Mathematics - 2017-01-12 (page 1 of 4)

Akiva Weinberger

12:05 AM

Apparently:$$\sqrt{x^2+1-\frac1{16x^4+8x^2+1}}=x+\frac{2x}{1+4x^2}$$

I wonder if this can be generalized

Mike Miller

@OneRaynyDay I know some with veeeery different tastes than yours.

Akiva Weinberger

We also have:$$\sqrt{x^2+1+\frac1{4x^2}}=x+\frac1{2x}$$

So I guess the question is, for what polynomials $P$ is $\sqrt{x^2+1+\frac1{P(x)}}$ a rational function.

Mike Miller

@Arrow That's definitely wrong. That would say that the fibers of a nowhere smooth homeomorphism $\Bbb R \to \Bbb R$ are smoothly varying. I disagree with that.

user97303

Hey, silly question, but if you have functions $u, v$ from/to $\mathbb{R}$ and you apply integration by parts $\int_a^b u(t) dv(t)$, do you have that $v(t) = \int_a^t dv(x) $?

user97303

Wait, nevermind

Akiva Weinberger

12:23 AM

Ah, OK, it does generalize. Pretty easily, in fact.

$P(x)=64x^6+64x^4+16x^2$ is the next one

The idea is that $\sqrt{x+1}-\sqrt x=\frac1{2\sqrt x+(\sqrt{x+1}-\sqrt x)}$, so you can use the continued fraction to approximate $\sqrt{x+1}-\sqrt x$.

Jessy Cat

12:45 AM

1

Q: Need help with change of variables in very strange heat equation problem

I'm asked to find the equilibrium solution of the heat equation $u_{t} = \nabla^{2} u$ in the ring $0< a<r<b$ if $u_{r}(a,t)=1$ and $u_{r}(b,t) + u(b,t) = 2$. Now, I know how to find equilibrium solutions of normal looking heat equations of the form $u_{t} = \alpha u_{rr}$, but $u_{t} = \nabla^{...

user228700

Hi, again.

user228700

Can anybody please help me to prove that if $f$ and $g$ are two functions which are inverse of each other, then $f^{'}[g(x)]=1/g^{'}(x)$).

user228700

$f^{'}(x)$ may be expressed as $$\lim_{h \to 0}\{f[g(h+x) - f[g(x)]\}/h$$

user228700

Since $f$ and $g$ are inverse of each other, we know that $f[g(x)]=x$.

user228700

...and I am looking to simplify the above expression to get $1/g^{'}(x) = h/\{g(h+x)-g(x)\}$

user228700

12:52 AM

Can anybody help me, please?

heather

1:14 AM

anyone around to help with conversion between coordinate systems and calculus and all that?

the hbar's empty.

arctic tern

@Kaumudi.H differentiating f(g(x))=x gives f'(g(x))g'(x)=1

user228700

@arctictern Ah, yes! Thanks!

heather

the problem is

in The h Bar, 10 mins ago, by heather

okay, so imagine that there is a field, and we are looking at a point $p$ in that field. we can use a coordinate system with a $x_1$ axis and a $x_2$ axis. there is also an altogether different coordinate system, with a $y_1$ axis and a $y_2$ axis. and then to find the change in height, $d\phi$, over that field, I have the equation $d\phi=\sum\limits_n\frac{\partial\phi}{\partial x^n}\,dx^n$. if i am trying to go from $\frac{\partial\phi}{\partial x^n}$ to $\frac{\partial\phi}{\partial y^n}$

user228700

@heather Over my head, what you've written over there.

heather

@Kaumudi.H, I probably have made it overly complicated =P I understand barely enough calculus to have gotten this far in the video, so.

in The h Bar, 9 mins ago, by heather

how would i get the conversion equation $\frac{\partial\phi}{\partial y^1}=\frac{\partial\phi}{\partial x^1}\frac{\partial x^1}{\partial y^1}+\frac{\partial\phi}{\partial x^2}\frac{\partial x^2}{\partial y^1}$?

Adeek

1:40 AM

hey @arctictern around ?

arctic tern

maybe

Adeek

I want to ask a quick question.

to make sure I understand this correctly.

Here $B_{b,E}$ is the matrix for the linear transformation ?

$\hat{b} : V \rightarrow V^{\star}$

is a linear transformation from V into its dual.

this is induced from the bilinear form b

i.e $\hat{b}(x)(y) = b(x,y)$.

so the matrix $\hat{b}$ represents the matrix which is obtained from the linear transformation $\hat{b}$ ?

arctic tern

looks like $B_{b,E}$ is the matrix of $b$ on $V$, and this equals the matrix of the map $\hat{b}:V\to V^\ast$

Adeek

oh ok so the matrix obtained from the bilinear form b should equal the matrix obtained from the adjoint transformation $\hat{b}$.

arctic tern

"adjoint transformation" is that what you call it?

Adeek

1:50 AM

I understand the part $(\hat{b}e_i)e_j = b(e_i,e_j)$ but why does that equal $\hat{b}e_i = \Sigma b(e_i,e_j)e_j^{\star}$ ?

@arctictern Yeah that is what the book calls it.

I really don't understand this proof.

arctic tern

$\sum b(e_i,b_j)e_j^\ast$ and $\hat{b}(e_i)$ do the same thing to every basis vector, hence are the same transformation

Adeek

I mean that matrix obtained from b has entries $b(e_i,e_j)$ for the $i,j$ entries.

ohhh I see

this is from the definition of $e_{j}^{\star}$ I guess to be the canocial delta function.

This book likes to have things on the right for some reason. Even though it is more natural to have them on the left.

I find it helpful to discuss things with people. Thanks a lot for providing me this opporunity. Especially for your knowledge in Algebra. @arctictern

@arctictern Btw I think I found an error in Michael Atiyah book.

@arctictern I think this definition of rings is wrong.

I think you can find rings which is closed under addition,multiplication, and contains the identity, but not closed under subtraction right ?

arctic tern

you mean find subsets, and yes

like N in Z

Adeek

yeah

Ramanujan

2:13 AM

Quick question: what is that thing which has 0 surface area but infinite volume?

Perturbative

@heather When you say field, are you referring to a vector field?

heather

@Perturbative, nope, the video i'm watching talks about a field with grass and stuff (though I guess it could be a vector field, or a field of just about anything)

arctic tern

@Ramanujan you mean infinite surface area and finite volume.

gabriel's horn

Ramanujan

@arctictern 0 volume and infinite surface area :P

Perturbative

@Ramanujan A plane?

Ramanujan

2:18 AM

@Perturbative no

Perturbative

@heather I'm sure the grass they're referring to is an analogy of some sort to try and explain what a field is

Ramanujan

A fractal

heather

@Perturbative i think they assume knowledge of basic linear algebra

they definitely are assuming knowledge of calculus.

Perturbative

@heather It looks like they are trying to express differentials from a multivariable calculus point of view (which is why they make use of multiple axes and so forth). See here : en.wikipedia.org/wiki/Total_derivative]

@heather What they are trying to define is the notion of a total derivative

heather

okay...

i'm sorry, but the video is a stretch for me in of itself; the wikipedia article is over my head.

Perturbative

2:26 AM

@heather Are you watching a KhanAcademy video?

heather

my single-variable calc is decent, but my multi-variable calc...well, it could be a bit better.

@Perturbative no, i'm watching the video

Einstein Field Equations - for beginners!

►

it's rather good, but occasionally, especially as the video progresses, i get lost in the math and forget what i'm doing.

case in point being understanding how he got the transformation equation.

Perturbative

@heather I'd recommend going through a bit of Multivariable Calculus, first before tackling those videos (not to discourage you at all, though). It's just that to understand the Einstein Field Equations you need to have a good understanding of Multivariable calculus, and Differential Geometry (that's where one usually learns tensors and the like)

heather

indeed, so i've been told =) the guy is explaining tensors as he goes, so that's not as big a deal, but a better understanding of multivariable calc would probably help.

Ali Caglayan

If you are going to learn about Einstein field equations I recommend Leonard Susskinds General Relativity lectures

Perturbative

@heather A good book (but definitely not a rigorous one) to look at would be Div, Grad Curl and All That :goodreads.com/book/show/703104.DIV_Grad_Curl_and_All_That

heather

2:41 AM

thank you @Perturbative, and @AliCaglayan

Perturbative

@heather No problem :)

Semiclassical

3:45 AM

hi chat

Ramanujan

Hi

TheGreatDuck

Slightly tangential question to this one. Using just 3 (and only 3) is it reasonable for me to conclude that x' = (x')^2

-2

Q: Are these alternate axioms of differential algebra (on real number functions) consistent?

In another question I asked, I asked about using a particular axiom to define the derivative so that I could negate it in general. However, I realized that there are two alternative systems in particular that I find interesting*, and that the core alternative truths that describe them as I desire...

derivatives logic axioms formal-systems differential-algebra

:p

TheGreatDuck

4:14 AM

I would think $x'=(x \circ x)'=(x' \circ x)x'=(x')x'=(x')^2$

any flaw in my reasoning?

I don't think 3 alone can prove x' = 1

but I believe it proves that $x' = (x')^n$, which I believe would suggest that x' is either 0 or 1 at any given point as when raised to an arbitrary power both 0 and 1 equal themselves.

Kaj Hansen

Hey chat

TheGreatDuck

I hope my reasoning is not flawed.

@KajHansen hello

@KajHansen I'm trying to prove that given only the identity $(f \circ g)'=(f' \circ g)g'$ that $x' = (x')^2$

no limits or anything. I'm looking at more as abstract rules that define derivation if that makes sense to you.

Kaj Hansen

Hmm. where is the composition of functions in $x' = (x')^2$ that will allow you to use the chain rule?

TheGreatDuck

x is composed with x

implicitly since x is the identity function

Kaj Hansen

Oh I see it now

TheGreatDuck

4:29 AM

I just wasn't sure if there was anything wrong with me concluding that $(x' \circ x) = x'$

Kaj Hansen

As long as you're willing to take $x' = 1$ for granted I think

TheGreatDuck

why?

isn't any function composed with x in that manner equal to said function?

(I am logical. I do accept x' = 1. However, I do not believe the chain rule is purely capable of proving that.)

@KajHansen for instance. for arguments sake, if I said only two things were given: the chain rule, and x' = 0. Could you at all prove that I have contradicted myself. Obviously it's false, but I'm not curious what is true but rather what can be proven true with just the chain rule.

and I just realized the purpose of axiom 4 in the list I found. :-)

Kaj Hansen

@TheGreatDuck Ok yeah, I see what you're saying. That seems fine then

Since $x$ is the identity function, anything composed with it will just give the thing back

TheGreatDuck

if x were 0 at any point, then for all other non-constant functions the chain rule would result in that point always being 0. However, axioms states that for any point there is a function that gives a non-zero derivative. Therefore, the derivative of x must not be 0 at any point. Therefore, since axiom 3 implies x is only either 0 or 1 at any point, it must be 1 at every point by elimination.

and I just worked out how to prove x' = 1. Interesting.

Kaj Hansen

There's something somewhere saying something along the lines of as soon as you have one inconsistency, like $x' = 0$ in this case, then you can "prove" anything you want

TheGreatDuck

4:38 AM

yeah but axioms 2-4 in my list give no inconsistency because they are true things of the derivative.

Kaj Hansen

Whether I can specifically, I'm not sure lol

TheGreatDuck

whether or not every property is provable from them is questionable

I think the "prove everything" is a more trivial thing. Basically you prove "FALSE" is "TRUE" and from there you just say "every statement is TRUE or FALSE. Therefore they are all TRUE."

Kaj Hansen

You're gonna need that limit definition at some point. I bet there are functions that cannot be derived like this.

TheGreatDuck

well of course.

this isn't a replacement

to be fair though, I'm only concerned about the smooth functions as (assuming consistency in my alteration) I think given these proving everything about smooth functions the alteration might prove anything relevant (in that system) regarding piecewise continuous. Anything beyond that is best left to... whatever Lebesgue integrals cover.

(those functions are important I'm sure but far too messy for what I would intend to look at)

probably one thing I want to try to reach ASAP in terms of proving would that $(e^x)' = e^x$

or that if $y' = y$ then $y = e^x$ as that would serve as a partial building block of a linear differential equation (and the ability to solve one).

Jessy Cat

0

Q: Applying difficult initial conditions to general form of a solution to get bounds on a solution

This question is related to one I asked earlier here This time, however, I have the same equation, $u_{t} = 2u_{xx} - 2u_{x} - u$, $0<x<1$ to solve, but this time the boundary conditions are different: $u(0,t) = 4t+1$, $u(1,t) = \cos t$, and the initial condition $u(x,0) = 1$. What I'm doing t...

pde heat-equation maximum-principle

One lat ditch effort to see if I can get any help with this last problem I have to do.

TheGreatDuck

4:46 AM

@KajHansen while I am starting with norma differentiation to get a foothold my actual intention is to later re-do the proofs using an alternate axiom and look at the results. That's why the limit definition cannot be explicitly used as it wouldn't apply to something that doesn't have the same identities.

It's like asking whether I am doing geometry on a torus or geometry on a plane. I cannot necessarily differentiate the two using Euclid's postulates. However, if I can prove I'm not on a sphere that is an interesting consequence but not directly relevant to whatever results the postulates give in and of themselves.

Kaj Hansen

That last submit makes sense. You'll have to flesh out the details more on what you're doing with re-axiomitizing derivatives for me to see things how you envision them

user228700

5:10 AM

Hi, again.

user228700

I have a quick-ish question about inverse trigonometric functions. The $\sin$ function isn't invertible as it is, since it's not one-one so we restrict the domain to $[-\pi/2, \pi/2]$, correct?

user228700

Then we can define an inverse function whose domain is $[-1,1]$ and range is $[-\pi/2,\pi/2]$, yes?

user228700

In this process, have we restricted the domain of the $\sin$ function just to be able to find an inverse function or is it that it's domain is still $R$, even when the conversation includes its inverse function? I'm a bit confused about this especially because some sources continue to take $R$ as the domain of the $\sin$ function just as soon as they're done defining the inverse function.

user228700

...I hope I've made my dilemma quite clear. Do ping me if u know an answer...

Kaj Hansen

5:27 AM

@Kaumudi.H, the domain is always R on the sine function. The inverse function has domain [-1,1] and will give the $x \in [-\pi/2, \pi/2]$ for which $\sin(x) = y$ for a given $y \in [-1,1]$.

user228700

@KajHansen I see. So even though we need to restrict the domain of the $\sin$ function to obtain a proper inverse, we go back to defining $R$ to be the domain just as soon as we're done defining the inverse function?

Kaj Hansen

Yeah @Kaumudi.H.

user228700

:-| That seems a bit...sketchy, actually.

Kaj Hansen

I mean technically you never have to redefine the domain. You just have to recognize that the sine wave repeats itself every $\2pi$...things.

user228700

Hmm, yeah, OK.

Kaj Hansen

5:36 AM

So if you want to know what $x \in \mathbb{R}$ maps to a $y$, we know we're gonna be able to write $x = z + 2\pi k$ for some $k \in \mathbb{Z}$ and $z \in [-\pi/2, \pi/2]$

user228700

Yes, alright, thanks very much :-)

Kaj Hansen

mhm; that's difficult for me to explain well given the relative simplicity of the concept.

user228700

What is? I think I do understand what you're saying.

Kaj Hansen

What's going on with that; idk why. I'm glad you understand though :D

user228700

Yeah :-| OK, thanks :-)

Ramanujan

6:20 AM

this question got 6 upvote in just 15 views and 5 minutes :D 👍

Jessy Cat

6:42 AM

I wish my questions got upvotes like that.

If I have the general form of a solution to a diffusion equation

How can I use the Maximum principle to put bounds on the exact solution without having to find it?

The b.c.'s are horrible.

So I'm pretty sure I'm not supposed to actually apply them.

Glorfindel

Morning all. Can anybody with 10k (or a moderator) tell me why this question was deleted?

Axoren

@Glorfindel I don't have that much rep, but was it something like find primes for which $p^2q^{2^{2017}}$ is a square?

I'm inferring from the URL

Glorfindel

Yep.

Axoren

6:58 AM

Well, that's probably because it's a national quiz question for one such quiz in progress.

Very likely, given that structure

But it's also super easy.

Do you care about the answer? Or the question more?

Glorfindel

I know the answer; I wrote one that got a +7 score. That's why I'm asking. No idea that that one was from an ongoing competition ...

Axoren

I don't know for sure, but it's likely.

That's the only thing I can infer without seeing the page myself.

Secret

0

Q: Rigged Hilbert Space in Quantum Mechanics and Generalized Notion of Sequence

I have been reading Arno Bohm and Israel Gel'Fand, on Rigged Hilbert Space ($\Phi \subset H \subset \Phi'$), which is an attempt to put the Quantum Mechanics Bra-Ket notation, of Dirac, on rigorous mathematical footing. This makes heavy use of linear topological spaces, which in turn makes heavy ...

quantum-mechanics mathematical-physics hilbert-space

Axoren

If it was a question around for longer than a day, unlikely.

Secret

That's a continuum and not a sequence ,right? A "sequence" where the index set is a superset of integers

Axoren

7:00 AM

@Secret Isn't the difference whether it's discrete or not?

Secret

Well, if the index set is $\mathbb{Q}$ it is still discrete, but it already differs a lot from the integers?

Jessy Cat

Anybody in here know how to use the maximum principle to find bounds on the solution to a diffusion equation?

Axoren

@Secret $\mathbb Q$ is just as big as the integers, though

Secret

yeah bad example, how about the index set being the set of transcendentals

Glorfindel

@Axoren no, it was from yesterday 6PM UTC, but it has been on the site until two hours ago.

Axoren

7:02 AM

@Secret Slow your roll, Satan.

Secret

then you get something that is not quite a continuum

Axoren

Indexing over the transcendentals is something I'll have to think about

And it's definitely not a superset of the integers.

@Glorfindel You'll have to ask a mod directly.

Or someone in the 10k Club

I think you can ask in the Mod chat

Glorfindel

Thanks!

Axoren

@Secret How is this any different from just having a function $f: \mathbb R \to S$?

Secret

It's kinda complicated

Axoren

7:06 AM

A sequence is just a function $f: \mathbb N \to S$

Where $f(n) = a_n$

Secret

@Axoren Hmm, that does sounds like that, so I guess functions already covered that

Axoren

But just thinking about the topology of a function over the transcendentals is scary.

Or worse yet, the transcendental plane.

It would be like looking at the fuzzy end of a microfiber cloth. There's technically holes and gaps everywhere, but it's just so damn soft that it feels smooth.

OneRaynyDay

Just confirming, the function $f(x) = a^tx = \sum_i a_ix_i$ is convex right?

Because $\nabla^2_xf(x) = 0$

Axoren

@OneRaynyDay I think there's another condition to go that route: that it must be smooth.

Luckily, there's an easier route: I'm pretty sure all linear transformations are convex.

Which is stronger than saying $f(x)$ is smooth and has 0 second derivative.

OneRaynyDay

@Axoren You mean $D(f)$ is convex?

I mean, I suppose - but then it'd be like a chicken-or-the-egg problem

because I'm trying to prove that a general linear transformation is convex. LOL

Axoren

7:15 AM

@OneRaynyDay Multiplication by $a^T$ is, and when you take $\nabla f$, what do you end up with?

OH

Well then just go back to the definition of convexity.

OneRaynyDay

There's multiple definitions though. Like Jensen's inequality, first order positive definiteness, second order 0 matrix, the classic $\theta x + (1-\theta)y \in$ convex set

Axoren

$f(ta + (1-t)b) < tf(a) + (1-t)f(b)$

OneRaynyDay

Just not sure what you mean by smoothness - you mean continuity?

Axoren

Continuity and differentiability.

OneRaynyDay

righty. Gotcha :)

Axoren

7:18 AM

For example, $f(x) = -|x|$

It has $\nabla^2f = 0$ everywhere it's defined.

But at $x = 0$, it's not defined.

And $f(x) = -|x|$ isn't convex.

Double check me on that, I'm partially retarded tonight.

OneRaynyDay

Yeah - you're right

-|x| isn't convex.

Alessandro Codenotti

If your function is nice enough, that is $C^1$ you can see that convexity is equivalent to the graph of the function being above the tangent plane to the graph at every point

Axoren

@AlessandroCodenotti Could you explain that?

I've never heard that before

OneRaynyDay

@AlessandroCodenotti Sorry - Not that far into the course yet - could you explain what $C^1$ is?

Oops I got ninja'd

Alessandro Codenotti

The equivalence is shown here @Axoren

OneRaynyDay

7:30 AM

Isn't that just the first order taylor approximation at point x

Alessandro Codenotti

$C^1$ means differentiable with a continuous derivative, but just differentiable should be enough here

OneRaynyDay

Ah gotcha. That makes sense. Thank you

Alessandro Codenotti

@OneRaynyDay The tangent plane where defined is the best linear approximation at a point. For convex functions you have a $\ge$ in there though

OneRaynyDay

Righty. Thank you :)

I have some basic mathematical knowledge, and am trying to go through a crash course through convex optimization

Since it's a pre-requisite to a machine learning class I'm taking right now.

Alessandro Codenotti

The point was that the tangent plane to the graph of a linear function is particularly nice

OneRaynyDay

7:39 AM

yes, because it's convex by satisfying the = in $\geq$

Ramanujan

Now trival answers are automatically converted to comments :(

Alessandro Codenotti

Yep. You could have also seen that directly from the usual definition of convex, that is $f(tx+(1-t)y)\le tf(x)+(1-t)f(y))

OneRaynyDay

Yep - worked it out :)

Alessandro Codenotti

Since $f$ is linear $f(tx+(1-t)y)=f(tx)+f((1-t)y)$ and you take out the constants

OneRaynyDay

learning about convex functions made me realize how restricted this class of functions actually are

yupyup

Tobias Kildetoft

9:10 AM

I was just told that there is a letter for me from MathReviews (sent to my university address, so they will forward it to me). I wonder what that could be.

s.harp

@TobiasKildetoft this: mathreviews.ca?

Tobias Kildetoft

@s.harp I assume the ones behind MathSciNet

s.harp

Looking at the website: When they say they store reviews, do they mean the reviews by referees for papers in consideration for publication?

Tobias Kildetoft

@s.harp You mean on MathSciNet?

s.harp

yes

I cant look at anything on the site because I don't have an id D:

Tobias Kildetoft

9:18 AM

Each paper there is accompanied by a review of it written specifically for the site by an expert in the field. This is non-anonymous and is more of an overview than a critical review (most of the time)

Ahh yeah, it requires a subscription

s.harp

So the papers are then also restricted to "classic" or impactful papers from some years ago?

Tobias Kildetoft

all papers are indexed as long as they are published in reputable journals

though the indexing can take a bit of time, and the review usually only appears after some further time (since they need to find someone to do it and that person has to actually do it)

s.harp

@TobiasKildetoft but there are lots of papers being published in reputable journals, surely they do not find a reviewer for every paper?

Tobias Kildetoft

They really do

Pissed off layman

Aka peer review

Tobias Kildetoft

9:22 AM

@Pissedofflayman No, this is not the same as peer review

s.harp

But then they do not just take every paper from a certain journal?

Tobias Kildetoft

@s.harp All papers from all journals that they have deemed good enough (which basically means that the journal should do proper peer review and publish only math)

Pissed off layman

Not even sort of @TobiasKildetoft?

:P

Tobias Kildetoft

journals that do peer review and publish both math and non-math are usually partially indexed (so they index the math papers)

@Pissedofflayman It is similar, but this happens after publication and will usually not include any suggestions for corrections (as those would be pointless)

Pissed off layman

I see.

s.harp

9:24 AM

@TobiasKildetoft so either I am severely underestimating the amount of math papers published or this is an absolutely gigantic thing :D

@s.harp It is huge

morning everyone

hi

But then, there are also a lot of mathematicians, and unlike peer review, they do get paid for this (though the amount is like a few bucks to be used at the AMS bookstore)

s.harp

@TobiasKildetoft how does one get a subscription? Do the libraries of universties usually have one?

Tobias Kildetoft

9:26 AM

@s.harp Yeah, basically all universities have one and going online through the university net will usually suffice to get access

s.harp

I'm in the network of a bakery next to the math department, so I'm not in the uni net. That might be why I can't access :)

Tobias Kildetoft

Using eduroam might also be enough (at least it is for me), and then one can pair the computer with the access so you keep access on that computer for 90 days, even without having to use the university net

I mainly use it to find the correct reference for papers (it can give you the BiBTeX code)

though it also has a nice search feature, as it can search the review text which is often more detailed than the abstract

(and of course, it can calculate collaboration distance which is invaluable for finding Erdos numbers)

Pissed off layman

@TobiasKildetoft when do you get to see the letter?

Tobias Kildetoft

@Pissedofflayman Probably in a few days, depending on the postal service

Fargle

@Semiclassical I'd watch this.

Secret

9:55 AM

Claim: $0\mathbb{Z}=\{0\}$

Tobias Kildetoft

@Secret Yes, clearly

Pissed off layman

0 is the identity element for addition.

Secret

We known in groups we can quotient any group by the trivial group $\{1\}$ to get the group itself, and that quotients also exist in the context of rings. However there's something not very clear here will I get different result if given a ring $R$, I quotient it by $\{1\}$ and $\{0\}$ given that they are regarded as identities for the multiplication and addition structure respectively?

Tobias Kildetoft

@Secret For rings you need to quotient by an ideal and $\{1\}$ is not an ideal

Secret

Ah I see

SteamyRoot

10:43 AM

If $1 \in I$ for some ideal $I$, then $I = R$.

Secret

12:04 PM

It's just a thursday and everywhere is quiet like space

Alessandro Codenotti

Lectures are starting again in the US I think

SteamyRoot

Exams are starting at my uni tomorrow

Alessandro Codenotti

12:22 PM

Here too, but I think they are before the winter holidays in the US

Steve DL

12:45 PM

Hi. Trying to remember the name of something in graph theory, could someone please lend a helping hand? :-)

I'm looking a measure of how many vertices can be reached from any vertex in a graph. I.e. something similar to the degree of a vertex, but applied to the entire graph, following edges to calculate the total number of other vertices reachable from any given vertex.

I'm trying to measure the changes in connectedness of a graph after running a community finding algorithm on it

Tobias Kildetoft

@SteveDL is the graph directed?

Steve DL

@TobiasKildetoft nope.

Tobias Kildetoft

then this is the size of the connected components

Steve DL

I could write the algorithm, I just know there's already a name for it which I forgot

@TobiasKildetoft oh, good call.

Thanks!

ah, turns out this property is called reachability.

PearlSek

1:55 PM

I'm sick

Time to do math :)

Fak I cought on my paper

I'm rooted into my bed

Mithlesh Upadhyay

Someone voting down my all network post in alternate days. Here is one the post, should deassociate from this? This was first post on SE network:

http://math.stackexchange.com/questions/1364274/let-g-be-finite-group-if-a-b-le-g-with-orders-4-5-respectively-then-a

Akiva Weinberger

Oh no

@PearlSek Feel better

SteamyRoot

@MithleshUpadhyay If you feel one and the same person is voting down all your posts, notify a moderator (this usually should get caught by the system, though).

Akiva Weinberger

I didn't realize you could still vote on a closed question

PearlSek

Last time $i \, 2^3 \, \sigma \, \pi$, maybe I was food poisoned

arctic tern

2:06 PM

\Sigma not \sigma

Akiva Weinberger

$2b\lor\lnot2b$

SteamyRoot

Hmmm... I actually wonder now

$\varsigma$

Oooh, that does work.

$\varkappa$ $\vartheta$ $\varrho$

Damn, I only knew about varepsilon and varphi for so long >.<

Akiva Weinberger

\varpi $\varpi$

SteamyRoot

Ugh, I hate correcting exercises the day before the exam

Giving the number of bijections between $X = \{a,b,c\}$ and $Y = \{1,2\}$ should not be difficult.

Semiclassical

I suspect I'm misunderstanding. How could you have a bijection between two finite sets of different cardinality?

SteamyRoot

2:21 PM

You're not misunderstanding.

Semiclassical

Ah.

SteamyRoot

The answer is just 0...

Semiclassical

Yeah, if people aren't getting that it's a bit disappointing.

Counting how many bijections there are between -subsets- of X, Y would require a bit of thought, but between X and Y themselves?

Secret

Turning most of my proofs into those handy category theoric looking diagrams do help to illuminate that they are not perfect and I miss some rather subtle cases. They are now under investigation

Bob

I have the following post on Stack Exchange Computer Science. I am thinking that I should have posted it to Mathematics instead. Please comment. cs.stackexchange.com/questions/68581/…

Semiclassical

2:25 PM

Hmm. Is there a name for that, actually: Given two finite sets, how many bijections are there between their subsets?

Bob

1

Q: Generating Pairs of Random Numbers

Please consider the following probability density function of two variables. \begin{eqnarray*} f(x_1,x_2) &=& \begin{cases} 2(x_1+x_2) & \text{for} \,\, 0 <= x_1 <= x_2 <= 1\\ 0 &\text{ otherwise } \\ \end{cases} \end{eqnarray*} I would like to generate two random values $X_1$ and $X_...

numerical-algorithms numerical-analysis random-number-generator

SteamyRoot

Hmm... I don't know if there is.

a-kile

Hi

:)

anyone here?

Semiclassical

Writing down that as a sum seems pretty straightforward: For any $k>0$, we pick $k$-element subsets of $X$ and $Y$. This can be done in $\binom{|X|}{k}$ ways for $X$ and $\binom{|Y|}{k}$ ways for $Y$. There are then k! (?) bijections between these two subsets.

So should be $$\sum_{k=1}^\infty \binom{|X|}{k}\binom{|Y|}{k} k!=\sum_{k=1}^\infty \frac{|X|!|Y|!}{k!(|X|-k)!(|Y|-k)!}$$ @SteamyRoot

@a-kile If you've got a question, ask it; if anyone is interested, they'll respond.

a-kile

@Semi

@Semiclassical nope

Semiclassical

2:33 PM

okay.

a-kile

I'm pretty new around here, so just checking out

Semiclassical

Ah, fair enough.

There's a lot of people who wander in here with questions and, as the room description states, "ask if they can ask a question." That tends to be a waste of time.

Akiva Weinberger

So it's the exponential generating function of $\frac{a!b!}{(a-k)!(b-k)!}$?

Mithlesh Upadhyay

@SteamyRoot , thanks for suggestion. I report this to SO Team.

Akiva Weinberger

(Cont'd) Don't know if that helps any.

Semiclassical

2:35 PM

@akiva Yeah, that works. I don't know if it helps either though.

Socrates

a linear equationsystem over $\mathbb{F}_3$ with 3 equations and 3 variables can have at most 27 solutions. This would be the case if all equations are 0=0, and therefore the variables can be chosen in 27 ways ($3^3$). But how do I show that this is the maximum number?

Akiva Weinberger

What's the formula for the product of two exponential generating functions, again?

Semiclassical

Can't remember myself.

Akiva Weinberger

Oh, never mind

a-kile

so

cool idk what i should talk here xd

Semiclassical

2:43 PM

@SteamyRoot It doesn't seem to be tabulated over at OEIS that I can tell.

Antonio Vargas

Hi my name is Antonio and I'm a procrastinator

Semiclassical

Hi Antonio.

SteamyRoot

I recall someone once saying this chat was "procrastinators unanonymous"

Semiclassical

Oh, I should mention: I may be running back into Riemann-Hilbert stuff again.

Fuckoffsittian

@AntonioVargas you are not alone friend

Antonio Vargas

2:50 PM

@Semiclassical Ooo neat

Semiclassical

More precisely, I'm running into stuff where the only analytical proofs seem to be either based on symmetry (for small matrix cases) or really painful asymptotics if you try to do it in a way that doesn't rely on that.

Antonio Vargas

Painful asymptotics is my jam.

Semiclassical

I figured. Let me sketch the simplest case; it's a building block for others.

Consider the first-order ODE system $$i\frac{d}{dt}\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix}=\begin{pmatrix} -\alpha t & 1\\ 1 & \alpha t \end{pmatrix}\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix}$$

More compactly, $i \frac{d}{dt}\Psi = (\sigma_1 -\alpha t\,\sigma_3)\Psi$ (names of the matrices $\sigma_1,\sigma_3$ are conventional.)

Antonio Vargas

ah yeah Pauli matrices

Semiclassical

Yup

GFauxPas

2:57 PM

math.stackexchange.com/questions/2094785/… I'm missing something obvious, why does symmetry make it trivial?

Semiclassical

If I act twice with $i\frac{d}{dt}$ and use that equation to simplify, I get $$-\ddot{\Psi}=-i\alpha \sigma_3 \Psi+(\sigma_1-\alpha t\sigma_3)^2\Psi = (1+\alpha^2 t^2)\Psi-i\alpha \sigma_3\Psi$$

This decouples the system, yielding the second-order ODE $-\ddot{\psi_1}=(1-i \alpha+\alpha^2 t^2)\psi_1$.

Antonio Vargas

I follow so far

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