Re: [sage-support] sagetex and stereographic plots
On Fri, 12 Mar 2010 at 01:27PM +0100, G. Damm wrote: > is it possible to make stereographic 3d-plots with sagetex? > I'd want to make a beamer-presentation with these plots and want to > help my students see the 3d. Can Sage make stereographic 3-d plots? If there's a way to get a PNG image of such a plot out of Sage, then you can get it into your presentation. What exactly do you mean by "stereographic"? Do you mean the kinds of image in which there's two images, from slightly different angles, and you project it using polarizing filters and the students wear polarized glasses? Or those "magic eye" images that look like static unless you cross your eyes just right? My guess, though, is that if you are already using a computer and a projector, it would be better to use an actual Jmol applet in your class. Then, instead of a three-dimensional static image, your students see something interactive, that can be twisted around, zoomed in and out, and so on. Dan -- --- Dan Drake - http://mathsci.kaist.ac.kr/~drake --- signature.asc Description: Digital signature
[sage-support] test if an expression is polynomial
Dear support the following two commands can be used to test, if a function is a polynomial or quotient of two polynomials in x variable. sage: (x^3+x+3+1/(x))._maxima_().polynomialp([x]).sage() 0 sage: ((x^3+x+3+1)/(x^4+3))._maxima_().polynomialp([x],"constantp", "integerp").sage() 1 Is there a solution how to do this in pure Sage, without calling Maxima? (I think that each call of Maxima takes a long time on my server, especially the first call.) Many thanks. Robert -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] Re: convert trigonometric/hyperbolic functions to exponentials
And I guess the answer to Paul's question is then: sage: (sinh(log(t)))._maxima_().exponentialize().sage() 1/2*t - 1/2/t sage: (cos(log(t)))._maxima_().exponentialize().sage() 1/2*e^(-I*log(t)) + 1/2*e^(I*log(t)) -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] Re: convert trigonometric/hyperbolic functions to exponentials
On 12 bře, 21:19, Guillaume wrote:
> > No, AFAIK, nothing other than explicit substitution with .subs().
>
> Hello,
>
> there are a few weird results. I'd like to solve this homogenous edo :
>
> $tx'=x+\sqrt{x^2+y^2}$.
>
> using x=tu
>
> sage: t=var('t')
> sage: x(t) = function('x',t)
> sage: id(t)=t
> sage: u=function('u',t)
> sage: d=diff(u*id,t)
>
Is this what you want?
sage: t=var('t')
sage: x= function('x',t)
sage: id(t)=t
sage: u=function('u',t)
sage: d=diff(u*id,t)
sage: assume(t>0)
sage: DE=(t*d==x+sqrt(t**2+x**2)).subs_expr(x==u*id)
sage: A=desolve(DE,u)
sage: C=var('C')
sage: A._maxima_().ev(logarc=true).sage().solve(u)[0].subs(c=log(C))
u(t) == C*t - sqrt(u(t)^2 + 1)
sage: eq = u == C*t - sqrt(u^2 + 1)
sage: ((eq-C*t)^2).solve(u)
[u(t) == 1/2*(C^2*t^2 - 1)/(C*t)]
The fact that the integral in A in not evaluated is probably a bug.
You may want to open trac on this and test, if this bug is inside
Maxima or in Sage interface to Maxima.
Robert
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[sage-support] Re: convert trigonometric/hyperbolic functions to exponentials
On 12 bře, 16:48, Burcin Erocal wrote: > On Fri, 12 Mar 2010 15:23:43 +0100 > > Paul Zimmermann wrote: > > is there a way in Sage to convert expressions involving trigonometric > > or hyperbolic functions to exponentials, like the convert/exp > > function of Maple? > > > > convert(sinh(log(t)),exp); > > 1 > > t/2 - --- > > 2 t > > > > convert(cos(log(t)),exp); > > 1/2 exp(ln(t) I) + 1/2 exp(-I ln(t)) > > No, AFAIK, nothing other than explicit substitution with .subs(). > You can use something like this desolve(t*diff(x,t)==x +sqrt(x^2+t^2),x)._maxima_().ev(logarc=true).sage() t == (sqrt(x(t)^2/t^2 + 1) + x(t)/t)*c desolve(t*diff(x,t)==x+sqrt(x^2+t^2),x) t == c*e^(arcsinh(x(t)/t)) Robert -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] Re: convert trigonometric/hyperbolic functions to exponentials
> No, AFAIK, nothing other than explicit substitution with .subs().
Hello,
there are a few weird results. I'd like to solve this homogenous edo :
$tx'=x+\sqrt{x^2+y^2}$.
using x=tu
sage: t=var('t')
sage: x(t) = function('x',t)
sage: id(t)=t
sage: u=function('u',t)
sage: d=diff(u*id,t)
apparently, you can't substitute inside diff
sage: DE=(t*d==x+sqrt(t**2+x**2)).subs_expr(x==u*id)
sage: desolve(DE,[u,t])
arcsinh(u(t)) == c + integrate(abs(t)/t^2, t)
sage: assume(t>0)
sage: desolve(DE,[u,t])
arcsinh(u(t)) == c + integrate(1/t, t)
sage: desolve(DE,[u,t],[1,0])
oups, first problem :
Division by 0
#0: ic1(soln=asinh(u) = 'integrate(abs(t)/t^2,t)+%c,xc=t = 0,yc=u = 1)
(ode2.mac line 297)
-- an error. To debug this try: debugmode(true);
sage: sol=desolve(DE,[u,t]).simplify_exp()
arcsinh(u(t)) == c + log(t)
OK
sage: desolve(DE,[u,t],[0,1]).simplify_exp()
but the problem reamins the same. Welle, let's make the substitution
by hand :
sage: solp=sol.subs(c=0)
sage: solp
arcsinh(u(t)) == log(t)
sage: Sol=solve(solp,u)[0]
sage: Sol
u(t) == sinh(log(t))
sage: x(t)=t*Sol.rhs()
sage: x(t)
t*sinh(log(t))
here comes our hyp2exp problem...
sage: sh(a)=(exp(a)-exp(-a))/2
sage: x(t).substitute_function(sinh,sh).simplify_full()
t*sinh(log(t))
argh
sage: x
t |--> t*sinh(log(t))
x is what we expect. Is it t ?
sage: x(a).substitute_function(sinh,sh).simplify_full()
a*sinh(log(a))
no...and now, weirder and weirder...
sage: a*sinh(log(a)).substitute_function(sinh,sh).simplify_full()
1/2*a*log(a) - 1/2
sage: t*sinh(log(t)).substitute_function(sinh,sh).simplify_full()
1/2*t^2 - 1/2
this 1/2*a*log(a) - 1/2 is a bit unexpected...
well, I'm discovering sage but these results look strange.
cheers,
Guillaume
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[sage-support] Re: possible error in manual
On Mar 12, 9:18 am, Michael Beeson wrote: > On the pagehttp://www.sagemath.org/doc/reference/sage/symbolic/expression.html > > the first example has eqn.subs(x==5) and I think it should be > eqn.subs(x=5) It seems to work either way, and there are examples in the documentation using both: type sage: eqn = (x-1)^2 <= x^2 - 2*x + 3 sage: eqn.subs? to see the docs. -- John -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] Re: norm or absolute value
Hi,
Is this helping?
sage: var('a,b,z')
(a, b, z)
sage: f=a*z+i*b*z^2
sage: f.norm()
b*z^2*conjugate(b)*conjugate(z)^2 - I*a*z*conjugate(b)*conjugate(z)^2
+ I*b*z^2*conjugate(a)*conjugate(z) + a*z*conjugate(a)*conjugate(z)
sage: f.norm().full_simplify()
b^2*z^4 + a^2*z^2
sage: f.norm().factor()
(b^2*z^2 + a^2)*z^2
On Mar 12, 6:46 pm, Michael Beeson wrote:
> sage: var('z'); var('a');var('b');
> sage: F = a*z + i*b*z^2
> sage:
>
> Now F.norm() or something should give me (a*z)^2 + (b*z^2)^2, but I
> can't find a command to do that.
> I want to do this when F is a polynomial of degree 6 with complex
> rational coefficients to eliminate i and produce
> a polynomial of degree 12 with rational coefficients.
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[sage-support] norm or absolute value
sage: var('z'); var('a');var('b');
sage: F = a*z + i*b*z^2
sage:
Now F.norm() or something should give me (a*z)^2 + (b*z^2)^2, but I
can't find a command to do that.
I want to do this when F is a polynomial of degree 6 with complex
rational coefficients to eliminate i and produce
a polynomial of degree 12 with rational coefficients.
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[sage-support] Re: plot3d and expression evaluation
I think I've got a vague idea what happens here... The same thing also happens when I define the max() function myself: sage: def my_max(x,y): sage: if(x>y): return x sage: else: return y sage: fermi2(x,y,d,L) = 1 - 1/( exp( ( my_max(abs(x),abs(y))-L) /d) + 1) sage: fermi2 (x, y, d, L) |--> -1/(e^(-(L - abs(y))/d) + 1) + 1 This looks a bit like max(x,y) gets passed the the objects 'abs(x)' and 'abs(y)' and decides, for whatever reason, that 'abs(x)' is not greater than 'abs(y)', so the else-clause gets executed and we are left with abs(y). Does this make sense to you? (I'm just thinking aloud, I got no idea how sage actually does it) Most likely the internal "max" function is defined in rather the same way (it just returns abs(x) instead of abs(y). ) The bottom line seems to be that it is a very dangerous game to mix symbolic expressions and python functions. Is there some way to control when (and how) Sage evaluates expressions and invokes python functions? Something like saying "first evaluate abs(x) and abs(y) and then invoke the python function max() with the result of the evaluation"? On Mar 12, 5:49 pm, Harald Schilly wrote: > Ah, I got it, the max function is evaluated during the definition of > the function, look: > > sage: fermi(x,y,d,L) = 1 - 1/( exp( ( max(abs(x),abs(y))-L) /d) + 1) > sage: fermi > (x, y, d, L) |--> -1/(e^(-(L - abs(x))/d) + 1) + 1 > > especially: > sage: max(abs(x),abs(y)) > abs(x) > > There is no y any more! > > H -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
Re: [sage-support] newbie to cython: wrap int func( int n, float x[] )
On Mar 12, 2010, at 9:05 AM, gerhard wrote: Trying to wrap an existing library. I managed to at least get started with a .spyx file as follows: cdef extern from "stdlib.h": void *malloc(size_t size) int free(void*) int sizeof() cdef extern from "func.h": int func( int n, float* x ) cdef double* d def __cinit__(self): self.d = malloc( sizeof(double)) self.n = 1 def __dealloc__(self): free( self.d ) def func( self, data ): cdef int i self.n = len(data) free( self.d ) self.d = malloc( self.n*sizeof(double) ) for i from 0 <= i < self.n: self.d[i] = data[i] func( self.n, self.d ) l = [] for i from 0 <= i < self.n: l.append( d[i] ) return l - but this leaves much to be desired. It should be a fairly standard operation for which sage has good solutions. Could someone please give me some pointers as to how to improve the above? Sounds like you want NumPy. - Robert -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] possible error in manual
On the page http://www.sagemath.org/doc/reference/sage/symbolic/expression.html the first example has eqn.subs(x==5) and I think it should be eqn.subs(x=5) -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] Re: plot3d and expression evaluation
On Mar 12, 5:47 pm, Harald Schilly wrote: > I'm not sure but there might be a bug or problem evaluating the > expression. Anyways, going the "pure" python way works: > > sage: def fermi(x,y,d,L): return 1 - 1/( exp( ( max(abs(x),abs(y))- > L) /d) + 1) > > sage: plot3d(lambda x,y : fermi(x,y,0.2,1), (x,-2,2), (y,-2,2)) > > Does this look fine? Yes, the "pure python" way works. But defining the function that way, I loose the ability to perform symbolic manipulations with it, don't I? -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] newbie to cython: wrap int func( int n, float x[] )
Trying to wrap an existing library. I managed to at least get started with a .spyx file as follows: cdef extern from "stdlib.h": void *malloc(size_t size) int free(void*) int sizeof() cdef extern from "func.h": int func( int n, float* x ) cdef double* d def __cinit__(self): self.d = malloc( sizeof(double)) self.n = 1 def __dealloc__(self): free( self.d ) def func( self, data ): cdef int i self.n = len(data) free( self.d ) self.d = malloc( self.n*sizeof(double) ) for i from 0 <= i < self.n: self.d[i] = data[i] func( self.n, self.d ) l = [] for i from 0 <= i < self.n: l.append( d[i] ) return l - but this leaves much to be desired. It should be a fairly standard operation for which sage has good solutions. Could someone please give me some pointers as to how to improve the above? -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] Re: plot3d and expression evaluation
On Mar 12, 5:39 pm, stefan wrote: > I have encountered a somewhat strange problem ... Ah, I got it, the max function is evaluated during the definition of the function, look: sage: fermi(x,y,d,L) = 1 - 1/( exp( ( max(abs(x),abs(y))-L) /d) + 1) sage: fermi (x, y, d, L) |--> -1/(e^(-(L - abs(x))/d) + 1) + 1 especially: sage: max(abs(x),abs(y)) abs(x) There is no y any more! H -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] Re: plot3d and expression evaluation
On Mar 12, 5:39 pm, stefan wrote: > Basically, I am trying to plot this expression: > > #sage> fermi(x,y,d,L) = 1 - 1/( exp( ( max(abs(x),abs(y))-L) /d) + 1) I'm not sure but there might be a bug or problem evaluating the expression. Anyways, going the "pure" python way works: sage: def fermi(x,y,d,L): return 1 - 1/( exp( ( max(abs(x),abs(y))- L) /d) + 1) sage: plot3d(lambda x,y : fermi(x,y,0.2,1), (x,-2,2), (y,-2,2)) Does this look fine? H -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] plot3d and expression evaluation
Hello group, I have encountered a somewhat strange problem in plotting a simple function. It seems to be related to the issues described in the tutorial section "Some Common Issues with Functions", but the lambda- function trick does not work here. I have uploaded a worksheet with what I have been trying so far here: http://www.sagenb.org/home/steja/1/ Basically, I am trying to plot this expression: #sage> fermi(x,y,d,L) = 1 - 1/( exp( ( max(abs(x),abs(y))-L) /d) + 1) as a function of (x,y) for some different values of d and L (nothing too useful, I invented it as a toy project to get familiar with Sage). I tried the following commands, but none of them works: #sage> plot3d(lambda x,y: fermi(x,y,0.2,1),(x,-2,2),(y,-2,2)) #sage> def plfermi(x,y): return fermi(x,y,0.2,1) #sage> plot3d(plfermi,(-2,2),(-2,2)) Both of these commands do plot something, but the result is wrong. I also tried this: #sage> pf(x,y) = fermi(x,y,0.2,1.0) #sage> plot(pf,(-2,2),(-2,2)) Here, the result is an error message (which looks like Klingon language to me). Could somebody send some light down my way? I am almost dying to get rid of Mathematica in favor of an open source CAS (because of their #*! § license that always seems to fail when one desperately needs it), but I'm afraid I've still got a long way to go with Sage when I'm not even able to plot a simple function... -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] sagetex and stereographic plots
Hello all, is it possible to make stereographic 3d-plots with sagetex? I'd want to make a beamer-presentation with these plots and want to help my students see the 3d. Thanks, Georg signature.asc Description: This is a digitally signed message part.
Re: [sage-support] convert trigonometric/hyperbolic functions to exponentials
On Fri, 12 Mar 2010 15:23:43 +0100 Paul Zimmermann wrote: > is there a way in Sage to convert expressions involving trigonometric > or hyperbolic functions to exponentials, like the convert/exp > function of Maple? > > > convert(sinh(log(t)),exp); > 1 >t/2 - --- > 2 t > > > convert(cos(log(t)),exp); > 1/2 exp(ln(t) I) + 1/2 exp(-I ln(t)) No, AFAIK, nothing other than explicit substitution with .subs(). Implementing these shouldn't be too hard given the rewriting capabilities of pynac. The problem is to provide an intuitive user interface. Francois Maltey had some good suggestions for this. Putting these up on the wiki has been on my todo list for quite some time, but I never managed to find the time. I'd be glad to forward the relevant emails to anyone interested in taking over this project. Cheers, Burcin -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] convert trigonometric/hyperbolic functions to exponentials
Hi, is there a way in Sage to convert expressions involving trigonometric or hyperbolic functions to exponentials, like the convert/exp function of Maple? > convert(sinh(log(t)),exp); 1 t/2 - --- 2 t > convert(cos(log(t)),exp); 1/2 exp(ln(t) I) + 1/2 exp(-I ln(t)) Paul Zimmermann -- 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 http://groups.google.com/group/sage-support URL: http://www.sagemath.org
Re: 回复: [sage-support] Re: 3D plot in sage
Actually I do need them to be plotted by color.it seems that sage is not
designed for this way.could anyone give some information aboutwhat open source
software can do this?Thanks a lot!YC - 原文 - 发件人: Jason Grout 主 题: Re:
回复: [sage-support] Re: 3D plot in sage时 间: 2010年3月12日 01:26:30On 03/12/2010
03:51 AM, Marshall Hampton wrote:> One option is several implicit plots,
colored by value, i.e. something> like:> > var('x,y,z')>
f=cos(x)*cos(y)+cos(y)*cos(z)+cos(z)*cos(x)> > imps = []> plot_range =
srange(0,3,.75)> color_range = [i/N(len(plot_range)) for i in
range(len(plot_range))]> > for i,q in enumerate(plot_range):> red
= color_range[i]> imps.append(implicit_plot3d(f-q==0, (x, -2, 2), (y,
-2, 2), (z,> -2, 2), opacity = .5, rgbcolor = (red,0,1-red)))> >
show(sum(imps))> If you didn't need them to be plotted by color, then you
could use thecontour option, like in the doc examples:sage:
implicit_plot3d((x^2 + y^2 + z^2), (x, -2, 2), (y, -2, 2), (z, -2,2),
plot_points=60, contour=[1,3,5])Jason-- To post to this group, send email to
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Re: 回复: [sage-support] Re: 3D plot in sage
On 03/12/2010 03:51 AM, Marshall Hampton wrote:
> One option is several implicit plots, colored by value, i.e. something
> like:
>
> var('x,y,z')
> f=cos(x)*cos(y)+cos(y)*cos(z)+cos(z)*cos(x)
>
> imps = []
> plot_range = srange(0,3,.75)
> color_range = [i/N(len(plot_range)) for i in range(len(plot_range))]
>
> for i,q in enumerate(plot_range):
> red = color_range[i]
> imps.append(implicit_plot3d(f-q==0, (x, -2, 2), (y, -2, 2), (z,
> -2, 2), opacity = .5, rgbcolor = (red,0,1-red)))
>
> show(sum(imps))
>
If you didn't need them to be plotted by color, then you could use the
contour option, like in the doc examples:
sage: implicit_plot3d((x^2 + y^2 + z^2), (x, -2, 2), (y, -2, 2), (z, -2,
2), plot_points=60, contour=[1,3,5])
Jason
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Re: 回复: [sage-support] Re: 3D plot in sage
One option is several implicit plots, colored by value, i.e. something
like:
var('x,y,z')
f=cos(x)*cos(y)+cos(y)*cos(z)+cos(z)*cos(x)
imps = []
plot_range = srange(0,3,.75)
color_range = [i/N(len(plot_range)) for i in range(len(plot_range))]
for i,q in enumerate(plot_range):
red = color_range[i]
imps.append(implicit_plot3d(f-q==0, (x, -2, 2), (y, -2, 2), (z,
-2, 2), opacity = .5, rgbcolor = (red,0,1-red)))
show(sum(imps))
-M. Hampton
On Mar 11, 2:21 am, wxu...@sohu.com wrote:
> I don't want to plot f(x,y,z)=0.Yes, it should a 4d plot as you said.can that
> be done in sage?Thanks!YC - 原文 - 发件人: John H Palmieri 主 题:
> [sage-support] Re: 3D plot in sage时 间: 2010年3月10日 07:04:29On Mar 10,
> 9:39 am, wxu...@sohu.com wrote:> Hi everyone,I want do this thing as
> follows in sage, is that
> OK?var('x,y,z')f=cos(x)*cos(y)+cos(y)*cos(z)+cos(z)*cos(x)and then I want to
> plot f. how can do it in sage?Thanks in advance!regards,YCIsn't this a 4d
> plot? Or do you want to plot f(x,y,z)=0? In thatcase, you could dosage:
> implicit_plot3d(f == 0, (x, -2, 2), (y, -2, 2), (z, -2, 2))--John-- To post
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