Re: sum of zetas


"Gottfried Helms" <helms@xxxxxxxxxxxxx> wrote in message
news:dfotto$d5i$05$1@xxxxxxxxxxxxxxxxxxxx
> Jon -
>
> I have a question concerning your definitions in:
>
> Am 07.09.05 17:10 schrieb Jon Slaughter:
>>
>> http://www.usna.edu/Users/math/meh/mult.html
>>
> Did I get it right, that Z(a,b) means the sum of the
> lower triangle, when the summands of Z(a) and
> summands of Z(b) are taken as vectors VA and VB and
> then the vectors are multiplied VA'*VB to form an
> infinite matrix?

yes, that is the basic idea..

Zeta(n_1,...,n_m) =

sum(....sum(1/i_1^n_1*...*1/i_m^n_m, i_m=i_(m-1)+1..oo),...),i_1-1..oo)

for m = 2

Zeta(n,m) =
sum(sum(1/i^n*1/j^m, j=i+1..oo), i=1..oo)

if you look at the matrix[i,j](n,m) it has (i,j)th entry of


1/i^n*1/j^m

the sum above is the same as summing over the upper part of the matrix.

i.e. it is true for all sums of the form sum(sum(f(i,j),j=i+1..oo),i=1..oo)
that if you write f(i,j) out as a matrix then the sum if equivilent to
adding up the upper part of the matrix

(ofcourse it maybe true that it happens to be the same as summing over the
lower too)

this is how you get the identity that

2Z(n,n) = 1/2*Z(n)^2 - Z(n)


Notice that if m = n then you have

1/(ij)^n and that adding up all the entries in that matrix is the same as
adding the lower half and adding it to the upper half - 1/2 the
diagonal(cause it gets counted twice because we go from

i.e. in general if we have a double sum over a function f(i,j,n,m) such that
f(i,j,n,m) = f(i,j,m,n) then the matrix for f(i,j,n,n) will be symmetric and
we will get a formula like the one above.

The unusal cases come when n != m. it turns out there are many relations
ships for specific powers but it is not known for the general case.




>
> I'm asking just to get things right.
>

Yes, I think so, but you are making it more complicated and also these
principles are not specific to the zeta function but apply to summing over
any matrix.

> Gottfried

Jon


.

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