## Factoring Zeta

```We've seen how Leibniz might have anticipated Euler by discovering
that the sum of the inverse squares is (pi^2)/6 purely on the basis
of the arctangent series for pi/4.  Another way of expressing that
approach is to regard it as essentially a factorization of the
product form of zeta(2).  In other words, beginning with the
expression

1     1     1     1                   /     1     \
1 + --- + --- + --- + --- + ...   =  PROD ( ----------- )
2^2   3^2   4^2   5^2             all  \ 1 - 1/p^2 /
primes

we factor the right hand side into two parts

/    1     \          /    1    \       /     1   \
PROD( ---------- ) =  PROD( --------- ) PROD( --------- )
\1 - 1/p^2 /          \ 1 - 1/p /       \ 1 + 1/p /

However, the two individual products on the right side aren't very
useful because the first is just the harmonic series, which diverges
to infinity, and the second converges on zero.  In order to get a
more useful factorization, we note that the above is really just a
special case of a more general form

/    1     \         /      1     \       /      1     \
PROD( ---------- ) = PROD( ------------ ) PROD( ------------ )
\1 - 1/p^2 /         \ 1 - a(p)/p /       \ 1 - b(p)/p /

where we've replaced the unit numerators with the functions a(p)
and b(p), which for each prime p are the roots of x^2 - 1, but any
permutation of those roots is allowable for any given prime.  For
example, we are free to define a(p) and b(p) in terms of the
Legendre symbol
/-1 \       /  +1  if p=1,2 mod 4
a(p) = -b(p)  =  (     )  =  (
\ p /       \  -1  if p=3 mod 4

in which case we have decomposed zeta(2) into the factors

/      1     \       /      1     \       pi  pi
PROD( ------------ ) PROD( ------------ )  =   --  --
\ 1 - a(p)/p /       \ 1 - b(p)/p /        2   3

where the first factor is simply 1/(1 - 1/2) times the arctan series
for pi/4, and the second factor is 1/(1 + 1/2) times the sign-reversal
of the Leibniz series (product) as discussed in the previous note.

Now, what happens if we take this same approach to zeta(3), i.e., the
sum of the inverse CUBES?  There is no known "simple" expression for
zeta with odd arguments, in contrast with the fact that all the even-
index zeta values are given by

(2pi)^(2n)
zeta(2n)  =  ---------- |B_{2n}|
2(2n)!

where B_{2n} is the (2n)th Bernoulli number.  The problem with this
formula relative to odd zeta arguments is that all the odd-index
Bernoulli numbers vanish.  (Of course, the odd-index Bernoulli
POLYNOMIALS do not vanish, so people have tried to generalize the
above expression in terms of the full polynomials rather than just
the constant coefficients, but nothing seems to work.)

Anyway, following our factorization of zeta(2), we might try the same
trick on the product-form of zeta(3), i.e., we might try to factor
the product
/     1     \
zeta(3)  =   PROD( ----------- )
\ 1 - 1/p^3 /

which can obviously be formally factored as

/     1     \         /    1    \       /         1       \
PROD( ----------- ) = PROD( --------- ) PROD( ----------------- )
\ 1 - 1/p^3 /         \ 1 - 1/p /       \ 1 + 1/p + 1/p^2 /

but again we have the problem that the first factor is the harmonic
series (divergent) and the second factor converges to zero.  We might
simply replace the "1/p" terms with a(p)/p and b(p)/p, but it seems
more in the spirit of our previous example to factor zeta(3) into
LINEAR factors.

To simplify the notation, let me use square brackets to denote a
product evaluated over all primes, so our putative factorization of
zeta(3) can be written as
_           _     _            _  _            _  _            _
|      1      |   |       1      ||       1      ||       1      |
| ----------- | = | ------------ || ------------ || ------------ |
|_ 1 - 1/p^3 _|   |_ 1 - a(p)/p _||_ 1 - b(p)/p _||_ 1 - c(p)/p _|

where a(p), b(p), and c(p) are now (some permutation of) the roots
of x^3 - 1.  Except for the prime 3, every prime falls in one of
the congruence classes {1,2}, {4,5}, {7,8} modulo 9, so if we let
r1,r2,r3 denote the cube roots of 1

r1 = 1      r2 = (-1+sqrt(-3))/2     r2 = (-1-sqrt(-3))/2

then one possible way of defining the functions a,b,c is presented
below, along with the resulting factorization of zeta(3).

p (mod 9)
-----------------
1,2   4,5   7,8,3               product
---   ---   -----      --------------------------
a(p)      r1    r2    r3        F1 = 1.484093 - 0.209977 i
b(p)      r2    r3    r1        F2 = 1.057036 + 0.224225 i
c(p)      r3    r1    r2        F3 = 0.740445 - 0.050781 i

Multiplying the factors together gives

zeta(3)   =   F1 * F2 * F3  =  1.20205...

Unfortunately, unlike the case of zeta(2), these "linear factors"
aren't immediately recognizable as fractions of pi (or anything
else). However, there were quit a few arbitrary choices made in
the above construction, mainly involving the definitions of the
functions a,b,c.  In the case of zeta(2) we essentially classified
the primes into two categories according to whether x^2 + 1 splits
or is irreducible in the field of integers modulo p (i.e., whether
or not -1 is a quadratic residue), so it would seem that for zeta(3)
we should classify the primes into three categories according to
whether some cubic polynomial splits completely, has one linear
factor, or is irredicible in Z_p.  However, we can't use x^3 + 1
because that ALWAYS has at least one linear factor.  It's not
clear to me how to come up with the "most natural" factorization
of zeta(3).

By the way, it's possible that we should really work with 3! = 6
factors, maybe using +- the cube roots of 1.  This may be related
to the fact that we can get zeta(3) from EITHER of the products
1/(1 - 1/p^3)  or  1/(1 + 1/p^3), because they are "conjugate"
factors of zeta(6), i.e.,
_           _       _           _  _           _
|      1      |     |     1       ||      1      |     pi^6
| ----------- |  =  | ----------- || ----------- |  =  ----
|_ 1 - 1/p^6 _|     |_ 1 - 1/p^3 _||_ 1 + 1/p^3 _|      945

So if we had a simple expression for either of the factors we would
automatically have one for the other factor.  Unfortunately the
second factor is just as intractible as the first, so this doesn't
seem to help much, except perhaps to suggest that the cube roots
of -1 should also be included in a six-term factorization.
```