Mathematics - 2018-06-17 (page 1 of 4)

12:02 AM

ok thanks

hi

@LeakyNun are you comfortable with renormalization as used in physics for a very simple case, say $phi^4$ theory or the general idea around this in qft?

Leaky Nun

???

random ping?

Cows

renormalization group in physics

I pinged you :P

I know you are very capable and good at math things, I thought may be you'd studied something like this in the past, and had some intuition about this type of stuff

@LeakyNun any hints ?

Leaky Nun

I don't know what you're talking about at all

Cows

oh

It is some physics thing

Leaky Nun

12:16 AM

sorry i'm busy

Cows

ok

no worries

Mary Star

12:35 AM

Hello!!

I want to calculate the limit when $n\rightarrow \infty$of the successive approximation \begin{equation*}y_{n+1}(x)=1+\int_0^xty_n(t)\, dt\end{equation*} with $y_0(x)=1$, $x\in [-1,1]$.

We have that
\begin{align*}&y_0(x)=1 \\ &y_{1}(x)=1+\int_0^xty_0(t)\, dt=1+\int_0^xt\cdot 1\, dt=1+\int_0^xt\, dt=1+\frac{x^2}{2} \\ &y_{2}(x)=1+\int_0^xty_1(t)\, dt=1+\int_0^xt\left (1+\frac{t^2}{2}\right ) \, dt+\frac{x^2}{2}+\frac{x^4}{8}\end{align*} Do we have to find a general formula for $y_{n+1}(x)$ or how can we calculate the limit ?

Leaky Nun

@MatheinBoulomenos I find it useful at times to view UMPs as statements about Hom sets

Alex

1:22 AM

@LeakyNun UMP?

Leaky Nun

universal mapping properties

Alex

Gotcha

GFauxPas

good evening everyone

math.nyu.edu/student_resources/wwiki/index.php/… Does anyone see any benefit to this answer invoking the standard basis? I don't see where it's used at all

to me, the answer is, you multiply a vector you want to act upon on the left by $V^{-1}$ such that $A$ acts on the orthonormal vectors by $UV$. Then, because $V$ is an orthonormal matrix, $V^{-1} = V^\intercal$

Thus $A = UV^\intercal$

Adam

hello

GFauxPas

Hi Adam!

Adam

1:36 AM

Hi GFaux! :)

GFauxPas

Do you see any flaw with my way of answering the question I linked to?

Adam

I haven't learned any Linear Algebra yet, sorry.

GFauxPas

oh okay

Adam

Are we allowed to joke here or is this for serious stuff only ?

GFauxPas

we'll make an exception for you. no one is allowed to joke except for you. lucky!

Adam

1:42 AM

Wow that's neat!

GFauxPas

Also, everyone else is allowed to joke also!

Adam

Yeah I realized that after looking at the starred messages

lol

does latex work here ?

$e^{i\pi\cdot 0}=1$

do I have to install something for it to work

oh hey it works now:)

Xander Henderson

2:00 AM

@MatheinBoulomenos I thought up another one: Yo mama so ugly, she scares the monster group.

Oh, hey... that is a strangely appropriate time for me to seagull into the conversation. Neat.

Adam

That is indeed funny

Xander Henderson

ARG!!! Stupid clickbaity titles that turn out to be shudder category theory. YOU TRICKED ME AGAIN, BAEZ!

Silent

Why is function $f:\Bbb (0,2)\to\Bbb R$ defined as $f(x)=x^2$ if $x$ is rational and $x\in(0,2)$, and $f(x)=2x-1$ otherwise; not differentiable at $x=1$?

Daminark

2:17 AM

Hmm, let's consider the difference quotient $\frac{f(x) - 1}{x-1}$. If $x\ne 1$ is rational, this is $\frac{x^2 - 1}{x-1} = x+1$. If $x$ is irrational, this is $\frac{2x-2}{x-1} = 2$

Hmm, so if you have a sequence $x_n \to 1$, you want to say $y_n = f(x_n) \to 2$. Take a subsequence $y_{n_k}$, then it has a subsequence which converges to $2$, namely take either the rational or the irrational sequence

I might be doing something wrong here but the function feels differentiable

I concur

Hello

@Daminark Thank you!

2:31 AM

No problem!

Silent

3:27 AM

Let $T$ linear transformation on $\Bbb R^3$ defined by $T(x_1,x_2,x_3)=(3x_1,x_1-x_2,2x_1+x_2+x_3)$. How to conclude $T^{-1}(x_1,x_2,x_3)=(\frac{x_1}3,\frac{x_1}3-x_2,-x_1+x_2+x_3)$, without taking inverse of matrix?

Akiva Weinberger

3:38 AM

Write it as a concatenation of simpler transformations

Leaky Nun

or just solve the equation directly

Akiva Weinberger

Right, yeah, $y_1=3x_1$, $y_2=x_1-x_2$, $y_3=2x_1+x_2+x_3$, solve for the $x$s

Balarka Sen

4:21 AM

\o @loch @Alessandro

Secret

@famesyasd yeah, your proof makes me curious about the reverse implication, which obviously fails to hold in general because nonassociative algebra is its own domain of study. I have not thought on how exactly that failed however

Balarka Sen

6:11 AM

COB (Alexi Laiho) cover Vivaldi's Four Seasons Summer Theme

►

Wait wait wait woah what

Alessandro Codenotti

6:40 AM

@BalarkaSen Hi @Balarka

AbhigyanC

6:53 AM

Is it always true that |a|*|b|=|ab|?

math.stackexchange.com/questions/633642/… No prob, thanks I got my answer...

jcora

7:30 AM

fellas need some help with another proof, it's not in english but I think it's legible:

what I had in mind was using the comparison test so that a(n) < 1/(n^p) where p>1, and then multiplying that inequality with n, gaining n*a(n) < 1/n^(p-1), where p-1>0, and then the limit would be 0

but I don't know whether that's correct it definitely sounds too informal sort of

Daminark

7:55 AM

It's that time of day again

Commentary!

(Really I'm just working through shit and need to talk out loud in a place where there's TeX so because doing it in the air is hard sometimes)

So let's say $G$ is a finite group and you have an exact sequence $0\to P\to M \to N\to 0$ of $G$-modules, where the first map is $\phi$ and the second is $\psi$

If $p\in P^G$, then $\phi(p)^g = \phi(p^g) = \phi(p)$, so $\phi(p) \in M^G$, and similarly for $\psi$. Obv $\phi$ is still injective on $P^G$, and the sequence is gonna be exact at $M^G$ because if $m\in M^G$ and $psi(m) = 0$, then $m = \phi(p)$ for some $p\in P$, but we see that $\phi(p^g) = m^g = m$ for all $g$, so by infectivity $p^g = p$ for all $g$, so $p\in P^G$

So this gives an exact sequence $0\to P^G \to M^G \to N^G$

Want to find an example where the last guy isn't surjective

Let's see if there's an easy one, $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$

And let's say $G = \pm 1$ acts on $\mathbb{Z}/4\mathbb{Z}$ where 1 acts trivially and $-1$ sends $1\to 3$. Then it sends $2 \to 2$, $3\to 1$, and $0\to 0$

So now $\mathbb{Z}/4\mathbb{Z}^G = \{0,2\}$, and the image of that in the quotient is $0$, but the induced action of $G$ on the quotient is trivial, so yeah ripperino

Hey @mercio!

mercio

8:17 AM

hi

Daminark

How's it going?

mercio

not too well

Daminark

Darn, what's wrong? :(

mercio

is it legal for a homeschooled kid in new york to have 0 contact with any school person for years

Daminark

School person meaning what?

mercio

8:21 AM

also your exact sequence is the start of a long exact sequence where the next term is $H^1(G,P)$

Daminark

Other kids?

mercio

meaning school intendant

so you would like the connecting map there to be nonzero if you want third induced map to not be surjective

Daminark

Hmm, I'm not aware of the law on this myself, a brief search + a naive guess on my part suggests it's more likely legal than not?

And yeah the next part here is building up $H^1$

So let's say $\xi$ and $\eta$ are 1-cocycles. Then $(\xi + \eta)(\sigma\tau) = \xi(\sigma\tau) + \eta(\sigma\tau) = \xi(\sigma)^{\tau} + \xi(\tau) + \eta(\sigma)^{\tau} + \eta(\tau) = (\xi + \eta)(\sigma)^{\tau} + (\xi + \eta)(\tau)$. If $\zeta$ is a 1-coboundary, then $\zeta(\sigma) = m^{\sigma} - m$ for some $m$, so $\zeta(\sigma\tau) = m^{\sigma\tau} - m = (m^{\sigma} - m)^{\tau} + m^{\tau} - m = \zeta(\sigma)^{\tau} - \zeta(\tau)$

mercio

looks like you have a sign mismatch

Daminark

Hmm?

mercio

8:32 AM

your last step it should bet $+\zeta(\tau)$

Daminark

Oh tru

Thanks!

mercio

since $G$ is $\Bbb Z/2 \Bbb Z$, the cochain conditions are $\zeta(1) = \zeta(0) +\zeta(1) = \zeta(1)+\zeta(0)^1, \zeta(0) = 2\zeta(0) = \zeta(1) + \zeta(1)^1$ ?

Daminark

These are just maps $G\to M$, not required to be homomorphisms

For reference, this is what I'm working with

mercio

the cocycles*

jcora

chat.stackexchange.com/transcript/message/45197042#45197042 - anyone?

Daminark

8:39 AM

Yeah I figured that's what you meant, the point is I'm not sure if the cocycle condition implies that the map is a homomorphism still, and it's not specified

mercio

so it looks like $\zeta(0) = 0$ and then $\zeta(1)^1 = - \zeta(1)$

and it's a coboundary if $\zeta(1) = m^1 - m$

also your group acts trivially on $P$

so I think the $H^1(G,P)$ should be $\Bbb Z/2\Bbb Z$

Daminark

Yeah it seems in general that if $G$ acts trivially on $P$, then $H^0(G,P) = P$ and $H^1(G,P) = Hom(G,P)$

Since coboundaries are $0$ and the cocycle condition becomes just that it's a group homomorphism

And this agrees with what you say

Hey @loch!

Hi @Dami @mercio

hi

Hey @Alessandro!

8:49 AM

@Daminark hey

Daminark

So now time to work through this proposition

How's it going for you two?

Alessandro Codenotti

"finite groups are the only groups" -Daminark, 2018

2

Daminark

I'm a group-ultrafinitist

Okay so first thing is we want to say that $\xi$ is well-defined. So if $\psi(m) = \psi(m') = n$, then $m^{\sigma} - m = m'^{\sigma} - m'$.

Alessandro Codenotti

I was looking at old questions I asked and noticed your comment here @mercio, I'm about an year late, but I don't see how to get those counterexamples (I'm also not sure about the precise statement of said conjecture because googling doesn't bring up much)

Drew Brady

We say a matrix is positive definite if its eigenvalues are positive

(assume real, symmetric)

can someone help me get a counter example to the statement : if A - B is positive definite then lambda_i(A) >= lambda_i(B) ?

we assume that eigenvalues are in sorted order

you can't get a counterexample if A and B commute because then this inequality is true

since lambda_i(A - B) >=0, and lambda_i(A- B) = lambda_i(A) - lambda_i(B) when these matrices commute.

so the counterexamples must be some non-commuting matrices

ÍgjøgnumMeg

8:57 AM

Morning @Daminark @Alessandro

Alessandro Codenotti

Hi @ÍgjøgnumMeg

Daminark

Hey @ÍgjøgnumMeg!

loch

@LeakyNun rep theory = studying how groups act on vector spaces. Since we know that group actions are important to studying groups I think it's not hard to convince someone that this is a good idea (now with linear algebra at our disposal!)

Maybe you're looking for applications - then eg you have character tables for finite groups / I think galois reps can be used to prove fermats two square etc

But also these are just things that show up in nature

mercio

@AlessandroCodenotti suppose your prime count is bounded by $n$, then if you take the first $n+1$ terms in your translated sequences, that would be a constellation, and unless there is a trivial objection to it (i probably haven't checked this), it is expected that there are infinitely many primes shaped in that constellation, and this would give a translated sequence with $n+1$ primes

loch

@Daminark not bad :)

Daminark

9:02 AM

Oh hmm maybe you don't need $\xi$ to be well-defined, but that two different $\xi$s have the same cohomology class. Because it's very much looking like $\xi$ depends on the choice of $m$ here

mercio

yes

Alessandro Codenotti

@mercio Hmmm, I see, thanks

mercio

Twin prime

A twin prime is a prime number that is either 2 less or 2 more than another prime number—for example, either member of the twin prime pair (41, 43). In other words, a twin prime is a prime that has a prime gap of two. Sometimes the term twin prime is used for a pair of twin primes; an alternative name for this is prime twin or prime pair. Twin primes become increasingly rare as one examines larger ranges, in keeping with the general tendency of gaps between adjacent primes to become larger as the numbers themselves get larger. However, it is unknown if there are infinitely many twin primes or if...

aw

I wanted to link to the section about the first HL conjecture

Daminark

Okay so then first thing is to verify that $\xi(\sigma) \in P$, so $\psi(\xi(\sigma)) = \psi(m^{\sigma} - m) = n^{\sigma} - n = 0$ since $n\in N^G$, so that's done. And since it's of the form of a 1-coboundary for $M$ it should satisfy the cocycle condition by what happened above, meaning it's a cocycle in $P$, so $\xi\in Z^1(G,P)$

Alessandro Codenotti

@mercio It does link to that section!

mercio

9:07 AM

but the preview isn't the preview of that section

Daminark

So now if $\xi$ is given by choosing $m$ and $\xi'$ is given by choosing $m'$, then $\xi-\xi'(\sigma) = m^{\sigma} - m - m'^{\sigma} + m'$

Alessandro Codenotti

yeah that's a weird beahaviour

Daminark

Which is $(m-m')^{\sigma} - (m-m')$. Now if $\psi(m) = \psi(m') = n$, then $\psi(m-m') = 0$, so $m-m'\in P$

mercio

yes

Daminark

Meaning $\xi-\xi'$ is a coboundary and the cohomology classes check out

mercio

9:10 AM

sometimes I answer things here and at the end I think "wait that construction felt just like the construcyion of the connecting homomo... oh yeah the whole thing is cohomology in disguise"

Daminark

So now time for exactness

Yeah I've kinda seen a similar idea because some time recently I worked through a proof that if you have a long exact sequence of cochain complexes (or as the source called them, "differential complexes", since they were focusing on De Rham), then this induces an LES of cohomology groups

Sorry I meant SES of cochain complexes

I guess I'm kinda redoing it in this context because we're only working with $H^0$ and $H^1$, and defining this out of an SES of groups instead of an SES of cochain complexes, also it's good practice

Though verifying exactness at this stage is probably gonna be completely identical now that the well-definedness step also looked the same

Maybe given that fact I'll call it a night since it's past 4AM

Actually one thing I will ask, here they're calling an abelian group $M$ on which some finite group $G$ acts such that $(m+n)^g = m^g + n^g$ a "G-module". Is there gonna be some canonically associated ring such that $M$ is a module module over that ring? Or is just kinda analogous?

mercio

$\Bbb Z[G]$

Daminark

Oh tru

Well, thank you for your help, and have a good night!

mercio

sleep well

ÍgjøgnumMeg

The product topology just has as a basis $\prod_{i \in I} U_i$ with each of the $U_i$ an open subset of $X_i$ for each $i \in I$, right?

($X_i$ top. spaces)

I think the restricted product topology is just the same but you take an open subset $Y_i \subset X_i$ and you have a basis $\prod_{i \in I} U_i$ with $U_i \subset X_i$ open subsets and $U_i = Y_i$ for all but finitely many $i \in I$

mercio

9:25 AM

in the product topology you need $U_i = X_i$ for all but finitely many $i$

ÍgjøgnumMeg

Ah okay

so this is the same but restricting to an open subset of $X_i$

Alessandro Codenotti

Equivalently the product topology on $\prod_{i\in I} U_i$ is the coarsest wrt which all of the projections $\pi_i\colon\prod U_i\to U_i$ are continuous

mercio

yep

Drew Brady

9:46 AM

I'm struggling to find a nice counter example to this statement: Suppose A and B are symmetric matrices and that A - B > 0 (i.e, A - B is positive definite), then lambda_i(A) > lambda_i(B) for all i.

I know that the counter example will come from A, B which are not commuting (else they're simultaneously diagonalizable, and this is true)

sfz

11:46 AM

hi there

Wouter M.

11:59 AM

hi @sfz, just looking for a place to discuss some generalities. Maybe this isn't where I should be.

Poline Sandra

hello

need help for this : math.stackexchange.com/questions/2821953/…

Wouter M.

hi Poline, that's not my cup of tea I'm afraid ;-)

Leaky Nun

@WouterM. you're in the right place

Secret

12:38 PM

[Random]

There are known knowns, unknown knowns, known unknowns and unknonwn unknowns

The known knowns are those you know that you know
The known unknowns are those you know that you don't know (unknown)
The unknown known are those you don't know that you know (forgotten knowledge, or known things with uncertainties)
The unknown unknown are those you don't know that you don't know (perfect surprise)

But it can go further:

13

Q: Can unprovability unprovable? Is there an $\omega$-fold unprovability?

I was just thinking about unprovability. I just wanted to know if it is possible to make a concrete boundary between provable problems and unprovable problems in a certain axiomatic system. We know that there is a statement that is true yet unprovable. Then is it possible that a statement is tr...

There are unknown to the power n and the first limit are countably unknown

The ultimate limit is the unknowable: Those that you know that you can never know

But we can generalise this further:

To forget is you forget

GFauxPas

just use transfinite induction

Secret

To obliviate is you have forgotten that you forgot

To overlook is you don't know that you forgot

And we can use explosive generalisation to go even further

Let A be a description

Then You A

GFauxPas

there is at least one way you can forget something, for example, I forgot my childhood email

Secret

Then define $A^2$ be You A that you A, and so on

GFauxPas

define the initial segment $S_a$ as forgetting how to forget $a$ but perhaps not forgetting $a$ itself

dot dot dot induction

Secret

12:44 PM

yeah

But I wonder, can we go even further than that. Attempt to do so yields the following:

GFauxPas

:O this is intense

Secret

The collection S under explosive generalisation transfinitely many times is complete

Interestingly, this is actually false

Because if it were true, we will have a problem:

Gödel's incompleteness theorems

Gödel's incompleteness theorems are two theorems of mathematical logic that demonstrate the inherent limitations of every formal axiomatic system containing basic arithmetic. These results, published by Kurt Gödel in 1931, are important both in mathematical logic and in the philosophy of mathematics. The theorems are widely, but not universally, interpreted as showing that Hilbert's program to find a complete and consistent set of axioms for all mathematics is impossible. The first incompleteness theorem states that no consistent system of axioms whose theorems can be listed by an effective procedure...

You cannot prove that A within A itself, so to speak

Thus the conclusion of this winding series of message is: (At least to our limitation as human beings) at least one unknowable exists

> and this is the answer to life, the universe and everything known to humans

:P

GFauxPas

good job

Secret

Now here's something more obvious, but otherwise fascinating at least to me:

There exists way to turn unknown unknown into known unknowns and known knowns

One of the ways is called experiment

GFauxPas

the problem with learning order theory is that sometimes you have to write things down like $\prec \, = \, <$

Secret

12:52 PM

Why is it fascinating: Because when you look at we we know about the concepts of unknown, you will not normally expect they have a relation. Thus the fact such relationship exists tells us there is something very special about the deep structure of real life

Secret

I tend to use ≺ for preorders and reserve < for linear orders and partial orders

GFauxPas

context: I'm trying to form equivalence classes of well ordered subsets of $\mathbb N$

I know that for every $A \in \mathcal P(\mathbb N)$, there is at least one pair $(A,\prec)$, taking $\prec \, = \, <$

maybe I should just write it out in words

"there is at least one well-order on each $A$, namely, the usual ordering on natural numbers"

less confusing I think

Secret

yeah it does

GFauxPas

it is you mean

k

proofwiki.org/wiki/User:GFauxPas/Sandbox trying to get an intuition onto this exercise in Munkres and I'm not sure how to interpret this theorem

recursive definition on well-ordered sets

Secret

so I am guessing $h$ induces a well ordering on $C$

GFauxPas

1:02 PM

en.wikipedia.org/wiki/Recursive_definition okay WP has the case for $\mathbb N$

let's see

WP: $A$ is a set, $a_0 \in A$, $f$ maps non-empty sections of positive integers into $A$

there exists a unique $h: \mathbb N_{>0} \to A$ with

forgot to define $\rho$ any function that sends an arbitrary $f$ to an element in $A$

$A$ a set, $a_0 \in A$, $f$ a function that maps non-empty sections of positive integers into $A$,$\mathcal F$ the collection of all such $f$s, $\rho: \mathcal F \to A$

there is a unique $h$ satisfying

$h: \mathbb Z_+ \to A$

$h(1) = a_0$

$h(i) = \rho(h \vert 1,2,\cdots,i-1)$ for $i > 1$

that's WP's version of Munkres's theorem for recursion on segments of $\mathbb N$. let's see

okay here's an example in Munkres for the version for segments of $\mathbb N$

$C \subseteq \mathbb Z_+$, $a_0$ smallest element of $C$, define:

oh it's too long to type the whole thing into here

Tug' Tegin

Hello! The picture shows an integral problem with the integral sign missing. Could some one explain to me how did '-' sign and an expression $\frac{1}{|x|}$ originate at the end?

Anonymous

Good morning/Good afternoon/Good evening, could someone help me with this simple problem? I am unable to find a flaw in my solution. Thankyou :)

Anonymous

1:17 PM

0

Q: Flaw in evaluating limit $\lim_{x\to \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x$

There was this question asked yesterday here: Link: Evaluating limit $\lim_{x\to \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x$ Consider $P(x)= ax^2+bx+c$ where $a,b,c \in \mathbb R$ and $P(2)=9$. Let $\alpha$ and $\beta$ be the roots of the equation $P(x)=0$. If $\alpha \to \infty$ an...

calculus algebra-precalculus limits functions limits-without-lhopital

GFauxPas

Tug it's integration by substitution but you have to think about it for a while

let $u = 1/\vert x \vert$, $du = -x^{-2} dx$, and $x^2 > 0$ so they drop the absolute value

Fargle

@Tug'Tegin Notice that $d\left(\frac{1}{|x|}\right) = d\left(\left|\frac{1}{x}\right|\right) = (\mathrm{sgn}\;x)\cdot\frac{-1}{x^2} dx$; that's where that step comes from.

GFauxPas

then $x^2$ in the denominator because $x^{-2}$

oh right forgot the $\operatorname{sgn}$ thanks Fargle

Fargle

The $\mathrm{sgn}$ being the derivative of the absolute value (technically it'd be $\mathrm{sgn}\;\frac{1}{x}$ but that's the same thing). That's just chain rule.

Silent

How do we get $\sum _{n=0}^{\infty \:}\:\frac{\left(n^2+3n+1\right)}{\left(n+2\right)!}=2$?

Mary Star

1:30 PM

Hello!! Does someone of you have an idea about my question that the image of each open subset is open: math.stackexchange.com/questions/2822587/… ?

Tug' Tegin

@GFauxPas @Fargle Thank you both very much! I am studying your answers.

Fargle

No problem

Silent

@Fargle, will u please look at my above question?

Fargle

Trying to work it out.

Wouter M.

When is a 'rare' mathematical object interesting? Take twin primes : I bet there is quite a bit of literature about them. Same with Latin Squares, Golomb rulers, Fermat primes etc. Say I come up with some new 'rare' object, what would make it 'interesting'?

ÍgjøgnumMeg

1:40 PM

Interesting is pretty subjective

Wouter M.

The safe bet is not in the rare object itself, but in its connection to a wider set of math relations.

Semiclassical

There’s also the degree to which the infinitude of twin primes has resisted proof or disproof.

Especially as compared with how elementary a claim it seems

Wouter M.

Example: there is a unique free necklace of 2 colours (0 and 1) with q beads where q=2p+1 with q and p prime, and p of them white. Would anyone care to collect them and search for 'the system' in this 'madness'?

Semiclassical

I wouldn’t, but a combinatorist might

Especially if there’s a bijection between such necklaces and another mathematical problem

Wouter M.

@Semiclassical: right. It's all in the interconnections. But isn't a true 'randomness' very unlikely in any well defined set of unique objects (function of that prime p)

Semiclassical

1:55 PM

Couldn’t tell you.

Something something finite fields, maybe

Wouter M.

funny you say that! it IS connected to Algebraic Number fields (via minimal polynomial)

Semiclassical

Sure.

Silent

2:29 PM

@AkivaWeinberger, will you please look at this:

59 mins ago, by Silent

How do we get $\sum _{n=0}^{\infty \:}\:\frac{\left(n^2+3n+1\right)}{\left(n+2\right)!}=2$?

Akiva Weinberger

$(n+2)(n+1)=n^2+3n+2$, right?

So $\dfrac{n^2+3n+2}{(n+2)!}=\dfrac1{n!}$

So $\dfrac{n^2+3n+1}{(n+2)!}=\dfrac1{n!}-\dfrac1{(n+2)!}$, right?

And so it's a telescoping series essentially

Everything cancels but $\dfrac1{0!}$ and $\dfrac1{1!}$

@Silent

Silent

@AkivaWeinberger Wow! Thank you very much!

Anonymous

2:45 PM

Do we always use two planes for representing functions in complex numbers? I mean, isn't that uncomfortable?

Anonymous

(I am just a lily pily beginner)

Fargle

@IceInkberry There are other ways to go about it--e.g. making a color wheel graph, where you draw points in the output planes with colors corresponding to the input points that land there (usually hue corresponds to angle, saturation corresponds to distance from 0). But the complex numbers basically are a plane, and so we do the two planes thing.

It's uncomfortable in the sense that we'd like to just be able to plot the function in 4D and be able to visualize it, but we can't, so we just kind of have to live with it.

Anonymous

I get it. We have to live with it. I wonder what having another dimension would seem like.

Fargle

Maybe geometry would be even more fun.

Plato would have six "solids" to his name instead of 5.

Secret

Domain coloring is actually rich enough to be not hard to read, so complex numbers are relatively tame compared to visualising generic 4D objects

Fargle

2:58 PM

I don't disagree

Akiva Weinberger

You know the usual proof of the five (3D) Platonic solids?

Where it's decided that there are no solids with, say, three hexagons to a vertex, since three hexagons would lie flat

Well, in hyperbolic space ($\Bbb H^3$), three hexagons do not lie flat.

So what happens? (I know the answer)

Fargle

I don't know

zany stuff I imagine

Secret

Hint: His avatar

Fargle

:|

nah that's pretty sweet though

Akiva Weinberger

You know what, I forgot that was there

But yeah, it's infinite-sided and looks like that^

Fargle

3:12 PM

now that's just friggin sweet

Akiva Weinberger

and if you do if right, so that the dihedral angle is 120 degrees, you can tile hyperbolic space with them.

Fargle

Hyperbolic space just seems goofy.

Secret

How round are these shapes, or they only look round in poncaire disk projections?

Akiva Weinberger

The vertices lie on a "horosphere", which is intrinsically the same as a Euclidean plane

Since the horosphere is not a geodesic plane, the edges lift off of the horosphere

Also, "similarity" is not a thing in hyperbolic space, which is why the dihedral angle can change (depending on the size of the hexagons).

Nick

3:30 PM

Hi, does anyone know the general equation for the roots of $a_0 x^{n} + a_1 x^{n-1} + \dots + a_n = 0$. As in what is x=?

Balarka Sen

There is no satisfactory answer to that question for $n \geq 5$.

Abel–Ruffini theorem

In algebra, the Abel–Ruffini theorem (also known as Abel's impossibility theorem) states that there is no algebraic solution—that is, solution in radicals—to the general polynomial equations of degree five or higher with arbitrary coefficients. The theorem is named after Paolo Ruffini, who made an incomplete proof in 1799, and Niels Henrik Abel, who provided a proof in 1824. == Interpretation == The theorem does not assert that some higher-degree polynomial equations have no solution. In fact, the opposite is true: every non-constant polynomial equation in one unknown, with real or comple...

Akiva Weinberger

Not in terms of radicals (square roots, cube roots, etc), anyway

Radicals combined with addition, subtraction, multiplication, division, and exponentiation

Eric Silva

3:45 PM

did someone say radical

Akiva Weinberger

@Nick Some degree 5 equations, like $x^5-2=0$, can be solved in terms of radicals, but others, like $x^5-x+1=0$, can't

@EricSilva No.

Eric Silva

:(

Fargle

Ironic that, as a revolutionary, Galois had a hand in showing the non-existence of certain radical solutions.

ÍgjøgnumMeg

lol

Eric Silva

@Fargle i cant believe u just ruined galois for me

user193319

3:48 PM

Why is the map $\Bbb{Z}_2 \times \Bbb{Z}_2 \to \Bbb{Z}_2$ defined by $(a,b) \mapsto ab$ being nonzero and linear in both $a$ and $b$ imply that the simple tensor $1 \otimes 1$ is not zero?

Balarka Sen

> going to die tomorrow in a duel for a woman
> sleeping with another woman the night before
> actually writing the thesis for starting an entire branch of mathematics instead

That's what I call a revolutionary

Poline Sandra

@AkivaWeinberger hello, i think that you can help me on topology math.stackexchange.com/questions/2821953/…

please

Fargle

@BalarkaSen Galois was an absolute unit

loch

@user193319 because (1,1) is not mapped to 0, and this is the image of $1 \otimes 1$ by the universal property

user193319

@loch Ah, okay. Thanks!

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$Mathematics$ Mathematics