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2024-02-20 18:14:55 +0100 edited question Ideals of $\mathbb Z[x]$

Ideals of $\mathbb Z[x]$ I think the following code should output True instead of False because we have $1 = (1-2x)+2x \

2024-02-20 18:14:03 +0100 edited question Ideals of $\mathbb Z[x]$

Ideals of $\mathbb Z[x]$ I think the following code should output True instead of False because we have $1 = (1-2x)+2x \

2024-02-20 18:13:04 +0100 edited question Ideals of $\mathbb Z[x]$

Ideals of $\mathbb Z[x]$ I think the following code should output True instead of False because we have $1 = (1-2x)+2x \

2024-02-20 18:11:24 +0100 asked a question Ideals of $\mathbb Z[x]$

Ideals of $\mathbb Z[x]$ I think the following code should output True instead of False: R.<x> = ZZ[] I = R.ideal

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2023-07-09 23:00:03 +0100 commented answer Power series with alternating exponent

Is there any way Sage can handle $\sum_{k=0}^\infty 4^{(-1)^k} x^k$ directly/automatically?

2023-07-09 22:59:25 +0100 edited question Power series with alternating exponent

Power series with alternating exponent Mathematically, we have $4^{(-1)^k} = \frac18 (17+15(-1)^k)$ for all integers $k

2023-07-09 22:58:47 +0100 commented answer Power series with alternating exponent

Is there any way Sage can handle $\sum_{k=0}^\infty 4^{(-1)^k} x^k$ directly?

2023-07-09 22:58:35 +0100 commented answer Power series with alternating exponent

Is there any way Sage can handle $\sum_{k=0}^\infty 4^{(-1)^k}*x^k$ directly?

2023-07-09 22:58:29 +0100 commented answer Power series with alternating exponent

Is there any way Sage can handle $\sum_{k=0}^\infty 4^{((-1)^k}*x^k$ directly?

2023-07-09 22:57:46 +0100 commented answer Power series with alternating exponent

Is there any way Sage can handle $\sum_(k=0)^\infty 4^((-1)^k)*x^k$ directly?

2023-07-09 22:57:37 +0100 commented answer Power series with alternating exponent

Is there any way Sage can handle $sum_(k=0)^\infty 4^((-1)^k)*x^k$ directly?

2023-07-09 16:46:16 +0100 edited question Power series with alternating exponent

Power series with alternating exponent Mathematically, we have $4^{(-1)^k} = \frac18 (17+15(-1)^k)$ for all integers $k

2023-07-09 16:44:53 +0100 asked a question Power series with alternating exponent

Power series with alternating exponent Mathematically, we have $4^{(-1)^k} = \frac18 (17+15(-1)^k)$ for all integers $k

2023-07-09 16:35:17 +0100 commented answer Compute power series

Thank you both!

2023-07-09 16:33:06 +0100 marked best answer Compute power series

I would like to compute $\sum_{k=0}^\infty (1+(-1)^k) x^k$. In Mathematica, one can do

In[1]:= Sum[(1+(-1)^k)*x^k,{k,0,Infinity}]                                     

          -2
Out[1]= -------
              2
        -1 + x

How does it work in Sage? I tried

sage: var('k')
k
sage: sum((1+(-1)^k)*x^k, k, 0, oo)
sum(((-1)^k + 1)*x^k, k, 0, +Infinity)

which doesn't give me any new information.

2023-07-09 14:42:59 +0100 asked a question Compute power series

Compute power series I would like to compute $\sum_{k=0}^\infty (1+(-1)^k) x^k$. In Mathematica, one can do In[1]:= Sum

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2023-05-09 15:27:14 +0100 marked best answer Recursive combinatorial species

The combinatorial species of rooted trees can be defined recursively via $A = X\cdot E(A)$, where $X$ denotes the singleton species and $E$ the set species. How can I make the following code work? The last command yields an error.

sage: X = species.SingletonSpecies()
sage: E = species.SetSpecies()
sage: A = CombinatorialSpecies()
sage: A.define(X*E(A))
sage: A.generating_series()
2023-05-09 00:30:02 +0100 edited question Recursive combinatorial species

Recursive combinatorial species The combinatorial species of rooted trees can be defined recursively via $A = X\cdot E(A

2023-05-08 23:19:28 +0100 edited question Recursive combinatorial species

Recursive combinatorial species The combinatorial species of rooted trees can be defined recursively via $A = X\cdot E(A

2023-05-08 23:19:01 +0100 asked a question Recursive combinatorial species

Recursive combinatorial species The combinatorial species of rooted trees can be defined recursively via $A = X\cdot E(A

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2023-04-13 00:58:10 +0100 commented question Accelerating for-loop

You are right in principle, and that would be the proper (elegant) mathematical way, but knowing the result in the case

2023-04-13 00:51:31 +0100 commented question Accelerating for-loop

You are right in principle, and that would be the proper (elegant) mathematical way, but knowing the result in the case

2023-04-13 00:24:44 +0100 edited question Accelerating for-loop

Accelerating for-loop I have a loop like n = 3 M = MatrixSpace(Integers(n),n) L = [] for m in M: if condition:

2023-04-13 00:24:13 +0100 edited question Accelerating for-loop

Accelerating for-loop I have a loop like n = 3 M = MatrixSpace(Integers(n),n) L = [] for m in M: if condition:

2023-04-13 00:23:43 +0100 edited question Accelerating for-loop

Accelerating for-loop I have a loop like n = 3 M = MatrixSpace(Integers(n),n) L = [] for m in M: if condition:

2023-04-13 00:21:33 +0100 asked a question Accelerating for-loop

Accelerating for-loop I have a loop like n = 3 M = MatrixSpace(Integers(n),n) L = [] for m in M: if condition:

2022-12-18 13:26:52 +0100 commented answer Linear Independence in Quotients of Polynomial Rings

Thanks, I've already played with that, too. In this case this should be equivalent to the fact that if we define all the

2022-12-18 13:16:42 +0100 commented answer Linear Independence in Quotients of Polynomial Rings

Thanks, I've already played with that, too. In this case this should be equivalent to the fact that if we define all the

2022-12-18 13:15:34 +0100 commented answer Linear Independence in Quotients of Polynomial Rings

Thanks, I've already played with that, too. In this case this should be equivalent to the fact that if we define all the