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Timing : are hurwitz_zeta values cached?

asked 2020-08-02 05:50:38 -0500

anonymous user


Dear all, It seems the values of hurwitz_zeta are cached in some way. This makes sense, but I couldn't find documentation on that issue, and in particular, on how to clear the cache: I want to time several instances of a script and need to start afresh each time. Here is an ECM:

import sys
from timeit import default_timer as timer
def ECM(prec):
    start, CF = timer(), ComplexField(prec)
    hurwitz_zeta(s = CF(2), x = CF(0.5)) 
    end = timer()    
return end-start

for i in range(0,20):


Then start afresh, get a similar answer. I am pretty sure this system-caching mechanism is explained somewhere but my morning queries drew a blank --

Pointers would be appreciated! Best, Olivier

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answered 2020-08-02 06:22:52 -0500

slelievre gravatar image

updated 2020-08-02 07:14:47 -0500

Regarding the timing issue, the timeit function could be a good choice.

sage: def ECM(prec):
....:     t = timeit(f'C = ComplexField({prec}); hurwitz_zeta(C(2), C(0.5))')
....:     print(f'prec = {prec:3d} --', t)
sage: for i in range(0, 4):
....:     ECM(3000)
prec = 3000 -- 25 loops, best of 3: 9.74 ms per loop
prec = 3000 -- 25 loops, best of 3: 9.8 ms per loop
prec = 3000 -- 25 loops, best of 3: 9.76 ms per loop
prec = 3000 -- 25 loops, best of 3: 9.81 ms per loop

See its documentation:

sage: timeit?

Regarding computing in arbitrary precision, using Arb could be a good choice.

Compare using ComplexField vs ComplexBallField (which uses Arb).

With ComplexField:

sage: C = ComplexField(20)
sage: hurwitz_zeta(C(2), C(0.5))

With ComplexBallField, using the zeta method:

sage: C = ComplexBallField(20)
sage: C(2).zeta(a=C(0.5))
[4.93480 +/- 2.40e-6]
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Thanks for the syntax in Arb to get the Hurwitz-zeta!

Olivier R. gravatar imageOlivier R. ( 2020-08-02 11:40:17 -0500 )edit

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Asked: 2020-08-02 05:50:38 -0500

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Last updated: Aug 02