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12:00 AM
Consider the map $f:A \to B$
 
i'm considering it. i have to say, it isn't very interesting.
 
What does this mean? $f_i : A \to B$ for $i=1,2,3,...$
 
@anakhro yes, over 10 years now.
unless subscripting has been given some kind of meaning, whereby subscripting an already-defined function does something goofy to the already-defined function, i'd assume that was just notation for a sequence of functions, each one of them from A to B.
 
I want to capture the set of linear maps between $A$ and $B$ and make a diagram of it
so if $V$ and $W$ are vector spaces then $L(V,W)$ denotes the set of linear maps between them
 
i dunno. i have never personally done that. it gets big even if V and W are of fairly small finite dimension so there is no hope of notationally capturing all of the individual maps. you have to choose what to abstract away.
"L(V,W)" is itself a pretty good abstraction of that set. i'm slightly kidding, but only slightly. it's the main one i've used.
when people have a small number of maps, each from the same V to the same W, they might draw a diagram with V and W as vertices, and then a family of oriented edges from V to W, perhaps labeled with letters or other symbols, each edge denoting an individual map within the family.
there are even fairly robust tex packages that help with stuff like that, although i have not used them.
 
12:13 AM
@leslietownes okay
 
12:25 AM
@anakhro Did you ever respond to my response?
@leslietownes 6 1/2 for me! Time flees (or is that fleas?).
 
@leslietownes How do you maintain a good breadth of knowledge on math? Did you find it never left, or do you have to keep reading, chatting on here, etc.?
@TedShifrin you mean your identification of the fact that I was trying to show a submanifold is totally geodesic?
 
Yes … There are ways to do that having nothing to do with following geodesics. Context matters.
 
Well I did provide my context. But it's okay, it was solved and I was being a BIG FAT LOSER.
 
I didn’t scroll enough. Sorry for interrupting.
 
It's okay, teddykins.
I love you regardless of how much you scroll.
 
12:38 AM
What the hell is the $L^2$ metric? But we have an open set of a linear subspace, so that surely has zero second fundamental form.
 
The metric tensor is given by $\langle\dot\varphi,\dot\varphi\rangle_\varphi = \int_{\mathbb R^n}\|\dot\varphi(x)\|^2\,\mu$ for a volume form $\mu$.
 
Are these compactly supported diffeos?
In which case $GL(n)$ ain’t a submanifold.
Makes zero sense to me.
 
No, I don't think they are compactly supported.
 
Why is that integral finite?
 
shrugs
All I know is Arnold does some weird stuff.
I am still but a stranger to this realm of infinite dimensional manifolds.
 
12:54 AM
Arnol’d in my experience omits most details but is correct.
 
I will download this 1966 paper of his and see if I can uncover the mysteries of this metric.
Did you ever meet Arnold, Ted?
 
@geocalc33 fancy seeing you here! :)
My internet was off earlier today
 
No. He was supposed to be at the conference I went to in honor of Elie Cartan’s 100th birthday, but the government wouldn’t let him go ….
Fabulous conference in Lyon,
 
ghetto fabulous, son
^_^
 
Huh?
 
1:04 AM
@TedShifrin in the actual paper, Arnold seems to do only the L^2 metric for Diff D for a closed domain D.
 
@anakhro sorry for delay in response, i was picking up a demon. i have an OK memory for the basics and a lot of old books/notes around.
 
Compact?
Munchkin is promoted to demon? Oh, we got the rains.
 
At one point he does say bounded domain, but I am not sure if that was just a special statement.
 
Well …. As I said, he’s sloppy with details. But my complaint still will hold.
 
Demons are pretty cool, Leslie. Does yours have wings and a tail?
 
1:09 AM
ted, the california theater on kittredge in berkeley is closing. may have operated under a different name in your time.
following the UC theater into oblivion by a few decades.
 
I remember it. The little arts theater on Telegraph (where I saw Bogart movie after Bogart movie, etc) closed ages ago.
Covid closed one of my favorite theaters in SD.
 
while i was there, a tiny arts theater on shattuck that had the projection "booth" in the aisle was demolished for expensive condominiums.
as a middle finger to history, the entryway was styled somewhat like that of a movie theater in the new building.
 
I forget if the Albany theater on Solano near @copper is still there.
 
SD is a nexus for MSE chatters
 
it was still there the last time i drove by it.
 
1:13 AM
does anybody know where algebra can come in useful in analysis
 
All over?
 
so for example?
 
i like my theaters old and either huge and ornate or really small. not new and mid-size. no cup holders.
 
Hilbert space is first some algebraic structure, I think a vector space
 
yeah not trivial examples
 
1:14 AM
Depends on the algebra and on the analysis. Some fields of analysis are all algebra.
 
a lot of the elementary theory of operator algebras is pretty algebraic. you can prove fairly 'hard' early 20th century analysis with it very quickly.
 
Lie groups is a good intermingling.
Lots of analysis in representation theory.
 
ANT is heavy analysis once you get to L-functions
 
SCV and residues overlaps AG and plenty of algebra.
 
there's the algebra of fourier series on the circle with absolutely convergent fourier coefficients. an element of that algebra without roots on the circle has a reciprocal in that algebra. norbert wiener, no slouch, proved it with a lot of tricky analysis. if you know the right algebra it's an exercise you could assign in week 3 of an operator algebras course.
 
1:16 AM
Not much algebra there, Space.
 
in complex analysis, some of runge's theorems on uniform approximation by analytic functions or polynomials can be proved with general algebra techniques.
you do need topology and limits and stuff. it's maybe not "algebra" algebra.
 
Lots of sheaf cohomology everywhere.
 
no True Analysis Statement can be proved with algebra.
 
Ha ha.
 
fighting words, finally
 
1:20 AM
@JoeShmo along the lines of what Leslie was saying, you should check out the Gelfand-Naimark theorem.
 
change my mind
 
Define “true analysis.”
 
ted, if you can prove it with algebra, that ain't it.
 
well apparently all of early 20th century analysis?
 
I like math that integrates different fields of math, not narrow shit.
 
1:24 AM
What's the toughest analysis? Complex analysis, real analysis, measure theory, functional analysis, fourier analysis, euclidean harmonic analysis, abstract harmonic analysis, etc....
 
lately I'm into concrete math, not abstract nonesense
harmonic analysis
 
analysis of the self.
 
lol
 
lmao
 
oy vey
this escalated quickly
 
1:26 AM
That’s super-narrow. Trompian.
Degenerated.
 
probabilistic proofs of analytic facts are black magic
 
There’s plenty of those proofs in combinatorial number theory. Several masters at UGA.
 
probabilistic combinatorics is really cool
 
my daughter is demanding "raspberries and cheese sticks and milk" for dinner. this is pure free association based on what she could see in an open refrigerator.
 
Yuck.
 
1:31 AM
you could game that to your advantage leslie
re: black magic. like try finding $\lim_{n \rightarrow \infty} e^{-n} \sum_{k=0}^n\frac{n^k}{k!}$ by hand. you could do it, after a long while, but the probabilistic proof is a couple of lines
 
she also wanted chocolate. that isn't visible, but she wants it. she had some on her birthday, over two weeks ago.
 
it feels like cheating
I always crave chocolate. always.
like you said chocolate. so now I want brownies. look what you did.
 
tosses Shmo a dark-chocolate covered walnut
 
none of that healthy option shit
I want fudge
I actually do appreciate walnuts in my brownies
 
i like bernstein's probabilistic-ish proof of the weierstrass approximation theorem.
 
1:35 AM
talk about math please
 
although there is a functional analytic proof that is even shorter.
 
Makes my teeth hurt. Nope.
 
and I don't indulge often either. hence I always crave.
yeah! bernstein's polynomials is another one
just popped up in the homework this week
 
a lot of baire category arguments have a probabilistic flavor, with genericity (or whatever it is called) in place of presence a.e. with respect to a measure.
 
@leslie Given a smooth curve γ in C parametrized by z : [a, b] → C, and f a
continuous function on γ, we define the integral of f along γ by
 
1:37 AM
yeah its not really a probabilistic proof though is it. the only pseudo-probabilistic feature there is the binomial distribution
 
Why do books say that?
 
i don’t know all these proofs to which y’all refer.
 
the bernstein polys one you should definitely check out
its 3/4 of a page, if that
 
the existnece of a lebesgue integrable function whose fourier series diverges pointwise everywhere is something you can do with BCT although i think kolmogorov initially did it with a clever construction.
 
Why not just a smooth curve $\gamma : [a,b] \to \Bbb{C}$?
 
1:38 AM
durrett has a proof
 
Because to some of us a curve is a set, not a mapping. Done.
 
i think they may want to distinguish between the curve as a geometric object and the parametrization. i would have to see more context for more.
 
I will get back to you on that :)
Reading from a $\Bbb{C}$ analysis book
 
I agree with the book.
 
I found the answer and it doesn't match Ted's
lol
 
1:40 AM
”the answer”?
 
The family of
all parametrizations that are equivalent to z(t) determines a smooth
curve γ ⊂ C, namely the image of [a, b] under z with the orientation
given by z as t travels from a to b.
 
yes, for line integrals we need oriented curves. Read Ted’s book.
 
So $\gamma$'s are equivalence classes, is the reason
Of a set of parameterizations
So $z$ is just one such rep
@TedShifrin where's your book?
 
This formalism is crap.
Published.
 
Is that it?
I thought you meant $\Bbb{C}$ analysis book though :)
Here's the book I'm reading:
 
1:45 AM
I prefer Palka's book
 
Where's a link to that?
 
It's more for dummies like me
 
Oh, cool, I got a copy of "Visual complex analysis". It's great
 
he explains with very much detail
 
however, I like the formalness of the link
 
1:46 AM
he talks a lot
but I like that
it fits with my type of learning
 
What type is that? Genius?
 
Math genii, plural
 
no, it's like complex analysis for dummies I think
 
@Twink oh you mean this free book here:
 
1:48 AM
he explains everything with a lot of details
yes, that one
that's my favorite book
 
Actually I couldn't find a free version. That link was just the TOC
 
I bought it and it didn't arrive to my address, so I made a complain, and they sent another copy. After I received it, I got the other one, so I have two books
 
Nice, amazon is good for that
 
I want to sell the other book
No, I bought it on springer's site
 
I mean for selling book
 
1:51 AM
it was cheaper than in amazon
should I return the other copy of the book or should I keep it?
and sell it?
 
I was thinking on that
 
xD
 
I think maybe have a nice copy and a copy you can treat badly
 
I love Palka, I can't treat it badly
 
Cuz I have old mathbooks that were just treated like shit (by me)
and even missing pages
 
1:53 AM
I never treat books badly
I treat them with love and care
 
You're a sait of bookdom
*saint
I even wrote in the covers of some books
 
I'm just very carefully with my things
that's a sacrilege
 
to write in the covers with pen
 
They are my Fermat notes
When I die, nothing will be conjectured though :|
 
1:55 AM
use post-it
 
The greats stole all the good maths from us
 
are take home tests the norm in the second half of a math undergrad?
 
Define norm
on the space of undergrads
 
actually it's not a norm
it's a seminorm
 
Hello! To locate item in the list, then possibility the item is there on average is $\Sigma_i (i)/n = 1/n + 2/n+ \cdots+n/n = (n+1)/2$. Possibility of element is not in the list is simply $n$ meaning search through the entire list. No on average, we need half the times the item is not there and half the time the item is there, which equals to $n/2 + (n+1)/4$. What do you think about last statement please?
i.e., n/2+(n+1)/4
 
1:57 AM
Did Bateman or Horn invent that?
 
@Derivative Assuredly this is COVID most places.
@Twink smack
 
and in the beforetimes? How was it like?
 
You spread germs like a petri dish and everyone got sick
 
I gave take-homes almost never. I learned that even with “honors students” cheating took place.
 
@TedShifrin Ouch!
 
2:01 AM
Cheating can be honorable in some circles, if you don't get caught :|
 
Great.
Sounds like one of our political parties.
 
Yes, probably them
 
@AbstractSpacecraft which circles?
 
Circles of cheaters
 
the unit circle?
hmm, I had never heard about that
 
2:02 AM
It's actually the whole world
the whole world cheats somewhere
in there lives
Little cheating never hurt no body less they got caught
lol
 
I don't think Ramanujan cheated
 
He cheated us of his great maths in older age
And he also was known to not give a proof
 
@TedShifrin did you ever cheat on an exam?
 
at all :)
 
Note to self: Ted does not like poop jokes.
 
2:05 AM
I have never cheated, I'm too anxious for that
 
For me it depends on what you consider cheating.
 
copying
 
I don't think so
 
I never cheated either. I think everyone in this room is cool
They would never cheat.
 
I have terrible handwriting, and at least once I did a problem the incorrect way and wrote random gibberish as my work because I didn't know how to do it the way we were supposed to; I just knew it involved summations.
I got full credit for the problem.
 
2:08 AM
Goin to hell, cheater!
^_^
j/k
 
I didn't consider it cheating but some might.
 
Yeah, so cheating is a relative phenomena
 
I got paid back slightly either way.
It was something to do with power series, and later when I went back to reteach myself calc cause I didn't remember much from school, I went over my exams, and I spent some time trying to reason through my work on the problem, before the incident all came back to me.
 
Nice, I know how it is now...
I can't even comprehend my notes later, but somehow other people in the world can comprehend theirs
So I usually just throw away my accumulated notes
 
Now that I'm self studying I take great notes.
When I was in school I was awful at it.
 
2:13 AM
Hmmm, too bad, I always understand and keep my notes
you should be more organized
I always take good notes, unless the teacher is disorganized
and I use different colors of pens and highlighters
 
2:50 AM
@Twink Yes, once. A 7th grade history class where we had to memorize pages of dates and names. I never cheated again. But I did hate classes like that
 
I don't take notes
I didn't take notes all grad school
or undergrad for that matter
can't do it. I like the book better anyway
 
Some of us don’t copy books on the board. For advanced courses there often is no book.
So you may have to grow up.
 
i'm not even a person, i'm a chatterbot trained on my designer's excellent notes.
 
Channeling munchkins.
 
3:09 AM
@geocalc33
 
3:56 AM
@AbstractSpacecraft
 
 
2 hours later…
5:58 AM
@TedShifrin Hi Ted, the landmark twin beside kains on Solano is still going strong!
 
copper did you have any memories of the california theater? just east of campus, shutting down now.
 
 
1 hour later…
7:00 AM
@leslietownes on kittridge?
yes
there used to be some veggie restaurant on the oxford corner. good something or other i think?
 
oh, yeah.
i forget what it was. i got food poisoning there, or near there, once.
 
 
2 hours later…
8:42 AM
hi, im trying to understand the conformal map $f$ in this image imgur.com/a/O1N4JIy , it maps the upper half plane to some polygon. I understand the authors argument saying that $\arg(f')$ jumps at the $z_k$ , but he later then says that since $f_k = (z-z_k)^{- \beta_k}$ jumps the same amount at $z_k$, $\frac{f'}{f_k}$ is holomorphic in a neighbourhood of $z_k$, this would make sense to me if we knew $|f'|$ did not jump at $z_k$, but im not sure how to see this
 
 
4 hours later…
12:50 PM
Schwarz–Christoffel mapping, yum
 
Let $U \subset \Bbb R^n$ be open and let $\vec a \in U$, and let $f:U \to \Bbb R^m$ be differentiable at $\vec a$. Prove that the Jacobian at $\vec a$ is identical to the derivative at $\vec a$.

We say $f$ is differentiable at $\vec a$ if there exists a linear map $f'(\vec a): \Bbb R^n \to \Bbb R^m$ such that $\lim \limits_{\vec h \to \vec 0} \frac{f(\vec a + \vec h) - f(\vec a) - f'(\vec a)\vec h}{\|\vec h\|} = 0$. Notice this can be rewritten as the equality $f(\vec a + \vec h) - f(\vec a)= f'(\vec a)\vec h + \epsilon(\vec h)$ if we suppose $\epsilon(\vec h)$ satisfies $\lim \limits_{\ve
Let $\vec e_j$ be a standard vector of $\Bbb R^n$ and let $\mu(t\vec e_j)$ satisfy both $\lim_{t \to 0} \frac{\|\mu(t\vec e_j)\|}{|t|}=0$ and $f'(\vec a)t\vec e_j + \mu(\vec e_j) = f(\vec a+t\vec e_j) - f(\vec a)$. With some algebra, we obtain $\frac{f(\vec a+t\vec e_j) - f(\vec a) - f'(\vec a)t\vec e_j}{|t|} - \frac{\| \mu(t\vec e_j) \|}{|t|} = 0$, and taking the limit as $t$ goes to $0$, we obtain: $\lim \limits_{t \to 0} \frac{f(\vec a+t\vec e_j) - f(\vec a) - f'(\vec a)t\vec e_j}{|t|} = 0$.
Using the linear map properties of $f'(\vec a)$, we obtain $\lim \limits_{t \to 0} \frac{f(\vec a+t\vec e_j) - f(\vec a)}{|t|} - f'(\vec a)\vec e_j =0$, i.e., $\lim \limits_{t \to 0} \frac{f(\vec a+t\vec e_j) - f(\vec a)}{|t|} = f'(\vec a)\vec e_j$, and therefore the jth column of $f'(\vec a)$ is the jth partial derivative of $f$ at $\vec a$, which is the definition of the Jacobian.
mein gott that is a bit longer than i expected
well, if anyone has the grace and charity to read it, is it believable?
argh i need to prove the limits exist
ignore the above, my bad for spamming chat
 
 
1 hour later…
2:04 PM
Hey,
$exp(x_{1},x_{2})$ - what does the author mean by that expression? I thought exp only has one argument
exp(x1, x2)

pdf page 159 https://mml-book.github.io/book/mml-book.pdf
ah nvm its not a comma but a multiplication
 
Hi folks, I'm working on my assignment for decision theory. I got iid random variables and need to show that the given estimator is inadmissible under the quadratic loss. Does it suffice to compare its risk (expected loss) to an unbiased estimator (sample mean) whose risk would be 0? I wrote my attempt in latex here: cdn.discordapp.com/attachments/554278933554659328/…
Could someone please let me know if that makes sense?
 
2:42 PM
Suppose that I want to prove that L=R, where L and R are some mathematical expressions that evaluate to something. I consider: L is finite and equals some value say m, then using some methods I show that R is also equal to m. Then can I say that L=R? Or should I do the same from R to L also?
 
3:12 PM
Is my question clear?
 
not sure what you mean: $L=m$, $R=m$ together imply $R=L$ and $L=R$
 
@wklm I'm no expert, but seems correct. You have shown the sample mean dominates your estimator, right? You need to know $(a - 1) \mu \neq b$
You have written $a \mu \neq 1/b$. Typo?
 
@Koro equality is transitive and commutes
 
@BalarkaSen I assumed that without the loss of generality we can set $\mu = 0$ and then it suffices to show that there are a and b for which $ (b + (a - 1)\mu)^2 > 0$ while for an unbiased estimator it is 0
 
@shintuku L=R is to be proven. Doesn't that mean that I have to prove that "$L=m\implies R=m$ and $R=m\implies L=m$"?
The confusion arose while trying to prove equality of limit of two sequences...
What am I missing?
nevermind, I think my confusion is clear now.
If L has say three possible values u,v,w then accordingly R will also have u,v,w if the equality is to hold.
Then, I think that my proof here is correct:
 
3:28 PM
it is faster to use instead $R = m \land L = m \implies R = L$ citing transitivity of equality. It is transitivity that is doing the work here
 
 
@wklm Why there exists $a, b$? Don't you want to show for all $a > 1$ and for all $b$?
 
@Koro also $R = m \implies L = m$ if and only if $R=L$, so you don't need $L=m \implies R = m$
 
@BalarkaSen imgur.com/QVm9B0R
this is the definition I am using
 
3:31 PM
@shintuku in my proof shared above, I used $\iff$ wherever possible.
 
$L = m \implies R = m$ is equivalent to $R = L$, and so is $R = m \implies L = m$
furthermore, $L = m \implies R = m$ if and only if $R = m \implies L = m$
 
hmm, right.
I got confused earlier.
 
it says that for a rule A(P) to dominate rule B(P) for every P, So if there's a P for which Risk(B(P)) < Risk(A(P)) then A doesn't dominate B
 
Thanks @shin :)
 
math.stackexchange.com/q/4288020/922120 I have a question on free $\Bbb Z_2$-space
 
3:36 PM
@wklm OK, but that means the quantifier is on $\mu$, not $a, b$.
I agree it suffices to take any $\mu$ s.t. $(a - 1)\mu \neq b$.
$\mu$ is the parameter. $a, b$ are some arbitrary constants someone fixed for you.
 
ahh!! correct! Many thanks for pointing this out!
thank you @BalarkaSen
 
Welcome.
Or "wklm"
 
:D
 
koro, as an exercise, you might try finding an argument that fits on one page and doesn't involve cases :)
 
I'm glad you saw my proof. Thank you! I think the cases may be reduced if I use Bernoulli's inequality instead of using log and exponent function.
 
3:46 PM
i mean, maybe some case analysis is necessary, but my instinct is that even if you use cases, you can wrap some of them up in a few lines, instead of repeating variations on the same thing
liminfs always exist, for example. in this way they are better than limits, and often one can avoid separating out the case where the limit does exist
a lot of the problem seems to be in relating limiting behavior of a_n log(1+1/n) to a_n/n (i'm not sure what you wrote fully explains the relationship, although i think everything you wrote is true). that might be the focus for the simplification, even if you still use cases
just thinking out loud
 
liminf always exists -yes so I had to consider cases on liminf only (liminf finite non zero, infinite and zero).
 
you really love your cases
 
I used $\lim \frac{\log (1+x)} x=1$ as $x\to 0$ to relate the two limiting behavior.
 
i guess maybe it's necessary depending on the setup. rudin (for example) spends what some would regard as too much time with the extended real number system to avoid this
 
I'll think about reducing no. of cases and let you know if there's a progress in that.
 
3:54 PM
The projection matrix $A(A^TA)^{-1}A^T$, where the matrix $A$ has as column vectors a basis for the $V$ onto which we want to project, is equal to the identity matrix if $A$ is invertible. does this have any geometrical significance? I'm trying to figure out why it collapses into the identity matrix
 
i mean, do whatever you want :) i just note a lot of analysis arguments can take up a whole lot of space if they use cases and it can be helpful to minimize them. for example the triangle inequality |x-z| <= |x-y| + |y-z| encodes a lot of true information regardless of the relative positioning of x, y, z on the line, which would come up if you think of |t| as a 'piecewise-defined function' and break down in terms of whether |t| is t or -t in each of those component | |s
shin, that's orthogonal projection onto CS(A)? orthogonal projection onto the full space is I? if you want to find the nearest vector in the full space to a vector itself, there's nothing to do? i dunno if any of this counts as 'geometrical'
 
right, it's orthogonal projection onto the column space of $A$
 
algebraically, if A is invertible just expand (A^T A)^(-1) as a product of the inverses of A and A^T
not what you asked for but worth keeping in mind at all times. the hypothesis is what lets you bring ^{-1} inside the product
 
right, if $A$ is invertible we get $A(A^TA)^{-1}A^T = AA^{-1}(A^T)^{-1}A^T = I$
 
@shin I revised it about two days ago or so. First find projection of vector a onto vector b. Then, project a onto space U (Leslie mentioned what is meant by it -nearest vector ...) and note that the formulae come almost similar.
 
4:01 PM
in terms of the arguments that realize that formula as a geometric projection, i think it's:if you want to find the closest vector in [a subspace of R^n] to v in R^n, that's sometimes some work, but if the subspace of R^n happens to be all of R^n, you just return v
 
i'm trying to interpret why the matrix would collapse, when we usually think of projection as getting the closest vector in the column space to a vector that is not in the column space "plane"
oh, right leslie
oh: if a matrix is invertible, the row space and column space are identical, no? i.e. have the same span
 
if the matrix is regarded as an operator on R^n, yes
sometimes people use matrices to represent operators on different spaces of the same dimension :) where that doesn't quite work out, but you get "whole space" for the row space and "whole space" for the column space, whatever those are, in any event
 
that means the column space spans the domain! so that explains that
yeah i'm thinking R^n
hm so this isn't always the case then
 
just think R^n. people hate on it but for finite dimensional linear algebra it (and C^n) can't be beat
 
right, thanks for the tips
 
4:25 PM
ah, we can also notice that if $A$ is invertible (which makes it square), then $P = A(A^TA)^{-1}A^T$ will also be invertible and square, meaning both $P$ and $A$ have the same column space
which is obvious when we set $P=I$
hm, but that's not exactly the fact that makes a projected vector not move
 
projected vectors move?
 
hm, its shadow instead
 
they can move, i'm not particular about this.
all sorts of interpretive narratives help or not. whether it's the thing changing or the label of the thing changing. that comes up a lot.
moving vs. casting shadows is different but also kind of the same.
 
ok we need to work on these interpretative narratives
this can become an allegory for the cold war kind of thing
$P$ is the soviet union in all of this
this will get people learning
$\vec x$ is trying to save democracy
 
there was a trend in the mid 20th century where some math publications under particularly authoritarian regimes would include front matter explaining how some vague quote from lenin or mao led to the work in question.
maybe in a few years we'll have that in the USA.
 
4:41 PM
$\vec y$ is trying to save $\vec b$, and the only way he can do it is by finding a projection onto $\vec A$
$\vec c$ is not quite right in the mind, in all of this. he goes on a long run and gets a big following, from vectors in $\vec A$
$\vec y$ has a fear of heights after having his work partner fall while pursuing a criminal
that complicates things
 
gosh it's weird to see low-rep accounts asking questions about stuff that most people don't see until halfway through graduate school if ever. particularly when the questions are, for lack of a better word, [the type of question that people say there's no such thing as].
 
the trap as a newbie is that you can understand the words since its in english
so there's an illusion of proximity and semblance
 
that's true. it's also fairly standard and even reasonable in many fields to simply skip over stuff that seems like it might be too involved on a first, second, or third pass. 'i can come back to it later if i need to.'
 
they need to work on their interpretative narratives
 
guess it depends on where and how you were raised, too - for at least a subset of environments I would guess that math education is shifting quite a lot younger, and so the topics are growing more compact (and content slough is high)
 
4:53 PM
that's an interesting point. i've seen that first hand a little, even a long time ago.
 
Seems chat has changed a lot
 
this was a criticism of the 'moore method' of teaching. it can produce knowledge along some narrow curated dimension, and maybe even novel work along that dimension, while also manufacturing ignorance of anything else.
 
did old members like Ted, Mike, Balarka Sen etc left ?
Leaky Nun is there
 
ted and balarka have both been around very recently although they are not present presently.
 
Okay thanks
 
4:55 PM
i don't know which mike you're referring to. you've been around fairly recently.
 
Mike Miller
 
@shintuku haha
 
We had conflicting names, thus I changed my name
 
i haven't seen him in a while.
 
@leslietownes Not familiar with the Moore method - not a math educator, as fun as I think it'd be. Skimming it, the conclusion makes sense (but eh, a naive perspective is worth the price).
 
4:58 PM
the moore method is problematic for other reasons. excluding black people from the classroom was part of the moore method.
 
...seriously?? good lord
 
Akiva Weinberger was another old member. I am talking about this Mike mathoverflow.net/users/40804/mme
He changed his username
 
Mr. Akiva has also been around lately
 
i don't think most math educators would be familiar with the moore method. it is too extreme for K-12 deployment and probably has never made much headway into ed schools.
 

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