Yeah it's a little bit stranger. I think it should still work

yea i just tried it out in mathematica

Taylor about any point should still have finite radius of convergence

and it's pretty simple

OK, great

i'm gonna try to find it analytically :

:d*

oh really?

it's not possible to have a taylor with radius of convergence $\infty$?

oh wai

t

i'm sorry

Sure, why not? Just e^x

i misunderstood your message:P

Ah ok

What's not possible is for it to have infinite radius of convergence around one point but finite around another.

hm okay

Hey guys, does anyone know why adding a l2 norm regularizer on a convex objective function gives the MAP estimate instead of the MLE?

Hey, does anyone know how can I graph $r(t) = (t,-t,2t)$ ??

I cant seem to figure it out, I'm suppose to know how the function look like only by looking at that

would someone care to explain how to find the taylor expansion for $\frac{1}{1+x^2}$? I tried writing out the first few derivatives, but I'm getting a lot of terms after the fourth derivative, which seem to keep expanding, and i don't see the pattern yet. i can see that a lot of them will become 0 at the end, but it still seems like quite some work. otherwise, i'll just ask on the forum!

ugh, I have a really amazing example but I'm having trouble proving its properties

intuitively it works, it's just that writing proofs is not easy

@TedShifrin Could you share the image from the solution to 4.5.6? I don't have it for some reason

Dropping signs on problems is the worst

Also, is that the only image in the solutions?

@ShaVuklia Try geometric series

So we were given 2 definitions of the tangent space of the manifold at a point

@Maks You can also write it as $t(1,-1,2)$, that is, $t$ times a vector

Hullo

hii

@AkivaWeinberger Hey ! The program is still running, 671088640 and counting

Wow

@AkivaWeinberger Ok... and how do I know the graph ?

That's never going to stop running, you know

I mean, the universe will stop before it does

One of them was that if you express it as a graph $H:\mathbb{R}^k\to \mathbb{R}^{n-k}$, then let $e_i$ denote the standard basis vectors of $\mathbb{R}^k$ and take $x_0$ such that $(x_0,H(x_0)) = p$, you have that $T_p(M) = span((e_1,\delta_1 H(x_0)), \ldots, (e_k,\delta_k H(x_0)))$.

@Maks Ackerman?

@Maks $t$ times a vector, as $t$ varies?

They should all be parallel to that vector, right?

@AkivaWeinberger Hahaha we will see

@Akiva I find that line of reasonning very pessimistic

@AkivaWeinberger To be honest I really dont fully understand this part of the mathematics

That is supposed to give me a graph right ?

Cause when I plot it in wolfram it gives me 3 separated lines

I hope the universe will not end before the maximal memory allocation is reached

The other is that if you express it as the 0 set of smooth functions $F_1,\ldots,F_{n-k}:\mathbb{R}^n\to\mathbb{R}$, that the tangent space is the intersections of their kernels

We had to prove that they were equivalent, and I was all like ?????

Right, @Daminark.

Because I dropped some signs

So I was getting $2\delta_j h_i(x_0)$ when I was supposed to get $0$

Then it hit me

@Maks It should be one line in 3D space

@AkivaWeinberger How can I use the geometric series? I know that $\sum s\cdot r^k=\frac{a}{1-r}$ for $|r|<1$, and $a\in \mathbb R$, but I don't know how to get the square?

@Astyx Oh, touché

Make r into r^2, and use an alternating series

@Sha Typo

@ShaVuklia $1/(1+x^2)$ is $1/(1-(-x^2))$

so that's $a=1$, $r=-x^2$

ahh, i was afraid to turn it into something negative:P

My new top comment on Reddit is about using LSD in cooking.

So it goes

great, thanks for the help! i learned a lot from you guys!

Torture

this is going to be torture to prove

What is ?

What are you trying to prove

Maybe "This sentence cannot be proven in fewer than googol words"

(A proof can be obtained by checking all possible strings of fewer than googol words.)

@AkivaWeinberger Do you know anything about probability theory? I have a proof that I have been trying all day to solve.

So ?

A little

You abandoned me :(

Here is the problem:math.stackexchange.com/questions/2129683/…

It should be one line in 3D space, I said…

@Maks

And how do I know how that line is ?

I'm being asked to graph it

It's the line containing (1,-1, 2) and going throught the origin @Maks

It's $\Bbb R\to\Bbb R^3$. I assume that by "graph it" they mean draw its image, i.e., the curve it parametrizes.

@AkivaWeinberger Exactly

So draw $\Bbb R^3$, plot (1,-1,2) and draw the line going from the origin to this point

I don't get the concept of how to do it

And my book doesnt really explain it much

That's all ?

A line that goes through the origin and (1,-1,2) ?

Yup

That's what I'd draw at least

Unless you know how to draw $\Bbb R^4$

On a piece of paper

@Astyx @AkivaWeinberger a highly technical example

in topology

the likes of which has never been constructed before

Right

How many dimensions @ForeverMozart

or are we doing weird point set stuff

yes set-theoretic topology

Aw

its a type of countable connected space

Weird non-Hausdorff stuff?

so non-metric

no its hausdorff

Countable, connected, and Hausdorff? Hmm

Though I'm guessing you have more conditions

yes. There are many examples of those

Genius

and I assume the Continuum Hypothesis as well

@ForeverMozart Like what?

you want a countable connected Hausdorff example?

Take $\mathbb Q\times \mathbb Q$.

rationals points in the plane

then define a topology as follows

a neighborhood of a point on the $x$-axis is an open horizontal interval

a neighborhood of a point $p$ elsewhere is defined as follows

there is an equilateral triangle with vertex $p$ and base along the x-axis

the two vertices on the x-axis are q and r

now the basic neighborhood of $p$ is the point $p$ together with horizontal intervals around $q$ and $r$

this is the nicest example I think

O_o

easy to visualize anyway

The vertice of the triangle are not in $\Bbb Q^2$ though, are they ?

@Astyx yes, and that's useful

when I say horizontal intervals, I mean intervals in $\mathbb Q$

Ah right

They're conjugates in $\Bbb Q[\sqrt3]$ I think?

I think I get it

The vertices, I mean

so since the base vertices are not in $\mathbb Q$, you can separate a point from it's projection onto the x-axis with disjoint open sets

it helps to get Hausdorff property

So it's essentially $\Bbb Q[\sqrt3]$ except we only require neighborhoods of a point to include rational numbers and the point, and we quotient out conjugates

or something weird like that

hi chat

"Look at my fanny, look at my fanny, look at my fanny"

Finally I can chat

I have a question.

Let $ A= (a_{i,j}) \in K^{m \times n}$ Then $\phi_A = K^n \to K^m, \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} \mapsto \begin{pmatrix} y_1 \\ \vdots \\ y_m \end{pmatrix}$ with $y_i = \sum_{j=1}^{n}a_{i,j}x_j$. So, we have a matrix. And a function.

This function maps an element of $n$-dimensional space to an element of $m$-dimensional space. That is good and clear. But how does this thing interact with the matrix, according to the last condition? If I had a linear system, I could say - the function takes the solution vector and maps it into the vector, which positions are the sums of the lines of the system.

But I have only a matrix and some tricky function, which sacral meaning I cannot get. Can somebody interprete it with human words? Thank you.

@Hardeep Yay

hi chat

@Semiclassical hello

hi @semi

@ForeverMozart Takes on a different meaning when you consider what fanny means as slang in the UK. (I'm not going to say it here.)

Could someone tell me what does this 1 displayed besides my pic near the chatbox indicate?

@Kirill Writing $x,y$ for the column vectors, you've got $\phi_A(x)=y$. But you also have $y=Ax$ since the sum you is the definition of matrix multiplication. So you can write $\phi_A(x)=Ax$.

That is, the action of $\phi_A$ on vectors in $K^n$ is represented by matrix multiplication on the left, sending vectors in $K^n$ to $K^m$.

@DanielFischer Hi, for the question about holomorphic function on a neighborhood of $0$ such that $\vert f^{(n)}(0)\vert \ge (n!)^2$ for all $n\in J$ where $J$ is an infinite subset of $\Bbb{N}?$ When you said Cauchy estimates, I write $\vert f^{(n)}(0)\vert\le \frac{n!}{\varepsilon^n}\sup_{\partial D(0,\varepsilon)}\vert f(z)\vert$ so that $n!\le \frac{f^{(n)}(0)}{n!}\le(\frac{1}{\varepsilon})^n\sup_{\partial D(0,\varepsilon)}\vert f(z)\vert$.

@Semiclassical Let us take a certain case. $ A= (a_{i,j}) \in K^{3 \times 2}, A=\begin{pmatrix} a_{1,1}&a_{1,2} \\ a_{2,1}&a_{2,2} \\a_{3,1}&a_{3,2}\end{pmatrix}$. Then $\phi_A = K^2 \to K^3, \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}$ with $y_i = \sum_{j=1}^{n}a_{i,j}x_j$. So, $y_1=a_{1,1}x_1 + a_{1,2}x_2$, for example.

now if $\varepsilon<1$, we have an absurdity by taking the limit, this is what you have in mind?

@Semiclassical so, what do we do?

$\phi_A(x)=y=Ax$.

Same as I said above. It is not any more complicated than that.

@Semiclassical Is $A$ a matrix?

Well, you just said it was.

So yes.

@Semiclassical So, $y$ ist a product of a matrix $A$ with an $x$ vector?

Yes.

@Semiclassical I do not know, how to multiplicate a matrix with a vector. Could you show it, please? On that example above?

It's literally just matrix multiplication.

A column vector is just a matrix with one column.

$y$ is the column vector you wrote above (with y_k components) and similarly for $x$.

If you don't know matrix multiplication, look it up. I'm not going to explain it here.

@Semiclassical I know :) I just do not know how to multiplicate them. But I can read it somewhere. What I still do not get, what this function does.

If you know what matrix multiplication is, then you should know what $Ax$ means.

@Semiclassical no, I said it above, I do not know it

Ah. Then, look it up. :/

@Semiclassical I will. Can I continue the questioning after reading it?

Sure.

@Semiclassical Thank you.

@Semiclassical Ok, I am reading at the moment. Strange, but the lecture script says nothing about the matrix multiplication, it just gives me this formel and says that this one is very important.

That's kinda goofy.

I mean, otherwise the matrix $A$ is literally just an array of numbers.

@Semiclassical I mean, "officially" we have not have matrix multiplication yet. That is why I am wondering that I nedd this to understand the formel.

Well, you can understand the mapping without knowing matrix multiplication: We take the vector $x=(x_k)$ and map it to $y=(y_j)$ where $y_j=\sum_{k=1}^n a_{j,k}x_k$.

hey, would anyone know why I don't get -1 as the value when I type in $e^{i\pi}$ in Mathematica?

@ShaVuklia Are you doing e or E? Mathematica uses capital E for the natural number.

ohhh! yea, I think that's it

@Kirill So you're creating a new vector in K^m whose components are particular linear combinations of those of the vector in K^n.

@Semiclassical So , I have read the basics. The multiplication is not commutative, that is good.

Yeah.

@Semiclassical ok. The direction is also good, I am starting to uderstand it. Slowly. That is why I need some word exchange, please. So, what does this function does? I have some vector $x$. Is it the same alle the time? Are the components fixed?

@JeSuis For $\varepsilon \geqslant 1$, the inequality becomes impossible even faster. The point is that for every $\varepsilon > 0$ and $K \in [0,+\infty)$ the inequality $$n! \leqslant K\varepsilon^{-n}$$ can hold only for finitely many $n$ (because the factorial grows faster than every exponential function).

The components of $x$ would presumably not be fixed, because you'd want to be able to apply $\phi_A$ to an arbitrary vector in $K^n$.

@ok. than, what is the domain of $x$? What could be a domain?

The domain is $K^n$.

They tell you that.

i.e. a vector of $n$ components, each an element of $K$.

So, $K \times K \times \ldots \times K$ $n$ times, right?

If you mean $n$ copies of $K$, then yes.

@Semiclassical Ok. Let $K$ be $\mathbb{R}$. Let us take $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ e.g.

@DanielFischer arf right! I have the same kind of question but now it's $\vert f(1/n)-\frac{\cos(\pi n)}{2n+1}\vert<1/n^2$, I get that $f(0)=0$ and so we have $f(z)=z^pg(z)$ where $g(0)\ne 0.$ But for I don't get $g(0)=0$ exploiting the inequality. So, I presume there is another argument here?

@Semiclassical what now?

Not sure what you're after. But, for instance, we could compute $\phi_A(x)$.

And by the definition of $\phi_A$, that'll be $y\in K^m$ such that $y_1=a_{1,1}(2)+a_{1,2}(3)$, $y_2=a_{2,1}(2)+a_{2,2}(3)$, and so forth.

@Semiclassical You said, $Ax$?

Same thing.

I'm avoiding talking in terms of matrix multiplication, though, since you're not familiar with it.

@Semiclassical I've read it in the time. Yes, it is $Ax$, now I see it.

Okay.

Now, one crucial aspect of all of this is that $\phi_A$ is a linear operator.

@Semiclassical So, we have station A where we take the vector, we send it to station B, where it combines with the matrix, and station C where the result comes out. Is that correct?

@Semiclassical linear operator $=$ linear function?

Yeah.

You've got the domain, which is $K^n$, and the codomain, which is $K^m$. A linear function from the former to the latter corresponds to an m-by-n matrix in $K^{m,n}$.

To check that it's linear, to what does $\phi_A$ map the vectors $(1,0)^T, (0,1)^T?$

@Semiclassical how should I read A unted the phi? Just "phi Ei?" Why we need this lower index?

One moment/

Eh, it's notation (read as "phi sub A")

Just telling you that this linear function corresponds to the matrix $A$.

ok

A reason to use it is if you had more than one such function, e.g. $\phi_A$ and $\phi_B$.

@Semiclassical $T=$ transpose?

right.

hey

Hi @Zach

studying vieta's formulas for the elementary cyclic sums of the roots of a polynomial

it appeared quite a bit in the problem sets from previous years of the competition im doing

@JeSuis Note $\cos (\pi n) = (-1)^n$. Consider even $n$ and odd $n$ separately.

Now that I'm beginning to get past the stage of being intimidated with the definition of a manifold and all, the stuff is actually really nice

@Semiclassical $(1,0) \mapsto (a_{1,1}, a_{2,1})$, $(0,1) \mapsto (a_{2,1}, a_{2,2})$?

Only if $K^m$ has $m=2$.

@Daminark Manifolds are life.

If not, you'll instead have a total of $m$ such entries in each output vector.

@Daminark A lot of the intro stuff is scary since it's all symbol pushing and abstract definitions. It's like point set topology: you gotta do it to get to the cool stuff. But the cool stuff is beautiful.

Also, the second one should be $(a_{1,2},a_{2,2})$.

@Semiclassical I am not sure that I read your $(1,0)^T$ right.

Yeah, we're only doing embedded submanifolds of $\mathbb{R}^n$, so not much going on with charts and all

You did.

just $(1,0)^T=\binom{1}{0}$.

@Semiclassical yes, that is why. On the paper it is clear. I have wondered why you used this notation.

Because I'm too lazy to write them as column vectors in Mathjax.

aha

The reason I noticed the above typo, I should note, is because the first index in each component should be the same as which component I'm looking at.

@Daminark The multivariable calculus with the embedded manifolds in R^n might be a bit tedious in the beginning but it pays off. The chart language eg is just a convenient local setup coming up from the inverse function theorem.

So I could see stuff like $a_{1,1}+2a_{1,2}$ in the first output entry, but not $a_{1,1}+a_{2,1}$.

Though the definition of a tangent space was scaring me first

So, for (1,0) it is the first column of the matrix, for (0,1) - the second one?

Right.

hi @Balarka long time no see

But one of our problems was to prove that the preimage of a regular value of a smooth function mapping to a lower dimensional space is a manifold

Later it says to prove the special linear group is a Lie group

Good!

Once it clicked, I was like WAIT

Hi @Zach

@Semiclassical how does it check the linearity? I know only those two conditions for linear functions - homomorphism and a multiplication with an element of the field

Well, now suppose I take a generic $x=(x_1,x_2)^T$.

ok

@Daminark Yup, it gives you a buckload of examples of manifolds

Computing tangent spaces of the group is a bit trickier, still wrestling with that

It's not hard to check that you get $$Ax=(a_{1,1}x_1+a_{1,2}x_2,a_{2,1}x_1+a_{2,2}x_2, \cdots ,a_{m,1}x_1+a_{m,2}x_2)^T$$

Which is great, because that's the same as

But at least now translating between the definitions as a graph and as a 0-set is getting better

$$(a_{1,1},a_{2,1}, \cdots ,a_{m,1})^Tx_1+(a_{1,2},a_{2,2}, \cdots ,a_{m,2})^Tx_2$$

And these vectors, as you just computed, are just $\phi_A(1,0)$ and $\phi_A(0,1)$ respectively.

Writing in terms of unit vectors and matrix multiplication to make things easer, we have $$Ax=A(x_1 e_1+x_2 e_2)=x_1 (Ae_1)+x_2(Ae_2)$$

To put this into context: We start with $x$, which can be understood as a linear combination of unit vectors with appropriate coefficients $x_1,x_2$.

And we end with $Ax$, which is a linear combination of vectors with those same coefficients.

@Semiclassical I got it. Let me think about it about 5 minutes long, please

:)

@Daminark Here's a thing to wonder about. A manifold is a subset of R^n which is locally 0-set of a smooth function where 0 is a regular value, right? Can there be a manifold which is not globally 0-set of any smooth function which has 0 as a regular value?

@Semiclassical we used a $m \times 2$ matrix, right?

Right.

I made a small edit, otherwise the question is way more nontrivial.

@Semiclassical I have to write it down in order to be sure that I get it

@Balarka My immediate inclination is to say no

Interesting.

Well, I'm not going to reveal this. Problem to be kept at the back of your head!

Alright

Well, I hope to get to it soon, but for now my pset beckons

You probably don't have the tools to answer it now, but it's still a question you might find appealing.

@Semiclassical stands a comma for a new row or a new position in $Ax$?

Yeah. I think I'll find some book on differential geometry to pursue it further

Because we're only really doing the very basics of manifolds here

In what I wrote above?

We started defining them by 0 sets, graphs, proved the equivalence of the two definitions, and some examples

I think these sort of things are more about differential topology than differential geometry (the latter deals with extra structures on the manifold - eg Riemannian metric, connections, complex structures, foliations, etc etc).

Guillemin-Pollack is my favorite diff. top. book.

I intended it to be (in conjunction with the transpose) the next row.

Monday we're gonna do a bit more, I think, but we're moving on to functional analysis

I have a pdf of Guillemin-Pollack

...I think?

I've also got a pdf of Lee's Intro to Smooth Manifolds

Hmm. Now I'm not sure I was careful enough in writing it out.

@Daminark Right. That's the gateway to manifolds.

This is why matrix multiplication is nicer, tbh.

@Balarka That = GP?

No, I mean, equivalence of those two definitions.

Ah, yeah

I haven't studied from Lee but it's ok. The theory is very well-explained; the problems, not so much.

G-P is quick and dirty and fun and fascinating

We hadn't done parametrizations since my professor said you need to be careful about generating manifolds with them, since you have to have the additional property that the topology generated by the parametrization needs to correspond to the inherited topology from $\mathbb{R}^n$

@Semiclassical but,, the result is $m \times 1$ matrix, or?

Should be, yes.

$(m \times 1)=(m\times n)(n\times 1)$ at the level of matrix multiplication.

In the side lessons from my physics TA, he defined it in the more abstract way, using charts, atlases, differential structures, all that.

@Daminark Well, the problem is if you have a manifold $M$ in R^n you can't parametrize that by a single map. A sphere is not image of an embedding of a subset of R^2.

You need multiple such parametrizations and "glue them up". That's what charts and atlases do

He said you have a set, nothing else, now do this. Then manufactured a topology and said we'd keep things second countable and Hausdorff

In order to get uniqueness of limits and partitions of unity

Right.

The cool thing is that the class of abstractly-defined manifolds and the class of submanifolds of R^n (for some n, not fixed) are not distinct.

Though I did hear that secretly all manifolds are embedded submanifolds of $\mathbb{R}^n$

Precisely.

Whitney's theorem

I don't like to call that Whitney's theorem but sure :)

I think you can double the dimension and guarantee an embedding

Why so?

Whitney's theorem is that you can get the dimension down to $2n+1$ (and the thing that he did that was really clever was that you can get it down to $2n$).

The fact that you can get it into $\Bbb R^N$ is less inspiring.

Yup, and it's much easier to prove every manifold embeds in some R^N

@Semiclassical and $e_i$-s are unit vectors?

Essentially a partition of unity argument; you embed each chart in some big dimension so you can glue them up without changing the manifold.

Lol I should actually find out about partitions of unity

Do they have them in GP?

I know they're in Spivak

Yes.

Though the argument that every abstract manifold gets into some R^N is carried out in Hirsch

GP uses partition of unity for different reasons; because their manifolds are already sitting in R^n you don't need to care much

Ah

@MikeMiller I don't actually know how to prove the Whitney's Clever Theorem. How's it done?

I can easily immerse in R^2n. Does one remove singularities by pushing along disks or something like that?

Yes.

Immerse w/ double points is what I want I guess

@Semiclassical I want to thank you very much for perfect explanations and hints! Now, at least, I can understand what they want from me asking about the kernel of $\phi_A$ and the basis of the image of the $\phi_A$ :) Thanks again!

Strange; I have no reason to believe that it actually works.

You can always eliminate double points with opposite sign by embedding a disc (as long as the manifold is of dimension at least 3). If you don't have any double points with opposite sign, you can easily add a single double point somewhere. (Not by a homotopy through immersions, but by modifying the immersion).

So add and cancel as appropriate.

I'm still working on this! I've been working on it for two days and I still have another problem on this assignment I haven't even tried yet.

I did post an answer where it shows that I got up to a point and got stuck.

Could somebody please help me finish it?