Did you read the documentation of the function? It makes it quite
clear:
Definition: lcalc.zeros(self, n, L='')
Docstring:
Return the imaginary parts of the first n nontrivial zeros of
the
L-function in the upper half plane, as 32-bit reals.
INPUT:
* ``n`` - integer
* ``L`` - defines L-function (default: Riemann zeta function)
This function also checks the Riemann Hypothesis and makes sure
no
zeros are missed. This means it looks for several dozen zeros
to
make sure none have been missed before outputting any zeros at
all,
so takes longer than ``self.zeros_of_zeta_in_interval(...)``.
You could always define your own function to return the complete zero:
sage: [1/2+I*y for y in lcalc.zeros(10)]
[0.500000000 + 14.1347251*I, 0.500000000 + 21.0220396*I, 0.500000000 +
25.0108576*I, 0.500000000 + 30.4248761*I, 0.500000000 + 32.9350616*I,
0.500000000 + 37.5861782*I, 0.500000000 + 40.9187190*I, 0.500000000 +
43.3270733*I, 0.500000000 + 48.0051509*I, 0.500000000 + 49.7738325*I]
!
On Jun 1, 4:13 pm, Anne Driver <[email protected]> wrote:
> Hello,
>
> I am new to this list, and relatively new to Sage. I'm puzzled by the logic
> of one part of Sage though.
>
> Although I don't have access to Mathematica at the minute on this computer,
> I know if I compute the first zero, I get something like
>
> In[1] = ZetaZero[1] //N (to get a numerical value)
> Out[1] = 1/2 + I*14.134...
>
> Trying this in Sage, I get:
>
> sage: lcalc.zeros(1)
> [14.1347251]
>
> Why does Sage not do the sensible thing like Mathematica and return the
> complex number 0.5 + I 14.1347251 ? It would seem much more logical.
>
> Of course, it is not proven that the real part is 1/2, so how would the case
> be handled if a root was not found to have a real part of 1/2 ?
>
> Anne
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