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12:07 AM
@DanDonnelly did you find an answer
also, on the subject of english
"Writing is inhibiting. Sighing, I sit, scribbling in ink this pidgin script. I sing with nihilistic witticism, disciplining signs with trifling gimmicks --impish hijinks which highlight stick sigils. Isn't it glib? Isn't it chic? I fit childish insights within rigids limits, writing schtick which might instill priggish misgivings in critics blind with hindsight. I dismiss nit-picking criticism which flirts with philistinism."
the guy only ever uses one vowel per chapter
so, this is a mixture of satisfying and profoundly dismaying
the good: our book is finally up for download on Springer! link.springer.com/book/10.1007/978-3-030-85939-8
woah, that's a big achievement congratulations
the bad: no matter how happy you are with what you did at the time, you always find stuff you'd have done differently
the ugly: read the chapter titles that Springer has listed :3
@shintuku ty
12:27 AM
@TedShifrin bough cough dough lough rough sough tough trough through thorough borough plough slough hiccough
(bau coff dou lock ruff sau tuff troff throo thorou buruh plau sloff hiccup)
@shintuku how monochromatic!
(three stanzas, one vowel per stanza)
12:44 AM
@LeakyNun nor do breach nor break nor bread
1:40 AM
semi: so tell us, how do you peneralize to singlet state of two
is this legal in all 50 states?
@shintuku nope, still an open question
1 hour later…
3:03 AM
Let $f:[0,1]\to\Bbb R$ be a function with $f(0)=0$ and $0<f(y)/y\leq f(x)/x$ for any $0<x<y\leq 1$. Then is it true that $\int_0^1f(x)dx<\frac{1}{2}f(1)$?
So far for polynomial $f$, I think I can prove it because $f(0) =0$ implies $f(x) = xg(x)$ for some $g$. I'm not sure I'm able to write for just function.
@PeterJohn bro it's malformated
Ignore the above question lots of typos
Let $f:[0,1]\to\Bbb R$ be a continuous function with $f(0)=0$ and $0<f(y)/y< f(x)/x$ for any $0<x<y\leq 1$. Then is it true that $\int_0^1f(x)dx>\frac{1}{2}f(1)$?
For polynomial $f$, $f(0) = 0$ implies $f(x) = xg(x)$ for some polynomial $g$. But for continuous $f$, I'm not sure I'm able to write in that way.
3:29 AM
@PeterJohn you still haven't fixed it
3:51 AM
@LeakyNun Anything more I need to fix?
@PeterJohn yes, you were missing an opening $
it'd be hard to point to where it is so I've just fixed it for you
@hyper-neutrino Oh right. Thanks. and I found that any function with $f(0) = 0$ can be written as $f(x) = xg(x)$ for some $g$.
such a g might not be continuous at 0, or even bounded near 0. e.g. f(x) = sqrt(x)
thinking geometrically it looks like the hypothesis implies that f is strictly 'concave down' and hence that its graph ought to lie strictly above any secant line to the graph. if that's true, then f(1)/2, being the trapezoidal rule approximation of the integral corresponding to partition [0,1], ought to be an underestimate of the integral
i haven't looked deeply into this
@PeterJohn do you want to assume $f$ Is differentiable?
4:07 AM
@PeterJohn $\int_0^1f(t)\, dt=\int_0^1\frac {f(t)}t.t\,dt\le \int_0^1 \frac{f(1)}1 t\,dt$
Wrong inequality but right argument, Koro?
Use continuity to get strict inequality. @PeterJohn
Read his hypotheses carefully, Koro.
you're right. My inequality is reversed hence not what the question asker desired.
Yup, so that proves what he wants.
4:24 AM
2 hours later…
5:55 AM
is there some name for the somewhat weird operation of minimizing the distance+squared distance metric
6:12 AM
Given a set of orthonormal basis B of a finite dimensional vector space V, is it possible to create an another orthonormal basis B' of V by using B?
Permute them?
For example: if $e_1,e_2$ are standard orthonormal basis of $R^2$ then is there a way to get: $\frac 1{\sqrt 2} (e_1+e_2), \frac 1{\sqrt 2}(e_1-e_2)$?
Just apply any orthogonal matrix.
That's necessary and sufficient.
i agree
I guess our work is done here for tonight.
6:17 AM
Geometrically, that would mean rotating x-y axis (treating $e_1$ along x-axis and $e_2$ along $y-$ axis) :)
Well, in 2D, that plus reflection is all you can do.
If you've never done that exercise, you should. Show that the set of $2\times 2$ orthogonal matrices is two copies of a circle.
I don't understand. What is "two copies" please?
orientation-preserving orthogonal matrices are rotations, parametrized by a circle
One circle for rotations. The other circle for a reflection composed with a rotation.
Ohh I see.
6:19 AM
Write actual formulas for these matrices and you'll see.
I have a matrix with entries like $\frac {\pm x (\text {or } y)}{\sqrt {x^2+y^2}}$ in mind.
Trigonometry is your friend.
Hi All...
ew, no, parametrize with t
ted has been going on about this geometry stuff, maybe there's something to it
which is like cos t and sin t
6:23 AM
Right, @Koro, with perhaps a minus sign or more sprinkled around.
hmm, I got it Ted.
Yeah, you really must think about this stuff geometrically. Even leslie admits it.
I do actually :)
Finally, the subtitle of my linear algebra book makes sense :D
Currently I am studying singular value decompositions. I learnt something about its use in transferring signals over long distances but don't remember that now.
So I'll soon see what SVD is geometrically or what its applications are.
6:25 AM
i find that yelling performs the same function with considerably less computational overhead
I understood that every matrix is diagonalizable :)
(if we allow two basis)
SVDs are an amazing tool.
spent the day at a trade show in moscone west
You mean $P^{-1}AQ$?
Oh yeah, @copper, you said you were gonna be there.
6:27 AM
had a lovely lunch funded by generous sales folks
Better than Mcdonalds?
one of my favourite Italian comfort dishes
Ah, yes, takes you back to your "puta" background.
plus a starter of smelt
not sure of the origin of the name.
i learned not to go for osso bucco in a work situation.
after i planted the entire dish in my lap once.
What I just said is the origin of the name, assuming you know what that is in Eye-talian.
6:30 AM
No Ted, I mean I don't know yet. Axler does SVD a bit differently (different than Strang's). There should be similarity between the two but I have not yet tried to see the similarity). If T is an operator on V then there exist orthonormal basis $e_i$'s and $f_i$'s such that $Tv= s_1\langle v,e_1\rangle f_1+...s_n\langle v,e_n\rangle f_n$ for every $v\in V$ and $s_i$'s are singular values of $T$.
The name comes, I think, from the garlic and capers.
dinner was palak paneer
@Koro That is what I wrote, actually. Different change of basis matrices for domain and range. Check it out.
But saying every matrix is diagonalizable is very misleading.
@copper.hat Ohh, did you try soan papdi?
6:32 AM
I don't recognize that name. What is it?
@TedShifrin I did say with respect to two basis.
@Koro i am not too fond of dessert in general
You said it quite a bit later after I said "huh?"
@TedShifrin it's a sweet :)
It's very tasty. :)
Every matrix can be written as $\left[\begin{matrix} I & 0 \\ 0 & 0 \end{matrix}\right]$, in fact.
6:33 AM
now i'm hungry. i love palak paneer
there is a place 3 blocks from home. fast, cheap, good. hits all my buttons.
especially the "cheap" :D
i prefer 'value conscious'
I merely quoted the source.
albeit i am trying to stop my daughter from being a penny pincher
6:35 AM
Oh, my matrix form disregards orthonormal bases, of course.
who does she think she is, scottish?
to be fair, my mum made an art form of being 'value conscious'.
my dad inherited that from his scottish grandmother.
ok. Lindt Lindor #3 and that's it
my aunt who died of covid used to send us those. you can stop at only 3?
6:37 AM
i'm not sure why that stereotype is confined to scots (among celts)
no, there is an inductive method of consuming them in operation
i would limit myself to one of each type in the package. i think there were four or five types.
i just buy the milk chocolate ones
as you know i am not a purist
milk, dark, hazelnut, and then a white chocolate one. if you were truly not a purist you'd eat the white chocolate one.
see, you do have standards.
6:39 AM
white chocolate is pure wrong
i actually turned down a drinking session tonight
it isn't chocolate, for one. it's chocolate the same way that herbal tea is tea.
'herbal tea'
copper, should we call some kind of help line?
i had a splitting headache when i turned it down
i reconsidered, but by the time lunch came around things improved
i'm not sure that counts. particularly if the splitting headache was from not turning down beer with dinner.
6:41 AM
but i had a nice italian red
and the waitress was a delight
and, get this, when they laid out the cutlery, they approached all four of us from the correct direction.
such attention to etiquette is remarkable
(in sf)
I'm not cultured. There is a correct direction?
always from the customer's right
my lunch companions, an international bunch just rolled their eyes are my delight
What if the table is against the right wall?
never eat there
I guess you never will eat there if you can't be served food.
6:44 AM
the waitress had certain characteristics that would have compensated for any lapse in etiquette
dining etiquette as copper understands it boils down to not getting any food on the walls.
i am not a stickler for etiquette, but i know what the etiquette is...
i try to wash hands after going to the bathroom
You only try?
Or by try to wash hands, do you mean 'try to wash other peoples hands'?
once i was introduced to a co-worker of my wife's. just before meeting him i was talking a leak. he was there and i noticed he did not wash his hands.
when we were introduced he reached out his hand...
i had to rush away to disinfect myself
6:47 AM
anyhooo, i am sure there are matters mathematical to be attended to.
just remember, serve from the right
the etiquette with proper use of cutlery and placement of cutlery in the plate is for another day...
and try not to get any on the walls.
i suspect it evolve so that guests could avoid,, god forbid, interacting with staff
in boarding school we used to throw those little lumps of foil wrapper butter against the ceiling
i think it's to complicate the project of attacking your server, on the assumption that most people are right handed and they have closed the distance. i'm left-handed, though.
they stick sometimes and when it heats up a little, someone gets a nice surprise.
i believe there are some such reasons behind the etiquette as it exists today
4 lindors is a bit much.
look at the wonderful buttons on offer at the trade show:
there should be a DON'T LOOK AT ME button
6:56 AM
always seemed like an unfortunate name, especially for a hand sanitiser give-away
oh goodness.
mentor graphics was less graphic
ohh, wait... a convex psq
2 hours later…
8:40 AM
What does this relation mean? i am not familiar with this symbol

$ \mathbb{C} \simeq \mathbb{R^2} $
does it mean they are isomorph?
more context needed
'isomorphic' has different meanings dependent on context, so even if 'isomorphic' is intended, more context needed
9:09 AM
Okay so this is where it is mentioned, i do not understand what is happening here. I am reading the proof of a the "Cauchy Riemann differential equations" theorem. let me type it:
$ a,h \in \mathbb{C}$ with the identification $ \mathbb{C} \simeq \mathbb{R^2}$ the complex multiplication with $ a $ represents a real linear function $ A :\mathbb{R^2} \rightarrow \mathbb{R^2}$
But $ a * h = a_1h_1 - a_2h_2 + i ( h_1a_2+h_2a_1) $ how is this real?
9:22 AM
$\mathbb C$ is a 2-dimensional real vector space, with basis $(1,i)$
so in your case, $a*(h_1, h_2) = (a_1h_1 - a_2h_2 , h_1a_2+h_2a_1)$ coordinate-wise
yes as a vector space isomorphic
9:46 AM
Hi folks! We can't obtain the closed form solution for the maximum likelihood estimator of the second parameter of the Gamma ditribution, because of the digamma
term in there. However we know that gamma distribution is a member of the expon
ential family and we know the MLE for it, however from it we would only get the
estimation of theta, right? I'm trying to fugure out if it's somehow possible to decompe calculate the MLE for gamma using that fact
But why is $ \mathbb{C} $ is considered as a real vector space? $i \notin \mathbb{R}$ !
10:05 AM
@MadSpaces just as neither $(1,0)$ nor $(0,1)$ are in $\mathbb{R}$, $\{1,i\}$ is a basis of $\mathbb{C}$ over $\mathbb{R}$.
What is meant by $ \{1,i\}$ ? Ordered pair?
@MadSpaces it is the basis consisting of $1$ and $i$, so yes, an ordered pair.
Why did you not write it as $ (1,i) $ ? is there a difference
10:09 AM
$(1,i)$ is the ordered pair, $\{1,i\}$ is the set of two vectors
@MadSpaces since we are writing $(1,3)$ as a vector, it seemed confusing
Oh yeah that's my bad
I used $(1,i)$ for the basis, then $(\cdot, \cdot)$ for the coordinates in that basis
It has been a while since i did Linear algebra. so i am out of touch.
How can a two dimensional vector space have one spanning vector as basis vector?
For me it is not clear why we use the describtion "real vector space" even though the element i is not in the real numbers...
Surely $ R^2 $ is real, however atleast the basis vectors consists of elements that are in the real numbers..
@MadSpaces there are two basis vectors $1$ and $i$
So $ (1,0), (0,i) $
10:13 AM
@MadSpaces Is $(1,0)$ a real number?
No but the elements $ 1, 0 $ are.
it is a vector
So we can say $i=(0,1)$
Thats just cheating..
absolutely not
it is simply two ways of writing the same thing
You might want to look at en.wikipedia.org/wiki/Vector_space
that a vector can be written as an n-tuple of reals is simply one representation of a vector.
a real vector space can have a basis $\vec x$, $\vec y$, where those are not pairs of reals or anything other than those abstract entities
What book introduces the complex set numbers and vector space in a formal algebraic way i can look at to study?
Apparently i need to reread this....
10:22 AM
@MadSpaces this is confused. These look like two elements in a two dimensional complex vector space
however, they could be basis elements of a real vector space as well.
You need to define the basis of your vector space as well as the underlying field
it is a real vector space when the underlying field is the reals
all two dimensional real vector spaces, with no further added structure, are isomorphic.
it doesn't matter what you call the basis elements
often, we denote the element $a\vec x+b\vec y$ as $(a,b)$, an ordered pair of two elements in the underlying field
$\mathbb{C}$ is a two dimensional real vector space with added structure for multiplication
10:41 AM
@MadSpaces A vector space V over field F is called a real vector space if $F=R$ and complex vector space if F=C, the set of all complex numbers.
If you take $V=\mathbb C$ and $F=\mathbb R$ then vector space definition is satisfied for V with the usual "complex" addition and usual scalar multiplication where the scalars come from $\mathbb R$.
That is if $z_1,z_2\in \mathbb C$, then for any scalars $a,b\in \mathbb R$ the vector $az_1+bz_2\in \mathbb C$ etc.
11:30 AM
writting more explicitely what I have before
Let $\phi:\mathbb C\to \mathbb R^2$ be the map $z\mapsto (Re(z), Im(z))$
Then this map is an linear isomorphism
The preimage of the canonical basis (1,0), (0,1) of $\mathbb R^2$ is $1, i$, which is a basis of $\mathbb C$ as a real vector space
the action of multiplication by $a\in \mathbb C$ through $\phi$ is precisely $\phi(a\phi^{-1}(h))$ where $h=(h_1, h_2)$. You can compute that $\phi(a\phi^{-1}(h)) = (a_1h_1 - a_2h_2 , h_1a_2+h_2a_1)$
where $a = a_1 + ia_2$
Now this formalism is very heavy, and to avoid it we often identify $z$ and $\phi(z)$, which is what robjohn did when he said that $i=(0,1)$
12:33 PM
$|\mathbb{Q}(\sqrt{-3},\sqrt[3]{2}.\sqrt[3]{5}:\mathbb{Q}|=18$ , right?
12:44 PM
@mathsresearcher sure
what would be a fast way to see it? @LeakyNun
@mathsresearcher Kummer theory lol
12:58 PM
Thank you for the multiple answers.
1:08 PM
also $|\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{5}):\mathbb{Q}|=9$, right?
it follows from the other one
1:20 PM
$Gal(\mathbb{Q}(\sqrt[3]{2}/mathbb{Q})$ has only two elements?
1:37 PM
@mathsresearcher do you have latex enabled or are you just seeing latex codes?
which two elements do you think it has?
just seeing latex code @LeakyNun. The identity and $\sqrt[3](2)\rightarrow -\sqrt[3](2)$
@mathsresearcher why do you think that is an element?
you can enable latex in chat by following the link in the room discription
oh sorry, just identity
since $\sigma(\sqrt[3](2))^3= 2$ and so $\sigma(\sqrt[3]{2})=\sqrt[3]{2},\sqrt[3]{2}\rho, \sqrt[3]{2}\rho^2$, but the last two are not in the the field extension
@mathsresearcher correct
1 hour later…
2:49 PM
when someone states : " Find the structure of the galois group Gal(L/K)" what do they mean? Size of Galois group? number of elements?
@mathsresearcher probably means make an isomorphism with some well-known group
why not just say that then?
@mathsresearcher Because it is faster to say "Find the structure" than it is to say "Find an isomorphism to a well-known group". And because mathematicians often use language in a specialized manner, and learning that specialized language is a part of learning mathematics.
(Of course, every profession has specialized language, and learning a profession includes learning that language.)
Oh, jeebus. I have a student who just sent me an email explaining to me how I should teach the class next time around. He is frustrated that I gave them too much homework in which I asked them to compute volumes using cross-sections and cylindrical shells.
Completely ignoring the fact that I have repeatedly indicated that homework is optional, and that I will give credit to any effort which is turned in (because the exams are where you are going to succeed or fail).
But, hey, 19 year old student, I am sure that you have much more pedagogical experience than I, and would run the class perfectly. :/
3:50 PM
Hey everyone, I believe I may have solved a question that I posted. Is it better to edit the original post, or reply to my own question with an "answer"? My solution was inspired by someone who initially answered my question but then deleted their post so I can't give credit :(
@TedShifrin i had an analysis quiz last week, and i've successfully proved that $\lim_{(x,y)\to(0,0)} \sin(xy)(x+y)=0$ with the simple fact that if $d((x,y)(0,0))<\delta$, then $x<\delta$ and $y<\delta$.
remember you taught me that fact!
@ChrisLangfield i believe an answered question is better than one without.
you can mention how you solved it and your inspiration.
@sevdaicmis ty
@sevdaicmis congrats!
such a small accomplishment, eh? but i missed to be happy, i've been failing for the past 1,5 year, because of pandemic. i hate remote education.
@LeakyNun but yeah, thanks :)
@LeakyNun what was your education status? were you a graduate student?
@sevdaicmis i'm studying phd
4:03 PM
oh, how nice. how is it going, do you enjoy it? :)
@LeakyNun i now remembered something else. i think i've started to be interested in proof assistants because of you. i remember you sent a formal proof of a simple proposition, here.
yeah it's fun
then i've started researching about it, and seen coq, lean and others.
that's nice
i can now prove simple set theoritic propositions with coq :)
but you were using... HOL, maybe? i'm not sure.
when i retire, i will implement my dissertation in coq
4:14 PM
If $G$ is any open set and $\gamma : [a,b] \to G$ is any rectifiable curve, why does it follow that $dist(\text{Im } \gamma, \partial G) > 0$? I know it has something to do with compactness, but I can't see the proof.
rectifiable curves are continuous by definition, no? so (as continuous image of compact set is compact, and compact = closed and bounded) im(gamma) is a closed subset of R^2. if that distance is 0, then, there's actually a point in im(gamma) on the boundary of G.
@leslietownes Typically, $C^1$ at least (or, at least, piecewise $C^1$).
which is ruled out by gamma having image contained in G, which is assumed open.
So sure, continuous.
i knew fractal guy would show up.
i put up the fractal bat signal.
4:25 PM
Yes, they are continuous, so their image is compact. That's the part I'm having trouble seeing. Why does having zero distance mean they have a common point.
user: for each n you can pick a point p_n on im(gamma) with d(p_n, partial G) < 1/n. by compactness p_n has a subsequence convergent to some p in the image of gamma. i think you can show d(p, partial G) = 0. but this subsequence is also a convergent sequence in the closed set G union partial G = the closure of G, so it's in the closure of G. p can't be in G (ruled out by d(p, partial G) = 0), so p is in partial G.
i may be circling around the same idea in one or two different guises there.
@leslietownes great!
4:45 PM
hooray my brain still works
5:05 PM
@leslietownes i am very familiar with that feeling.
@sevdaicmis I'm glad you actually understood and learned :)
@leslietownes What brain?
huh? why are the lights on my screen blinking at me
5:33 PM
go away, lights
Yup, munchkin has taken over.
Thanks again for the guidance the other night, Ted. Your course notes were really helpful in understanding what was going on in the Euler-Lagrange. Everywhere else I was looking was missing just that little bit
You're most welcome, @Axoren. No big deal :)
Hi, can you tell me if the standard topology on R^n (or R) and the usual topology are the same thing?
Does the fundamental theorem of line integrals only apply to conservative vector fields?
5:47 PM
@Under Yes.
Ah, ok.
@antonio "standard" = "usual" ...
@TedShifrin Thanks
@Under It's like asking whether the fundamental theorem of calculus $\int_a^b f'(x)\,dx = ...$ applies if you integrate something other than a derivative.
@TedShifrin :)
5:49 PM
The usual topology on $\mathbb{R}^n$ is the standard topology. By contrast, the standard topology is not different from the usual topology.
Yeah, just took a second to reread the definition. I'm not sure why I asked now, lol
smacks @Xander
Not definition, theorem
@Under Generalizations (for closed curves) to non-conservative fields will be Green's and Stokes's Theorems, soon to be seen.
Hm, that'll probably be today then. Green's Theorem is next on the syllabus.
This is all under the heading: Generalizations of the FTC ...
@Xander: Sugar is an evil thing.
@TedShifrin Lies. I need simple carbohydrates to live!
Well, a few non-sugar smacks will get you back in order.
I need sugar. I just got a pack of trail mix from the vending machine.
Ah, over-priced stale stuff from vending machines.
5:53 PM
This is true.
I don't have time to get proper food before class, lol.
clucks like a mother hen
Coffee counts as proper food, right?
In which case, I've had plenty of proper food today.
But I need to teach now.
At this point, yes.
Vending machine coffee does not remotely count as coffee. It is vile.
But it says starbucks on it ;-;
5:59 PM
Ah, so you paid $8.00 for it.
The more I read through math literature, the more I feel like I was never introduced to some key foundational concepts and their notations, and now there's so much behind the points I'm at that I don't know where to even go back and review.
That's called life, @Axoren.
It was $5
It's bad enough that the book I was reading on Diff Geo years had me contemplating a religious war with the author over whether or not 0 was a natural number.
6:02 PM
We've had that argument here uncountably many times.
"Now, here's the exact same argument but for 0." "Consider $n \in \mathbb N \cup \{0\}$"
Why should that be remotely relevant in a differential geometry setting?
That's something I thought was weird when I started my proofs class. I screwed up a few problems because I assumed $0 \in \mathbb N$, until my professor clarified.
In Europe they say $0\in\Bbb N$. In the US we do not.
It added a whole lot more pages of nothing between all the good parts.
6:03 PM
Dang, I thought it was universal.
I always confirmed with professors what our standards were, if we used NIST or some other standard for the class.
I've only recently gained the kind of foresight that would lead me to do things like that at the beginning of a course.
One professor was very much in the $0 \in \mathbb N$ camp that he told the whole class "This class will not be any easier if you start working with fewer numbers. 0 is natural and if you disagree, tough ****."
Who the h*** really cares?
Some inductive statements won't make sense at $0$, but ...
As long as it's clear and doesn't waste my time reading proofs and theorems, I don't mind.
Textbooks have started including standard definitions at the beginning of the book, as well. Which helps a lot in this regard.
6:08 PM
My textbooks don't do that.
I recently moved to another state, so a lot of my textbooks are back home, but a couple of them had a pre-chapter 1 section dedicated to notation.
Oh, you're no longer in MA?
TX now.
No comment.
I know.
But it's for work.
6:10 PM
Well, I'm sure there must be a handful of civilized people with a brain left somewhere in TX.
It's cheaper to live here by a lot and I'm getting paid very well. Austin is like that to an extent.
Austin and Houston are probably relatively safe.
I do not recommend ever living in Houston.
Did it for a few months. The people were nice, but the environment was sad and depressing.
When I interviewed for a job at Rice in 1979, the city was under feet of water.
(I exaggerate only slightly.)
I lived down there during a drought for like half a year. It got so hot that it melted my shoes to the pavement. I was supposed to borrow a friend's car while I was down there, and the heatwave melted the engine belt apart.
6:13 PM
Well, between climate change and the dissolution of a democratic society, we're in for disaster, regardless.
The core problem I had with Houston was that it just felt like people living there were either doing very well for themselves or struggling to just get by. Very high income disparity and it showed in how the city was laid out.
Well, that's true in Boston, Atlanta, NY, SF, ... hmm ... every large city.
It was a very different experience from Boston, by far. Having come from there to Houston, it was stark.
Maybe you never went to Jamaica Plain or other less well-off parts of Boston. Or maybe things have changed since I lived there.
Definitely overt Irish/black racism in the years I lived there.
Even Roxbury is doing well now, I don't know about the racial tendencies since I've left.
Before I moved to TX, I moved north towards NH, but before then, Boston proper and the Greater Boston area were overall safer and happier places.
So I can't really claim much about from 2018 onward.
6:19 PM
@Axoren tough luck? what do the asterisks stand for?
Well, my data on Boston go back to the late 70s, early 80s.
smacks Leaky
@LeakyNun Defecation.
i've never heard of that
like those two words put together that way
Oh back then? Terrible everwhere. That was the Bulger era.
@TedShifrin ouch
6:21 PM
@LeakyNun It's a slang idiom meaning "I have no sympathy for you."
I see
Often used in a situation where nothing can be done about someone's problem, like having to pass a difficult (tough) ____.
@TedShifrin time to revive the us labor party
@Axoren my own tack is: if I’m enumerating things, zero is natural. If I’m multiplying, it’s not.
I would just use positive integers if I wanted to exclude zero.
I’m particular, zero is natural for power series. But it’s not natural for Dirichlet series
I don't like the idea of a number being natural-fluid. Makes definitions and theorems situational and subjective instead of objective claims and conclusions.
I vaguely remember that someone here was working on something-like-a-field in which division by the zero element was well-defined. Did anything come from that?
I can't find the chatroom anymore.
6:52 PM
@Axoren its called the extended real number line
I think it was something different, I believe it was a finite field personal research topic.
But yeah, the extended reals does that, too.
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