@user10478 I'm not seeing the error. In fact I am very confident your things satisfy the ODE.

(For all points of their domain of definition, aka when $t \neq 0, \sqrt{C}$.)

Why do some authors seem to prefer the phrasing "coprime" and others "relatively prime"? I'm curious as to the history of this terminology.

@Fargle co- and relatively both mean relative to each other, so I doubt there is any significance to the matter beyond that.

I was more curious about whether the etymologies are informed by different branches or perspectives in math, or by different languages, or by something entirely other.

The meaning is transparent, but I'm curious about how those terms became competing standards, and which perspectives prefer which terms.

I see. Might you ask the same of commutative group and abelian group?

I prefer using commutative group myself, because it is more descriptive, and applies to other structures too, like rings.

@Fargle in fact in a lot of number theory texts you'll often just see "prime to"

@ÍgjøgnumMeg Interesting.

probably another shorthand, eventually number theory texts will just be strings of contextless symbols

Hey. I'm trying to read Natural operations in differential geometry, by Kolar, Michor and Slovak.

In page 64 they present the exterior derivative.

Oh, hell.

I'll just paste a pic

Is this right?

(In my last expression, $L$ is the Lie derivative).

Is there any difference between Vector form and parametric vector form?

@GaussianElimination Hello, these terms have no fixed meaning. We need the context. vector form just means involving vectors. Parametric vector form just means involving vectors and parameters, that is all.

Sorry, I mean in terms of linear algebra. Look the answers to the question on this page. math.stackexchange.com/questions/28051/…. I have heard 'vector form' and 'parametric vector form' used interchangeably to refer to solution sets that use vectors.

@MikeMiller, @Fargle Sorry had to run for work. Not sure if you're still here, but the book's answer is $y = \frac{1}{t} + \frac{2Ct}{1 + Ct^2}$.

It seems to make more sense than mine, since $C = 0$ yields the given particular solution.

any idea of this notation ?

$5_{\frac{1}{2}}$

@BAYMAX We need the context. Otherwise notation is just notation. One can define cat to mean dog and dog to mean cat.

nice

it was actually

find the sum of the sequence $4, 5_{\frac{1}{2}}, 6_{\frac{1}{2}} \ldots $ till 37 terms

The notation has to be defined somewhere in the paper you are reading, I guess

Guys can anybody help me with this question?

I have this inequation: $+-\sqrt{5-x}<5-x$. How do I solve this? Can I square both sides and find the correct solution?

Can you help with $\sum_{0}^{\infty} {r^n sinn\theta}$ if r =0.5 $\theta = \pi/3$?

@Nobodyrecognizeable geometric series

@LeakyNun actually n goes to 0 to infinity.

sure

$\sin2\theta$ not equal to $sin^2\theta$

sure

@LeakyNun how's that geometric series then ?

because you can write $\sin(n \theta)$ as $\frac1{2i} [\exp(in\theta) - \exp(-in\theta)]$

@LeakyNun what will be the ratio of successive terms ?

$$\sum_{n=0}^\infty r^n \sin(n\theta) = \frac1{2i} \left[ \sum_{n=0}^\infty (r\exp(i\theta))^n - \sum_{n=0}^\infty (r\exp(-i\theta))^n \right]$$

is that better?

@LeakyNun you are a genius man... thanks

Thoughts: the diagonals of a rectangle are distinguishable

There is no symmetry of the rectangle that takes one diagonal to the other, at least if orientation-reversing symmetries are disallowed

This is unlike the case of a square

(We can specify one by saying, "take the long axis of the rectangle and rotate it counterclockwise")

(or "clockwise" for the other one)

Hi

The group $< a, b|a^q = b^{p^2}= 1, bab^{−1 }= a^i, ord_q(i) = p >$ can be represented as $\mathbb{Z}_q \rtimes \mathbb{Z}_{p^2}$. How to represent the group $< a, b|a^q = b^{p^2}= 1, bab^{−1} = a^i, ord_q(i) = p^2 >$?

(Also, in a rectangular cuboid, all the space diagonals are distinguishable)

@BuddhiniAngelika What's ${\rm ord}_q(i)$?

means that the multiplicative order of i modulo q is p

For the second group the multiplicative order of i modulo q is p^2

Is it correct if I represent the second group as $\mathbb{Z}_q \rtimes_2 \mathbb{Z}_{p}$

?

I don't know what $\rtimes_2$ is, sorry

I don't know a lot of group theory

Ok, thanks anyway @AkivaWeinberger. $\rtimes _2$ is like having theta represented with semidirect product sign. It's a notation. I found this notation in some paper, but I just wanted to know it's exactly correct, that's why

in a cayley graph of a group there are generator sets for which some subgroups will be apparent in the graph as subgraphs right?

Suppose in a cayley graph of a semidirect product the normal subgroup and its cosets are apparent in the cayley graph.

Is it ok to refer the cosets apparent in the graph by using the word "coset" as usual?

Is there better terminology to identify the subgraphs representing the cosets ?

@Buddhini what paper did you see $\rtimes_2$ in?

Oh my, Nash's ghost was here.

math.stackexchange.com/questions/318299/… please see the answer given by Nathan A S , wiki community. what is the guarantee that $I_a$ contain $x$.?

What's I_a? I don't see it in the answer

Aren't you all excited about how, in a little more than a year, we can all start living in a decade again

Unknown x has commented on the answer.

What does it mean for a vector field to point inward/outward?

@user193319 divergence negative/positive

Suggestion: the forward slash (/) should be called the Z slash. The backslash (\) should be called the N slash.

So, if $v$ is a vector field that points directly outward at $x$, this means that $v(x) = ax$ for some $a > 0$? I'm encountering this idea for the first time in Munkres, and he didn't provide any definitions.

Z N

@AkivaWeinberger Nonono, the N-slash is frowned upon

@BalarkaSen an interesting fact after my question yesterday on geodesicity and isometries: If $X$ and $Y$ are quasi-isometric and $Y$ is quasi-geodesic then so is $X$! This is one of the steps needed to show that Gromov hyperbolicity is a quasi-isometry invariance among geodesic spaces

@Alessandro quasi-geodesic metric space means that there is a quasi-isometrically embedded interval between any two points?

between any two quasi-points

@BalarkaSen Yes

Very cool

Ah, wait, you want a $(c,b)$-quasi geodesic between any two points, the same constants for every pair of points

Why isn't that clear from composing with the quasi-isometry $Y \to X$?

well you need to quasi-compose them

@Alessandro Got it

The details are kinda ugly because you need to fiddle with the constants a bit, but the idea is indeed "just push forward and pull back the quasi geodesics everywhere"

Aha

Which makes sense, because it's exactly what you do without all the quasi- prefixes

Yup

@BalarkaSen Also the fact that the quasi isometry needs to be neither surjective nor injective is very quite annoying and definitely a complication from the actual isometry case

it's just a quasi-complication

That makes sense

@ÍgjøgnumMeg in this paper arxiv.org/abs/1504.00801

In page 6, in Case 1c

Do u think it's ok?

What did the mathematician pirate say?

The Rrrrr language should walk the plank!

:P

"Aye aye is minus one"

Sorry

Arrrrrrg(z), me mateys!

"X" marks the spot, to within +/- Rrrrr

haha

Obligatory sea shanty:

Consider the cayley graph of a semidirect product between Z3 and Z5 X Z5, where Z5 X Z5 is the normal subgroup. Then we can view it as 3 copies of 5 X 5 square grid (torus) structures arranged along a 3-cycle. Between these 3 copies the vertices are mapped based on the homomorphism $\theta: Z3 -> Aut(Z5 X Z5). The elements (0,0,0

The elements (0,0,0), (1,0,0), (2,0,0) are connected by a 3-cycle. (0,0,0) is the identity element. Does anyone know what we call (1,0,0) and (2,0,0)? Identity elements of cosets of Z5 X Z5?

Hi chat.

Hi @Lucas

@Secret This font seems to work on a similar principle to Truchet tiles

clanbadge.com/knots.htm

Actually @BalarkaSen Löh's proof that being quasi-geodesic is a quasi-isometry invariant uses AC (I don't know if it is necessary though)

Maybe @PaulPlummer knows?

In Commutative Algebra by Atiyah and Macdonald it's stated that for a field $k$, a $k[x]$-module $M$ is a $k$-vector space with a linear transformation, I understand that $M$ is a $k$-vector space but what is the linear transformation?

$m\mapsto xm$ I guess?

@AlessandroCodenotti yes

in fact let's form a category $\mathscr C$ whose objects are $k$-vector space $V$ with an endomorphism $T:V \to V$, and let a morphism $(V,T) \to (W,U)$ be a linear map $\varphi : V \to W$ such that $\varphi \circ T = U \circ \varphi$

I claim that $\mathscr C$ is equivalent to the category of $k[X]$-modules

@AlessandroCodenotti what do you think?

Seems both right and needlessly complicated :P

So did you decide wether Jech's example works or not in the end?

eh it doesn't really matter

it isn't part of any argument

@AlessandroCodenotti but my category can be generalized can it not

for any category I can do the same thing to get a new category

does my category have a name?

I don't know

In general your new category is a category where objects are pairc $C$ and an element $f\in\mathrm{Mor}(C,C)$ where $C$ is an object in the original category, right?

call it the category of dynamical systems in $\mathcal C$ if you like

I suppose the name would make more sense to me if I knew what a dynamical system is :P

Some object with some dynamics on it

Dynamics move things around

So I'm thinking of something like a topological space $X$ with a function $f:[0,1]\times X\to X$?

@AlessandroCodenotti yes

I am thinking of a topological space with a self-homeomorphism, or maybe just a self-map :P

You're thinking of the idea of dynamical systems as being about flows

and then we have a forgetful functor from the new category to the original category

and I believe it has a left adjoint

In which case, ok, take $\mathcal C$ to be smooth manifolds where a morphism is a smooth map $f: X \to Y$ and a section of $f^*TY$ (a map and a direction to move the map in). Then in Leaky's category, $(X, (\text{id}, V))$ is an object, where $V$ is some vector field on the smooth manifold $X$

I see, that makes sense

Why do mathematicians want to count the number of closed orbits and fixed points

why not

what else were we doing with our time?

why do painters want to paint

i guess its like that

but after you identify a fixed point what do you do with that knowledge

Hi there. Does anyone know of something to help e to calculate the following: Sum of all (1^a)(2^b)(3^c)(4^d)(5^e) such that a+b+c+d+e=n

@Ultradark Sometimes finding that fixed point solves some other problem you cared about (solving some equation, maybe).

okay

Hello, I am working on representing an epicycle as an epitrochoid, and to do so I need only relate the radii of the fixed circle and rolling circle to the given values for the epicycle. However my working out seems to be wrong, and I am not sure why. My question has not received many views, link is here: math.stackexchange.com/questions/3056183/…