2024-07-18 15:07:13 +0200 | commented question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Thanks, I'm taking a look. |

2024-07-18 12:58:21 +0200 | edited question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) I ha |

2024-07-18 12:49:07 +0200 | edited question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Alternative to changing the __add__ of Integers class (need $0 + x = x$, $x$ belonging to custom class) I have a class c |

2024-07-18 11:24:36 +0200 | edited question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Alternative to changing the __add__ of Integers class (need $0 + x = x$, $x$ belonging to custom class) I have a class c |

2024-07-18 10:16:23 +0200 | edited question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring $R$, which moral |

2024-07-18 10:16:20 +0200 | edited question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring $R$, which moral |

2024-07-18 08:36:31 +0200 | edited question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring $R$, which moral |

2024-07-18 08:22:52 +0200 | edited question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) |

2024-07-18 08:22:03 +0200 | edited question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring which morally sh |

2024-07-18 08:19:39 +0200 | edited question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring which morally sh |

2024-07-18 08:19:20 +0200 | asked a question | Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring which morally sh |

2024-07-18 08:09:31 +0200 | answered a question | Tensor product of Clifford algebra with another (non-Clifford) The main question was already answered in the comments. Of course, both tensor factors should be a module of the same ri |

2024-07-18 08:07:06 +0200 | edited question | Tensor product of Clifford algebra with another (non-Clifford) Tensor product of Clifford algebra with another (non-Clifford) For a given quadratic form $Q$, I'd wish to symbolically |

2024-07-18 08:05:28 +0200 | commented question | Tensor product of Clifford algebra with another (non-Clifford) Thanks for the link |

2024-07-12 17:52:31 +0200 | commented question | Tensor product of Clifford algebra with another (non-Clifford) Many thanks. I should say that forming a triple product sometimes work, sometimes it doesn't (by running the same code) |

2024-07-12 12:36:56 +0200 | edited question | Tensor product of Clifford algebra with another (non-Clifford) Tensor product of Clifford algebra with another (non-Clifford) over a multivariable polynomial ring For a given quadrati |

2024-07-12 12:36:52 +0200 | edited question | Tensor product of Clifford algebra with another (non-Clifford) Tensor product of Clifford algebra with another (non-Clifford) over a multivariable polynomial ring For a given quadrati |

2024-07-12 10:36:23 +0200 | edited question | Tensor product of Clifford algebra with another (non-Clifford) Tensor product of Clifford algebra with another (non-Clifford) For a given quadratic form $Q$, I'd wish to symbolically |

2024-07-12 08:05:19 +0200 | commented question | Tensor product of Clifford algebra with another (non-Clifford) Many thanks, (moral: I should think in the math first for next time). I should say that forming a triple product sometim |

2024-07-12 08:04:48 +0200 | commented question | Tensor product of Clifford algebra with another (non-Clifford) Many thanks, (moral: I should think in the math first for next time). I should say that forming a triple product sometim |

2024-07-12 08:04:33 +0200 | commented question | Tensor product of Clifford algebra with another (non-Clifford) Many thanks, (moral: I should think in the math first for next time). I should say that forming a triple product sometim |

2024-07-12 08:00:05 +0200 | commented question | Comparing two symbolic expressions without sympy It surely does, thanks. |

2024-07-11 23:33:31 +0200 | asked a question | Comparing two symbolic expressions without sympy Comparing two symbolic expressions without sympy From diff(sin(x),x) == cos(x) one gets just trivial equality cos(x) |

2024-07-10 09:39:32 +0200 | commented question | Tensor product of Clifford algebra with another (non-Clifford) Many thanks, (moral: I should think in the math first for next time). |

2024-07-10 09:33:56 +0200 | edited answer | complex expand Hi, welcome. Don't know how far the next works, but while nobody answers, consider: var('x,y',domain=RR) def csimp(comp |

2024-07-10 08:39:44 +0200 | answered a question | complex expand Hi, welcome. Don't know how far the next goes, but if nobody answers consider: var('x,y',domain=RR) def csimp(complex_p |

2024-07-10 08:01:19 +0200 | edited question | Radical numerical expressions not handled by simplify_full ? Rational numerical expressions not handled by simplify_full ? To simplify rational sqrt(21)*sqrt(6) to, 3*sq |

2024-07-10 08:01:15 +0200 | edited question | Radical numerical expressions not handled by simplify_full ? Rational numerical expressions not handled by simplify_full ? To simplify rational sqrt(21)*sqrt(6) to, 3*sq |

2024-07-09 23:45:13 +0200 | edited answer | partial derivatives The next: x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) z = function('z')(x) (((f(y,q).diff(q)).s |

2024-07-09 23:44:53 +0200 | edited answer | partial derivatives The next: x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) z = function('z')(x) (((f(y,q).diff(q)).s |

2024-07-09 23:44:33 +0200 | edited answer | partial derivatives x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) z = function('z')(x) (((f(y,q).diff( |

2024-07-09 23:44:19 +0200 | edited answer | partial derivatives x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) z = function('z')(x) (((f(y,q).diff( |

2024-07-09 23:43:41 +0200 | answered a question | partial derivatives x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) fy1 = f.diff(z) z = function('z')(x) (((f(y,q).diff(q |

2024-07-09 21:58:30 +0200 | edited question | Tensor product of Clifford algebra with another (non-Clifford) Tensor product of Clifford algebra with another (non-Clifford) For a given quadratic form $Q$, I'd wish to symbolically |

2024-07-09 21:54:21 +0200 | edited question | Tensor product of Clifford algebra with another (non-Clifford) |

2024-07-09 21:53:19 +0200 | asked a question | Tensor product of Clifford algebra with another (non-Clifford) |

2024-06-21 17:43:08 +0200 | commented answer | Question on tensor products (of the free algebra) sure I forgot to specify $F\otimes F$ has non-homogeneous elements, was a little bit sloppy of me. Thanks |

2024-06-21 17:40:14 +0200 | edited question | Question on tensor products (of the free algebra) Question on tensor products (of the free algebra) If $F$ is the free algebra over the rational numbers with generators |

2024-06-21 17:36:42 +0200 | marked best answer | Question on tensor products (of the free algebra) If $F$ is the free algebra over the rational numbers with generators ${a,b,c,\ldots}$, then $F\otimes F$ is implemented Typical element are (sums of) A first issue I have is that -- despite the And more importantly, how read off each factor of this type of expressions? That is, how to map to, say, or (who gets the numerical factor I don't care) ? |

2024-06-21 14:59:15 +0200 | asked a question | Free algebra 'inside the trace' Free algebra 'inside the trace' In the free algebra of generators $a,b,c,\ldots$, is there a way to do additive simplifi |

2024-06-21 12:03:55 +0200 | edited question | Question on tensor products (of the free algebra) Question on tensor products (of the free algebra) If $F$ is the free algebra over the rational numbers with generators |

2024-06-21 12:01:50 +0200 | edited question | Question on tensor products (of the free algebra) Question on tensor products (of the free algebra) If $F$ is the free algebra over the rational numbers with generators |

2024-06-21 12:01:01 +0200 | edited question | Question on tensor products (of the free algebra) Question on tensor products (notation) If $F$ is the free algebra over the rational numbers with generators ${a,b,c,\ld |

2024-06-21 11:59:18 +0200 | edited question | Question on tensor products (of the free algebra) Question on tensor products (notation) If $F$ is the free algebra over the rational numbers with generators ${a,b,c,\ld |

2024-06-21 11:58:40 +0200 | edited question | Question on tensor products (of the free algebra) Question on tensor products (notation) If $F$ is the free algebra over the rational numbers with generators ${a,b,c,\ld |

2024-06-21 11:56:41 +0200 | asked a question | Question on tensor products (of the free algebra) |

2024-06-19 23:08:16 +0200 | marked best answer | Free algebra identity (empty word) and involution I didn't find the way to define an involution on the free algebra (say in two generators) yet. Specifically I want unitary elements. Thus, I added two variables, $y$ and $z$ which serve as $w^\star,x^\star$ and force a relation so that $w y=1$, the empty word, and similar relation $x z=1$ . But then typing yields an error (#This is dirty, coercion is broken). That is: Question: which is the free algebra empty word here? Of course, I tried to add a symbol $E$ which then satisfies $gE=g=Eg$ for all the generators $g$, But then yields also an error. Second question: does somebody know an easier way to to implement the involution here? |

2024-04-07 22:04:51 +0200 | received badge | ● Notable Question (source) |

2023-09-06 21:10:07 +0200 | commented question | Free algebra identity (empty word) and involution It doesn't but thanks anyway, good to know. |

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