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 Group Title: Department of Computer and Information Science and Engineering Technical Reports Title: B-splines and the Riemann's zeta function on integers
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 Material Information Title: B-splines and the Riemann's zeta function on integers Alternate Title: Department of Computer and Information Science and Engineering Technical Report ; 2009-474 Physical Description: Book Language: English Creator: Entezari, Alireza Publisher: Department of Computer and Information Science and Engineering, University of Florida Place of Publication: Gainesville, Fla. Publication Date: November 8, 2008 Copyright Date: 2008
 Record Information Bibliographic ID: UF00095733 Volume ID: VID00001 Source Institution: University of Florida Holding Location: University of Florida Rights Management: All rights reserved by the source institution and holding location.

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B-splines and the Riemann's Zeta Function on

Integers

Alireza Entezari
entezari@cise.ufl.edu
Technical Report: REP-2009-474
CISE Department, University of Florida

Nov 08, 2008

1 Background and definitions

The Poisson's summation formula is:
1 2x
f(kT) Z- (n ), (1)
kCZ nCZ
where f is the Fourier transform of f:

f(wu) I f(x) exp (-iwx) dx.

The Poisson's summation formula holds for all integrable functions, given the
series on the right hand side is absolutely convergent.
The simplest (centered) B-spline, 3o, is the characteristic function of [-1/2, 1/2).
Higher order B-splines 3, are defined recursively by 3m = Pm-1 po. Fourier
transform of a B-spline is:
(w) = sinc"m+(w), (2)
where sinc(t) := sin(t/2)/(t/2).
Lemma 1 (Partition of Unity).

m3,(k) 1
kcZ
Proof. Using Poisson's summation formula, setting T = 1, we have:

S3,m(k)= since + (2)7n) 1,
kCZ nCZ
since sin(rn) 0, for all n E Z and the only non-zero term in the right hand
side comes from n = 0: sinc(0) := 1 that agrees with the actual limit of sine at
0. O

1.1 Zeta function

The Riemann's Zeta function is defined as:

((x) (3)
nl
n 1

2 Relationship between /3m and (m n+ 1)

By playing around with the parameter T, one can recover parts of the series.
The linear B-spline can be related to the ((2) and generally, ,3m can be related
to ((m + 1). For the case of linear B-spline, by setting T = 2 in (1), we have:

3 Bi(2k) s- sinc2(n7)

1C
kCZ neZ

2 + since nr)
n-l
_1 + 0 sin2(n7)
2 (7/2)2 n2
1 4 sin 2(n)
_2 (/2 n2
n-l
The last series is nearly ((2) except that sin2(n-) zeroes out the even terms.
But it is, surely, convergent, since it is a sub-series of ((2), so as the series with
remaining terms of ((2). The last series can be resolved, however, using ((2):

Ssin2(n) 1 17
n2 W2 2 (2n)2
n=l n=l n=l

n=1
((2) ((2)
4

4
4((2).
Therefore, we can relate 31 to ((2) by:
1 3
31(2k) 2 + ((2). (4)
kcZ
Since the support of 31 is [-1, 1), the left hand side is 1 and we have:

((2) 2
6
((2) has been used in the so called Basel problem. Its inverse is the probability
that two randomly selected integers are relatively prime.

2.1 Result
Using the above approach, one can derive the exact values for all even integers
of C using evaluation of B-splines on even integers. Wikipedia lists C(2), ((4)
and C(6), but, apparently, evaluation of C, in general is difficult and there are
papers for evaluation such as [1].
Is there a closed-form solution for all B-splines of odd order:

S 32m +(2k) =? (5)
kCZ

3 The difficult, but interesting case

The odd values of C, are interesting; for instance, C(3), known as Ap6ry's
constant [2] is a curious number that occurs in various physical problems. The
exact value of this constant is not known and it is an open problem whether
this number is transcendental.
To derive Ap6ry's constant using the B-spline approach, we shall focus on
the quadratic B-spline; employing the Poisson's sum (1), and T = 2, we have:

1 32(2k) -1 sincs(rn)
kcZ ntZ
1 sin3 (n")
2 (,(n)3

1 8 sin3(n )
2- + s3 ( ) even terms are zero.
2 + 8 n3
n= 1

The last series in the above derivation is more difficult to tackle since sin3 (n )
is alternating its sign for the non-zero terms.

sin n(n) \ sin ((2n +1)2)
S n3 (2n + 1)3
n=1 n=0
(-1)
= w (6)
Y-0(2n + 1)3 (6)
f 1 1
S(4n+ 1)3 (4n+ 3)
n=0 n=0

On the other hand:

((3) t 7
n=l
1 00
(2n + 1)3 (2n)3

(2n + 1)3 + (
n=0 n1
1 1
E + -((3).
n=-(2n + 1)3 8

Hence, we have:

7 1
((3) 1 (2n + 1)3
n=0
00 00
1 1 +1O (7)

O(4 + )3 + (4n + 3)3

(7) and (6) are different and hence we can not resolve ((3) using this approach.

4 S.O.S

The question is: is there a more appropriate choice than T = 2? By choosing
different values like T 3/2, we get different sub-series of ((3). Can we build
((3), perhaps, from multiple choices for T?

References

[1] Jonathan M. Borwein, David M. Bradley, and Richard E. Crandall. Compu-
tational strategies for the Riemann zeta function. J. Comput. Appl. Math.,
121(1-2):247-296, 2000. Numerical analysis in the 20th century, Vol. I, Ap-
proximation theory.

[2] Wikipedia. Apery's constant -wikipedia, the free encyclopedia,
http://en.wikipedia.org/wiki/Ap%C3%A9ry%27sconstant, 2008. [On-
line; accessed 12-November-2008].