Mathematics - 2022-07-21

shintuku

12:01 AM

i'm waiting on the phone for my phone company to answer

i wonder if it actually costs something to offer more minutes

or more internet

Lemon

1:13 AM

\nabla_(e_i) e_j = \Gamma^k_{ij} e_k , is there an Einstein summation going on in this equality? That is, do they mean \nabla_(e_i) e_j = \sum_k \Gamma^k_{ij} e_k ?

shintuku

@Lemon add $ around your latex so we can read it

Lemon

$\nabla_(e_i) e_j = \Gamma^k_{ij} e_k$ , is there an Einstein summation going on in this equality? That is, do they mean $\nabla_(e_i) e_j = \sum_k \Gamma^k_{ij} e_k$ ?

$...$ doesn't compile for me

shintuku

check out the info box on the top right

there's a link

it gets you to a page with instructions to get latex to render

takes 2 sec

Ted Shifrin

Yes. It can’t make sense any other way. You need subscript for the $e_i$.

Lemon

@shintuku i don't know if i did it right because it isn't loading still. do i have restart the page? i just copied the link and manually added it into Safari

shintuku

1:21 AM

as a bookmark?

Lemon

yeah. but it doesn't matter if i can't read it. as long as others can read it i guess is fine. i can just copy and paste the math stuff into other places.

shintuku

Ted Shifrin

You have to click on the bookmark while you’re on the chat page.

Lemon

oh i have to click on it

shintuku

you have to drag it onto your bar

and then click on it, on your bar

Lemon

1:22 AM

let me just remove all the 10 start chatjax i have

Ted Shifrin

It’s another Secret Service scandal.

Lemon

Let me correct question:
$\nabla_{e_i} e_j = \Gamma^k_{ij} e_k$ , is there an Einstein summation going on in this equality? That is, do they mean $\nabla_{e_i} e_j = \sum_k \Gamma^k_{ij} e_k $?

Ted Shifrin

Good. And I already answered.

shintuku

@TedShifrin did i miss an important recent piece of news

Lemon

@TedShifrin You mean to say $\nabla_{(e_i)_r}$?

Ted Shifrin

1:28 AM

Yes, shin. It’s been news several days. SS mysteriously wiped all their text messages from 1/6 after they’d been asked for them.

@Lemon What!??

shintuku

@TedShifrin jesus

Lemon

@TedShifrin You said I need subscript on $e_i$.

Ted Shifrin

No, I didn’t. You originally had not typeset it as a subscript.

Lemon

Oh you mean to say there is an Einstein summation (hidden) and they did mean it is a summation over the k indices and your subscript comment was on the notations I used from the beginning.

shintuku

theguardian.com/us-news/2022/jul/18/…

maybe some good news?

apparently the 1/6 committee will receive the texts

Ted Shifrin

1:31 AM

I personally write summation symbols. People are sloppy and “use” Einstein summation sometimes without upper/lower.

@shin They are “missing.”

We shall see. Criminals everywhere in Tromp world.

shintuku

@TedShifrin i bet they're "missing" in the sense of, "being pasteurized by lawyers"

Ted Shifrin

Not at all clear from my watching of the news.

shintuku

hehe, that's worrying

Ted Shifrin

You bet.

Everything is these days.

Everyone has God with their AR-15s.

shintuku

in your opinion what were good years in terms of stability

Ted Shifrin

1:44 AM

This destruction of democracy by the GOP started with Newt Gingrich in the 90s. But the election of Obama lit these fires. We never dreamed what McConnell and Tromp would do, but it was clearly the plan for many decades.

shintuku

didn't know about this Newt Gingrich guy, am wikipedia-ing

Ted Shifrin

From my former home state. Such a breeding ground of vile people.

shintuku

argh i'm failing to upload picture

oh well, it was a screenshot with the sentence Political scientists have credited Gingrich with playing a key role in undermining democratic norms in the United States and hastening political polarization and partisanship.

followed by like 6 citations on wikipedia

found funny the high the number citations

Ted Shifrin

Well, yeah, I'm not the only one to have this opinion. People with far more expertise :P

shintuku

do you have more expertise if it is not sustained by the sound structure of mathematical knowledge?

smh these liberal arts "scientists"

Akiva Weinberger

2:35 AM

Beware Viking Era

fun with anagrams

Ted Shifrin

@shin I do not dismiss the humanities and history, etc. Far from it.

shintuku

i rate history 10/10, literature 9.27/10, philsophy 9.2/10

philosophy gets +0.2 because of community volunteering

hm no

history 10/10, literature 9.67/10, philosophy 9.8/10

philosophy gets +0.02 because of community volunteering

copper.hat

3:18 AM

the problem with history is that there are always $\ge 2$ sides, all of whom think their's is the one.

runway44

3:32 AM

anybody see a copy of $S_4\times S_4$ in $A_8$, and/or a copy of ${\rm Aff}_2\Bbb F_2\times{\rm Aff}_2\Bbb F$ in ${\rm Aff}_4\Bbb F_2$?

leslie townes

4:22 AM

gingrich saw sooner than many that if you don't have a majority you can just act as though you do and if you have control over a small number of choke points there is basically nothing to stop you in the US political process

in a way, it's weirder that this moment didn't happen sooner in US history

copper.hat

5:19 AM

tyranny of the mass

user4539917

12:08 PM

Let's not leave out the catalyst known as covid-19.

The great global anti-robinhood.

in English Language & Usage: Multi-Layered Discourse Room, 2 hours ago, by CowperKettle

Plato's Cave

and for those interested in a geometric approach to things:

in This Is Fine, 6 hours ago, by Elise

Cool visualization of temperatures from 1880 to 2021 in relation to global average temperature

Sources: tweeter

shintuku

1:35 PM

@user4539917 that is such an optimal way to expose that information

wow

Akiva Weinberger

1:54 PM

So here's a neat question

Let $P$ be a lattice polyhedron, and $L_P(n)$ refer to the number of lattice points in $nP$ (the dilation of $P$ by a factor of $n$, $n$ a positive whole number)

A neat theorem is that $L_P$ is always a polynomial

Since $L_P$ is a polynomial, we can extend its domain to more than just positive whole numbers. We can compute, for example, $L_P(-n)$.

What is the interpretation of $L_P(-n)$?

(There's a relatively straightforward answer)

Jakobian

I only know lattices as in orders in which we can take infima and suprema

Akiva Weinberger

I mean the set of points with integer coordinates @Jakobian

A lattice polyhedron is a polyhedron whose vertices have integer coordinates, that is, whose vertices are on the integer lattice

(My question applies to arbitrary dimension, so perhaps I should have said "polytope" rather than "polyhedron" (which implies dimension 3))

Jakobian

2:16 PM

@AkivaWeinberger is it convex

Akiva Weinberger

Yes

Pretty sure we must have that

Akiva Weinberger

2:47 PM

@Jakobian For example. If $P$ is the tetrahedron whose vertices are $(0,0,0)$, $(1,1,0)$, $(1,0,1)$, and $(1,1,0)$, I'm pretty sure the number of lattice (that is, integer) points in $nP$ are $\frac16(n+1)(n+2)(n+3)$

which is interesting enough in itself, I suppose

If $P$ is the standard cube (vertices are $(0\text{ or }1,0\text{ or }1,0\text{ or }1)$) then the number of all lattice points in $nP$ are $(n+1)^3$

In two dimensions, if $P$ is the triangle with vertices $(0,0)$, $(1,0)$, and $(0,1)$, then the number of lattice points in $nP$ are $\frac12(n+1)(n+2)$.

If $P$ is a unit square, the number of lattice points in $nP$ are $(n+1)^2$.

14 mins ago, by Akiva Weinberger

@Jakobian For example. If $P$ is the tetrahedron whose vertices are $(0,0,0)$, $(1,1,0)$, $(1,0,1)$, and $(1,1,0)$, I'm pretty sure the number of lattice (that is, integer) points in $nP$ are $\frac16(n+1)(n+2)(n+3)$

^This is wrong

Gwyn

4:30 PM

I am a little confused but what $o(f)=o(g)$ means. I know that $f=o(g)$ as $x \to 0$ if $\lim_{x \to 0} \frac{f(x)}{g(x)}=0$, so $o(f)=o(g)$ as $x \to 0$ simply means that $\lim_{x \to 0} \frac{o(f)}{g(x)}=0$? For example, $o(x^2)=o(x)$ as $x \to 0$ is true because $\lim_{x \to 0} \frac{o(x^2)}{x}=\lim_{x \to 0} \frac{o(x^2)}{x^2} \cdot x =0 \cdot 0=0$

However, this doesn't work as an equality because it is not true that $o(x)=o(x^2)$ as $x \to 0$, because $\lim_{x \to 0} \frac{x}{o(x^2)}=\lim_{x \to 0} \frac{x^2}{o(x^2)}\cdot\frac{1}{x}$ is indeterminate.

So, should I say that "$o(x^2)$ is a $o(x)$ as $x \to 0$" without using equality sign and prove it evaluating the limit as I have done in my first message, or what I have done up to now is incorrect?

Ted Shifrin

Where are you seeing someone write such an expression?

What you should say is: If $f(x)=o(x^2)$, then $f(x)=o(x)$ as well — assuming we're talking about $x\to 0$.

Gwyn

I have seen in "Olver - Asymptotics and Special Functions": a problem asks to show that $O(\phi)O(\psi)=O(\phi \psi)$ as $x \to \infty$. It is big O, but the nature of the doubt is the same.

Ted Shifrin

Yeah, it's sloppy mathematics.

Gwyn

@TedShifrin This is way more understandable!

@TedShifrin Ok so it is indeed imprecise. Thank you Ted for the help!

Ted Shifrin

In the case you just wrote, it's certainly true that if $f = O(\phi)O(\psi)$, then $f = O(\phi\psi)$.

The other inclusion is not immediately obvious to me.

It's not clear that every $f$ satisfying $f=O(\phi\psi)$ can be written as a product of appropriate functions.

Akiva Weinberger

4:38 PM

Whenever you see ${}=O(\cdot)$, you should think ${}\in O(\cdot)$

as it is a set of functions, and the $=$ sign is not symmetric

In fact, they should just write $\in$, but tradition is tradition

Gwyn

5:03 PM

@AkivaWeinberger This clarifies a lot. Thank you!

Mad Spaces

6:59 PM

Dumb question,
i am having a problem with the rigor of applying a function twice,
I have
$\varphi : \mathbb{R}^{\mathbb{R}} \rightarrow \mathbb{R}^{\mathbb{R}} \\ \varphi(f)(x) \to 0.5 (f(x)+f(-x))$
First of all, how come the notation says we send functions to functions, but the result is a scalar. me confused.
i want to show that $\varphi \circ \varphi = \varphi$ but as i said, the notation above is confusing me, so i am not sure how $\varphi$ looks like, is it taking a function to a scalar or a function to function

leslie townes

bad use of $\mapsto$ there. they mean to say that $\varphi(f)$ is the function whose rule is $x \mapsto 0.5 (f(x) + f(-x))$.

Mad Spaces

okay, i just solved it by defining $ f^{'} $ to be the function which sends $ x $ to $f(-x)$ then writing $\varphi(f) = f/2 + f^{'}/2$

leslie townes

so your investigation might begin, $\varphi(\varphi(f(x)) = 0.5 (\varphi(f(x)) + \varphi(f(-x)) = 0.5 (0.5 (f(x) + f(-x)) + 0.5 (f(-x) + f(-(-x))) = \cdots$

sounds like you got there.

Mad Spaces

Yes that was clear, but your notation implies f(x) is a scalar, and then phi can not take that as argument, that was my issue...

leslie townes

well, to compare two real-valued functions of a real variable, it's very much OK to look at what they both do to the same real variable $x$.

it is, essentially, a scalar calculation.

Mad Spaces

7:05 PM

You are right, i am not arguing with that, i am just saying, the function is defined on R^R not R, so we can not put x before, it is just a rigor thing.

leslie townes

i agree that this can be written confusingly.

as i should have put $\varphi(f)(x)$ instead of $\varphi(f(x))$ above.

but, it is best not to get too caught up in this kind of thing.

once you're at the point of caring about it's $\varphi(f)(x)$ or $(\varphi(f))(x)$ you are down the path where 'rigor' is not what it seems to be.

a computer parsing most math textbooks would, with some justification, be throwing type exception errors left and right.

Mad Spaces

I would like to think my brain works like a program to a certain degree.

An thats the reason i am asking this question, because it throw an error.

leslie townes

well, it is certainly good practice not to write things that invite this kind of discussion. but people inevitably do. so i'd try to have the reaction of "oh, they're failing to notationally distinguish X and Y" and not just "type exception." it's a good question, though.

and worth asking if you get confused.

Mad Spaces

You see this is leading me to even more confusion, because

how can i show in the previous question with the described notation that the image are the even functions?!, for that it must apply that for the image $\varphi (f) = \varphi(-f)$ which is not true.

leslie townes

mm, there i think the issue is notational ambiguity about what "-f" is.

Mad Spaces

7:19 PM

Because you would get $ 0.5 (f + f') \neq -0,5 (f +f') $

leslie townes

you need a notation for the map $x \mapsto -x$. if you called it $i$ then the question is whether $\varphi(f) = \varphi(f \circ i)$ for all $f$.

the point being it's precomposing with that map, and not post-composing with it.

and buried within that i guess is the fact that $(\varphi(f)) \circ i = \varphi (f \circ i)$.

but whatever it is, it isn't $-f = i \circ f$.

Mad Spaces

With your previously mentioned notation, it would suffice to show $\varphi(f)(x)= \varphi(f)(-x)$ which is true

Anyway, thank you for taking time for my madness.

When i used the name mad spaces couple of years ago, i never thought it would truly fit well, it was more of a meme.

leslie townes

surely. there's a good general notion in here about notation for composition operators. if $T_f(g)$ means $g \circ f$ then $T_f(T_h(g))$ is $g \circ h \circ f$ or $T_{h \circ f}(g)$, where i am using the associativity not to write parenthesis around $g \circ h \circ f$. in other words $T_f \circ T_h = T_{h \circ f}$, which feels like a reversal of the natural order of things.

sometimes people deal with this by writing some applications functions on the left, and other applications of functions on the right.

something like that is kind of going on with $\varphi$, which is about precomposing (not post composing) with the function i was calling $i$ above.

Xander Henderson

7:46 PM

Notation for composition is a pain in the ass.

Shaun

Please may I have some feedback on the following?

-1

Q: The outcome of part of Exercise 7.1.4 of one of Robinson's books.

This is concerning part of Exercise 7.1.4 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to this search of "similarity type" in the group-theory tag, it is new to MSE. The Details: (This can be skipped.) Arguments are written on the left of functions. A permutation ...

Xander Henderson

If one has $f: X\to Y$ and $g : Y \to Z$, I would imagine that one would really like to write the composition from left-to-right, i.e. $f g$ or $f \circ g$ or $f \otimes g$ or whatever symbol you like.

But that is wrong! (Or so I have been told).

It makes me want to cry. :(

Ted Shifrin

8:06 PM

I boycott functions on the right. Perhaps OK in Hebrew.

Mad Spaces

i agree with xander!

thats how i initially thought of it as i was a layman...

Xander Henderson

8:37 PM

@TedShifrin I used to know a bit more than an epsilon's worth of Hebrew.

Akiva Weinberger

@MadSpaces $\phi(f)(x)$ can be tough to read if you don't keep track of what everything's type is, but you find that sort of notation all the time (especially in, say, category theory)

$f$ is type $\Bbb R^{\Bbb R}$, or in other notation, $\Bbb R\to\Bbb R$; it eats a scalar and spits out a scalar. $\phi$ is type $(\Bbb R^{\Bbb R})^{\Bbb R^{\Bbb R}}$, or in other notation, type $(\Bbb R\to\Bbb R)\to(\Bbb R\to\Bbb R)$; it eats a function (type $\Bbb R\to\Bbb R$) and spits out a function (same type)

$\phi(f)$, then, is a function. It's type $\Bbb R\to\Bbb R$.

Therefore, $\phi(f)$ eats a scalar and spits out a scalar. So $\phi(f)(x)$ refers to the application of the function $\phi(f)$ on the scalar $x$.

(Sidenote: the "exponential notation" gives us a hint that the type $(\Bbb R^{\Bbb R})^{\Bbb R^{\Bbb R}}$ is equivalent to the type $\Bbb R^{\Bbb R\times\Bbb R^{\Bbb R}}$. What does that mean?

It means that we can view $\phi$ as taking in a scalar and a function, and returning a scalar… which is exactly what it does in $\phi(f)(x)$.)

(That is, we can define $\tilde\phi$ by $\tilde\phi(f,x):=\phi(f)(x)$, and $\tilde\phi$ will contain the same amount of information as $\phi$.)

Technically, $\Bbb R\times\Bbb R^{\Bbb R}$ (the type of the pair $(f,x)$) is equivalent to $\Bbb R^{\Bbb R\sqcup\{ *\}}$, where the exponent is the type of something that is a scalar or an extra point, but I'll let you think through why that makes sense…

Jakobian

8:54 PM

@Gwyn Here what they really mean is $O(\phi)O(\psi)\subseteq O(\phi\psi)$

Akiva Weinberger

($\sqcup$ is disjoint union; $*$ is just some random object that isn't a real)

(I suppose more suggestive notation would be $\Bbb R^{\Bbb R+1}$.)

@XanderHenderson I've seen $f;g$ to mean $g\circ f$, albeit with a "bubbled" semicolon that I think isn't (by default) in LaTeX

Jakobian

It doesn't matter if we write composition from right to left or left to right, it's just a matter of getting used to a convention.

Akiva Weinberger

Sure, but it can be useful to have both options.

@Jakobian Here's a neat fact. Say $f$ means rotate the plane by angle $\theta_A$ about point $A$; $g$ means rotate the plane by angle $\theta_B$ about point $B$; and $h$ means rotate the plane by angle $\theta_C$ about point $C$

I claim that $f\circ g\circ h$ equals the result of rotating the plane about $A$, then about where $B$ now is, then about where $C$ now is (by the appropriate angles)

and that $f;g;h$ equals the result of rotating the plan about $A$, then about where $B$ originally was, then about where $C$ originally was

where $f;g;h$ is defined to by $h\circ g\circ f$

Bob

10:07 PM

@TedShifrin Hello Ted

I am thinking this problem would interest you:

1

Q: Unable to find my error in calculation of the minimum net worth of a family needed to be in the Top 1% by using the equation $n t^{c_1} = c_2$

Problem: a) How large does a family's net worth have to be to be in the top $1$ percent in 2019? b) a) How large does a family's net worth have to be to be in the top $2$ percent in 2019? Answer: I will work in units of thousands of dollars. My answer is based upon data from the following website...

statistics

maybe not

Ted Shifrin

10:48 PM

I don’t see why your model is remotely plausible. You wrote it weirdly, but the percentage is inversely proportional to some power of wealth. Maybe that’s plausible, maybe not.

Why not do a least squares regression with more data points?

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$Mathematics$ Mathematics