Mathematics - 2022-10-04 (page 1 of 2)

Ted Shifrin

12:00 AM

For a loop?

copper.hat

like when someone says "no, no, no...., no", i am busy in my head working out the parity...

indeed:-)

i'm going to go and pretend to exercise...

dang, missed that, whatever it was

Ted Shifrin

@copper.hat I think you’re missing the emphatic point?

It was a repeat. Chat messed up.

wol

2:57 AM

Is it true that if $S$ is a dense subset of $\mathbb{R}$, $-S$ is also dense? Not sure if I am overthinking here . . .

Ted Shifrin

You have a proof?

Akiva Weinberger

Me and a friend working on the same p-set

He writes that $C^\infty(M_1\times\cdots\times M_k)$ and $C^\infty(M_1)\oplus\cdots\oplus C^\infty(M_k)$ are isomorphic by the "universal property of the direct sum"

($M_i$ are smooth manifolds)

That's... not true, is it?

Ted Shifrin

Sure is crap.

Is $xy$ a sum of a function of $x$ and a function of $y$?

Akiva Weinberger

That's exactly the example I gave

@TedShifrin Delicate as always

Ted Shifrin

So you didn’t need us :)

I wonder what universal property this is.

Akiva Weinberger

3:13 AM

The exercise is to show that $T_{(p_1,\dots,p_k)}(M_1\times\cdots\times M_k)\cong T_{p_1}M_1\oplus\cdots\oplus T_{p_k}M_k$

Ted Shifrin

Well, that of course is correct.

There are projections that we don’t have for functions.

Akiva Weinberger

My proof uses the finiteness of their dimension as a crucial element @TedShifrin

Is it still true in the infinite-dimensional case (technically not called manifolds anymore but we can generalize)?

(I found a map from left to right called $f$ and a map from right to left called $g$ and show $f\circ g$ is the identity, which combined with the fact that they're linear maps between spaces of the same dimension is enough)

Ted Shifrin

Yeah, I think it’s true for Banach manifolds as well.

Akiva Weinberger

How would you prove it? My proof method doesn't work

Actually, is it true for infinite-dimensional real vector spaces $V$ that $T_vV\cong V$?

I feel like if so you could use that

Ted Shifrin

I sure hope so.

I haven’t thought about even the definition of tangent space for a long time in the Banach case.

Akiva Weinberger

3:25 AM

If we have charts to vector spaces $V$ and $W$ I think we can do $T(M_1\times M_2)\cong T(V\times W)\cong V\times W\cong T(V)\times T(W)\cong T(M_1)\times T(M_2)$

Ted Shifrin

Still have charts and the product chart on a product .

Right. That’s what I was thinking.

Akiva Weinberger

Still gotta show that the composition of isomorphism maps is the map you want

The exercise says "prove that this specific map is an isomorphism" and then gives a formula

Problem 3-2 is "prove Prop. 3.14" of course

Ted Shifrin

That’s the projections I mentioned a while ago.

Akiva Weinberger

I suppose it specifies at most one of the Ms is a manifold with boundary because the product of two of those has corners and that's not a valid sm.man.w/bd

robjohn

@AkivaWeinberger is that Prop. $\frac{22}7$?

Akiva Weinberger

3:29 AM

I think in Indiana it's just Prop 3

22/7 is Alabama.

Hrm?

I thought so.

1 Kings 7:23

(I don't get the reference @ ted)

Ted Shifrin

3:31 AM

They passed a law years ago.

Akiva Weinberger

I think it's Indiana

Oh, apparently there is a (false) hoax stating that the same thing happened in Alabama

snopes.com/fact-check/alabamas-slice-of-pi

The Indiana one actually happened though: en.wikipedia.org/wiki/Indiana_Pi_Bill

@robjohn Fun fact

In the Hebrew Bible, there are several words that are written one way and pronounced another

The word for "and a line" in that verse is written וקוה but pronounced וְקָו

This leads to a bit of fun numerology: the numerical value of קוה (adding up the values of the letters) is 111, and the numerical value of קָו is 106

(I've deleted the first vav, which means "and", to make the result nicer)

The ratio 111/106 works out to be really close to pi/3

robjohn

$\frac{333}{106}$ is the convergent before $\frac{355}{113}$

Akiva Weinberger

In other words, 3*111/106 gives you pi to four decimal places

So it's kinda like the text sneakily gives you a correction term

Ajay

Here is a philosophical sort of question, I think. If a theorem is proved by a computer, how can we claim that it is true?

Akiva Weinberger

There are three other instances of this (words being "miswritten" but pronounced correctly) in that chapter

@Ajay We can prove mathematically that, if a mathematically ideal computer ran that code and produced that output, then the theorem is true

We cannot prove mathematically that any given physical computer behaves like a mathematically ideal one

but it certainly inspires confidence

There's a more thorny philosophical issue with "probabilistic proofs". There's a certain primality testing algorithm,

that works like this: every composite number has certain "witness numbers" to its compositeness

For a composite number C and a witness W between 1 and C, there's some function you can compute that if it gives a certain answer means C is a composite

Primes have no witnesses, while 3/4 of the numbers between 1 and a composite number are witnesses

So here's the algorithm: select a whole bunch of candidate witnesses at random and check them in turn. If none of them work, then the number is probably prime

Fargle

3:43 AM

@Ajay It seems that it would depend upon whether you're confident that whatever axiom set the system is using corresponds with the one(s) people use, and that whatever inference rules the system is using are valid, and that the system is sufficiently error-free.

Akiva Weinberger

You can make that probability as high as you want by choosing more candidates, but you can only get certainty by checking literally all numbers between 1 and P

@Ajay If I run this test 100 times, the odds of it falsely claiming a composite number is prime is (3/4)^100 which is like 3*10^-13

Fargle

My understanding is that the current formal verification state-of-the-art is still a decent clip away from being able to formally verify a lot of things, and certainly a long way away from producing its own theorems, but there also appears to be quick progress in the former regard, at least.

Akiva Weinberger

(even if the number being tested is in the trillion trillions)

Would you call that a proof?

@Fargle It doesn't need to be a hypothetical. This is what happened with the four-color theorem in the 70s.

Fargle

(And here, what I'm talking about is distinct from the sort of "lots of case-checking" proofs that are the typical examples of computer-aided proofs, like the one for the four-color theorem. That's mostly just an issue of "do people trust the people who gave the computer the inputs and the algorithm to have done that properly")

Akiva Weinberger

I'm not sure I understand the distinction.

In any case, the four-color theorem has also been formally verified in computer proof languages.

Fargle

3:48 AM

At least to my eyes, there's a difference between "formally verifying Perelman's proof of the geometrization conjecture in Lean" and the original four-color theorem computer-assisted step.

Ajay

Yeah, I have heard about the case for the four-color theorem.

Fargle

We're 100% at the point of being able to do the second. It's not clear when we'll be at the point of being able to do the first.

Akiva Weinberger

Ah, I see - the former was all comprehensible to humans (or at least one human) but we'd like to automatically check for errors, whereas the latter was a massive search

These lead to different philosophical questions

Fargle

Yeah, that's basically what I mean. It would be helpful to be able to take some kind of research preprint and slap that into Agda or Coq or Lean or whatever, but as I understand it, the state of the art is such that that's only possible in narrow cases, usually ones that are already relatively easy to "algorithmize" (for lack of a better word).

Akiva Weinberger

Four-Color: "How can you call it a proof when humans can't understand it?" Perelman: "How can you call it a proof when a computer hasn't checked it?"

Both are valid questions but in very different directions

@Fargle There have been some high-profile formal verifications. A year and a half ago, they formally verified in Lean the proof that the Continuum Hypothesis is independent from ZFC

(The type theory of Lean is as strong as ZFC+countably infinitely many universes, I think?)

Fargle

3:53 AM

Something like that.

Akiva Weinberger

(so it can prove that ZFC is consistent)

Fargle

And yeah, the progress is very fast. FVS was not even close to being able to do that five years ago.

Akiva Weinberger

Does the S stand for software?

Ajay

ZFC, FVS, Lean....what?

Fargle

Yeah, I just didn't want to type it all out

ZFC = Zermelo-Fraenkel set theory, with the axiom of choice. The typical axiom set for working mathematicians, I guess

Akiva Weinberger

3:55 AM

@Ajay Lean is a computer language for writing proofs

Fargle

FVS = formal verification software

Akiva Weinberger

You can learn some of it here: ma.imperial.ac.uk/~buzzard/xena/natural_number_game

Fargle

And yep

Akiva Weinberger

Here's a neat puzzle: Given that computer proof verification exists, and given that it's possible to write computer programs that have access to their own code (it's a neat puzzle to show that you can do this, by the way),

Fargle

To be clear, my claim that "we're not there yet" is highly contingent, and may just straight up not being true in the next decade.

Akiva Weinberger

3:57 AM

part 1: Suppose we write a program R that searches for proofs in Lean that R doesn't halt, and if it finds one it halts. What does R do?

part 2: Suppose we write a program F that searches for proofs in Lean that F does halt, and if it finds one it halts. What does F do?

@Fargle Yeah I can't give you a timeline but I feel like it's on its way

Basically we need to (a) fill Lean up with enough libraries that it has the standard definitions of nearly every field built-in and (b) teach undergrads Lean

We teach undergrads coding already, (b) isn't too difficult

(a) is a legit multi-year research project but it's happening super quickly I think

copper.hat

i don't trust computers

Ajay

Another question: How would I find the number of solutions to the equation $$4^x = y^2 + 15$$ By plugging in values i've found that there are 2 pairs of solutions, but that doesn't seem so efficient.

copper.hat

Surely $({\log (y^2 +15) \over \log 4}, y)$ provides lots of solutions?

Ted Shifrin

4:12 AM

Did Ajay forget to say integers?

Ajay

Yes Ted, I did.

Positive integers too

copper.hat

get real

Ted Shifrin

I can’t imagine why.

copper.hat

i always find it bothersome that discrete problems are harder than continuous ones.

Ted Shifrin

I think modular arithmetic and unique factorization do it.

Ajay

4:18 AM

Unique factorisation I can do. Modular arithmetic, I struggle to do.

Ted Shifrin

$y$ has to be of the form $4k+1$. Now $(4k+1)^2+15$ has to beca power of $4$.

Ajay

Why does it have to be 4k +1?

Ted Shifrin

Work mod 4.

Akiva Weinberger

y is either 0, 1, 2, or 3 mod 4. What are the possible values of y^2 mod 4? What are the possible values of y^2+15 mod 4?

Ted Shifrin

Oh, also $4k-1$ … same analysis.

Akiva Weinberger

4:22 AM

I suppose it's easier to say 2k+1

Ted Shifrin

Yeah. I can’t do math typing on my ipad.

Ajay

0 mod 4 is 0

1 mod 4 is 1

2^2 mod 4 is 0

3^2 mod 4 is 1

15 mod 4 is 3

1 + 15 mod 4 is 4

2^2 + 15 mod 4 is 7

Akiva Weinberger

7 is the same as 3

Ajay

3^2 +15 mod 4 is 12

Akiva Weinberger

How do you get 12?

You can use your previous result (3^2 mod 4 is 1) to add 15 to both sides (3^2+15 mod 4 is 16, which mod 4 is 0)

or alternatively do it from scratch (3^2+15 is 9+15=24, which mod 4 is 0)

Ajay

4:29 AM

Oh yeah, oops.

But why was mod 4 taken though?

copper.hat

number folks like mod 4

Akiva Weinberger

I'll be honest, I'm not sure… I'm also not sure if it will work in the end

I think it was 'cause the powers of 4 are all zero mod 4

Ajay

Then why?

Akiva Weinberger

so it's convenient there

Ajay

That does make sense

So if I had like 9^x = 2^y + 3, what mod would I take?

mod 9, or mod 2, or both?

Akiva Weinberger

4:32 AM

Try a few and see what works

I don't think any are guaranteed to work

Ajay

Wow... so modular arithmetic is pretty much like guessing.

Ted Shifrin

@Ajay Because $4^x=0$ mod $4$.

Oh, Akiva said that.

Akiva Weinberger

$25^2+15=640=4^3\cdot10$

That's the first time it's a multiple of 4^3 I think

Yeah I don't think doing this mod 4 will work. I found some (large) ys with $y^2+15$ a multiple of 1024=4^5

4^5 might be the limit though

Never mind, eventually you get a multiple of 4^6

3289^2+15 is a multiple of 4^6

This is a dead end

That might be the highest power of 4 you can get to though

@Ajay Where did this question come from?

Can you solve this "old-style" problem from 1986?

►

@Ajay This seems similar

Akiva Weinberger

5:04 AM

Wolfram Alpha tells me 1241^2+15 is a multiple of 4^7

one potato two potato

math.stackexchange.com/questions/420642/… @robjohn I understand that the radius of convergence is greater than $1$. But how that implies $|\sum_{j=0}^n\binom{n}{j}(-1)^{n-j}z_0^{n-j}a_{k-j}|\leq cr^k$ for some $r$?

Akiva Weinberger

@Ajay Here's an idea:

$4^x-64=y^2-49$

$64(4^{x-3}-1)=(y-7)(y+7)$

Not sure where to go from there

PM 2Ring

5:39 AM

Let $z=2^x \implies z^2=4^x$
$z^2-y^2=(z+y)(z-y)=15$
$z+y=15,z-y=1\implies z=8,y=7,x=3$
Or
$z+y=5,z-y=3\implies z=4,y=1,x=2$

PM 2Ring

5:49 AM

The key insight is that $4^x$ is a perfect square, so we can transform the problem from being about some unknown power to a quadratic. And then see if the solutions to that quadratic give us any $z$ which are powers of 2.

Ajay

@AkivaWeinberger ukmt.org.uk/sites/default/files/ukmt/…

Overtherainbow

@copper.hat but wouldn't this mean that in our case $S$ is $\Bbb{N}^N$ since $\Bbb{Q}$ is countable. So $S$ is countable?

copper.hat

6:11 AM

@Overtherainbow was there something ambiguous in what i wrote earlier?

${\{0,1\}}^\mathbb{N}$ has the same cardinality as $(0,1)$.

$\mathbb{Q}^n$ is countable, so $\cup_n \mathbb{Q}^n$ is countable.

Overtherainbow

7:22 AM

@copper.hat no sorry my mistake there was nothing abigous.

I made a thinking error according to the union since I did not take the union over $n$ but I wanted to take the union over $x_n$ in $\Bbb{Q}$ therefore I got confused with the cardinality but now it is clear. Sorry for the circumstances:(

robjohn

8:17 AM

@onepotatotwopotato $\sum\limits_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}$ is the coefficient of $z^k$ in $f(z)(z-z_0)^n$, and since the radius of convergence of $f(z)(z-z_0)^n$ is greater than $1$, the root test shows that there is an $r\lt1$ so that $\left|\sum\limits_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right|\le cr^k$.

Xander Henderson

1:29 PM

Note to self: $10+40=50$. Not $40$. :/

anak

@XanderHenderson blunder during your lecture?

Xander Henderson

@anak Nope. A blunder in a conference program I sent out yesterday.

I had a 40 minute talk start at 1:10, and end at 1:40.

And, of course, this propagated to all the following start times. :/

anak

And if I know talks, they all end 5 minutes late.

Xander Henderson

@anak Technically, it is a 35 minute talk, with five minutes of transition time.

(Technically, it is two 15 minute talks, back-to-back, with five minutes between talks. Which is why it is 35 minutes, which is a slightly weird amount of time.)

anak

Can always speed-run the transition times until you make up the 10 minutes.

Alessandro Codenotti

1:53 PM

I'm familiar with Ehrenfeucht-Fraisse games @Akiva, I don't do much of it, but officially I work in the model theory group!

Xander Henderson

@anak Easier to just correct the schedule.

It is a small conference (community college mathematics folk in Arizona), so no one should be too put out.

Alessandro Codenotti

Anyway a sentence like $\exists A_1,\ldots,\exists A_4$ such that $\bigcup A_i$ is everything and $\forall B_1,\ldots,\forall B_4$ if $\{B_i\}$ is a cover refining $\{A_i\}$ there are $i_1,i_2,i_3\in\{1,2,3,4\}$ with $B_{i_1}\cap B_{i_2}\cap B_{i_3}\neq\varnothing$ can be written in your language, holds in $\Bbb R^2$ and fails in $\Bbb R$

Or in words you can cover $\Bbb R^2$ with four sets, such that any refinement of that cover also consisting of $4$ sets is such that three of them have nonempty intersection (so witnessing that $\dim\Bbb R^2\geq 2$)

anak

@XanderHenderson What kind of stuff gets talked about by community college math folk usually? I am quite ignorant of how the community college system works, and I didn't even realize they would have conferences for that.

Alessandro Codenotti

Now if you can figure out the smallest size of a cover of $\Bbb R^3$, such that any subcover of the same size witnesses that $\dim\Bbb R^3\geq 3$, you can distinguish $\Bbb R^3$ and $\Bbb R^2$ @Akiva

Xander Henderson

@anak Talks are usually about pedagogy.

anak

2:10 PM

@XanderHenderson are most community college math classes trending towards active learning classes there?

Xander Henderson

@anak I think that is the impression that advocates of active learning would like you to have. I am not sure that is the reality.

anak

You say that like you aren't an advocate for active learning. :P

Xander Henderson

But that is a comment made out of ignorance. I genuinely don't know the statistics.

@anak I'm not convinced that active learning is much of an improvement.

I am convinced that it is a useful technique for those people for who it works (which is tautological), but it is not very scalable, works poorly in remote / hybrid environments, and requires a lot of buy-in (both on the part of the instructor and on the part of the students).

anak

So I guess you have seen all the studies that suggest it increases performance, but have concluded this is probably because the surrounding factors were just right for it to happen?

leslie townes

sounds like another ed school theory that might founder when the instructor or the students lack sufficient knowledge of the subject matter

"if the buy-in is there, it works" yeah cool

Xander Henderson

2:17 PM

@anak In part. It is also worth noting that the studies are conducted in environments where the instructors have a lot more autonomy with respect to course content and practices than a lot of us have.

anak

Are community colleges well known for being rigid with regards to content and practices?

Xander Henderson

There are certain topics which I must cover (because of our articulation agreements with the universities). If I don't cover those topics, the classes don't transfer. But active learning techniques usually take more time to cover the same material.

@anak Depends.

But there is also a common effect in studies regarding education (and other social programs): the people who are engaged in the study are highly motivated. They want their techniques to be successful, hence they are likely to put a lot of time and effort into making the program successful.

The kinds of effect reported in such studies tend to shrink significantly when you try to scale them, or when you ask others to implement the same ideas.

leslie townes

"does it scale?" in education often means more frankly "would it scale to people who don't know and maybe don't give a shit"

anak

Heh.

Xander Henderson

@leslietownes This.

anak

2:23 PM

So you mean "scale" by implementation by all teaching staff, rather than scaling up/down class sizes?

Xander Henderson

And, to be fair, I have been interested in "active learning" since it was called "the Moore method". There are things to be taken from it. But it is not a panacea.

@anak Both.

anak

I've found both large and small classes to be effective for active learning. I do agree about your online sentiment, though, most students taking online courses don't want to be engaged.

leslie townes

a lot of K-12 stuff fails because there are lot of really good general ed research notions that founder when the instructor doesn't know math, and the instructor generally doesn't know math. it's like trying to teach a language that you can't speak. the latest and greatest techniques will not help

Xander Henderson

@anak I said "remote". Don't confuse "remote" with "online". :D

ThunderGlove

I am strictly adhering to the rule that I should not cancel any terms in a trig equation, but I just looked up a solution where terms were cancelled though there was a possibility of them being zero

leslie townes

2:26 PM

community college is kind of a middle ground where the subject matter knowledge is there but often the students lack motivation, maybe more than high school students do

i couldn't explain why math X is a prerequisite for so many things in college, it shouldn't be

anak

@XanderHenderson Do they mail-in their homework?

Xander Henderson

@anak They used to.

We have been teaching classes remotely here for 50 years.

leslie townes

thunder i think often there is an unstated rule that a trig identity is an identity if it holds at all but a discrete set of points. once tan or sec get in there for example you are going to have those points.

Xander Henderson

The college owns 6 radio transmission towers for instruction.

And we used to have couriers who would transport VHS tapes between our different locations so that students could watch recorded lectures.

ThunderGlove

Not identity, its an equation

Xander Henderson

2:29 PM

About 15 years ago, we contracted with Cisco to create "connected classrooms", which are classrooms with a lot of cameras and screens, which enable remote (but not "online") courses.

@ThunderGlove An identity is an equation.

Not all equations are identities, but all identities are equations.

But it would be helpful to have an example which exemplifies your confusion.

anak

@XanderHenderson it doesn't, by any chance, direct the camera/audio signals through the internet, does it?

ThunderGlove

How do I share a picture?

anak

By the "send" button is "upload..."

leslie townes

pasting the url should work, if there is one. or 'upload.' don't know how it looks on mobile.

Xander Henderson

@anak Yes, but via a more-or-less private WAN.

anak

2:31 PM

@XanderHenderson I am not quite sure about your distinction between online and remote anymore.

Xander Henderson

@anak In an "online" course, students are expected to attend courses remotely, from whatever location they choose, using their own equipment, over the interwebs.

"Remote" just means that students and instructors are not necessarily in the same room.

All online classes are remote, but not all remote classes are online.

anak

What are remote classes like that aren't online?

(modern ones)

I get that pre-internet revolution, things were different.

Xander Henderson

Again, "online" doesn't mean "it uses the internet". It has a specific meaning of every student connecting to a course one their own, possibly asynchronously.

ThunderGlove

Xander Henderson

For the remote classes that I teach, I am in one classroom with a group of students, and am simultaneously being broadcast to other classrooms with their own groups of students.

ThunderGlove

2:35 PM

I have highlighted the portion where the term has been cancelled

Xander Henderson

The fact that the signal may (or may not) be transported over the internet is not a relevant part of being "online".

anak

@XanderHenderson "online course", in the most common context I hear it, means a course where classroom content (assignments, lectures, etc.) is distributed and accessed primarily (if not solely) online.

@XanderHenderson This makes sense now, though, thanks!

Xander Henderson

@anak And our remote classes don't fall into that definition.

anak

Agreed.

Xander Henderson

@ThunderGlove The author starts by making the assumption that $2x \ne n\pi$ (among other things).

ThunderGlove

2:38 PM

Thanks!!

Oh god how could I not notice that

Thank you so much

Xander Henderson

@ThunderGlove To be fair, it could have been made more clear.

ThunderGlove

Also, how is 3x=n(pi)

Xander Henderson

If I were writing that solution, I would have started with something like "The equation is only meaningful if... . Therefore, assume that these conditions hold. The equation becomes...".

ThunderGlove

Doesnt that make cot infinity

Xander Henderson

@ThunderGlove I think that is a typo.

In the context of the other conditions, the author meant $3x \ne n\pi$.

ThunderGlove

2:41 PM

So does he mean 3x is not equal to npi?

Xander Henderson

@ThunderGlove That is my assumption.

ThunderGlove

I mean it does make sense since he mentions "the eqn is meaningful if...:

ILikeMathematics

3:14 PM

Amongst mathematicians, reading math books is considered the best way to learn, right? (I mean the best medium of course, there's no substitute for practice)

anak

No, most people think you have to do actual math to learn.

"math is not a spectator sport" is the common adage

So watching + reading someone else doing math isn't ideal.

leslie townes

subject to anak's qualifiers, i do think that a well organized book is the best input to math learning.

Ted Shifrin

@ThunderGlove Look at what is needed to make sure there is no zero in the denominators.

ILikeMathematics

@anak I meant best medium, like preferring books over lectures

anak

The mediums are not equivalent, and tend to have different goals in mind.

Ted Shifrin

3:26 PM

Besides, different people learn in very different ways. Some people absorb lecture material and interact in class in a lively fashion; others just find the classroom experience overwhelming and would rather read on their own.

leslie townes

good point. i've seen maybe five good lectures. everything i know i learned from a book

for others i imagine it is the opposite experience

ILikeMathematics

@anak Do lectures tend to not go as much into detail for a topic as books?

Ted Shifrin

I always liked learning from lectures, getting insights from the prof (assuming a good prof). The occasional disastrous lecturer (such as Irving Segal teaching out of his own book for graduate real analysis) was not a good experience.

leslie townes

ted i did click into one of your lectures one time and enjoy it. way better than my multivar prof.

Ted Shifrin

@ILikeMathematics This varies widely, depending on the level of the course and the style of the lecturer.

@leslie But you're not exactly a struggling MATH 51/52 student at this point.

robjohn

3:31 PM

@ThunderGlove the second is equivalent to $2\tan(x)=\tan(2x)$

leslie townes

that's the best way to experience lectures, after you know the material.

ILikeMathematics

@TedShifrin If you had to choose between books and lectures, you would definitely go with books though, right?

Ted Shifrin

Damn, this is scary.

anak

@ILikeMathematics Generally speaking, you can pack more details in a book without boring learners than a lecture, since readers can skip details, while lecturees fail to be able to escape the narrow corridor of time.

Ted Shifrin

@ILike I just finished saying I always liked learning from lectures.

leslie townes

3:32 PM

oh wow, yikes.

Ted Shifrin

I know lots of calculus students are still making that mistake, @leslie, but a complex variables student?!!

leslie townes

i will praise the OP for one thing, which is that they wrote their misunderstanding very clearly.

Ted Shifrin

Once one gets to advanced mathematics, some texts are so dense and sophisticated that one basically has to know the material before one can read the text.

@leslie Yes. Dare I tell the OP about the geometric series? Yikes.

robjohn

Well, $\frac1{1-x}=1-\frac1x$

anak

Another freshman's dream, huh.

robjohn

3:37 PM

Extrapolating from the post Ted referenced

Ted Shifrin

@robjohn Yes, that's an immediate corollary.

ILikeMathematics

Oh, so in those cases you would first have to attend a lecture before being able to read the text (or pick a different one ig)

Ted Shifrin

Yup.

Xander Henderson

@leslietownes There is a professor at UCR is generally considered to be a great lecturer. I went to a lot of his talks, and always felt like I was really understanding everything as he went along. And then I would go home and try to consolidate my notes. At that point, I would learn that I had understood nothing, that most of the lecture was entirely vacuous, and that I had several hours of reading ahead of me. :/

But the lectures were energizing! And you felt like you were learning something!

leslie townes

the best of those folks like to sit at the intersection of two disciplines. i know an econ guy who made a career out of this.

each half of the audience thinks that what he's saying must make sense to the other half.

Ted Shifrin

3:41 PM

I had that same experience with Atiyah and with Bott when I heard them lecture. Both stupendously clear and provided great intuition. I think the moral of the story is that one needs to take notes in math lectures; saying "oh, I'll never forget a word of this" is hopeless.

Xander Henderson

@TedShifrin Oh, yikes! I just came across that on the main page and though "Oh, this just came up in chat. Maybe I should share it?"

But it seems that it was the instigating post.

@leslietownes Ha!

Ted Shifrin

Indeed. And I did give away the main hint (although I suspect it's like casting pearls before swine).

Xander Henderson

This comment does not fill me with hope:

Please explain the first comment, I don't understand and the only thing I've done is reversing the sin(z) expansion? Cant see that I've broken any rules? — zzz__ 5 mins ago

As you say, pearls, meet swine.

Ted Shifrin

@Xander The OP just responded. All I can say is, OY.

Oh, you just did that.

I must have gotten out of bed on the wrong side of the world today.

@leslie This is an interesting one I haven't seen before.

leslie townes

that's a cool question.

Ted Shifrin

3:56 PM

I realize it's too geometric for you, but I don't yet see an argument.

anak

leslie, despiser of all things geometric

leslie townes

although it involves an inner product, it strikes me as something that might appeal to banach space weirdos.

Ted Shifrin

Inner product and very finite-dimensional....

I bet @copper likes it ... anti-convexity all over.

I guess we're intersecting a bunch of half-spaces?

Xander Henderson

Why do quantifiers seem to cause so much difficulty?

Ted Shifrin

That's the thing that makes math so challenging, I think. It's also the reason that I think that linear algebra and algebra are easier than baby analysis — fewer quantifiers!

I wonder also if writing sentences like that OP using symbols is partly the problem. Writing out for all and there exists makes one understand a little better what is being said.

Xander Henderson

4:09 PM

@TedShifrin Very likely true.

@TedShifrin I remember that, when I took analysis, I wanted to write everything in symbols. That's how real mathematicians write, right?

$>$ symbols = $>$ good

copper.hat

i think most folks go through that phase

Ted Shifrin

I've said this numerous times here, but I was taught sophomore year never to use the symbols. Munkres was very strict.

copper.hat

its the bourbaki phase

Ted Shifrin

@copper Any thoughts on that anti-convexity problem?

Xander Henderson

@TedShifrin I'm anti-(anti-convexity).

Note that I am not convexity, either. The law of the excluded middle can take a long hike of a short pier. While wearing cement undergarments.

copper.hat

4:12 PM

@TedShifrin lookin at it (on a boring call atm)

Ted Shifrin

It's a nice question.

I see the half-space argument vividly in $\Bbb R^2$, but not in $\Bbb R^{10}$.

PM 2Ring

For the sin reciprocal guy:

How do you even get to Taylor series when you're oblivious to the properties of simple fractions?

anak

A combination of anxiety, confusion, and not being able to take a step backwards.

Ted Shifrin

4:28 PM

@PM2 As I said, I experienced that phenomenon with plenty of standard calculus students, numerous times.

PM 2Ring

And there's also that curious aversion to just plugging in some numbers to test stuff.

I guess that people are less likely to do that if they aren't comfortable with mental arithmetic.

anak

Or if they feel comfortable with what they are thinking. They'd probably only check it by accident, or if they are not confident with it.

But I think someone in their situation would feel much more confident with arithmetic relative to the material they are trying to work through.

PM 2Ring

Fair point. There's the tendency to not worry about simple stuff when you're doing advanced stuff. OTOH, you can't build a solid advanced structure if your foundations aren't solid.

Akiva Weinberger

@PM2Ring Ohh that's so simple

In my defense it was past midnight when I was writing that

PM 2Ring

:)

It gets a bit trickier when the coefficients of z^2 &/or y^2 aren't 1. Or you have a zy term.

Ted Shifrin

4:40 PM

@PM2Ring Nice solution. I actually assigned a few questions like this to my algebra students many years ago. Shame on me for not thinking of it.

Akiva Weinberger

@AlessandroCodenotti Four probably?

I don't know how to prove it

PM 2Ring

Solving quadratic Diophantines can get a bit messy. But it is kinda satisfying. I can't remember all the tricks. But I know where to find them when I need them. :)

Ted Shifrin

DogAteMy, you're clever. Any ideas?

PM 2Ring

Thanks, Ted.

More Anonymous

@TedShifrin Hey Ted sorry for the late reply. Given 2 curves how does one construct geodesic congruences

Congruence (general relativity)

In general relativity, a congruence (more properly, a congruence of curves) is the set of integral curves of a (nowhere vanishing) vector field in a four-dimensional Lorentzian manifold which is interpreted physically as a model of spacetime. Often this manifold will be taken to be an exact or approximate solution to the Einstein field equation. == Types of congruences == Congruences generated by nowhere vanishing timelike, null, or spacelike vector fields are called timelike, null, or spacelike respectively. A congruence is called a geodesic congruence if it admits a tangent vector field...

Ted Shifrin

4:45 PM

What exactly are you starting with?

Akiva Weinberger

@TedShifrin Interesting. I suppose we want to show that they map to some linearly independence set of vectors in $\Bbb R^{2n}$ somehow

Ted Shifrin

That's an interesting notion. I was trying to intersect half-spaces.

ThunderGlove

Akiva Weinberger

Here's an idea

ThunderGlove

In the highlighted portion shouldn't it be 6pi + pi/2 instead of 6pi + 3pi/2??(first line is the question)

Akiva Weinberger

4:48 PM

Pick a random point in $S^n$. How many $\alpha$s do we expect there to be that have positive dot product with it?

$m/2$, yeah?

More Anonymous

@TedShifrin Lets assume I have 2 curves in flat space time as an example how do I construct geodesic congruences?

Ted Shifrin

@ThunderGlove Where does that one value come from? Are we still supposed to have $\sec x = 3$?

@MoreAnonymous Those curves have to be geodesics to start with.

ThunderGlove

Yes we are counting the number pf possible solutions for secx=1/3

Akiva Weinberger

Hm not sure where to go from here

copper.hat

i think the intersection of half spaces seems like the moist fruitful approach

Ted Shifrin

4:51 PM

@ThunderGlove It would help if you gave the actual question first. This just doesn't make sense.

More Anonymous

@TedShifrin yes. So lets say I have $s_1^\mu = x^\mu$ and $s_2^\mu = x^\mu + \delta^\mu$

where $\delta^\mu$ is a spacelike vector

Ted Shifrin

@MoreAnonymous I don't do relativity, so let's forget about that. Just do the question in $\Bbb R^3$ to start with. You're starting with two skew lines and asking how you fill up space with lines? This has nothing to do with differential geometry.

ThunderGlove