12:01 AM
1/x has a multiplicative structure - e^-x doesn't have a nice structure I can see

@geocalc33 with the definition you've given $f(ab)=e^{ab}\ne e^{a+b}$ in general

yeah sorry about that
all I can think of is that exponential changes addition into multiplication...

The exponential is a group homomorphism from the additive reals to the multiplicative positive reals. Functions do not have multiplicative “structure.”
You still haven’t taken a basic algebra class?

Hi folks. Not sure if this is more of a physics question, but I'd really appreciate if someone could at least point me to a resource that can help me understand.
In a Phase Diagram, why is the gradient $dT_d/dP$ equivalent to the rate of change of the volume wrt Entropy $\Delta V/\Delta S$?

fyi, there's a chat room for the physics site: chat.stackexchange.com/rooms/71/the-h-bar

12:14 AM
Hm, doesn't seem particularly active

also, for googling purposes you may want the name of that identity: Clausius-Clapeyron relation

physics to me sounds like mathematics to someone that doesn't know mathematics

@Semiclassical That's it. This is what I was looking for. Marvelous, thank you!

@Jakobian At my undergraduate institution, physics was the major for people who failed out of the mathematics program.
At my phd institution, mathematics was the program for people who failed out of the physics program.

1 hour later…
1:17 AM
@robjohn Yes, the dual of the hypercube exists in higher dimensions. From math.ucr.edu/home/baez/platonic.html "In higher dimensions, the n-simplex is self-dual, and the dual of the n-cube is the n-dimensional cross-polytope".

1:55 AM
dual is a hugely overused word in mathematics.

It makes sense. Just like projective duality.
Vertices dual to faces, edges dual to edges.
And projective duality really is literally duality. Dual vector space, etc.

ahh, i see your point, you had to draw the line somewhere.

smack

:-) sry

I need to get your daughter on my side.

2:00 AM
I have sets with the property that if $F_1\cap F_2 = \emptyset$ then $\overline{F_1}\cap\overline{F_2} = \emptyset$.
Can I get the same property for finite amount of sets from this one?

:-) her friend tested +ve unfortunately, hopefully her already tested immunity will suffice
i am certain in the context of puns she will take your side

@Jakobian why not?

why yes

@PM2Ring Yeah, I thought about it today, and realized that that was a valid regular solid in any dimension.
Thanks.

closure isn't really closed under intersection
I mean that $F_1\cap ...\cap F_n = \emptyset \implies \overline{F_1}\cap ...\cap\overline{F_n} = \emptyset$
I think this is false

2:17 AM
Not the question, @robjohn.

Okay, sorry, I just saw the last posts.

is intersection closure under closed?

Huh?

i was just trying to see how confusing i could make something.
i did look twice when i saw closure not being closed under intersection, wanted to help someone else go on a surrealistic journey.

Flexing those redundant muscles?

2:23 AM
jazz odyssey also would have been accepted.

Maybe we didn’t miss you after all.

2:42 AM
@Jakobian $F_1 = (-1,0)$, $F_2 = (0,1)$.

@XanderHenderson I tried that, and was rebuffed
I didn't read back far enough

2:53 AM
I excel at rebuffing.
My floors are quite polished.

mine are irished.

Carpet gang.

More in the news with the Irish fishing fleet! They got fleeced by the Russian ambassador, presumably.

@TedShifrin it is on the border of the economic zone, i don't know why they didn't go a little

Yup, Putin can’t be outdone by fishermen!

3:48 AM
@robjohn Well, that's annoying.
I'm going to bed.

4:03 AM
Hello @leslietownes, glad to see you back. :) How are you feeling now?

better, thanks.

:-)

3 hours later…
6:59 AM
@TedShifrin Finnegan's Fleeced Fishing Fleet: Fun Fishing for the Family

7:21 AM
How come $D_{2n}$ acts on the set consisting of pairs of opposite vertices of a regular n-gon, if n is even?
How do I define such action?
All I know about $D_{2n}$ is that $D_{2n}:=\langle r, s: r^n=s^2=1, rs=sr^{-1}\rangle, \langle r\rangle$ is the subgroup of rotational symmetries, whereas $\langle s\rangle$ represents subgroup of reflection symmetries.

that's maybe not the most helpful way of thinking of D_{2n}
the presentation, i mean
or maybe it is, usually when people think of 'symmetries' you're already acting in something and there it's just symbols acting on symbols
good think ted isn't here to read this

I say if $x\in D_{2n}$ then
case 1: x is rotation
Let (a,a') be pair of opposite vertices of n-gon and let $L_{aa'}$ be a vector connecting them. Then, $x.(a,a'):= |L_{aa'}|(\cos (r+\theta), \sin (r+\theta))$.
Something like this.
But with this, verifying that . is actually an action seems complicated.

4 hours later…
11:05 AM
@XanderHenderson I am looking for a finite subfamily of sets such that their intersection is empty, but closures aren't, assuming that the original family has properties 1) disjoint sets have disjoint closures, 2) they're closed under union and intersection

I am not sure I understand what condition 2 means. What are closed under union and intersection?
Oh, $F_i\cup F_j$ is in the collection of sets, too.

11:44 AM
yes

11:59 AM
In the context where $(X,d)$ is a metric space, what does $Ox$ mean?

12:23 PM
@Derivative No idea. Where are you seeing that?

12:43 PM
Open neighbourhoods of x maybe

@XanderHenderson an assignment. We asked the professor, it was the X axis

1:03 PM
for this question
-2

The quesstion: What does the following series converge to: \sum_{}^{} \frac{1000 + 20n + sinh(5n) arcsin(ln(12n^{2000}-\pi{2n^{25}}+34))^{arccos\frac{1}{n}}+n^{10^{9^{8^{7^{6}}}}} - tan(n!) + 6^{n} - \sqrt{n+3}}{arccsc(ln(cos(n^{5}+\frac{12}{n}))) + sin^{2}(e^{n})(csc^{2})(6e^{2n^{2n}})(n^{24...

Was it wrong for me to give an upvote to encourage the OP to improve his post and add more context to the problem?
I also think it's a good question with a challenging problem, thought I don't like the OP's tone.
*though

1:26 PM
Hi there, question on opinion relating to notation
I'm working with a bunch of elementary symmetric polynomials, how should I denote what I'm doing similar to say, a big product, summation or repeated fraction (K notation)?
Is there something similar for a set?
So for example, e_5({}^_{i=1}^5 x_i) or something?
My proof will contain elementary symmetrics of a variable number of arguments so I'm just not sure how to write it down nicely
Or should I define a helper variable? Meh
Oops, that won't compile, remove the first hat ^ :p

1:54 PM
When evaluating $\lim_{x \to 0-}\frac{x+1}{x^2}$, am I allowed to use substitution $t = \frac{1}{x}$ ? What about if $\lim_{x \to 0}$? Since $t \to \pm \infty$ when $x \to 0$, I'm not sure.
Also, if there is a simpler way to solve this, I would surely like to know.

2:30 PM
@mohan10216 Vote however seems best to you. I would have rewarded, however.

what happend to the comments?
I saw that the user got suspended, is everything ok?
@robjohn

3:17 PM
how to prove that x^(n-1)=n-1-x,if (^) is xor function, n is positive power of 2,x is non negative integer less than n?

@mohan10216 On the fence about it. Could be a language barrier

I'm kind of stuck at an exercise but I don't think anyone here will understand what I'm talking about ;_;

@1010011010 It doesn't seem like a language barrier, as much as an attitude problem.

Based on what exactly?
The alias is not english for one
The use of "useless" seems indicative of a non-native speaker as well

@1010011010 the comments that were left.

3:22 PM
well, you can't guess someone's attitude on anything else other than your observations

3:37 PM
Suppose that H is a subgroup of a finite group G and that H acts on G. Let $x\in G$ and suppose that O is an orbit of x under the action of H. Then the mapping from H to O given by $h\mapsto hx$ is a bijection?

How does H act on G

The problem that I have here is $hx$. The mapping is indeed a bijection if we define $h.x:=hx$ (composition under the operation of G) then . is a bijection. However, if . is any generalized action then I don't know how to show the mapping to be 1-1.

Ok, so $H$ acts by left multiplication say, $h\mapsto hx$ on its orbit

yeah, by left action.

multiplication?

3:42 PM
$.:H\times G\to G$ is not given.
Only that . is an action as described above.
:(
And by orbit of $x\in G$ under the action of H, I mean the set $\{h.x: h\in H\}$.

I don't think it's true at all

yeah, I would love to agree with you on this.
But apparently, this is true.

Define $h.x = x$ for all $h$
It's not true
you have to have a specific action for this to work out
like by left multiplication

This is an exercise problem from Dummit &Foote.
yeah, you are right.

they have lots of exercises, probably lots of errors
but here it was just misunderstanding most likely

3:47 PM
I think I misunderstood-We'll write $h.x$ as $hx$ if no ambiguity.
I think I misunderstood definition of orbit also.

Orbit is simple
If $G$ acts on $X$ then orbit of $x$ is just $Gx = \{gx : g\in G\}$
so everything you can get from x

Here gx is $g.x$, where . is action. Right?

sure
actions are actually important in topology

I see. :) I am studying group theory this time using actions.

yeah, I mean they're important in group theory itself too
$p$-Sylow stuff
the most fundamental family of finite groups, the symmetric groups
I know some group theory

3:55 PM
Jakobian: the exercise is correct but I misunderstood it.
(H acts on G by left multiplication).
This was given there :). And you also said that earlier.
Jakobian: Thank you so much :).

Np.

1 hour later…
5:13 PM
@TedShifrin that makes sense. I came across the dual number and I've seen this $\epsilon^2 = 0, \epsilon \neq 0$ which I do believe similar in some sense.

5:27 PM
I'm trying to find the maximum likelihood estimate of $P_\theta (X_1 = 0)$ for $n$ iid Bernoulli trials $X_1,...,X_n \sim Bern(\theta)$.
My intuition just tells me it should be $\theta = 0$, but that seems very wrong
If I follow the method of finding the MLE for $1-\theta$ I get $\theta = 1 - \bar{x}$ as my MLE
Since $P_\theta(X_1 = 0) = 1 - \theta$
Which answer is right?

Relating to "$\sqrt{-1}$"
Or, rather, relating to "define $i$ to be the number satisfying $i^2=-1$"
In the reals there is none; in the complex numbers there are two; in the quaternions there are continuum many
(note that $(ai+bj+ck)^2=-(a^2+b^2+c^2)$ in the quaternions so you have a sphere of solutions)

Right. Which is why context matters when you write the definition.

The proper way is to view it structurally. Don't define $i$; define $\Bbb C$
$\Bbb C$ is, up to isomorphism, the field whose underlying set is $\Bbb R^2$ with operations $+$ and $\times$ defined by $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b)\times(c,d)=(ac-bd,ad+bc)$
$i$, then, is $(0,1)$.

@AkivaWeinberger I disagree about what is "proper". I agree that, in a context where you have to distinguish between the reals, the complex, the $p$-adics, the quaternions, and so on, it is best to define things structurally.
But this is, I think, bad pedagogy for folk who have never taken calculus, let alone a course which deals with structures (groups, rings, fields, etc).

5:43 PM
Perhaps

Or do you believe that we should teach third graders that a rational number is an ordered pair of integers satisfying certain properties? :P

I see your point

Which is not to say that I disagree with you---I actually agree entirely that it makes more sense to define the complex numbers first if the concern is rigorous mathematics. But pedagogically, this can be troublesome.

I think it should still be acknowledged that, the fact that we can "add in" this number $i$ satisfying $i^2=-1$ without breaking things like associativity, is nonobvious
We create this new structure, called set of the complex numbers, out of the set of reals, and a priori we have no reason to expect that it still has all of our old properties. And then we show that it does
Well, if you believe associativity for real polynomials, then you should believe associativity for complex numbers. But you can say that explicitly
Incidentally, it's fun to see what happens if we "add in" a number $j$ satisfying $j^2=1$, but $j\ne1,-1$
What ends up happening is that $1+j$ has no multiplicative inverse
$(1+j)(1-j)=1^2-j^2=0$, so you have the product of two nonzero numbers being zero, and therefore either you have to give up that property or you have to give up the $j\ne1,-1$ condition
(In fact, a better notation for $j$ is $\pm1$. And then I've just written $(1+\pm1)(1-\pm1)=0$, or perhaps better, $(1\pm1)(1\mp1)=0$)
($e^{jx}=\cosh x+j\sinh x$ is mysterious until you realize it's $e^{\pm x}=\cosh x\pm\sinh x$)

6:02 PM
@CroCo No, this is ostensibly a totally different notion. That is working in an algebraic construction called a quotient ring in which you introduce an object $\epsilon$ with the property that its square is $0$. Whereas I was just talking about how a general polynomial of degree $n$ (in our case $2$) has $n$ different roots (in the complex numbers, say). Now you can phrase this as quotient ring construction, too.

Does my question make sense, maybe I'm misunderstanding what the MLE is

Too much statistics for this room, @Govind.

6:27 PM
sorry @TedShifrin I'll create a thread

6:40 PM
I think what I want to prove, is somehow show that my family of sets satisfies $\overline{F_1}\cap\overline{F_2} = \overline{F_1\cap F_2}$
Hmm... $p\in \overline{F_1}\cap ...\cap\overline{F_n} \implies F_1\cap ...\cap F_n\neq\emptyset$ would be good too

7:09 PM
What are $F_i$'s?

7:40 PM
If I get that some random variable $X \overset{d}{\longrightarrow} N(0,\sigma ^2$, can I say anything about $\frac{X}{\sqrt{{S_n}^2}}$
I know the sample variance is a consistent estimator of $\sigma ^2$, but how can I use this

$F_i\in\mathcal{F}$ where $\mathcal{F}$ is something called a Wallman base @Koro

Subsets of what space, @Jakobian?

$F_i$ are subsets of $X$ and we take closures in its compactification

Do we have properties of $X$?
And what sort of compactification it is? One-point, e.g.?

Well, $X$ is only assumed to be separable and metrizable
it can be literally any kind of compactification that's metrizable and separable
I say literally because it belongs to a class called Wallman compactifications and any metrizable and separable compactification is Wallman
I am trying to prove that $f(p) = \{F_1\cap ...\cap F_n : p\in\overline{F_1}\cap ...\cap \overline{F_n}\}$ is a $\mathcal{F}$-ultrafilter
maybe I chose wrong thing for $f(p)$, I don't know, it seems like a reasonable choice
$\mathcal{F}$-ultrafilter means maximal centered family of subsets of $\mathcal{F}$ here
So the goal here is to show $F_1\cap ...\cap F_n \neq \emptyset$

8:05 PM
It's so exhausting to even look at

Hi folks
I'm trying to computationally implement the unfolding algorithm described in Lemma 2.8 of the Schwartz (2009) paper on Triangular Billiards (projecteuclid.org/journals/experimental-mathematics/volume-18/…)
I'd be very grateful for any suggestions.

@Jakobian i haven't studied those yet :(.

I'd be surprised to know someone else than me studied them here
it's not something you get out of your typical general topology curriculum I feel like

I've already created a simulation of billiards bouncing in a triangular table.
^ You can see a screenshot above of my simulation.
But I'm not quite sure how to go about implementing the unfolding algorithm.
Here's an example of the unfolding algorithm (Credit to @Intelligenti pauca for the image):
In essence, it turns a periodic path into a straight line by repeatedly reflecting the triangle itself over one of its sides (in a particular order)
I would really appreciate any advice.
And if anyone needs any clarification, I'd be glad to provide it.

8:31 PM
@Jakobian There are a few people here (@Alessandro, for example, and @Thorgott, I think) who love that stuff. I do not and did a whole career never having to think about that.

@Jakobian why don't you ask it on mse as well?

9:18 PM
yeah, I actually consulted this with a topology professor, he seems to have no idea
and me neither

9:48 PM
Hello, i have a small question how to choose x to get that $[nx]-n[x]=n-1$
[] is the floor functions

10:07 PM
@TedShifrin I've thought about those kinds of things, but only in cases where the "billiard table" has a fractal boundary. :D

P = NP

@PenAndPaperMathematics That's what she said.

to pee or not to pee

Really, I think if you transform everything into matrices. The matrices can represent the worst case of operations involved in an algorithm. Dynamic array allocation is simply changing vector spaces.

the solution is an empty bladder

10:09 PM
Whether tis nobler to suffer the slings and arrows of a full bladder, or take up arms against a sea of urine...

and then hamlet let loose

I would do it using an "adding machine" a machine that can only add or subtract constants or variables, has conditionals, loops, functions, and everything is stored in a global vector of memory. I.e. any Turing-complete language is such that its programs can be converted to this form.

Good ol' Willy Shakes a Spear. Hamlet is one of the few good ones. :D
But it doesn't make up for Romeo and Juliet or Titus Andronicus.
Ugh...

When you use an adding machine, everything becomes matrix multiplication

Some snakes have real trouble with higher math. They can't multiply for heavens sake!
2

10:14 PM
There will be a slow-down for already efficient algorithms, but a speed-up for supposedly "hard" algorithms
This is because computing $A^k$ can be done with smart exponentiation (exponentiation by squaring)
Every conditional in a Turing complete language with $\gt$ can be converted into $y \gt 0$ or $y = 0$. You put your conditional expression into a variable before testing it against $0$.
I have the proof in my head, but P = NP is the conclusion.

@XanderHenderson in my time in Ireland we had to read an inordinate number of spokeshave plays. coriolanus (Kerry's ole' anus) was my most detested.
i would liken it to having to study calculus using Newton's fluxions
i found a nice convex psq, but it was a comment answer unfortunately, so no risky behaviour there

@copper.hat That's another bad one.
Christopher Marlow FTW.
@copper.hat I have put effort into reading Newton. It is hard, but not as bad as R&J.

if i recall correctly, ullyses was banned in ireland at the time
not that i ever want to see it again, i scanned it looking for the dirty bits
nothing