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2015-12-02 10:11:52 +0100 | commented answer | Latex name of generators of number fields extending other number field This is now trac ticket 19657 |
2015-12-01 17:26:31 +0100 | answered a question | Latex name of generators of number fields extending other number field The problem is caused in the first place by a mistake in Unfortunately, you cannot get round this by using The inconsistencies here indicate that there are other things that need correcting in this part of the code. |
2015-05-21 09:31:44 +0100 | answered a question | get the coefficients from the Polynomial or |
2015-05-20 11:20:52 +0100 | commented question | Elements in the lattice $A_n$ Perhaps |
2015-05-16 11:28:22 +0100 | answered a question | TypeError for hom between multivariate Laurent polynomial rings? It seems that the root of the problem is : |
2015-05-15 11:54:46 +0100 | commented question | TypeError for hom between multivariate Laurent polynomial rings? I note that |
2015-05-10 07:45:28 +0100 | commented answer | How to build a matrix thought of as an array of smaller matrices? This is mostly fairly basic python syntax: (1) d is not an empty list; it is a dictionary. (2) m is a list of four lists of four 2 by 2 zero matrices. It is not a matrix, but (after it is modified) it is used to create a (block) matrix. The command could have been written as (3) |
2015-05-09 14:14:59 +0100 | answered a question | How to build a matrix thought of as an array of smaller matrices? The I first set up a dictionary containing data of the kind you discussed (I'm assuming that the final entry in your data set should be $(1,3,D)$ if you want a $4k$ by $4k$ matrix ) Then I defined a 4 by 4 array of zero matrices and put the data matrices in the appropriate positions Now All the usual matrix methods are available, e.g., |
2015-05-02 18:47:09 +0100 | answered a question | Factor base of class group computation The following does involve calling PARI's bnfinit. But for a number field K, K.class_group() calls bnfinit, the output of which is cached. The factor base used can be recovered as follows: |
2015-02-01 18:57:36 +0100 | answered a question | Units in number fields. The problem is that your |
2014-12-17 14:55:42 +0100 | commented answer | Computations in a Quotient Ring But |
2014-12-03 09:35:33 +0100 | commented question | Bug in roots()? Seems to be a bug in the Singular interface. For |
2014-11-19 11:26:43 +0100 | answered a question | Possible coefficients for a given discriminant You can do: and to get the coefficients: |
2014-11-02 10:47:51 +0100 | commented answer | How to compute this exponential generating function? Probably worth noting that print oeis('Narayana')[0].programs()[0] gives a very simple formula for the coefficients. |
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2014-10-27 11:11:09 +0100 | answered a question | what matrix.plot meanings? The lighter shades correspond to the larger entries of the matrix. In your example the black square are in positions (0, 1) and (2, 1), where the entries are -1, and the white square is in position (0, 0), where the entry is 3, which is the largest entry in the matrix. Typing will give more information. |
2014-10-25 18:21:28 +0100 | commented answer | Homomorphisms for relative number fields The problem should be solved by trac #10843. |
2014-09-18 17:49:41 +0100 | commented answer | Homomorphisms for relative number fields Now I see. Your F_pol was a symbolic expression, but mine was a polynomial. Unfortunately the root methods for each are not compatible. |
2014-09-18 13:52:59 +0100 | answered a question | Homomorphisms for relative number fields First you need to write Then the following works Typing explains the syntax. There should be an easier way to do this. On the second questionThis is a bug, the check parameter does not get passed on as it should be. I will raise a ticket to deal with this, but in the meantime you can get the result you want as follows. This is taken, with some slight modifications, from the code for |
2014-08-16 17:43:59 +0100 | commented question | Plane curve genus: new minimal polynomial No, I did look briefly at the Singular documentation, but couldn't find anything helpful. Yes, I also found that the minimal polynomial varies. I tried calling genus(I) 32 times, and got 17 distinct polynomials. However, there were only 5 distinct leading coefficients. |
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2014-08-15 09:57:38 +0100 | commented question | Plane curve genus: new minimal polynomial The calculation seems to be performed by Singular. Unwrapping the code, what C.genus() does is: sage: I = C.defining_ideal() sage: genus = sage.libs.singular.ff.normal__lib.genus sage: genus(I) // new minimal polynomial: 2097152a6+2359296a5-239616a4-518400a3+37800a2+13203a+54 11 |
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2014-06-20 03:46:53 +0100 | answered a question | polynomial list, array But maybe you meant: or |
2014-04-12 06:15:27 +0100 | commented question | QQ.extension() with embedding: incorrect modulus This is definitely a defect in Sage. If z is an element in a number field K, then z.abs(i) returns the absolute value of the the i-th embedding of K into CC , and if i is omitted, the zero-th embedding is used. Thus any specified embedding used to construct K is ignored. I have raises a ticket on this issue; see http://trac.sagemath.org/ticket/16147. |
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2013-12-11 04:46:46 +0100 | answered a question | interger-ring() and maximal_order() is the same in numberfield I think you must mean |
2013-11-30 06:23:09 +0100 | answered a question | Yet another linear combination This is just standard linear algebra: form a 21 by 20 matrix whose rows consist of the linearly dependent vectors and the 20 linearly independent vectors, and find the one-dimensional left kernel (nullspace); alternatively transpose and find the right kernel. The coefficients of the single element in a basis of the kernel provide a linear combination of all 21 vectors which is zero. Hence the dependent vector can be written as a linear combination of the 20 independent vectors. A small scale version in Sage: Confirming that the vectors are linearly independent, otherwise repeat. Then form a random linear combination, and create the matrix: Now take the first (and only) generator of the kernel, and verify the result: |
2013-05-27 13:39:45 +0100 | commented answer | referencing polynomial variables It is worth mentioning that R = PolynomialRing(ZZ, n, 'x') and P = sum(c\*R.gen(i) for i, c in enumerate(mylist)) are perhaps a little less cryptic, and that R.inject_variables() allows one to write things like Q = 3\*x1\*x2^3 - x3^4 |
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2013-02-25 06:09:05 +0100 | answered a question | berlekamp massey Yes, there is a problem with the Sage implementation. But other implementations such as the one at bma.bozhu.me may give varying results. The point is that for your sequence $s_0,s_1,\dots,s_{35}$ there does not exist a unique sequence $c_1,c_2,\dots,c_k$ in $GF(2)$ such that $$s_i+\sum_{j=1}^{k}c_js_{i-j}=0,\qquad\text{for $i=k,\dots,35$.}$$ The result that Sage produces comes from the fact that $$s_i+s_{i-2}+s_{i-3}+s_{i-4}+s_{i-5}=0,\qquad\text{for $i=16,\dots,33$.}$$ But this fails (generally) if $i=5,\dots,15$ or $i=34, 35$. The degree 19 result indeed says that $$s_i+\sum_{j\in\lbrace1, 2, 5, 6, 9, 10, 11, 13, 16, 17, 19\rbrace}s_{i-j}=0,\qquad\text{for $i=19,\dots,35$.}$$ However, this is not unique (it cannot be because the input sequence only has length 36), for example we could add in $s_{i-2}+s_{i-4}+s_{i-5}+s_{i-6}+s_{i-7}$, which, as we have seen, is zero for $i=18,\dots,35$. Thus the degree 19 polynomial has no right to be called the minimal polynomial. Of course, Massey's 1969 paper "Shift-register synthesis and BCH decoding" is clear on this point; the algorithm he describes produces "one of the [linear feedback shift registers] of [minimum length] which generates $s_0,s_1,\dots,s_N$" [ยง III]. |
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2013-02-16 07:29:17 +0100 | answered a question | How does one find solutions to a polynomial over a finite field? Alternatively |
2013-02-06 06:22:52 +0100 | answered a question | polynomial evaluation For polynomial substitution there are several possibilities: To handle symmetric polynomials it may be best to work in a polynomial ring. For example: We can then check symmetry and do other sorts of substitution |
2013-01-16 02:43:40 +0100 | answered a question | extracting digits in p-adic expansion But note that here the coefficient of $3^9$, which is zero, yields [] as its list of coefficients. Similarly, for the coefficient of $3^4$, which is $2$, we get [2]. |
2012-12-30 18:05:47 +0100 | commented answer | Problem with infinite sum The remark that $|B_{2k}|\to\infty$ was intended only to justify divergence for $n=1$. The final sentence implies divergence for $n>1$. |
2012-12-30 12:10:19 +0100 | commented answer | Problem with infinite sum The series certainly doesn't converge for n=1, since abs(bernoulli(2*k)) tends to infinity as k tends to infinity. In fact I don't believe that the series converges for any integral value of n, for I think $sum_{k=1}^{\infty}B_{2k}x^{2k}$ has zero radius of convergence. This should follow from knowing that $sum_{k=1}^{\infty}B_{2k}x^{2k}/(2k)!$ has radius of convergence $2\pi$; the value is $x/(e^x-1)-1-x/2$. |