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Yes, the absolute value is omitted, as seen from integrating without limits, then differentiating the result: sage: ass

If we start with an integer $n>0$ with known (or computable) factorization $n=\prod_{p\in P} p^{r(p)}$, for some fini

What about also indenting input ?!

For the first question, on can use the $k$'th power of the adjacency matrix of the graph. (With or without the diagonal

Here is the same code as posted on MO. However, here i is the place to say some words about the sage part of the impleme

Modular Forms What is the most efficient way to determine the coefficients $$a_{1}, a_{2}, ..., a_{21}$$ such that $$\D

Let us see what is exactly CC. For this, we compare: sage: CC(0.13816890584139213) 0.138168905841392 sage: CC Complex F

I hoped to get at least $\zeta_L(3)$ as a linear combination with rational coefficients of values of the trilogarithm co

I hoped to get at least $\zeta_L(3)$ as a linear combination with rational coefficients of values of the trilogarithm co

Let us get an explicit field $L$, i will ignore any restriction on $a,b$. a, b = 7, 3 # chosen so that we have a nic

Here is the same answer, some nuances are shown. The function / method numerical_approx(...) has the short hand n(...)an

Ausland-Reiten Quiver - knitting Given a quiver representation A3 = DiGraph({1 : {2 : ['a']}, 2 : {3 :

Let us take a look at the constructor called in order to build the instance QM2. First of all, it is mathematically a

Let us use names for the objects, i am working with particular examples. P = Permutations(7) Q = Permutations(9) s =

Make sure that x is a variable when defining f. In my case, the following worked: sage: var('x'); sage: f = heaviside(x

.... continued: But Jordan form will require that the eigenvalues of the matrix can be represented within Sage, r

Which is the field / domain of definition for the entries of the two matrices K_1 and K_2 - that i will denote by A and

You may want (next time) to rearrange the code, so that it gets readable. Also, respecting PEP8 and PEP40 may let the re

Here is a way, i tried in hurry: N = 1000 R = range(N) def myrgbcolor(k): # you may try also some other functions

Which operations are needed for these matrices over the octonions? Just make an own class if the wanted mathematical app

Please give us the reason why this question is relevant to sage. (The code does in sage specifically only the story wit

Please give us the reason why this question is relevant to sage. (The code does in sage specifically only the story wit

Please always mention the own effort to solve the problem. Well, in this case it is simpler to deliver the solution then

As reference i am using https://doc.sagemath.org/html/en/reference/groups/sage/groups/matrix_gps/orthogonal_gap.html. T

I will use $\tau$ instead of $z$, and assume that we have the nome-convention: $$ q = q(\tau)=e^{2\pi i\; \tau}\ . $$ Th

See also https://doc.sagemath.org/html/en/reference/spkg/rubiks.html

Here is a first brute force solution. I will use q instead of b. Let $F=\Bbb F_q=$GF(q) be the field with $q$ elements.

No trick could force the present code (in my hands) to use FLINT. From https://fredrikj.net/math/lyon2023flint.pdf is se

Let us take a look at the doc string... sage: ?PolynomialRing After some lines, we get the block: * "implementati

Version 10.1 does the job. To see if a method accepts some arguments, ask for its doc string. For instance, with the ab

After the comment of FrédéricC i took at the generalized hexagon graphs that sage may construct for some small values of

Incidence graphs of generalised quadrangle and hexagon For an introduction to generalised polygons, see https://cage.uge

Please give some references to the mathematical definition of the involved objects. Or some sample minimal code initiali

I will use $p$ for $13$ for short. Let $F$ be the field with $p=13$ elements, $F=\Bbb Z/13=\Bbb F_{13}$. Let $R=\Bbb Z/1

I will use $p$ for $13$ for short. Let $F$ be the field with $p=13$ elements, $F=\Bbb Z/13=\Bbb F_{13}$. Let $R=\Bbb Z/1

I took a look to the code of the VectorPartitions and there was a line that i could not understand immediately... @stat

The following works for me: # parameters for BLS12-381 z = -0xd201000000010000 # this is z = -15132376222941642752

Posting as anonymous has some drawbacks... it may be a good idea to get please a user name... The question starts by int