[sage-support] Re: using existing functions with vectors / lists / arrays
On 05/22/2011 07:52 AM, mankoff wrote: I've never yet figured out in Python when to use lists, arrays, or vectors. I have something like: y = x^2 I can evaluate it at one point: y(2) # 4 I'd like to evaluate it at a group of points: y( x ) # 2, 4, 9, 16, ... Can this just work like in IDL or MATLAB? How would one define x? What is the quickest syntax if it can't just be a variable? Perhaps: for xx in x: y(xx) Something simpler? Clearer? You can use something like that ys = map(y, xs) or ys = [y(x) for x in xs] where xs is a list of the x values. Kind regards -- 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] Re: using existing functions with vectors / lists / arrays
I have something like: y = x^2 I can evaluate it at one point: y(2) # 4 I'd like to evaluate it at a group of points: y( x ) # 2, 4, 9, 16, ... Can this just work like in IDL or MATLAB? How would one define x? What is the quickest syntax if it can't just be a variable? Perhaps: For purely numerical work a la Matlab, you may want to use numpy arrays (you will be using Python and not any Sage functionality): sage: import numpy sage: x = numpy.array([1, 2, 3, 4]) sage: def y(x): : return x*x : sage: y(x) array([ 1, 4, 9, 16]) -- 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] Easy operations on Cyclic groups...
Hello everybody ! I want to enumerate the k different translations of [0,1,2] in Z/kZ. I thought there should exist in Sage something like that (don't try it it does not work) : g = CyclicGroup(k) S = g([1,2,3]) for i in g: print S+i Is that possible ? It looks like the CyclicGroup is Sage natively contains permutations, and as I had no idea in which direction to look I wrote a custom script :-) Thanks for your help !!! Nathann -- 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] imposing commutation relation
Hi, I wish to simplify some calculation that appear in quantum mechanics. To begin we use non-commutative variables as - sage: R.a,b = FreeAlgebra(QQ, 2) sage: (a+b)^3 + a*(a+b) a^2 + a*b + a^3 + a^2*b + a*b*a + a*b^2 + b*a^2 + b*a*b + b^2*a + b^3 I want to impose the commutation relation [a,b]=1 and bring the expression to normal form (i.e. in all terms b appears before a, e.g. a*b gets replaced by b*a + 1). Is it possible to do this? If not then can I get the expression such that a*b^2 appears as a*b*b? If the later can be done then I can use some string manipulation hack and get my job done. Thanks in advance. Regards, Rajeev -- 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] Re: using existing functions with vectors / lists / arrays
On Sun, May 22, 2011 at 4:45 PM, Berkin Malkoc malk...@gmail.com wrote: I have something like: y = x^2 I can evaluate it at one point: y(2) # 4 I'd like to evaluate it at a group of points: y( x ) # 2, 4, 9, 16, ... Can this just work like in IDL or MATLAB? How would one define x? What is the quickest syntax if it can't just be a variable? Perhaps: For purely numerical work a la Matlab, you may want to use numpy arrays (you will be using Python and not any Sage functionality): sage: import numpy sage: x = numpy.array([1, 2, 3, 4]) sage: def y(x): : return x*x : sage: y(x) array([ 1, 4, 9, 16]) -- 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 if the function of the single variable is somewhat more complicated then one can use vectorize from numpy - sage: import numpy as np sage: x = np.array([1,2,3,4]) sage: x array([1, 2, 3, 4]) sage: def y(x): : if x 3: : return 2*x : else: : return x*x : sage: y_vec = np.vectorize(y) this fails - sage: y(x) --- ValueErrorTraceback (most recent call last) while this works - sage: y_vec(x) array([ 2, 4, 9, 16]) hope it helps. Rajeev -- 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: imposing commutation relation
Hi Rajeev, On 22 Mai, 20:25, Rajeev rajs2...@gmail.com wrote: I wish to simplify some calculation that appear in quantum mechanics. To begin we use non-commutative variables as - sage: R.a,b = FreeAlgebra(QQ, 2) sage: (a+b)^3 + a*(a+b) a^2 + a*b + a^3 + a^2*b + a*b*a + a*b^2 + b*a^2 + b*a*b + b^2*a + b^3 I want to impose the commutation relation [a,b]=1 and bring the expression to normal form (i.e. in all terms b appears before a, e.g. a*b gets replaced by b*a + 1). Is it possible to do this? Well, if you start with a free algebra generated by a and b, and you mod out the commutator of a and b, then you obtain the polynomial ring generated by a and b, up to isomorphism. Hence, you could create a polynomial ring and convert your free algebra elements into a polynomial like this: sage: R.a,b = FreeAlgebra(QQ, 2) sage: p = (a+b)^3 + a*(a+b); p a^2 + a*b + a^3 + a^2*b + a*b*a + a*b^2 + b*a^2 + b*a*b + b^2*a + b^3 sage: P = QQ['a','b'] sage: P(p) a^3 + 3*a^2*b + 3*a*b^2 + b^3 + a^2 + a*b You say that you want to appear b before a, but that should merely by a choice of order: sage: P2 = QQ['b','a'] sage: P2(p) b^3 + 3*b^2*a + 3*b*a^2 + a^3 + b*a + a^2 Both P1 and P2 have a degree lexicographic order. If you want pure lexicographic, it can be done as well: sage: P3 = PolynomialRing(QQ, ['b','a'], order='lex') sage: P3(p) b^3 + 3*b^2*a + 3*b*a^2 + b*a + a^3 + a^2 sage: P4 = PolynomialRing(QQ, ['a','b'], order='lex') sage: P4(p) a^3 + 3*a^2*b + a^2 + 3*a*b^2 + a*b + b^3 Cheers, Simon -- 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: imposing commutation relation
On Sunday, May 22, 2011 10:44:50 PM UTC-7, Simon King wrote: Hi Rajeev, On 22 Mai, 20:25, Rajeev rajs...@gmail.com wrote: I wish to simplify some calculation that appear in quantum mechanics. To begin we use non-commutative variables as - sage: R.a,b = FreeAlgebra(QQ, 2) sage: (a+b)^3 + a*(a+b) a^2 + a*b + a^3 + a^2*b + a*b*a + a*b^2 + b*a^2 + b*a*b + b^2*a + b^3 I want to impose the commutation relation [a,b]=1 and bring the expression to normal form (i.e. in all terms b appears before a, e.g. a*b gets replaced by b*a + 1). Is it possible to do this? Well, if you start with a free algebra generated by a and b, and you mod out the commutator of a and b, then you obtain the polynomial ring generated by a and b, up to isomorphism. The poster wants to impose the relation [a,b]=1, not [a,b]=0. -- 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
