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2017-02-11 20:31:11 -0600 | asked a question | solution to homogeneous system of linear equations with coefficients over field $\mathbb{F}_2$ Solve the following homogeneous systems of linear equations with coefficients over field $\mathbb{F}_2$: I want to output on the sage following: I want to solve a general homogeneous systems with coefficients over field $\mathbb{F}_2$ and output same as above. I hope that someone can help. Thanks! |

2017-02-11 20:08:47 -0600 | asked a question | Code error in sagemath My code in sagemath but it was an error as below |

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2017-02-11 06:40:43 -0600 | asked a question | Find all non-negative integer solutions of $a+b+c+d+e = 8$ in Sagemath Find all non-negative integer a, b, c,d, e such that $$a+b+c+d+e = 8$$ Is there any method for this? I have no idea. I can just fix the limit. |

2017-02-10 21:01:50 -0600 | commented answer | Construction of formula in Sagemath program I use the code above but it error. File "<ipython-input-1-b39e5d3b522e>", line 3 Multivariate Polynomial Ring in x1, x2, x3, x4 over Finite Field of size Integer(2) SyntaxError: invalid syntax |

2017-02-10 09:24:29 -0600 | asked a question | Construction of formula in Sagemath program Let $P_k:= \mathbb{F}_2[x_1,x_2,\ldots ,x_k]$ be the polynomial algebra in $k$ variables with the degree of each $x_i$ being $1,$ regarded as a module over the mod-$2$ Steenrod algebra $\mathcal{A}.$ Here $\mathcal{A} = \langle Sq^{2^m}\,\,|\,\,m\geq 0\rangle.$ Being the cohomology of a space, $P_k$ is a module over the mod-2 Steenrod algebra $\mathscr{A}.$ The action of $\mathscr{A}$ on $P_k$ is explicitly given by the formula $$Sq^m(x_j^d) = \binom{d}{m}x_j^{m+d},$$ where $ \binom{d}{m}$ is reduced mod-2 and $\binom{d}{m} = 0$ if $m > d.$ Now, I want to use the Steenrod algebra package and Multi Polynomial ring package and using formular above to construction of formula following in Sagemath program $$ Sq^m(f) = \sum\limits_{2^{m_1} + 2^{m_2} + \cdots + 2^{m_k}= m}\binom{d_1}{2^{m_1}}x_1^{2^{m_1}+d_1}\binom{d_1}{2^{m_2}}x_2^{2^{m_2}+d_2}\ldots \binom{d_k}{2^{m_k}}x_k^{2^{m_k}+d_k}.$$ forall $f = x_1^{d_1}x_2^{d_2}\ldots x_k^{d_k}\in P_k$ Example: Let $k = 5, m = 2$ and $f = x_1^2x_2^3x_3^2x_4x_5\in P_5.$ We have $$ Sq^2(x_1^2x_2^3x_3^2x_4x_5) = x_1^4x_2^3x_3^2x_4x_5 + x_1^2x_2^5x_3^2x_4x_5 + x_1^2x_2^3x_3^4x_4x_5 +x_1^2x_2^3x_3^2x_4^2x_5^2 + x_1^2x_2^4x_3^2x_4x_5^2 + x_1^2x_2^4x_3^2x_4^2x_5^1.$$ I hope that someone can help. Thanks! |

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