Mathematics - 2021-03-04 (page 2 of 3)

leslie townes

16:00

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.

Matthew Christopher Bartsh

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.

leslie townes

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.

Matthew Christopher Bartsh

At ELU some are saying it doesn't belong in ELU.

Leaky Nun

@MatthewChristopherBartsh you're just posting it in the wrong place

Astyx

I don't know what the usual answer length is for ELU, but that's a long post

Leaky Nun

16:03

also you're lying if you say it is the first feedback

Matthew Christopher Bartsh

Yes, I was wondering if it was too long.

Leaky Nun

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 ^

leslie townes

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.

Matthew Christopher Bartsh

I meant the responses from ELU were the first feedback. Not that one comment.

What would be the right place?

leslie townes

can you provide a link? i am at sea here.

Leaky Nun

16:06

your own blog

@leslietownes https://english.stackexchange.com/a/561666/235777

Matthew Christopher Bartsh

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.

leslie townes

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.

Matthew Christopher Bartsh

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.

leslie townes

binary 10 should not be pronounced ten. we can all agree on that. that's as close to objective truth as anything gets.

Matthew Christopher Bartsh

I notice that few were saying that in Mathematics.

Few to none.

leslie townes

16:11

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.

Matthew Christopher Bartsh

Agreed.

Astyx

The highest upvoted answer on the ELU post agrees with that

Jakobian

well, I disagree. With words being in base 10, we have to always switch, which might impair our thinking

leslie townes

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

Jakobian

wait, let me rethink what I'm disagreeing with

Astyx

16:16

What do you want me to do? LEAVE? Then they'll keep being wrong!

Edward Evans

lmao

Matthew Christopher Bartsh

That looks like an XKCD cartoon. Am I right?

Astyx

it is one

user2103480

@EdwardEvans got something for yer

Edward Evans

don't want it

user2103480

16:18

be quick to cash it

Edward Evans

aight got it

Matthew Christopher Bartsh

I was wondering whether I should ask and answer my own question to prevent accusations that I am straying from the question.

Jakobian

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

Jeff

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]?

Mike Miller

I think that depends on whether there is anyone else who wants to see said question and answer.

Edward Evans

16:19

This is so slavic

Mike Miller

But I mean, ultimately, you can do whatever you want to do.

Astyx

:57236492 sounds like little big

user2103480

@EdwardEvans yeh i love it

leslie townes

has anyone figured out what counting in french is all about? like, there's 20s in it? and other weirdness? whose idea was that

user2103480

russian hardbass is a permanent mood

Edward Evans

16:22

well that was ghastly

user2103480

@Astyx I'd almost say little big is more unique than that lol every russian hardbass song sounds the same

Astyx

lol

leslie townes

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.

user2103480

@EdwardEvans you're only envious that they dont have kontaktbeschränkung there

Alessandro Codenotti

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

Mike Miller

16:22

It's not a typo, it's just rearrangement as you say.

Alessandro Codenotti

This is very not natural/canonical though

Jeff

@MikeMiller Oh? Can you show me?

leslie townes

i mean, it's certainly also an approximation for the area on [n, 2n]. whatever else it might be

Leaky Nun

> 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).

source

@leslietownes ^

leslie townes

that's just goofy. they must have been on fermented goat milk

Matthew Christopher Bartsh

16:24

Am I allowed to post two different answers to the same question in the one forum, ELU say?

Leaky Nun

you'd ask ELU for that

Matthew Christopher Bartsh

What about in Mathematics then?

Astyx

"four-twenty sixteen eff-ten nine" is how we say "96g9" in french

Leaky Nun

I suppose you can, if you have two different approaches

I don't really know

leslie townes

russians sure do like casual athletic wear

Jeff

16:25

@leslietownes That's why I figure it's a typo

Mike Miller

I mean, I can show you, but this sounds like an exercise.

user2103480

@leslietownes rightly so

Mike Miller

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.

Jeff

@MikeMiller Is there a formal definition? I just know the idea.

Mike Miller

Yes, it was covered in your calculus textbook when you learned the definition of integral.

leslie townes

16:27

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.

Matthew Christopher Bartsh

@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.

Jeff

@MikeMiller It is. But I can't figure it out. Looking at your other post now.

Thorgott

@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).

Matthew Christopher Bartsh

It invites you to "add another answer" at ELU.

I am not here to play matador to anyone's bull.

Mike Miller

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.

Matthew Christopher Bartsh

16:31

So should I just go ahead and post it in Mathematics?

Mike Miller

@Thorgott I don't have TeX on so I can't really read that but it all smells right.

leslie townes

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

Mike Miller

@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!

Thorgott

only the part about arguing why the flow is actually defined where we want it is surprisingly awkward

Jeff

@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.

Matthew Christopher Bartsh

16:34

I could shorten it I suppose.

Thorgott

>(2) Prove that if M is connected every embedding ψ:Dn→M is isotopic to one lying in a single chart
just shrink it

Mike Miller

That is truly bizarre. Are you actually using a math textbook?

leslie townes

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

Mike Miller

Or rather, what's the textbook you're using?

Matthew Christopher Bartsh

Did my reputation drop from eleven to one just because my binary code post dropped from zero to minus two, do you think?

Mike Miller

16:36

(My previous phrasing was rather presumptuous. It's not like calculus textbooks are universally well-written. The opposite is more true.)

leslie townes

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.

Jakobian

Mike Miller

@Thorgott :) Aye. Notice that your previous argument didn't require that M is closed, just compact.

Jakobian

I don't know why would z_k be defined on nbd of p in M

user2103480

@leslietownes there's an astounding variety

Mike Miller

16:39

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.

Jeff

@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.

leslie townes

@user2103480 i'm also thinking of bleaching my hair. it would go with the look.

Mike Miller

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

Jeff

@MikeMiller Thanks!

Matthew Christopher Bartsh

I posted it in Mathematics just now.

Mike Miller

16:44

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

Jeff

@MikeMiller I'm not sure what 'awkward' phrasing you mean. I didn't sense any rudeness. The book is Thomas' Early Transcendentals.

leslie townes

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.

Thorgott

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

Mike Miller

@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.

leslie townes

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.

Mike Miller

16:51

@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.

Astyx

@leslietownes The book's canceled, transcendentals just called, they can't make it

Mike Miller

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)

leslie townes

Calculus: The Possibility of Transcendentals. They Are Totally Going To Make It

Mike Miller

@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?

Thorgott

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

Mike Miller

16:54

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

user2103480

@leslietownes oh yeah, I mean, if now isnt the time to try out a new look, when is it?

Thorgott

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 $p\in M$

leslie townes

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.

Thorgott

I take a point $q$ in my arbitrary disk embedding, choose a path from $q$ to $p$, apply isotopy extension

user2103480

@leslietownes amen!

Matthew Christopher Bartsh

16:56

Thanks for all your help.

Mike Miller

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)

Thorgott

ok nice

time to analyze the top dimension

Mike Miller

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

leslie townes

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.

Jeff

@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.

Jakobian

17:00

actually, things like tracksuits are more of a symbol of social pathology in Russia

of course, normal people wear them as well

Astyx

Mike is having a hard time leaving :p

leslie townes

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?

Jeff

@leslietownes I think so. I'm using left endpoints as (k-1)/n and right as k/n. Is that what you mean?

leslie townes

yeah, that seems right

Jeff

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))

Astyx

17:09

Brain check: a line bundle not having nonzero sections is not something that happens in a topological setting right ?

It's a geometric phenomenon

Thorgott

huh?

wait, do you mean a nowhere zero section or a section that's just not literally the zero section

user2103480

@leslietownes will that also be the signature outfit for a music genre

Astyx

not what I wrote

A section that is not literally the zero section

Thorgott

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

leslie townes

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.

Astyx

17:13

Right, My brain has been going in circles about this for a month now

Leaky Nun

@Astyx triple negative...

Astyx

double

Leaky Nun

"not" having "non"zero ... "not" something

Astyx

Ah yeah

Every line bundle in a topological setting admits a nonzero section

Leaky Nun

Tietze extension theorem

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

Astyx

17:17

I think it's easier than that

Thorgott

my argument assumes the base is nice enough

Astyx

Take a bump function on a trivial subset

Thorgott

yeah, but to have bump functions, the base needs to be nice enough

normality should suffice, I think

Leaky Nun

are you in the setting of a C0 manifold?

Astyx

yes, or even $C^{\infty}$

Thorgott

17:19

ok, then no issues whatsoever

Leaky Nun

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|)

Mike Miller

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

Jeff

@MikeMiller OK. Is Geogebra difficult to learn? I could never get started quickly, like in Desmos.

Astyx

I feel you

Mike Miller

@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

Jeff

17:27

Is how I learned LaTeX and Desmos

Sorry, I mean "learned".

Mike Miller

I am probably not going to be teaching any LaTeX classes, but I would say I've learned it well enough for practical use.

Thorgott

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

Mike Miller

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

Thorgott

17:44

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 $f,g$.

Edward Evans

17:54

@user2103480 ich bin diesbezüglich verdammt jealous of the tracksuit wearing badboiz

leslie townes

i think it's pretty clear that russians know how to party. under any circumstances

Edward Evans

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."

leslie townes

if anyone needs me i'm going to do a bowl haircut and then wear a tracksuit

Edward Evans

Ehrenmann

@user2103480 ever tried marmite?

user2103480

@leslietownes proper madlad

@EdwardEvans i think I tried a variation once but I think I can live without it

Edward Evans

18:02

lol what

I had my mum send over a kilo to Germany because I couldn't live without it

user2103480

Br*ton

Burned taste buds

Mike Miller

@Thorgott I love this!

Edward Evans

Unit** ******* ** ***** ***tain

user2103480

@EdwardEvans shhht dont say that word here

Mike Miller

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

Edward Evans

18:04

The language is that of Great Britain, which I will not utter here

Mike Miller

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!

user2103480

But tbh british food isnt that bad. Dutch food, on the other hand...

Edward Evans

Headcheese

Mike Miller

I will be yoinking that idea for my own materials. I can credit you it you want and you tell me your name

Jeff

(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 $\ln(2)$.

Mike Miller

18:09

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"

leslie townes

ow

Jeff

@MikeMiller Here it is programmed. But we probably have to use partial rectangles or something.

Mike Miller

Also, int_1^2 dx/x = int_{n*1}^{n*2} [1/n du] /[1/n u] by taking u = nx

Jeff

desmos.com/calculator/kfq05d2zal

Mike Miller

Which you can rewrite as int_n^{2n} du/u

leslie townes

18:12

that's what i was missing

it seems silly now

Mike Miller

I didn't do the u sub at first

Jeff

@MikeMiller Gotcha. But I guess we knew they were the same anyway since they evaluate to $\ln 2$.

Mike Miller

I just knew the answer was ln(2n) - ln(n) = ln(2n/n) = ln(2).

Jeff

@MikeMiller I got the u-sub right on paper. (i think you just made a little typo).

Mike Miller

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.

Jeff

18:16

Thanks for help @MikeMiller, @leslietownes.

Mike Miller

🗿👍

Jeff

BTW, I kinda like these pics I created pats self on back

Mike Miller

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

leslie townes

i love algorithmic symbol pushing. i wish i could 'see.' people who can see seem to grasp things. i just push symbols.

Ted Shifrin

GRR.

leslie townes

18:19

ted's here and he's angry. let's all step back

Mike Miller

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

Ted Shifrin

Even my abstract algebra textbook is full of pictures.

leslie townes

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.

Ted Shifrin

Well, we all need symbols to do mathematics.

leslie townes

i pay careful attention to the symbols that i push. i am not an automaton.

Ted Shifrin

18:20

You cannot teach multivariable calculus as symbol pushing.

leslie townes

well, i did. malpractice i guess

Mike Miller

most people cannot teach multivariable calculus!

Ted Shifrin

Indeed. And even plenty of calculus textbooks do it horridly.

Mike Miller

[Note that I have not asserted I can]

leslie townes

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.

Ted Shifrin

18:21

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.)

Mike Miller

You may have seen above or previously that Jeff is learning out of his book

Ted Shifrin

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.

Mike Miller

It is easier to teach topology.

leslie townes

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.

Ted Shifrin

I first learned epsilonics from Thomas's book when I was in high school. It was, in retrospect, horribly executed epsilonics. Cube roots of $\epsilon$.

leslie townes

18:24

if you're doing more than epsilon over two, that's the wrong epsilonics.

Ted Shifrin

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.

leslie townes

how. goodness.

Ted Shifrin

@leslie: Just to make you mad :) The proof of the product rule should be a picture of an expanding rectangle. :)

Mike Miller

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.

leslie townes

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.

Mike Miller

18:26

I find it implausible that Bob Hope has no visual sense.

Ted Shifrin

ROFL

Mike Miller

You'd need it for the choreography, I feel.

leslie townes

he does have a little bit if you need a song and dance.

i can go there.

Ted Shifrin

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.

Mike Miller

If someone understands what's going on I don't care how they got there.

Ted Shifrin

18:28

Maybe even third edition.

Mike Miller

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...

Ted Shifrin

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

leslie townes

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.

Ted Shifrin

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.

leslie townes

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

Ted Shifrin

18:35

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

Mike Miller

@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.

Ted Shifrin

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.

Mike Miller

Sure

I should get back to the work I keep saying I am doing. I just don't want to write. :(

Ted Shifrin

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!!

Mike Miller

@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.

Ted Shifrin

18:39

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.

Mike Miller

Of course we all agree a connection on G -> P -> B is a continuous homomorphism Hol from the loop-space of B to G.

Jeff

@TedShifrin Co-authored by Hass, Heil, Weir

Ted Shifrin

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.

leslie townes

what a strange commitment they seem to have had to having four-letter last names

Ted Shifrin

LOL ... I'm sure that's what the publisher had in mind :P

Thomas & Finney matched, too.

Jakobian

18:57

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

dc3rd

19:22

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

Ted Shifrin

@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.

Ted Shifrin

19:51

Salut, @Astyx. Qu'est-ce qu'on a mangé comme dîner?

Astyx

Salut ! Risotto ce soir

Ted Shifrin

Yum. Aux écrivisses? :)

Astyx

Non, tout simple

Ted Shifrin

Eh bien.

Tu peux quand même m'en envoyer tout de suite.

Astyx

Ah, il n'en reste plus ;)

Ted Shifrin

19:59

Merde.

C'est pas gentil!!

Astyx

haha

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$Mathematics$ Mathematics