Mathematics - 2022-09-22

Ted Shifrin

00:37

@RE60K No clue. But just to clarify: No repetitions allowed.

Akiva Weinberger

01:16

Suppose you want to cut a square into pieces and reassemble them to form the same square but rotated 45 degrees. The pieces are not allowed to rotate. How many pieces do you need?
I can do it with five pieces, which I conjecture is the smallest amount possible.

Ted Shifrin

Number, not amount.

Xander Henderson

@AnthonySaint-Criq One of the standard texts on this topic is Lee's Intro to Smooth Manifolds. The definition he gives of a topological manifold is one without boundary (a manifold with a boundary is actually not a manifold at all).

Munkres also gives a definition of "manifold" which does not have a boundary.

Akiva Weinberger

…which I conjecture is the fewest possible.

Xander Henderson

@TedShifrin I do that all the time.

Ted Shifrin

01:36

Only cuz it’s DogAteMy did I make the correction.

copper.hat

02:24

so fussy, maximal atlases and all

leslie townes

the real question is how does bourbaki do it.

copper.hat

the immortal Nick

Akiva Weinberger

03:20

Can we cut the square into finitely many pieces and rearrange the pieces into a circle, if we allow ourselves to dilate the pieces as well as simply translate them? Surprisingly, yes!

From here: link.springer.com/article/10.1023/A:1025944327967

BAYMAX

$M=\begin{pmatrix} 1 & 1 & 1 & 0 & \cdots &\cdots\\
0 & 1 & 1 & 1 & \cdots&\cdots \\
0 & 0 & 1 & 1 & \cdots&\cdots \\
0 & 0 & 0 & 1 & \cdots&\cdots \\
\vdots & \vdots& \vdots& \vdots& \ddots&\vdots\\
0&0&0&0&\cdots&1\end{pmatrix}

and inverse

$$M^{-1}=\begin{pmatrix}
1&-1&0&1&-1&0&\cdots\\
0&1&-1&0&1&-1&\cdots\\
0&0&1&-1&0&1&\cdots\\
0&0&0&1&-1&0&\cdots\\
0&0&0&0&1&-1&\cdots\\
0&0&0&0&0&1&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\
0&0&0&0&0&0&\cdots&1
\end{pmatrix}$$

Does $M$ and $M^{-1}$ have any special name?

$M=\begin{pmatrix} 1 & 1 & 1 & 0 & \cdots &\cdots\\
0 & 1 & 1 & 1 & \cdots&\cdots \\
0 & 0 & 1 & 1 & \cdots&\cdots \\
0 & 0 & 0 & 1 & \cdots&\cdots \\
\vdots & \vdots& \vdots& \vdots& \ddots&\vdots\\
0&0&0&0&\cdots&1\end{pmatrix}$

@AkivaWeinberger I wonder how they make these diagrams and insert math symbols?

copper.hat

04:11

other than Toeplitz, i don't think so...

leslie townes

BAYMAX: none that i know of. these are examples of toeplitz matrices (which are often defined to be matrices whose i,jth entry is constant along all 'diagonals' i-j=k, k fixed).

people who do certain types of signals processing recognize toeplitz matrices as corresponding to 'time-invariant' filters (because you can characterize toeplitz-icity in terms of commutation relations with a shift). lower triangular toeplitz matrices get more attention than upper triangular ones because they correspond to 'causal' time-invariant filters.

so you're definitely not writing on a blank slate here, but i don't think the general theory of these things is so good that the recognition of "oh boy! this is a toeplitz matrix!" necessarily leads to a lot of generally applicable and fruitful insights. a lot of diversity and difficulty is present within the world of toeplitz matrices

math.stackexchange.com/questions/786108/… is a bare-handed discussion of how to invert toeplitz matrices

it's sometimes helpful to think of them as the result of compressing multiplication operators to a space of polynomials. if you think of L^2(circle) as a hilbert space in the usual way, multiplication by sum c_n z^n followed by compression to the span of [1, z, ..., z^n] is an nxn toeplitz matrix, where the entry on the diagonal i-j=k is c_k. that perspective might make that MSE post make more sense.

hi copper.

copper.hat

04:59

Hi @leslietownes!

leslie townes

what's new?

Koro

05:38

Suppose that R is the ring of all real valued continuous functions on [0,1]. Define M={f in R: f(0)=0}. Then, is M = principal ideal generated by the identity map i (i(x)=x for all x in [0,1])?

leslie townes

05:55

koro consider something like sqrt(x)

Koro

perfect! thanks a lot!

Instead of f(0)=0, if one considers f(c)=0 for some c in (0,1), then |x-c| works just fine.

1

A: $R$ local ring, $I$ maximal ideal then $x\notin I$ implies $x$ unit

Here's a different idea. Suppose $x \not \in I$. Then since $R$ is local, $x$ must generate $R$ (unless $x=0$). Hence there is some $z \in R$ such that $xz=1$.

Does this answer make any sense?

Why must (x) generate R?

leslie townes

06:12

see the other answers.

it makes sense in that it doesn't say anything false, but it doesn't explain what's going on.

Koro

If (x) generated a prime ideal, then also there exists a maximal ideal containing (x) (by Zorn's lemma) and hence (x)= R.

But not every prinicpal ideal is prime. All that seems to be correct is (x) is principal.

leslie townes

any proper ideal is contained in a maximal ideal. primality, principality, etc. aren't needed for this

Koro

06:32

oh yes. Thanks a lot.

Meanwhile, I also applied Zorn's lemma to {set of all proper ideals containing a given non unit element a}

to deduce that there is a maximal ideal containing a.

I got mixed up in 'proper' and 'prime'.

leslie townes

06:56

might be worth thinking about whether you can give a zorn-free proof of that fact for R = C[0,1] up above.

Not Tfue

08:56

Poisson Distribution doesn't make sense...

I get all the distribution but this one isn't intutive.

Is there a good source for Poisson distribution? Or should I just memorize it.

robjohn

09:45

@NotTfue Does this answer help?

Jakobian

10:00

My favourite application of axiom of choice is existence of Bernstein sets

Not Tfue

10:38

@robjohn Sir yes sir!

I think when I have problem I can literally search from your profile.

ZaWarudo

11:41

to show that a differentiable function with $f'(x) \ne 1$ for any $x \in \mathbb{R}$ has a unique fixed point, it is valid to say that assuming $x_1$ and $x_2$ fixed points by the mean value theorem it is $x_1-x_2=f(x_1)-f(x_2)=(x_1-x_2)f'(x) \implies [f'(c)-1](x_1-x_2)=0 \iff f'(c)=1 \vee x_1=x_2$ and, since $f'(x)\ne 1$, it is necessarily $x_1=x_2$?

Thorgott

11:58

you seem to write $c$ and $x$ slightly inconsistently, but otherwise yes, your argument is correct

I should stress, though, you only show that if there is a fixed point, then it is unique

there might very well not be any fixed point, however

so the claim "has a unique fixed point" is not true

one potato two potato

0

Q: Generator of $H^1(S^1)$ via integration of a bump $1$-form on $S^1$

I have a statement in Bott & Tu's Differential form in Algebraic topology (p.36) that I can't understand. We say previously that a generator of $H^1(S^1)$ is a bump $1$-form on $S^1$ which gives the isomorphism $H^1(S^1)\simeq\Bbb R^1$ under integration. It says integration but I can't see the ...

Thorgott

you misunderstand

integration has nothing to do with arguing what a generator of $H^1(S^1)$ is

one potato two potato

12:14

Yeah I'm sure I misunderstand this. Then how the word integration should be interpreted here?

Thorgott

integration should be interpreted in the regular sense, I think you're just misunderstanding the structure of the sentence

what they are saying is that integration gives an isomorphism $H^1(S^1)\rightarrow\mathbb{R}$

this is a separate claim from the claim that a bump $1$-form generates $H^1(S^1)$

Monty

12:55

if i have a function $f:\mathbb{R}^d\to \mathbb{R}$ I write $\Delta f=\sum \partial^2_{x_i}f$ as the laplacian, i.e sum of 2nd derivatives

Say I have a function $f:\mathbb{R}^d\times \mathbb{R}^d \to \mathbb{R}$, calling the variables $x,y\in \mathbb{R}^d$. What is the appropriate notation for $\sum_{i}\partial_{x_i}\partial_{y_i}f$, maybe something like $\Delta_{x,y}f(x,y)$?

robjohn

13:42

You would have to define that that is what you mean. I don't think it would share many properties with the Laplacian; the extension Laplacian would seem to be $\sum\limits_{j=1}^d\partial_{x_j}^2f+\sum\limits_{j=1}^d\partial_{y_j}^2f$

Tachytaenius

15:31

Does anyone have any ideas for getting a series of integers that add up to integer n from a series of arbitrary numbers from 0 to 1?
Really broad question I know, I'm not sure how to better define it.
So, {0,0,0,1} would be {0,0,0,n}, and {1,1,0,0} would be {n/2,n/2,0,0} etc.
Accounting for rounding error by adding to or subtracting from the last integer or something, I'll figure that part out afterwards.

Koro

Dimension of a subset S of R^m is defined as $k$ if S is homeomorphic to an open subset of R^k.

Tachytaenius

15:44

I have noticed that "{1,1,0,0} would be {n/2,n/2,0,0}" is division of n by the sum of the starting numbers.

What would {1, 0.5, 0, 0} be? Maybe it would be {2*n/3, 1*n/3, 0, 0}
i noticed that 2*n/3 is the same as n/1.5, which is the sum of these starting numbers. Clues, but I haven't figure it out yet.

Jakobian

16:02

@Koro and what if it isn't?

Akiva Weinberger

Question (complex analysis homework): "Suppose that $\lim_{z\to z_0}$ exists. Show that the limit is unique." Student: "Let $f:\Bbb C_1\to\Bbb C_2$, where $z_0$ is a limit point of $\Bbb C_1$."

My note: "We have two complex numberses?"

Otherwise the rest of the answer was correct

Koro

@Jakobian It's a definition that I learnt recently.

This is in the context of understanding implicit function theorem.

The definition makes sense by "invariance of domains", which I don't yet know in detail as it seems to be from algebraic topology.

Akiva Weinberger

This means that if $S$ is the closed ball, it has no dimension

@Koro

There are several nonequivalent topological definitions of dimension. Here's one: https://en.wikipedia.org/wiki/Lebesgue_covering_dimension

(@Jakobian also)

They all agree for open sets, though

Invariance of domain basically says open subsets of $\Bbb R^m$ are not homeomorphic to open subsets of $\Bbb R^n$, $~m\ne n$ EDIT: No, the statement is something else, but this is a consequence

Koro

@AkivaWeinberger yes using this the dimension definition makes sense.

Akiva Weinberger

16:18

The algebraic topology thing in general uses things called (singular) chains. An $n$-chain is a collection of maps from $\Delta^n$ (the $n$-dimensional equivalent of a triangle, or an $n$-simplex) to your manifold

and by collections I mean a formal sum and difference of a bunch of them

It turns out it makes sense to talk about the "boundary" of an $n$-chain, which will be an $n-1$-chain

and then basically your thing is $n$-dimensional at $p$ if there's an $n$-chain in your space whose boundary does not contain $p$, and which is not the boundary of an $n+1$-chain minus an $n$-chain that does not contain $p$

Xander Henderson

Right. This is, essentially, a notion of "inductive dimension".

Inductive dimension

In the mathematical field of topology, the inductive dimension of a topological space X is either of two values, the small inductive dimension ind(X) or the large inductive dimension Ind(X). These are based on the observation that, in n-dimensional Euclidean space Rn, (n − 1)-dimensional spheres (that is, the boundaries of n-dimensional balls) have dimension n − 1. Therefore it should be possible to define the dimension of a space inductively in terms of the dimensions of the boundaries of suitable open sets. The small and large inductive dimensions are two of the three most usual ways of capturing...

To be three dimensional is to have a two dimensional boundary, more or less.

Of course, if you are playing with manifolds, most of the complications about "dimension" go away. Manifolds are very simple from the point of view of dimensions.

Akiva Weinberger

Is it the same? I'm basically doing relative homology and $H_n(X,X\setminus\{p\})$

Xander Henderson

@AkivaWeinberger It is not the same. But it is a similar idea.

Anywho, I need to go teach a few classes. And then drive an hour to Show Low to teach an evening class.

user 726941

Cya

Thorgott

hmm, not sure if cohomological dimension works that well for top spaces

Koro

16:43

@AkivaWeinberger I don't understand why. Can you please explain?

I think that S is open/closed in itself in subspace topology of R^n

So I don't see a contradiction if S is homeomorphic to an open subset U (of some R^k).

Ted Shifrin

Being open/closed in itself is totally not relevant.

Shinrin-Yoku

If we have a random polygon in the plane (need not be convex) a linear function would be optimised at the vertices?

Ted Shifrin

Yes.

Shinrin-Yoku

Yes to me? Is that cause a linear function would just stretch out the lines ?

Ted Shifrin

You’re talking about a real-valued function if you’re talking optimization, so that comment makes no sense.

Shinrin-Yoku

16:52

I mean if you visualise the plane we get when you apply the function

Ted Shifrin

You’re not listening.

Koro

youtube.com/… Here, at around 10:00, I think the Jacobian matrix of Df (a) is written incorrectly.

Ted Shifrin

Koro: I refuse to look. This is not rocket science. Figure it out.

Shinrin-Yoku

@TedShifrin a function from R^2 to R can be visualised as a plane in 3d space?

Akiva Weinberger

@Koro Closed balls are not homeomorphic to any open subset of R^n for any n

Ted Shifrin

16:57

OK, now you’re talking about the graph. Since it’s a region in a tilted plane, then ….

Koro

Suppose that $f=(f_1, f_2,..., f_k)$, then Jacobian matrix for f should have partial derivatives of $f_i$ in the $i$ the row (and not in the ith column).

Akiva Weinberger

(continued) I mean, you'd have to prove that, which uses invariance of domain, but it is a true fact

Shinrin-Yoku

@TedShifrin Yes then it would be optimised at vertices

Koro

I see. So I'll need invariance of domain. I'll take the statement for granted for now then.

Shinrin-Yoku

Am I correct?

Ted Shifrin

16:58

@Shinrin-Yoku Have you learned multivariable calculus? Do you know about gradients? That’s the way I think about optimization.

@Koro No, totally wrong.

Shinrin-Yoku

Yes, but isn’t that overkill? When it could be proved purely geometrically

Ted Shifrin

Sorry, Koro. Too crazy here. You’re right. The function is a column. Derivatives go across rows.

copper.hat

@Koro The Jacobian must satisfy $f(x+h) = f(x) + J h + o(\|h\|)$, so you can figure it out from that.

Ted Shifrin

@Shinrin-Yoku If you state it correctly, sure. Your original statement was wrong.

Koro

yeah, copper and Ted: exactly!

Shinrin-Yoku

17:02

@TedShifrin which original statement?

Koro

I was just saying that the video at that time-stamp (around 10:00) had made an error.

Ted Shifrin

Koro: In fact, this is the reason I wrote everything with column vectors throughout my multivariable book, despite the typographical nightmare.

Talking about how a linear map maps the plane to the plane.

Shinrin-Yoku

Ok I apologise

Ted Shifrin

Linear programming is quite nice. The simplex method is a nice application of very simple linear algebra.

It finds the optimal vertex immediately.

Jakobian

17:19

@AkivaWeinberger I was asking for education purposes, not because I don't know

robjohn

@AkivaWeinberger: I hope you don't mind I fleshed out your comment.

@TedShifrin either direction can prove nightmarish.

Ted Shifrin

How so, robjohn?

Koro

$f(x,y)=x^2+y^2$ and g(x,y)=$x+y-1$, then by implicit function theorem $g^{-1}(0)=\{(x, 1-x), x\in \mathbb R\}$.

hmm, not really.

robjohn

@TedShifrin row vectors can take up a lot of horizontal space, though much less vertical space.

Before LaTeX, it would be horrible to align vertical vectors.

Ted Shifrin

Oh, but writing the input and output of vector functions both as columns is a pain to typeset. Not so bad writing on the board.

I’m talking calculus, not just linear algebra. But it helps to stay consistent so as not to have Koro’s issues.

robjohn

17:32

Are you talking about pre-LaTeX, with a typewriter? With LaTeX, it is just a space consideration.

@TedShifrin yes

I, too, pretty much use column vectors (vertical vectors)

geocalc33

I have a vector field question

Ted Shifrin

@robjohn Yes, and lots of macros.

Williamson, Crowell, Trotter was typeset that way back in the late 60s, early 70s. Only other book I’d seen.

I think Hubbard and Hubbard, which drove me to write mine after 2 months of teaching, also did it.

robjohn

Wrote a book that could be taught from

geocalc33

Has the space of 3d vector fields $X$ on $D^3$ s.t. $X$ vanishes at the poles $p_0$ and $p_1$ and the flow maps the boundary, $S^2,$ to itself, been discussed in some book or is anyone here familiar with it?

Ted Shifrin

Hubbard is very smart and very idiosyncratic. That said, his book is in its nth edition, so way more popular than mine. Famous author rule.

You don’t need to mention flow, geocalc. This is just vector fields on the disk tangent to the boundary. I don’t know what more to say.

leslie townes

17:45

start with "Dianetics." the later books make a lot more sense after that

Koro

Suppose that $U\subset R^n\times R^k; f: U\to R, g:U\to R^k$ are $C^1$. Suppose $(a,b)\in U, g(a,b)=0$ and $Dg(a,b)$ has rank $k$.

Ted Shifrin

Ha ha … L Ron not quite at the right level.

Koro

My question is: why can I apply implicit function theorem on g? Please note that Dg has rank k and that the hypothesis does not say that either $D_1g(a,b)$ or $D_2(a,b)$ has rank $k$.

Ted Shifrin

Reorder the variables.

This is totally standard.

And I have no idea why you put $f$ in there.

Koro

Suppose that after re-ordering, I have $k_i$ linearly independent columns in $D_ig(a,b)$, i=1,2 such that $k_1+k_2=k$, then implicit function theorem is not applicable.

Actually, I'm trying to understand Lagrange's multipliers and hence $f$ was also put.

Ted Shifrin

17:50

I have no idea what you’re doing. Put the $k$ good columns at the end.

Koro

I see. That should not create any problem.

Reordering of variables function is diffeomorphism, I think.

i.e, the map $(x_1,x_2,...,x_n)\mapsto (x_{\sigma 1},..., x_{\sigma n})$, where $\sigma \in S_n$, the symmetric group of order n.

Thanks a lot, Ted :).

Ted Shifrin

It’s a linear isomorphism!

Maybe watch my lectures instead of the random ones that are wrong?

Koro

Ohh. The lectures I'm watching are not random ones. The professor teaching in the video lectures is also an author of several texts on mathematics. The error I pointed above is just a typo.

I have watched some portion of your lectures on inverse function theorem.

and I have also watched your lectures on surface integrals.

:)

Ted Shifrin

18:05

I’m surprised you haven’t caught mistakes and complained!

copper.hat

everyone makes musytakes

Koro

on surface integrals ? (I watched them months ago and didn't find anything suspicious). On IFT ? (started watching it few days ago. I'm yet to finish watching the complete lectures on this. So far I have watched till - given s, p, discussing x and y such that x+y=s, xy=p).

geocalc33

I'm asking about vector fields in the (ball) tangent to the boundary, not disk

Koro

@copper.hat you are raight :).

Ted Shifrin

@geocalc33 you’re misinterpreting. Most of us call a closed ball the disk (in whatever dimensions).

geocalc33

18:16

gotcha

Ted Shifrin

@Koro I made a small error discussing the meaning of flux, but caught and corrected it a bit after.

Koro

:-)

copper.hat

18:31

what the flux?

robjohn

shut the flux up

Ted Shifrin

covers eyes and ears

copper.hat

stoking up some divergence in the chat room

Ted Shifrin

Curl up with a good book?

schn

19:00

On Wikipedia the Fourier transform (FT) of a function has a minus sign in the exponent and a plus sign for the inverse FT. In a book I'm reading, the author has swapped those signs. Does it matter much?

leslie townes

not particularly, no, although that's an unusual choice. a more common potential point of difference from one book to another is fiddling with where factors of 2pi, or sqrt(2pi), go

schn

Ok, thank you @leslietownes.

leslie townes

is it a book with prob/stat applications?

schn

More engineering I would say, for which I have seen that it is common to define $j=-i$, which would explain the sign change.

geocalc33

okay, instead, define a diffeomorphism from the open ball to the open cube with a prescribed vector field in the open ball. I think this would let you get an explicit form for the vector field in the open cube...right??

leslie townes

19:15

wow, j is -i? never heard of that. i thought j was just a funny name for i

the positive sign for the transform would agree with a prob/stat convention about the 'characteristic function' of a random variable, but in that context they usually never call it the fourier transform at all

schn

interesting

Ted Shifrin

I thought so, too, @leslie. So engineers are reverse-oriented.

@geocalc How explicit is your diffeomorphism?

Is it really $C^1$?

copper.hat

20:02

i have never seen j defined as -i

robjohn

Nor have I

leslie townes

i think we've caught the probabilist trying to masquerade as the author of an engineering textbook

copper.hat

probably true

am i missing something here? math.stackexchange.com/q/4536867/27978

kinda want to write "do some work" but perhaps i am missing something

politeness always trips me up

geocalc33

20:18

diffeomorphism from the cube to the ball could be $\psi_{1}(x_{1}, \dots, x_{n}) = \bigl(f(x_{1}), \dots, f(x_{n})\bigr)
\quad\text{and}\quad
\psi_{2}(y) = \frac{y}{\sqrt{1 + \|y\|^{2}}}$ for $f(x) = \tanh^{-1} x$

leslie townes

copper: you're not missing something, but i don't see value in engaging further. there seem to be several levels of confusion there. my reaction is "yikes" if this is a functional analysis class.

geocalc33

with $\psi_1 \circ \psi_2$

copper.hat

i know, scary right!

leslie townes

definitely helps to have a grasp of R^2 before considering the infinite dimensional case. at least, even i found that handy.

copper.hat

:-)

Koro

20:20

Leslie: Re: our earlier discussion, the set $M_c=\{f \in C[0,1], f(c)=0\}$ is a maximal ideal. This can be shown without Zorn's lemma.

leslie townes

koro: yes, i meant: "any proper ideal in C[0,1] is contained in a maximal ideal of C[0,1]."

Koro

ohh, I see. I'll think about this.

leslie townes

your example is a pretty good example in functional analysis. "multiplication by x" on C[0,1] being a classic example of a non invertible operator that does not have 0 as an eigenvalue (it clearly fails to be surjective). its range not being all of what you called "M_0" is an example of a proper ideal sitting properly within a maximal ideal.

if you topologize C[0,1] with the sup norm it is also an example of a bounded operator that does not have closed range.

Ted Shifrin