Mathematics

12:36 AM

anyone any good at graph theory?

@Abcd i have never seen a good high school math textbook pretty sure its a does not exist thing.

1:49 AM

@Faust nodes, edges, directed lines, paths, circuit, hamiltonian path, 3 houses problem, konisberg bridge, traveling saleman, tours, knight's tour, connected graphs, weighted binary trees, rings... whatever it is, shoot your question.

Faust

0

Q: Cycle edges cut of a 3 regular graph

Denote $\lambda_c(G) $to be the smallest integer k (if it exists) such that there exists a set Y with k edges such that $G-Y$ is disconnected and each of its components has a cycle. let $G \notin \{K_4,K_{3,3}\}$ be a connected cubic graph such that $|Y|=\lambda_c(G)$ and $G-Y$ is disconnected an...

graph-theory

fucked up 3 regular edge cut of cycles

i think im close with the ideas i have but cant quite finish it

its driving me nutz i found an exceedinly complicated solution of 2 switchs

to reduce to the case wher ei have 1 2 or 3 edges of diffrent colours in Y but i want a diffrent solution

@Nick

Faust

3:21 AM

woot woot got the Socratic badge

user 2646

congrats

Faust

first gold badge actually proud of

and its something most people will never get cause they know things, lol

user 2646

> knowing is half the battle

Faust

i ask alot of questions when i am learning something but i usually dont forget it

user 2646

the answers to your questions are memory hooks

2

mercio

7:27 AM

ugh I caught a cold and I didn't sleep

user 2646

i guess, trying to sleep it off is not an option?

mercio

but I want to do things

user1732

the story behind "the royal road"

twitter.com/mathematicalsvs/status/1042220572292243456?s=21

Rudi_Birnbaum

@mercio Gute Besserung!

mercio

goedemorgen

Leaky Nun

7:40 AM

@mercio goe'e morge'

Rudi_Birnbaum

@Mercio I have some metal picture about those $E_i$ reps all gving $Alt^2(E_i)=Rotz$.

mercio

metals have lots of moving electrons

Rudi_Birnbaum

Its like all consist of two vectors from the same plane, but with different angles.

mercio

my guess is that faithful representatoins into GL(2) * GL(1) is just too restricting

Rudi_Birnbaum

the angles correlate somehow with the divisors of the order of the highest rotation in the elements

In chemistry we see these IRREPs all the time in the symmetry of the orbitals

mercio

7:43 AM

that's because

of the molien series

Rudi_Birnbaum

In the solid state the $i$ from $E_i$ gets a continuous parameter. Because

you can imagine a chain as an infinite circle

mercio

do you have a nice description of all the Ei of those families of groups ?

Rudi_Birnbaum

For the solid its old and well known; for point groups I would not be aware of such a thing, but that doesn't mean there are no people who do.

In the solid the parameters are called "reciprocal lattice vectors"

vectors since the solid is 3D periodic

but in our infinite circle analogy it would be just a scalar.

actually they are not really "continuous" I guess but I don't know.

Strictly they are in $\Bbb Q$, I guess.

The reciprocal lattice is a kind of FT of the chain of atoms.

(and in my mental picture the infinite periodic chain is like a circlular pattern with circle radius = $\infty$)

here I have drawn three such $E$-orbitals

ignore the straight lines in $E_3$

mercio

it looks like it's parametrized by N instead of Q

Rudi_Birnbaum

@mercio maybe in 1D its N and in 2D it gets Q ?

mercio

7:58 AM

is this 1D ?

Rudi_Birnbaum

yes

the infinite circle is the picture for a chain

but we here stop at $i=3$

mercio

what is the circle equivalent for 2D ?

Rudi_Birnbaum

So that really would be irreps of $C_7$ or something

a torus

mercio

mhm

Rudi_Birnbaum

gernot-katzers-spice-pages.com/character_tables/C7.html

Leaky Nun

10:35 AM

@MatheinBoulomenos hab ich ein dank beweis, das Splitkorper sind normale

frage mir wann du hier seist

actually I’ll just type it now

ÍgjøgnumMeg

@Leaky ich habe einen DaNkEn Beweis, dass Zerfaellungskoerper normale Koerpererweiterungen sind

Leaky Nun

danke

Alessandro Codenotti

You have a thankful proof?

ÍgjøgnumMeg

@Alessandro na, dank

as in

DaNk

hahaha

Leaky Nun

that’s right :p

let F be the base field, and let f, g in F[X] irreducible. Let K1 be a splitting field for f; let K2 similarly for g with roots b1 ... bn. Let b1 in K1, and we need to show that the other roots are also in K1. For each root bj of g, create K2j a copy of K2. We now have the following maps: phi:F(b1)->K1 and psi_j:F(b1)->K2j where psi_j(b1)=bj.

consider the big ring K1 (x) K2_1 (x) ... (x) K2_n where all tensors are based with F(b1). quotient by a maximal ideal to make it field. We treat this field as a common F(b1)-extension of K1, K2_1, ... K2_n

Let B be the big field, and we now have maps alpha : K1 -> B, beta_j : K2_j -> B

Leaky Nun

11:14 AM

being F(b1)-algebra maps, they satisfy alpha(b1) = beta_j(psi_j(b1)) = beta_j(bj)

Leaky Nun

11:25 AM

I’ll continue this later

Leaky Nun

11:53 AM

well since K1 is a splitting field and b1 in K1, then (beta_j o psi_j)(b1) is in alpha(K1), i.e. beta_j(bj) in alpha(K1), so I believe we’re done, lol

Perturbative

12:40 PM

Hey everyone

Jacob P.J

what is integral 1/3+ x^2

is it tan^-1 or log

Perturbative

Quick question, is the definition for the sign function for a permutation $\alpha \in S_n$, the following: If $\alpha = \beta_1\beta_2 \dots \beta_k$ is a representation of $\alpha$ in terms of disjoint cycles then we define $$\sigma(\alpha) = \left(\sum_{i=1}^k \text{length}(\beta_i)\right) - k$$ and $$\operatorname{sgn}(\alpha) = (-1)^{\sigma(\alpha)}$$

Or is $\sigma(\alpha)$ defined differently as $$\sigma(\alpha) = \left(\sum_{i=1}^k \text{length}(\beta_i) - k\right)$$

Julius

12:57 PM

Hi. Let $X$, $Y$, $Z$ be random variables with $0<c_1<Z<c_2<\infty$ a.s. I'm dealing with $|E[X\exp(i Y/Z)]|$, where $i^2=-1$. Is there some way to get a nontrivial upper bound for it that would still contain $X$ and $Y$ but not $Z$?

mercio

2:26 PM

@Julius like E[|X|] ?

user193319

Problem: Suppose that $V$ is a vector space and $S,T \in \mathcal{L}(V,V)$ are such that $S(V) \subseteq \ker T$. Prove that $(ST)^2 = 0$....By definition $T(V) \subseteq V$, so $S(T(V)) \subseteq S(V) \subseteq \ker T$, where the last set inclusion follows from the hypothesis. Thus we have $S(T(V)) \subseteq \ker T$, but taking the image of both sides under $T$ yields $T(S(T(V)) \subseteq T(\ker T) = \{0\}$, proving that $T(S(T(V))) = \{0\}$ and therefore $S(T(S(T(V)))) = \{0\}$.

This says that, for every $v \in V$, we have $0 = STSTv = (ST)^2v$ so that $(ST)^2=0$.

Does this sound right?

It seems that I was able to prove something stronger, namely that $TST=0$, which of of course implies $(ST)^2 =0$.

Do I have that terminology right? If proposition $p$ implies $q$, then we say $p$ is stronger than $q$, right?

Or do we say that $p$ is weaker? Anyone?

Secret

2:47 PM

math.ucdenver.edu/~langou/5718/nelsen/HW3.pdf

Actually, you don't need the T(V) step, all you need is showing TS=0 then when composing STST, the TS bit vanishes and the conclusion is concluded

So TST=0 is true because TS=0 is true

jdavison

Man, it'd be nice if this room rendered MathJax :^/

Secret

tinyurl.com/cfqcvpc

user193319

@Secret Ha, I actually just proved that. I was thinking to myself, I wonder if I can prove something weaker (stronger?) than $TST=0$.

Would you happen to know the correct terminology?

Secret

let me check...

jdavison

thank you secret!

user193319

2:51 PM

I can't remember if $p$ is said to be weaker or stronger than $q$.

Secret

15

Q: Precise definition of "weaker" and "stronger"?

If I say that $A$ is stronger than $B$, do I mean that $A \Rightarrow B$, or that $B \Rightarrow A$? (Or something else?) I feel like I have seen both usages in literature, which is confusing. Thoughts based on intuition: $A \Rightarrow B$ means $A$ is a special case of $B$ -- $B$ is more gen...

logic terminology definition

So, the thing that has more implication, and the converse is not true, then it is stronger

So $P \implies Q$ and $Q \not\implies P$ means $P$ is stronger

user193319

@Secret Ah, very nice! That's what I suspected. Thanks for the link

jdavison

@jaco

@JacobP.J it would be $ {x^3\over 3} + {1\over3}x + c $

Julius

4:14 PM

Hi. Suppose $a_{n,N} \geq 0$. Is there some relatively weak condition on $\sum_n a_{n,N}$ that would imply $a_{n,N}\to 0$ as $N\to\infty$ for each $n$? Of course $\lim_{N\to\infty}\sum_n a_{n,N}=0$ works but it's too strong. Is there anything weaker?

Poline Sandra

@LeakyNun bonsoir, s'il vous plait connaissez vous une suite de Cauchy dans $(\mathcal{C}(\mathbb{R},\mathbb{R}),||.||_{\infty})$ mais pas convergente

Leaky Nun

4:35 PM

@PolineSandra quel est le norm de $e^x$?

Poline Sandra

$+\infty$

Leaky Nun

ca c'est permet?

Poline Sandra

je cherche une suite qui est de cauchy

Ultradark

$K{_\phi}= \phi(s,t)\phi(1-s,t)=\zeta(t)^{log(s)}\zeta(t)^{log(1-s)}=\zeta(t)^{log(s)+log(1‌-s)}$

support for the solution space for $K_\phi=0?$

Poline Sandra

j'ai trouvé dans un livre que $(\mathcal{C}(\mathbb{R},\mathbb{R}),||.||_{\infty})$ est de banach. @LeakyNun

Leaky Nun

4:47 PM

ta livre permet que ||quelque chose|| = infinite?

Ultradark

$\zeta(t)$ is the riemann zeta function by the way

Poline Sandra

@LeakyNun non il dit rien il l'a juste utiliser dans un exercice

Leaky Nun

comment definit-il ta livre un norm?

Poline Sandra

positive, inégalité triangulaire, $N(\lambda x)=|\lambda| N(x)$, $N(x)=0\Longleftrightarrow x=0$

Alessandro Codenotti

$C(\Bbb R)$ with the $\|\cdot\|_{\infty}$ norm isn't even a normed vector spacce with the usual definitions @PolineSandra

Ultradark

4:56 PM

$K{_\phi}= \phi(s,t)\phi(1-s,t)=\zeta(t)^{log(s)}\zeta(t)^{log(1-s)}=\zeta(t)^{log(s)+log(1‌‌-s)}$. $s,t \in \Bbb C$

where does $K_\phi=0$ ?

specifically, where do the non trivial zeros live?

do they live along the vertical y-axis?

Poline Sandra

@AlessandroCodenotti where i can found the proof please

Alessandro Codenotti

If $f(x)=x$ for all $x$ what's $\|f\|_{\infty}$?

Poline Sandra

$\infty$, but i dont ses that the norme must ne finite

Secret

$$\sqrt{\frac{f+1}{f+2}}\lvert ⟲\rangle + \sqrt{\frac{1}{f+2}}\lvert ⟳\rangle$$

there's always something funny about seeing squareroot expressions like these

Ultradark

@Secret does my question make sense

Alessandro Codenotti

5:01 PM

@PolineSandra Norms are functions $V\to\Bbb R$ usually

Poline Sandra

yes

Secret

I am the wrong person to ask about anything riemannian zeta

Ultradark

okay but

Secret

I cannot even get my head around irrational proof of pi yet

Alessandro Codenotti

And there's no $\infty$ in $\Bbb R$

Poline Sandra

5:04 PM

@AlessandroCodenotti but one time i find an exercise of metric space for let E,d be a metric space where d is not bounded

Alessandro Codenotti

Sure, but not bounded and infinite are very different!

For example the usual absolute value on $\Bbb R$ isn't bounded, but there is no $x\in\Bbb R$ with $|x|=\infty$

Ultradark

I'm just trying to understand these solutions

I don't think one needs to necessarily be an expert on the riemann zeta function to answer this

just trying to understand the support of the solution space

Secret

I don't think you can say much about where in the real axis, other than in (-1,1) where the nontrivial zeros lie as shown in the 2nd and 4th line with a squareroot sign

the inequalities of $s$ meanwhile suggests the imaginary part lies some distance away from the real strip (-1,1)

Ultradark

5:21 PM

that's the solution space for only $\phi(s,t)=\zeta(t)^{log(s)}$

so I'm thinking if you do the following multiplication: $K{_\phi}= \phi(s,t)\phi(1-s,t)=\zeta(t)^{log(s)}\zeta(t)^{log(1-s)}=\zeta(t)^{log(s)+log(1‌‌-s)}$ the square roots might cancel or the resulting solution space might be nicer to work with

Poline Sandra

si de cant even find a Cauchy sequence which do not converge in this space @LeakyNun

@AlessandroCodenotti

Secret

how will $\zeta(t)^{\log (s(1-s))}$ be any simpler

Ultradark

yeah maybe not, it's more complicated probably

Poline Sandra

@AlessandroCodenotti de can't construct a cauchy sequençe

Secret

5:38 PM

brainly.in/question/1862483