« first day (3865 days earlier)      last day (1150 days later) » 

4:00 PM
i do like linking to other things. the point of all the rules, to me, is making the site a useful resource. a lot of the time that means being nonduplicative of other things, but not always, if there is some connection, making the connection explicit can be of value.
 
My post at ELU is at -2. One person called it bullsh*t with the full spelliing.
I've done a lot of thinking for many years and this is the first feedback I've had. It's one of my favorite ideas.
 
something about these kinds of sites brings out the worst in people. i see a lot of downvotes on math.SE that seem to be motivated by something other than increasing the quality of posts on the site. but i don't know. everybody has their own frame of reference. maybe in some world, calling bullshit is a helpful hand.
 
At ELU some are saying it doesn't belong in ELU.
 
@MatthewChristopherBartsh you're just posting it in the wrong place
 
I don't know what the usual answer length is for ELU, but that's a long post
 
4:03 PM
also you're lying if you say it is the first feedback
 
Yes, I was wondering if it was too long.
 
This could do with a little editing to make it easier to read. At the moment it's several big blocks of text. — KillingTime 2 days ago
this is the first feedback ^
Please realise that ELU deals with established usage. It is certainly not intended as a platform for the suggestion / championing of DIY 'words', unrecognised 'naming conventions' (the scare-quotes indicating that these are not words / naming conventions). — Edwin Ashworth 2 days ago
and this is the second feedback ^
 
sometimes i think people act reflexively and don't just say the simplest thing. for example, not the right place, is a simple thing to say, and useful for the poster.
 
I meant the responses from ELU were the first feedback. Not that one comment.
What would be the right place?
 
can you provide a link? i am at sea here.
 
https://english.stackexchange.com/questions/52494/how-do-you-pronounce-numbers-written-in-different-bases/561666#561666

https://math.stackexchange.com/questions/65760/how-do-you-say-10-when-its-in-binary
These seem to be about the same question.
 
it's certainly a long answer, and maybe the length alone set people off. in bases other than base 10 i just list the 'digits' (scare quotes because the origin of the word might have something to do with base 10). i don't see much of a good solution here. a shorter answer might not have produced the same response.
 
Certainly the answers were more similar to mine. At ELU many were saying that binary 10 should be pronounced 'ten' which would be the one way I would not pronounce it.
 
binary 10 should not be pronounced ten. we can all agree on that. that's as close to objective truth as anything gets.
 
I notice that few were saying that in Mathematics.
Few to none.
 
4:11 PM
it's an interesting topic, but maybe too subjective to avoid downvotes based on subjective feeling. 10 should be one-zero or maybe what it actually is, two. not ten. ten is chaotic evil.
 
The highest upvoted answer on the ELU post agrees with that
 
well, I disagree. With words being in base 10, we have to always switch, which might impair our thinking
 
everyone on this subject is wrong except for me. time to join ELU and tear this question a new one.
actually maybe i'll just enjoy some herbal tea and not embrace a life of conflict
 
wait, let me rethink what I'm disagreeing with
 
4:16 PM
 
lmao
 
That looks like an XKCD cartoon. Am I right?
 
it is one
 
@EdwardEvans got something for yer
 
don't want it
 
4:18 PM
be quick to cash it
 
aight got it
 
I was wondering whether I should ask and answer my own question to prevent accusations that I am straying from the question.
 
I think both using zero-one word system and ten, eleven etc. are good, but both of them you have to get used to, and since word system for numbers in English is so weird and has a bias towards base 10, ultimately I agree
 
My boss insists this is not a typo: For the series $\sum_{n=1}^{\infty} a_n$ has terms $a_n = \sum_{k=n+1}^{2n} \frac 1k$, "Show that $a_n$ is equal to the right-hand approximation for the area under $\frac 1x$ over the interval $[1, 2]$". Shouldn't the interval be $[n, 2n]$? How could I compare sum 1/k to interval [1,2]?
 
I think that depends on whether there is anyone else who wants to see said question and answer.
 
4:19 PM
This is so slavic
 
But I mean, ultimately, you can do whatever you want to do.
 
:57236492 sounds like little big
 
@EdwardEvans yeh i love it
 
has anyone figured out what counting in french is all about? like, there's 20s in it? and other weirdness? whose idea was that
 
russian hardbass is a permanent mood
 
4:22 PM
well that was ghastly
 
@Astyx I'd almost say little big is more unique than that lol every russian hardbass song sounds the same
 
lol
 
jeff: looks like a typo to me. maybe there is some way of rearranging the sum. but stuff like like 1/(1 + 1/n) is equal to n/(n+1) and not that.
 
@EdwardEvans you're only envious that they dont have kontaktbeschränkung there
 
You can build a set $A\subseteq[0,1]$ such that $A\cap[0,x]$ has Hausdorff dimension $x$ for all $0\leq x\leq 1$. I suppose it generalizes to $\Bbb R^n$ with no issues
 
4:22 PM
It's not a typo, it's just rearrangement as you say.
 
This is very not natural/canonical though
 
@MikeMiller Oh? Can you show me?
 
i mean, it's certainly also an approximation for the area on [n, 2n]. whatever else it might be
 
> Numerals
The French counting system is partially vigesimal: twenty (vingt) is used as a base number in the names of numbers from 70 to 99. The French word for 80 is quatre-vingts, literally "four twenties", and the word for 75 is soixante-quinze, literally "sixty-fifteen". This reform arose after the French Revolution to unify the counting systems (mostly vigesimal near the coast, because of Celtic (via Breton) and Viking influences. This system is comparable to the archaic English use of score, as in "fourscore and seven" (87), or "threescore and ten" (70).
@leslietownes ^
 
that's just goofy. they must have been on fermented goat milk
 
4:24 PM
Am I allowed to post two different answers to the same question in the one forum, ELU say?
 
you'd ask ELU for that
 
What about in Mathematics then?
 
"four-twenty sixteen eff-ten nine" is how we say "96g9" in french
 
I suppose you can, if you have two different approaches
I don't really know
 
russians sure do like casual athletic wear
 
4:25 PM
@leslietownes That's why I figure it's a typo
 
I mean, I can show you, but this sounds like an exercise.
 
@leslietownes rightly so
 
Like, did you write down the definition of right-hand approximation?
Use boxes whose feet are between k/n and (k+1)/n where n+1 <= k <= 2n.
 
@MikeMiller Is there a formal definition? I just know the idea.
 
Yes, it was covered in your calculus textbook when you learned the definition of integral.
 
4:27 PM
if someone is posting two genuinely different answers to the same problem, i wouldn't personally see it as an issue. if there is conceptual overlap people might regard it as spamming or who knows what. a priori, personally, i would regard it as waving a red flag in front of the internet forum bull.
 
@Leaky Nun, what a good point. Three score years and ten is analogous to soixante- dix. I never noticed. Wow. Four score is analogous to quatre-vingt. Wow. I can't believe I never noticed that.
 
@MikeMiller It is. But I can't figure it out. Looking at your other post now.
 
@MikeMiller Ok, let $F\colon M\times I\rightarrow N$ be an isotopy, $M$ compact. The track map $\tilde{F}\colon M\times I\rightarrow N\times I$ is an embedding. Taking time derivative gives a map $M\times I\rightarrow T(N\times I),\,(p,t)\mapsto(F_p^{\prime}(t),1)$. Taking $M\times I$ to be an embedded submanifold of $N\times I$ via $\tilde{F}$, we extend this to a vector field $X$ of $N\times 1$, which is $1$ in the second component (first extend to a tubular neighborhood by pre-composing with a retraction, then smooth first coordinate off using a bump function).
 
It invites you to "add another answer" at ELU.
I am not here to play matador to anyone's bull.
 
There are very few rules on these sites other than things about harassment. Most moderation is done via upvotes and post closures / user deletions, which are communally decided. I do not post on ELU, nor does most anyone here, so they won't have insight about that site.
If you can handle some downvotes and possible deletion post whatever you want.
If you can't, don't post. Seems simple.
 
4:31 PM
So should I just go ahead and post it in Mathematics?
 
@Thorgott I don't have TeX on so I can't really read that but it all smells right.
 
just browsing around ELU it reminds me quite strongly of what i didn't like about elementary school. i would avoid it
but that's just me
 
@MatthewChristopherBartsh If you want. I wouldn't read it or upvote it myself because it's way too long.
It's not a sufficiently interesting topic for me to read pages of text.
But that's my taste, maybe others will benefit, I dunno.
@Thorgott Nicely done!
 
only the part about arguing why the flow is actually defined where we want it is surprisingly awkward
 
@MikeMiller Responding to just this now (then going to look at your other message): There are two sections of the text describing partitioning, but there is no formal definition of right-hand approximation. It's also not in the index.
 
4:34 PM
I could shorten it I suppose.
 
>(2) Prove that if M is connected every embedding ψ:Dn→M is isotopic to one lying in a single chart
just shrink it
 
That is truly bizarre. Are you actually using a math textbook?
 
the right hand approximation involves splitting an interval [a,b] into a number of pieces, often but not always intervals of equal length, and using the right endpoint of each interval as the thing you plug in to compute the Riemann sum. i presume Riemann sums to be known
 
Or rather, what's the textbook you're using?
 
Did my reputation drop from eleven to one just because my binary code post dropped from zero to minus two, do you think?
 
4:36 PM
(My previous phrasing was rather presumptuous. It's not like calculus textbooks are universally well-written. The opposite is more true.)
 
i'm still watching this incredibly weird russian video and deciding what adidas tracksuit to purchase
calculus instruction comes in two types: very good, or very ungood. the calculus instructor at my high school was mostly a volleyball coach. i don't think he could have gotten an A in his own class.
 
@Thorgott :) Aye. Notice that your previous argument didn't require that M is closed, just compact.
 
I don't know why would z_k be defined on nbd of p in M
 
@leslietownes there's an astounding variety
 
4:39 PM
Interestingly this is probably false topologically. The right condition on $\psi: D^n \to M$ to guarantee that there is a, in your language, homeotopy taking $\psi$ into a coordinate chart should be that $\psi$ is a tame embedding (extends to an embedding of a slightly larger disc).
Since then you can shrink on the image of the smaller disc and "slow down the shrinking" on the extra thickened part.
Until on the boundary of the extended disc you do nothing.
 
@MikeMiller OK, when n=2, then k= 3, 4 and the first box's feet are 3/2, 4/2. The second box' feet are 4/2, 5/2. I get the idea, but those feet aren't within [1,2], so maybe we need a sligtly different formula.
 
@user2103480 i'm also thinking of bleaching my hair. it would go with the look.
 
I meant to say (k-1)/n and k/n.
You're using the right-hand approximation so I want the right-hand term to look nicer.
I am spending too much time in here, this is supposed to be a work day... Ciao
 
@MikeMiller Thanks!
 
I posted it in Mathematics just now.
 
4:44 PM
For sure. I am still curious about your textbook, and hope my awkward phrasing about it seeming like a bad text did not parse rudely
 
@MikeMiller I'm not sure what 'awkward' phrasing you mean. I didn't sense any rudeness. The book is Thomas' Early Transcendentals.
 
is it a rule that every new post on math.se has to be from a new account who wants homework done? i used to be very open minded but things are getting crazy over there.
 
Alright, let $\psi\colon D^n\rightarrow\mathbb{R}^m$ be an embedding. By translation, you can isotope it to an embedding fixing the origin. Then $D^n\times I\rightarrow\mathbb{R}^m,(x,t)\mapsto\psi(tx)/t$ is an isotopy between $\psi$ and $d\psi_0$, which is injective. If $n=m$, two linear maps are isotopic iff they act the same way on orientation.
If $n<m$, two linear injections are always isotopic (by the previous case, it only remains to argue that the identity and a reflection are isotopic one dimension higher and this can be done by duplicating a coordinate in the extra dimension and th
ah no, that's only half the argument
 
@Jeff Seemed like I might have sounded condescending, glad to see I did not. I do not know Thomas' book but it sounds unfortunately written.
 
i'm going to corner the market with a book involving Later Transcendentals. Transcendentals that didn't bother to show up until chapter 10. Ooops, make that 11. Sorry. We had a lot of transcendental sh*t to do.
 
4:51 PM
@leslietownes I deleted my account some two years back for similar reasons. I used the site for amusing and moderately interesting small questions which I could solve in a few hours or a day (because research does not exactly provide those, but old textbook exercises become quite dull very quickly). But eventually all the questions on MSE became dull homework exercises instead of someone being genuinely curious about a subject, so my interest was killed.
 
@leslietownes The book's canceled, transcendentals just called, they can't make it
 
Now I come here for said moderately interesting questions (though now reframed, often I know the answer and it is instead an exercise in pedagogy)
 
Calculus: The Possibility of Transcendentals. They Are Totally Going To Make It
 
@Thorgott Seems like you've successfully proved that there are at most two ambient isotopy classes of discs, yeah? And you just have to finish the job by analyzing when there is 1 and when there are 2?
 
no, I'm still missing an argument
I need to argue that if $M$ is connected, the isotopy class doesn't depend on which chart I land in
 
4:54 PM
I don't think so
Oh I see
Rephrase your first argument
Don't show that every disc is isotopic to one in a chart depending on that disc
Fix a single chart once and for all and show that every disc is isotopic to a linear disc in that chart
This requires like one extra step
Good catch though
 
@leslietownes oh yeah, I mean, if now isnt the time to try out a new look, when is it?
 
Hmm, I can shrink towards any point in the disk, not just the center, so it remains to argue that every disk embedding is isotopic to one containing an a priori fixed $p\in M$
 
i've been cutting my own hair for about a year now. i think it's time i branch out into bleached hair and tracksuits. as in all things, russia points the way.
 
I take a point $q$ in my arbitrary disk embedding, choose a path from $q$ to $p$, apply isotopy extension
 
@leslietownes amen!
 
4:56 PM
Thanks for all your help.
 
Yup that's it
If q is in the chart I would have picked a path from psi(0) to q
(This is a good catch --- I should verify that I haven't made this error in any of my written exposition)
 
ok nice
time to analyze the top dimension
 
I did catch that one needs to get every disc centered at a fixed point q first, but I am not certain that I carefully wrote that we are using a single chart for every disc
You're killin it
OK but really I need to stare at the ceiling until I understand gluing parameters
Goodbye
 
i think i need an older car and a vacant lot where i can drink liquor in public. i'm almost all the way there.
 
@MikeMiller I'm making a picture (because I'm in capable of visualizing anything until I draw it AND program a picture). But it's pretty clearly going to work out nice. I thought you were going to work, but if you're utterly bored, just change the value of $n$
https://www.desmos.com/calculator/x56jpttjbn
TBF, you might be bored by the pic, too.
 
5:00 PM
actually, things like tracksuits are more of a symbol of social pathology in Russia
of course, normal people wear them as well
 
Mike is having a hard time leaving :p
 
i don't want to appropriate or parody anything. there's something wholesome and refreshing about not giving a f*ck, which seems to be the tracksuit ideal. that's what i'm going for. it's OK if i don't do that by imitating people in bad situations.
sometimes i let my daughter, who is 2.5 years old, dress me. i look insane. it's a new style that we will introduce to the universe. it will replace the tracksuit
jeff are you seeing the algebraic manipulations that turns the right endpoint approximation on [1,2] into the sum you're talking about?
 
@leslietownes I think so. I'm using left endpoints as (k-1)/n and right as k/n. Is that what you mean?
 
yeah, that seems right
 
Mike gave it to me. I was sort of thinking along those lines when he posted it (I almost wish he'd waited about a few minutes -- but then I also need to get this done (which does not explain why I'm busy making a pretty picture in Desmos))
 
5:09 PM
Brain check: a line bundle not having nonzero sections is not something that happens in a topological setting right ?
It's a geometric phenomenon
 
huh?
wait, do you mean a nowhere zero section or a section that's just not literally the zero section
 
@leslietownes will that also be the signature outfit for a music genre
 
not what I wrote
A section that is not literally the zero section
 
yeah ok, that always exists
find a compactly supported section that is not the zero section in one local trivialization and extend it by the zero section
 
my daughter's latest thing is dressing me in a felt crown. it has leaves on it. she made it in day care a few months ago in celebration of fall, but we're using it all year. it will become the outfit for a subgenre of metal. i've determined that there aren't enough subgenres of metal.
 
5:13 PM
Right, My brain has been going in circles about this for a month now
 
@Astyx triple negative...
 
double
 
"not" having "non"zero ... "not" something
 
Ah yeah
Every line bundle in a topological setting admits a nonzero section
 
In topology, the Tietze extension theorem (also known as the Tietze–Urysohn–Brouwer extension theorem) states that continuous functions on a closed subset of a normal topological space can be extended to the entire space, preserving boundedness if necessary. == Formal statement == If X is a normal topological space and f : A → R {\displaystyle f:A\to \mathbb {R} } is a continuous map from a closed subset A of X into the real numbers carrying the standard topology, then there exists a continuous map...
maybe something like this
 
5:17 PM
I think it's easier than that
 
my argument assumes the base is nice enough
 
Take a bump function on a trivial subset
 
yeah, but to have bump functions, the base needs to be nice enough
normality should suffice, I think
 
are you in the setting of a C0 manifold?
 
yes, or even $C^{\infty}$
 
5:19 PM
ok, then no issues whatsoever
 
yeah sure use a bump function then
I mean if you only need continuous then you can just use something like max(0, 1-|x|)
 
In fact there always exist smooth sections transverse to the zero section
@Jeff Pictures are great. I didn't look, but once you have fully understood this part, try to understand (a) why this is also gives what "leslie" suggested, the right-side approximation to the area under y = 1/x for x in [n, 2n]; use intervals of length 1, and (b) why the area under y =1/x is the same above [1,2] and [n, 2n] (thing u-sub).
This is another picture you can program; get geogebra to show you how the area changes as n varies continuously.
@Astyx I have never left my office, I just keep not working in it :8
 
@MikeMiller OK. Is Geogebra difficult to learn? I could never get started quickly, like in Desmos.
 
I feel you
 
@Jeff You can probably do this in desmos as well. To be honest, I only learned geogebra by taking other people's programs in it and then hacking together a new program out of the old one. Do that enough and you know how to use it.
Standard way to learn TeX as well Amusingly enough
 
5:27 PM
Is how I learned LaTeX and Desmos
Sorry, I mean "learned".
 
I am probably not going to be teaching any LaTeX classes, but I would say I've learned it well enough for practical use.
 
Ok, one direction is easy. If $M$ is oriented and $f,g\colon D^n\rightarrow M$ are isotopic embeddings and $\varphi\colon D^n\times I\rightarrow M$ is an isotopy, the differentials $df,dg\colon TD^n\rightarrow TM$ are bundle-homotopic via $TD^n\times I\rightarrow TM,\,((x,v),t)\mapsto d\varphi_t\vert_x(v)$, hence equioriented. Thus, we have exactly two isotopy classes of embeddings $D^n\rightarrow M$ when $M$ is oriented, one orientation-preserving and one -reversing.
In the unoriented case, I probably want to isotope two embeddings along an embedded higher-dimensional Möbius strip or something along those lines, but I'll have to think for a bit on how to make that work
 
I have an approach in mind which is very dependent on thinking of non-orientability a certain way
I am curious what you come up with, if you get hopelessly stuck I'll give pointers
 
5:44 PM
Actually, can't we cheat as follows: Let $\tilde{M}\rightarrow M$ be the oriented double cover and take two embeddings $f,g\colon D^n\rightarrow M$. Since $D^n$ is simply connected, these lift to embeddings $\tilde{f},\tilde{g}\colon D^n\rightarrow\tilde{M}$.
$\tilde{M}$ possesses an orientation-reversing deck transformation, so assume WLOG that $\tilde{f},\tilde{g}$ are equioriented. Since $M$ is non-orientable, $\tilde{M}$ is connected, so that $\tilde{f},\tilde{g}$ are isotopic by the previous argument. Then project down to get an isotopy between $f,g$.
 
5:54 PM
@user2103480 ich bin diesbezüglich verdammt jealous of the tracksuit wearing badboiz
 
i think it's pretty clear that russians know how to party. under any circumstances
 
truth
As Gorbatschow once said: "He drinks a Whiskey drink, he drinks a Vodka drink. He drinks a Lager drink, he drinks a Cider drink. He sings the songs that remind him of the good times."
 
if anyone needs me i'm going to do a bowl haircut and then wear a tracksuit
 
Ehrenmann
@user2103480 ever tried marmite?
 
@leslietownes proper madlad
@EdwardEvans i think I tried a variation once but I think I can live without it
 
6:02 PM
lol what
I had my mum send over a kilo to Germany because I couldn't live without it
 
Br*ton
Burned taste buds
 
@Thorgott I love this!
 
Unit** ******* ** ***** ***tain
 
@EdwardEvans shhht dont say that word here
 
My proof is to observe that a manifold is oriented iff it has no orientation reversing loop, where an orientation reversing loop is one that pulls back the tangent bundle to the non-trivial bundle over the circle
 
6:04 PM
The language is that of Great Britain, which I will not utter here
 
So drag your disc along an orientation reversing loop
Of course that's what your proof does in the end but it's much cleaner!
 
But tbh british food isnt that bad. Dutch food, on the other hand...
 
Headcheese
 
I will be yoinking that idea for my own materials. I can credit you it you want and you tell me your name
 
(Unless you finally made it to work) How can $n$ vary continuously?
Also, what is the comparison using u-sub? From this exercise I get that
$$\int_1^2 \frac 1x dx \leq a_n \leq \int_1^2 \frac 1x dx$$
and from the other exercise (over the interval from [n, 2n]) I found that
$$\int_{n+1}^{2n+1} dx \leq a_n \leq \int_{n}^{2n} dx.$$
All the integrals resolve easily to $\ln(2)$.
 
6:09 PM
Ehh, just pretend nobody told you it's a natural number. So look at pictures of integrals over [t, 2t] if you prefer to use t for a real variable
I get the sociological pain of saying "let n be a real number"
 
@MikeMiller Here it is programmed. But we probably have to use partial rectangles or something.
 
Also, int_1^2 dx/x = int_{n*1}^{n*2} [1/n du] /[1/n u] by taking u = nx
 
Which you can rewrite as int_n^{2n} du/u
 
6:12 PM
that's what i was missing
it seems silly now
 
I didn't do the u sub at first
 
@MikeMiller Gotcha. But I guess we knew they were the same anyway since they evaluate to $\ln 2$.
 
I just knew the answer was ln(2n) - ln(n) = ln(2n/n) = ln(2).
 
@MikeMiller I got the u-sub right on paper. (i think you just made a little typo).
 
I trust you.
BTW, some people choose to define ln as ln(a) = int_1^a dx/x
So what I just suggested is the standard proof that ln(ab) = ln(a) + ln(b) I think.
 
6:16 PM
Thanks for help @MikeMiller, @leslietownes.
 
🗿👍
 
BTW, I kinda like these pics I created pats self on back
 
I think visual understanding is very important and it's good to see you learn things that way
It's closer to actual understanding than any algorithmic symbol pushing
 
i love algorithmic symbol pushing. i wish i could 'see.' people who can see seem to grasp things. i just push symbols.
 
GRR.
 
6:19 PM
ted's here and he's angry. let's all step back
 
I get the impression you picked up a PhD at some point so I don't buy that your understanding of calculus stops at algorithm
 
Even my abstract algebra textbook is full of pictures.
 
they used to make me teach multivariable calculus. i taught it as symbol pushing. i can't draw a circle. what chance do i have with these cylinders or whatever.
 
Well, we all need symbols to do mathematics.
 
i pay careful attention to the symbols that i push. i am not an automaton.
 
6:20 PM
You cannot teach multivariable calculus as symbol pushing.
 
well, i did. malpractice i guess
 
most people cannot teach multivariable calculus!
 
Indeed. And even plenty of calculus textbooks do it horridly.
 
[Note that I have not asserted I can]
 
i just felt completely incompetent at it. so the theorems were symbol pushes. i don't think i was giving value for money. i was a better linear algebra instructor.
 
6:21 PM
George Thomas was a nominal number theorist. He had no feeling for analysis or multivariable calculus.
It was really bad.
(Yes, he actually "taught" me multivariable calculus the semester before he retired.)
 
You may have seen above or previously that Jeff is learning out of his book
 
No, I didn't see. Of course, Thomas died 40 years ago or so. The publishers are on their second round of "coauthors." But I even bitched at Joel Hass for a mistake he made in the multivariable.
Finney didn't improve it much, even though I literally gave him a list of conceptual errors that needed to be fixed.
 
It is easier to teach topology.
 
i just don't have a visual understanding of anything. i tried to get people through the process without having to acquire such a thing. coauthors do ruin the soup.
 
I first learned epsilonics from Thomas's book when I was in high school. It was, in retrospect, horribly executed epsilonics. Cube roots of $\epsilon$.
 
6:24 PM
if you're doing more than epsilon over two, that's the wrong epsilonics.
 
I wonder if we can dig up his proof of the product rule for limits from the early editions of his book.
Cube roots of epsilon. I promise you.
 
how. goodness.
 
@leslie: Just to make you mad :) The proof of the product rule should be a picture of an expanding rectangle. :)
 
I don't know what epsilonics are but in these arguments one just needs a continuous function of epsilon as the bound which vanishes at 0. This phrasing is often useful when it is difficult to get things as explicit as you want. Of course it sounds like this is a student's first pass at eps-delta.
 
it's just goofballing with the triangle inequality to me. i can't see anything. i had a student who was blind once, i think we had more common ground than i had with my sighted students.
 
6:26 PM
I find it implausible that Bob Hope has no visual sense.
 
ROFL
 
You'd need it for the choreography, I feel.
 
he does have a little bit if you need a song and dance.
i can go there.
 
I acknowledge that some students are not visual thinkers, either, of course. And so when I taught I tried to balance pictures with algebraic formulas. One needs both.
I have understood and taught the "correct" proofs since I was 18, so it's hard to recreate the wrong proof at this stage, but I swear ... Find Thomas from the 60s.
 
If someone understands what's going on I don't care how they got there.
 
6:28 PM
Maybe even third edition.
 
But I have to get them there in the way I am most competent.
Same with Bob. If he got them where they needed to be...
 
Well, @MikeM, we're supposed to be competent to help them get there the way they're most competent :P
I swear I had a student in my multivariable years ago who said she couldn't see/draw anything. A few months later, she was drawing multi-colored pictures on my blackboard better than I could.
As I recall, she transferred to Yale to be an art major. :D
 
one time i did nap and have a dream that solved a problem in something that became a paper. i have been slightly mystical since then. i have still been unable to visualize anything.
 
In all seriousness, I do believe that many linear algebra and abstract algebra courses are taught as symbol-pushing. That does the field (pun intended) gross disservice.
 
that's right, i think. almost everything i did could be a matrix calculation. i tried to convey the underlying ideas of the calculation even if it ultimately became mechanical.
a lot of books are just, 'here you go, do this.' not helpful
 
6:35 PM
On the other hand, many geometric topology arguments are far too geometric for me, I confess. I certainly could never "see" all the stuff that Kirby and his students do, and often Balarka's and DogAteMy's topological machinations are too geometric for me. So we all have our limits :P
 
@TedShifrin Sure. I just mean that one can teach with pants on their head if it works. I agree that singleminded focus on one pedagogical approach will not work.
 
Sure. Actually, I think it's good to aim to expand the students' skill/thinking set, too, of course. Hence my being so thrilled with the student I mentioned earlier.
 
Sure
I should get back to the work I keep saying I am doing. I just don't want to write. :(
 
I guess I should complain about a lot of graduate differential geometry courses, too. I think a lot of times they're taught very much in the symbol-pushing style. If a student hasn't developed intuition for a connection and parallel translation in an undergraduate course, the student is unlikely to develop it in a formalistic graduate course.
Go work, @MikeM!!
 
@TedShifrin The handleslides are the hardest part. Once those are understood the rest is possible. Can I understand a Kirby calculus argument? Yes. Can I produce one? ...
I strongly agree about this, we have talked about this extensively.
 
6:39 PM
John Harer taught me to understand some of that stuff in grad school — we shared an office and also shared beers numerous times. But I could never do it.
 
Of course we all agree a connection on G -> P -> B is a continuous homomorphism Hol from the loop-space of B to G.
 
@TedShifrin Co-authored by Hass, Heil, Weir
 
Yup. UGA was using that for many years. Doesn't much resemble the original Thomas. But still not my favorite.
Hass and I went to grad school together.
 
what a strange commitment they seem to have had to having four-letter last names
 
LOL ... I'm sure that's what the publisher had in mind :P
Thomas & Finney matched, too.
 
6:57 PM
I've already asked this question here, but would it be fine if I repeat myself?
I have a problem with a book I'm reading
so f^k, how I'm understanding it, comes from x^k o (x_1, ..., x_m)^-1 on W
but after that I just don't get it
 
7:22 PM
Recovering symbol pusher checking in 🙋🏾‍♂️....................what's a good free internet app I could use that is similar to Geometer's Sketchpad to fiddle around with some shapes?
no worries.....geogebra.org seems to be satisfactory
 
@Jakobian: Here's the idea. You reduce to an $m$-dimensional submanifold of (an open subset of) $\Bbb R^n$, which you have as a graph over the $(x^1,\dots,x^m)$-plane. Thus, it is given by $y=(y^{m+1},\dots,y^m) = f(x^1,\dots,x^m)$. Now you just want to change coordinates to straighten this out to be $y=0$ instead.
So you use as coordinates $(x^1,\dots,x^m,x^\mu-f^{\mu}(x^1,\dots,x^m))$, $\mu=m+1,\dots,n$. Check this is a coordinate system (no problem) and that the previous graph is now given by the slice $x^\mu = 0$, $\mu=m+1,\dots,n$. Done.
 
7:51 PM
Salut, @Astyx. Qu'est-ce qu'on a mangé comme dîner?
 
Salut ! Risotto ce soir
 
Yum. Aux écrivisses? :)
 
Non, tout simple
 
Eh bien.
Tu peux quand même m'en envoyer tout de suite.
 
Ah, il n'en reste plus ;)
 
7:59 PM
Merde.
C'est pas gentil!!
 
haha
 

« first day (3865 days earlier)      last day (1150 days later) »