t;
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On May 21, 8:55 am, Maarten Derickx
wrote:
> Good you found this. I already noticed that the modular symbols code
> some times gave me a different awnser then before, but I couldn't
> reproduce this so I guess this is the underlying reason.
That is very plausible: it also causes occasional and
code in sage/ext/multi_modular.pyx is not handling a
corner case properly.
The relevant routine (mpz_crt_vec_tail) involves both Cython and GMP, so
if someone with Cython and/or GMP experience could give me a hand
debugging it then I would be very grateful.
This is trac #11358.
Yours,
Tom
-
On 16 June, 07:48, rjf wrote:
> By your reasoning, and for other domains we would have the following
> behavior:
>
> 1-2 --> error. 1 and 2 are both positive integers. In order to
> provide the answer -1, one must
> expand the domain to include negative integers.
>
> 1 / 2 --> error..
As ddrake pointed out in Trac #9248, even when x is in the symbolic
ring, factorial(x) is not simply calling gamma(x+1):
sage: x=I; factorial(x)
0.498015668118356 - 0.154949828301811*I
sage: gamma(x+1)
gamma(I + 1)
So something strange is going on. I think that the first example here
is probabl
> My vote is to have factorial(n) = n(n-1)...2.1 whenever n is integer.
>
> Cheers,
>
> Florent
We certainly need to allow for symbolic input too, so that Sage can
simplify expressions involving factorials and binomial coefficients
such as:
sage: var('k,n')
(k, n)
sage: f = factorial(k)*factoria
I will change the
documentation of factorial() to make this clear.
Best,
Tom
On 15 June, 01:14, David Kirkby wrote:
> On 15 June 2010 05:21, Tom Coates wrote:
>
>
>
>
>
> > Dear sage-devel,
>
> > I believe that the following:
>
> > sage: gamma(x).ful
ching functionality in pynac handle
this? Or is there another way to do it?
Best,
Tom
---
Tom Coates
Royal Society University Research Fellow
Reader in Pure Mathematics
Imperial College London
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):
http://trac.sagemath.org/sage_trac/ticket/9240
and uploaded a patch to fix it.
Best,
Tom
---
Tom Coates
Royal Society University Research Fellow
Reader in Pure Mathematics
Imperial College London
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h.org/sage_trac/ticket/9217
and uploaded a patch to fix it.
Best,
Tom
---
Tom Coates
Royal Society University Research Fellow
Reader in Pure Mathematics
Imperial College London
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It turns out that for my problem it suffices to compute the
polynomials
1, f, f^2, ..., f^K
modulo N for some large but known number N, so I suspect that it may
be faster to work modulo p for various primes p and then use the
Chinese Remainder Theorem.
How large, roughly speaking, can I take my
On May 15, 9:03 pm, William Stein wrote:
> 1. On what hardware?
This was on 64 bit GNU/Linux (Fedora release 12) running on a dual
processor machine with two Intel Core 2 CPUs (each 2.4GHz, 4Gb
cache). I have included the contents of /proc/cpuinfo at the bottom
of this reply.
> 2. Can you po
Thank you (everyone!) for the many extremely helpful comments and
links.
Recall that I need to compute
1, f, f^2, ..., f^K
for f in ZZ[x,y,z] and K known but large. (In fact I only need
certain coefficients of the f^i, but this does not seem to help very
much.)
I have implemented the most naiv
Dear Georg,
Many thanks! Running "make" again solved the problem. I updated the
trac ticket.
Best,
Tom
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