Re: Correctness of derivnum(X=1/2,zeta(X),41)

On 26 Sep 2021, at 08:55, Georgi Guninski <gguninski@gmail.com> wrote:

On mathoverflow [1] there is conjecture about the nearest integer to
[zeta^(k)(1-1/B)]=-B^(k+1)*factorial(k)

Answer of controversial theoretic result claims the explicit
numerical counterexample k=41,B=2

I couldn't compute the counterexample on mpmath with high precision
due to internal error,

that is wierd; using python as an interface to mpmath it gives

languasc@languasc-macmini Desktop % python3 conj.py 70

------

precision 70

value = 3.0

languasc@languasc-macmini Desktop % python3 conj.py 80

------

precision 80

value = 3.0

The point is that these quantities are large (their order

of magnitude is about 10^62) but their sum is tiny.

So you need to choose a sufficiently large precision

to perform the computation.

The python-mpmath script I wrote is at the bottom of this file.

An alternative code for gp is the following

? \p100

realprecision = 115 significant digits (100 digits displayed)

? K=41

%2 = 41

? B=2

%3 = 2

? round(zetahurwitz(1-1/B,1, K))+B^(K+1)*factorial(K)

%5 = 3.000000000000000000000000000000000000000000000000000000000

Best regards

Alessandro Languasco

but pari agrees the counterexample is correct:

K=41;B=2;T=derivnum(X=1-1/B,zeta(X),K);(round(T)+B^(K+1)*factorial(K))
%16 = 3.00....

Is it plausible that the pari computation is correct with high (what?)
precision?

What other CAS say about it?

[1]: https://mathoverflow.net/questions/404779/on-the-nearest-integer-to-zetak1-1-b-b-ge-2

import sys

defaultprecision = sys.argv[1]

import mpmath as mp

# function z

def z(s,a,k):

y = mp.zeta(s,a,k)

return y

# print("Give me the desired accuracy (# of decimal digits): ")

# defaultprecision = input();

defaultprecision = int(defaultprecision);

#print(defaultprecision);

mp.mp.dps = defaultprecision;

mp.mp.pretty = True

print("------ ");

print("precision ", defaultprecision);

B=2

K=41

print("value = ", mp.nint(z(1-1/B,1,K))+ mp.power(B,K+1)*mp.fac(K));