Hi Michael,
On 2019-03-11, Michael Beeson wrote:
> I tried various simplification functions.I suppose I could start over,
> not using "symbolic expressions" but
> declaring K to be a suitable field or ring, maybe a quadratic extension of
> the field of rational functions in a.
> That is
Le lundi 11 mars 2019 19:05:04 UTC+1, Michael Beeson a écrit :
>
> I appreciate Eric's post, and I do use subs sometimes, but it makes me
> nervous since
> it will happily substitute any old thing you tell it to, even an
> incorrect thing. So, if your idea
> is to check a computation, it
On Monday, March 11, 2019 at 11:05:04 AM UTC-7, Michael Beeson wrote:
>
> I appreciate Eric's post, and I do use subs sometimes, but it makes me
> nervous since
> it will happily substitute any old thing you tell it to, even an
> incorrect thing. So, if your idea
> is to check a
I appreciate Eric's post, and I do use subs sometimes, but it makes me
nervous since
it will happily substitute any old thing you tell it to, even an incorrect
thing. So, if your idea
is to check a computation, it is a dangerous thing. True, if you put only
correct equations in,
What about something like
sage: var('p,q,r,a')
(p, q, r, a)
sage: b = sqrt(1-a^2)
sage: eq = (p*a+r*b+q)^2
sage: eq = eq.expand(); eq
a^2*p^2 - a^2*r^2 + 2*sqrt(-a^2 + 1)*a*p*r + 2*a*p*q + 2*sqrt(-a^2 + 1)*q*r
+ q^2 + r^2
sage: b = var('b')
sage: eq.subs({sqrt(1-a^2): b})
a^2*p^2 + 2*a*b*p*r -
On 2013-11-22, John Cremona john.crem...@gmail.com wrote:
1. It is a fact that if a^2+b^2=1 then arcsin(a)+arcsin(b)=pi/2, but
how can I get this to happen automatically? I have an expression
which comes out as
arcsin(1/3*sqrt(3)*sqrt(2)) + arcsin(1/3*sqrt(3))
which I would like to get
On Feb 22, 2:37 am, JamesHDavenport j.h.davenp...@bath.ac.uk wrote:
Canonical form and simplify aren't the same thing (necessarily).
See Carette,J., Understanding Expression Simplification. Proc. ISSAC
2004 (ed. J. Gutierrez), ACM Press, New York, 2004, pp. 72-79.
I don't have access to that
Try
http://www.cas.mcmaster.ca/~carette/publications/simplification.pdf
The real point is that GiNaC's canonical form has different goals from
a 'simplify' command in the sense of minimal complexity.
One really needs to separate the two, and I don;t know how easy that
is with the current design.
So once it comes back to Sage, its internal representation goes back to the
Ginac one.
Doh! So much for a possible workaround involving maxima.
Dox, I was using full_simplify() and also the handful of simplify
methods it invokes. Evaluate a.full_simplify? to see their names.
Nils, trivial
On Feb 21, 9:36 pm, Mark Rahner rah...@alum.mit.edu wrote:
So once it comes back to Sage, its internal representation goes back to the
Ginac one.
My initial problem was the severe obfuscation that resulted when extra
factors added by the canonical form refused to cancel and then
replicated
On Feb 18, 8:24 pm, Mark Rahner rah...@alum.mit.edu wrote:
I appreciate that background info. I hadn't tried invoking maxima
because I read somewhere that simplify() used maxima. I must've been
reading outdated material. As you stated, maxima does the correct
thing. Because Sage can
On Feb 18, 5:24 pm, Mark Rahner rah...@alum.mit.edu wrote:
converts 1/sqrt(5) to 1/5*sqrt(5) so I suspect that this issue can be
traced to the GiNaC canonical form.
Yes, it does so for a very good reason: By simplifying expressions
this way, you're sure to recognize equal expressions. Compare
I appreciate that background info. I hadn't tried invoking maxima
because I read somewhere that simplify() used maxima. I must've been
reading outdated material. As you stated, maxima does the correct
thing. Because Sage can invoke maxima, perhaps I have a work around.
You're right that this
In some previous incarnation, where Sage used Maxima for things like
this, your simplification happened.
(%i1) 1/sqrt(5);
1
(%o1) ---
sqrt(5)
(%i2) sqrt(5)/5;
As far as I understand from your previous comments, a way to extract the
exponential functions from the expression is all you need. You don't
really need to walk through the tree. Here is one way to do this:
sage: t = exp(x+y)*(x-y)*(exp(y)+exp(z-y))
sage: t
(e^(-y + z) + e^y)*(x - y)*e^(x
On 6/24/10 6:15 AM, kcrisman wrote:
Right. This crops up in the middle of a more complicated
expression. If I could figure out how to break the expression
up in the right way, then I guess I could search for parts
that are exponential functions, take the log of those, and
then simplify the
On 06/26/2010 03:26:06 PM, Jason Grout wrote:
On 6/24/10 6:15 AM, kcrisman wrote:
Right. This crops up in the middle of a more complicated
expression. If I could figure out how to break the expression
up in the right way, then I guess I could search for parts
that are exponential functions,
I believe that Ticket #9329 was generated in response to my original
post, before I understood that there was a Latex issue involved.
I believe that Ticket #9329 should be deleted (closed or whatever).
But part of your question was also to try to simplify more complicated
expressions, and it
On 06/26/2010 05:21:21 PM, kcrisman wrote:
I believe that Ticket #9329 was generated in response to my original
post, before I understood that there was a Latex issue involved.
I believe that Ticket #9329 should be deleted (closed or whatever).
But part of your question was also to try to
Dear Mike,
Just to follow up:
There is further discussion at http://trac.sagemath.org/sage_trac/ticket/9329
if you are interested in saying exactly what sort of data structure
would enable you to perform the simplifications you would like to
without having to create a custom Maxima
On 06/25/2010 06:07:02 AM, kcrisman wrote:
Dear Mike,
Just to follow up:
There is further discussion at
http://trac.sagemath.org/sage_trac/ticket/9329
if you are interested in saying exactly what sort of data structure
would enable you to perform the simplifications you would like to
sage: n=var('n')
sage: f=e^(i*x*pi*n-i*2*pi*n)
sage: f.simplify_full()
e^(I*pi*n*x - 2*I*pi*n)
# Is there a way I can get this to simplify?
This apparently isn't even that easy in Maxima.
Maxima 5.21.1http://maxima.sourceforge.net
using Lisp ECL 10.4.1
Distributed under
On 06/24/2010 06:15:52 AM, kcrisman wrote:
sage: n=var('n')
sage: f=e^(i*x*pi*n-i*2*pi*n)
sage: f.simplify_full()
e^(I*pi*n*x - 2*I*pi*n)
# Is there a way I can get this to simplify?
This apparently isn't even that easy in Maxima.
Maxima
I've noticed too about how maxima continues to ask things that
(it would seem) you have already told it. I guess it would be
in my best interests to learn more about maxima.
If you are serious about doing symbolic manipulation that you can
control from within Sage, yes. That said, various
On 06/22/2010 12:41:17 PM, kcrisman wrote:
sage: n=var('n')
sage: f=e^(i*x*pi*n-i*2*pi*n)
sage: f.simplify_full()
e^(I*pi*n*x - 2*I*pi*n)
# Is there a way I can get this to simplify?
This apparently isn't even that easy in Maxima.
Maxima 5.21.1 http://maxima.sourceforge.net
using Lisp
sage: n=var('n')
sage: f=e^(i*x*pi*n-i*2*pi*n)
sage: f.simplify_full()
e^(I*pi*n*x - 2*I*pi*n)
# Is there a way I can get this to simplify?
This apparently isn't even that easy in Maxima.
Maxima 5.21.1 http://maxima.sourceforge.net
using Lisp ECL 10.4.1
Distributed under the GNU Public
On May 14, 3:57 am, Laurent moky.m...@gmail.com wrote:
Hello
Hi Laurent,
x,y=var('x,y')
s = x*y2 + x*(-y2 - x2 + 1) + x3 - x
print simplify(s)
Answer :
2 2 2 3
x y + x (- y - x
Dear Laurent,
On May 14, 12:57 pm, Laurent moky.m...@gmail.com wrote:
Btw, the function simplify_full does not exist ... so I suppose that I
*do* miss something.
Yes. simplify_full is not a function but a method (after all, python
is object oriented).
So, you can do:
sage: x,y=var('x,y')
On May 14, 2009, at 3:57 AM, Laurent wrote:
Hello
x,y=var('x,y')
s = x*y2 + x*(-y2 - x2 + 1) + x3 - x
print simplify(s)
Answer :
2 22 3
x y + x (- y - x + 1) + x - x
On May 14, 12:57 pm, Laurent moky.m...@gmail.com wrote:
Btw, the function simplify_full does not exist ... so I suppose that I
*do* miss something.
Yes. simplify_full is not a function but a method (after all, python
is object oriented).
Thanks all.
Indeed, simplify_all(s)
On Nov 24, 2008, at 8:45 PM, William Stein wrote:
On Mon, Nov 24, 2008 at 4:46 PM, Tim Lahey [EMAIL PROTECTED]
wrote:
Hi,
Is there a specific way to add rules (and apply them) to rewrite
expressions in Sage?
Such as, log(a)-log(b) = log(a/b)
I need this (and others) in order to
On Mon, Nov 24, 2008 at 6:03 PM, Tim Lahey [EMAIL PROTECTED] wrote:
On Nov 24, 2008, at 8:45 PM, William Stein wrote:
On Mon, Nov 24, 2008 at 4:46 PM, Tim Lahey [EMAIL PROTECTED]
wrote:
Hi,
Is there a specific way to add rules (and apply them) to rewrite
expressions in Sage?
Such as,
On Nov 24, 2008, at 9:05 PM, William Stein wrote:
I can easily run timing comparisons between maxima and FriCAS, but
because
of how sympy does things (with its separate variables), I'll have
to run
them separately. Comparing maxima and FriCAS, the timings are pretty
close on
both for
On Nov 24, 6:17 pm, Tim Lahey [EMAIL PROTECTED] wrote:
On Nov 24, 2008, at 9:05 PM, William Stein wrote:
Hi,
I can easily run timing comparisons between maxima and FriCAS, but
because
of how sympy does things (with its separate variables), I'll have
to run
them separately.
On Nov 24, 2008, at 9:21 PM, mabshoff wrote:
We have a timeit doctest framework that is supposed to hunt for speed
regressions. It is merged in 3.2, but we need infrastructure to
compare the output from several runs.
But I guess you are asking if timeit('foo') could return the time so
On Mon, Nov 24, 2008 at 6:31 PM, Tim Lahey [EMAIL PROTECTED] wrote:
I know I could parse the output, but I thought someone might have done
it and it sounds like the timeit doctest framework might do it.
Where can I find this in the source so I can see how it is doing it?
You can do this in
On Nov 24, 7:03 pm, Tim Lahey [EMAIL PROTECTED] wrote:
On Nov 24, 2008, at 9:51 PM, Mike Hansen wrote:
You can do this in 3.2:
sage: s = timeit.eval(2+3)
sage: s
625 loops, best of 3: 942 ns per loop
sage: s.stats
(625, 3, 3, 942.230224609375, 'ns')
The code is in
On Nov 24, 2008, at 10:07 PM, mabshoff wrote:
You should consider creating one or a couple large files with the
integrals for doctesting and stuff them into $SAGE_ROOT/devel/tests.
Hopefully it can be arranged to feed the input into Maxima/Axiom/
Maple/
MMA/sympy and so on and compare the
On Mon, Nov 24, 2008 at 6:17 PM, Tim Lahey [EMAIL PROTECTED] wrote:
On Nov 24, 2008, at 9:05 PM, William Stein wrote:
I can easily run timing comparisons between maxima and FriCAS, but
because
of how sympy does things (with its separate variables), I'll have
to run
them separately.
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