Mathematics - 2018-07-15 (page 1 of 2)

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12:00 AM

Hello. Does anyone have a computer handy? I need to factorise 8wxyz+(w²yz+w²xz+w²xy+x²yz+x²wz+x²wy+y²xz+y²wz+y²wx+z²xy+z²wy+z²wx)+w²xyz+wx²yz+‌wxy²z+wxyz²+xyz+wyz+wxz+wxy=0 and Wolfram Alpha gives something daft. Please help :)

Good luck, @rschwieb :)

12:42 AM

$8wxyz+w^2(yz+xz+xy) +x²(yz+wz+wy) +y²(xz+wz+wx)+z²(xy+wy+wx)+wzxy(w+x+y+z)+w(xy+wy+wx)+wxy=0
$

3 hours later…

Jackozee Hakkiuz

3:24 AM

Hey.

Anyone who knows about connection regarded as Lie algbra valued 1-forms?

4:04 AM

Working on the command timeline feature of the assistive diagram feature (undo / redo ability)

It's not trivial

to code this part

4:15 AM

ZoomspaceScreenRecording1

2 hours later…

5:58 AM

Why I am not interested in $\gamma$ stuff (because I have yet to fully comprehend it), that question by mick sounds like a useful framework to explore all sophomore dreams expressions

1 hour later…

6:58 AM

@Shaun well, Z is UFD so Z[w,x,y,z] is also UFD. I believe it is also decidably a UFD.

but the decidable part is much harder, because it includes determining whether an arbitrarily large polynomial is reducible

7:19 AM

i think you can use SAGE to factor things, and it seems to suggest that that is irreducible (I'm slightly confused because for some reason SAGE only has .is_irreducible() for univariate polynomials, but .factor() works for multivariate polynomials lol)

@loch what do you think about the "hints" under the exercises

I think that's Atiyah saying "it's too hard, might as well just give you the answer"

i.e. it's actually another proposition

oh

yeah

lol

so I should read them?

i would

it's up to you lol you can think about the Q first too

and maybe you'll figure everything out

lol I probably won't

7:25 AM

anyway im off

ok

7:48 AM

user image

8:38 AM

36

Q: Examples of "miraculous" proofs

Concerning the proof that $\zeta(3)$ is irrational, Van der Poorten famously noted that "Apéry's incredible proof appears to be a mixture of miracles and mysteries". Indeed, many ideas introduced in Apéry's proof such as $\zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {\binom {2k}{...

emptyset - Collapse [Official HD]

2 hours later…

10:18 AM

4 hours til game time

In The Air Tonight - Phil Collins [Lyrics on screen]

for all my life

10:54 AM

34

Q: Has anyone ever drawn a football field in latex?

Has anyone ever drawn a football field in latex? I know that could seem like a strange request. But I'm looking for a package that allows me to draw a football field.

LegionMammal978

11:05 AM

So what is the difference between a polynomial in a field (i.e., $p\in\Bbb R[x,y,\dots]$) and a polynomial function (i.e., $p:\Bbb R^n\rightarrow\Bbb R$)?

@LegionMammal978 a polynomial is defined formally, i.e. the polynomial $X^2 + Y^2 - 1$ is just $X^2 + Y^2 - 1$, not a function

every point $p \in \Bbb R^n$ gives an evaluation homomorphism $\Bbb R[x_1, \cdots, x_n] \to \Bbb R$

it is a ring homomorphism

LegionMammal978

Ah, okay, that makes sense. I've never been able to understand all this field theory too well.

the difference is clearer if the field is finite, e.g. $\Bbb F_2$ the field of two elements, where $X^2 - X + 1$ and $1$ are two distinct polynomials, but the same function

11:35 AM

sucessive values in a sequence, each with a long dash followed by a comment about how it seems related to previous values in the sequence

1
2 — *1
2 — *1
4 — *2 or +p2
12 — *3
24 — *2
40 — +p3
80 — *2 or +p4
176
352 — *2
6721

+pn means the sum of the previous "n" values

*n means the previous value multiplied by "n"

this is reached as a sequence of digits in a calculation and is infinite, this is just the first few terms

other sequences generated in similar ways relate to pascals triangle (n choose k), exponentials, and the fibinarchi sequence, but this one seems odd

framed another way

1 — 0th power of 2
2 2 2 — 1st power of 2
4 — 2nd power of 2
8 — 3rd power of 2
12 — "3.5th" power of 2
16 — 4th power of 2
24 — "4.5th" power of 2
40 — don't you mean 48? +p2
64 — 6th power of 2
96 — "6.5th" power of 2
144 — 96 * 1.5
224 — factors are just 2s and a 7
etc… 1984 — no, 1024!

only between powers of two linearly, not 2^3.5 etc…

sorry, that last set is a similar one, not the same sequence, similarly confusing though

there made of adjoining "2s complement" form (but writtain here in base 10) numbers (like truncated p-adics) together of the same length and taking the reciprocal. Values are equally spaced in the resulting digits, segments can be made arbitrarily long for the same sequence to be able to continue for further.

in this case using square brackets to show how the calculation is constructed 1/[-2][1][-4] being 1/999800019996 and with longer segments for longer of the same sequence

they work the same way in any base

for ones that I can model as recurrence relations, summing all the previous terms seems occur very similarly to using just the last term, as does adding an additional 1 every N steps a adding aucessive points along a mod like sawtooth wave

12:42 PM

@MatheinBoulomenos can I just skip Exercise 4.23

MatheinBoulomenos

12:58 PM

:P

@MatheinBoulomenos 4real

MatheinBoulomenos

do what you want

the proofs are not that different

so you don't lose out by skipping them

brilliant

MatheinBoulomenos

I wonder why A-M didn't just prove it for modules in the first place

I've advanced more than I could in the previous few months

MatheinBoulomenos

1:00 PM

chapter 5 is really important for ANT

I think I'm more familiar with chapter 5 than chapter 4

somehow

MatheinBoulomenos

ah, you don't do everything in order?

deep down I've always known that x is integral in A iff A[x] is finite over A

I've never read chapter 5

MatheinBoulomenos

I liked the exercises in chapter 5

I'm still reading :)

I'm on Proposition 5.7 now

I think I proved 5.7 myself before

MatheinBoulomenos

1:04 PM

yeah it's pretty straightforward

user image

I already spotted a mistake lol

I type this on May 16

Does this always hold: Let $D_{\vec v}f(\vec a)$ be directional derivative of $f$ at $\vec a$ in the direction $\vec v$. Is this always true that: If $\vec v\dot \vec a=0$, then $D_{\vec v}f(\vec a)=0$?

MatheinBoulomenos

@LeakyNun what's a characteristic polynomial of an endomorphism of a module that is not free?

@MatheinBoulomenos :)

!enter image description here

1:16 PM

@MatheinBoulomenos let's list out rings that we know

(don't go to that site :P

where by "rings" I mean commutative unital rings

I realized that I don't really know that many rings

MatheinBoulomenos

you get to know a lot of rings when do ANT or classical Alg Geo

• Let A1,...,An be independent events each with the same probability of occurring: p = P(Ai), i = 1,...n. What is the probability that exactly m ≤ n events will occur?
A pm
B 1−􏰀n􏰁pn
m
C 􏰀n􏰁pm(1−p)n−m m
D 􏰀n􏰁pm m

@MatheinBoulomenos enlighten me

MatheinBoulomenos

you really want to list me all rings I know?

Could take a while.

1:19 PM

sure

Can someone help me with my question?

MatheinBoulomenos

$\Bbb Z/1\Bbb Z$, $\Bbb Z/2\Bbb Z$, $\Bbb Z/3\Bbb Z$, $\Bbb Z/4\Bbb Z$ ... am I doing this right?

@MatheinBoulomenos that's a class of rings, go on :)

the class $\Bbb Z/n\Bbb Z$

aka surjections from $\Bbb Z$

btw hail Urysohn

MatheinBoulomenos

really impotant class $\mathcal{O}_K$ where $K$ is a finite field extension of $\Bbb Q$ this is the integral closure of $\Bbb Z$ inside $K$

alright, go on

MatheinBoulomenos

1:22 PM

smooth irreducible affine curves over $\Bbb Z$

do the same thing with $\Bbb Q$ replaced by $\Bbb Q_p$ and $\Bbb Z$ by $\Bbb Z_p$

then you get the same class that you get if you start with some $\mathcal{O}_K$ for some $K$, pick a maximal ideal $\mathfrak{p}$, and take $\varprojlim \mathcal{O}_K/\mathfrak{p}^n$

ok, ring of integers of local/global fields, go on

MatheinBoulomenos

you can also do this for infinite extensions

in the local case you get non-discrete valuation rings that way (if you have infinite ramification)

right

add all the roots of the uniformizer

omega-bar

MatheinBoulomenos

if you take the integral closure of $\Bbb Z$ in $\overline{\mathbb Q}$, you get a Bezout domain (so all f.g. ideals are principal)

(that's non-trivial)

isn't that normal basis theorem

MatheinBoulomenos

1:27 PM

no

ok

MatheinBoulomenos

it's the finiteness of the class number

also the normal basis theorem doesn't exend to rings of integers

I don't really know what kinds of rings you want

rings that are used a lot or rings with some weird properties

consider $\mathbb{Q}[[x,y]]^{\Bbb N}$

that's a pretty weird ring

it's coherent, but the polynomial ring over it is not coherent

then you also have the adele ring

really weird non-Noetherian ring, but quite important

I'm not really sure what you want honestly

you can write down a lot of rings

just take any random ring, take the polynomial ring, quotient localize at some random stuff

you really like Z and Q

MatheinBoulomenos

yes

conjecture: all rings you know are constructed from Z using the operations that include but are not limited to: quotient, completion, localization, polynomial, direct product, coproduct, extension

MatheinBoulomenos

1:37 PM

lol

every ring is the quotient of a polynomial ring over Z

you only need two operations

come on

MatheinBoulomenos

what?

you know what I mean

Sanity check: is the proof that $fg$ is primitive iff $f,g$ are primitive as easy as it looks?

is that AM ex.1.2?

1:42 PM

Yeah.

I don't know what you mean by "as easy as it looks"

I can't read the solution off your mind

I'm typing it

give me a second

MatheinBoulomenos

Let $\Gamma \subset \mathrm{SL}_2(\Bbb Z)$ be a subgroup such that $\Gamma$ contains $\mathrm{ker}( \mathrm{SL}_2(\Bbb Z) \to \mathrm{SL}_2(\Bbb Z/N \Bbb Z))$ for some $N$. $\Gamma$ acts on $\Bbb H$ by Möbius transforms. Fix $k \in \Bbb N_0$, then let $M_k(\Gamma)$ be the set of holomorphic functions $f:\Bbb H \to \Bbb C$ such that: $f$ extends holomorphically onto $\Bbb H \cup \{\infty\} \cup \Bbb Q$.
For all $\begin{pmatrix} a & b \\ c & d\end{pmatrix} \in \Gamma$, we have $f(\frac{az+b}{cz+d})=(cz+d)^{k}f(z)$

27 mins ago, by Leaky Nun

where by "rings" I mean commutative unital rings

MatheinBoulomenos

this ring is commutative unital

1:45 PM

you win

MatheinBoulomenos

$1 \in M_0(\Gamma)$

but of course that's just a quotient of a polynomial ring over $\Bbb Z$

that's still very number-theoretic :P

@MatheinBoulomenos no, you took a direct sum

MatheinBoulomenos

?

hmm

nvm

MatheinBoulomenos

yeah I just take the multiplication as functions

I mean just any function $\Bbb H \to \Bbb C$ that is a finite sum of elements from $M_k(\Gamma)$s

it's actually true that $M_k(\Gamma) \cap M_l(\Gamma) = 0$ if $k \neq l$

so the additive group is a direct sum

1:50 PM

ok

If $fg$ is primitive, then some linear combination of the coefficients of $fg$ will be $1$, but then you have a linear combination of the coefficients of $f$ that gives $1$, and similarly for $g$. Conversely, if $f$ and $g$ are primitive, then no maximal ideal of $A$ can contain all the $a_i$, or all the $b_j$. So if $\mathfrak{m}$ is a maximal ideal of $A$, we have that $f$ and $g$ are non-zero in $(A/\mathfrak{m})[x]$, whence $fg$ is non-zero there too--thus $fg$ is primitive.

@MatheinBoulomenos so you're doing it in the wrong category and lifting it to the right one?

MatheinBoulomenos

@LeakyNun I don't understand

I'm just multiplying functions

which category are you taking direct sum in?

MatheinBoulomenos

it's an internal direct sum if you wish

1:52 PM

in which category?

MatheinBoulomenos

abelian groups or $\Bbb C$ vector spaces

@Fargle why is $fg$ primitive?

MatheinBoulomenos

I don't understand your confusion, have you seen graded rings before?

fair enough

I didn't recognize graded ring

I'm old

MatheinBoulomenos

I don't know the category-theoretic interpretation, but when you have a sequence of abelian groups $(A_n)_{n \in \Bbb N}$ and operations $A_n \times A_k \to A_{n+k}$ satisfying some obvious axioms, then you can make $\bigoplus A_n$ into a graded ring in a canonical way

1:54 PM

@MatheinBoulomenos do two series of groups have a common refinement?

@LeakyNun Because $fg$ is non-zero in all $(A/m)[x]$, so no maximal ideal can contain all coefficients of $fg$; therefore the coefficients of $fg$ generate $(1)$.

@Fargle fair enough

my answer is similar

user image

MatheinBoulomenos

Schreier refinement theorem

In mathematics, the Schreier refinement theorem of group theory states that any two subnormal series of subgroups of a given group have equivalent refinements, where two series are equivalent if there is a bijection between their factor groups that sends each factor group to an isomorphic one. The theorem is named after the Austrian mathematician Otto Schreier who proved it in 1928. It provides an elegant proof of the Jordan–Hölder theorem. It is often proved using the Zassenhaus lemma. == Example == Consider Z / (...

that's isomorphic refinements

MatheinBoulomenos

if you want common refinement, then obviously no

1:56 PM

how obvious is that

Cool. Just wanted to make sure I wasn't missing a detail.

MatheinBoulomenos

think about two different series for $\Bbb Z/2\Bbb Z \times \Bbb Z/3 \Bbb Z$

@MatheinBoulomenos I see

@MatheinBoulomenos I mean, there's common refinement in Lebesgue integral

why isn't there common refinement in group theory

@MatheinBoulomenos how'rt thou doing

thou'st eaten?

MatheinBoulomenos

2:14 PM

ah I'm busy

I need to go now

but the only ring I know is $\Bbb Z$ anyway

2:34 PM

Hey guys

user image

@MatheinBoulomenos I don't think this shows $bt \in C$

why not?

because you're arguing inside the localization

the homomorphism doesn't need to be injective

2:50 PM

oh hm

it's not injective when $S$ contains a zero-divisor right

3:26 PM

anyway i think your problem is fixed by multiplying by something in $S$, so you get $ubt\in C$ and you can deduce the same conclusion

@loch that isn't the point

the point is that the proof is wrong

er

sure lol

The final is not going to be 1-0 after all

Thank god

lol

I support neither team

a draw?

3:29 PM

Yeah both are kinda meh tbh

so 1-1 is good for me

well I just don't have a preference

But I do think France plays markedly worse

I guess I’m nominally in support of Croatia, since it’s a team from a smaller nation and therefore seems like an underdog. Can’t say I really care though

can we watch the match without support either of the teams?

Alessandro Codenotti

Hi @Balarka

3:34 PM

A friendly reminder to all chat inhabitants in maths: h bar is currently stuck in a toxic state. Do not enter unless otherwise specified

@Alessandro Hey

Disastercode: Penrosediagram.conformal.primenumber.obssession.et

looool penalty

stop making fun of me secret

ok fine I'll stop

But really, put some more effort on your questions, I have been reading the transcript and those conversations are not close to clarifying them

that's why many lost patience of you

chat.stackexchange.com/search?q=geocalc&room=36

3:43 PM

I just asked if penrose diagrams have been used as tensor networks

but yeah, I am not joking when I pinged @Slereah, I knew he knew about GR and penrose diagrams, though I don't know if he knew tensor networks

records suggests the literature of tensor network is very hard to read though...

Aug 23 '15 at 14:49, by Kevin Driscoll

Anyone know of a good reference for learning tensor networks?

4:11 PM

Subheim - Arktos

lelelel 3-1

4:37 PM

0

Q: A method to solve "non continuous differential equations"?

Question Given a "non continuous differential equation"(see example) is the below technique (see conjecture) known for "integration"? Is it possible to make any of this more rigorous? (for example what is the set of $f$ for which the conjecture holds true) Example Consider a non-continuous di...

real-analysis measure-theory mp.mathematical-physics stochastic-processes

@TheGreatDuck you might be interested in this

1 hour later…

5:54 PM

@Secret Hey! thanks for sharing a post of mine :D

Daniel Pietrobon

Hi all. Q about Shark's answer math.stackexchange.com/questions/2852712/… here. Can someone explain how the second equality is obtained?

I.e why -- $e^{ix} + 1 + e^{-ix} = e^{3ix/2} - e^{-3ix/2}/e^{ix/2}-$e^{-ix/2} ?

6:21 PM

@BalarkaSen say what ?

lol

well its true

their win against belgium was a bad fluke

even today i thought their game was mediocre, but so was croatia's

international football is ded m8

2 hours later…

8:33 PM

hey crowd. who's a differential geometer in the crowd?

i guess one person makes a crowd

two's a crowd

thou'rt a crowd

you know what? it's time i installed tex on this thing. brb

there.

Define $S \coloneq \Set{(x, y, z) \in \mathbb{R^3}}{x^2 + xy + y^2 + z^2 = 1}$.

What's the volume of this manifold?

\coloneq... more like :=

8:43 PM

hmmm

$S := \{ (x,y,z) \in \Bbb R^3 \mid x^2 + xy + y^2 + z^2 = 1 \}$

S := \{ (x,y,z) \in \Bbb R^3 \mid x^2 + xy + y^2 + z^2 = 1 \}

lets try this again

$S := \{ (x,y,z) \in \Bbb R^3 \mid x^2 + xy + y^2 + z^2 = 1 \}$

different standards may have different preferences for \Bbb R vs \mathbb R

\coloneq don't work

anyway this is a quadratic form that you can diagonalize

coloneq isn't even a command

8:45 PM

ok. whats special about quadratic forms?

they can be diagonalized

and turned into ellipsoids

who's volume is well known

?

right

ok. suppose i don't know quadratic-form theory

that means, express $x^2+xy+y^2+z^2$ as the sum of three squares

8:47 PM

i may have gotten one of the signs here wrong

ok

presumably you want $(x+y)^2 + z^2$ stuffs

right, those stuff

hast thou gotten it right?

i am missing an $xy$ term

I mean, the signs that thou saidest were wrong

orbit-stabilizer

8:53 PM

Question

Actually nvm

let me try something, ill get back to you

ok

MatheinBoulomenos

@LeakyNun why dost thou quethe as in the days of yore?

9:11 PM

@MatheinBoulomenos the glorious past, alas, is nevermore

quoth the raven

QUA

@MatheinBoulomenos would it be not more natural if thou say'st, why quethest thou as in the days of yore

MatheinBoulomenos

right

thou hast broken the iambic pentameter!

hello

someone speak French please

orbit-stabilizer

9:27 PM

bonjour

bonsoir ca va?

orbit-stabilizer

comme ci comme ca

et toi?

ca va merci

@Vrouvrou felicitations

MatheinBoulomenos

bonjour, je suis une baguette

Am I doing this right?

9:29 PM

yep, absolutely, thou'rt doing this right

orbit-stabilizer

oui

s'il vous plait savez vous traduire cette phrase :covering spaces that are path
connected est ce que c'est recouvrement d'espaces connexe par arcs ou bien espaces de recouvrement... ?

@orbit-stabilizer

recouvrement > revetement

orbit-stabilizer

je ne cais pas - my french is tres mal

mon francais est tres mal?

ma francais est tres mal

ok merci bien

9:32 PM

je ne sais pas, mon francais est tres mal

MatheinBoulomenos

@LeakyNun mein Französicsh ist dreimal?

orbit-stabilizer

guten morgen

@MatheinBoulomenos lol, drei (deutsch) = trois (franzosisch) = tres (spanisch)

MatheinBoulomenos

I understood lol, just making bad jokes

I heard someone say "Je suis une baguette" in a French bakery by the way

thou'st some pretty bad jokes ymakt

(Old English) macian / ġemacod

btw maken / machen (High German consonant shift)

@MatheinBoulomenos I think we should revive the thou/thee/ye/you distinctions

9:37 PM

i know some old english as well

10 bucks to anyone who can guess the word i am thinking of

MatheinBoulomenos

@LeakyNun Ay!

hwaet!

@MatheinBoulomenos du/dich/ihr/euch am I doing this right

@BalarkaSen is it a meme?

yes

whom'st

no fail

9:40 PM

whom'st'd've

i had cuckold in mind lol

I see

my memery likenth not to thine

an insult you might encounter in a bar in the middle ages

and in 2018

meme revival :clap: :clap:

orbit-stabilizer

35

A: Position of least significant bit that is set

Most modern architectures will have some instruction for finding the position of the lowest set bit, or the highest set bit, or counting the number of leading zeroes etc. If you have any one instruction of this class you can cheaply emulate the others. Take a moment to work through it on paper ...

I don't get the answer by Andrew Grant

oh got it

which bit?

orbit-stabilizer

9:43 PM

I don't know what magic is at play... but somehow, the answer just appears after posting

hah

@BalarkaSen stareth

orbit-stabilizer

It's like I'm trying to guard against being stupid, so I think extra hard

@MatheinBoulomenos art thou here?

MatheinBoulomenos

Forsooth, I am

user image

I present thee, more follies from Atiyah Macdonald

Alessandro Codenotti

9:51 PM

That looks more normal than a lot of stuff in AM to me

by "folly" I mean "error"

hath nobody found the error?

MatheinBoulomenos

The set could be empty

bingo

example: $K = \Bbb F_2$, $\Omega = \Bbb C$

MatheinBoulomenos

I noticed it immediately, but I couldn't find a way to say it in archaic english

lmao

maths in archaic english

MatheinBoulomenos

9:56 PM

@LeakyNun 'Tis a folly, but it hurteth not the overall argument

hurteth not

es schmerzt nicht?

MatheinBoulomenos

Kann "hurt" nicht auch "schaden" heißen?

jawohl

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