Mathematics - 2021-08-10 (page 1 of 2)

leslie townes

00:04

that looks OK to me. maybe i'm missing something.

err, the coefficient of x^2 when you expand that is B-A. so if A = 1 (which makes sense) shouldn't that mean B = 2b+1?

when i see repeated roots i think derivatives but maybe that's not the way to solve the problem.

Ted Shifrin

Something that is titled "FOR ALL APPLICANTS" makes it seem awfully suspicious to be getting help ...

leslie townes

i'm an applicant. aren't you?

Ted Shifrin

You're more of a supplicant.

leslie townes

i'm a replicant.

Ted Shifrin

I recant that.

Anyhow, far be it from me to say anything categorical, but I like the idea of saying bijection in the category of sets and embellishing to get homeo/diffeo in the respective categories.

leslie townes

00:13

bijection does strike me as very set theoretic.

LearningCHelpMe

@leslietownes Oh yes, the derivative was a better way to approach this question. thanks. The value of the coefficients is the thing that's confusing me. Because if you use $B = a^2+b^2$ you get the correct coefficient for the $x^2$ term, but you get the wrong coefficent for the $x$ term.

dc3rd

Hello Math World.....one of the only places in my current life where it seems common sense prevails......for the most part......

LearningCHelpMe

@TedShifrin I'm not cheating on a test or anything haha. it's a maths admission test. a university admission test only runs on the 5th november each year. im just practicing old past papers

Ted Shifrin

No common sense here, @dc3rd.

dc3rd

most folks here are vaccinated or planning on doing it once availble in their respective countries...........a fairly decent proportion of my friends and families circles on the other hand..............

leslie townes

00:19

i can confirm that it appears to be a test from 2014.

Ted Shifrin

OK, detective leslie, thank you.

leslie townes

i'm sorry to hear that, dc3rd. it is a mystery why anybody with the good fortune to have access to a vaccine would decline it. i mean, it isn't a mystery and i could say more about how i feel about that, but i won't.

Ted Shifrin

Darwinism in practice, @dc3rd. I actually don't know the status of most people in the chat, other than those of us who blab about everything.

dc3rd

I just finished reading your higher order derivatives section.....Where did you find the kindness not to put any monstrosities of higher order derivatives where one always loses track of what is getting differentiated?

leslie townes

i got vaccinated the minute i had the opportunity. it was surprisingly effiicient and weirdly inspiring. about 200 people scheduled to get something medical, for free, in the united states. you almost forget that government can work, but they did a very good job in long beach.

dc3rd

00:21

Unfortunately it is Darwinism in practice. Just makes it frustrating when it is loved ones and people of the sort...

but can't expend all my energy worrying about how they choose to live their life with the "iron clad" immune systems

leslie townes

oh boy. yeah, you can't argue against that.

Ted Shifrin

I was quite pedantic in that section, actually, @dc3rd, doing the standard change of variable to solve the wave equation.

The selfishness astounds me more than the stupidity.

leslie townes

one of my friends isn't healthy enough to receive the vaccine, which also means she's very much at risk for the virus. i don't know what's going on with people who just have it offered to them and could take it and say no thanks.

dc3rd

Oh I read that.....and that is involved in and of itself.

It's the stupidity that is drivng me mad about the whole situation....all of a sudden a bunch of folks who failed 9th or 10th grade science are now virus pathologists 10 yeas deep studying these phenomena.......

leslie townes

the presumption that random people can 'do their own research' and reach their own conclusions could not be more wrong. i don't know what the solution is. it's disappointing. there are reasons to doubt the medical establishment but the vaccines are some of the most-researched things in existence. everybody jumped on it at once.

Ted Shifrin

00:26

I don't know where this notion came from. Newsmax and FOX?

leslie townes

tucker carlson is a big fan of doing your own research. easy for him to say. born rich. probably vaccinated and won't tell anybody.

dc3rd

the other thing that frustrates me goes beyond the covid situation.....this scenario has just brought to light the lack of critical thinking in a good portion of the population and that can't be corrected over night....it will involve people wanting to make the effort to adjust their thought processes......but that is another can of worms of egos, self centredness, and personal insecurities...

leslie townes

what bothers me is that there are parts of the world who don't have access to the vaccine. people can't get it even if they want it. and you have people in the USA just saying, no, not for me, thanks.

spoiled brats.

Ted Shifrin

Totally, because Tromp told them science is bad.

leslie townes

HIV was the same way. absolutely devastated parts of the earth.

i hated george W bush but he did do some funding for HIV stuff in africa.

Ted Shifrin

00:28

Reagan signed the death warrants of hundreds of thousands.

leslie townes

stupid virus, killed so many people. fauci was at the lead of that too. what a life.

dc3rd

it a convergence of things like this.....I am all for people expressing their opinions, but the democratization of opinions has gone too far where an ordinary person thinks they can question a researchers credentials in a field and somehow find it valid

leslie townes

ronald reagan had blood on his hands.

b---ard.

dc3rd

I was a kid during the Reagan years so I was a sucker for the propaganda......talk about reading history....

Ted Shifrin

Democracy is a yuge mistake, all around the world. The truly stupid have the right to derail decency.

leslie townes

00:30

my mom was a nurse in the bay area during the early AIDS epidemic and had the worst stories. most of her stuff was end of life care. nobody was paying attention or taking it seriously, or happy that it was happening.

Ted Shifrin

Obviously, as a gay man, I took the AIDS crisis very seriously/personally. Very good friends died ... numbers of them.

dc3rd

Was the amount of deaths due in part to people not taking it seriously sort of like now?

leslie townes

there was definitely a 'do your own research' crowd then, too. people were suspicious of public health officials who wanted to shut the bathhouses down because it seemed homophobic.

Ted Shifrin

It took a long time to get decent drug treatments. But, yes, there were the young'uns who insisted they were immortal and were stooopid.

leslie townes

it traumatized my mom. nurses weren't really equipped to get in between patients and their families who wouldn't let loved ones into the hospital. she was fired from a job because she let someone's husband in.

dc3rd

00:33

simple cliche sayings like "study your history to not repeat mistakes"....comes to my mind

Ted Shifrin

And the idiots who insisted unprotected sex (both gay and str8) was their birthright.

Yes, disgusting self-righteous homophobic parents of gay children.

leslie townes

they'd come in right at the end and want to dictate everything.

Ted Shifrin

OK, I'm going back to watching tennis. It's more fun.

leslie townes

pieces of garbage.

i might do the same.

dc3rd

and now we are in an age where it is almost automatic that the first thing you do is protect yourself for sex.....it took a tragedy for the behaviour to change....I don't like how I'm forecasting this.

oh.....one quick question......why don't we define/prove smoothness in your course? do we need more tools?

leslie townes

00:35

it seems like the coronavirus will just be a thing now. i don't see it going away like SARS or other random illnesses. it's too much out there.

dc3rd

tennis?......olympics just finished there is already another contest?

leslie townes

there's always tennis.

shintuku

Prove $\mathbb{N}$ is infinite, i.e., there is no bijection $f : n \to \mathbb{N}$, where $n \in \mathbb{N}$.
Suppose there is such a bijection $f: n \to \mathbb{N}$. Since $f$ is bijective and $n \in \mathbb{N}$ , for each $y \in n + 1$, there is exactly one $x \in n$ s.t. $f(x) = y$. But this implies the cardinality of $n$ is identical to the cardinality of $n+1$, which is a contradiction. Therefore there is no such bijection.

is everyone here compelled by the power of logic to believe me

dc3rd

yeah it does seem like it will be a thing here on out.....and it could be ok if folks would get vaccinated....anyways....I'm not trying to get irate again....

why not the well ordering principle?

shintuku

the author introduces it later :'(

dc3rd

00:38

lol......interesting

well learning it this way will give you some technical proving skills that's for sure....

leslie townes

how is mathbb N defined? do you have weak induction?

dc3rd

you must

leslie townes

there's a maximum for the map from n to mathbb N. and every natural has a successor.

but formalizing it is very much dependent on what you are 'allowed' to use.

dc3rd

that's the only other thing I could think of....even if you don't show the two statements equality

shintuku

uh, $\mathbb{N}$ is defined as the intersection of all inductive subsets of any inductive set

leslie townes

00:41

which book, shintuku? this sounds a bit like enderton's set theory.

shintuku

Goldrei

Classic set theory

dc3rd

welp...this went above my pay grade........I still got to do Munkres Topology.....then I might consider Halmos book.....

leslie townes

i worked with a student of halmos and have tried to imitate his writing style, even in my non mathematical job.

he was a great writer.

shintuku

using the successor, we might also say that $n^+$ has a greater cardinality than $n$

geocalc33

Is there a way to prove that $X$ and $Y$ have the same properties?

leslie townes

00:45

he also liked cats, and nobody who likes cats is all bad. maa.org/sites/default/files/images/spotlight/images/Halmos2.jpg

shintuku

@geocalc33 in which sense?

dc3rd

THis man got a robe on sitting back in a lazy boy and just working on his art......I'm keeping this picture.....such a G.....lol

leslie townes

i would like to know more about the garment he is wearing.

geocalc33

@shintuku for example, say $X$ and $Y$ are isomorphic groups. I would like to convince myself that every true statement for $X$ has a corresponding true statement in $Y.$

leslie townes

if it's a sentence in a finite number of symbols you can just move the map to move everything over.

i dunno about infinite stuff.

dc3rd

00:51

that was their version of rockin' sweatpants leslie...

leslie townes

he's just such a boss. i want some of that swagger.

dc3rd

I've been noticing it a bit more and more with the "famous" mathematicians......they weren't "nerdy" they were just really so comfortable in their skin and went about things in their way.....I love it

leslie townes

my advisor was the least stereotypically mathematicianly person of anyone i've known. i have known a few who did fit the type.

i fit it too, i'm weird.

an extroverted mathematician is someone who looks at your shoes while talking to you.

dc3rd

I'm in the same boat...if you saw me you would not be thinking anything math related was coming from me....even though I'm still in the "aspiring" category.....

leslie townes

my advisor was very comically profane. he cursed all the time.

Ted Shifrin

00:57

So that’s where your daughter acquired it?

leslie townes

possibly. that may have been where i got it.

she was born too late to meet him but i may have carried over some of his traits.

Ted Shifrin

@leslietownes Nonsense. We’re not all so pathological.

leslie townes

one thing that's interesting about mathematicians is that a lot of the stereotypes are really about math students, not people who go on. you can't be a member of a department and fulfill all of those stereotypes.

Ted Shifrin

Well, some do.

leslie townes

there was a guy during my time at berkeley who did that throat singing thing. where you sing multiple notes at the same time. that was pretty weird.

one time someone dropped something on my foot at target and i involuntarily let loose with a stream of profanity. it terrified the person who had done it and i wound up apologizing to them although they were the ones who dropped something on me.

Ted Shifrin

01:02

You have Tourettes.

leslie townes

something like that, maybe.

it must have been a little scary. i try to hold my tongue more now because of the kid.

can't have her cursing at people.

Ted Shifrin

And yet she already does. Well done.

leslie townes

yeah. the other day she told the cat to f--- off.

probably my DNA there.

dc3rd

that's comedic and cute....but probably not too good for a lil one who isn't aware of what she is doing.

leslie townes

she isn't even three years old and will need to learn lessons later about what is and is not appropriate.

Ted Shifrin

01:06

Probably.

leslie townes

she's too young to understand very much, she's just in parrot mode.

which means she says a lot of cuss words.

Koro

01:58

What is $\lim (^{nk}C_n)^{\frac 1n}$

where k>1 is a fixed integer and $^nC_r=$ no. of ways of choosing $r$ objects out of $n$ distinct objects

I’m getting the answer as $ke^{1+\frac 1{2k}-\frac 1{6k^2}}$

which feels wrong.

love_sodam

02:15

@TedShifrin Thanks

Koro

03:09

$(^{nk}C_n)^{\frac 1n}=\frac n{n!^{\frac 1n}}(\frac kn)^{\frac 1n}((k-\frac 1n)(k-\frac 2n)…(k-\frac{n-1}n))^{\frac 1n}$

Inside parentheses, we’ll not get negative number as $k>1$. By taking log, inside parentheses is $e$ to the power $\frac 1n ((n-1)\log k+\sum_{r=1}^{n-1}\log (1-\frac r{nk}))$

o.9

Sometimes I wonder if the people who ask questions on this site are 90% bots

Koro

03:28

The expression simplifies to: $\frac{n-1}n \log k -\sum_{r=1}^{n-1}(\frac r{nk} -\frac 12 \frac{r^2}{k^2n^3})+o(\frac 1{n^2})$

and all this simplifies to $\log k+\frac 1{2k}-\frac 1{6k^2}$

so we get:

2 hours ago, by Koro

I’m getting the answer as $ke^{1+\frac 1{2k}-\frac 1{6k^2}}$

I doubt this step:

4 mins ago, by Koro

The expression simplifies to: $\frac{n-1}n \log k -\sum_{r=1}^{n-1}(\frac r{nk} -\frac 12 \frac{r^2}{k^2n^3})+o(\frac 1{n^2})$

I can however compute this limit using other method: simply by noting that if $\lim\frac {a_{n+1}}{a_n}=b$ then $\lim |a_n|^{\frac 1n}=b$ if $a_n>0$ for all $n\in \mathbb N$

Koro

04:14

I have posted my complete question here: math.stackexchange.com/questions/4221026/…

copper.hat

04:35

@TedShifrin Much appreciated!

dc3rd

the product of the conjugate and conjugate transpose matrix is not necessarily the identity correct?

I did a counter example and showed it wasn't the case. I just feel I'm buggin' because a solution applies it as if the identity matrix is obtained.

Ted Shifrin

04:58

@dc3rd Huh?

Akiva Weinberger

@Koro First instinct is to use Sterling's approximation

(second instinct is to look up the spelling)

I was incorrect, it's "Stirling"

$n!\approx\sqrt{2\pi n}(n/e)^n$

(This is related to a great question: An n-dimensional cube and n-dimensional sphere have the same (hyper)volume, for large n. What is the ratio of the cube's diagonal and the sphere's diameter? Does this approach a limit as n goes to infinity?)

Akiva Weinberger

You ever notice that the two diagonals of a (nonsquare) rectangle are distinguishable? There's no rigid motion (disallowing reflections) that takes one to the other. I could define the "A diagonal" and "B diagonal" of a rectangle without knowing how you're oriented relative to it.

> Definition: take one of the longer edges, rotate it a bit counterclockwise, that's the A diagonal. The other one's the B diagonal.

This is not true of the diagonals of a cuboid (aka rectangular prism aka brick). Those are all indistinguishable.

Russian Bot 2.0

Akiva Weinberger

On the other hand, I think the two tetrahedra in the cuboid (from half the vertices each) are distinguishable?

Basically for one, draw the A diagonals of the smallest and largest faces (as defined above) and then the B diagonal of the middle-sized face. Reverse these for the other

Semi-relatedly, the names for the two ways to fold a piece of paper in half are called "hamburger style" and "hotdog style", which I love

Koro

05:16

@AkivaWeinberger hi Akiva, how are you? I have received an answer on my post which uses Stirling's approximation

rather, a series form of the approximation which I have never used to solve any limit before

Also, I don't understand what is wrong in my method

Akiva Weinberger

See also the terms for orientations of a cuboid (in this case, apple boxes): en.wikipedia.org/wiki/Apple_box#Position

Koro

I have a strong feeling that step involving little o has some mistake but I can't pinpoint the exact mistake right now

Akiva Weinberger

Maybe I'll check it out later

Koro

thanks.

@AkivaWeinberger I never heard of n-dimensional sphere before

Akiva Weinberger

Set of all points in n-dimensional space that are a certain distance away from a given point

Also called a hypersphere

Graph of $w^2+x^2+y^2+z^2=1$ for example

Koro

05:23

ahh

i know that equation but never really tried to find volume/surface area of graphs associates with this.

Akiva Weinberger

You can find formulas here:

Volume of an n-ball

In geometry, a ball is a region in space comprising all points within a fixed distance from a given point; that is, it is the region enclosed by a sphere or hypersphere. An n-ball is a ball in n-dimensional Euclidean space. The volume of a unit n-ball is an important expression that occurs in formulas throughout mathematics; it generalizes the notion of the volume enclosed by a sphere in 3-dimensional space. == Formulas == === The volume === The n-dimensional volume of a Euclidean ball of radius R in n-dimensional Euclidean space is: V n...

Strangely, you get an extra "pi" every other dimension

Koro

strangely, not from dim 4 to dim 5

Akiva Weinberger

Yeah every even dimension gives you one

Super surprising that they decrease with enough dimensions, too

Koro

thanks I didn't know that @Akiva. What branch of mathematics is this taught in?

Akiva Weinberger

I guess this is geometry

High-dimensional geometry

Same sort of thing as "area of a circle is $\pi r^2$", but more than 3 dimensions

There's this great quote from Geoff Hinton, a famous AI researcher

He was giving a talk about neural networks or something

One of the slides in his talk was this:

Relevant stuff:

Perfect Shapes in Higher Dimensions - Numberphile

►

dc3rd

06:33

figured it out @TedShifrin, I just followed your advice from awhile ago and trusted my solution cause what I read as a solution was wrong.

porridgemathematics

06:49

are these all the images of continuous injections $\mathbb{R} \rightarrow \mathbb{R}^2$? A wedge of two circles, a O-O shape, a O- shape, a capital theta , $\mathbb{R}$ itself

(up to homeomorphism)

P-addict

07:58

hi! i've been trying to understand the algebraic nature of restrictions of functions on smooth manifolds. in particular if we have a smooth manifold $M$ and an open set $U\subseteq M$, how does $C^\infty(U)$ relate to $C^\infty(M)$? clearly the latter is a subring of the former, but what else is there? i've been trying to figure this out using $M=\mathbb{R}$, $U=\mathbb{R}\setminus\{0\}$ as an example but it's not clear to me how to understand all the additional functions

Balarka Sen

@porridgemathematics Yes

porridgemathematics

how would one prove this ?

Balarka Sen

@P-addict $C^\infty(M)$ is not apriori a subring of $C^\infty(U)$, there's a map $C^\infty(M) \to C^\infty(U)$ given by restriction of a function to $U$.

The image is, of course, a subring.

This map is a split epimorphism of rings.

I don't know what else to say

P-addict

shoot! it's not necessarily injective, don't know why i thought that

Balarka Sen

@porridgemathematics Well, you have to find a way to trace these figures without lifting your pen

That's basically what image of a continuous injection from R means

porridgemathematics

08:10

sure, in my head I arrived at those figures by considering whether I loop at some point or continue straight, etc.

i guess this can be formalized?

Thorgott

@BalarkaSen hows it split

Balarka Sen

Take a bump function supported on a compact set containing U?

Oh U is not a small chart

I mean doesn't matter

Thorgott

take $U$ to be $M$ punctured

Balarka Sen

yeah fair enough

you want it to be contained in a compact set for split

Thorgott

that works, but we can do better, right

we want existence of a tubular nbhd

Balarka Sen

08:16

haha yeah but whatever man

you're right of course

thats what i was thinking when i said doesn't matter

P-addict

sorry i'm kind of lost... is that map being split equivalent to saying every map on $U$ extends to one on $M$?

Balarka Sen

yeah

Thorgott

in a homomorphic way, yes

but it's not true in general regardless of whether we postulate homomorphy or not

@Balarka I caught myself recently accidentally assuming that every open disk in a manifold is contained in a slightly larger disk lol

then I realized that's bs

Balarka Sen

hah

P-addict

i guess my concern is with functions that behave weirdly at endpoints, like $\frac{1}{x}$ for the example above on $\mathbb{R}\setminus\{0\}$. i'm not sure how to characterize these

blowing up probably isn't the only way it can fail either, something that oscillates a lot sounds problematic

Balarka Sen

08:21

yeah thats what thorgott said, if the open set U is too big you can't do that

Thorgott

yeah, to extend something on $\mathbb{R}\setminus\{0\}$, you need well-behaved limits at 0

what you want for extension is that there is a bit of room about $U$ to wiggle

P-addict

sorry, i'm not sure what you mean, do you mean something like $\mathbb{R}\setminus[a,b]$?

nvm this would have the same problem

oh but what if we think locally, like each point will have a neighborhood which is really the restriction of a global function? using bump functions probably

Thorgott

yeah, you're right, I'm being imprecise

P-addict

idk, not sure where i'm going with this

Thorgott

when I say "wiggle", I really want a retraction from a neighborhood

the actual reality is that $C^{\infty}(M)\rightarrow C^{\infty}(U)$ is almost never surjective (only if $U$ is a connected component of $M$)

P-addict

08:28

yeah, makes sense

porridgemathematics

that thinking locally thing you said pretty much leads to the definition of germ right

Thorgott

I think a true statement along these lines would be that $C^{\infty}(M)\rightarrow C^{\infty}(\overline{U})$ is surjective if $U$ has smooth boundary and the complement has non-empty interior

(the ultimate statement should be that $C^{\infty}$ is a soft sheaf and not a fine sheaf, but who cares)

Balarka Sen

It is a fine sheaf, not a flabby sheaf.

It is soft and fine.

Thorgott

wait

you're right lol

Balarka Sen

lol

Thorgott

08:32

too many adjectives

Balarka Sen

who cares

P-addict

haha

Thorgott

exactly

the first statement I gave is actually better than the sheaf nonsense

Balarka Sen

yes lol

Thorgott

or, well, they're almost the same, but also kinda not

it's not as artificial, let's put it like this

P-addict

08:34

@porridgemathematics sorry, not sure i understand the relationship to germs here

i might be missing something

porridgemathematics

$\hat{C_{x}^{\infty}}(M) \rightarrow \hat{C_{x}^{\infty}}(U)$, $[h] \rightarrow [h \restriction U]$ is an isomorphism , because of what you said about bump functions

but you need to fix a point for that

robjohn

@AkivaWeinberger $\Omega_n=\frac{2\pi}n\Omega_{n-2}$

P-addict

@porridgemathematics gotcha, makes sense

Thorgott

yes, bump functions are a somewhat germinal concept

they tell you any smooth function agrees with a global smooth function on a slightly smaller open set

porridgemathematics

why is math obsessed with agriculture

P-addict

08:39

i have also wondered this lol

Thorgott

so the canonical $C^{\infty}(M)\rightarrow C_p^{\infty}(M)$ is surjective even though none (say $M$ connected) of the maps in the codirected system is surjective

P-addict

what is a codirected system?

oh do you mean the collection of restriction maps to open sets containing $p$

or rather $C^\infty$ of those sets

porridgemathematics

i need to learn category theory properly at some point :/

P-addict

haha same

i guess there's a way to check if a function should belong to $C^\infty(U)$ by examining it locally and checking it agrees with a global function but i'm not sure how (or if there is a way) to turn this into something $C^\infty(M)$ sees

...well the brute force way would be to reconstruct $M$ from $C^\infty(M)$ but i was hoping for something a little more straightforward haha

Thorgott

what do you mean by "if a function belongs to $C^{\infty}(U)$"? where is your function defined to begin with?

P-addict

08:54

sorry, not really being precise, i meant given some function on $U$ and given the smooth global maps we can check it actually is smooth by asking if it locally agrees with a smooth global map

porridgemathematics

i.e. to see if a function is smooth, you check if its smooth at all points where its defined?

at/around

Thorgott

sure, you could do that, though it sounds a bit antithetical

smoothness is a germinal concept, so you usually want to check it locally

P-addict

yeah, that makes sense

PM 2Ring

09:27

@Koro Also see this (and the other answers on that page):

118

A: What's new in higher dimensions?

In high dimensions, almost all of the volume of a ball sits at its surface. More exactly, if $V_d(r)$ is the volume of the $d$-dimensional ball with radius $r$, then for any $\epsilon>0$, no matter how small, you have $$\lim_{d\to\infty} \frac{V_d(1-\epsilon)}{V_d(1)} = 0$$ Algebraically that's o...

A cute illustration of "almost all of the volume of a ball sits at its surface" is given by the well-known limit for $e$: $\left(\frac{n+1}n\right)^n$. This one converges a little faster: $\left(\frac{2n+1}{2n-1}\right)^n$. — PM 2Ring Apr 23 '18 at 14:19

Russian Bot 2.0

09:53

A person is trying to prove me that evolution is bs

All his claims are baseless

shintuku

12:38

what axiom do I refer to in order to state $\{x_1\} \neq \{x_1, 0\}$?

the second set is a pair

ah extensionality

leslie townes

13:23

@RussianBot2.0 all of evolution? some people believe it for agriculture, because you kind of have to, but disbelieve it for people.

the time scale involved makes it harder to see modern evidence of it for people. people have gotten somewhat taller in living memory, i think because of better childhood nutrition, but natural selection doesn't really operate on people anymore.

robjohn

Medicine cancels natural selection

Someone can't see, give them glasses

someone can't walk, give them a wheelchair

leslie townes

i'm for it. the history of the 'eugenics' movement is not a pleasant one.

robjohn

in the animal world, these afflictions would most likely be deadly

leslie townes

you can see it in domestic animals. ill-advised breeding practices have created breeds of animals that certainly could not survive in the wild and sometimes can barely survive as domestic animals.

robjohn

yep

leslie townes

13:31

but here they are.

robjohn

some bulldogs cannot breed without medical aid

leslie townes

it's actually frightening. if you look at the 'standard' for persian cats 100 years ago, the cats still have normal faces. now they have inverted faces.

shintuku

What's the axiomatic definition of a set $\{a, b, c\}$?

is it $\{ x: x = a \lor x = b \lor x = c\}$?

and similarly for bigger sets?

leslie townes

many pugs can barebly breathe. the cavalier king charles spaniel is prone to seizures. they had one win a pet show that could barely stand up.

i dunno about 'the' axiomatic definition, but that's definitely what that set is.

shintuku

great

now i need to find what $\{ x: x=1 \lor x=2 \lor \cdots \lor x=n \}$ means

leslie townes

13:34

in some formulations of ZF i think you'd define it in terms of two iterations of the pair set axiom. there is also probably an axiom schema that you could fit into.

i was really into set theory for about a year of college. i lost interest, but it is cool stuff. maybe not at this level. i don't need axioms to know what {1,2,...,n} is.

shintuku

i just need to prove this stuff once and I'm satisfied

i'll look into iteration, thanks

leslie townes

i'm happy with not proving it at all, but i understand the impulse. you do have to be careful about what your axioms are. maybe it's the union axiom or axiom of replacement.

the axiom of replacement renders a bunch of the other axioms redundant, i think. it's a high powered schema of axioms.

it's fun to write proofs in purely symbolic language. what i don't get is people mixing a human language like english with symbols. just pick one.

no accounting for taste, i suppose.

vitamin d

can anyone help me to find a sequence of continuous functions f_n so that \int_0^1 f_n dx=0 for all n but f_n does not converge as n->inf for all x\in[0,1]?

shintuku

i'll check those too, thanks!

leslie townes

something like a shifted version of the haar basis for [0,1] might work.

vitamin d

13:46

isn't it discontinous

leslie townes

eh, make it continuous with little diagonal lines.

how about sin(2pinx)?

just thinking out loud.

copper, coming to us live from the emerald isle.

vitamin d

@leslietownes the limit exists for x=0

leslie townes

maybe a version of something we used to call the roving spike. its graph's main feature looks like /\/ and it moves around, e.g. enumerate the rationals in [0,1] and center the nth one on the nth rational.

that might have the same problems at 0 and 1, now that i think about it.

maybe add little spikes that alternate in sign at 0 and 1.

that could work for sin(2pinx) too. add narrow little spikes at 0 and 1 and flip them.

vitamin d

14:05

could you please be more precise? not really sure if I've understood you correctly

leslie townes

a function g_n whose graph is the union of line segments from (0,1) to (1/n,0), (1/n,0) to (1-1/n,0), and (1-1/n,0) to (1,1). then do something like sin(2pinx) + (-1)^n g_n.

maybe that doesn't work.

leslie townes

14:25

to (1,-1) i should say.

robjohn

14:48

@vitamind you want $\lim\limits_{n\to\infty}f_n(x)\ne0$ for all $x$?

vitamin d

I want $\lim\limits_{n\to\infty}f_n(x)$ does not converge

Koro

for any $x\in [0,1\rbrack $?

@vitamind

vitamin d

for any x\in[0,1], yes

robjohn

Then the convergence cannot be in $L^1$.

Because then the $f_n(x)\to0$ for almost every $x$ or at least a subsequence

vitamin d

This is the exercise verbatim: Construct a sequence of continuous functions f_n : [0, 1] → [−1, 1] such that \int_0^1 f_n(x) dx = 0 for all n and f_n(x) does not converge for any x ∈ [0, 1].

Are you saying it's not solvable meaning there is no such sequence?

robjohn

14:59

That is not what I said.

vitamin d

I already tried thing like sin or x-1/2, didn't work out. Could you give me a hint?

robjohn

Have you tried $\sin\left(n\!\left(x+\frac12\right)\right)-\frac2n\sin(n/2)\sin(n)$?

vitamin d

@robjohn the integral is not zero or am I missing something

Russian Bot 2.0

He was referring to all of evolution

shintuku

how do they explain the Advent of Dogs

robjohn

15:16

@vitamind the integral over $[0,1]$ of the latest one is $0$

Since the limit of the additional constant is $0$, it should display the same convergence properties as $\sin\left(n\!\left(x+\frac12\right)\right)$

leslie townes

@RussianBot2.0 not a lot can be done about that, i suppose.

vitamin d

ok thanks robjohn

leslie townes

15:34

i just checked my earnings history with the social security administration. they seem to have lost my wage data from 2004-2007, which is weird, because they used to have it.

thankfully i was a grad student at the time and the net effect on my lifetime benefits is probably 50 cents per month.

i don't know where i'd dig that information up. i shred anything tax related after five years.

also, social security probably won't exist when i retire. if i do retire, which i probably won't.

Christoffer Ringtved

Hi does anyone know what
f(x) = ( x+2, for x < 2
x-2, for 2 < x)
is called?

shintuku

15:50

@leslietownes how come social security won't exist?

leslie townes

no specific reason, just pessimism.

shintuku

comrades, it is time we say we've had enough

robjohn

@shintuku It requires funds, and the funds will have dried up by then.

shintuku

well, there's always the chance the US will finally create a social-democratic party in a period of lesser economic prosperity

leslie townes

there has been so much money spent on convincing americans that the entire project of governance is futile. i doubt it.

shintuku

16:01

yeah something weird happened with the crumbling of the new deal coalition, it's on the top of my reading list

but cardinal numbers first

robjohn

@ChristofferRingtved $x+2-4H(x-2)$ where $H$ is the Heaviside function

Christoffer Ringtved

@robjohn Great thank you just what I was looking for

robjohn

Wolfram calls it HeavisideTheta

John_Krampf

16:17

I LIKE MATH

robjohn

I HATE MILK <-- anagram

John_Krampf

@robjohn D:<

robjohn

My wife really does hate milk. I just suggested to her that we could get shirts.

leslie townes

EMAIL KITH [and kin]

John_Krampf

Maybe she'd like soy milk? I usually drink soy milk, though not because I hate milk but because there's less cholesterol

leslie townes

16:26

my wife had a milk allergy. no bad effects other than huge dark circles under her eyes. switched to soy milk, the circles went away but had digestive trouble. soy allergy too.

i don't know what the 'milk' she gets now is made of.

it's called soylent something.

that's a joke. she can do almond milk.

robjohn

@leslietownes Soylent white is pimples!!

leslie townes

hahahaha

my daughter might have the same allergy. or sensitivity. i don't know what the right word is.

geocalc33

Does $(e^x,e^y) \mapsto (e^{ax},e^{y/a}) $ preserve area?

leslie townes

i can eat basically anything but if i have too much cheese, particularly aged cheese, i break out in hives. i have to manage that. pizza on monday means no leftover pizza on tuesday.

robjohn

@geocalc33 The Jacobian has determinant $1$, I believe

leslie townes

16:28

or i just take hives medication and go for it.

robjohn

My wife is allergic to gluten and milk. Milk is not so bad, she still eats ice cream.

I think soy is okay

leslie townes

the first time i broke out in hives i had to take a bus 90 minutes to the dermatologist i was referred to on the hottest day of the summer. he said, you really shouldn't be out in the heat. try to avoid the heat. thanks doc.

robjohn

my son inherited the gluten allergy and so I have to eat all the bread and cookies that come to us over the holidays

leslie townes

what a sacrifice. sadly my daughter can eat gluten so i have to share with her. it used to be all mine.

geocalc33

@robjohn so the Jacobean determinant equaling 1 implies that the transformation $(e^x,e^y) \mapsto (e^{ax},e^{y/a})$ preserves area?

robjohn

16:37

yes. That is the ratio of the infinitesimal pieces of area (volume, etc) in integrals.

geocalc33

@robjohn and if $x,y \in \Bbb Z$ this area preservation property would still hold you think?

robjohn

and all you're doing is $(x,y)\mapsto(ax,y/a)$

@geocalc33 if you divide, you probably won't stay in $\mathbb{Z}$

leslie townes

also a little weird to think about area for just a map on a lattice.

robjohn

and points have measure? so boundaries are not negligible

@leslietownes my wife makes gluten-free things, and has found some decent gf foods in the market. She found some GF Oreos that she says are just like the original.

geocalc33

@leslietownes why?

leslie townes

16:43

tate's make some ok GF cookies. not the same by a long shot, but better than most.

robjohn

She has the ginger cookies and chocolate chip cookies from Tates

leslie townes

the ginger ones are really good.

geocalc33

@leslietownes can you help me understand whether the lattice of points $(e^x,e^y)$ for $x,y \in \Bbb Z$ is a discrete subgroup of $\Bbb C$?

leslie townes

subgroup under what operation? pointwise multiplication i guess? let me think about the topology.

geocalc33

I might need that x,y \ne 0

@leslietownes for pointwise multiplication

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$Mathematics$ Mathematics