# Hadamard product for the Riemann zeta function

What is the best way to compute numerically with Sage the product over the first n nontrivial zeros of the Riemann zeta function?

p(s,n) = product((1-s/rho(k))*exp(s/rho(k)) for k in (1..n))

where zeta(rho(k)) == 0 and Im(rho(k)) != 0.

Wikipedia advises: "To ensure convergence the product should be taken over 'matching pairs' of zeroes, i.e. the factors for a pair of zeroes of the form rho(k) and 1-rho(k) should be combined."

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First you need to install Odlyzko's database of zeta zeros. In a terminal, type:

\$ sage -i database_odlyzko_zeta


The command

sage: zeta_zeros()


will then give you a list of the imaginary parts of the first 100,000 non trivial zeros of zeta. Note that as usual, the list is indexed from 0.

The information page about this package also warns: Note that only the first 9 digits after the decimal come from the database. Subsequent digits are the result of the inherent imprecision of a binary representation of decimal numbers. So you should have that in mind and check how this affects precision in your product.

Then you can do the following:

sage: def rho(k):
....:     return CC(0.5,zeta_zeros()[k])
....:
sage: def a(x):
....:     return (CC(1.,0.)-x)*exp(x)
....:
sage: def p(s,n):
....:     return prod(a(s/rho(k))*a(s/(CC(1.,0.)-rho(k))) for k in (0..n))
....:
sage: p(0.5,2)
1.00221605640651
sage: p(0.5,20)
1.00403653183532
sage: p(0.5,200)
1.00527298745769
sage: p(0.5,2000)
1.00567828272459


If you plan to use large values of n, you might want to use Cython to speed up computation, but you should probably first check what precision the computations really give you.

more

Nice. Thanks!

( 2013-07-27 08:12:01 -0500 )edit

If you need more precision but not too high n, [Odlyzko's tables](http://www.dtc.umn.edu/~odlyzko/zeta_tables/index.html) include one of imaginary parts of the first 100 zeros of the Riemann zeta function, accurate to over 1000 decimal places.

( 2013-07-28 11:04:56 -0500 )edit