On Tuesday, September 15, 2020 at 3:20:13 PM UTC-4 Emmanuel Charpentier
wrote:
> sage: L[1].n()
>
> fails because L1 is an equation, i. e a symbolic expression whose operator
> is the built-in “eq”, which has no n() method.
>
> However,
>
> sage: PP=-625/1000*t^4 + 2355/100*t^3 - 264051/1000*t
sage: L[1].n()
fails because L1 is an equation, i. e a symbolic expression whose operator
is the built-in “eq”, which has no n() method.
However,
sage: PP=-625/1000*t^4 + 2355/100*t^3 - 264051/1000*t^2 + 10269/10*t - 8538/10
sage: PP.parent()
Symbolic Ring
sage: L=solve(PP,t)
sage: L[1].rhs()
Thanks! I've already learned more.
What I first did was this:
sage: PP
-0.625*t^4 + 23.55000*t^3 - 264.0510*t^2 +
1026.900*t - 853.8000
sage: L=solve(PP==0,t)
sage: L[1]
t ==
-1/1250*sqrt((390625*(4/1953125*I*sqrt(37468876945450884598)*sqrt(5) -
39
> I still don't know my way around the Sage documentation... Sorry for the
> elementary question.
>
> Yeah, we are sorry that it never has gotten more organized (though it is
actually quite thorough!). You may want to try the French (now in
English) Sage book, or Greg Bard's AMS (but free on
H = (Mod(20192834, 20876441)^-1 * 5260862).nth_root(17)
On Thursday, May 14, 2020 at 3:26:09 PM UTC+2, Madison Adams wrote:
>
> Solve this Equation for H: 5260862 = M*H^y % 20876441
>
> Does anyone happen to know the steps to solve for h:
> Equation: 5260862 = M*Hy % 20876441
>
> Where y = 17, and
Have you had a look at
https://doc.sagemath.org/html/en/reference/numerical/sage/numerical/mip.html?
Seems to be what you want to do essentially.
julian
On Sunday, October 20, 2019 at 5:34:37 PM UTC+2, Santanu wrote:
>
> Hi,
> I have inequalities like these:
>
> 3 x1 + 5 x2 + 2 x3 + 5 x4 + 7 x
On Tue, 30 Jul 2019 at 05:56, Kwankyu wrote:
>
>
> On Thursday, July 25, 2019 at 12:08:20 AM UTC+9, chandra chowdhury wrote:
>>
>> I have matrices B and C of size (m,n) over integer with m>n.
>> I know there is matrix A of size (m,m) such that
>> AB=C. How to find A efficiently in Sage?
>>
>
Try
On Thursday, July 25, 2019 at 12:08:20 AM UTC+9, chandra chowdhury wrote:
>
> I have matrices B and C of size (m,n) over integer with m>n.
> I know there is matrix A of size (m,m) such that
> AB=C. How to find A efficiently in Sage?
>
I guess there is no special way in Sage to solve your kind
after solving an equation (or not) for x, I can check if the answer still
contains x by ans.has(x).
That should weed out any non-explicit solutions.
But still: am I guaranteed for any class of equations, e.g. polynomial
equations of degree <= 4,
that if solve produces an empty list there re
On Tuesday, February 19, 2019 at 8:56:50 AM UTC-8, Michael Beeson wrote:
>
> When I try to reproduce Eric's post, I get an error message about an
> unexpected keyword argument
> (maybe my version of Sage is too old.) But look at this:
>
> sage: solve(*2**(x+sqrt(*1*-x^*2*))-*7*,x,explicit_so
When I try to reproduce Eric's post, I get an error message about an
unexpected keyword argument
(maybe my version of Sage is too old.) But look at this:
sage: solve(*2**(x+sqrt(*1*-x^*2*))-*7*,x,explicit_solutions=True)
[1/4*I*sqrt(41) + 7/4 == -1/2*sqrt(7/2*I*sqrt(41) + 2), 1/4*I*sqrt(41)
Eric's post shows me how to get that particular example solved. But my
real
concern is, when my code (inside some deep loop) calls solve, I want to
know
(a) if it returns an answer, that answer really is a solution, and (b) if
it
returns an empty list, there really is no solution.
S
Hi,
Le lundi 18 février 2019 21:56:56 UTC+1, Michael Beeson a écrit :
>
> sage: solve(*2**(x+sqrt(*1*-x^*2*))-*7*,x)
>
> [x == -sqrt(-x^2 + 1) + 7/2]
>
>
> sage: version()
>
> 'SageMath version 8.0, Release Date: 2017-07-21'
>
>
> That doesn't look like a solution to me because x still appears on
yes exactly what I wanted. Sorry again for the bad explanation of my
problem, by the way I hadn't found the wiki you gave the link
Thanks
Henri
Le vendredi 5 octobre 2018 13:05:06 UTC+2, HG a écrit :
>
> I would like to solve these equations but I don't know how?
> >
> > t_0=t_p==gamma*(t-V*x/c^2
It seems you want to revert the Lorentz transformation:
https://en.wikipedia.org/wiki/Lorentz_factor#Occurrence
In this case you have a system of equations relating
the various quantities c, ga, t, tt, x, xx, v, where
- ga is the gamma factor, c is the speed of light,
- v is the relative speed of
sorry for the bazar !
I don't know how to make latex in the mail
I will make it with sage
Le vendredi 5 octobre 2018 13:05:06 UTC+2, HG a écrit :
>
> I would like to solve these equations but I don't know how?
> >
> > t_0=t_p==gamma*(t-V*x/c^2);show(t_0)
> > x_0=x_p==gamma*(x-V*t);show(x_0)
> >
>
Sorry for not being clear :)
I wonder if I can calculate
x'′=γ(x−vt)
t′=γ(t−vxc2)
γ=11√−v2c2
Le vendredi 5 octobre 2018 13:05:06 UTC+2, HG a écrit :
>
> I would like to solve these equations but I don't know how?
> >
> > t'=gamma*(t-vx/c^2);show(t')
> > x'=gamma*(x-v*t);show(x')
> >
> > solve(t_0
On Wed 2018-10-03, 18:34:51 UTC+2, HG wrote:
> I would like to solve these equations but I don't know how?
>
> t_0=t_p==gamma*(t-V*x/c^2);show(t_0)
> x_0=x_p==gamma*(x-V*t);show(x_0)
>
> solve(t_0,gamma*(t-V*x/c^2))
> desolve(gamma*(t-V*x/c^2)==0,x)
>
> error desolve() takes at least 2 arguments (
You are doing inexact computations, and imaginary errors creep in. Indeed,
let's first run without assume()'s:
sage: s=solve(f(x) == 0,x)
sage: [t.rhs().n() for t in s]
[0.464973569257452 - 1.11022302462516e-16*I,
0.0267727617252126 + 1.11022302462516e-16*I,
2.00825366901734]
so you see a tin
> >
> > Interesting. If you have ideas on how to improve it so that you might
> do
> > so, please start a thread on sage-devel about that.
>
> I don't. I have expressed my dislike of AskSage as soon as it was
> available. It is just that I don't like all these badges and awards and
> I do
Sage has two sources of support, the mailing list sage-support and the
AskSage site. Users and helpful developers can use whichever they
choose, and I don't think that any of us should be telling anyone
which one they ought to use, either as an asker of questions or as a
helpful answerer. Of cour
Hi Karl-Dieter,
On 2016-11-09, kcrisman wrote:
>> > This is a very good question for Ask Sage, would you ask it there?
>>
>> Why should he? He did ask here. And I, for one, dislike the Ask Sage
>> pages to the extent that I wouldn't answer questions there.
>>
>
> Interesting. If you have ide
> > This is a very good question for Ask Sage, would you ask it there?
>
> Why should he? He did ask here. And I, for one, dislike the Ask Sage
> pages to the extent that I wouldn't answer questions there.
>
Interesting. If you have ideas on how to improve it so that you might do
so, plea
A bit of numerical analysis (see enclosed Jupyter notebook) proves that
this polynomial has at least two real roots, and quite probably four), one
of them being positive.
This triggers the question : does Sage have built-in facilities for
uncertainty computation ("calcul d'erreurs" in French, b
On 2016-11-08, slelievre wrote:
> This is a very good question for Ask Sage, would you ask it there?
Why should he? He did ask here. And I, for one, dislike the Ask Sage
pages to the extent that I wouldn't answer questions there.
Cheers,
Simon
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Sun 2016-11-06 16:00:30 UTC+1, Francis Banks:
> I am solving a polynomial which arises from plotting titration cures
> in chemistry. The rule of signs suggests it has one positive root.
> Find_root seems to find it. Solve with poly_solve=true does not.
> Instead it gives 4 complex roots, which don
On Sunday, November 6, 2016 at 3:00:30 PM UTC, Francis Banks wrote:
>
> I am solving a polynomial which arises from plotting titration cures in
> chemistry. The rule of signs suggests it has one positive root. Find_root
> seems to find it. Solve with poly_solve=true does not. Instead it gives 4
thank you, Simon
>
> many computational aspects, but still human insight is needed to break a
> problem up into sub-problems of chewable size, or transform the original
> problem into one that is more accessible to automatic solutions. And
> even if a computer algebra system (no matter which o
> > In fact, the solution is: w=t+t^2
>
> Are you sure? Assuming some value for t, plotting the expression
> doesn't seem to show a solution at w = t + t^2.
>
I am sorry. I made a mistake.
> best
>
> Robert Dodier
>
>
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Hi!
On 2015-03-09, platane wrote:
> Thanks to Emmanuel Charpentier for the solution.
> But the trick is S1[0]^2 that needs human works.
Sure. But that's a very common situation. Computers are very good in
many computational aspects, but still human insight is needed to break a
problem up into su
Thanks to Emmanuel Charpentier for the solution.
But the trick is S1[0]^2 that needs human works.
Le samedi 18 octobre 2014 11:58:37 UTC+2, Emmanuel Charpentier a écrit :
>
> Well, after a bit of sleep, the solution was (semi-)obvious :
>
> sage: var("w,t")
> (w, t)
> sage: E1=-(1/2*sqrt((4*w+1)+1
Hello guy,
Your "question" looks like a modeling math problem, ...and is not related
to SAGE.
I wouln't try to solve "quadratic" problem using "linear" solvers : LP
means "Linear Programming"
for definition of LP, read
http://www.sagemath.org/doc/reference/numerical/sage/numerical/mip.html
Hello Dima,
Thank you for your response. It gave me some ideas :)
On Thursday, December 11, 2014 3:19:24 PM UTC+1, pegah Ali wrote:
>
> Hello everybody,
>
> I am new to sage and try to solve the following LP with
> MixedIntegerLinearProgram:
>
>
>
> Maximization:
> 3.0 x_0 + 2.0 x_
On 2014-12-11, pegah Ali wrote:
> --=_Part_141_146171712.1418307564716
> Content-Type: multipart/alternative;
> boundary="=_Part_142_2134155063.1418307564716"
>
> --=_Part_142_2134155063.1418307564716
> Content-Type: text/plain; charset=UTF-8
>
> Hello everybody,
>
> I am new to
Well, after a bit of sleep, the solution was (semi-)obvious :
sage: var("w,t")
(w, t)
sage: E1=-(1/2*sqrt((4*w+1)+1))*t+w==0
sage: S1=E1.solve(w)
sage: S1
[w == 1/2*t*sqrt(4*w + 2)]
sage: S2=((S1[0])^2).solve(w)
sage: S2
[w == 1/2*t^2 - 1/2*sqrt(t^2 + 2)*t, w == 1/2*t^2 + 1/2*sqrt(t^2 + 2)*t]
A n
Le vendredi 17 octobre 2014 16:37:55 UTC+2, vdelecroix a écrit :
>
> 2014-10-17 10:09 UTC, Emmanuel Charpentier >:
> > Ahem !
> >
> > On one machine :
> >
> > sage: sage.version.version
> > '6.4.beta4'
> > sage: var("w,t")
> > (w, t)
> > sage: solve(-(1/2*sqrt((4*w+1)+1))*t+w==0,w)
> >
2014-10-17 10:09 UTC, Emmanuel Charpentier :
> Ahem !
>
> On one machine :
>
> sage: sage.version.version
> '6.4.beta4'
> sage: var("w,t")
> (w, t)
> sage: solve(-(1/2*sqrt((4*w+1)+1))*t+w==0,w)
> [w == 1/2*t*sqrt(4*w + 2)]
This solution is implicit. This is the problem.
Vincent
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Ahem !
On one machine :
sage: sage.version.version
'6.4.beta4'
sage: var("w,t")
(w, t)
sage: solve(-(1/2*sqrt((4*w+1)+1))*t+w==0,w)
[w == 1/2*t*sqrt(4*w + 2)]
sage: maxima.version()
'5.34.1'
Another one :
sage: sage.version.version
'6.4.beta1'
sage: var("w,t")
(w, t)
sage: solve(-(1/2*sqrt((4*w
>
> > sage: solve(-(1/2*sqrt((4*w+1)+1))*t+w==0,w)
> > [w == 1/2*sqrt(4*w + 2)*t]
>
> Well, if I'm not mistaken, Sage punts to Maxima's 'solve' function,
> which, I'm sorry to report, is not very strong (it can solve a
> relatively narrow range of problems). But I find that Maxima's
> 'to_pol
On 2014-10-15, platane wrote:
> solve does not solve ?
>
> sage: solve(-(1/2*sqrt((4*w+1)+1))*t+w==0,w)
> [w == 1/2*sqrt(4*w + 2)*t]
Well, if I'm not mistaken, Sage punts to Maxima's 'solve' function,
which, I'm sorry to report, is not very strong (it can solve a
relatively narrow range of probl
Have zeros of order 2 too, so the sign change doesn't help in general, but
may work for some zeros. Thank you!...
On Saturday, June 21, 2014 1:21:06 AM UTC-7, Dima Pasechnik wrote:
>
> On 2014-06-21, David Ingerman > wrote:
> > Thank you, that's helpful. Is there a way to get all roots of a P
On 2014-06-22, David Ingerman wrote:
> Thank you, that makes sense. My Python function is not continuous though,
> has poles, so I'll probably have to plot it to find its zeros...
you might perhaps try finding a pole p by solving 1/f(x)=0, and
regularise f by multiplying f with (x-p)i^m, for a
Thank you, that makes sense. My Python function is not continuous though,
has poles, so I'll probably have to plot it to find its zeros...
On Saturday, June 21, 2014 1:21:06 AM UTC-7, Dima Pasechnik wrote:
>
> On 2014-06-21, David Ingerman > wrote:
> > Thank you, that's helpful. Is there a wa
On 2014-06-21, David Ingerman wrote:
> Thank you, that's helpful. Is there a way to get all roots of a Python
> function on an interval?
I never heard of robust procedures for such a task, and doubt they are
even possible (think about roots of sin(1/x) on [0,1]).
Certainly you can partition t
Thank you, that's helpful. Is there a way to get all roots of a Python
function on an interval?
On Friday, June 20, 2014 2:10:38 AM UTC-7, Dima Pasechnik wrote:
>
> On 2014-06-20, David Ingerman > wrote:
> > Thank you, so what to do for Python function? Matlab had general
> purpose
> > 'opt
On 2014-06-20, David Ingerman wrote:
> Thank you, so what to do for Python function? Matlab had general purpose
> 'optim(f)' if my memory is right...
you can e.g. use find_root(); this is a numerical thing that accepts
Python functions. Here is an example:
sage: def f(x):
return x-cos(x)
Thank you, so what to do for Python function? Matlab had general purpose
'optim(f)' if my memory is right...
On Wednesday, June 11, 2014 1:50:10 AM UTC-7, Dima Pasechnik wrote:
>
> On 2014-06-10, David Ingerman > wrote:
> >
> > How to solve([f(x)==0],x) for a function "f(x)" defined in a
On 2014-06-10, David Ingerman wrote:
>
> How to solve([f(x)==0],x) for a function "f(x)" defined in a .sage file?
>
> The error message: TypeError: Cannot evaluate symbolic expression to a
> numeric value.
what is f(x) ?
solve() won't work for a Python function, it needs a symbolic
expressi
I've opened http://trac.sagemath.org/sage_trac/ticket/14738 for a lot of
this stuff. It's all related to keywords not being sufficiently happy.
> There is a ticket for adding a lot of that to the main 'solve?' but Trac
> seems to be down right now so I can't find it...
>
http://trac.sagemath
>
>
> Sorry to keep on here, but I've
> got three other related queries:
>
> (1) "solve?" gives me " solve(sin(x)==x,x,explicit_solutions=True)"
> as an example which returns an empty list of solutions.
> But x=0 surely counts as an explicit solution? I guess my
> interpretation of an empt
On Thu, Jun 13, 2013 at 12:45 PM, kcrisman wrote:
>
>> Smells like a bug...
>
>
> Not exactly. The documentation for solve makes it clear (I hope!) in the
> examples that these are "generic" variables generated by Maxima which we do
> not make "Sage" variables. They just mean, "any old integer"
kcrisman wrote:
On Wednesday, June 12, 2013 8:09:02 PM UTC-4, leif wrote:
robin hankin wrote:
> hello, sage 5.9
>
> If solve() gives an unspecificed integer, how do I substitute a
> particular value into the expression?
>
> subs() does not work as expected/desi
On Wednesday, June 12, 2013 8:09:02 PM UTC-4, leif wrote:
>
> robin hankin wrote:
> > hello, sage 5.9
> >
> > If solve() gives an unspecificed integer, how do I substitute a
> > particular value into the expression?
> >
> > subs() does not work as expected/desired because the free variable
leif wrote:
robin hankin wrote:
hello, sage 5.9
If solve() gives an unspecificed integer, how do I substitute a
particular value into the expression?
subs() does not work as expected/desired because the free variables
don't seem to be defined.
sage: a=solve(sin(x)==0,x,to_poly_solve='force')
robin hankin wrote:
hello, sage 5.9
If solve() gives an unspecificed integer, how do I substitute a
particular value into the expression?
subs() does not work as expected/desired because the free variables
don't seem to be defined.
sage: a=solve(sin(x)==0,x,to_poly_solve='force');a
[x == 2*pi
We are talking about the Boolean polynomial ring here, right? So an F4 style
algorithm is used by default (subject to some heuristics). To emphasise you'd
have to construct your ring using the BooleanPolynomialRing constructor.
On Saturday 08 Dec 2012, john_perry_usm wrote:
> On Saturday, Decemb
On Saturday, December 8, 2012 11:07:31 AM UTC-6, Santanu wrote:
>
> Thank you. But when I try to solve
> f1=x1 + x2 + x4 + x10 + x31 + x43 + x56 ,
> f2=x2 + x3 + x5 + x11 + x32 +x44 + x57,
>
> it becomes very slow. Is there any faster approach like
> F4 algorithm available in Sage?
>
F4 is not
Thank you. But when I try to solve
f1=x1 + x2 + x4 + x10 + x31 + x43 + x56 ,
f2=x2 + x3 + x5 + x11 + x32 +x44 + x57,
it becomes very slow. Is there any faster approach like
F4 algorithm available in Sage?
On 8 December 2012 17:25, Martin Albrecht wrote:
> Or compute a Gröbner basis:
>
> sage:
Or compute a Gröbner basis:
sage: P. = BooleanPolynomialRing()
sage: Ideal(x^2 + y^2).groebner_basis()
[x + y]
sage: Ideal(x^2 + y^2).variety()
[{y: 0, x: 0}, {y: 1, x: 1}]
On Saturday 08 Dec 2012, Volker Braun wrote:
> I take it you mean polynomial equations:
>
> sage: AA. = AffineSpace(
I take it you mean polynomial equations:
sage: AA. = AffineSpace(GF(2),2)
sage: S = AA.subscheme(x^2+y^2)
sage: S.point_set().points()
[(0, 0), (1, 1)]
On Saturday, December 8, 2012 6:14:19 AM UTC, Santanu wrote:
>
> I have a system of non linear equations over GF(2). How to solve
> them in Sa
On Jul 21, 9:03 pm, Marshall Hampton wrote:
> Ticket 8553 doesn't fix this, so while it is a related issue this will
> need a ticket of its own. Looking at the source its not clear to me
> why your last example doesn't return a list.
>
> I createdhttp://trac.sagemath.org/sage_trac/ticket/11618t
Thanks, this at least gives a reasonable answer and will work in the
short-term. I'm still confused about why solve() didn't work, though.
On Thu, Jul 21, 2011 at 11:44 PM, achrzesz wrote:
>
> Try something like this:
>
> from mpmath import *
> mp.dps = 30; mp.pretty = True
>
> f=[lambda s00, s
Ticket 8553 doesn't fix this, so while it is a related issue this will
need a ticket of its own. Looking at the source its not clear to me
why your last example doesn't return a list.
I created http://trac.sagemath.org/sage_trac/ticket/11618 to address
this. Hopefully someone from 8553 can fix t
Try something like this:
from mpmath import *
mp.dps = 30; mp.pretty = True
f=[lambda s00, s01, s10, s11, k, p:0.55*k*s00 + 0.6*k*s01 + 0.6*k*s10
+ 0.6*p*s01 + 0.6*p*s10 +0.55*p*s11 + 33*s00 + 33*s01 + 33*s10 +
33*s11 - 33.0,
lambda s00, s01, s10, s11, k, p:0.55*k*s00 + 0.6*k*s01 + 0.6*k*s10 +
3
On 06/29/11 16:28, Jason Grout wrote:
> On 6/29/11 3:22 PM, Michael Orlitzky wrote:
>> This is probably just a case of "don't do that," but I thought I'd check:
>>
>>sage: c = [ var('c[0]') ]
>>sage: system = c[0]*x == 1
>>sage: solve(system, c[0])
>>
>>...
>>TypeError: unab
Thats
http://trac.sagemath.org/sage_trac/ticket/7496
and its scheduled for sage-4.7.2
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On 6/29/11 3:22 PM, Michael Orlitzky wrote:
This is probably just a case of "don't do that," but I thought I'd check:
sage: c = [ var('c[0]') ]
sage: system = c[0]*x == 1
sage: solve(system, c[0])
...
TypeError: unable to make sense of Maxima expression '[c[0]==1/x]' in
Sag
So, with 4.7 it doesn't work either.
But if I do some transformation by hand, and square both sides (I'm
not completely sure if that's allowed), then I get the solution.
f_2 = ((1/4*a^2*t^4 + a_x*t^3*v_x + a_x*t^2*x + t^2*v_x^2 +
t^2*v_y^2 + 2*t*v_x*x + 2*t*v_y*y + x^2 + y^2 -m)/(t^3*v_y +
t^2
Normally solve(e == m, a_x, to_poly_solve='force') works, but interestingly,
not for your equation.
Joal Heagney
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Btw. I use sage 4.6.2. I wasn't aware that 4.7 allready exists.
I will try out if the issue still exists in 4.7.
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On May 18, 9:17 am, tvn wrote:
> so is there someway we can do so that this patch or temporary solution be
> pushed into the next Sage release ?
Well, *you* can do that yourself - see, for instance,
http://www.sagemath.org/doc/developer/walk_through.html#reviewing-a-patch
Unfortunately, unti
so is there someway we can do so that this patch or temporary solution be
pushed into the next Sage release ?
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On May 18, 2:21 am, zsharon wrote:
> This is already mentioned
> here:http://www.mail-archive.com/sage-support@googlegroups.com/msg20832.html
> (It seems from that thread that this may not be a problem with Sage
> itself.)
>
> It should be reported as a bug if it hasn't been already.
It is htt
This is already mentioned here:
http://www.mail-archive.com/sage-support@googlegroups.com/msg20832.html
(It seems from that thread that this may not be a problem with Sage
itself.)
It should be reported as a bug if it hasn't been already.
On May 17, 12:20 pm, tvn wrote:
> shouldn't this be take
On Oct 19, 5:23 am, Robert Samal wrote:
> Speaking about solve(), is there a place to report equations it cannot
> solve
> (and I believe it should?).
>
> I suppose putting it on the same Trac ticket is wrong practice? But
> should
> it be another ticket, or some yet other place?
>
Since for no
Speaking about solve(), is there a place to report equations it cannot
solve
(and I believe it should?).
I suppose putting it on the same Trac ticket is wrong practice? But
should
it be another ticket, or some yet other place?
Robert
> Yup. solve() probably needs a general overhaul (and has for
On 13 okt, 03:39, kcrisman wrote:
> On Oct 12, 6:37 pm, Robert Samal wrote:
>
>
>
> > I observed that solve behaves inconsistently in the following regards:
>
> > sage: solve([x==1,x==-1],x)
> > []
>
> > (this is as expected)
> > However:
>
> > solve([x==1,x==-1],x, solution_dict=True)
>
> > pr
On Oct 12, 6:37 pm, Robert Samal wrote:
> I observed that solve behaves inconsistently in the following regards:
>
> sage: solve([x==1,x==-1],x)
> []
>
> (this is as expected)
> However:
>
> solve([x==1,x==-1],x, solution_dict=True)
>
> produces an error message. Easy to live with, but I was sca
Hello everyone
thanks for the help here.
In Mathematica, Reduce[] works like Solve, except that it returns a
Boolean list of possible solutions. I use it to check what
the necessary conditions for the"real" solution to work:
MMA> Reduce[a*x == b, {x}]
MMA> (b == 0 && a == 0) || (a != 0 &&
On Aug 19, 5:39 pm, robin hankin wrote:
> Hello Simon
>
> thanks for this. One problem
> with the solution you mention is that I can't do the
> general case. What I need is the sage equivalent
> of mathematica's Reduce[] function.
>
I think that solve() is the closest that comes to this (thoug
Hello Simon
thanks for this. One problem
with the solution you mention is that I can't do the
general case. What I need is the sage equivalent
of mathematica's Reduce[] function.
Is there one?
rksh
On Thu, Aug 19, 2010 at 10:06 PM, Simon King wrote:
> Hi!
>
> On 19 Aug., 22:41, robin hankin
Hi!
On 19 Aug., 22:41, robin hankin wrote:
> sage>solve([a*b==15*I-5,a*conjugate(b)==-13*I+9],[a,b])
> []
>
> So, from the first two lines I know that a=2+I, b=1+7I should
> be a solution to the system in the third, yet solve() returns empty.
Admittedly I am no expert for symbolics, and I don't
On Fri, Apr 16, 2010 at 9:57 AM, Owen wrote:
>> Is there another method approach I could take? I'd like to reach 6 at
>> least. My homework depends on it! :)
There's also the to_poly_solve argument to solve which cause non-exact
answers to be returned:
sage: solve([x^5 - x^4 - x^3 - x^2 - x -
Gak! I found it a work around: find_root() seems fine. Sorry for the
interruption!
On Apr 16, 10:47 am, Owen wrote:
> I'm solving recurrence relations for a k-SAT algorithm, and have run
> up against an apparent limit in solve. I'm running:
> solutions = solve([x^3 - x^2 - x - 1 == 0], x, solu
The problem related to conversion of h1into maxima has been reported
at
http://trac.sagemath.org/sage_trac/ticket/8634
Robert
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For more op
You can also use find_root:
sage: h1 = var('h1')
sage: eq = (6000*(h1/2)/(((1/12)*0.1125*(h1^3)) -
((1/12)*(0.1125-0.012)*((h1-(2*0.012))^3 == 8928880.28799800
sage: eq.find_root(-1.2, -0.8)
-0.96594148395195312
sage: eq.find_root(-0.8, 0.2)
0.00033054212748907948
sage: eq.find_root(.2, 1)
0.3
Mike wrote:
Thanks for the reply. Using the exact same setup I changed "h1" to "p"
and it worked out fine. Well not fine exactly but at least I'm getting
some kind of answer. It's giving me some kind of "I" variable. Anyway
this seems like kind of a major bug. Is there a way to add this into
the t
Try to_poly_solve=True
sage: h1=var('h1')
sage: solve ([(6000*(h1/2)/(((1/12)*0.1125*(h1^3)) -
((1/12)*(0.1125-0.012)*((h1-(2*0.012))^3 == 8928880.28799800],
h1,to_poly_solve=True)
[x^k*binomial(n, k) == -0.965941485371, x^k*binomial(n, k) ==
0.000330542127524, x^k*binomial(n, k) == 0.3626109
On Mar 2, 8:10 pm, Sharpie wrote:
> So I guess at this point my question is: is there another way to
> convert from a symbolic polynomial equation to a Polynomial Ring? I
> The methods I used feel very hacky and I don't trust them.
Ok, so I think I found something in the manual page "Conversion
On Mar 2, 2:34 pm, Alex Ghitza wrote:
> The way I see it, it is not actually a question about the variable
> representing a real number; it is more a question of using polynomials
> and their specialised built-in roots() method rather than symbolic
> functions and the general-purpose solve().
>
>
On Tue, 2 Mar 2010 05:27:45 -0800 (PST), Sharpie wrote:
> Thanks for the reply Alex. I think I understand that by choosing a
> variable of the appropriate type, in this case one that is restricted
> to the real numbers, the roots can be determined in a straight-forward
> manner.
The way I see it,
On Mar 1, 3:02 am, Alex Ghitza wrote:
> On Sun, 28 Feb 2010 23:02:08 -0800 (PST), Sharpie
> wrote:
> > However, tonight I have been trying to solve an open channel flow
> > problem which requires me to find the roots of:
>
> > y^3 - 1.39027132807289 * y^2 + 0.090610488164005 == 0
>
> > find_ro
Should we create a ticket for this ? I'd have done it if not for my
doubt on the section I should pick for this... :-)
Nathann
On Feb 10, 1:42 pm, Harald Schilly wrote:
> On Feb 10, 11:16 am, Simon King wrote:
>
> > sage: f = y(n+2) - y(n+1) - y(n)
>
> ahh ... ok. now i get it ^^
> When I look
On Wed, Feb 10, 2010 at 8:50 AM, Walking Randomly
wrote:
> Regarding namespace pollution. Mathematica has thousands of function
> names in it's global namespace but it never causes programmers a
> problem because they have a convention. All mathematica functions
> start with a capital: Integrate
On Feb 10, 2:48 pm, Nathann Cohen wrote:
> I mentionned having to import simpy.*...
nonono, rsolve should be a new sage function, inside of it it only
imports the really necessary sympy classes. how is
integrate(algorithm='sympy') done? i think that's a good starting
point. see calculus/calculus.
I think you always mean
from sympy import *
Don't get it confused with simpy, which is another software package for
simulation (versus symbolic computation).
On 02/10/2010 02:48 PM, Nathann Cohen wrote:
> I mentionned having to import simpy.*, which included classes like
> Symbol or Function, a
Regarding namespace pollution. Mathematica has thousands of function
names in it's global namespace but it never causes programmers a
problem because they have a convention. All mathematica functions
start with a capital: Integrate, Plot, ListPlot etc. So, stick to
lower case variables and you d
I mentionned having to import simpy.*, which included classes like
Symbol or Function, and having to use those types. I wouldn't
personally mind if I had to import rsolve from some Sage class... Why
should it necessarily imported by default (namespace kept clean)?
Nathann
--
To post to this grou
On Feb 10, 11:16 am, Simon King wrote:
> sage: f = y(n+2) - y(n+1) - y(n)
ahh ... ok. now i get it ^^
When I look into sympy/solvers/recurr.py right the first thing rsolve
does is to compute lhs - rhs. So, f = y(n) == y(n-1) - y(n-2) *should*
work. But it doesn't, because that equation doesn't ge
Hi Nathann,
On Wed, Feb 10, 2010 at 9:25 PM, Nathann Cohen wrote:
> Well, for example :
>
> - You need to import simpy to use it
One either directly import the functionalities of a third-party
package, or have them imported automatically at Sage startup. But
consider for a moment, the implicatio
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