@AlessandroCodenotti Ahh cool I'll scroll up and check

@Perturbative you have $\varnothing=\bigcap V(f_i)=V(\sum (f_i))$, by taking $I$ on both sides you get $A=\operatorname{rad}(\sum(f_i))$ so $1$ is in the rhs, but then $1\in\sum (f_i)$ as well

It was actually a set theory question rather than an AG one we were discussing

@AlessandroCodenotti Thanks for that response, let me just go through what you said carefully

I hope it's clear because I'm about to leave to sleep!

So just quick check, $\sum_{i \in I} (f_i)$ is the ideal generated by $\bigcup_{i \in I} (f_i)$, then $\emptyset = \cap V(f_i) = V(\cup(f_i)) = V(\sum (f_i))$ but since $V(A) = \emptyset$ it follows that $A = \sum(f_i)$

Actually wait that last part is probably nonsense

Anyway thanks for the help @AlessandroCodenotti, I should get some sleep too

@Bob It doesn't really matter which tbh. It's like asking if you write on a whiteboard or paper etc

Depends on how you feel and what you want to do with it

There is no problem with tex'ing it straight away if that's what you're thinking

If $f(z)$ is analytic, then is it true that $f(-z)$ is analytic?

@user330477 How would you differentiate the latter function if you knew the derivative of the former? That should answer your question.

Well, it really depends on where $f(z)$ is analytic, but since you didn't specify a particular region, I assume that the function is analytic everywhere on the complex plane. Given that, what can you say about $-z$?

@Fargle I know what you mean. But still to able to differentiate you should that the derivative must exist. I see this true when $f(z)$ is analytic over $\abs{z}<r$, but what if it is analytic over an arbitary open set.

@Rithaniel That case is really simple. See my comment to @Fargle.

Indeed---what if it is analytic over an arbitrary open set? There's a detail here, but it's not too hard to work out.

So finished the problem I complained about earlier, turns out it simplifies faster than I thought but it was still non-trivially annoying to expand out a bunch of stuff

But you would never whine, Demonark.

Heya @Ted

Oh of course

@Fargle This is exactly what my question is? What detail are you talking about?

hi @Fargle

@user330477 I would be giving it away at that point. That's no fun. Here's a leading question: if f(z) is analytic for z in U an arbitrary open set, then where would f(-z) be analytic?

@Fargle Is it the set $-U$, which is also open.

Indeed.

@Fargle How does this help?

This has answered your question: if f(z) is analytic in a place, then you can also say that f(-z) is analytic in a (different) place.

I think.

(Ted can slap me if I've led you astray)

Note: That different place might also be the same place as f(z) is analytic

that doesn't mean entire

Right---as is the case for the whole complex plane, or for $|z| < r$.

it's just vague

If I could be more specific, there's an answer that's almost a non-answer: if f(z) is analytic for z in U, then f(-z) is analytic for (-z) in U.

And this is just because symbols really do be like that.

Or any region centered at the origin with reflective symmetry along the line x-ix. (right?)

@Fargle Thank you for your help. This question came in context of this problem: Suppose $f(z)$ is analytic with power series representation $\sum a_k z^k$. If $f(z)$ is an even function, then $a_n=0$ for $n$ odd.

@Rithaniel Not along a line---"reflective" symmetry about the origin.

This is just happening on some disk centered at the origin @user330477

@TedShifrin @Fargle This question is a bit vague as it does not mention anything.

I know. I'm just telling you context. :P

Ah, right, just worked it out a little bit more in my head and I see what you mean.

Yeah. To borrow terms from other memelords in here, you just kinda, uh, flooop

@TedShifrin Ok, but then how do I show that the power series corresponding $(z/2) \cot (z/2)$ only contains even terms. I was planning on using this theorem, but this function has singularities at $n \pi$, where $n \in \mathbb{Z}$.

Are you sure about where the singularities are?

But, anyhow, the power series only converges on a disk of radius $R$, where the first singularity occurs at a point $z$ with $|z|=R$.

@TedShifrin Oops, sorry the singularities are $2n \pi$, where $n \in \mathbb{Z}$.

Agreed.

Oops. $n\ne 0$.

Is it true that $\sup_{x \in X} f(x) - \sup_{x \in X} g(x) = \sup_{x,y \in X} (f(x)-g(y))$?

@TedShifrin The problem is that we have a singularity at $0$.

No, it's removable, @user330477.

@user193319: Hardly ever.

Oh wait.

(There's a sin x/x limit hiding in the weeds, @user330477)

If you change the $-$'s to $+$'s, that's true.

But subtracting a sup is adding the negative of an inf, isn't it?

Yes.

So can you work it out from that?

Well, $\sup_{x \in X}f(x) - \sup_{x \in X}g(x) = \sup_{x \in X} f(x) + \inf_{x \in X} -g(x) = \sup_{x \in X} (f(x)-g(y))$...Is that right?

You'd better work on that last equality.

I'm not sure I follow.

No, it's not right.

Oh, it should be $\sup_{x,y \in X} (f(x)-g(y))$.

Why?

Why is the equality valid?

I'm not sure...I'm just trying to show that the uniform limit of Riemann integrable functions is Riemann integrable, where $X \subseteq \Bbb{R}^p$ is a Jordan set.

I don't want to hear about that. Focus.

is the $0 \times 0$ matrix diagonalizable?

Is it empty?

$\forall \text{ stuff } \in \varnothing \implies \text{stuff is true}$

Hi

If $T$ is a linear transformation between finite-dimensional euclidean spaces, what can we say about $\lim_{\mathbf{h}\to \mathbf{0}}\frac{T\mathbf{h}}{\|\mathbf{h}\|}$ ?

Lets say, $T: \mathbb{R}^n \to \mathbb{R}$.

any thoughts? M I bitin way more than I can chew?

Can anyone tell me how to simplify this $|x+y|+|x-y|$so that I can plot a graph of $|x+y|+|x-y|\leq 4$

wolfram alpha it? and its a square

@MoreAnonymous I want to learn to graph such functions

@LoopBack You know how to plot $|x| + |y| = c$ right? now rotate $x$ and $y$ coordinates by 45 degrees and you get the same thing!

@MoreAnonymous Yes I know how to plot $|x| + |y| = c$ . But how would I know, when to rotate coordinate axis to plot a particular graph.

I've been at this simple proof for hours and I have no idea how to prove it except setting up many cases... is there any better way? $|a+b|+|a-b|=|a|+|b| \implies |a|=|b|$. Is there something obvious I'm missing? I've tried applying the triangle inequality but it's an equality I want.

Someone should work on more $E_{10}$

@BalarkaSen Hey

I'm reviewing my probability and came across this statement. "We need to estimate P(Y| X1, X2, …, Xn). How many quantities to estimate? Ans - 2^n" Can someone explain how this is true?

Another paper: arxiv.org/pdf/math/0309407.pdf

Ok test test $x+1=y$

Okay that didn't work :D

Ok I'm good to go, ok can anyone hook me up with a good suggestion for "inverses of functions", something that is tractable for graduate/post-graduate students... I've hit a bit of a snag in my research and texts are either too simple (basic calculus) or they're way too formal (endless theorems that don't give me insight on how to approach it, only how to prove new statements related to function inverses)

Big ups and bonus points if you have a text relating specifically to non-homogeneous integral equations and their related inverses :-)

@user55789 "inverses of functions" seems absurdly broad

Hmm, okay so inverses of 1st kind Fredholm eqns, what's a good read? :D

Because I was thinking of just transforming the whole thing by tacking a heaviside function between my integral boundaries onto the kernel and then takign the Fourier transform, but I'm not sure within what constraints that would be legal

To make it more concrete what should I look at to investigate the behaviour of equations of the form $M(q) = \int_{-\infty}^\infty dx \left[\Theta(a-\lvert x\rvert)d(x,q)\right]\rho(x)$ ?

Where $a$ is some previously specified parameter, ok technically I should put it inside the arguments of $M$ as well

Please someone let me know why we can use 'invertible continuous map' here?

If $\gamma [a,b] \to \Bbb{R}^n$ continuously differentiable, show that the arc length $\ell (\gamma)$ equals $\sup \{ \sum_{i=1}^n ||\gamma (t_i)-\gamma (t_{i-1})|| \mid a = t_0 < t_1 < ... < t_n = b \}$.

I'm trying to solve the above problem. I've reduced it to showing that $\int_{a}^{b} ||\gamma ' (t)|| dt \ge ||\gamma (b) - \gamma (a)||$ holds. Intuitively/geometrically it's obvious, but I having trouble mathematizing my intuition. I could use some help.

How do you define the arc length ? @user193319

$\int_a^b \Vert \gamma'(t)\Vert dt$ ?

@Astyx I actually just decided to ask this on the main site. Here's the link with the definition of arc length math.stackexchange.com/questions/3015743/…

Yes, that's the definition.

You want to use the fact that $\gamma$ is uniformely continuous

Oh, really? Okay. I'll try that. Thanks!

Let me know if you're having trouble

I haven't gone into the detail but I think that's the way to do it

Oh, wait. I did try this, but i wasn't sure where to go with it.

is 1/(x^(-1/2)) Riemann intergable ?

I meant $\gamma'$, not $\gamma$

@neraj that's x^(1/2)

@Astyx I'm still having trouble seeing how uniform continuity of $\gamma '$ helps...sorry.

Wait I think I have misled you

Give me a minute

Yeah, forget what I said

Remember that $\int \Vert f\Vert \ge \Vert \int f\Vert$

Does this happen to rely on some mean value theorem for vector valued functions?

Because there is no mean value theorem for vector valued functions.

No this is just a consequence of the triangle inequality when you define the integral

Because you have $\Vert \sum_{i=1}^N h f(ih)\Vert \le \sum_{i=1}^N \Vert h f(ih)\Vert$

And you take the limit as $h = {1\over N} \to 0$

Yes, I understand that ∫∥f∥≥∥∫f∥ follows from the triangle inequality.

Okay, what are $h$ and $f$?

and why are you taking a limit?

Do you know how one defines the integral ?

f is a function

Yes, with upper and lower darboux sums and supremums and infimums

So you have sums of terms of the form $h_if(x_i)$ in your definition of the integral ? And you make the $h_i$ go to 0 right ?

I mean I'm just arguing that $\int \Vert f\Vert \ge \Vert \int f\Vert$, but you said you understood that so I guess there it's okay

But I don't see how the problem follows from $\int \Vert f\Vert \ge \Vert \int f\Vert$.

Instead of $f$, take $\gamma'$

Yeah, I did. And it seems like your relying on a MVT for vector-valued functions, but there is no such theorem.

Oh....

The FOTC, which does hold for vector valued functions, tells us that $\int_{a}^{b} \gamma ' (t) = \gamma (b) - \gamma (a)$.

I thought you were using the MVT for some reason, but you're just using FTOC.

Right

Can you tell me how you prove that the sup is indeed the arc length (and not lower than it) ?

Oh, because I can find a sequence of points in the set converging to $arc(\gamma)$. But I had to use a relatively deep theorem to conclude that.

Which one is that ?

I can't type it up at the moment (busy typing up HW at the moment), but when I finish I can ping you with the theorem.

Sure, thanks

if i am in a group of 20 people, and there's a 3/20 chance that there's a criminal in the group

knowing that i am not a criminal

does that mean there's a 3/19 chance that there's a criminal in the group?

if there are two teams of 10, and there's a 1% chance of someone in the group of 20 being a criminal

knowing that i am not a criminal, is it more likely for the other team of 10 to have a criminal?

Are you saying: If I had an arbitrary group of 20 people, not necessarily including myself, then the probability of the group containing a criminal is 3/20.

And then you want to see how the probability changes if you know that one person in the group (yourself) isn't a criminal?

Hi chat.

Hi @Albas

What's up

hi chat

here's a boring question about convex polytopes

suppose I intersect a (bounded) convex polytope with a hyperplane. (In my case of interest, it's a convex hull of 14 points in R^6 and I intersect it with a 3D hyperplane)

the result should presumably be some 3D convex polytope

It is

(i mean, you could probably find some non-generic cases where the dimension is less than 3, analogous to intersecting the closed disk with one of its tangents. but that's too annoying)

Yeah fair. For most hyperplanes it's dimension 3

Right.

What I'm trying to figure out is how to characterize the vertices of this 3D polytope

My thinking was to look at all the edges in the 6D polytope (i.e. convex combinations of two vertices), intersect them with the hyperplane, and take the convex hull of that

That seems like a plausible approach, but I can't decide if it's airtight

(For instance, is it possible that one of the convex corners would lie on one of the faces of my 6D polytope rather than one of the edges? I don't think so but I'm not certain.)

Mostly I'm hesitant because the family of cross-sections I get (if I generalize the hyperplanes slightly) is seemingly just a set of simplices

@Semiclassical I think that's possible. Note that a 2-dimensional and a 3-dimensional affine subspace of R^6 can intersect at a point very easily.

yuck

I mean, x1x2 and the x3x4x5 subplanes of R^6

My mental picture is that your R^3, the slicing plane, hits one of the 2 dimensional faces of the convex set and just picks up a point from it's interior when it intersects it

But I am too lazy to think about it seriously

lol

Having trouble figuring out how to show that the intersection of a closed and open subset of a Hausdorff, locally compact space is also locally compact. I think it's because I'm having trouble seeing why "the intersection of a closed and open subset" doesn't just cover "all subsets."

Because the whole space is clopen, so any open or closed subset intersected with it counts. If the space is not open or closed, then you need to require that it can be written as the intersection of an open subset and closed subset, but why does it fail to be locally compact if you can't do that?

For a real orthogonal 3x3 matrix it’s gotta have 3 eigenvalues

I can get that $\lambda = 1,-1$ but can’t figurre out what the other will be?

i was considering roots of unity but that’s just the same

I’m only getting 2 :/

ahyone got an idea

@JakeRose Consider the three-by-three identity matrix.

Yes

Is it real and orthogonal?

Yes

What are its eigenvalues?

1

-1

0?

Remember the definition of eigenvalue. You're looking for whether there's a nonzero vector $v$ such that $Mv=\lambda v$

and in this case, $M=I$

So you're looking for solutions to $v=\lambda v$.

Not a lot of options there

Is it just 1?

That's the only one possible, yeah.

Moreover, what's the characteristic polynomial of $I$?

The paradox is that a real orthogonal 3x3 matrix need not have 3 distinct eigenvalues.

shhh

$(1-\lambda)^3$

@JakeRose right. so $\lambda=1$ is a root of multiplicity 3

Mhm

Hence the identity matrix has eigenvalues 1,1,1

so in this question I’m doing

it says something along the lines of, show the 3 eigenvalues are $1,-1, e^{\alpha i}$

hmm

so is alpha $n\pi$

I don't think that's true in the obvious way.

If you had a matrix with eigenvalues $1,-1,e^{\alpha i}$, then their product would be $-e^{\alpha i}$

But the product of the eigenvalues should be the same as the determinant

and the determinant of a real matrix had better be real

(Complex roots of a real cubic comes in conjugate pairs, else all the roots are real)

Right.

$\alpha = \pi$?

What is your matrix anyway?

(Could I ask you a small question about special relativity afterwards Semi ?)

havent got one

What's the question

Word by word

@Astyx go ahead

@Astyx I’ve done some SR I also may be able to help

How did things turn out with you sen

@BalarkaSen would you prefer a screenshot?

Sure

m.imgur.com/6FuNDMU

Done the first bit

Wait actually I'll give it more thought on my own, otherwise I won't be able to make sense of my question

Note that their wording allows you to have an eigenvalue set like $1,e^{i\alpha},e^{-i\alpha}$

In which case the product of the roots is 1 and you don't run into issues with reality

Is the point of this being

you have 3 roots

$1, -1 $ and an exponential which will be either of the first two eigenvalues?

no

If two of the eigenvalue are 1 and -1, then the last eigenvalue had better be real as well

thats what I mean

okay

as in $\alpha = \pi$

do you follow?

Or 0 I suppose

No. You can have real orthogonal matrices with complex eigenvalues.

but for 3x3 they’re not

What are some examples of real orthogonal 3x3 matrices that you know, @Jake?

bold of you to assume I know any

As an example, take $$M=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1\\ 1 & 0 & 0\end{pmatrix}$$

identity as previously said

@Jake You should. Rotations.

we honestly don’t use them that much yet

Actually I have another very basic question : The referential you define does not care where you are, but only what speed you're going right ? My referential is the same the one of someone who isn't moving according to me

more of next terms stuff atm I think

@Astyx eh. Your choice of reference frame determines your origin

In any case, without even writing down the matrix, most rotations have exactly one real eigenvalue (the eigenspace is the axis of the rotation)

The two other eigenvalues are complex.

OkY

So if you translate your reference frame without changing the velocity, you'll shift all your position coordinates accordingly

But appart from that, distances and durations will be the same

@Astyx same reference frame, different coordinates in said frame

@Astyx in one frame yes

Right. And only changes in position/time are measurable anyways

So in my case how do I show that the 3rd eigenvalues is the alpha term?

brb lemme show you guys what I did

to be clear: you can have complex eigenvalues even in the 3-by-3 case.

imgur.com/Gnpt4L9

I've read that $c^2t^2 - x^2 - y^2 - z^2$ does not depend on the reference frame for an event, what does that mean for changes of coordinates ? Does the time of the event change ?

I can see that you could , but I don’t get how you would work them out from what I’ve done if gym?

@Astyx $c^2 (\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2$ is frame-independent

Note you really should write it with $\arcdelta$

Otherwise you could shift coordinates by a constant amount and it'd no longer hold

i have no idea how to do a capital letter

\Delta

why infinite copies of field has dimension 0?

What's $\Delta x,y,z,t$ ?

$\Delta t = t_f-t_i$

more obvious than it seems apparently thanks

Now what does frame independant mean ?

@Astyx SR deals with intervals. Without specifying that it is $\Delta$ you’re implying there’s an absolute time, or absolute reference frame which there isn’t. Also, everything you talk about is an interval, so you should write it accordingly.

Three things. First, the above is obviously invariant under translations of positions/time

@Semiclassical you can probably say that better than me

@BalarkaSen (if you’ve looked at my work) is anything immediately wrong?

Second, it's also obvious that the above expression is unchanged if you rotate your position coordinate system

since $\Delta x^2+\Delta y^2+\Delta z^2$ is just the square of the distance between the two points

@Semiclassical Will that be invariant in non-inertial frames, too?

Ok so it's just that your $\Delta x$ are my $x$ etc

@Astyx I mean yes. But you really should denote them like semi did

I believe you, it's just not what my teacher did

Well, you can also assume that $x_i=y_i=z_i=t_i=0$

@Astyx I had the same problem before undergrad

@Semiclassical seems like a slippery slope

What are $t_f$ and $t_i$ reffering to ? Is $i$ where the event starts ?

$i$ is the initial event, $f$ is the final event

@Semiclassical could you expand on how you get the third eigenvalues?

I really don't get what you're asking.

If the first two eigenvalues are $\pm 1$, then so is the third.

But that's not the only case allowed here.

I don’t understand where the exponential term really comes from

Right I get what I missunderstood now

Is that a real orthogonal matrix?

Because we assumed that the event started at the origin, and it kinda confused me

Yeah. In general you don't want to do that

Anyways. The third property that expression has is that it's preserved under boosts i.e. Lorentz transformations which change the velocity of your reference frame

Yes

@JakeRose Is that in reply to me?

Right it all makes sense now

Cool, thanks a lot

@Semiclassical yes

Okay. What is its characteristic polynomial, and what are its eigenvalues?

@Semiclassical I get the idea generally. I think the root of my problem is that I took a transpose instead of a hermitIan conjugate, but I don’t fully understand why this makes a difference

What?

Hermitian conjugate and tranpose are identical for a real matrix.

Exactly

but if you use hermitIan conjugate it gives modulus =1 straight away and includes complex eigenvalues

if you look at what I’ve done, then the resulting expression for the eigenvalue is just $\lambda^2=1$

i don’t know how to get that third exponential term

I guess? I think complex conjugation is sufficient there tho: $Av=\lambda v\implies (Av)^* =Av^* = \lambda^* v^*$

So if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $v^*$ is an eigenvector of $A^*=A$ with eigenvalue $\lambda^*$

Oh fair

I see you've got $Q^\top Q=I$, which is fine

How are you getting to $\lambda^2$ tho?

set y=x

what

just let the other eigenvector be equal to x

what?

So I’ve set it up with two eigenvectors

x and y

if you let y = x then you get the expression

(And hence $\mu = \lambda$

Write it out.

write what out?

On here?

Your argument. Don't use words when symbols are appropriate.

I’ve wrote it on my work

it’s just cumbersome to type latex on iPad

oh, I missed the second link

So you've got $Qx=\lambda x$ and $x^\top Q^\top = \lambda x^\top$

@Semiclassical sorry I should have linked you

And therefore $\lambda^2 x^\top x = x^\top Q^\top Q x = x^\top x$

Yes

Okay. Let's go back to the example I gave before with $M$

that was a real orthogonal matrix. What were its eigenvalues?

1

just 1?