMANDA KNECHT
Proof: Suppose f (z) is analytic at z0 and has an mth -order root at z0 . Then
f (z0 ) = f 0 (z0 ) = f 00 (z0 ) = . . . = f (m−1) (z0 ) = 0 but f (m) (z0 ) 6= 0.
2
f (m+1) (z0 )
f (m+2) (z0 )
Letting g (z) = (m+1)f
(m) (z ) (z − z0 ) + (m+2)(m+1)f (m) (z ) (z − z0 ) + . . .,
0
0
f (z) = (z − z0 )m
f (m)(z0 )
(1 + g (z)) , which is valid in a neighborhood of z0 .
m!
Now let
θ0 = lim arg (z − z0 )
z→z0
φ0 = lim arg (f (z))
z→z0
Then,
f (m) (z0 )
(1 + g (z)) .
φ0 = lim arg (z − z0 )
z→z0
m!
And because arg (xyz) = arg (x) + arg (y) + arg (z) for any x, z, y ∈ C,
1 + g (z)
m
(m)
φ0= lim arg (z − z0 ) + arg f (z0 ) + arg
.
z→z0
m!
g(z) → 0 as z → z0 , so
φ0→ m arg (z − z0 ) + arg f (m) (z0 ) as z → z0 .
m
We will now let α denote the angle between two arcs C1 and C2 passing through
z0 and let β denote the angle between the arcs Γ1 = f (C1 ) and Γ2 = f (C2 ) . Then
α = θ1 − θ2 and β = φ1 − φ2 . And so,
β = mθ1 − arg f (m) (z0 ) − mθ2 + arg f (m) (z0 ) .
= m (θ1 − θ2 ) .
= mα.
Letting the arc C1 be an arc in the level set < (f (z)) = 0 and C2 be an arc in the
level set = (f (z)) = 0 the function f maps C1 and C2 to the real and imaginary
axes, respectively. We know that the angle between the real and imaginary axes
.
on the complex plane is π2 +2kπ radians. So β = mα = π2 +2kπ implies α = π+4kπ
2m
Thus the angle between the level sets < (f (z)) = 0 and = (f (z)) = 0 is a multiple
π
of 2m
radians.
Now let c =
f (m)(z0 )
m!
6= 0 and write
m
p
m
f (z) = (z − z0 ) c (1 + g (z)) = (z − z0 ) c + 1 + g(z) .
m
Define a new function h(z) by
p
h(z) = (z − z0 ) m c (1 + g (z)).
p
Observe that since g(z0 ) = 0, m c + 1 + g(z) is well
√ defined in a neighborhood
of z0 . Because g(z) → 0 as z → z0 , h0 (z0 ) = m c 6= 0. Thus by the Inverse
Function Theorem, h−1 (z) exists in a neighborhood of z = z0 . Now we know
kπ
the angle between the curves C1 = z|arg(z) = 2m
and C2 = {z|arg(z) = 0}
kπ
−1
is 2m radians. Because h preserves angles, the angle between h−1 (C1 ) and
kπ
h−1 (C1 ) is 2m
radians. Knowing f (z) = (h(z))m , the angle between f (h−1 (C1 )) =
INVESTIGATING ZETA FUNCTIONS
13
z|arg(z) = kπ
and f (h−1 (C2 )) = {z|arg(z) = 0} is kπ
radians. Since the images
2
2
of these curves are the real and imaginary axes for every k ∈ Z, 4m angles are
created.
3. Consequences of the Slope Conjecture
The proof of the following statements rests entirely on the previously stated
lemmas and the proof of the Slope Conjecture.
Consequence 1. The only non-trivial roots of the Riemann Zeta Function, ζR
can be found on its critical line, < (w) = 0.
1
2
+ 2w ,
Proof: If roots did exist outside the critical line, they would be the result of
the level sets of the real and imaginary parts of ξ intersecting. But the Slope
Conjecture tells us that on either side of the line
<(z) = 1/2 the slopes of the level
sets < ξR 21 + 2w = 0 and = ξR 21 + 2w = 0 have the same sign. Thus,
outside of this line the level sets do not intersect. Therefore, the only simple roots
of ξR 12 + 2w occur when < (w) = 0.
We know that
1
1
1
− 12 −2w
ξR
+ 2w = π
Γ
+ 2w ζR
+ 2w .
2
2
2
And thus
ζR
1
+ 2w
2
=
ξR
1
1
2
π − 2 −2w Γ
+ 2w
1
2
+ 2w
.
1
Realizing that π − 2 −2w is never zero and that Γ only has simple roots at w =
−1 −3 −5
, 4 , 4 , . . ., where we know the trivial zeros of ζ lie, we conclude that ζR 21 + 2w
4
has non-trivial roots exactly when ξR 21 + 2w = 0. And thus, ζR 21 + 2w = 0
only when < (w) = 0.
Consequence 2. All of the roots of the Riemann Zeta Function are simple.
Proof: We know from the Slope Conjecture that the only roots of ζR are on
the line <(z) = 1/2. Now if these slopes were not simple, then for <(w) > 0 either
the slope mRe ( 21 + 2w) or mIm ( 21 + 2w) would have to approach a number greater
than or equal to 1 as w → 0. But this would contradict the Slope Conjecture; thus
all roots are simple.
4. The Epstein Zeta Function
The Epstein Zeta function is a generalization of the Riemann Zeta function and
is defined as follows,
14
AMANDA KNECHT
Definition 2. [Te] Epstein’s zeta function of Y , an n × n, positive symmetric
matrix, and s ∈ C, with < (s) > n/2 is
Z (Y, s) =
X
Y [v]−s .
v∈Zn −0
Here v is a column vector and Y [v] =t vY v with t v = transpose of v. Thus
P
Y [v] = ni,j=1 yij vi vj . In the special case where n = 1 and Y = 1, we find that
Z (1, s) = ζR (2s) . The general Epstein zeta function has an analytic continuation
similar to that of ζR (s). Terras [Te] gives the following theorem for the analytic
continuation.
Theorem 1. The Analytic Continuation of Epstein’s Zeta Function Let
R∞
γ (s, x) be the incomplete gamma function defined by γ (s, x) = x us−1 e−u du, and
set
−s
Z
G (s, x) = x γ (s, x) =
∞
u(s−1) e−xu du.
1
Then Epstein’s zeta function can be analytically continued to all s ∈ C with its
only pole a simple one at s = n/2. Here |Y | denotes the determinant of Y .
The analytic continuation comes from the incomplete gamma expansion:
Λ(Y, s) = π −s Γ(s)Z(Y, s)
P
|−1/2
1
− 2s
+ 21 v∈Zn −0 G (s, πY [v]) + |Y |−1/2 G
= |Y2s−n
n
2
− s, πY −1 [v] .
Thus Epstein’s Zeta function satisfies the functional equation
Λ(Y, s) = |Y |−1/2 Λ(Y −1 , n/2 − s).
Now for a 2 × 2 matrix Y, we define a new function ξE (Y, S) that is defined on
the entire complex plane and shares its roots with the nontrivial roots of Z(Y, s).
To do this we merely alter the function Λ(Y, s), which is defined in Theorem 4.1.
But before defining our new function, we first simplify our symmetric 2 × 2
matrix. Say Y = cM where M is a symmetric matrix and c is a nonzero conP
−s
stant. Then Z(Y, s) = Z(cM, s) =
= c−s Z(M, s), and c−s
v∈Zn −0 (cM [v])
INVESTIGATING ZETA FUNCTIONS
is never zero. So given a matrix Y ,let c =
1
|Y |1/n
and then M =
15
1
Y
|Y |
and
1
n |Y | = 1. Thus we see that every symmetric matrix can be written
||Y |1/n |
as a product of a nonzero constant and a matrix with determinant 1, and every
|M | =
Epstein Zeta function can be written as the product c−s Z(M, s) with |M | = 1.
So we will restrict our study to matrices with determinant 1. Earlier discov
t 0
eries of nontrivial zeros of Z(Y, s), [Te], when Y is of the form
and
0 1/t
t = 10, 100, 1000 . . . prompts us to simplify even further to matrices of this form.
1/t 0
−1
Now with this form of matrix we see that because Y =
,
0 t
X
X
X
X
1
1 2
Y −1 [v].
a + tb2 =
Y [v] =
ta2 + b2 =
t
t
v∈Zn −0
v∈Zn −0
a,b∈Z
a,b∈Z
not both zero
not both zero
Finally we can define our new function
ξE (Y, s) =
= s(2s − 2)Λ (1 − s)
X
= 1 + s (s − 1)
(G (s, πY [v]) + G (1 − s, πY [v])).
v∈Z2 −0
Letting s =
1
2
+ x + ıy since the line of symmetry is <(z) = 1/2 for a 2 × 2 matrix
1
ξE (Y, + x + ıy) = 1+
2
X 1
1
1
2
2
−1
− + x − y + 2ıxy
G
+ x + ıy, πY [v] + G
− x − ıy, πY [v] .
4
2
2
2
v∈Z −0
Because our column vector has two elements which are not both zero, say a, b ∈
Z, we can define a new function f (a, b) = πY [v] = π ta2 + 1t b2 and rewrite
t 0
1
ξE
, 2 + x + ıy as
0 1/t
1
t 0
ξE
, + x + ıy =
0 1/t
2
Z ∞
1
X
1
1
2
2
= 1 + − + x − y + 2ıxy
e−uf (a,b) u− 2 +x+ıy + u− 2 −x−ıy du
4
1
a,b∈Z
not both zero
1
= 1 + − + x2 − y 2 + 2ıxy
4
X
a,b∈Z
not both zero
Z
1
∞
2
e−uf (a,b) √ cosh ((x + ıy)(ln u)) du.
u
16
AMANDA KNECHT
Knowing that cosh(x + ıy) = cos(y) cosh(x) + ı sin(y) sinh(x), we write the real
and imaginary parts of ξE
1
t 0
< ξE
, + x + ıy
=
0 1/t
2
Z ∞
X
2
1
2
2
e−uf (a,b) √ cos(y ln u) cosh(x ln u)du
= 1+ − +x −y
4
u
1
a,b∈Z
not both zero
−2xy
1
a,b∈Z
not both zero
t 0
0 1/t
X
= ξE
= 2xy
∞
Z
X
2
e−uf (a,b) √ sin(y ln u) sinh(x ln u)du.
u
1
=
, + x + ıy
2
Z ∞
2
e−uf (a,b) √ cos(y ln u) cosh(x ln u)du
u
1
a,b∈Z
not both zero
1
+ − + x2 − y 2
4
Z
X
a,b∈Z
not both zero
1
∞
2
e−uf (a,b) √ sin(y ln u) sinh(x ln u)du.
u
Again we take the partial derivatives of these functions.
∂
1
t 0
< ξE
, + x + ıy
=
0 1/t
∂x
2
Z ∞
X
1
2
2
2
=
− +x −y
e−uf (a,b) √ cos(y ln u) sinh(x ln u)(ln u)du
4
u
1
a,b∈Z
X
+2x
Z
e
1
a,b∈Z
not both zero
X
−2y
∞
Z
a,b∈Z
not both zero
−2xy
not both zero
∞
−uf (a,b)
X
a,b∈Z
not both zero
1
Z
1
2
√ cos(y ln u) cosh(x ln u)du
u
2
e−uf (a,b) √ sin(y ln u) sinh(x ln u)du
u
∞
2
e−uf (a,b) √ sin(y ln u) cosh(x ln u)(ln u)du.
u
INVESTIGATING ZETA FUNCTIONS
17
∂
1
t 0
< ξE
, + x + ıy
=
0 1/t
∂y
2
Z ∞
X
2
1
2
2
e−uf (a,b) √ sin(y ln u) cosh(x ln u)(ln u)du
= − − +x −y
4
u
1
a,b∈Z
not both zero
X
−2y
Z
−2x
∞
2
e−uf (a,b) √ sin(y ln u) sinh(x ln u)du
u
Z
a,b∈Z
not both zero
X
−2xy
2
e−uf (a,b) √ cos(y ln u) cosh(x ln u)du
u
1
a,b∈Z
not both zero
X
∞
1
∞
Z
1
a,b∈Z
not both zero
2
e−uf (a,b) √ cos(y ln u) sinh(x ln u)(ln u)du.
u
1
∂
t 0
= ξE
, + x + ıy
=
0 1/t
∂x
2
Z ∞
X
1
2
2
2
=
− +x −y
e−uf (a,b) √ sin(y ln u) cosh(x ln u)(ln u)du
4
u
1
a,b∈Z
X
+2y
e
1
a,b∈Z
not both zero
X
+2x
Z
a,b∈Z
not both zero
+2xy
not both zero
∞
−uf (a,b)
Z
X
a,b∈Z
not both zero
∞
1
Z
1
2
√ cos(y ln u) cosh(x ln u)du
u
2
e−uf (a,b) √ sin(y ln u) sinh(x ln u)du
u
∞
2
e−uf (a,b) √ cos(y ln u) sinh(x ln u)(ln u)du.
u
18
AMANDA KNECHT
∂
1
t 0
= ξE
, + x + ıy
=
0 1/t
∂y
2
Z ∞
X
2
1
2
2
e−uf (a,b) √ cos(y ln u) sinh(x ln u)(ln u)du
=
− +x −y
4
u
1
a,b∈Z
Z
X
+2x
e
1
a,b∈Z
not both zero
a,b∈Z
not both zero
1
Z
X
−2xy
∞
Z
X
−2y
not both zero
∞
−uf (a,b)
a,b∈Z
not both zero
2
e−uf (a,b) √ sin(y ln u) sinh(x ln u)du
u
∞
1
2
√ cos(y ln u) cosh(x ln u)du
u
2
e−uf (a,b) √ sin(y ln u) cosh(x ln u)(ln u)du.
u
As before we can define new functions A, B, C and D.
Z ∞
X
2
e−uf (a,b) √ cos(y ln u) sinh(x ln u)du(ln u)
A =
u
1
a,b∈Z
not both zero
B =
X
a,b∈Z
not both zero
C =
X
a,b∈Z
not both zero
D =
X
a,b∈Z
not both zero
Z
∞
2
e−uf (a,b) √ cos(y ln u) cosh(x ln u)du
u
∞
2
e−uf (a,b) √ sin(y ln u) sinh(x ln u)du
u
∞
2
e−uf (a,b) √ sin(y ln u) cosh(x ln u)(ln u)du.
u
1
Z
1
Z
1
With these new functions we can more easily express the partial derivatives and
show that ξE
∂
< ξE
∂x
∂
< ξE
∂y
∂
= ξE
∂x
∂
= ξE
∂y
is an analytic function.
1
t 0
, + x + ıy
= − 14 + x2 − y 2 A + 2xB − 2yC − 2xyD
0 1/t
2
1
t 0
, + x + ıy
= −2xyA − 2yB − 2xC − − 14 + x2 − y 2 D
0 1/t
2
1
t 0
, + x + ıy
= 2xyA + 2yB + 2xC + − 41 + x2 − y 2 D
0 1/t
2
1
t 0
, + x + ıy
= − 14 + x2 − y 2 A + 2xB − 2yC − 2xyD.
0 1/t
2
Now we turn our study from an investigation of the change of
t 0
1
ξE
, 2 + x + ıy as x and y change to a study of how the function
0 1/t
INVESTIGATING ZETA FUNCTIONS
19
changes with respect to t by taking the derivative with respect to t of ξE .
∂
1
t 0
ξE
, + x + ıy
=
0 1/t
∂t
2
1
2
2
= π − + x − y + 2ıxy ·
4
Z ∞
1
X
1
b2
2
−ue−uf (a,b) u− 2 +x+ıy + u− 2 −x−ıy du
a − 2
t
1
a,b∈Z
not both zero
1
2
2
= π − + x − y + 2ıxy ·
4
Z ∞
X
2
b2
2
−ue−uf (a,b) √ cosh((1 + x) ln u)du.
a − 2
t
u
1
a,b∈Z
not both zero
This expression can be simplified to
∂
1
t 0
=
ξE
, + x + ıy
0 1/t
∂t
2
X
1
2
2
= −π − + x − y + 2ıxy
4
a,b∈Z
b2
a − 2
t
2
·
not both zero
b2
1
b2
3
2
2
+ x + ıy, π ta +
+ G − − x − ıy, π ta +
.
G
2
t
2
t
Similarly to the investigation of ξR , we studied the images of the graphs for
different values of t, which are included in the appendix. The numerical data
produced by these graphs suggests that when t is sufficiently large the analogue
of the Slope Conjecture fails. However the conjecture does seem to be valid when
0 < t < 7. Also the analogue of the Riemann Hypothesis fails for larger t. The
non-trivial roots of ξE we encountered are results of the slopes of the level sets
< (ξE (s)) = 0 and = (ξE (s)) = 0 being both positive and negative for <(s) > 1/2.
The first five images were created by the program Fractal Explorer written
by Peter Stone. The first image is the four color graph of the complex plane.
Points colored blue have positive real and imaginary parts, while red points have
negative real and positive imaginary parts. Green and yellow are similar. When
this program graphs a function f (z), the location of the point is the value z and
24
AMANDA KNECHT
the color represents the quadrant f maps z to.
The last images, those graphing ξE , were created by a graphing applet written by
Ken Richardson. For these images, purple represents the first, yellow the second,
blue the third, and red the fourth quadrant.
References
[Ed]
[Pa]
[Te]
[Ti]
H.M. Edwards. Riemann’s zeta function. Dover Publications, Mineola, New York, 1974.
S.J. Patterson. An introduction to the theory of the Riemann zeta-function. Cambridge
University Press, Cambridge, 1989.
Audrey Terras. Harmonic analysis on symmetric spaces and applications.I. SpiringerVerlang, New York, 1985.
E.C. Titchmarsh. The theory of the Riemann zeta-function. Oxford University Press,
London, 1951.