2022-11-01 13:30:27 +0200 | received badge | ● Popular Question (source) |

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2022-02-26 22:14:14 +0200 | commented answer | recognize a rational function is a polynomial Thank you! |

2022-02-26 22:14:07 +0200 | marked best answer | recognize a rational function is a polynomial Is it possible to make sage recognize that a certain rational function is a Laurent polynomial, and treat it as such? A simple example: Here, Is this possible to achieve? |

2022-02-26 15:53:12 +0200 | commented answer | recognize a rational function is a polynomial the error it gives is: fraction must have unit denominator. |

2022-02-26 15:51:53 +0200 | commented answer | recognize a rational function is a polynomial Thanks, that's also what I thought, but why is it not working e.g. here? R.<q1,q2> = LaurentPolynomialRing(ZZ,2) |

2022-02-25 17:29:13 +0200 | edited question | recognize a rational function is a polynomial recognize a rational function is a polynomial Is it possible to make sage recognize that a certain rational function is |

2022-02-25 17:28:49 +0200 | asked a question | recognize a rational function is a polynomial recognize a rational function is a polynomial Is it possible to make sage recognize that a certain rational function as |

2022-02-25 16:09:08 +0200 | commented answer | factorization and multiplication in polynomial ring Thanks, although I thought symbolic computations would be slower here, right? |

2022-02-22 10:39:29 +0200 | commented answer | factorization and multiplication in polynomial ring Thanks a lot! it seems to work well. |

2022-02-22 10:39:14 +0200 | marked best answer | factorization and multiplication in polynomial ring I'm looking for an efficient (i.e. fast enough for $k=4$ and higher) way in Sage to build large rational functions (ratio of polynomials with integer coefficients), and then take residues at simple poles (which I know by construction), by multiplication and taking limit (via subs). I can do that in the symbolic ring, which I believe uses maxima or sympy, but at some point it becomes slow in factorization (say compared to mathematica), so I tried to work with polynomial rings, hoping that singular would be faster. However, at present I have issues: since by default it expands whatever factorized polynomial I define, then it has a hard time factorizing it back (see how long it takes even just to return $\chi$). Moreover, I cannot operate with factorization objects, e.g. I need to multiply them, but it seems this is not allowed.
This is my current attempt: Any help/suggestions? Thanks. |

2022-02-19 21:54:30 +0200 | commented answer | factorization and multiplication in polynomial ring Yes, this assumption is correct. Let me run a few more checks. Note that k=1 is still broken as I pointed out. |

2022-02-18 17:16:08 +0200 | commented answer | factorization and multiplication in polynomial ring Thanks for helping me. Two comments: it seems that R = PolynomialRing(QQ, is better here, e.g. k=1 does not complain. Wh |

2022-02-18 11:29:55 +0200 | commented answer | factorization and multiplication in polynomial ring Sorry, but this does not look right: cancellations between numerator and denominator of chi will happen along the way (i |

2022-02-18 11:22:44 +0200 | commented answer | factorization and multiplication in polynomial ring Simultaneous substitution is not allowed here: we always compute an iterated residue, for which order matters. |

2022-02-18 11:13:35 +0200 | commented question | factorization and multiplication in polynomial ring I've also reversed rho, to get correct ordering. |

2022-02-18 11:12:56 +0200 | edited question | factorization and multiplication in polynomial ring factorization and multiplication in polynomial ring I'm looking for an efficient (i.e. fast enough for $k=4$ and higher) |

2022-02-18 11:11:57 +0200 | commented question | factorization and multiplication in polynomial ring You're right, that's not correct. |

2022-02-17 21:37:29 +0200 | edited question | factorization and multiplication in polynomial ring factorization and multiplication in polynomial ring I'm looking for an efficient (i.e. fast enough for $k=4$ and higher) |

2022-02-17 21:36:29 +0200 | commented answer | factorization and multiplication in polynomial ring in case I did not make it clear: the order in which residues are taken is crucial |

2022-02-17 21:35:45 +0200 | commented answer | factorization and multiplication in polynomial ring Thanks, let me study your suggestions. Is there a reason why you changed the order from invlex? the actual ordering shou |

2022-02-17 21:27:35 +0200 | commented question | factorization and multiplication in polynomial ring You're right. I never use X[p] with p \geq k, I just didn't know how to code it correctly. Thanks. |

2022-02-17 15:09:54 +0200 | commented answer | fast factorization of ratios of polynomials I wrote a new question, trying to explain what I want and what goes wrong https://ask.sagemath.org/question/61151/factor |

2022-02-17 15:07:07 +0200 | asked a question | factorization and multiplication in polynomial ring factorization and multiplication in polynomial ring I'm looking for an efficient (i.e. fast enough for $k=4$ and higher) |

2022-02-11 15:19:36 +0200 | commented answer | fast factorization of ratios of polynomials the change is that what I called chi, is then used to compute a residue (see the last 3 lines), so one needs to be able |

2022-02-10 16:24:13 +0200 | commented answer | fast factorization of ratios of polynomials do you think using the dictionary could work for the above? |

2022-02-10 16:23:39 +0200 | commented answer | fast factorization of ratios of polynomials But see the edited question, as there will be cancellations. |

2022-02-10 14:25:51 +0200 | commented answer | fast factorization of ratios of polynomials I added a full explanation, sorry for being not complete. I'm not sure using a dictionary is the fastest way here? |

2022-02-10 14:24:55 +0200 | edited question | fast factorization of ratios of polynomials fast factorization of ratios of polynomials I consider polynomials of the form $P=\prod_i (1-1/y_i)$, where each $y_i$ i |

2022-02-10 14:24:10 +0200 | edited question | fast factorization of ratios of polynomials fast factorization of ratios of polynomials I consider polynomials of the form $P=\prod_i (1-1/y_i)$, where each $y_i$ i |

2022-02-10 14:20:16 +0200 | edited question | fast factorization of ratios of polynomials fast factorization of ratios of polynomials I consider polynomials of the form $P=\prod_i (1-1/y_i)$, where each $y_i$ i |

2022-02-10 14:14:52 +0200 | commented question | fast factorization of ratios of polynomials Sure, but let me be more precise. What I asked is only part of the story, and I'd like to understand if using Singular i |

2022-02-07 10:19:34 +0200 | commented question | fast factorization of ratios of polynomials @tmonteil I added a code example, does this clarify my question? |

2022-02-07 10:18:44 +0200 | edited question | fast factorization of ratios of polynomials |

2022-02-06 10:09:03 +0200 | received badge | ● Notable Question (source) |

2022-02-06 10:07:52 +0200 | commented question | fast factorization of ratios of polynomials Well, numerator and denominator are factored separately, but I want to cancel common factors when taking the ratio, name |

2022-02-06 01:55:36 +0200 | received badge | ● Popular Question (source) |

2022-02-05 19:04:30 +0200 | edited question | fast factorization of ratios of polynomials |

2022-02-05 19:02:55 +0200 | asked a question | fast factorization of ratios of polynomials |

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2021-02-14 10:05:48 +0200 | commented question | prove an identity for any integer Thanks, I did see that, although in my case it just prints the sums and does not simplify them to zero. |

2021-02-13 21:40:36 +0200 | asked a question | prove an identity for any integer Let $n$ be a positive integer and $m = (m_1, \ldots, m_n)$ an $n$-dimensional vector of real numbers. Let $g$ be a real number. I want to prove, for any $n$ and $m$, an equality of the form $$ \sum_{i=1}^n f_i (m,g) = 0 $$ where the function $f_i$ is a rational function of $m$ and $g$. Of course it's easy to check this by substituting finite values of $n$, but is there a way in Sage to prove it for any integer? |

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