sage: s=maxima('1/2*sqrt(%pi)*sqrt(2)*sqrt(x)*bessel_j(1/2,
x),besselexpand:true')
sage: s.sage().full_simplify()
sin(x)
On Saturday, September 15, 2012 6:53:34 AM UTC+2, Renan Birck Pinheiro
wrote:
The command sum( ((-1)^k*(x^(2*k+1))/factorial(2*k+1)),k,0,oo)
For example:
sage: a=[[QQ((x2).rhs()),+oo],[-oo,QQ((x1).rhs())]]; a.sort(); print a
[[-Infinity, 1], [2, +Infinity]]
On Saturday, September 15, 2012 7:32:22 PM UTC+2, aurelie...@yahoo.fr wrote:
I stumbled upon the following problem: the following code does not work as
it should:
The equations you provided are vector-valued and Sage expects scalar ones
Is that what you are expecting:
a,b,c,d,e,f,t = var('a,b,c,d,e,f,t')
v(t)=(t,t)
u(t)=(2+t,1-t)
B(t)=(a*t^2+b*t+c,d*t^2+e*t+f)
eq10=v(t).substitute(t==1)[0]==B(t).substitute(t==0)[0] # v(1)==B(0)
Sometimes a workaround is possible:
sage: x = var('x')
sage: y = function('y', x)
sage: C = var('C')
sage: f = desolve(diff(y,x) + y, y, ics=[0,C]); f
C*e^(-x)
sage: c = var('c')
sage: f = desolve(diff(y,x) +c*y, y,ivar=x, ics=[0,C]); f
C*e^(-c*x)
On Thursday, July 26, 2012 12:37:56 PM UTC+2,
l1=[(cos(phi),sin(phi)) for phi in srange(n(pi)/4,3*n(pi)/4,0.01)]
l2=[(2*cos(phi),2*sin(phi)) for phi in srange(n(pi)/4,3*n(pi)/4,0.01)]
l2.reverse()
p=polygon(l1+l2)
c1=circle((0,0), 1,rgbcolor=(0,0,0))
c2=circle((0,0), 2,rgbcolor=(0,0,0))
l1=[(3*cos(phi),3*sin(phi)) for phi in srange(n(pi)/2,3*n(pi)/2,0.01)]
l2=[(6*cos(phi),6*sin(phi)) for phi in srange(n(pi)/2,3*n(pi)/2,0.01)]
l2.reverse()
figure4=polygon(l1+l2)
figure4.show(xmin=-6,xmax=6)
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from
#You probably need the right-hand sem-iring:
l1=[(3*cos(phi),3*sin(phi)) for phi in srange(-n(pi)/2,n(pi)/2,0.01)]
l2=[(6*cos(phi),6*sin(phi)) for phi in srange(-n(pi)/2,n(pi)/2,0.01)]
l2.reverse()
figure4=polygon(l1+l2)
figure4.show(xmin=-6,xmax=6)
--
To post to this group, send email to
On Monday, July 16, 2012 12:49:32 AM UTC+2, The Doctor (Michael) wrote:
Yes,that is correct Jason.Thanks.
i have another question along the same vein. I need to plot a polar region
between 2 circles of radius 1 and 2, from pi/4 = theta M= 3*pi/4. Here is
what I tried, but I got a bunch
Notice that roots gives you the multiplicity
sage: f=(x-1)^2
sage: f.roots(ring=RealField())
[(1.00, 2)]
sage: f=x^3-x-1
sage: f.roots(ring=RealField())
[(1.32471795724475, 1)]
sage: f.roots(ring=RDF)
[(1.32471795724, 1)]
sage: f.roots(ring=CDF)
[(-0.662358978622 - 0.562279512062*I,
sage: import sympy
sage: sympy.rsolve?
On Wednesday, June 20, 2012 6:43:21 PM UTC+2, Daniel Krenn wrote:
Is there something to solve recursions (e.g. linear recursions, but also
others) in Sage? Or, formulated in another way: Is there something in Sage
like RSolve in Mathematica?
--
To
On Saturday, April 14, 2012 6:04:44 PM UTC+2, Laurent Decreusefond wrote:
Dear all,
here is my problem
sage: var('t')
t
sage: integrate(e^(-2*t)/sqrt(1-e^(-2*t)),t,0,infinity)
-1
... a negative value for the integral of a positive function.
On the other hand
sage:
On 1 Lut, 10:49, Ajay Rawat ajay.rawa...@gmail.com wrote:
Hi,
I want to integrate
(int x*exp(-x^2)/(x-1) , -inf, inf)
how to do this
Thanking you
--
Ajay Rawat
Kalpakkam, IGCAR
-
Save Himalayas
On Jan 25, 6:21 pm, William Stein wst...@gmail.com wrote:
On Wed, Jan 25, 2012 at 7:25 AM, kcrisman kcris...@gmail.com wrote:
Probably that should be implemented inside the trigonometric functions
code itself, instead of in any of the simplify's.
What do you think?
+1
It's
On Jan 25, 2:26 pm, Peter Luschny peter.lusc...@googlemail.com
wrote:
Hi, with Sage Notebook Version 4.7.2:
Z0 = 1118557440*x^5 + 180204024*x^4 + 15195180*x^3 +
523250*x^2 + 4095*x + 1
Z1 = 4115105280L*x^6 + 1118557440*x^5 + 180204024*x^4 + 15195180*x^3 +
523250*x^2 +
On Jan 9, 6:39 pm, Renan Birck Pinheiro renan.ee.u...@gmail.com
wrote:
Hi,
I'm trying to solve a differential equation with unit step, e.g. the
equation y'(x) = U(x-5) - where U is the unit step, and the inicial
condition y(0) is 0.
The result is 0 for 0x5 and x-5 for x5 (it's a simple
On Dec 31, 11:21 am, Christophe BAL projet...@gmail.com wrote:
Hello,
the following command has a very bad output.
plot(sqrt((x - 2)*(x - 1)), (-4, 5)) + plot(-sqrt((x - 2)*(x - 1)), (-4, 5))
How can I ameliorate this ?
Best regards.
Christophe
Your function is comlex-valued on the
On Dec 28, 8:13 pm, akm andrew.mussel...@gmail.com wrote:
Hi all, can anyone recommend a way of generating random numbers via a
Poisson distribution?
I'm trying to wrangle scipy.stats.poisson to get something centered at
10, say, with a nice long tail out to the right, but I can't figure
On Dec 28, 6:26 am, shreevatsa shreevatsa.pub...@gmail.com wrote:
Hi,
I'm trying to use Sage to find the asymptotics of binomial
coefficients. Specifically, I wanted to find out the rate at which
binomial(n, n/2)/2^n goes down to 0 as n goes to infinity.
See Wolfram
On Dec 28, 1:54 pm, achrzesz achrz...@wp.pl wrote:
On Dec 28, 6:26 am, shreevatsa shreevatsa.pub...@gmail.com wrote:
Hi,
I'm trying to use Sage to find the asymptotics of binomial
coefficients. Specifically, I wanted to find out the rate at which
binomial(n, n/2)/2^n goes down to 0
On Dec 28, 6:14 pm, achrzesz achrz...@wp.pl wrote:
On Dec 28, 1:54 pm, achrzesz achrz...@wp.pl wrote:
On Dec 28, 6:26 am, shreevatsa shreevatsa.pub...@gmail.com wrote:
Hi,
I'm trying to use Sage to find the asymptotics of binomial
coefficients. Specifically, I wanted to find out
sage: (cos(x)^3).reduce_trig()
1/4*cos(3*x) + 3/4*cos(x)
On Dec 13, 7:45 pm, Christophe BAL projet...@gmail.com wrote:
Hello,
what is the easiest way to linearize cos(x)^3 ?
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
On Dec 12, 7:03 am, Santanu Sarkar sarkar.santanu@gmail.com
wrote:
Sorry I meant to write
But it does not work
apologies for the typo
On 12 December 2011 07:49, Santanu Sarkar sarkar.santanu@gmail.com
wrote:
I have a set of Boolean functions like
A[0]=x1*x2+x3*x4
On Dec 12, 2:39 pm, v...@ukr.net wrote:
On Mon, 12 Dec 2011 07:16:17 -0600Jason Grout jason-s...@creativetrax.com
wrote:
sage: round(-0.02800200,12)
-0.028
See
http://www.sagemath.org/doc/reference/sage/misc/functional.html?#sage...
Thanks a lot! :)
It looks like I
sage: N(pi^2/6, digits=17)
1.6449340668482264
sage: numerical_integral(x/(exp(x)-1),0,oo)
(1.6449340668482264, 5.9356452836178026e-10)
On Dec 11, 6:45 pm, andres.ordonez andres.felipe.ordo...@gmail.com
wrote:
That'll do. Thanks
On Dec 10, 8:19 pm, Renan Birck Pinheiro renan.ee.u...@gmail.com
On Dec 11, 7:30 pm, achrzesz achrz...@wp.pl wrote:
sage: N(pi^2/6, digits=17)
1.6449340668482264
sage: numerical_integral(x/(exp(x)-1),0,oo)
(1.6449340668482264, 5.9356452836178026e-10)
On Dec 11, 6:45 pm, andres.ordonez andres.felipe.ordo...@gmail.com
wrote:
That'll do. Thanks
On Dec 11, 10:59 pm, achrzesz achrz...@wp.pl wrote:
On Dec 11, 7:30 pm, achrzesz achrz...@wp.pl wrote:
sage: N(pi^2/6, digits=17)
1.6449340668482264
sage: numerical_integral(x/(exp(x)-1),0,oo)
(1.6449340668482264, 5.9356452836178026e-10)
On Dec 11, 6:45 pm, andres.ordonez
On Dec 12, 12:10 am, juaninf juan...@gmail.com wrote:
Hi everybody
I want choose different minimal polynomial to build a Galois Field
2^m, how?
For example: m = 8
sageF.a=GF(2^8)
sage:print a.minpoly()
I get ...
x^8 + x^4 + x^3 + x^2 + 1
but I want now other polynomial for example
On Dec 12, 12:31 am, achrzesz achrz...@wp.pl wrote:
On Dec 12, 12:10 am, juaninf juan...@gmail.com wrote:
Hi everybody
I want choose different minimal polynomial to build a Galois Field
2^m, how?
For example: m = 8
sageF.a=GF(2^8)
sage:print a.minpoly()
I get ...
x^8 + x^4
On Dec 12, 12:41 am, Juan Grados juan...@gmail.com wrote:
but when I put this polynomial x^8+x^7+x^4+x^3+x+1 I get ValueError:
finite field modulus must be irreducible but it is not
2011/12/11 achrzesz achrz...@wp.pl
On Dec 12, 12:10 am, juaninf juan...@gmail.com wrote:
Hi
On Dec 12, 12:47 am, achrzesz achrz...@wp.pl wrote:
On Dec 12, 12:41 am, Juan Grados juan...@gmail.com wrote:
but when I put this polynomial x^8+x^7+x^4+x^3+x+1 I get ValueError:
finite field modulus must be irreducible but it is not
2011/12/11 achrzesz achrz...@wp.pl
On Dec 12
On Dec 12, 1:08 am, achrzesz achrz...@wp.pl wrote:
On Dec 12, 12:47 am, achrzesz achrz...@wp.pl wrote:
On Dec 12, 12:41 am, Juan Grados juan...@gmail.com wrote:
but when I put this polynomial x^8+x^7+x^4+x^3+x+1 I get ValueError:
finite field modulus must be irreducible
On Dec 10, 2:48 pm, v...@ukr.net wrote:
Hello!
Today I have noticed that if I cannot use the legend_label plot
parameter while joining several plots in one grahpics_array object.
For example, the following code works fine:
plot1 = plot(sin, xmin=0, xmax=8*pi)
plot2 =
On Dec 9, 12:18 pm, David Joyner wdjoy...@gmail.com wrote:
This is a question for sage-support, which I am ccing.
On Fri, Dec 9, 2011 at 5:04 AM, Jean-Patrick Pommier
jeanpatrick.pomm...@gmail.com wrote:
Hi,
Is it possible to solve equation with complex variable such 1/z=1+i?
I try
On Dec 8, 3:48 pm, Eric Kangas eric.c.kan...@gmail.com wrote:
thanks i tried it with N(pi, digits = 100) and sage didn't like that
format.
On Dec 8, 6:39 am, Jason Grout jason-s...@creativetrax.com wrote:
On 12/8/11 7:16 AM, Eric Kangas wrote:
Hi,
I remembering reading one of the
On Dec 8, 3:58 pm, achrzesz achrz...@wp.pl wrote:
On Dec 8, 3:48 pm, Eric Kangas eric.c.kan...@gmail.com wrote:
thanks i tried it with N(pi, digits = 100) and sage didn't like that
format.
On Dec 8, 6:39 am, Jason Grout jason-s...@creativetrax.com wrote:
On 12/8/11 7:16 AM, Eric
On Dec 8, 4:35 pm, achrzesz achrz...@wp.pl wrote:
On Dec 8, 3:58 pm, achrzesz achrz...@wp.pl wrote:
On Dec 8, 3:48 pm, Eric Kangas eric.c.kan...@gmail.com wrote:
thanks i tried it with N(pi, digits = 100) and sage didn't like that
format.
On Dec 8, 6:39 am, Jason Grout jason-s
On Dec 8, 9:36 pm, robin hankin hankin.ro...@gmail.com wrote:
hello
I have been playing with assumptions(). I want to assume ab
but solve() gives me a solution which is not consistent with this:
sage: var('a b')
(a, b)
sage: assume(ab)
sage: assumptions()
[a b]
sage:
On Dec 6, 3:26 pm, kcrisman kcris...@gmail.com wrote:
The problem is that the integral should not depend on the center of
the circle
containing the pole. It looks like maxima bug (?)
I've reported this
athttps://sourceforge.net/tracker/?group_id=4933atid=104933
Dan, if you want to
On Dec 5, 5:31 am, Dan Drake dr...@kaist.edu wrote:
I keep wondering whether Sage is making a mistake, or I'm not
understanding complex analysis. I'm a little afraid to learn the answer.
:)
Take f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3)). It's analytic everywhere
except at -1/2-I/3, where it has
On Dec 5, 10:04 am, achrzesz achrz...@wp.pl wrote:
On Dec 5, 5:31 am, Dan Drake dr...@kaist.edu wrote:
I keep wondering whether Sage is making a mistake, or I'm not
understanding complex analysis. I'm a little afraid to learn the answer.
:)
Take f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3
On Dec 5, 3:18 pm, Subhadeep Banik monsieurlel...@gmail.com wrote:
If I have a boolean expression of a certain number of variables, how
do I define a polynomial ring with them.. For example, let
(x0,x1,x2,x3,x4,x5,x6,x7) be a set of variables, and I have an
expression S = x0 + x1*x3 + x7
On Dec 5, 4:56 pm, kcrisman kcris...@gmail.com wrote:
So it sounds like you should file a ticket, Dan. Maybe we're just
sending it to Maxima wrong.
(%i9) f(z):=(z-%i)*(z-1)^2/(z-(-1/2-%i/3));
2
(z - %i)
On Dec 5, 5:31 am, Dan Drake dr...@kaist.edu wrote:
I keep wondering whether Sage is making a mistake, or I'm not
understanding complex analysis. I'm a little afraid to learn the answer.
:)
Take f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3)). It's analytic everywhere
except at -1/2-I/3, where it has
On Dec 2, 2:24 pm, Julie juliewilliams...@googlemail.com wrote:
Hi all,
I am attempting to obtain coefficients of a generating function to
obtain probabilites, but in order to obtain the coefficients, I first
need to expand a power series, which is necessary for my paricular
function.
Is
On Nov 30, 8:17 pm, Th btho...@nexus.hu wrote:
Dear All,
I am new to sage (converting from mathcad), so please forgive my basic
question.
I am trying to solve system of equations in a symbolic way, as a
simple example:
var('a,b,c,d,d1,d2,d3,x,x1,x2,x3,y,y1,y2,y3')
symsys=[
On Nov 27, 2:16 am, kcrisman kcris...@gmail.com wrote:
Here's this in Sage. See the end of this post for a solution you
might like better than the ramblings in between :)
--
| Sage Version 4.7.2, Release Date: 2011-10-29
On Nov 24, 5:55 pm, Julie juliewilliams...@googlemail.com wrote:
I have a generating function programmed with two variables, for which
I wish to find the coefficients of each polynomial in the series
expansion, i.e. coefficiecnt of xy, x^2y, x^2y^2,...
For example, using a simple function
Workaround:
sage: f=[f1,f2,f3,f4,f5,f6]
sage: sum([implicit_plot(g,(x,-8,8),(y,-5,5),plot_points=num) for g in
f])
(The eq. prod(f)==0 is really complicated)
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage: pl=sum([plot(lambda x:bessel_J(x,n),(x,0,1),color=hue(n/
5.0),legend_label=str(n)) for n in [1..4]])
sage: pl.show()
sage: numerical_integral(lambda x:bessel_J(x,2),1,2)
(0.48267531157352617, 5.3587724432705002e-15)
Andrzej Chrzeszczyk
--
To post to this group, send email to
sage: maxima('integrate(bessel_j(1,x), x)')
-bessel_j(0,x)
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more options, visit this group at
sage: import sympy
sage: sympy.integrate(1/(sqrt(x)*((1+sqrt(x))^2)),(x,1,9))
1/2
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more options, visit this
Hello Urs
Here is an example with 6 variables
Unfortunatly I dont remember the source or Author
import scipy
from scipy import optimize as opt
def f(x):
s00=x[0];s01=x[1];s10=x[2];s11=x[3];k=x[4];p=x[5]
eq1 = 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 +
Hello Urs
fsolve from scipy does solve systems (look at scipy reference)
All functions I have mentioned solve nonlinear systems of *equations*
Maxima has also function mnewton solving nonlinear systems
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
For larger problems:
from scipy.optimize import newton_krylov
newton_krylov?
(examples in scipy reference)
A Ch
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more options,
1)
solve?
(some nonlinear systems are solved numerically)
2)
from scipy.optimize import fsolve
fsolve?
3)
from mpmath import findroot
findroot?
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
In Sage you need more steps.
Assuming that you need the exact and the approximate value:
var('x y')
s=solve([x^2+y^2-4==0,x+y-1==0],[x,y])
x0,x1=[sol[0].rhs() for sol in s]
assume(abs(x)2)
ii=integrate(integrate(1,(y,1-x,sqrt(4-x^2))),(x,x0,x1))
#print ii
print ii.n()
#3.31871699805750
(The
I must correct myself
The region you asked has the area
sage: (pi*2^2-ii).n()
9.24765361630167
A Ch
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more options, visit this
If you realy need the exact solution then you can tray something like
that
var('x,y,lam,x4,y2')
assume(x40)
assume(y20)
L= x+2*y -lam*(16-2*x^(1/4)*y^(1/2))
eq1=numerator(factor(L.diff(x)))
eq1=eq1(y=y2^2,x=x4^4).full_simplify()
eq2=numerator(factor(L.diff(y)))
sage: maxima('batch(solve_rec)')
/home/andy/Pobrane/sage-4.7.1/local/share/maxima/5.23.2/share/contrib/
solve_rec/solve_rec.mac
sage: maxima('deq: u[n+2]=2*u[n + 1] + 8*u[n];')
u[n+2]=2*u[n+1]+8*u[n]
sage: maxima('sol:solve_rec(deq,u[n],u[0]=2,u[1]=7);')
u[n]=11*4^n/6-(-2)^(n-1)/3
Andrzej
Also:
sage: from sympy import *
sage: n = Symbol('n', integer=True)
sage: u = Function('u')
sage: f=u(n+2)-2*u(n+1)-8*u(n)
sage: rsolve(f,u(n),{u(0):2,u(1):7})
(-2)**n/6 + 11*4**n/6
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from
On Oct 8, 10:05 am, achrzesz achrz...@wp.pl wrote:
The exact integral
1/2*(erf(pi - 1/2*I) + erf(pi + 1/2*I))*sqrt(pi)*e^(-1/4)
contains *real* expression
(erf(pi - 1/2*I) + erf(pi + 1/2*I))
but maxima/sage can't check that it is real
In WolframAlpha one can check that
Im[Erf[Pi-1/2
On Oct 23, 11:33 pm, Eric Kangas eric.c.kan...@gmail.com wrote:
Hi,
I have one function that I need to plot in an array with different
values for two constants. I would like to only have to repeat the
function once, and go off of a string of different variables to
produce this array. Is
On Oct 22, 4:57 am, Gary Church gary.chur...@comcast.net wrote:
Hello all,
How can I evaluate
integral(sin(x^2),x,0,2)
to get a real value instead of the nasty expression involving erf() and I
that it spits out at me?
Thanks much,
Gary
One can check that the nasty expression is
Is that integrate_numerical or numerical_integral?
sage: numerical_integral(sin(x^2),0,2)
(0.80477648934375612, 1.0932458096886601e-14)
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
This version works for me:
r,z = var('r,z')
gro = 1.0
kro = 3.0
def f(r,z):
term1 = (1+(bessel_J(0, gro)/bessel_J(2,gro)))*(r/
kro)*(bessel_J(1, r)/bessel_J(1,gro))*cos(z)
term2 = -(bessel_J(0,r)/bessel_J(2, gro))*(r**2/kro**2)
return term1 + term2
p = contour_plot(f , (r, 0, 2.5),
We need to work on the numerical approximation stuff for the error
function!
Dan
More generally
in my opinion n() in complex domain needs improvement
For example (as I have mentioned)
sage: ((-1)^(1/3)).n()
0.500 + 0.866025403784439*I
leads to wrong solution
sage:
The exact integral
1/2*(erf(pi - 1/2*I) + erf(pi + 1/2*I))*sqrt(pi)*e^(-1/4)
contains *real* expression
(erf(pi - 1/2*I) + erf(pi + 1/2*I))
but maxima/sage can't check that it is real
In WolframAlpha one can check that
Im[Erf[Pi-1/2*I]+Erf[Pi+1/2*I]]
iz zero
Andrzej Chrzeszczyk
--
To
The *indefinite* integral of zero is an arbitrary constant (not only
zero)
Compare more 'concrete' version
sage: y = function('y',x)
sage: a = 1 + diff(y,x) == 0
sage: var('t')
t
sage: assume(t0)
sage: integrate(a,(x,0,t))
t - y(0) + y(t) == 0
Andrzej Chrzeszczyk
--
To post to this group, send
This also works:
sage: k=var('k')
sage: sum('binomial(8,k)',k,0,8)
256
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more options, visit this group at
It seems unreasonably annoying to plot a bunch of Bessel functions
together. How can I work around this?
Dan
sage: pl=sum([plot(lambda x:bessel_J(x,n),(x,0,1)) for n in [1,2]])
sage: pl.show()
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
To
On Sep 28, 2:48 pm, flyingsquirrel coskun.e...@yahoo.com wrote:
I have a symmetric matrix that I want to diagonalize, such as
x y z
y 0 xy
z xy xyz
x, y, z being variables, and the base field is CC (complex numbers). I
typed in the following:
R.x,y,z=CC[]
For trivial cases, yeah, but consider
parametric_plot((f(t-0.5j).real, f(t-0.5j).imag), (t,tmin,tmax))
parametric_plot(ReIm(f(t-0.5j)), (t,tmin,tmax))
--
For example:
sage: f(z)=[z.real(),z.imag()]
sage: t=var('t')
sage: parametric_plot(f(exp(I*(t-5*I))),(t,0,2*pi))
A colon is missing after range(20).
The following version works for me:
(but the indentation in this post may be broken)
x,y,z,t = var('x,y,z,t')
P = implicit_plot3d(x^2 +y^2 -z^2 ==1, (x,-3.2,3.2),(y,-3.2,3.2),
(z,-3,3),opacity=.2,color='blue')
for k in range(20):
P +=
Also:
sage: T = Cylindrical('height', ['radius', 'azimuth'])
sage: r, theta, z = var('r theta z')
sage: plot3d(r*sin(1/r), (r, 0.0, 0.2), (theta, 0, 2*pi),
transformation=T,adaptive=True)
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe
Also:
sage: n=var('n')
sage: assume(n,'integer')
sage: maxima('rectform(exp(%i*n*%pi))').sage().simplify_full()
(-1)^n
Andrzej Chrzeszczyk
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage: (integrate( exp(-x^2/2)/sqrt(2*pi) * sign(x-1), x, -oo, 1 )
+integrate( exp(-x^2/2)/sqrt(2*pi) * sign(x-1), x, 1,
oo )).simplify_full()
-erf(1/2*sqrt(2))
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
On i5-661:
sage: timeit('ainv=LA.inv(a)')
5 loops, best of 3: 54.3 ms per loop
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more options, visit this group at
sage: MS=MatrixSpace(RDF,512,512)
sage: A=MS.random_element()
sage: timeit('B=A.inverse()')
5 loops, best of 3: 41 ms per loop
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more
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 +
sage: assume(t,'real')
sage: solve(diff(myHH,t),t)
[t == 1/10*log(31)]
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
sage-support+unsubscr...@googlegroups.com
For more options, visit this group at
t=var('t')
taum, tauh, m = var('taum, tauh, m')
f(t) = (1-e^(-t/taum))^m*(e^(-t/tauh))
d=diff(f(t),t)
d1=(d==0).full_simplify()
print d1.solve(t)
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
It seems that Maxima has problem here but mpmath has not:
sage: from mpmath import *
sage: mp.pretty=True
sage: quad(lambda x:(x^2)*exp(x)/(1+exp(x))^2,[-inf,+inf])
3.28986813369645
sage: n(pi^2/3)
3.28986813369645
A Ch
--
To post to this group, send email to sage-support@googlegroups.com
To
``list_plot`` takes either a single list of data, a list of tuples,
or a dictionary and plots the corresponding points.
sage: list_plot(data.tolist())
A Ch
--
To post to this group, send email to sage-support@googlegroups.com
To unsubscribe from this group, send email to
Example; intersection of an elipsoid and sphere:
var('x,y,z')
solve([x^2 +y^2+z^2 ==1,x^2+y^2+2*z^2 ==1],[x,y,z])
#[[x == r1, y == -sqrt(-r1^2 + 1), z == 0], [x == r2, y == sqrt(-r2^2
+ 1), z == 0]]
var('r1')
p1=parametric_plot3d([r1,-sqrt(-r1^2 + 1),0],
(-1,1),thickness=10,color='red')
Maybe a series of implicit_plot3d with w=w0,w1,...
will be helpful?
On 5 Maj, 19:12, kcrisman kcris...@gmail.com wrote:
On May 5, 12:25 pm, ObsessiveMathsFreak
obsessivemathsfr...@gmail.com wrote:
I currently have a function of three variables w=f(x,y,z), which I
would like to plot in 3D
Artifical workaround:
sage: R.x,y=QQ[]
sage: w=3*x^2+4*x
sage: w.monomials()
[x^2, x]
On 29 Kwi, 00:28, tvn nguyenthanh...@gmail.com wrote:
Hi, the monimals function you suggested doesn't work for the below case
where I want to extract the terms [x^2,x] from the given expression
3*x^2+4*x
Hello nkulmati
If you want to get rid of virtualization
maybe you should try Linux version without instalation
burning Live-CD first
The present version fits into memory and after (rather long)
start process the CD can be removed
It works quite satisfactory (if you have =2GB RAM)
The native
http://ask.sagemath.org/question/335/multiple-3d-plots-in-one-panel-graphics_array-and
On 17 Kwi, 02:46, ObsessiveMathsFreak obsessivemathsfr...@gmail.com
wrote:
Currently in sage 3D I can graph 3D plots in the same output, but not
side by side.
p1=plot3d(lambda x, y: x^2 + y^2, (-2,2),
sage: numerical_integral(lambda t:imaginary(gamma(1-I*t)),-15,15)
(0.0, 5.3925521851144085e-15)
On 18 Kwi, 00:02, Ian Petrow ianpet...@gmail.com wrote:
Also, imaginary(gamma(1.+I*5)), say, works fine as long as it's not
inside a numerical integral
Ian
On Apr 17, 2:57 pm, Ian Petrow
One can discuss if in limits of f(x,n) as n--oo
x may depend on n or not but in the following version:
sage: assume(x-0.99,x0.99)
sage: n=var('n')
sage: sage: limit(x^(n+1)/(1-x), n=infinity)
-limit(x^(n + 1), n, +Infinity)/(x - 1)
sage: assume(x0)
sage: sage: limit(x^(n+1)/(1-x), n=infinity)
In W... alpha
Assuming[x-0.99,x0.99];Limit[x^(n+1)/(1-x),n-+Infinity]
remains unevaluated, so Maxima, Sage are nol alone
On 16 Kwi, 08:18, achrzesz achrz...@wp.pl wrote:
One can discuss if in limits of f(x,n) as n--oo
x may depend on n or not but in the following version:
sage: assume(x-0.99
sage: R.x,y,z=PolynomialRing(QQ,'x,y,z')
sage: w=3/5*x*y+5*y+3*z
sage: w.monomials()
[x*y, y, z]
sage: w1=3*x^2
sage: w1.monomials()
[x^2]
On 16 Kwi, 07:00, tvn nguyenthanh...@gmail.com wrote:
given an expression f of the form c1*t1 + c2*t2 + .. +cn*tn, I want to
extract from f the list of
I must correct myself
W... alpha:
Assuming[x-0.99,Assuming[x0.99,Limit[x^(n+1)/(1-x),n-+Infinity]]
gives correct answer 0
On 16 Kwi, 09:27, achrzesz achrz...@wp.pl wrote:
In W... alpha
Assuming[x-0.99,x0.99];Limit[x^(n+1)/(1-x),n-+Infinity]
remains unevaluated, so Maxima, Sage are nol alone
In the last post one can replace 0.99 by 1
but I wanted to exclude the following situation:
sage: n=var('n')
sage: x=var('x')
sage: x=(1/2)^(1/(n+1))
sage: limit(x^(n+1)/(1-x),n=+oo)
+Infinity # OK
On 16 Kwi, 15:36, achrzesz achrz...@wp.pl wrote:
I must correct myself
W... alpha:
Assuming[x
Without additional assumption that x is constant
the limit is not zero (take for example x=(1/2)^(1/(n+1))
(W... alpha:
Assuming[x=const,x1,x0];Limit[x^(n+1)/(1-x),n-+Infinity] 0 OK,
Assuming[x1,x0];Limit[x^(n+1)/(1-x),n-+Infinity] unevaluated OK)
On 15 Kwi, 06:00, Dan Drake dr...@kaist.edu
Or:
sage: import sage.libs.mpmath.all as mpmath
sage: V=mpmath.call(mpmath.legenp,2.1,0,-2);V
5.83105230126368 + 1.89579005740338*I
sage: type(V)
type 'sage.rings.complex_number.ComplexNumber'
On 13 Kwi, 23:45, Fredrik Johansson fredrik.johans...@gmail.com
wrote:
On Apr 13, 7:48 pm,
In 64bit 4.6.2 fedora13 Dell Vostro 1720
sage: numerical_integral(lambda x: cos(2*x)*cos(x), 0, pi)
(4.4478052108155282e-17, 1.3516940761795953e-14)
sage: plot(cos(2*x)*cos(x), (x, 0, pi))
Maxima and Wolfram alpha: 0
sage: integral(cos(2*x)*cos(x), x, 0, pi)
4/3
On 15 Kwi, 05:27, Dan Drake
To be more precise:
Maxima 5.23.2 gives 0
Sage 4.6.2 has 5.22.1 (and gives 4/3)
On 15 Kwi, 06:00, achrzesz achrz...@wp.pl wrote:
In 64bit 4.6.2 fedora13 Dell Vostro 1720
sage: numerical_integral(lambda x: cos(2*x)*cos(x), 0, pi)
(4.4478052108155282e-17, 1.3516940761795953e-14)
sage: plot(cos
Is your function usable?
def mplegp(n,m,x):
V=mpmath.legenp(n,m,x)
return float(V.real)+I*float(V.imag)
sage: time plot(lambda x:mplegp(2.1,0,x).real(),(x,-1,1))
CPU times: user 0.87 s, sys: 0.00 s, total: 0.87 s
Wall time: 0.87 s
On 13 Kwi, 19:48, ObsessiveMathsFreak
Maxima functions are OK
but mpmath is not as slow as one can think
sage: time list_plot([(x,mpmath.legenq(2,0,x)) for x in
srange(-0.99,1.01,0.05)],plotjoined=True)
+list_plot([(x,mpmath.legenp(2,0,x)) for x in
srange(-0.99,1.01,0.1)],plotjoined=True)
CPU times: user 0.20 s, sys: 0.00 s, total:
1 - 100 of 154 matches
Mail list logo
