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2020-07-31 10:22:10 +0200 | commented answer | lambda calculus Thank you for this link which I will examine carefully. But I was rather thinking of a possibility of handling lambda terms, of developing a beat-reduction, all these complicated things to experiment with by hand ... |

2020-07-30 17:39:02 +0200 | asked a question | lambda calculus Is there an implementation of lambda-calculus with sagemath (or python)? thanks in advance |

2020-01-27 16:04:20 +0200 | answered a question | power_mod strange... thank you so much. I will study this more closely |

2020-01-27 11:01:37 +0200 | asked a question | power_mod strange... where is the error? x = 15 z = power_mod(x,3,23) #z=15^3 mod 23 inverse = inverse_mod(3,23) #1/3 mod 23 assert mod(3*inverse,23)==1 #ok y = power_mod(z,inverse,23) # y = 15^(3/3) = 15 print(y) 18 Thanks a lot to everyone |

2020-01-01 16:16:21 +0200 | answered a question | How to get the canonical prime factorization, with zero exponents too something like this: [(7, 2), (83, 1), (101, 1)] [(7, 2), (83, 1), (101, 1), (11, 0), (13, 0), (17, 0), (19, 0), (23, 0), (29, 0), (31, 0), (37, 0), (41, 0), (43, 0), (47, 0), (53, 0), (59, 0), (61, 0), (67, 0), (71, 0), (73, 0), (79, 0), (89, 0), (97, 0)] happy new year |

2020-01-01 16:11:28 +0200 | received badge | ● Commentator |

2020-01-01 16:11:28 +0200 | commented question | How to get the canonical prime factorization, with zero exponents too something like this: facteurs = list((410767).factor()) print(facteurs) [(7, 2), (83, 1), (101, 1)] solution = facteurs deja =[p for p,e in facteurs] for premier in primes(facteurs[0][0]+1,stop=facteurs[-1][0]): print(solution) [(7, 2), (83, 1), (101, 1), (11, 0), (13, 0), (17, 0), (19, 0), (23, 0), (29, 0), (31, 0), (37, 0), (41, 0), (43, 0), (47, 0), (53, 0), (59, 0), (61, 0), (67, 0), (71, 0), (73, 0), (79, 0), (89, 0), (97, 0)] happy new year |

2020-01-01 10:20:30 +0200 | commented answer | Using piecewise-defined functions for recursive integer sequence define F(x)=10(x%n) so n(k)=F(F(F....(F(r//m)...) with k iterations. What do you mean by a piecewise function , when the function is defined for three variables k, r and m ? happy new year |

2020-01-01 09:48:08 +0200 | commented answer | Using piecewise-defined functions for recursive integer sequence it is not a sagemath problem but an interesting mathematical question, right? I write two or three equations, just to see if the answer is within pen's reach |

2019-12-31 14:45:22 +0200 | received badge | ● Teacher (source) |

2019-12-31 09:49:06 +0200 | answered a question | Using piecewise-defined functions for recursive integer sequence r = 10 m = 2 def fonction_n(k): def fonction_q(k): It is possible to derecursify this functions happy new year |

2019-12-30 14:58:43 +0200 | commented answer | not evaluating expression Great!! Happy new year to you |

2019-12-29 15:13:33 +0200 | asked a question | not evaluating expression var("x",domain=RR)
e =2 reply: 5x+5=5x+3 of course. But I would like to have 2x+3+5x+1=5x+3 Is there somethink like e.unevaluate()? big thanks and happy new year to the whole sagemath community |

2019-12-24 10:09:41 +0200 | commented answer | equality with integer exponents Indeed. But here I know the answer; in general, it will be necessary to show a lot of experience in the manipulation of symbolic expressions before concluding. |

2019-12-23 17:22:48 +0200 | asked a question | equality with integer exponents n = var("n",domain = ZZ) assume(n>0) bool(3^(2*n) == (3^2)^n ) reply: False Why? Big thangs |

2019-12-23 16:53:36 +0200 | commented answer | unicode coding You are absolutely right ... but it does not work better, while in Python 3 it works very well Big thangs |

2019-12-20 15:35:49 +0200 | commented answer | unicode coding not better unfortunately |

2019-12-20 10:49:42 +0200 | commented answer | unicode coding sorry but utf-8, utf_8, UTF-8 ou UTF_8 does not work on windows... (# -- coding= -- on the first line) something else to had? big thanks |

2019-12-19 10:35:23 +0200 | asked a question | unicode coding
does not work ? Big thanks |

2019-12-19 09:42:52 +0200 | commented answer | factorize symbolic expression ok I undersatnd.... |

2019-12-18 10:33:25 +0200 | commented answer | factorize symbolic expression Thanks for the replay. A little bit heavy for a simple factorization of an expression... Any particular reason for this restriction of the format method? Big thanks anyway |

2019-12-17 17:16:04 +0200 | received badge | ● Student (source) |

2019-12-16 10:22:28 +0200 | answered a question | factorize symbolic expression that's the problem... Thank to you |

2019-12-15 22:35:31 +0200 | asked a question | factorize symbolic expression (ax+bx).factor()=x(a+b) but (2a+2b).factor()=(2a+2b) how to obtain 2(a+b) ? Thanks in advance... |

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