A Balarka is always helpful.

(i didnt even look at the recursion, i picked a random number and said youre wrong because either way youll have to do the hard work to defend it)

bah

lol

@leslietownes excellent proof

I'll steal this trick

It’s not so tricky blatantly to assume what you’re trying to prove whilst proving it.

@BalarkaSen god

Thing is, you were right, so I didn't even suspect it

all's well that ends well

In any case, context:

(see the comment to the answer)

Moo.

Or, maybe, $\mu$.

Boo who mu?

do you mean the universal thom spectrum for complex vector bundles?

So when a cheater deletes a post with my help/answer and then a new cheater posts the identical question with identical wording and diagram, poof, I cannot prove there’s competition or exam cheating going on. My flag was, I guess, ignored.

No, the Milnor number, @Lukas.

Consider the sequence of functions $f_n(x)=x/n$ for odd $n$ and $f_n(x)=1/n$ for even $n$. It is simple to show that $f_n$ tends to $0$ pointwise on $\mathbb{R}$. Is the following reasoning correct to show that it is not uniformly convergent to $0$ on $\mathbb{R}$? Reasoning: let $n_0$ be a odd positive integer, it is $\sup_{x \in \mathbb{R}} |f_{n_0}(x)-f(x)|=\sup_{x \in \mathbb{R}} (|x|/n_0)$.

Since $|x|/n_0$ is unbounded above as a function of $x$ for fixed $n_0$, the sup is $\infty$ and so $f_n(x) \ge \epsilon_0$ for any given $\epsilon_0>0$ for infinite odd indexes big enough. This makes the definition of uniform continuity impossible to satisfy.

Indeed.

uniform convergence, not continuity

Oh yes, typo, thanks for the help @TedShifrin

No help given!

@TedShifrin I doubt that it was ignored. It may be that no one has seen it yet. I've been offline for the last several hours, and I believe that most of the rest of the moderation team is in Europe or father east (thus, probably, sleeping).

Oh, I flagged many hours ago.

Very annoying when they delete…. Cheaters.

@TedShifrin According to the super secret moderator statistics page, average flag handling time is 16 hours, 28 minutes.

Your flag was raised 2 hours ago.

I should send you some noodle kugel, Xander. Probably won’t last the trip, though.

I'm working on it now.

Though I am having trouble tracking down the earlier post.

Ah, tanx. It was under precalculus. I don’t remember the OP’s name. It was within the past week.

@TedShifrin Did you post an answer? or just comments?

Ah... found it.

math.stackexchange.com/questions/4589354/…

Just comments, with a link to another question on envelopes and ultimately the answer. But not worked out in detail.

Ah.

The identical picture makes it suspicious, of course.

Thanks for telling me mean time is >16 hrs. I had no idea.

@Xander a question for you as a moderator: should I use the "in need of moderator intervention" flag only for severe things or also for mundane things that just don't fit any other flag type, such as making a post community wiki

Gute Frage.

@LukasHeger If a standard flag applies, please use a standard flag. Also note that we are very unlikely to mark anything CW.

not even a big-list?

I thought these should be CW

of course if a standard flag applies I'll use the standard flag!

@LukasHeger Even for a "big list" question, I see no compelling reason to mark the post as CW.

oh I see, I thought this was a standard here for some reason

Not really.

I think I even raised a "please make this CW because it's a big-list" flag before and it was accepted

It is certainly possible.

There is not 100% agreement on this across the SE network, nor on Math SE in particular, and things are typically handled on a case-by-case basis.

okay I see

i don't know anything about the policies or standards, but i have a vague impression of there being more CW, and perhaps more conversions to CW, when i started using MSE (10+ years ago) than now.

Part of the thinking has to be "why should this be CW?" What is the intent?

@leslietownes You are correct. CW used to be more common.

I thought the argument was that it has no clear-cut answer for a big-list question and for some reason that implies it should be CW

not sure I can reconstruct the whole argument

Oh, I hate it when Leslie is right.

so my more general question was if a "in need of moderation" flag means "this is something really important that can't be handled by a standard review queue" or more something "this didn't fit any of the other flag types, so I chose this". I'm tending more towards the former

@LukasHeger The only real benefit that marking a post as CW has is that it allows low XP users to make edits.

@LukasHeger I think the latter. If you believe that some action needs to be taken, and none of the standard flags apply, flag it. Just keep in mind that there are only a handful of us that can handle custom flags, while most other flags are handled through community moderation.

And the current average wait time is nearly 16.5 hours. :D

okay thanks I see

In any event, I am calling it a night now. Laters.

later

Nighty night!

why/how is this pattern in a liquid being described as a lattice? youtube.com/watch?v=WSjL4jgpJCI

Suppose we have a vector space $P\oplus Q$ so that $P$ and $Q$ are orthogonal and span the space. Choose a linear map $X:P\to Q$ and let $S=\{v+Xv:v\in P\}$.

I wanted to find the formula for $\pi_S$, the projection map onto $S$, assuming we have access to $\pi_P$ and $\pi_Q$.

After some fiddling, I got$$\pi_S=(I+X)(I+X^\top X)^{-1}(\pi_P+X^\top\pi_Q).$$

Is this a known formula? Otherwise, is there a quick way to derive this?

Do you know the standard approach/formula for projection onto a subspace? It’s in my book, for example.

Not off the top of my head.

If the columns of $A$ give a basis for the subspace, it’s $A(A^\top A)^{-1}A^\top$.

See the last section in chap 5.

Ah-hah.

So in our situation, $A=I+X$, I suppose. EDIT: Oh, not quite

yeah, i'd try something like the usual formalism, maybe using block matrices with respect to that orthogonal decomposition. you want projection onto the range of $\begin{pmatrix} I & 0 \\ X & 0 \end{pmatrix}$.

Sure, because you’re doing the graph of $X$.

I mean I guess it's just $\begin{bmatrix}I\\X\end{bmatrix}$.

No need to make it square.

$I\oplus X$

i prefer squares but you do you.

So then it's $\begin{bmatrix}I\\X\end{bmatrix}(\begin{bmatrix}I &X^\top\end{bmatrix}\begin{bmatrix}I\\X\end{bmatrix})^{-1}\begin{bmatrix}I &X^\top\end{bmatrix}$

I don’t want square. I want lin. Indep. columns.

Now how do I do that product...

Leslie flunks.

$\begin{bmatrix}I\\X\end{bmatrix}\begin{bmatrix}I&X^\top\\ X &X^\top X\end{bmatrix}^{-1}\begin{bmatrix}I &X^\top\end{bmatrix}$ maybe?

This hasn't lessened the "oh god" factor

mainly because I have no idea how to invert that

No.

Cuz it’s the wrong shape.

Rethink block mult.

Second attempt is $\begin{bmatrix}I\\X\end{bmatrix}\begin{bmatrix}I&X\\ X^\top&X^\top X\end{bmatrix}^{-1}\begin{bmatrix}I &X^\top\end{bmatrix}$

but that's a braindead "well try the other one" approach

Crap. Think.

Oh duh

$\begin{bmatrix}I\\X\end{bmatrix}(I+X^\top X)^{-1}\begin{bmatrix}I &X^\top\end{bmatrix}$

Yeah?

Yup. Block mult and shapes.

Yup, I guess I was thinking $\begin{bmatrix}I\\X\end{bmatrix}\begin{bmatrix}I &X^\top\end{bmatrix}$ or something for some reason

Different!

Now this takes in a vector $\begin{bmatrix}v_P\\v_Q\end{bmatrix}$

and outputs $\begin{bmatrix}I\\X\end{bmatrix}(I+X^\top X)^{-1}\begin{bmatrix}I &X^\top\end{bmatrix}\begin{bmatrix}v_P\\v_Q\end{bmatrix}$

And if we work in $P\oplus Q$, this agrees with the earlier formula of $(I+X)(I+X^\top X)^{-1}(\pi_P+X^\top\pi_Q)$.

And incidentally, $I+X^\top X$ is always invertible since it's the sum of a positive definite and a semipositive definite.

Positive semidefinite?

Right.

Nonnegative definite. Whatever the word is EDIT: Positive semidefinite.

Positive semidefinite

Probably your honors class skipped this stuff, cuz “too applied.”

Grr.

Nah I skipped an honor class

In any case the problem is to show that the tautological bundle over the Grassmanian, $T=\{(S,v)\in\mathrm{G}_k(V)\times V:v\in S\}$, is a smooth rank-$k$ subbundle of the product bundle $\mathrm{G}_k(V)\times V\to\mathrm{G}_k(V)$

and we have a theorem saying smooth vector bundle homomorphisms of constant rank have smooth images

Huh, I’m talking about your freshman class.

so I basically just wanted the map $F:\mathrm{G}_k(V)\times V\to\mathrm{G}_k(V)\times V$ defined by $F(S,v)=(S,\pi_Sv)$

but to do that I needed to show that this actually is a smooth map

Oh, Grassmannians are my favorite.

and to do that the easiest thing I could think of was to use a formula, via the charts they defined for the Grassmanian

Or work with frames instead.

which were basically for all spaces that intersect $Q$ trivially ($Q$ a dim $n-k$ subspace) the coords for $S$ are the entries of $X$

That's probably doable... I'll have to review that section of the book @TedShifrin

I like frames and Stiefel manifolds.

Projections are easier with orthonormal bases.

I'm not sure how to use those here tbh

Stiefel manifolds aren't in the book (Lee's Introduction to Smooth Manifolds)

Just need to choose a local smooth section.

Of the Grassmanian?

Oh, of the tautological bundle.

Of the Stiefel manifold, which is the bundle of frames over the Grassmannian

I guess for each $k$-subspace $S$ of $V$ (for each element of $\mathrm{G}_k(V)$), we need a neighborhood of $S$ in $\mathrm{G}_k(V)$ whose inverse image maps perfectly onto that neighborhood times $S$

I think using those clunky charts obscures the easy geometry, but maybe it’s needed.

which basically means I need to map each nearby subspace onto $S$.

Which means I need the projection map I'd think.

Which means I need to prove smoothness still

Yeah, I really think about local triviality of the frames.

which I don't think I know a good way to do other than via coordinates because I need to refer to the smooth structure of the Grassmannian somehow?

@TedShifrin Is this a theorem about the Stiefel manifold?

I think of the structure from the homogeneous space $O(n)/(O(k)\times O(n-k))$.

That’s cuz I do geometry with moving frames and Lee doesn’t :)

This is isomorphic to the Grassmannian?

It is.

Ah, I see.

Oriented, put $S$ everywhere.

All ways of rotating $V$ up to symmetries that preserve $S$.

Right. And hence $S^\perp$.

And the dimension is right (I'm visualizing a triangular array of dots minus two triangles equals a rectangle) so that's everything you need to quotient by

Yup.

$T_n=T_k+T_{n-k}+n(n-k)$ basically

Ayup.

Does $\{(S,v):v\in S\}$, the total space of the tautological bundle, have a nice description like that?

It’s an associated vector bundle to a representation.

Awkward, bit you can edit.

I'm not sure I'm familiar with that.

Which representation, actually?

where can I learn about boot manifolds?

What's a boot manifold?

Stiefel means boot in German

Ah

In Israel once I needed to buy gloves but had no idea what the word was. Having vaguely remembered that in German they were literally "hand shoes", I asked someone if they knew where the 'naalei yadayim' were. She laughed but understood me

(The correct word is 'kfafim')

(Well, I said something to the effect of, "Excuse me, my Hebrew isn't so good, I don't know the word for it, but do you know where the 'hand shoes' are?")

no, the correct word is kfafot

feminine

Wait really? Crap

really

You're right

Damn

you must have given a bunch more Israeli hand shoe saleswomen a reason to laugh

I did not have many more occasions to buy gloves.

right, come to think of it. It's Jerusalem, not Wyoming

fun fact, we must have adopted the German "earth apple" for potato

I think French does it too

pomme de terre

youre right!

In fact, Kartoffel is more common in Germany I think

it must be regional

within Germany

does anybody have a fun analysis puzzle?

$$ \frac{1}{x}= \int_0^\infty \int_0^1 \varphi(s,t,x)~dt~ds $$

what's this type of problem called?

i guess some obscure type of integral equation

looks like some sort of wave equation with time and displacement idk what s is though

in Gauss' wiki "His breakthrough occurred in 1796 when he showed that a regular polygon can be constructed by compass and straightedge if the number of its sides is the product of distinct Fermat primes and a power of 2." what does this mean?

wikipedia has a pretty good page on it, en.wikipedia.org/wiki/Constructible_polygon

.. which states that he only showed constructibility of the 17-gon in 1796 and the more general theory came a little later

do better, random internet strangers working for free

Eh, maybe the proof came after his "breakthrough"

yes, but apparently not in the same year. at least, both pages can't be right. i don't care, i'm just quibbling.

What amazes me is he developed a way to compute the date he was born through only knowing it was 8 days before "The feast of the ascension" which sounds like a crazy holiday

the (completely understandable and inevitable) lack of consistency from one math wiki page to the next is one of my favorite hobbyhorses

yeah if it was incentivized with money though things would be much worse :P

At least they were inconsistent with good intent !

Smells like a contest problem, but it isn't on the USAMTS list. The Putnam was two days ago. math.stackexchange.com/questions/4592545/…

can't get more contesty than a problem with the year in it

@LukasHeger It’s named after a man, not a boot.

well technically ted, it's named after a man who was named after a boot.

Correct.

Why does $\int 2x + y dy = 2xy + \frac{y^2}{2} + C(x)$ and not $C(x) + C$ or $C(y)$ as well

traditionally $\int y \text{d}y = \frac{y^2}{2} + C$ but in this video they attribute the constant only to the other variable.

$C(x)$ includes $C(x)+C$

$C(x)$ refers to any function of $x$, such as $x^2+4$ or whatever

You integrate fixing $x$?

The extra term is something whose derivative is zero. In the single-variable case, that's a constant. In several variables, it's anything which doesn't depend on the variable you're antidifferentating

and $C(x)+C$ refers to any function of $x$ plus any constant... such as $x^2+4$ or whatever

Ohh..

You should work through doing an explicit 2-step path. This makes it more convincing. See my videos :)

You know what, I will @TedShifrin I think they will be condensed enough that if I can follow along with those I can pass this test on wednesday.

starting with 3510 day 1: intro to integration on YT

No, no, just pick topics you’re doing. Most of the lectures have more theory than you need. But setting up

integrals in different orders, polar coords, etc. Line integrals and finding potential functions. Seems like you waited until

the final to start studying.

You have to do

lots of problems. No substitute.

I was blindsided by a lab report and quiz for physics which I should have known was coming..

But I still have twice as much time as I did before, so I project twice the grade than my last test.

I still think your lectures are like a set containing subsets of the material covered in other courses lol.

@leslietownes

Any reference for modules over pid for linear algebra?

oh, take your pick. i think it's in all of the 'big' algebra books.

Axler doesn't cover that. I think it is somehow also used for rational canonical form.

i don't have enough exposure to algebra to have a favorite.

Leslie, yes it is in module theory book by TS Blyth.

Artin is very efficient on this.

@leslietownes okay thanks. I was thinking there would be some book taking the approach through 'more linear algebra' and less abstract algebra.

The abstract algebra is the right way. Avoiding it is clunky.

yeah, you will just reinvent the wheel if you avoid it.

it's not a good match for axler's "let F denote R or C."

Ad hoc torture

being sick is awesome. instant get out of preparing dinner, bathtime, reading bedtime stories. it's like i'm single again.

I wonder how many mathematicians there were across the history of civilizations compared to scientists and the distribution of the "types" even though types probably didn't even exist for a long time.

Also whether mathematicians were even considered separate from "inventors" and such

Like the further back I go I see on wiki their list of titles and endeavors gets longer lol. Like ancient Greek polymaths and maybe thats partly due to there being very little to work with so you have to kinda invent everything yourself.

yeah. 'pure mathematics' is a relatively new thing.

if you click back far enough on the mathematics genealogy project you see scholars doing dissertations on how to tell if somebody is a witch. wild times.

I wonder what their methods were.

@leslietownes Wait til it’s your wife’s turn to be sick!

@Obliv simple! (1) is it a woman (2) does she look smart

ask if she's a witch, and if she says no, well, isn't that what a witch would say?

or the old standby, throw them into the river and see if they use witch magic to escape.

ted: it'll never happen.

I like the monty python bit where they drown the witch to see if she would float, confirming she'd be a witch, but killing her if she were innocent lol.

it's historically accurate

horrifyingly enough

@leslietownes uh huh

@shintuku huh i just checked, the drowning part might have been a myth

my life is a lie

fear not people, for this was only one banner among my many

What do you mean a myth? Witches were burned at the stake, paraded through towns covered in tar, I don't think drowning would have been above them. 'Twas a dark time when people turned on each other due to many reasons.

blogs.loc.gov/law/2022/02/…

it says .gov in the address and it has references

i'm sold

would be pretty easy to fabricate evidence and to make me believe it on the basis of those two things

everything you heard is just something that hasn't been referenced maaan

that article is so informative

like, of course it would make sense that the practice was used by non-trained local clergy

we were idiots but not to that extent

it was all just ugly fear and fear mongering. When everyone you knew believed in witches it was hard not to trust your community and hear out the other side.

uh what is this notation $R = [0,1]\times [0,1]$

A x B is popular notation for the set of ordered pairs (a,b), with a taken from A and b taken from B.

so here, ordered pairs of elements of [0,1]. if you think in terms of the cartesian plane, the 'unit square' in the first quadrant.

yeah I'm mixing up interval notation with set notation I think

It looks like the interval of real numbers between 0 to 1 but it also looks like a matrix/set

a generally fine and not uncommon thing to do.

maybe there's a potential for confusion or multiple meanings that i'm not seeing.

@TedShifrin The $A(A^\top A)^{-1}A^\top$ formula, that only works when $A$ has full rank, right?

Yes. Otherwise there’s not a unique solution of the normal equations, and you need the pseudoinverse to get the shortest one.

There still is a unique closest point, though.

The pseudoinverse is $A^\top(AA^\top)^{-1}$?

Well that's still for full rank

I guess there isn't really a formula, just delete columns until you have a basis

and use that

Yeah?

I suppose it makes sense. This isn't continuous

Projecting onto the span of $\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}1\\t\\0\end{bmatrix}$ is the matrix $\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$ for $t\ne0$ and $\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}$ for $t=0$

If the rank drops, you don’t expect continuity?

Sure

And therefore no formula

(made out of nice pieces)

The pseudoinverse isn't a continuous operator so you can use that

but no polynomial formula like you get for the injective case

I don’t know why you’re worrying about this…

(surjective is trivial lol)

Just making sure I understand.

Oh, but not relephant to your situation.

I'm to find volume bounded by $z=4-x^2 - 2y^2$ and the xy-plane. I'm guessing I have to develop some type of $\int A(x,y) dz$ so it's areas added up for the full interval of z=0 to z=4 do I just have to rewrite this as a double integral?

Not quite $dz$.

oh right

should be z-oriented slices so dydx

or dxdy

Did your course do the change of variables formula?

It might have, I'm not entirely sure.

You want elliptic polar coordinates or it’s a mess.

Start with $z=4-x^2-y^2$.

I am having difficulty showing that every finitely generated free R-module can't have an infinite basis.

Breaking down the definition, we know that there exists a finite basis $v_1, v_2,...,v_n$.

Suppose there exists $\mathfrak B$ -an infinite basis of the module. Then $v_i$'s can be written as linear combination of elements of $\mathfrak B$. Collect these finitely many elements of $\mathfrak B$ and then clearly they span the module.

Let's call the set of these finitely many elements F. Then $\mathfrak B$ $-F$ is infinite.

How to show that $\mathfrak B$-$F$ is not linearly independent?

okay so when $z = 0 = 4 - x^2 - y^2$, the paraboloid touches the xy-plane. So $x = \sqrt{4-y^2}$ and $y = \sqrt{4-x^2}$. Ohh and when it touches the plane it's on the ellipse

so the bounds would be $\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}dxdy$

that looks awful

You need polar coordinates. Otherwise it’s beyond a mess.

nvm, I got it now.

oh change of variables is covered in next section, maybe shouldn't have tried the example problem blindly

should have been $\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(4-x^2-y^2)dy dx$ which you'd need to change $x=2\sin \theta$

what's the latex command to center expressions again?

If $u$ is a harmonic and bounded on the punctured disk $0<|z-z_0|<R$ then $\lim_{z\to z_0}u(z)$ exists.

I actually know a stronger one: Every bounded harmonic function on a punctured disk can be harmonically extended to the whole disk. But the argument I have uses a Dirichlet problem (poisson integral formula). I think it's overkill?

Guys how to show a photo in this chat

Have a doubt with a (quite basic) prof

Suppose $u_1(z)$ and $u_2(z)$ are harmonic in a simply connected domain $D$, with $u_1(z)u_2(z)\equiv 0$ in $D$. Prove that either $u_1(z)\equiv 0$ or $u_2(z)\equiv 0$ in $D$.

Any help on this problem? I don't know how to use the harmonic conjugates of $u_1$ and $u_2$ in $D$.

@nickbros123 you need a certain amount of reputation

Ok so basically i was trying to prove how to get infimum property (greatest lower bound) from the assumption that every set which is a subset of R which is bounded above has a supremum

that there exists an infimum?

what's the statement you're trying to prove

Iyes that there exists an infimum for lower bound sets

Basically using the assumption that upper bound sets have an infimum

Sets which are Subset of R,

Supremum*

Imy line of thinking was that if i define a new set s`= {-s : for s€S} then i say for set s to be lower bound, m≤s for all s€S, which gives me -m≥-s for all s€S, -m≥u for all u€S'

Can I write this

what's m

the least upper bound of s'?

@nickbros123

Yeah

I just wanna know if I can write that shrewd step, where i convert -x where x€S to u where u€S'

In my mind it was logical, but how can I show it mathematically

you have to show that the least upper bound of S', m, gives the greatest lower bound of S when it is -m

Ah I wish I could upload a photo or something

@shintuku can i type it out in latex notation

yes this chat can support latex, check the link in group descriptio

Ah not working on Android

.'ve come to the point where we find out S has an upper bound -m. Which means we can use the completeness axiom and so on.

I have a question here) how to mathematically show that the step "....m≥-s, for all s€S→ -m≥u for all u€S'....." Can be written. I can intuitively tell it is true, but how to prove that we can replace ".....-s where s€S ..... " into ".....u where u€S' ....."

give me 5 min i will read that

it is a lower bound, but why would it be the greatest lower bound? @nickbros123

We can prove it by some tricks. We got -m≥x for x€S', which means S' has a least upper bound let's say it's L. I can write= L≤n for all n: x≤n for all x€S' (n is any upper bound, L is least upper bound).

-L≥-n for all n:-x≥-n where x€S' which means -L≥k for all k: -x≥k where x€ S' ; -L≥k for all k: u≥k where u€S. We see k is any lower bound, and the term -L is greater than k always

I have a doubt regarding the step where we use S and S' interchangeably as we change -x and x

am reading

where did you get k from, what is k

I replaced -n with k

also when you write "for all n", this is incorrect, because it is not for all n, it is for those n that are in the real numbers, and that are upper bounds of S'

same with for all k

@shintuku no you are mistaking me, the way I've written is basically a condition, I've written that for all n for which -x≥n

Sorry, -n

The : symbol, i used it for putting a condition

This is my first intro to proving something rigorously and using axioms, i am a proper novice here

you are changing the names of variables, why? this makes it very confusing to read your proof

you're naming elements of S' using x and u

why not just use one of those?

@shintuku i am changing the variables so that I can negate the negative signs and make the term look better

k = -n?

can you rewrite the proof without changing names?

@shintuku ok, give me some time

we see m' is any lower bound of S, and we see L' is greater than any lower bound of S. Therefore L' is the greatest lower bound, hence the existence of infimum is proved.

hm, you need an elementary logic reading

you aren't properly using quantification

your proof should look like this

Let S be a subset of R. Let s be an arbitrary element of S.

Let S be bounded below. Then, there exists an m such that m is less or equal than s.

you only have a SINGLE s

it is an arbitrary one, yes

but you can only manipulate a single one in your proof

that is, if that's the way you want to proceed

@shintuku you mean a book specifically for logic?

Can I get the same rigor for proofs from set theory

Reading set theory*

some real analysis books come with an introduction to proofs

like Steven R. Lay's "Analysis with an introduction to Proof"

yeah a set theory book works well too