Mathematics - 2019-02-26

user193319

12:07 AM

Let $Q$ be the Quaternion group of order $8$, $N$ a a normal subgroup of order $4$. The reason why $1 \rightarrow N \rightarrow Q \rightarrow Q/N \rightarrow 1$ is not a split sequence is because $Q$ cannot be written as the semidirect product of any its subgroups in a nontrivial way, right?

Are there two notions of semidirect product? My book speaks of one where the group $G$ is the semidirect product of $N \triangleleft G$ and $H \le G$ if $N \cap H = \{1\}$ and $G = NH$. And then my book also speaks of a semidirect product as the cartesian product together with a homomorphism from one factors into the automorphism group, together with particular group multiplication defined on the order pairs. Are they related in any way?

Ted Shifrin

12:22 AM

@user193319: Every subgroup of $Q$ is normal. So there's no semidirect product, only direct product.

The semidirect product structure usually involves giving the homomorphism $H\to \text{Aut}(N)$.

user193319

And that semidirect product is different from the first one I mentioned?

Ted Shifrin

The first one is not complete information.

user193319

Hmm...really? I just typed in what was in my book.

Ted Shifrin

As I said, most books/people will say that to give the semidirect product structure, you must say what the homomorphism is.

Thus, the first description is not giving complete information.

user193319

Sorry. I actually took that from wikipedia: en.wikipedia.org/wiki/Semidirect_product

Ted Shifrin

12:25 AM

Well, we all know that wiki is perfect.

user193319

Haha

Ted Shifrin

I've said the same thing twice now, so I'm not going to say it again.

Zee

Yo @ted

You feeling the bern?

topologicalmagician

@TedShifrin, I'm watching your lecture, the only thing I don't understand is why that when you applied the one form on the basis vector you get 1 or 0

Ted Shifrin

That's how the $\phi_i$ are defined. $\phi_i(e_j) = 1$ when $i=j$ and $0$ otherwise.

Zee

12:29 AM

dx is the dual basis vector

topologicalmagician

Ahhhhhhh that's how they're defined

I've been trying to work it out for a while and it turns out that's how they're defined lol

Ted Shifrin

"It turns out"? That was clearly stated.

You have to change your approach to learning mathematics. You have to start by writing down and learning definitions, above all else.

Érico Melo Silva

pshh writing is for wimps

topologicalmagician

Yeah, true

Ted Shifrin

smacks Eric

topologicalmagician

12:33 AM

Anyways I'm gonna go now im really tired

Érico Melo Silva

tbh i deserve the smack ngl

topologicalmagician

Take care guys

Ted Shifrin

bye

Alessandro Codenotti

@user193319 The former is enough to conclude that there is some morphism such that $G$ is a semidirect product of $N$ and $H$ wrt that morphism

The former is giving you sufficient conditions to say that a group splits as a semidirect product, the latter is telling you how to combine two unrelated groups with a semidirect product, I've heard those called internal and external products before, but I don't think it's common (or useful) terminology

Ted Shifrin

@Alessandro: If you don't think of $N$ and $H$ as subgroups of a given $G$ in the second case, then it's "external."

Alessandro Codenotti

12:37 AM

Yes, but it's really the same construction so I'm not sure about giving it a different name!

Ted Shifrin

People use the same distinction for direct products.

Zee

That’s the only things Grothendieck requires of you , write the definitions and continue

user193319

1:35 AM

In Lebesgue measure theory, is there an example of a sequence of nonmeasurable functions converging pointwise to a measurable function?

Rajesh Dachiraju

2:03 AM

0

Q: Combining a coupled minimization and maximization problem. Is there a better approach?

$$A_{\lambda}(f) = B(f) + \lambda C(f)$$ $\lambda \in (0,\infty)$. $A_{\lambda},B,C$ are non negative, quadratic and convex functionals. Let $f_{\lambda}$ be the minimizer of $A_{\lambda}(f)$ over a convex set $S$. Now I need to maximize $D(f_{\lambda})$ over $\lambda \in (0,\infty)$, again $D$ ...

MatheinBoulomenos

2:32 AM

@user193319 let $X$ be any set that is not Lebesgue measurable. Let $\chi_X$ be the indicator function of $X$ (i.e. $1$ on $X$ and $0$ outside of $X$), then consider the sequence $f_n=\frac{1}{n} \chi_X$, this converges pointwise to $0$ and each $f_n$ is non-measurable

Secret

@AkivaWeinberger Make sense, permutations are either odd or even, so sums of terms and their permutations will going to do some kind of cancellations along the way. I have not investigated much on the permutahedron because I only learnt it from you yesterday. It seems it could be useful to understand how some cancellations can happen in the context of number theory because these sums of pairs, triplets etc. are very common in the context of polynomials of multiple variables

Ultradark

-2

Q: Is there a correspondence between these sets of areas?

Given are two functions, $p(x)$ and $\phi(x)$, defined as follows $$ \begin{align} p\colon [3, +\infty[ &\to [0, 1],\\ x &\mapsto \frac{1}{\ln x}, \end{align} $$ and $$ \begin{align} \phi\colon \mathbb{R}^{\geq 0}\setminus\{1\} &\to \mathbb{R},\\ x &\mapsto 2^{\frac{1}{\ln x}}. \end{align} $$ Se...

probability number-theory probability-theory soft-question prime-numbers

Is $f(x)=f(-x)$ a good property? Why?

Secret

@MatheinBoulomenos Good point (which I have not get back to last night). Actually, I am not sure if there is a way to avoid that result. One interesting thing about the expression $ad+bc-ab-cd$ and $ad+bc-ac-bd$ is that even if you allow associativity and commutativity in both + and x, they can never be made equal, and that is a desirable point because by the definition, the only difference between the two are the position of b,c,d get cyclically permuted, thus if it is nonzero, it means there is some

"noncyclicality" in the structure

This result suggests we need another way to define the cyculator so that it can play its role properly to measure the cyclicality of an algebraic structure

MatheinBoulomenos

2:54 AM

@Secret this might interest you, as it is slightly related:

Polynomial identity ring

In mathematics, in the subfield of ring theory, a ring R is a polynomial identity ring if there is, for some N > 0, an element P other than 0 of the free algebra, Z⟨X1, X2, ..., XN⟩, over the ring of integers in N variables X1, X2, ..., XN such that for all N-tuples r1, r2, ..., rN taken from R it happens that P ( r 1 , r 2 , … , r N ) ...

Secret

3:08 AM

Hmm... so that means in this terminology, the only PI ring that has $P(a,b,c,d)=ad+bc-ab-cd$ is trivial

MatheinBoulomenos

right

Secret

It is weird that I often end up picking these near counterexamples when doing maths

near counterexample: Examples where it seemed to obey a pattern, only to fail to satisfy a desired property the end

Secret

3:23 AM

Moufang loop

In mathematics, a Moufang loop is a special kind of algebraic structure. It is similar to a group in many ways but need not be associative. Moufang loops were introduced by Ruth Moufang (1935). Smooth Moufang loops have an associated algebra, the Malcev algebra, similar in some ways to how a Lie group has an associated Lie algebra. == Definition == A Moufang loop is a loop Q that satisfies the four following identities for all x, y, z in Q (the binary operation in Q is denoted by juxtaposition): z(x(zy)) = ((zx)z)y; x(z(yz)) = ((xz)y)z (zx)(yz) = (z(xy))z (zx)(yz) = z((xy)z).These identities...

hmm...

The generalisation of Moufang identities are n-ary versions of associators

so nothing inherently new

Secret

4:07 AM

@Ultradark what does eveness has to do with that?

Ultradark

4:25 AM

nothing that was a separate question

Secret

4:39 AM

2nd attempt at defining a useful quaternery operator:

Let $⫿a,b,c,d⫿$ be defined as follows:

$$⫿a(b(cd))⫿+⫿b(c(da))⫿+⫿c(d(ab))⫿+⫿d(a(bc))⫿=0$$

Secret

4:55 AM

fail

Proper substring(abc) = a,b,c,ab,bc,ac

discard all singles: ab,bc,ac

(ab)c,a(bc),(a|b|c)=(ac)b

typo: ignore ac

Proper substring ignore singles (abcd) = ab,bc,cd,abc,bcd

(ab)(cd)
(a(bc))d,a((bc)d),a(b(cd))
((ab)c)d,a(b(cd)),((a(bc))d,a((bc)d)

hmm...

((ab)c)d,(a(bc))d,(ab)(cd),a((bc)d),a(b(cd))

ok, so there's clearly an extra kind of ordering in here, that only become significant when there are 4 terms. Now to figure out how to formalise it

If there is a way to use a quadruple to represent the ordering on how the substrings of a given string is taken, then that will be a good 4-nary operator

ncatlab.org/nlab/show/identity+of+Bol-Moufang+type

Secret

5:53 AM

Definition: Define the Moufangator $⁅a,b,c,d⁆$ to satisfy the following equations:

$(((ab)c)d) = ((a(bc))d) ⁅a,b,c,d⁆$

$(((ab)c)d) = (ab)(cd) ⁅a,b,c,d⁆$

$(((ab)c)d) = (a((bc)d) ⁅a,b,c,d⁆$

$(((ab)c)d) = (a(b(cd))) ⁅a,b,c,d⁆$

Then, a magma is a moufang loop if $⁅a,b,c,d⁆ = e$ where $e$ is the identity

(Actually, need a subscript as a Moufang loop does not need to satisfy all 4 of them. But then the problem of this is then it is technically not a 4-nary operator, but a 5-nary one)

But then, n-ary associators explode in numbers already when n>3, so maybe that is not much of a deal

Tobias Kildetoft

8:56 AM

What sort of thing should I read into a company using the word "awesome" a lot? It appeared in both the job ad and 2 out of three mails from their HR manager.

Secret

I think the first step is to delete all instance of "awesome" and see if the remaining stuff still make sense

Tobias Kildetoft

@Secret well, some places you would need to replace it with another adjective, since one of them was something like "it is awesome that you have applied"

but sure, it makes sense without the words or with suitable replacements.

Alessandro Codenotti

Is this written by a native English speaker?

Tobias Kildetoft

It was written in Danish (but the word really is "awesome" rather than a Danish equivalent)

Alessandro Codenotti

Oh, that's weird

Tobias Kildetoft

9:02 AM

It is not that uncommon for people to use the word

(though mostly spoken)

The specifications for the assignment they gave me to show I can make android apps was written in English, but not by a native speaker (and it did not say "awesome" anywhere).

The assignment turned to be harder than I thought, but at least I managed to implement all the required features.

Leaky Nun

what would you call the opposite of leading coefficient? i.e. the coefficient of the smallest degree with non-zero coefficient?

Alessandro Codenotti

the led coefficent

Tobias Kildetoft

trailing might actually be a good word for it

both because it is opposite to leading and because it gives the right picture

Secret

meanwhile, "tail" is also a common word to describe the last of something

tail coefficient sounds strange though

Tobias Kildetoft

tailing could work

(stalking would probably be inappropriate)

Secret

9:18 AM

In other news:

If I understand the conditions correctly, both of the examples of $3$-ary operators you give (the associator and the l.h.s. of the Jordan identity) do not satisfy condition (2), so one might first try to construct a $3$-ary operator satisfying (1)-(3). — Travis 2 hours ago

that will probably mean I need a 3-nary operator that can only be defined in terms of identities and cannot be wrote in a more explicit form

meanwhile this is strange use of algebra in chemistry: match.pmf.kg.ac.rs/electronic_versions/Match65/n2/…

ChoMedit

9:46 AM

Here is the question. I've known that to prove Young's theorem, we need $C^2$ condition but Pugh(and maybe Dieudonne) proved it without $C^2$ assumption. Did I miss something? See the link below.. If you are interested in.. sites.math.washington.edu/~morrow/334_15/…

Secret

10:08 AM

Wikipedia mentioned that it is sufficient to have the weaker condition where the partial derivatives to be differentiable to establish Young's theorem, which is exactly the condition given in the notes at the line when it says Df=[f_x,f_y] is differentiable

ChoMedit

But does differentiability of $Df$ imply the continuity of second partial derivatives? Because wikipedia also mention with continuity of them.

Secret

if the partial derivatives f_x and f_y are differentiable by both x and y, then it means both f_x and f_y are continuous, which means they cannot exhibit jumps as shown in the counterexample, and hence f is twice differentiable at that point

user193319

$Mathematics$ Group Theory

Let's discuss group theory!

Secret

lol I have not gone that deep into short exact sequences yet. Leaky is a better person

user193319

10:25 AM

Okie dokie. Perhaps you should start learning about them so you can help me! :P

Secret

projecteuclid.org/download/pdf_1/euclid.ndjfl/1093958336

ChoMedit

Yes that's right but it doesn't imply that $f_{xy}$ and $f_{yx}$ are also continuous.. since there is no guarantee that $f_x, f_y \in C^1$.

Secret

Hmm.. this suggests quadrilaterals are primitive elements in many geometries

but if f_xy and f_yx are not continuous at that point ,it means f_x, f_y had a jump there. Then f_x,f_y will not be differentiable at the jump

o wait nvm

I have no idea

The part about sufficiency stated that differentiability of first derivatives is a weaker condition than continuity of second derivatives, so it will not be surprising that continuity of secondary derivatives is not implied from it

It basically means Young's theorem can hold for some non $C^2$ functions

Secret

10:42 AM

In other news:

Separation relation

In mathematics, a separation relation is a formal way to arrange a set of objects in an unoriented circle. It is defined as a quaternary relation S(a, b, c, d) satisfying certain axioms, which is interpreted as asserting that a and c separate b from d.Whereas a linear order endows a set with a positive end and a negative end, a separation relation forgets not only which end is which, but also where the ends are located. In this way it is a final, further weakening of the concepts of a betweenness relation and a cyclic order. There is nothing else that can be forgotten: up to the relevant sense...

A set with 4 points on a circle is the first example where any two points can be separated by other points

You cannot do that for 3 points because some pairs will always be next to each other

Secret

11:22 AM

Three possible way forward in finding a good quaternary operator:

1. Define when an object is chained to another

2. Define abstract quadrilaterals

3. Define separation

One difficulty is suppose the two triangles shown are elements from a ring, what does "hinging them together" means algebraically

Bijayan Ray

0

Q: What is the significance of sign of Jacobian determinant of a function?

In Tom Apostol calculus volume 2 (page 402) I found that the non zero jacobian determinant of a function whose components have continuous partial derivatives on a set, is either positive or negtive in the whole set. Why does the determinant has same sign in the whole set? In the text I fou...

real-analysis linear-algebra multivariable-calculus determinant jacobian

0

Q: What are Winding number?

I came across the term in Tom Apostol Calculus volume 2 (page 390) as a line integral , which seemed to be equivalent to the wikipedia definition in terms of polar coordinate. The line Integral in Apostol text simplifies to somewhat :- $$\tan^{-1}(\frac {Y(a')-y_0}{X(a')-x_0})-\tan^{-1}(\frac {...

real-analysis calculus multivariable-calculus vector-analysis winding-number

Secret

hmmmmmmmmmm....

how about:

Crossings only become possible in 3D, and need at least 2 line segments (hence 4 endpoints) to define.

Inspired from knot theory, we can thus have the first axiom:

$C(a,b,c,d)C(c,d,a,b)=e$

Likewise, if lines are unoriented, then

$C(a,b,c,d)=C(b,a,c,d)=C(b,a,d,c)=C(a,b,d,c)$

'

Thus a left handed trefoil can be denoted as follows:

$C(e,d,a,b)C(b,c,e,f)C(a,f,d,c)=0$

and a right handed trefoil:

$C(a,b,e,d)C(e,f,b,c)C(d,c,a,f)=0$

Now we can go further and generalise this quaternary operator:

Secret

12:20 PM

Define $⌘a,b,c,d⌘$ the tanglelator as follows:

$$⌘a,b,c,d⌘ = C(a,b,c,d)C(c,d,a,b)$$

in the context of knots the tanglator vanishes (i.e. = +1 and -1 crossing cancelled out into the unknot)

Akiva Weinberger

@Secret This feels like knot coloring… if you impose $a=c$ and $a+c=b+d\pmod m$ then you recover it I think

(If $m=3$ then this is the same as saying that you color every strand in one of three colors and at every crossing you have either all three colors or only one)

Secret

12:41 PM

well... I am not very familiar with knot theory yet, I just briefly googled some stuff since wikipedia give very little details on how knot colorings are computed

In particular, I don't know how to interpret the + sign, are we gluing the specified ends together?

Linking number

In mathematics, the linking number is a numerical invariant that describes the linking of two closed curves in three-dimensional space. Intuitively, the linking number represents the number of times that each curve winds around the other. The linking number is always an integer, but may be positive or negative depending on the orientation of the two curves. (This is not true for curves in most 3-manifolds, where linking numbers can also be fractions or just not exist at all.) The linking number was introduced by Gauss in the form of the linking integral. It is an important object of study in...

hmm...

Akiva Weinberger

1:27 PM

@Secret In $a+c=b+d$?

These are numbers you're drawing on the strands

Hold on let me draw an example

user193319

@Secret Why do you learn things so haphazardly? Why not pick up a book on a particular topic, instead of reading random wiki articles?

3

Secret

@user193319 Well the main reason is because I often read or play with concepts that first came to mind, and these concepts are like chains of memories. So when I learn with some spare time, I often end up jumping haphazardly from wikis to books, to slides and so on

Akiva Weinberger

So like basically it's as if you're assigning every segment a number (or element of $\Bbb Z/m\Bbb Z$) according to the rules on the right

and the reason we do this is because it plays well with the Reidemeister moves

Secret

1:43 PM

I see

so a=c defines a strand, and a+c=b+d assign colors mod m to the strands

Akiva Weinberger

And I guess if you add a constant number to all the strands then it still works

or if you multiply them all by a constant

Secret

@user193319 I did once read a book in detail before, and that is Munkres. But in order to continue on the book, I will need to complete my chemistry PhD first

There's slightly a bit of free time recently because the computational resources will not recharge until the end of March

user193319

I see. Well, it doesn't seem that your current method is particularly productive. A lot of your rambles in this chatroom are nonsensical.

Secret

Well, the amount of rambles have cut down alot, and some users can still follow my ideas and thus contribute to the discussion. I also don't ask questions alot, and instead try my best to dig for the answers

Most of the rambles are now in the Star Wars room

user193319

Haha, true. You use post a lot of drawings and sketches in the chatroom.

Secret

1:51 PM

@AkivaWeinberger While I can only continue the investigation much later because I felt I will need a lot more knot theory understanding to get it right, but basically the motivation of the operator $C$ is it tries to generalise knots into something which positive and negative crossing no longer cancel, and thus you have really tangled knots that can only be untied in specific ways, and the "strands" will be elements from rings or something more general

Akiva Weinberger

Have you heard of quandles

$user image$

Secret

yeah, briefly came across that from a user back in 2015, and then came across again when I lookup knot coloring rules

can't say I fully understood it, but it encodes coloring somehow

Akiva Weinberger

The Wikipedia page is good, as is this page I found a minute ago

Secret

wow, that page is short and sweet, a lot easier to read than wikipedia's

One thing that quandles are interesting to me is the self distributive law

the second axiom of quandles reminds me of quasigroups

Akiva Weinberger

Test

$\triangleleft$ $\triangleright$

$\vartriangleleft$ $\vartriangleright$

Secret

2:03 PM

$\vartriangleleft$ cross under from the left
$\vartriangleright$ cross under from the right

A lot less clumpsier than my $C$ operator. Probably stick with quandles and build stuff from there

Akiva Weinberger

You do need an orientation on the knot, I think

or, at least, on the overcrossings

(so you can decide which way is "left" and which way is "right")

It's strange that it doesn't take into account orientation on the undercrossing

Secret

true

well the orientation of the undercrossing is taken account by the $\vartriangleright, \vartriangleleft$ of the quandles

because the undercrossing has to be computed in reference to the overcrossing strand, whose orientation is known

that actually justify the name rack

Wish fields actually look like a field and rings look like rings (well ok, cyclic rings do look circular)

> The French term for a field is corps and the German word is Körper, both meaning "body." A field with a finite number of members is known as a finite field or Galois field.

33

Q: Origins of names of algebraic structures

Consider the names of basic algebraic structures: 'group', 'ring', 'space', 'field', 'Körper', even the name 'structure' itself - all of them time-honoured terms, deeply rooted in our history and culture. But what has an algebraic field to do with an acre? What has an algebraic group to do wit...

soft-question ho.history-overview

Akiva Weinberger

Note that the "Takasaki kei" is essentially the same as the coloring rule $a+c=b+d$

(and also that it satisfies $\triangleright=\triangleright^{-1}$ aka $\triangleleft$ and so it doesn't care about the knot's orientation)

Akiva Weinberger

2:39 PM

@Secret Hey, you know what video game I think you'd like? Antichamber

(Here's some gameplay to get an idea of what it is https://youtu.be/fXty2JYIjPo)

Secret

Ah, already knew that

They will probably need an Antichamber 2 because I have already watched all the walkthroughs

The position dependent perspective is one of its big selling points

I am not sure how one will model that geometry mathematically

Ultradark

Hi

What is the line of best fit for the prime counting function up to a given number

Tobias Kildetoft

@Ultradark isn't that essentially given by the prime number theorem?

abenthy

Hi

@TobiasKildetoft Hey Tobias, do you maybe happen to have some knowledge of quaternion algebras over number fields? I would love to discuss a question about these with someone.

Secret

3:00 PM

Hard to imagine this thing has no pattern, there are clearly those white stripes where there aren't many prime number hits

Akiva Weinberger

@Secret Unfortunately I think the creator isn't making games anymore

I mean, Antichamber was his only game

but he spent 7 years making it so I think he said he's a bit burned out

Ultradark

So the line of best fit is then approximately $x/\ln(x)$? @TobiasKildetoft

Akiva Weinberger

Also if you look at the reactions online, it seems to be really polarizing. Some people really like it, other people hate it

Weird is polarizing

It made a ton of money though so I guess it succeeded

Secret

Typical

Akiva Weinberger

3:16 PM

I just downloaded the Mandarin Pimsleur course from my library. I'm probably not gonna listen to more than an hour of it or so, so I won't actually learn the language, but hopefully I'll get a better idea of how the pronunciation works

which has been something I was wondering about

Oh, it's not Pimsleur, it's Mango

Same thing basically

Bijayan Ray

0

Q: What is the significance of sign of Jacobian determinant of a function?

In Tom Apostol calculus volume 2 (page 402) I found that the non zero jacobian determinant of a function whose components have continuous partial derivatives on a set, is either positive or negtive in the whole set. Why does the determinant has same sign in the whole set? In the text I fou...

real-analysis linear-algebra multivariable-calculus determinant jacobian

I would be thankful if some clarification is given regarding my doubt

Jake Rose

3:43 PM

Can anybody see where I’m going wrong in getting that first identity?

I actually cut out the rest of my working by mistake.

Essentially I get stuck with this

Nevermind

maths student

6:20 PM

@LeakyNun is R/I is submodule of field of fraction?

Leaky Nun

context?

maths student

If M is a finitely generated torsion module, then M ~ M* and there exists a canonical isomorphism M ~ M**. I want to prove this

Leaky Nun

context as in, give me the types of the variables that appeared in your question

i.e. hypotheses

maths student

encounterd with this math.stackexchange.com/questions/3127546/…

Zee

7:22 PM

Can you get an STD by kissing?

Akiva Weinberger

7:44 PM

@Zee The math chat is not the best place for this, but if they have a cut on their mouth, then yes it is possible

Zee

8:13 PM

Oh ok thx

Anadactothe

8:41 PM

why would you ask this in the mathematics room

maths student

@LeakyNun may I ask you question?

Leaky Nun

?

maths student

Have you seen this question ?math.stackexchange.com/questions/3127546/…

and last comment by jgon ;

As for why b−1/R≅R/b, you'll notice that the text is citing a result saying that a/ab≅R/b.

I want to ask why a/ab is isomorphic to R/b ; I know all this ideals are fractional ideals so all are invertible in dedekind domain but how do we get this result from this?

topologicalmagician

WTS:The 1-form $dx_i(v)=v_i$ is a basis for the space of all one forms, Hom($\mathbb{R}^n,\mathbb{R})$.

The proof for linear independence:

Let $c_1dx_1+c_2dx_2 +......+c_ndx_n$ $=0$ (where it is the zero function, because we are considering the space of all one forms and the 0 function is a one form). We would like to show that $c_1=c_2=....=c_n =0$. Since the LHS and RHS are both function, and are in the dual space, and $dx_i$ is in the dual space, it follows that if we take the the standard basis $e_i$ for some i $\in \mathbb{N}$ then the LHS will give us $c_i=0$.

Leaky Nun

@mathsstudent no idea

@topologicalmagician $i$ was arbitrary, so they're all zero

maths student

8:55 PM

fine no problem

topologicalmagician

@LeakyNun yeah

all of them are zero

but can't I have another input which gives me a non-zero c, but a 0 on the rhs?

Leaky Nun

@topologicalmagician do you understanding that plugging in $e_1$ to both sides tells you that $c_1 = 0$?

topologicalmagician

yeah

Leaky Nun

so similarly you can show that $c_i = 0$ for all $i$

topologicalmagician

@Leaky ahh I see, and that's the only solution because c cannot vary since its a constant, is my reasoning correct?

Ted Shifrin

9:12 PM

Oh oh ... hides from the magician :P

topologicalmagician

LOL

maths student

0

Q: About Dedekind domain and its results

Let $I,J$ be ideal of Dedekind domain $R$ ; then show that $I/IJ$ $\cong$ $R/J$. I know that every fractional ideal in Dedekind domain is invertible how to use this statement here to prove above result of isomorphism. Is this result true for only $I,J$ fractional ideal or for all ideals of $R$

abstract-algebra ring-theory commutative-algebra modules dedekind-domain

maths student

9:53 PM

@LeakyNun see comment on above question if curious to know answer

topologicalmagician

10:08 PM

guys, if I have two differential euqations dx/dt and dy/dt, why is it that when dx/dt=0 we have a vecrtical trajectory in the xy plane?

and when dy/dt=0 we have a horizontal trajectory?

user193319

Is the lift of a concatenation of two paths the concatenation of lifts the individual paths?

That is, $\widetilde{\alpha \cdot \beta} = \widetilde{\alpha} \cdot \widetilde{\beta}$?

Ultradark

Any idea how I could find the curve traced out by the blue?

The best I could come up with is $y=x/2$

Oh it's $kx/\ln(x)$

for some constant $k$ approximately equal to $2.7$

oh $k=e$

gotta prove that though

Zee

10:31 PM

Did anybody have any luck on Tinder ?

Akiva Weinberger

Tendril perversion

Tendril perversion, often referred to in context as simply perversion, is a geometric phenomenon found in helical structures such as plant tendrils, in which a helical structure forms that is divided into two sections of opposite chirality, with a transition between the two in the middle. A similar phenomenon can often be observed in kinked helical cables such as telephone handset cords.The phenomenon was known to Charles Darwin, who wrote in 1865, The term "tendril perversion" was coined by Goriely and Tabor in 1998 based on the word perversion found in the 19th Century science literature...

@Zee You are really asking the wrong crowd

Zee

Lol this crowd is exactly the crowd that needs it

That’s a cool article , thanks for posting that

Akiva Weinberger

Ultradark

I love drawing DNA

user193319

10:56 PM

Does anyone know of an online permutation calculator (permutations in the sense of $S_n$)? I found one a while ago (the website had the word "banana" in the title), but I can no longer find it.

Leaky Nun

@user193319 how about pari/gp

user193319

What is that?

Is it online?

Leaky Nun

it's a calculator / programming language with an extensive algebraic library

I don't know if there's an online demo

user193319

I just need to do something simple like $(12)(23)(12)(23)$.

Akiva Weinberger

${}=(23)(12)$