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12:07 AM
Picky @robjohn, and patient @TedShifrin, I feel I might be X-Y'ing huh. The ultimate goal us to get an expression for the time derivative of the expectation of momemtum. Ie $d/dt \int{\Psi^* [-i\hbar\frac{\partial}{\partial x}]\Psi}dx$ (the solution to this has been given to me, my task is to try get the method)
 
seems to be a bit of dirt on the $h$.
 
12:22 AM
So Schrödinger tells us relevant things?
 
just about cats
 
just looked inside her bed. alive.
 
you just collapsed her state
 
You persist doggedly.
 
this is getting too entangled
 
12:51 AM
@copper I thought they called them intertwining operators.
 
1:06 AM
@AndrewMicallef As Ted said, you have to use that $\Psi$ satisfies the Schrodinger equation. Of course, the answer is what you expect (quite literally; expectation recovers classical equations)
 
1:17 AM
Hi, a @Balarka!
 
Hi Ted
 
1:28 AM
@TedShifrin That would be putting a different spin on it :-).
 
Not to mention twistors.
 
i like twizzlers too
 
You would!
 
1:54 AM
Is anything interesting known about A082525 and A082526?
That is, the coefficients of the power series of $x^x$ at $x=1$?
 
To me, nothing about such things is interesting.
 
we agree again
 
that just takes the fun out of maths
like numbering poems or something.
 
2:13 AM
@AndrewMicallef So in that integral, $\hbar$ is Planck's constant, correct?
So we can pull the $-i\hbar$ out in front of the integral
 
is this not just Leibniz?
 
@copper.hat the coefficients?
 
the time derivative?
 
Yes, of course.
 
2:29 AM
i'm sometimes interested in formal analysis of coefficients of power series when it's integer coefficients. rational coefficients is ramanujan territory.
 
@copper.hat the limits are not time-dependent, so that makes things easier.
 
@copper.hat Are you around? I had a question regarding spectral theorem
 
@BalarkaSen i'm a square. i am no expert...
 
Took a second to get the pun
 
my working knowledge originated from finite dimensions, hopefully your questions lies there...
 
2:42 AM
Let $T$ be bounded self-adjoint on $H$. There are two different formulations for the spectral theorem, one says there is a projection-valued measure $\mu$ such that $T = \int_{\sigma(T)} \lambda d\mu(\lambda)$, the other says there is a real-valued measure $\nu$ such that $H = \int^{\oplus}_{\sigma(T)} H_\lambda d\nu(\lambda)$, and $T$ acts on this direct integral as multiplication by $\lambda$.
I would like it if for Borel subsets $B \subset \sigma(T)$ if $\int_B^{\oplus} H_\lambda d\nu(\lambda)$ was equal to the spectral resolution $E(\mathcal{B}) = \text{im} \mu(B)$.
Is that true?
I am guessing no but I am also having a hard time pinning down the precise relation between the two formulations above
 
2:55 AM
Maybe this is uniqueness of the spectral resolution? We can define from the second formulation a new spectral resolution for $T$ which associates to each borel subset $B \subset \sigma(T)$, the projection to the subspace (maybe closure of this) $\int_B^{\oplus} H_\lambda d\nu(\lambda)$ of $H$.
That seems to be a new spectral measure, but spectral measure is unique
 
3:07 AM
what are some mathematical facts that seem true but are not
no need for "trivial" examples
 
this is a sensitive subject for me, balarka. i knew all of this stuff. now i can't even remember the names of the books i learned it from.
or the stuff itself. i haven't just forgotten the names of the books. it's all gone.
 
@BalarkaSen out of my range right now, sorry!
 
it's probably in reed and simon, although i don't remember liking much of the way they did things.
 
 
1 hour later…
4:33 AM
i didn't know that...
 
ha, cool
 
maybe the logo creators were inspired by that
is "affine groups on small binary spaces" a book or an article?
 
angelfire.com/pa4/steve/copy_of_index.html which is of unknown provenance (angelfire?) has a link to a page of typewritten manuscript by that title
maybe the first page of a preprint
it looks like he didn't publish it, but finitegeometry.org/sc/gen/coord.html has more
lots of results, many unpublished, finitegeometry.org/sc/map.html
 
nice
 
4:52 AM
finite geometry is frightening
 
yeah that's true
 
You're frightened of anything with geo in it.
 
in (very, very weak) defense of this question, that integral seems way harder to do with Stokes than it does to do the line integral directly @TedShifrin
 
@TedShifrin if you are talking to me, then yes, that is true
i don't like geography, geometry, etc.
 
@Semiclassic I don't disagree, but your comment is wrong.
No, Euler, talking to Leslie.
 
5:06 AM
eh? his second term should be $xj$
 
ok
 
not $xi$
 
Oh, I looked at the $x$. It was in the right slot.
Sorry.
 
np
i find the wording of that problem confusing too
 
Of course. I complained.
 
5:08 AM
in what sense is the curve C shown the "borders of the surface of the cylinder"
but translation to English might be part of that
 
normal geometry is better than finite geometry. it's not frightening, it's just better for me if i don't think about it
 
if $(h,k)=1,k=0$, prove that there is a constant $A$ depending on $h$ and $k$ such that if $x\ge 2$
$$\sum_{p\le x, p\,\equiv\, h(\text{mod}\, k)}\frac{1}{q}=\frac{1}{\phi(k)}\log\log x+A+O\left(\frac{1}{\log x}\right)$$
can someone help to prove
 
The Kempner series is a modification of the harmonic series, formed by omitting all terms whose denominator expressed in base 10 contains the digit 9. That is, it is the sum ∑ ∑ ′ n...
I feel like "$\sum_{n=1}^\infty' 1/n$ where the prime indicates that n takes only values whose decimal expansion has no nines" is about the most useless application of summation notation i've seen
 
5:23 AM
oh no my kitten is typing random things on the keyboard lol
 
the second sentence communicates nothing that the first doesn't already
 
lol
as much as i like that strip, i feel like the kempner series isn't a great example
 
there is a typo in my question in the summation $1/q$ should be $1/p$
 
"The harmonic series over all integers diverges, but most large integers contain the digit 9 in their decimal expansion."
being able to prove it is neat, but ultimately the real fact is just "There's a lot more ways to create a long integer using 0-9 than 0-8"
 
5:28 AM
In that case, this trick works for any old integer?
 
Cool
 
nothing special about 9
and base 10 isn't special either
being able to actually compute Kempner's series, though, that's pretty crazy
 
seems like something for someone else to do
 
wolfram mathworld states that a kemnper series can be formed by removing all terms containing the digit $d$, not just 9
for $d=9$ it is approximately 23
 
5:35 AM
yeah, there's quite a few generalizations
which all come down to "if you exclude enough integers, then the harmonic series converges."
(i suspect there's no simple criterion for "enough integers", though.)
 
i would love it if there was
 
proving its convergence is an exercise in a lot of analysis books, it's easier than the other classic about sum 1/p, p prime
 
i mean, there's probably some conditions. like if you remove this much, it definitely converges, and similar criterion to guarantee divergence
but probably no one-size-fits-all criterion
 
definitely converges if you remove all of them
 
5:40 AM
i was also thinking the same lel
 
don't be greedy, just remove all but, like, 10 of them
 
here's one i have no idea about
Sample a Poisson random variable (with some rate constant) and use that to pick your first integer. Then generate another sample and add that to your first integer to get your second. Repeat ad infinitum. What can be said about convergence of the sum of reciprocals?
more mathematical statement of it: Consider a sequence $X_1,X_2,\cdots$ of iid Poisson random variables, and let $Y_k$ be the $k$th partial sum. What is the probability that $\sum_k 1/Y_k$ converges?
 
oof ask me in 3 months and maybe I'll have an answer
 
i'm mostly curious if this has been considered before
hmm, here's a different take on "random harmonic series": thatsmaths.com/2016/07/28/random-harmonic-series
the harmonic series diverges, but the alternating harmonic series converges. what if you pick each sign according to a coin flip?
 
5:55 AM
yeah I just read this lmao
 
lol
problem is then characterizing the distribution of the $Y_n$s
 
jp kahane has a book, some random series of functions, i think i saw some results of this type in there
probably has papers on it too
 
 
1 hour later…
7:10 AM
Hi Ted.
@TedShifrin Got a minute?
 
7:26 AM
@leslietownes I should've excluded the $\frac{1}{k!}$ factor, in which case it is A005727, which are integers.
 
Is there any mathematical term for a function g that satisfies g(g(x)) = f(x) for a given function f?
 
Isn't finite geometry just finite group theory? Or are there things in finite geometry that group theory can't really encapsulate?
 
 
2 hours later…
9:24 AM
@robjohn I took me a while, and it wasn't intuitively obvious, but I see how that works. Not sure why I would use that strategy though, but hey that was fun
I feel like someone played a prank on me
 
9:42 AM
Ohh!! that isn't a prank, that last line is a usefull simplification
So I guess, this would follow?
$$
\begin{align}
\int\limits_{-\infty}^{+\infty}{\left(\Psi^*\left[\frac{\partial}{\partial x}\right]\Psi\right)}\,\mathrm{d}x
&= \left[\Psi^* \cdot \Psi\right]_{-\infty}^{+\infty} - \int\limits_{-\infty}^{+\infty}
{\Psi\frac{\partial \Psi^*}{\partial x}
}\,\mathrm{d}x\\
&= \frac{1}{2}|\Psi|^2 \Big|_{-\infty}^{+\infty}
\end{align}
$$

And yeah $i \hbar$ was a constant term that I have put aside for the moment
I don't know enough about complex functions to know if that actually works or not.
 
 
2 hours later…
11:39 AM
does the infinite series $\sum 1/(r^n-1)$ for r>1 converges?
 
12:04 PM
@love_sodam i don't think i could write it properly here so i gave an image
 
12:25 PM
@love_sodam the summing term is equivalent to $1/r^n$, and $\sum 1/r^n$ converges, so yes
 
1:08 PM
Can anyone please help me understand why $\mathbb{R} ^2$ minus $n$ points deformation retracts onto a wedge sum of circles? This has been bothering me for 2 days
 
1:20 PM
What is meant by bottoms-up approach and what is meant by top-down approach?
 
 
1 hour later…
2:22 PM
Guys , any recommendations for book on 3D Geometry that is self explanatory.
 
2:48 PM
i think im missing something really obvious, if $f_n \rightarrow f$ uniformly on $\mathbb{R}$, and $f_n$ converges to something in $L^2(\mathbb{R})$, then it must be the case that this function is $f$, right?
is there a way to prove this besides using that there is a subsequence of $f_n$ converging a.e. to its limit in $L^2$, and therefore its limit in $L^2$ is equal a.e. to $f$?
like a 'more obvious' way
 
uniform convergence implies L^2 convergence on bounded sets
 
ah yes of course
and that means the L^2 limit agrees with $f$ on bounded sets a.e. so also a.e. on the whole line
 
yup
 
Hello
 
@GeorgeRevingston it doesn't matter where the $n$ points are, up to homeomorphism, so you can fan them out around the origin, now draw a wedge of $n$-petals , each petal around one of the points
now there is an easy choice of deformation retraction
inside each petal, you project radially outwards from the point onto the petal, and outside the petals you split the plane into $n$ regions, and within each region you move along the line between a point and the point in the petal corresponding to that region
of course this isn't the formal answer in terms of some actual function, but you can probably turn it into one
and on the lines that separate the region, you move towards the wedge sum along that line, so you'll eventually reach the wedge point from there
 
3:46 PM
why are proofs written in such an obtuse, ugly way.
$S(a,b)$ is not a sequence. "admits some integer". ffs.
 
@TedShifrin
 
4:05 PM
@feynhat You keep pinging me when I'm not even here!
 
I got a last day offer from IU Bloomington.
 
Oh, really down to the wire. Congrats.
 
I am very confused right now
 
Life is a bunch of decisions.
 
4:53 PM
you make the best decision given the info available. if you are having a hard time deciding you either need to get more info or there is no clear answer and either decision is good at that time.
 
if you've studied something several times before in math, its usually pretty easy to remind yourself of the material right? Like say you go to a uni where students do eight courses a year and exams are at the very end of the year, if you learned the first four courses (taught in the first half of the year) well at the time, it should be doable to review that material within a few days right?
assuming you're not relearning stuff, just reviewing material
maybe ive phrased my question poorly, its normal to not be able to recall little details of maths you've learnt before but haven't kept fresh in your memory right? does that count as 'forgetting' that piece of mathematics? but how could it if you can refresh your memory much faster than it would take to learn it the first time
 
@Thorgott this is just riemann integration right
 
@porridgemathematics that depends on the individual.
 
it's $\int_Af\le\int_A\lVert f\rVert_{\infty}=\mu(A)\lVert f\rVert_{\infty}$
 
oh yeah for sure, this will depend on the speed at which someone can parse mathematics, and lots of other personal differences
lets say you've been able to review material at the rate of three or four days before, and its material of a similar difficulty for you (or slightly harder)
 
5:04 PM
for me, it really depends on my intuition for a particular area.
 
hm
i sort of factor that into parsing time
if something is intuitive to you, you basically have the ability to skip certain details because they click on some lower level in your mind
at least thats my theory
(skip at no cost, you are doing something equivalent to the computations subconsciously)
anyway i shouldn't have asked, its a bit more philosophical than mathematics related a question
 
well, i studied in ireland back on the early 80's so this may not be relevant. but all of our (engineering) exams were right before summer (no semester/quarter thingy) so it was a different sort of problem. i would made detailed notes of the issues that i needed refreshing on and basically look at them frequently and just before the exam.
 
thats pretty relevant, its the exact same system im currently in actually
I see, yeah I happen to have made those notes, and I find myself doing a similar thing
in retrospect I wish I looked at notes for my first four courses more in the second half of my year, but at the same time i literally couldnt find the time because learning the latter four courses were hard enough
 
it is hard to find the time/discipline to do it. you need a life as well.
 
fortunately they give us a month to prepare for exams, but thats basically four days for each course, which is why I asked - im basically just able to finish the course content in those four days so it seems doable
i really get stuck on using the same word over and over again , i wonder what that says about the way I think
 
5:16 PM
for me, managing the feeling of impending doom/panic was the biggest contributor to moving ahead :-). i'm serious.
 
hah that is very relatable
why do exams have to be this way :-(
 
if it is any consolation, i am 60 and have worked in industry and from time to time encounter similar situations of total internal panic :-).
 
i guess everyone just deals with it
thats life
 
easy to say, but focus on what you need to do and do not let the unhelpful feelings get in the way. i am definitely not a ahhhoooommmm sort of person, but i found some simple mind settling sort of focus/meditations to help.
never thought i would be advising meditation...
 
@Thorgott ah yuh so we get le convergence in all the p's
 
5:25 PM
yeah
of course, L^infty convergence need not imply L^p convergence for any other p globally, but if that convergence is given, the limits have to agree a.e. by this argument
actually, now I'm unsure
what's a sequence of L^2 functions converging to something non L^2 in L^infty
 
@Thorgott uhh
 
How about $x^{-1/2+1/n}$ on $[1,\infty)$?
 
ah clever
 
(Not that I've ever thought about this before.)
 
@Thorgott doesn't convergence in L^infinity imply that the sequence converges pointwise to it a.e?
so then convergence in some L^p to something, that something would also be the pointwise limit of a subsequence
in which case they should coincide
 
5:37 PM
I've been trying to approximate a constant function but that didn't work due to the L^2 functions vanishing. A nonintegrable, vanishing function sounds like the way to go
/not square integrable
 
maybe im misinterpreting something, I thought all @Thorgott was saying is if you have convergence in L^infinity and L^p for some finite p, the limits coincide a.e.
which still seems reasonable to me
 
anyone have objections to defining identical vector spaces as: if the subspaces of a vector space have as possible dimensions the same number of dimensions as the subspaces of another vector space, then the two vector spaces are the same?
 
Well, we need to check whether my example works. What's the max of $|f-f_n|$?
@shintuku What does "the same" mean?
 
@TedShifrin that's the definition, I'm trying to see if it works out intuitively: two vector spaces are the same if their subspaces have the same number of possible dimensions
 
All your talking about is same dimension, conceivably — this is nothing like "the same" vector space.
So any two lines are the same, any two planes are the same? That's a horrible notion of "same."
No one in the world will agree with you.
 
5:42 PM
@porridgemathematics that's definitely true, the additional question I just raised is whether converging in L^p is implied by converging in L^infty or whether that is a genuine hypothesis
 
hm you're right
 
it clearly should be a genuine hypothesis, but I don't have an immediate example
 
You can say "isomorphic" — which in the case of finite dimension means that they have the same dimension.
 
@TedShifrin either I'm being silly or those are not in L^2
 
ah ok got it
no its not
 
5:43 PM
Oh, I meant $-1/n$, of course, @Thor.
 
@porridgemathematics you can create a 'comb' function that converges to zero in $L_\infty$ but converges nowhere pointwise.
 
Maybe easier to do on $(0,1]$ with mine.
 
@TedShifrin do you know if 'isomorphic' is a much used definition of identity in linear algebra?
will read up on it
 
No, identity means identity.
Literally the same.
 
oh yeah, makes sense
 
5:45 PM
Isomorphic is the meaningful concept, yes .... If you relabel everybody, the two spaces become identical.
 
I don't immediately see that this is uniform, will have to come up with an estimate
my analysis is rusty
 
@copper.hat I dont think thats true
 
ignore my last comment. my reasoning was wrong.
sry about that.
 
no worries
 
@Thorgott I'm too lazy to do the derivatives
I find a critical point at 1/(1-2/n)^n
of the difference
 
5:47 PM
I did screw up a negative sign, as Thor pointed out.
 
yeah no, of course it works out asymptotically for n>2
 
ugh really. if I didnt do a sign mistake then the critical point is < 1
 
the difference converges to 0 as x->infty for any fixed n>2
 
that'd be shitty. But the difference is 0 at 1 and 0 at infinity right
 
@porridgemathematics $f_n \to f$ in $L_\infty$ iff there is some null set $E$ such that $f_n \to f$ uniformly on $E^c$.
 
5:49 PM
yeah
 
so that this should have a critical point
 
yeah, but who cares about the critical point
 
almost uniform convergence
 
it's literally the supremum
 
5:50 PM
I mean, it obviously converges pointwise, which implies uniformly on compacta
so the only thing to worry about is what happens as x->infty
 
yeah yeah okay I see sorry for making hard estimates
 
@TedShifrin Would you say CUNY is strong for Topology/Geometry?
 
@Thor: I let Mathematica do the nontrivial limit. It is uniform.
 
let's face it
we would all have been very shocked if this werent uniform
 
@porridgemathematics so if $f_n$ converges in $L_\infty$ and some other $L_p$ then the limits must be the same ae. The $L_\infty$ convergence dominates the $L_p$ convergence on finite measure sets.
 
5:51 PM
@feynhat: I am out of touch at this point. It's certainly not on my radar screen. But nor is Indiana at the top of the heap.
@user2103480: I know. That's why I guessed it!!
 
good guess indeed
 
Hmm... thanks.
 
I'm looking at the faculty list. I assume you've researched all the faculty and their publications and emailed some?
 
@porridgemathematics so i guess i need some form of $\sigma$ finiteness.
 
I know Kent Orr and Dylan Thurston at IU, but most of their geometry is ergodic theory dynamical systems, and I know none of that.
 
5:55 PM
@copper.hat well since it dominates the L^p convergence on finite measure sets, the limits are a.e. equal on all bounded intervals
@copper.hat which implies a.e. equal on R in fact
so you're done
 
I haven't spoken to anyone at IU except the graduate director.
 
@TedShifrin what do you think of this one: two vector spaces with dim=n are the same if they can be generated from the same subset of vectors from a vector space of dim>n
 
I wrote to Mandell but didn't get a response. (except an auto-response saying that he was on a leave).
 
but yeah, we are implicitly using sigma finiteness as you say
 
I did speak to a few people at CUNY.
 
5:56 PM
I know several people at CUNY, but most of the ones I know are surely retired. And saying geometry/topology is awfully broad. Unless you truly are interested in anything/everything in those fields.
 
Isn't that how most grad students start out (in the US)?
 
@shintuku: You shouldn't make it a question of living inside someone bigger. Why are you doing that? Just say that you can choose identical bases. Period. And yes, that's the same. Isomorphic, of course, but also same.
Some do, some don't know fields at all. If you were committing to a particular adviser at that Canadian university, that is damn specific.
I basically knew I wanted to do complex geometry, and I did, although I was slightly tempted by one person in dynamical systems from whom I took a course first year.
 
@porridgemathematics in general, i think you need something like if $A$ is measureable non null set then there is some non null $B \subset A$ of finite measure.
 
@Euler2 did you mean $\frac1p$? Oh, nevermind.
 
Hi, demonic @Alessandro
 
6:00 PM
Hi
 
Hi, picky @robjohn.
 
I want to do algebraic topology. (But I am not dead set on this).
 
i've forgotten what that is called, non atomic?
 
@feynhat: I suspect either one would be fine for you. Big difference between living in NYC and Bloomington. Expense, culture, ambiance. Ignoring all things Covid.
 
im not sure, are you sure thats strong enough though?
 
6:01 PM
Oh, classical algebraic topology is not so popular these days.
 
@TedShifrin same bases seems like a good one, thanks! trying to make sense of some proofs so I'm working on the definition of identity
 
CUNY has more people in that direction, if they're still active.
 
ROFL @robjohn
henceforth, "blurry Ted."
 
like you could take the counting measure on borel sets, and that satisfies this, but its pretty far from sigma finite
 
6:02 PM
algebraic topology? you mean derived infty-model categories?
 
People nowadays claim they are doing homotopy theory while they are actually doing higher category theory
 
What is 'classical' algebraic topology?
 
I mean that as opposed to geometric topology and Seiberg-Witten topology.
 
Or at least that's the impression I got in Bonn/Münster
 
More like classical homotopy theory questions ... which have never appealed to me. Obviously.
 
6:03 PM
classical algebraic topology is the algebraic topology in which the word "category" only appears without any additional prefixes or adjectives
3
 
Well, that's "modern classical," still.
Poincaré didn't think categorically.
Not so far as I know.
 
@porridgemathematics relax it to any measureable set $A$ of infinite measure has a non null measureable subset $B$ of finite measure.
 
yeah, they also thought of homology as a collection of betti numbers and torsion coefficients back then
 
He didn't even believe in Cantor's ideas of sets, IIRC.
 
there's a difference between "classical" and "outdated"
 
6:06 PM
@feynhat Set theory was a mistake
 
insert misattributed set theory is a disease quote here
 
@Thorgott "A spectre is haunting Europe..."
 
I like those parts where you compute stuff.
 
Geometric topology is full of "computing stuff."
That's why people study invariants — characteristic classes in topology, geometry are a prime example.
Sounds like you probably don't know enough of geometry/topology to make this a thorough decision.
 
This semester I will compute the Euler class of a certain bundle over a certain grassmannian which will tell me the number of lines in a quintic threefold.
I like this stuff.
 
6:13 PM
does anyone know if pointing out that a subspace U of $R^1$ has the same standard basis constitute a proof of identity between U and R^1?
if U is nonzero
 
Ah, yes, that's beautiful stuff. Enumerative geometry is far more a part of algebraic geometry than of algebraic topology. My papers are full of Chern class computations.
 
if it is a non trivial subspace of $R^1$ it must be $R^1$. There are only two subspace of $\mathbb{R}^1$, one of them is trivial.
 
@shintuku The right thing to prove is this. If $W\subset V$ is a subspace and $\dim W = \dim V$, then $W=V$.
(I refer to finite-dimensional spaces.)
 
dang, got there before me
 
@copper For some reason shintuku is obsessed with "sameness."
 
6:18 PM
hahahah, I'm having a terrible time with sameness since I was asked to show $\W \subset \mathbb{R}^1$ are identical
 
OK, I'm gone for a few hours. Everyone misbehave without me.
 
@TedShifrin certainly there are many aspects of algebra that will never by 'natural' to me.
enjot!
 
thanks for the help @TedShifrin!
 
@TedShifrin I listened to a seminar talk on Chern-Weil theory and its applications to Kähler manifolds earlier. I understood very little, but it was quite interesting.
 
so, how about this: Proof. Notice that the dimension of any subspace of $\mathbb{R}^1$ is either 0 or 1. Let $U$ be a nontrivial subspace of $\mathbb{R}^1$. Since it is nontrivial, its dimension is 1. Therefore, since $dim(U) = dim(\mathbb{R}^1)$ and $U \subset \mathbb{R}^1$, they are identical.
 
6:25 PM
not wrong, but too complicated
what are the scalar multiples of a non-zero real number?
 
there's the one that uses only real numbers, but it bothers me terribly since there is no distinction between scalars and vectors
yeah there's the x/c = y proof, but it is outrageous
 
why's that a bother
 
it is scandalous! how dare people treat vectors like scalars
 
in R, they're literally the same thing
 
@shintuku that is fine.
 
6:30 PM
what bothers me is I think of a vector as an arrow, and a scalar as a distance only when it is a difference
but yeah, they behave very similarly in R^1
or literally the same, as you say
 
a distinction with no difference.
 
those are valid ways to visualize ways these concepts, but sometimes (and this is one of those times), there are distinct ways of visualizing the same mathematical object
 
Good morrows to all.
 
think of a real number as having a length and a direction $\pm 1$.
 
this to me @copper.hat ?
 
6:35 PM
yeah, I don't have any arguments against that
 
@dc3rd no :-)
 
@dc3rd we are discussing the scandalous treatment of vectors as scalars
 
scalars (as in fields) can be pretty odd objects.
 
let's not discuss C as a 2-dimensional real vector space but a 1-dimensional complex vector space, shall we?
 
never know....you may be pop quizzing me......well you could say said real number is bounded.....my physics is also very rudimentary....

the field of scalars? or I'm guessing you can construct various fields with select scalars
 
6:38 PM
@leslietownes wait how does this work? does it fix the scalar/vector distinction?
 
forget about physics. field has other meanings in math
also, hi chat
 
also very guilty of treating vectors as scalars until recently
 
meromorphic functions for example.
 
that is a "google" term for me or better put a term to be put in a search engine. because there is more than one.
 
@dc3rd what made you change? i'm building my case for the distinction but copper.hat has better arguments
@copper.hat very cool!
 
6:41 PM
fields are very nice objects
 
Hello all, suppose we have unknown distribution of data, say Gaussian distribution, what does it mean to choose the cut-off based on Chi distribution?
 
@shintuku what made me change?...one word: Ted :p...............but being more serious about it, looking at things in higher dimensions. you can't describe a position with just one value in a higher dimension
was such a convoluted term needed to describe a complex differentiable function at a point @copper.hat?
 
that's not what it means
 
i said let's not discuss that. also, don't think of an elephant.
 
I'm going on the very brief caption from wikipedia @Thorgott, I will definitely defer to you on this
 
6:50 PM
eventually you get to a place where for the same set of things, there might be one or more choices of what your 'scalars' are going to be. that's all. it doesn't have to be today.
 
@leslietownes is it ok if I think of Johnstone's Sketches of an Elephant, the classical treatise of topos theory
 
and now i'm thinking about topos theory. i guess it's fair to get a taste of my own medicine. well played.
 
the idea of a scalar is problematized in topos theory? @leslietownes
 
everything is problematized in topos theory. i don't know about linear algebra, but i'd guess so.
 
so is topos theory a thing for algebraic geometers or algebraic topologists?
i will not search that on ncat.
 
6:56 PM
i have that site blocked
 
@dc3rd meromorphic means more than that. the are particularly useful in algebraic methods for control system design.
 
i think topos theory is for people who are into categorical logic, if anything
@leslietownes you should be ashamed of yourself
 
> A quick formal definition is that an elementary topos is a category which

has finite limits,

is cartesian closed, and

has a subobject classifier.
(i lied)
 
Grothendieck topois are cool, though
 
@Thorgott ??
Topois
Topoi is already the shitty plural
 
7:08 PM
@dc3rd my comment on meromorphic functions reminded me that there is a nice, relatively informal discussion of the pitfalls of modern control system design called "Respect the unstable" by Gunter Stein. not entirely apropos, but a good reminder that algebraic methods applied without understanding the underlying dynamics/physics can have fairly disastrous consequences.
 
how many times do I have to cite mclarty explaining why toposes is the correct english plural
 
sorry mein english no good
 
surely it should be topoi?
it is greek after all.
 
7:23 PM
anyone know of complex functions that cause a rotation away from the imaginary axis? so first quadrant imaginary numbers get closer to positive real axis, second quadrant imaginary numbers get closer to negative real axis
 
@user2103480 is that the thermoi thing?
 
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