bk322's profile - activity

Why sage would give me polynomial representation of GF(8), but not GF(7)?

sage: G = GF(8, 'x')
sage: G.list()
[0, x, x^2, x + 1, x^2 + x, x^2 + x + 1, x^2 + 1, 1]
sage: G = GF(7, 'x')
sage: G.list()
[0, 1, 2, 3, 4, 5, 6]

Maybe there's no such thing as polynomial represenation of GF(7)?

How do I show in sage that

 (1234)=(12)(13)(14)

I tried make use of:

 PermutationGroup([1,2]) * PermutationGroup([1,3])

but it doesn't do what is needed.

Here's elements of Symmetric group of 6th order: S3:

I want to get the same in Sage. So I do:

sage: G = SymmetricGroup(3)
sage: G.list()
[(), (2,3), (1,2), (1,2,3), (1,3,2), (1,3)]

Now I can't find (1,3,2) element in the book. As far as I understand:

P1 -> ()
P2 -> (1,2,3)
P3 -> (2,3)
P4 -> (1,2)
P5 -> (1,3)
P6 -> ??? also (1,2,3) ???

So my question is to set the correct map from sage to my book...

So I've compiled sage from sources. The sage folder is now 3Gb. I guess I've no need for some files... What can I delete? Is there any reason I should not delete these things?

Suppose I'd like to compute

prod(1/x^4, x, 1, oo)

How can this be done?

I found an old thread, but with no answers.

Take the natural logarithm of your product and you get a sum which can be evaluated:

$$\ln\left( \prod_{x=1}^k \frac{1}{x^4} \right) = \sum_{x=1}^k \ln\left(\frac{1}{x^4}\right)$$

... now take the limit as $k \to \infty$:

sage: sum(ln(1/x^4), x, 1, oo)
-Infinity
sage: e^sum(ln(1/x^4), x, 1, oo)
0

See [here](http://www.sagemath.org/doc/faq/faq-usage.html#i-have-type-issues-using-scipy-cvxopt-or-numpy-from-sage).

Am... Great, but why post it here?

The problem is that you are comparing lists, not Sage objects. achrzesz gives an answer which compares two Sage group elements, so they are equal. The lists you give are simple Python ordered lists, and as ordered things, certainly aren't the same. Why didn't you just compare P[3]*P[6] and P[4]?

By the way, they're not the same. But that's a different issue. They're using the notation of the second row of your notation in your original question, not cycle notation.

sage: P[3] * P[6]
(1,3)
sage: P[4]
(1,2)

But mine are also the same. [3, 2, 1] is same as [2, 1, 3], isn't it?

Here's elements of Symmetric group of 6th order: S3:

I want to check that P3*P6=P4.

G = SymmetricGroup(3)

BookNumbers = [1, 4, 2, 3, 6, 5]

P = [0]
for i in BookNumbers:
    P.append(sorted(G.list())[i-1])

print (P[3] * P[6]).list(), P[4].list()
print (P[3] * P[6]) == P[4]

it gives:

[3, 2, 1] [2, 1, 3]
False

so they are the same actually. But how do I make sage say True?.

The results of

sage: G = SymmetricGroup(3)
sage: G.list()
[(), (2,3), (1,2), (1,2,3), (1,3,2), (1,3)]

are given in cycle notation. So the book's equivalent of (1,3,2) is one where 1 becomes 3, 3 becomes 2, and 2 becomes 1, which is P6, if I'm reading correctly.

If you want to match the bottom three elements of your matrix, you can simply convert each to a list, or maybe a dict would make the mapping more explicit:

sage: for g in G:
....:     print g, g.list(), g.dict()
....:     
() [1, 2, 3] {1: 1, 2: 2, 3: 3}
(2,3) [1, 3, 2] {1: 1, 2: 3, 3: 2}
(1,2) [2, 1, 3] {1: 2, 2: 1, 3: 3}
(1,2,3) [2, 3, 1] {1: 2, 2: 3, 3: 1}
(1,3,2) [3, 1, 2] {1: 3, 2: 1, 3: 2}
(1,3) [3, 2, 1] {1: 3, 2: 2, 3: 1}

That's very cool. I did this: `for e in sorted(G): print '{0:>8s}{1:>10s}{2:>19s}'.format(e, e.list(), e.dict())` (I type more then, but it inserts >...)