On 10/28/19 12:03 PM, Nils Bruin wrote:
> On Monday, October 28, 2019 at 3:30:03 AM UTC-7, Michael Jung wrote:
>
> Nice example! I see your point.
>
> However, I wonder. The matrix inversion should cause the same
> problem, but it is implemented for the symbolic ring. What is differen
On Monday, October 28, 2019 at 3:30:03 AM UTC-7, Michael Jung wrote:
>
> Nice example! I see your point.
>
> However, I wonder. The matrix inversion should cause the same problem, but
> it is implemented for the symbolic ring. What is different?
>
I don't think it is really implemented *for* the
Nice example! I see your point.
However, I wonder. The matrix inversion should cause the same problem, but
it is implemented for the symbolic ring. What is different?
Best wishes
Michael
Am Sonntag, 27. Oktober 2019 05:28:12 UTC+1 schrieb vdelecroix:
>
>
>
> Le 24/10/2019 à 09:53, Simon King a
Hi Vincent,
On 2019-10-27, Vincent Delecroix <20100.delecr...@gmail.com> wrote:
> This was an easy one. The following shows that SR is just
> broken pi is rational!
>
> sage: q = continued_fraction(pi).convergent(100)
> sage: q
> 8736149038303113005348154524599771853409352442745266/27808026060
Le 24/10/2019 à 09:53, Simon King a écrit :
On 2019-10-24, Michael Jung wrote:
Do you have an example where SR fails to be exact?
One can convert a float to SR. The result is in SR, but still behaves
like a float:
sage: a = SR(2.)^(1/500)
sage: a^500
2.05
sage: a.pa
Hi Emmanuel,
On 2019-10-24, Emmanuel Charpentier wrote:
> Writing *correctly* this decomposition is, IIRC, a numerical analysis
> bitch... You are, IIRC, led to compute differences of large products, where
> underflows can easily slip into... Definitely not an amateur's problem.
>
> While I agr
Dear Michael
Contrast
sage: A=matrix(QQ,[[1, 2], [3,4]])
sage: G,M=M.gram_schmidt()
sage: A=matrix(QQ,[[1, 2], [3,4]])
sage: G,M=A.gram_schmidt()
sage: M*G==A
True
with
sage: Ar=matrix(RDF,[[1, 2], [3,4]])
sage: Gr,Mr=Ar.gram_schmidt()
sage: Mr*Gr==Ar
False
sage: Mr*Gr-A
[-1.1102230246251565e-16
I see. Maybe it is possible to decompose/split the matrix in SR into an exact
and inexact part, convert the inexact part to RDF and apply the Gram-Schmidt
algorithm appropiately? I don't know, maybe it's too naive?
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On 2019-10-24, Michael Jung wrote:
> Do you have an example where SR fails to be exact?
One can convert a float to SR. The result is in SR, but still behaves
like a float:
sage: a = SR(2.)^(1/500)
sage: a^500
2.05
sage: a.parent()
Symbolic Ring
Best regards,
Simon
--
You
Do you have an example where SR fails to be exact?
Am Donnerstag, 24. Oktober 2019 18:20:36 UTC+2 schrieb Simon King:
>
> Hi Michael,
>
> On 2019-10-24, Michael Jung > wrote:
> > Maybe, I did get something wrong. But what's the problem about
> Gram-Schmidt
> > on SR? There are just sums and di
Hi Michael,
On 2019-10-24, Michael Jung wrote:
> Maybe, I did get something wrong. But what's the problem about Gram-Schmidt
> on SR? There are just sums and divisions (and probably roots to normalize)
> in Gram-Schmidt which should not lead to problems in SR.
>
> By the way, what does "exact"
Maybe, I did get something wrong. But what's the problem about Gram-Schmidt
on SR? There are just sums and divisions (and probably roots to normalize)
in Gram-Schmidt which should not lead to problems in SR.
By the way, what does "exact" actually mean?
Am Mittwoch, 23. Oktober 2019 22:41:53 UTC
Well... the error message is pretty explicit: since
sage: SR.is_exact()
False
M.gram_schmidt() wont work if M.base_ring is SR.
Creating a second special case for SR may not be as simple as for RDF,
since a lot of other cases (beyond RDF) can happen in this case
Le mardi 22 octobre 2019 17:03
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