A BASIC CALCULUS APPROACH TO ESTIMATING
THE ZEROS OF THE RIEMANN ZETA FUNCTION
By
JOAN A. PHARR
A Research Paper Submitted to
WAKE FOREST UNIVERSITY
In Partial Fulfillment of the Requirements
for the Degree of
BACHELOR’S OF SCIENCE
In the Department of Mathematics
May 2008
Winston Salem, North Carolina
John V. Baxley, Ph.D., Advisor
Abstract
Joan A. Pharr
A BASIC CALCULUS APPROACH TO ESTIMATING THE ZEROS OF
THE RIEMANN ZETA FUNCTION
Research under the direction of John V. Baxley, Ph.D., Professor of Mathematics.
Let p be a complex number with Re(p)=
1
2
ζ(p) =
∞
X
1
p
k=1 k
We are interested in enlarging the domain of this function and finding an approximation
for the value of its zeros using methods familiar to students of basic calculus.
1
The Zeta Function
To define the Riemann Zeta function we begin with the p-series from calculus. This
P
1
is the series ∞
k=1 kp , which converges for p > 1. The most common way to verify this
convergence is to use the integral test. For our definition we will need to consider complex
values of p = r + it. For such p we can use Euler’s definition of
eiθ = cos θ + i sin θ
and thus k −p = e−(r+it) ln(k) = e−r ln(k) e−it ln(k) . Then we simplify
1
1
= r [cos(t ln(k)) − i sin(t ln(k))].
p
k
k
Giving us | k1p | = k1r so we see that the p-series for a complex p converges absolutely for
r = Re(p) > 1. Real series which converge absolutely are also convergent. This is valid for
complex series as well. Therefore the p-series converges in the right half plane Re(p) > 1.
The sum of this series defines the Riemann Zeta Function, ζ(p), for such values of p.
Riemann’s Hypothesis
This Zeta function is the key feature of Riemann’s Hypothesis. Riemann (1859) used
complex analysis to extend the domain of ζ(p) to include all complex numbers, with the
exception of p = 1. This extended function has real zeros at each negative even integer.
These are called the trivial zeros of the Zeta function. Riemann realized that there were
many other zeros, all non-real, and showed that they were located in the critical strip, 0 ≤
Re(p) ≤ 1. Based on empirical evidence, Riemann conjectured that all these zeros actually
satisfy Re(p) = 12 , and called this line the critical line. To this day the conjecture has yet
to be proven theoretically, although there is now a wealth of computations supporting his
hypothesis.
Riemann stated he had given some “vain, fleeting attempts” to verify his conjecture,
but put these efforts aside because it was not necessary for his goal of understanding the
distribution of prime numbers. In 1896 his ideas were used to prove the prime number
theorem, separately and simultaneously by Jacques Hadamard and Charles de la Vallee
Poussin, which only depended on knowing that all the non-trivial zeros have Re(p) < 1.
Enlarging the domain of the Zeta Function
There are many ways to enlarge the domain of the Riemann Zeta function. We are
interested in the zeros which lie within the critical strip. Therefore it is necessary only
to enlarge the domain to include the critical strip, 0 ≤ Re(p) ≤ 1. We must ensure that
after enlarging the domain we still have the same function. Luckily, thanks to a wonderful
theorem in complex analysis, there is only one way to extend the function if we insist
on keeping the extended function differentiable, i.e. analytic. If our extended function is
analytic we can be sure that it is the same as Riemann’s extension, no matter how different
our expression may appear from the original or Riemann’s extension.
2
Using some advanced calculus, we can show that the Zeta function is analytic for
Re(p) > 1. We know that if the differentiated series converges uniformly then the sum of
the series is differentiable and can be differentiated term-by-term. The differentiated series
P
− ln(k)
≤ ln(k)
. If > 0 and r = Re(p) ≥ 1 + , then ln(k)
≡ Mk .
for the zeta function is ∞
k=1
kp
|kp |
k1+
P∞
Since k=1 Mk converges by the ratio test, then by the Weierstrass M-test, our differentiated series converges uniformly for r = Re(p) ≥ 1 + , for any > 0.
A very easy way to enlarge the domain is by changing the p-series to the alternating
series
∞
X
(−1)k+1
A(p) =
.
kp
k=1
By the alternating series test we know that this series converges when p > 0. This result
also holds for complex p for Re(p) > 0, thus includes the critical strip in the interval of
convergence. Considering Re(p) > 1 we see
ζ(p) − A(p) = 2
1
1
1
+
+
+ ...
2p 4p 6p
Which can be simplified to
ζ(p) − A(p) =
and thus
ζ(p) =
1
2p−1
ζ(p)
A(p)
1
1 − 2p−1
This method was first developed by Euler, but it converges exceedingly slowly because the
rate of convergence is that of the first term omitted. The error is really high. Adding
up over 10,000 terms yields only two decimal places of accuracy, making it a relatively
ineffective method for computation. Since absolute convergence is not true in the critical
strip and the Weierstrass M-test cannot be used, it is not obvious that A(p) is differentiable
for Re(p) > 0.
Our method described below is not nearly as effective as the Riemann-Seigel formula.
Although it lacks in sophistication, only knowledge of elementary calculus and some complex numbers is needed for the explanation. To enlarge the domain we will use Taylor’s
formula from calculus. This method was first used by Dr. Baxley to enlarge a domain
for computational purposes in a December, 1992, Mathematics Magazine article on Euler’s
Constant, Taylor’s Formula and Slowly Converging Series. This method via Taylor’s formula is effective for computation, if the Im(ζ(p)) is not too large. It seems to be roughly
as effective as a well-known method using the Euler-Maclaurin summation formula, which
we will not describe here.
Keeping f (x) = x1p in mind, Taylor’s formula provides us with
f (x, p) = f (k, p) + f 0 (k, p)(x − k) + · · · +
f (n) (k, p)
(x − k)n + Rn (x, k),
n!
(1)
where k is a positive integer. Note our function depends on p as well, the reasoning behind
this will be explained shortly. Since our f (x) is complex for real x, we must be careful
3
about the value of the remainder term Rn (x, k). Writing f (x, p) = g(x, p) + ih(x, p), with
g(x, p) and h(x, p) real, we can write down the Taylor formula for g(x, p) and also for
h(x, p). Multiplying the formula for h(x, p) by i and adding we get
Rn (x, k) =
g (n+1) (c, p) + ih(n+1) (d, p)
(x − k)n+1
(n + 1)!
for some numbers c and d between k and x. Thus
q
|Rn (x, k)| =
((g (n+1) (c, p))2 + (h(n+1) (d, p))2 )
|x − k|n+1
(n + 1)!
Remember for any complex number z = a + ib, we have |a| ≤
|b| ≤
q
q
(a2 + b2 ) = |z| and
(a2 + b2 ) = |z|. Therefore
q
|Rn (x, k)| ≤
|f (n+1) (c)|2 + |f (n+1) (d)|2 )
|x − k|n+1
(n + 1)!
Now letting Mn (f, k, x) be the maximum value of |f n+1 (t)| for all t between k and x, we
have
q
((Mn (f, k, x))2 + ((Mn (f, k, x))2 )
|Rn (x, k)| =
|x − k|n+1
(n + 1)!
√
2Mn (f, k, x)
|x − k|n+1 .
=
(n + 1)!
Equation (1) along with the above bound is a statement of Taylor’s theorem for such
complex valued functions with a real variable x.
Equation (1) is derived with p held constant and all derivatives are with respect to
x. For our function f (x, p) = x−p , one can check that f (x, p), f 0 (x, p), ..., f (n) (x, p) are all
differentiable, hence analytic, with respect to p and with x held constant. The resulting
expression when solving for Rn (x, p) is clearly an analytic function of p. Furthermore
this shows that the expression we get for the Zeta function after using Taylor’s formula is
analytic in the critical strip excluding p = 1.
Using n = 1 in (1) and integrating from k − .5 to k + .5 gives us
Z k+.5
k−.5
f (x)dx = f (k) + Ck ,
where
Ck =
Z k+.5
k−.5
(2)
R1 (x, k)dx.
Since |f 00 (x)| = |p(p − 1)xp−2 | = |p(p − 1)|xr−2 is decreasing for 0 ≤ r ≤ 1, it is straightforward to verify that
√
Z k+.5
√ 00
2 00
|Ck | ≤ 2|f (k − .5)|
(x − k)2 dx ≤
|f (k − .5)|.
24
k
4
Summing (2) from n + 1 to ∞ (for Re(p) > 1) gives
Z ∞
∞
X
f (x)dx =
n+.5
f (k) + En
(3)
k=n+1
where
√
√ Z
√
∞
2 X
2 ∞ 00
2|p(p + 1)|
00
|En | ≤
|f (k − .5)| ≤
|f (x − .5)|dx ≤
.
24 k=n+1
24 n
36(n − .5)Re(p)+1
Performing the integration in (3) with f (x) = x−p and replacing the sum in (3) by ζ(p) −
Pn
1
k=1 kp , we then solve for ζ(p) and get the 1st Taylor’s approximation
ζ(p) =
n
X
1
(n + .5)1−p
−
− En
p
1−p
k=1 k
(4)
This expansion yields an effective approximation for zeros when the Im(p) was not large.
We will provide some computational results later.
We increased the number of terms in Taylor’s formula with the hope of increasing
accuracy. Returning to Taylor’s formula (1) with n = 3, integrating from k − .5 to k + .5,
we get
Z k+.5
f 00 (k, p)
+ Dk
f (x, p) = f (k, p) +
24
k−.5
where Dk =
R k+.5
k−.5
R3 (k, p). Summing from n + 1 to ∞, we get
∞
X
Z ∞
1
f (k) +
f (x)dx =
24
n+.5
k=n+1
Z ∞
n+.5
00
f (x)dx −
En00
+ En0 ,
where we have used Equation (3), applied to f 00 (x) rather than f (x) and used the approximation for En above for the second derivative, while En0 is the sum of the Dk ’s. Note that
|f (4) (x)| is decreasing and the required convergence
occurs for f (x) = x1p , with Re(p) > 1.
R∞ 1
1
Thus we substitute f (x) = xp starting with n+.5 xp dx and Re(p) > 1 to obtain the 3rd
Taylor’s approximation
−
∞
X
(n + .5)1−p
1
p
=
+
+ Qn
p
1−p
24(n + .5)p+1
k=n+1 k
After substitution and some rearranging we have
ζ(p) = −
where
n
(n + .5)1−p X
1
p
+
−
− Qn
p
1−p
24(n + .5)p+1
k=1 k
√
13 2|p(p + 1)(p + 2)(p + 3)|
|Qn | ≤
.
5760(−Re(p) − 3)(n − .5)Re(p)+3
5
(5)
For real values p the calculus student will find the above calculations familiar. Fortunately,
the calculations remain valid for complex p. Note that all terms on the right side of (5) are
analytic if Re(p) > −3 and p 6= 1. Thus we can extend ζ(p) using (3) to include the critical
strip.
Computing ζ(p) on the Critical Line
1
Remember how we defined the zeta function earlier: ζ(p) = ∞
k=1 kp which can be
Pn
P
1
broken into ζ(p) = k=1 k1p + ∞
k=n+1 kp . By rearranging for the sum found in (5) and
solving for ζ(p), we now have an approximation for the Zeta function. To find the value of
ζ(1/2) using the 3rd Taylor’s approximation we substitute p = .5
P
ζ(1/2) ≈
√
1
1
√ − 2 n + .5 −
.
48(n + .5)3/2
k
k=1
n
X
According to Edwards’ book on Riemann’s Zeta Function, the correct value of ζ(1/2) to
six decimal places is 1.46035496; our approximation differs from the correct value by less
than five in the seventh decimal place. With p = .5 our error estimate yields
√
39 2
.
|Qn | ≤
9200(n − .5)7/2
For n = 10, we have error less than 1.725461240 ∗ 10−7 ; for n = 17, we have error less than
2.498872064 ∗ 10−8 . This appears to indicate that to increase our accuracy we only need
to increase n. However, as the imaginary part of .5 + it gets larger, the error estimate for
Qn grows rapidly. For example, for t ≈ 50 our error estimate yields roughly
√
13 2(50)4
.
|Qn | ≤
1560(3.5)(n − .5)7/2
For n = 50, 100, 200, 500, we get Qn ≤ .0067, .00058, .000051, .0000020. As p moves
further up the critical line, it becomes necessary that n becomes uncomfortably large. This
can be partially mitigated by taking more terms in Taylor’s formula. Attempting to attain
greater accuracy we extend using Talor’s 5th approximation formula
Z ∞
n+.5
f (x)dx =
∞
X
n+1
f (k) −
1 000
1 0
f (n + .5) −
f (n + .5)
24
1920
Using
f (x) = x−p , f 0 (x) = −px−p+1 ,
f 00 (x) = p(p + 1)x−p+2 , f 000 (x) = −p(p + 1)(p + 2)x−p+3 ,
we obtain
−
n
X
(n + .5)1−p
1
p
p(p + 1)(p + 2)
= ζ(p) −
+
+
p
p+1
1−p
24(n + .5)
1920(n + .5)p+3
k=1 k
6
Rearranging for ζ(p)
ζ(p) =
Let p =
1
2
n
X
1
p
(n + .5)1−p
p(p + 1)(p + 2)
−
−
−
p
p+1
1−p
24(n + .5)
1920(n + .5)p+3
k=1 k
+ it, thus
n
X
.5 + it
(n + .5).5−it
−
.5+it
.5 − it
24(n + .5)1.5+it
k=1 k
(.5 + it)(1.5 + it)(2.5 + it)
−
1920(n + .5)3.5+it
ζ(.5 + it) =
1
−
k −it = e−it ln(k) = cos(t ln(k)) − i sin(t ln(k))
The third Taylor’s doubling “ n” results in an additional decimal place of accuracy.
However, lengthening the terms in Taylor’s formula is not enough. The “n” in the denominator does aid in sharpening the estimate, but the additional “p” terms in the numerator
quickly detract from this improvement as the Re(p) increases. Consider the case were
p = .5 + i50. Clearly, ours is not the best method to compute ζ(p) for values far up the
critical line. The experience here is similar to that using the Euler-Maclaurin summation
formula. Serious computations require much more sophisticated methods. In this light, the
Riemann-Siegel formula is all the more impressive.
7
References
[1] John V. Baxley,(1992) Euler’s Constant, Taylor’s Formula and Slowly Converging Series
Mathematics Magazine 65.5. 302–313.
[2] John Derbyshire Prime Obsession 2003: Joseph Henry Press, Washington, D.C..
[3] Harold M. Edwards Riemann’s Zeta Function 1974: Dover Publications, INC., Mineola,
New York.
8