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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 How does it work in Sage? I tried 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. |
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 |