Mathematics - 2023-07-29

Xander Henderson

00:15

@D.C.theIII Well, the contraction mapping theorem says that a contraction mapping on a compact set has a unique fixed point. So if you know that $x \mapsto \sqrt{x}$ is a contraction mapping on $[1,2]$, and you know that it has a fixed point at $x=1$, then you are done, no?

(Compactness isn't even necessary---I'm thinking of the Brouwer fixed point theorem---metric completeness is the right hypothesis, I think).

(But $[1,2]$ is complete, so... meh.)

D.C. the III

@XanderHenderson Only reason I "know" that the fixed point is $1$ is because of the nice picture I drew. But I thought about it a bit and I could define a function as $g(x) = 1 - \sqrt{x}$ on my closed interval and apply Newton's Method to $g(x)$, Newton's method will also be a contraction map but at least I can solve for the root of $g(x)$ and that could show it

Ted would be proud of me in this moment for thinking about it this way...

Xander Henderson

@D.C.theIII $\sqrt{1} = 1$. QED

D.C. the III

Lol

Fair

THis is what happens when I get too caught up in formalism

Xander Henderson

I mean, in general, you are looking for a point $x$ such that $f(x) = x$. Use basic algebra if you can: $x = f(x) = \sqrt{x}$. So $x^2 = x \iff x^2 - x = x(x-1) = x \iff x \in \{0, 1\}$. So your function has two fixed points, only one of which is in the interval you care about.

D.C. the III

00:30

My take away from what you just wrote is to try to simplify more my ideas instead of trying to bring out the big guns unnecessarily

Xander Henderson

@D.C.theIII Sometimes all you need a flyswatter. Keep the nukes in reserve.

2

Xander Henderson

01:06

Man, mantis shrimp are metal af.

Mantis Shrimp Destroys Clam

►

geocalc33

I wonder what gometrical questions one can ask about noise

say the noise is gaussian distributed - one maybe pose a question such as: what is the largest distance between two points in the noise

or, what is the largest simply connected region with no points contained in it?

Shaun

04:09

Why the Hell was this deleted?

math.stackexchange.com/q/4361739/104041

Ted Shifrin

Click on the word deleted and it will tell you.

leslie townes

only the OP and fairly high rep users can see deleted posts, shaun. so the audience for that question is somewhat narrower than the audience for this chat. (which is not to say that i would have anything to say if i could see the post, which i can't)

Shaun

I have made a request to undelete:

0

A: Requests for Reopen & Undeletion Votes (volume 01/2022 - today)

The following was closed and deleted, presumably because is contains two exercises, and a cursory reading might place it, therefore, as needing focus; it does not: it's about an apparent contradiction in Robinson's book on group theory between the two exercises. https://math.stackexchange.com/q/4...

leslie townes

again, non high rep users cannot see what you are linking to there.

Shaun

I know, @leslietownes

Please would you help by voting to undelete, @TedShifrin, if you see fit?

It doesn't give details on why it was deleted, @TedShifrin; it just gives general information.

MathCrackExchange

04:32

@Shaun hey

SO Teams sucks for a math platform, wish SO/SE would get their shit together

I guess we're limited to MSE / the chat for now for study group

Shaun

Hi. Yeah, it's disappointing.

I had my hopes up.

MathCrackExchange

I noticed too that the MSE editor stopped working a year ago

I have to edit my posts on stackedit.io

So this book is really complicated, but we don't have to understand every example

We just want the guts of it

For example here definition of sheave is really unlike I've ever seen, but probably for a reason i.e. more useable in the theory she develops

Ted Shifrin

@Shaun A bot deleted it. It won’t have a human reason.

MathCrackExchange

I say we write our own book, after parsing this one

One that undergrads could pick up and actually read

Would cover the basics of category theory, adjoint functors, etc

Ted Shifrin

It seems to me that the post didn’t serve a purpose. You realized after comments that you were confusing two notions of product. Since no one posted an answer, that’s the end of it.

MathCrackExchange

04:39

@Shaun what parts of category theory do you need brushing up on?

Shaun

@MathCrackExchange It would be similar to this, I'm guessing. (I think it was self published.)

@TedShifrin I see. Thank you :)

@MathCrackExchange Everything.

MathCrackExchange

@Shaun not necessarily like that book. I have Visual Complex Analysis but it's not formal enough for my tastes

Shaun

That's an exaggeration.

. . .on my part.

MathCrackExchange

I say we write the book that college students will be using for the next 50 years

on the topic of categories / sheaves / topoi / homology

All that arrow shit

Shaun

@MathCrackExchange That's ambitious!

MathCrackExchange

04:42

Yes, but we're both really motivated and smaht

You could write the Group Theory chapters

Shaun

Why would anyone want to read such a book by amateurs in the area?

MathCrackExchange

Because it would be better than the professionals'

Written by amateurs because they know what amateurs need to learn this stuff

Shaun

@MathCrackExchange I doubt it.

MathCrackExchange

It seems like TST says "only requires basic category theory". Then she writes as if we're all as smart as her

on the subject

Shaun

But there's nothing stopping us from trying to write a book.

MathCrackExchange

04:44

Don't underestimate yourself

These jokers who write these treatise, are not doing it right

It's a cosmic joke I think

Shaun

I appreciate your confidence in me.

MathCrackExchange

Well, think on it.. We also have at our advantage more time than the current authors who are mainly teaching all the time so they're too busy to write a good book

I think preschoolers should be taught at the most fundamental level, which is not the integers

Ted Shifrin

Some of us jokers with 40 years of knowledge, teaching experience, and explaining things to students think you’re full of s***.

3

MathCrackExchange

Maybe you wrote a good book, and I'm not reffering to your book

Shaun

It's 05:49 here. I should be asleep. See you later :)

MathCrackExchange

04:49

Later!

Maybe we could incorperate or make it a digital book with learning software

@TedShifrin look inside of Theories, Sites, Toposes, and I think you might find yourself wondering the same things we are in regards to how it was written

I can't find one gd good topos theory book. It's a shame

And I'm actually a good writer, got A's in English, make great MSE posts, etc.

123

Hello Everyone...

leslie townes

hi 123

Ajay

05:50

@leslietownes Thanks for the response earlier

Also, where can I find a proof of the triangle inequality in the two dimensional case?

leslie townes

two dimensional case of what? R^2? C^2? even at that low level of generality it is often proved via a route (such as the schwarz inequality) that generalizes to any R^n or C^n. see the 62-upvoted answer in math.stackexchange.com/questions/91181/…

Ajay

R^2

Is there a way to do it by avoiding the schwarz inequality

leslie townes

well, you don't have to prove the general inequality. you could prove it just for R^2. i don't know if it's any easier.

Ajay

I need to prove that d(p,r) ≤ d(p,q) + d(p,r)

leslie townes

speaking of "avoiding" things at this level is, uh, not natural to my mind, but there are certainly lots of proofs that maybe would not route through the schwarz inequality.

in R^2 with the usual operations you would simplify that a little by setting x = p-q and y = q - r and proving that ||x + y|| <= ||x|| + ||y|| for any x and y. note that this reduces the number of R^2-valued things to consider by one. you have gone from p, q, r to just x, y.

math.stackexchange.com/questions/2122350/… is one example that is specific to R^2.

i don't find looking at that any more congenial than looking at slightly more abstract proofs of things like the schwarz inequality, but stuff like that is out there.

Thomas Finley

09:06

3

Q: Can this solution of a Ring Theory problem be improved?

If $R$ is the ring of integers, let $U$ be the ideal consisting of all multiples of $17.$ Prove that if $V$ is an ideal of $R$ and $U\subset V\subset R$ then either $V = R$ or $V = U.$ I tried, solving this problem as follows: First of all, it's given $R$ is the ring of integers. We assume the op...

Any suggestions in here ?

Jakobian

Horrible notation. Reminds me of open sets too much

(17) is a non-zero prime ideal, hence maximal

Because R is a PID

QED

PNDas

I am reading Calculus of variations from Evans' PDE. Can anyone recommend any supplement text? I mean a CoV book which is more focused on PDE and is easier to undertsand.

For example Evans let $i(\tau)=I[u+\tau v]$ and then found $i''(0)$ which is a big expression and then he took cut off function did some magic to get a nice compact expression $\sum L_{p_1p_j}(Du,u,x)\xi_i\xi_j\geq0$. Obviously this looks nice, $D_p^2L$ becomes positive semi definite, but why?

Oren Diskin

10:57

hello :)
https://math.stackexchange.com/questions/4743895/continuous-map-sending-dense-subset-via-a-homeomorphism/4743921?noredirect=1#comment10063944_4743921

asking about the lemma used in this question
I feel like we are showing that we can not extend a continous function $f':D\rightarrow Y$ to $f:X\rightarrow Y$ and it would still stay continous

Jakobian

The function that we are applying the lemma to has different codomain, namely not $Y$ but $f[D]$

what the lemma says is that we can't extend $f':D\to f[D]$ to a function $f:X\to f[D]$

or even, by one point

Oren Diskin

so how come in the question they say such a function exsits?

Jakobian

that is, all points of $X\setminus D$ are being sent to $Y\setminus f[D]$

Oren Diskin

ohhhhhh

Jakobian

well, $f[D]$ doesn't have to equal $Y$

Oren Diskin

11:01

okay, very cool, thank you :)

Jakobian

np, have a nice day

Oren Diskin

11:34

I'm having a bit of problem proving
Then 𝑓↾𝐷∪{𝑥}
is a continuous extension of 𝑓↾𝐷
to 𝐷∪{𝑥}
is there a simple way to see it?

its at the end (second line from last)

Jakobian

> Associated with Math.SE; for both general discussion & math questions alike. Just ask; don't ask to ask. Rarely if ever expressible as a ratio of integers. Chat guidelines: tinyurl.com/hzl2955 | $\LaTeX$ in chat: tinyurl.com/cfqcvpc

see here for LaTeX in chat

@OrenDiskin $f\restriction_A$ is just $f$ but on a set $A$

so since $D\subseteq D\cup \{x\}$, we have $f\restriction_{D\cup \{x\}}(y) = f(y) = f\restriction_D(y)$ for $y\in D$

this is precisely what it means for $f\restriction_{D\cup \{x\}}$ to extend $f\restriction_D$

it's a continuous extension because $f$ is continuous

Oren Diskin

Dont I need to prove for any open set A in $D\cup {x}$ $f^{-1}\restriction_{D\cup \{x\}} (A) is open in D\cup \{x\}?

Jakobian

restrictions of continuous functions are continuous

This follows from $(f\restriction_A)^{-1}[U] = f^{-1}[U]\cap A$ being open in $A$ whenever $U$ is open

you probably done an exercise like that, and if not, well here you go

Oren Diskin

11:51

excellent, thank you very much :)
and have a great day

one potato two potato

12:06

taking shower strangely wakes you up

Nick

12:20

I have produced some graphs today: https://imgur.com/a/07izsOm
Is there anything interesting to take note from them?

Jakobian

@onepotatotwopotato I think it depends on the temperature

it can't be too hot

冥王 Hades

12:58

Prove that $\Phi=30^\circ$

No calculators

The Empty String Photographer

B is 84 (I’ll drop the degree symbol)

D is 126

Draw line AD

actually don’t

Draw line between D and AB midpoint

And between B and CD midpoint

Those last two lines have equal length.

DC midpoint then has angle of 12

So AB midpoint is also 12

By scaling, we get that A is also 12.

No I’m wrong!

NVM

sunny

13:26

> Proposition. Suppose that $f:A\to\mathbb{R}$ and $c$ is a limit point of $A$. If $\lim_{x\to c} f(x)$ exists, then $f$ is locally bounded at $c$.

> Proof. Let $\lim_{x\to c} f(x)=L$. Taking $\epsilon=1$ in the definition of the limit we get that there exists a $\delta>0$ such that $$0<|x-c|<\delta \text{ and }x\in A \text{ implies that }|f(x)-L|<1.$$ Let $U=(c-\delta,c+\delta)$. If $x\in A\cap U$ and $x\neq c$, then $$|f(x)|\leq |f(x)-L|+|L|<1+|L|,$$ so $f$ is bounded on $A\cap U$. (If $c\in A$, then $|f|\leq\max\{1+|L|,|f(c)|\}$ on $A\cap U$.)

I don't understand the last part. Why, if $c\in A$, is $|f|$ bounded by the max of $1+|L|$ and $|f(c)|$ on $A\cap U$?

I guess simply because $x=c$ is the only point that is not covered in the proof. Cheers.

one potato two potato

@Jakobian you like cold showers?

冥王 Hades

@TheEmptyStringPhotographer Yeah this doesn’t really work

Jakobian

13:56

@onepotatotwopotato no, but I like when they're cold enough to be refreshing

Thomas Finley

14:18

A scalar is a mathematical object that can be represented by a real number. For eg: 5 is a scalar quantity. Scalar opeartions are those operations we use from our childhood like addition, subtraction, and so on. - Am I correct?

I am speaking from a mathematical point of view

Jakobian

@ThomasFinley not really

we usually say scalar when we're in the context of vector spaces

and it can refer to an element of the field the vector space is over

this could just as well be a complex number

one potato two potato

@Jakobian any personal tips for refreshing except cold shower?

冥王 Hades

Ice bucket challenge

I’ve done it

Jakobian

@onepotatotwopotato sitting in the window

brushing your teeth

washing your face

drinking cold water

The Empty String Photographer

14:36

@onepotatotwopotato Get someone to scare you every morning

Jakobian

14:58

I didn't think we were talking about waking up in the morning

But about this, around hour after waking up, drink a coffee

I don't think there's any way to deal with morning tiredness though, you'll always have to endure a little bit of it

coffee can deal with some of it, but it'll only be effective after ~hour after waking up

one potato two potato

Coffee transfusion.

冥王 Hades

15:32

Bingo

MathCrackExchange

I apologize for talking shit about authors last night. Turns out the book is actually awesome / mesmerizing / a great book. I will definitely be reading it this year (or next 2 years lol).

user 858770

15:56

Yeah, I was wondering what that was about. Writing is a difficult task.

@冥王Hades good work.

Thomas Finley

@Jakobian I am reading from a book by Hoffman and Kunze. It doesnot match with your terminology. Instead, it says, this:

In here, it's just an element of a certain set with two specific operations following some rules.

Jakobian

@ThomasFinley ? It says exactly what I said

Thomas Finley

2 hours ago, by Jakobian

and it can refer to an element of the field the vector space is over

@Jakobian You said this.

But my point is,

it refers to an element of a certain set with two specific operations following some rules. Here one of the field is F, but a vector is not in F.

It's in a different field

It's in V. It is also a field

But to avoid ambiguity, it's better to state which field, isn't it?

Jakobian

16:19

@ThomasFinley we call this a field

wdym different field I didn't specify

I only said the vector space is over that field

here the vector space V is over the field F

this is precisely what I said

unambiguous

Ted Shifrin

$V$ is NOT a field.

Jakobian

oh that's what you meant

$V$ is not a field like Ted said

well either way, I said field the vector space is over so there shouldn't be any confusion

Thomas Finley

@Jakobian to end all ambiguity: Yes, what you said maybe true. But, in my book, it defines a vector precisely to be an element of a set $V$ endowed with two operations satisfying certain criterias, in a Vector Space.

@TedShifrin Umm..can you clarify that what I wrote in bold is a standard definition or at least correct defn or not?

user 858770

Have you ever read this quotation before @copper.hat? "The faculty of voluntarily bringing back a wandering attention, over and over again, is the very root of judgment, character, and will."

Jakobian

16:34

@ThomasFinley ?

I never said anything about vectors

you're just not reading what I'm saying

Prithu Biswas

16:47

@XanderHenderson To me, math and science are two old friends whose collaborative work has enchanted the world around us.

copper.hat

@user858770 I have not, but I like it :-)

Ted Shifrin

@Thomas The confusion is yours, not Hoffman & Kunze's. The field gives you the scalars by which you multiply vectors. The set of real polynomials is a vector space $V$ over $\Bbb R$. The polynomials are elements of $V$. You never refer to $V$ as a field. It is not.

You should look up the definition of a field — separate from this.

sunny

18:41

> Definition. Let $f:A\to \mathbb{R}$, where $A\subseteq \mathbb{R}$. If $c\in\mathbb{R}$ is a limit point of $\{x\in A: x>c\}$, then $f$ has the right limit $$\lim_{x\to c^+}f(x)=L,$$ if for every $\epsilon>0$ there exists a $\delta>0$ such that $$c<x<c+\delta \text{ and }x\in A \text{ implies that }|f(x)-L|<\epsilon. $$

Why does $c$ have to be a limit point of the set $\{x\in A: x>c\}$? Specifically, why require $x>c$? What would be incorrect about $x\geq c$?

Ted Shifrin

You can't take limits as you approach an isolated point. There are no nearby $x$. And you can't allow $x\ge c$ because we never allow $x=c$ when we discuss $\lim_{x\to c}$.

sunny

ok

MathCrackExchange

Rasp Pi updating so I can chat for a few

Or study :)

MathCrackExchange

18:59

$\text{Mor}(C)$ sounds a lot like Morrisey

I know how to remove for all from my notes

Jakobian

raspberry pie

MathCrackExchange

Simply assume all unbound variables are foralls, and then special notation for when they're not

Jakobian

updating my cheesecake

MathCrackExchange

Foralls everyehwere is ugly and complicates things, we can assume them because usually in category theory it's $\forall \forall \forall ... \exists$

tfph = the following properties hold

Jakobian

cheesecake successfully updated into my mouth

MathCrackExchange

19:03

Jakobian, my pi updated when you said that

Maybe your entangled with my Raspi chip

Jakobian

I think so

Ted Shifrin

19:29

@Jakobian Make mine apricot or lemon, please.

peek-a-boo

19:44

@TedShifrin do you know/does there exist a catchy name (like the regular-value theorem) for the theorem about preimages of submanifolds?

MathCrackExchange

Preimages of submanifolds is descriptive

Ted Shifrin

@peek-a-boo Nope. But it is equivalent to the regular value theorem. :)

By the way, I have 4 volumes of Diedonné in French, but I’ve never looked at that section. I agree with Didier that the language is unexpected.

Avv

20:23

Hello,

For the following problem, the red-green-blue color "#AABBCC" can be written as "#ABC" in shorthand.

For example, "#15c" is shorthand for the color "#1155cc".
The similarity between the two colors "#ABCDEF" and "#UVWXYZ" is -(AB - UV)2 - (CD - WX)2 - (EF - YZ)2.

Given a string color that follows the format `"#ABCDEF"`, return a string represents the color that is most similar to the given color and has a shorthand (i.e., it can be represented as some "#XYZ").

Any answer which has the same highest similarity as the best answer will be accepted.

the math based solution is to increment the target color based by 17 for each of the 3 pairs of RGB color and then subtract the corresponding RGB color in the input color to find the similarity. I tried to increment the RGB color by 1 for each R, G, B and then subtract it from the corresponding R, G, B in the input color, but my answer is wrong!

Let $$ v $$ be a decimal value which is the result of parsing the hexadecimal string $$ s $$.

$\[ v' = \frac{v}{17} \]$

Then, an adjusted value $ v'' $ is defined as follows:
$$
\[ v'' = \left\{
\begin{array}{ll}
v' + 1 & \mbox{if } v \mod 17 > 8 \\
v' & \mbox{otherwise }
\end{array}
\right. \]
$$
Finally, the output hexadecimal string $$ s' $$ is obtained by converting $$ 17 \times v'' $ $to hexadecimal.

$\[ s' = \text{{toHexString}}(17 \times v'') \]$

robjohn

21:17

Hey, @Ted. How is the weather there? We were in 102° just a bit ago.

Rosie got a gash in her right shoulder and we took her in to get some stitches. We just got home.

We have no idea how she got it, but we just noticed it this morning.

Ted Shifrin

Oh, poor thing! Weird how you don’t know how! … It’s humid but low 80s here. I just walked over to CVS to get a prescription but had forgotten they close at 2 for lunch,

Xander Henderson

21:34

@robjohn We went to Springerville today for the hummingbird festival, then drove home the long way through the Sitgreaves. We got some rain in the mountains, and the temperature got down to 70°F. 3000 ft in elevation lower, at home, the temperature is over 90°F. Not as bad as where you are, but annoying nevertheless.

leslie townes

high 70s and not humid here. munchkin says it's too sunny.

Xander Henderson

@leslietownes Bright kid.

冥王 Hades

21:53

Today I learned that punching a computer screen and breaking it can also draw blood

Don’t ask me how I learned that though I just did

Jakobian

self-explanatory

I'm just worried about your anger issues

leslie townes

some guy kept posting general topology questions that nobody born after 1880 cares about and i just lost it

shintuku

might as well go study history smh

Jakobian

couldn't be me

冥王 Hades

It was more stress than anger. Trust me, competitive gaming can get really, really, really stressful

I was gonna purchase a new monitor anyway, I need a monitor with a refresh rate of 240Hz

Jakobian

22:02

I think you need a break

@leslietownes By the way I think the issue whetever or not Bing's example G is submetacompact for $P = \mathbb{R}$ is very important

leslie townes

i was gonna post a general topology question on here because i thought i'd found a friend for you, but it turns out when i clicked into it that it was actually your question and some other user was just editing it

doesn't like your prose stylings, apparently

shintuku

lol i didn't realize we were talking about jakobian my bad

Jakobian

@leslietownes I was my own friend all along

2

Xander Henderson

@冥王Hades "need".

冥王 Hades

@XanderHenderson yes, high refresh rate is key for competitive gaming. It is a need.

Xander Henderson

22:08

@冥王Hades "need".

Jakobian

@leslietownes that user that was editing it is the guy reponsible for pi-base database

I contribute a lot there

冥王 Hades

@XanderHenderson Okay fine you can do it at 60Hz but still, you really should have 240Hz if you’re serious about competitive gaming

Xander Henderson

@冥王Hades I disbelieve. The human eye can't perceive a difference much past 100 Hz. 240 is overkill.

冥王 Hades

@XanderHenderson I assure you that myth has been dispelled within the gaming community, the difference between 144Hz and 240Hz is very real and had notable impact on gaming performance. You can check Linus Tech Tips’ detailed video on it

Even having frame rates that go that high reduce input lag and latency which results in better reaction time

Xander Henderson

Yeah, I don't buy it.

And I am fairly sure that I have seen the video you are referencing.

冥王 Hades

22:13

@XanderHenderson Well once you do game on a 240Hz panel (like I have), you’ll see what I mean. Frame rate is one of those things that you won’t know you “need” it until you’ve tried it

60 feels perfectly fine and smooth, until you try 144, and so on

Xander Henderson

@冥王Hades I barely see the difference between 60 and 120. I couldn't care less about anything higher.

冥王 Hades

@XanderHenderson Some people are less sensitive to the difference yes, but it does exist. I can absolutely never go back to 60 after playing on 165Hz

Xander Henderson

I could possibly see an argument in favor of the very highest level gamers noticing some difference, in the same way that professional baseball players can feel the difference between more tightly wound balls or corked bats.

But for anyone who is at a level where they are still buying their own gear, I think that they are being taken for a ride if they shell out for such an expensive monitor which isn't likely to actually give them any real edge.

It is like gold plated monster cables.

冥王 Hades

@XanderHenderson It’s less about “highest level gamers” and more about the games you play. If you’re playing a story based games 240Hz is useless. However if you’re playing something like a fast paced FPS it can really help

Ted Shifrin

@robjohn I may have lied. I suspect it's closer to the neighborhood of 90 now.

Xander Henderson

22:17

@冥王Hades Again, past 100 Hz or so, I don't buy it.

冥王 Hades

@XanderHenderson Xander, you may not notice the difference, but it is very real. The frame-time graph (aka latency) alone shows it.

Xander Henderson

My claim is that the vast majority of humans will see no real difference past 120 Hz. I don't know what you mean by "frame-time graph", but that doesn't seem to be about human perception.

冥王 Hades

@XanderHenderson well yes most people will not notice much beyond 100Hz, but competitive gamers will and it has its advantage even beyond just perception. Frame-time graph just shows how much time passes between each frames. If you’re rendering 60 frames per second, there’s a delay of 16.5 milliseconds between each frame aka your latency

its done by inverting the frames per second formula to seconds per frame.

For 240 frames per second, your latency is 4.1 milliseconds, a quarter of that. It makes a huge difference in competitive gaming

Xander Henderson

@冥王Hades I already granted your first point: "I could possibly see an argument in favor of the very highest level gamers noticing some difference" chat.stackexchange.com/transcript/message/64093541#64093541

But at that level, someone else is paying for your gear.

冥王 Hades

@XanderHenderson for most yeah, but I’m paying for mine

Xander Henderson

22:24

@冥王Hades Which suggests that you aren't on that level.

Let's double blind your playing, and see if you do any better (or worse) at 240 Hz.

冥王 Hades

@XanderHenderson I mean I’ll swallow my pride and admit I’m not as good as the highest level players, but I do play against them and I’ll take any advantage I can get.

Xander Henderson

Putting on a pair of Air Jordans aren't going to get you into the NBA.

冥王 Hades

Well I certainly can’t perform any worse, and the little advantage it does have (regarding latency that has nothing to do with perception) is helpful nonetheless

@XanderHenderson As far as I know the average NBA Height is 6’3”. I’m 6’4”, just barely above average

Xander Henderson

@冥王Hades I mean, if it makes you happy to spend $1000+ on a monitor which has negligible impact, great. Buy you expressed it as a "need". Be honest with yourself: it is a want.

@冥王Hades I fail to see the relevance. I was making a point about equipment not making the competitor.

冥王 Hades

@XanderHenderson Well, okay you’re right. Even my gaming computer is just a want (I can do most of my work ona far less powerful machine that certainly doesn’t need a $1,600 GPU), but I would be really bored without it when I’m not doing math

Xander Henderson

22:31

When I was fencing competitively, I went through a lot of blades. I sometimes bought really expensive Uhlmann blades, and sometime bought really cheap non-name Chinese blades. I was good enough to feel the difference (and to actually prefer the noodly Chinese blades to some extent, but they favored a flick, and there are expensive blades that supposedly do the same thing), but the blades themselves made little difference at my level.

Man, I miss fencing. :/

But it trashed my knees. I can't really do it anymore.

冥王 Hades

@XanderHenderson Would it do any good to tell you I’ve managed to defeat those really high level players a few times?

I barely managed to do it but a victory is a victory

Xander Henderson

@冥王Hades At the top of my game, I scored touches against Sean McClain. Doesn't mean I was on his level.

(For what it's worth, the first time I came up against him was at a North America Cup event in 2002, not long after he won the US championship).

冥王 Hades

Welp

At least I’m near the top in the gaming club

robjohn

22:50

@XanderHenderson Hummingbird Festival! Sounds like a nice thing. My wife and I would like to go to something like that.

Xander Henderson

@robjohn It was fun.

And took that when I got home.

Not the bird I've been trying to photograph for the last week, but pretty nonetheless.

robjohn

@TedShifrin That's pretty high for there. I hope you survive the heat.

In the nest is the best shot I've been able to get of a hummingbird.

I might get better if I hung around our feeder.

Xander Henderson

@robjohn I got a couple of nest pictures last year, but I can't figure out where they are nesting this year.

The branch that they used last year fell off over the winter. :/

Ted Shifrin

Make sure you hum when you hang around. Reminds me of Winnie the Pooh and the bees.

robjohn

@TedShifrin No, but I flap my arms really fast.

Oh, bother.

sunny

23:09

Suppose we have a statement like $$\forall \epsilon>0, \exists \delta>0, \forall x \in D \Big(P(\epsilon,\delta,x)\text{ and }Q(\epsilon,\delta,x)\Big).$$ Can we then just distribute the quantifiers over the conjunction and obtain an equivalent statement?

Xander Henderson

@sunny I really don't like to think about things in this kind of symbolic manner, but I think that you will have problems with $\delta$.

As it is now, there is a $\delta$ which satisfies both $P$ and $Q$ (whatever those statements are). If you "distribute" that quantified doohickey, you may end up with different $\delta$s for $P$ and $Q$.

That is, $\exists \delta \text{ st }(P(\delta) \land Q(\delta))$ seems to be a different statement from $(\exists \delta \text{ st }P(\delta)) \land (\exists \delta \text{ st }Q(\delta))$

But, again, I don't really like to get into the nitty-gritty of symbolic logic all that much, so I could be way off base.

sunny

Ok. I can give some context, actually. It is in regards to this post (see my comment).

The answerer goes from $$\forall\varepsilon{>}0\,\exists\delta{>}0\,\forall x{\in} D\,\Big(\big(c{<}x{<}c+\delta\implies|f(x)-l|{<}\varepsilon\big) \text{ and
}\big(c-\delta{<}x{<}c\implies|f(x)-l|{<}\varepsilon\big)\Big)$$ to $$\forall\varepsilon{>}0\,\exists\delta{>}0\,\forall x{\in}
D\,\Big(c{<}x{<}c+\delta\implies|f(x)-l|{<}\varepsilon\Big) \text{ and
}\forall\varepsilon{>}0\,\exists\delta{>}0\,\forall x{\in}
D\,\Big(c-\delta{<}x{<}c\implies|f(x)-l|{<}\varepsilon\Big).$$

Xander Henderson

Yeah, okay, I'm out. I don't have the energy to try to parse that.

sunny

We're trying to prove that if the limit exists, then the right and left limits equal :)

Xander Henderson

@sunny Yeah, that symbol pushing looks like absolute overkill to me.

If the limit exists, then if $x$ is "close" to $a$, then $f(x)$ is close to $L$. Adding the restriction that $x < a$ or $x > a$ doesn't really do anything---if $x$ is close to $a$, then it doesn't matter if it is larger than $a$ or smaller than $a$. So that direction is relatively quick (you can, of course, zhuzh that up a little and make it a bit more formal).

Jakobian

23:21

they mean that they split a quantifier into two

or well, three quantifiers

Xander Henderson

Going the other direction, if $x$ is close to $a$, then either it is bigger than $a$, and you can use the fact that the right limit exists, or it is smaller than $a$, and you can use the fact that the left limit exists. And the two are equal, so you get a limit. Again, zhuzh as required.

Jakobian

So the issue is determining the logical laws $\forall (\phi \land \psi) \implies \forall \phi \land \forall \psi$ and $\exists (\phi \land \psi)\implies \exists \phi \land \exists \psi$

sunny

@Jakobian the first one is an equivalence, right?

Jakobian

yeah

the second isn't

so you just apply those laws three times in total

sunny

that makes sense, but

then we end up overall with an implication rather than equivalence

Jakobian

23:25

I thought that was the intention ?

sunny

because the second is just an implication

yeah, I guess

CroCo

Hello everyone, I have a question regarding abstract algebra. What is the reason for defining a group only on a binary operator? Can a group be defined on an unary operator?

Jakobian

how would that work

CroCo

For example, let $S={0,1,2}$ and $*a = |a|$ then if $a\in S$, then $*a \in S$.

Jakobian

@CroCo you need to clarify your question

CroCo

23:34

@Jakobian is it possible to define a group as a non empty set with an unary operator?

Jakobian

that wouldn't be a group then

CroCo

what is it?

Jakobian

I don't know. How am I supposed to translate the axioms of a group to a unary operation?

CroCo

I know a group is defined that way. It is a definition after all. What I'm trying to understand why we define it that way.

The unary operator is handy as well.

Jakobian

Okay let me ask you this... what properties do you expect for such "group" to have

CroCo

23:39

Associativity, identity and inverse elements

Jakobian

associativity doesn't have definition for unary operations

CroCo

@Jakobian true. Do we have any notion associated with unary operations on sets?

Xander Henderson

@CroCo Because that is the definition of the word "group".

I suppose that one could define some kind of structure on sets with unary operators.

Jakobian

yeah. A set with unary operation

Xander Henderson

For example, a valuation could, in principle, be defined on a set with no further structure, I suppose.

CroCo

23:44

@XanderHenderson true. It is defined that way. But why unary operations on sets are ignored in some abstract algebra textbooks.

Xander Henderson

Though isn't an "unary operation" just a map from the set to itself?

shintuku

it's not a very interesting object algebraically i guess

Xander Henderson

@CroCo Because they aren't relevant to what abstract algebra is seeking to study.

Jakobian

main objects of study in abstract algebra are groups, rings, field etc.

Xander Henderson

I mean, I'm having profound difficulty coming up with an example of an unary operation $f : X \to X$ which is of general interest.

Generally, we ask about functions which preserve some kind of structure (e.g. homomorphisms, automorphisms, homeomorphisms, etc).

CroCo

23:46

@XanderHenderson how so? Let $S=\{0,1,2\}$ and $*a=|a|$. It is clearly $*: S \rightarrow S$.

Xander Henderson

But those presuppose some other kind of structure already in place (a group, or ring, or topology, or whatnot).

CroCo

@shintuku this is the only guess I have.

Xander Henderson

@CroCo What does $|a|$ mean for an element of your three element $S$?

Jakobian

there's no reason to study unary operations

CroCo

@XanderHenderson good question. I will say the absolute value in the usual way. But I think you're going to ask me how you define it then?

shintuku

23:48

yeah then your set needs to be bigger if it means anything at all, otherwise it is just identity

Xander Henderson

Presumably, an "unary operator" is just a function from a set to itself. I see nothing intrinsically interesting about such functions, unless they preserve some other structure. The unary operator itself is not interesting---how it interacts with other structures is what is interesting.

@CroCo If you regard $S \subseteq \mathbb{R}$ (or the integers, I suppose), then $|a|$ makes sense, but is just the identity on $S$. Seems boring...

Jakobian

Groups in universal algebra where they care about such things as n-ary operation, are actually equipped with an important unary operation $x\mapsto x^{-1}$

but this is not the only operation on a group

CroCo

@XanderHenderson "seems boring" doesn't confuse me.

Xander Henderson

@Jakobian Sure, that is an operation you can do on a set which already has some other structure. If you don't have a multiplicative structure of some kind, what does $x^{-1}$ even mean?

It isn't the unary operation that is interesting---it is the effect it has on the other structures.

CroCo

So let me ask the question in more lucid way. Can we define a set with an unary operator? I think you're going to say you can define whatever you want, but is it useful?

Please forget about groups.

Jakobian

23:52

You can define a set with unary operator

shintuku

yeah, consider the shift operator on the set of permutations of 1,2,3 @CroCo

Xander Henderson

@CroCo Again, "a set with an unary operation" is just a set $X$, and a function $f : X \to X$.

Jakobian

those are precisely universal algebras with one function symbol corresponding to a unary operator

Xander Henderson

Of course you can define such a thing. But what about such a function makes it interesting to study?

shintuku

@CroCo 312 -> 231, 321 -> 132, etc.

CroCo

23:55

@XanderHenderson It's this I'm trying to explore.

Jakobian

I'd say having only a unary operation on your structure is too simple to be useful

Xander Henderson

@CroCo And I am asking you the question you need to answer in order for that exploration to be meaningful.

Jakobian

nothing will come out of studying such structure

unless you're studying Peano arithmetic...?

Xander Henderson

Mathematicians tend not to ask "Can you do [thing]?" Rather, they as "What can you do with [thing]?"

Suppose you define a function $X \to X$. Now what? What do you get from that?

What's the next step?

CroCo

@XanderHenderson merely an innate curiosity but I appreciate you illuminating me.

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$Mathematics$ Mathematics