• Definition
• Rough calculation
• Figures
• Source Code

Definition
   The Riemann zeta function zeta(s) of the complex variable s=sigma+it
is written as an infinite series:-
                n=oo
                 __   1
     Zeta(s) =  \    ____
                /__   s
                n=1  n


Euler investigated this function in the 1700s and studied the special
values s=1 and s=2.

The zeta function is important in number theory. One of the greatest
outstanding problems in mathematics concerns the locations of its
zeros. Hardy proved that it has an infinite number of zeros on the
half-line sigma = 1/2 [1]. The real problem is to prove that all its
zeros lie on this line, excepting known zeros at the values s= -2n : n
> 0. This statement about zeros is the famous RIEMANN HYPOTHESIS.

    Erica Klarreich [2] gave an interesting summary of recent work in New
Scientist. A superficial observation shows that the zeta function is
related to calculations used in quantum mechanics. Energy levels
of electron orbits follow a series like  E=k/n^2 for n=1,2,..etc.
All of this was known around 1908 when Niels Bohr invented quantum
theory.

    In fact the zeta function crops up all over the place in numerical
analysis because the values Zeta(2n) are related to Bernoulli numbers
used in the Euler-MacLaurin summation formula which connects sums and
integrals. Classical physics describes the universe in terms of infinitely
differentiable functions, whereas quantum numbers require summation
of series rather than integration of functions.

    There is a whole website devoted to Zeta(2).
    It's value is (PI*PI)/6, pisquaredoversix where PI=3.14159.


Calculations

    Zeta(k) is the sum 1/n^k for integers n >=0.

    The series does not converge at all for k=1.

    The numbers n>1 split into even numbers 2n, and odd numbers 2n+1 so
the sum Zeta(k) can be split:-

     Zeta(k) = sum 1/(2n)^k + sum 1/(2n+1)^k
             = E(k)         + O(k)

     E(k) is in fact 1/2^k * Zeta(k)

     Define  A-zeta(k) =  1 - 1/2^k +1/3^k - 1/4^k ....

     The sum O(k)-E(k) = Zeta(k)-2*E(k) = (1-1/2^(k-1))* Zeta(k)

     so  Zeta(k) = A-zeta(k) / (1-1/2^(k-1))

     The series for A-zeta(k) has terms with alternating sign, so it is
more convergent. There is no reason to expect very good convergence for
values of 0< k <1 but using a few thousand terms of the series with
complex values for k shows results compatible with textbooks[3].

    In APL try the expressions

$ 12 8 z #sqrt 6* +/1%N*N<-1.0+i60000
  3.14157674
$
$ 12 8 z #sqrt 12*+/((rN)r1 _1)%N*N<-1.0+i30000
$ 3.14159265


The first few zeros on the line 1/2+it have the approximate t-values

14.13 21.02 25.01 30.42 32.93 37.58 40.91 43.32 48.00 49.77 52.96 56.44
59.34 60.83 65.11 67.07 69.54 72.06 75.70 77.14 79.33 82.91 84.73 87.42
88.80 92.49 94.65 95.87 98.83 101.3 105.4 107.1 111.0 111.8 114.3 116.2
118.7

Figures

• |A-zeta(0.5+it)| t=0 .. 40
• |A-zeta(0.5+it)| t=40 .. 80
• |A-zeta(0.5+it)| t=80 .. 120
• |A-zeta(0.5+it)| t=120 .. 160
• |A-zeta(0.5+it)| t=160 .. 200
• |A-zeta(0.5+it)| t=200 .. 240
• |A-zeta(0.5+it)| t=240 .. 280
• |A-zeta(0.5+it)| t=280 .. 320
• |A-zeta(0.5+it)| t=320 .. 360
• |A-zeta(0.5+it)| t=360 .. 400