Hello!! I have posted a question about probabilities, does someone have an idea?

Could someone of you explain to me if it is the probability of intersection or theconditional probability?

Hi! What does it mean by from "Note that the bounds for the relative error when using k-digit rounding arithmetic" to "is constantfor all integers $n$"?

It says k digit arithmetic is independent of number being represented but what number?

I didn't understand result due to manner in which machine number are distributed but what distributed?

Exponential form of characteristics ? ??

I am confused by this description is there simple way to explain this with examples...

and thanks for attention in advance

umm...

;_;

alright

what do you do when you are learning math topics and every topic stops one by one?

meaning every topic makes you feel hard

@BalarkaSen oi

(^ from Hatcher)

$\alpha \cup \beta$ has been deleted from $D^2 \times I$.

Go on

Forget about $\gamma$ for a while.

Is the space homotopy-equiv to a 3-ball minus two disjoint diameters?

...which is then homotopy-equiv to a solid double torus.

What does this homotopy do to the $\gamma$? I think it takes $\gamma$ to a loop around one of the holes. Does this make sense?

lol

I agree with the double torus part, but where $\gamma$ goes depends on the homotopy you have in mind

I mean it has to be around one of the holes right? I can't see it going around both of the holes or going around a hole twice.

What I mean is, if I further def. ret. it to a figure 8, then $\gamma$ will either be $a$ or $b$. It can't be something like $ab$ or $ba$ or $a^2$ or $a^2b$ etc.

But clearly gamma has zero linking number with alpha and beta

The word it goes to should be a commutator

Why should it be a commutator?

It is not

$\gamma$ goes to a loop around both of the holes.

The point being that linking number in complement isn't well behaved since this isn't a closed manifold?

Oh no maybe you're right I was hasty

I dunno man draw a picture

Mike is probably right, linking number is zero, there's something to think about @feynhat

I am not going to think though

Oh it's clearly null homologous right?

Push it to the right, you bound a disc with two holes around the beta strands

Then fill in that pair of discs with a tube going around the beta loop

Ya

A torus bounds it

So gamma bounds S_{1,1} disjoint from alpha and beta

So it's a commutator

I mistranslated $\gamma$ in my head while doing the homotopy to the genus 2 handlebody

It winds around both the holes, in opposite orders

$aba^{-1}b^{-1}$

I got a Wacom tablet

Best thing I have ever spent on

I was expecting to see some sick drawings of that homotopy

If $A$ is a set with two elements, is there a way to know how many members are in the sigma algebra generated by $A$?

Seems not obvious

what about the set $A=\{ $\{$ $4,5,6$ $\}$ , $\{$ $5,7,8 $\}$ $\}$

16, right?

over $\mathbb{N}$ and over $\mathbb{R}$

Does anyone know about any online study/reading groups in elementary number theory or in abstract algebra?

@orientablesurface Every set they generate is a union of {5}, {4,6}, {7, 8}, and X - {4,5,6,7,8}, and you can write it as a union of those in a unique way, so I do get 2^4 = 16

So that's probably the answer for an arbitrary pair of sets with non-trivial intersection

If they intersect trivially you can only generate 8 things

This would hold for any two 3 element sets with a singleton as their intersecition, right?

@MikeMiller

The fact that there's a singleton as intersection is irrelevant

Just that $A \cap B$ is nonempty

If you're generated by two sets $A$ and $B$, then every set in the sigma-algebra is a unique union of unions of the four disjoint sets $A \cap B, A \setminus B, B \setminus A, X \setminus (A \cup B)$

Unique at least unless one of those guys is empty

Oh alright, I see

If $A \cap B$ is empty then you only have the other three generators (8 elements), or if $A \subsetneq B$ or $B \subsetneq A$, or if $X = A \cup B$

@BalarkaSen Too hard. But I think I get it now. I will try to make a cartoon of that homotopy and post it here.

Yo Bal. Put that Wacom to use.

(I actually don't know what it really is. I just assumed that its something that lets you draw stuff with a stylus.)

f(x)+f(y)=f(xy +sqrt((x^2-1)(y^2-1))) find f(x)

anyone remember how to approach these kind of things?

im trying to maybe complete a square algebraicly

maybe x=cost , y=sint gets me something

Is the Poisson bracket as a $\Bbb{R}$- bilinear map over $C^{\infty}(M)$ a non-degenerate bilinear form?

@SayanChattopadhyay What sense of non-degenerate?

Also I assume $M$ is symplectic?

Nope just a Poisson manifold

I mean $\{-, -\} = 0$ satisfies the axioms

Non degenerate in the bilinear form sense. If (X,Y) = 0 for all Y then X = 0

Yeah my point is that there are two natural notions of "nondegenerate bilinear form" on say a topological vector space

One is that the map $F: V \to V^*$ given by $F(v)(w) = (v,w)$ is injective (your notion). One could also demand that $F$ is an isomorphism

Anyway the zero bracket seems to give a counterexample and I'm sure I could cook up similarly trivial ideas

Oh not the isomorphism one. That would be overkill for my purposes.

Hmm I see.

Also doesn't the Poisson bracket vanish on constant guys

So the point is that I have the thing that $\{ \{H,K\}, L \} = 0$ for all L and I want to conclude something about the bracket of H and K

I think that's only true for like the "usual" Poisson bracket which is defined on a symplectic manifold (the non trivial Poisson bracket)

To explicitly state the problem, I have two vector fields $X_H = \{H, \cdot \}$ and $X_K = \{ K, \cdot \}$ and the lie bracket of $X_H, X_K$ is given to be 0. I have to show that the Poisson bracket of the functions are 0

Maybe I am taking some wrong approach and arriving at this

This should really be true for arbitrary Poisson structures? Surprising

Hello. Given $n \ge 2$, I am trying to define a short exact sequence $0 \rightarrow \Bbb{Z}_n \rightarrow \Bbb{Z}_{n^2} \rightarrow \Bbb{Z}_n \rightarrow 0$, but I am having trouble. I first tried finding a map between $\Bbb{Z}_n \rightarrow \Bbb{Z}_{n^2}$ and from this find the other map, but I was unsuccessful. I then tried the same for the map from $\Bbb{Z}_{n^2} \rightarrow \Bbb{Z}_n$.

Yeah it's surprising to me as well.

For example, for the first map I tried $z \mapsto z$ and $z \mapsto mz$, but neither worked.

I guess think of the bracket in terms of its action on functions

I can't seem to also find a counterexample

Yeah this should work

$$[X_H, X_K]f = X_H(X_K f) - X_K(X_H f) = \{H, \{X_K f\}\} - \{K, \{X_H f\}\} = \{H, \{K, f\}\} - \{K, \{H, f\}\}$$

Last term by antisymmetry is the same as $\{H, \{K, f\}\} + \{K, \{f, H\}\}$ and then by the Jacobi identity this sum is equal to $-\{f, \{H, K\}\}$

Exactly what I did

How do you conclude after that?

Ah I see your point

I think all you can conclude is that $X_{\{H, K\}} = 0$

Which is true by definition

Yep

If I make the manifold symplectic, does this hold then?

I'm unsure but skeptical. I bet that you can cook up examples of nonconstant functions whose Poisson bracket is constant

Hmm yeah, this is very weird then.

Are you sure it wasn't just a misstatement

The correct statement should just be that $[X_H, X_K] = X_{\{H, K\}}.$ That seems true to me in general by the proof we just gave

It was part of a homework problem set. I should maybe check once again, even though I have done it before.

@SayanChattopadhyay I do notice that {1,f} = 0 in general though. Constants are always in the kernel

Apply Leibniz to 1 = 1*1 to get that {1,f} = {1,f} + {1,f}

Oh yes

Here's the problem @Mike, the definitions here are that by a hamiltonian dynamical system they mean that $\{H, \cdot \}$ is complete and infinitesimal symmetry means that $L_{\xi_H}(\nu) = 0$

Any input on the problem I am working on? It should be simple, but I just don't see how to define the maps.

@user193319 Note that $\bar n = \bar 0$, and so $\bar n$ has to map to $\overline{n^2} =\bar 0$.

@TedShifrin So, are you saying that the first map is indeed $z \mapsto nz$?

Yes (you had written $m$ earlier).

To be more pedantic, we're saying $z\pmod n$ maps to $nz\pmod{n^2}$.

Oh, whoops. So $z \mapsto nz$ is injective; I must have made a dumb error. Let me recheck my work.

@SayanChattopadhyay This is correct though

$L_{\xi_H}(\xi_K) = \xi_{\{H, K\}}$ as we said

@user193319: It is totally important to work with equivalence classes, not integers.

Hi @MikeM and @Sayan.

So $\xi_K$ is an infinitesimal symmetry iff $\xi_{H, K} = 0$

Oh no I see the problem again, maybe $\{H, K\} \neq 0$ but $\xi_{\{H,K\}} = 0$. Whatever

I'm content to not worry about this irritating edge case

Hi Ted

Yeah

Hi @Ted

@SayanChattopadhyay It's possible that $\{H, K\}$ is never constant nonzero but I dunno how to prove it nor do I care

Yeah, thats cool. Let me see if I can cook something up. Thanks for the help though :)

This is a relatively logistical question. Say if I have to explain my thoughts to someone through an email and it involves mathematical equations. What is a better thing to do, link a pdf file with whatever I want to say (its not a lot in this case) or explain it on the mail itself (can LaTeX be enabled)?

Do you use chrome

Yeah

Use the browser extension TeX for gmail. It turns your tex into images and attaches them

Recipient doesn't need any extensions and it usually works great. Sometimes people can't see the images but I'd say only one every 20 that happens.

Neat, didn't know that. That's pretty good

I don't read my math email (or even my gmail, generally) on Chrome.

Doesn't cause a problem for people sending you messages using this extension, though, since it just turns the TeX into pictures.

I'm enough of a TeX expert that if people send me a moderate amount of LaTeX code in an email, I can interpret it with my brain. If it's long and convoluted, I'd rather paste it into LaTeX or have someone send me a pdf.

It does mean you can't use it but that's OK. :)

I agree with that ted

Unfortunately, relatively few of my students learned LaTeX so they might not be able to interpret my answers to their emails in LaTeX :D

Just look at the number of people who post in MSE (or in the chatroom) who don't know basics.

I tell my kids they either need to write legibly and scan into a PDF or they need to learn to TeX. Either is ok with me.

Do they need to write more legibly than you do? :D

I prefer TeX, but people seem to prefer scanned stuff here. Don't know why? Maybe nobody wants to learn

Probably! That's why I TeX.

I have office hours in 10, but it's the two classes back-to-back, and I'm never sure which office hours I'm doing until someone pops in.

@Sayan: That would have made sense to me 10 years ago, but no longer.

@MikeM: I used to have students from both my classes simultaneously almost all of the time. I don't know how I would deal with that on Zoom.

It was really fun the few times I taught 3 classes in one quarter.

Yeah

That's ideal for having more people (since sometimes one of my classes doesn't have a lot to ask about) but it would be hell in Zoom OH.

I'd have to set up breakout room type things. Awful.

Ah, Zoom allows for that? Makes sense.

It does, but I think it's not worth the time.

Are the unis in US planning to open anytime?

Here they seem to be planning for even the next year

My ex-university is open (i.e., having in-person classes, although not all), and the faculty (and most of the students) are up in arms about it. The government of the US is trying to kill off the citizens.

Yeah, I did see how Trump chums up with anti maskers. Whereas ours seems to be busy feeding peacocks.

Do you see a simple way to show that $\Bbb R^{\Bbb R}$ (product/pointwise convergence topology) is Baire?

I might have had an answer for you the last time I taught topology, demonic @Alessandro, but certainly not now.

I hope this question makes sense. If (v_1,\dots,v_n) is a basis for a free module M over a comm R with 1, then it can be shown that (v_1,\dots,v_n)C is also a basis if C is invertible matrix over R. Now my question is, what exactly are we "doing" (e.g. geometrically) to the set of basis (v_1,\dots,v_n) when we multiply it by an invertible matrix? If we expand out the product, we are just summing up the basis elements, but I feel like we are a bit more than that.

Interesting, demonic @Alessandro. Much more is true. Munkres has an exercise that $\Bbb R^J$ is Baire for any index set $J$ in all of the box, product, and uniform topologies.

I mean it's neither locally compact Hausdorff nor completely metrizable so the BCT doesn't apply

$J$ is irrilevant because any product of Cech-complete spaces is Baire, but was looking for a less nuke proof for this specific space

Well, the box topology surprises me, doesn't it?

@Hawk: It's essential that $C$ be invertible. Understand why.

@TedShifrin no i know it has to be invertible. i know the reason for the statement, but i want to know what is it doing

Work out some specific examples if you don't see it.

You should see that linearly independent sets map to linearly independent sets AND back again precisely when you use an invertible matrix.

i just realized this is just a change of basis

that's what the invertible matrix is doing

@TedShifrin Yes, I think I have an idea for showing that it is Baire, but I'm not sure whether it works

I wrote that the empty set is not the union of countably infinite many intervals of $\mathbb{R}$ and i've been told that is wrong?????

It's the empty union

But is the empty set countably infinite?

I take the index set to be countably infinite :P

votes with orientable

$\{\}=\bigcup\limits_{k=1}^\infty(0,0)$

The empty interval is not an open interval :D

Oh good grief; technically, it is.

yep

clopen

This is why students hate mathematics.

When I took point-set topology from Munkres, he appointed in each class a custodian of the empty set. He was writing his book and that person was in charge of making sure that the empty set was not a counterexample to each statement as it came along.

Null watch

We had a class with something like 75-80 students (shocking, I know), and I escaped the assignment.

each day a new custodian?

No, no, although my comment suggested that. There was a fixed custodian. I have no recollection who was in the class or who was custodian.

Ah, so not so surprising you escaped.

Although I'm sure my students thought me pedantic, I think that I probably only earn a C grade in being a mathematical stick-in-the-mud.

What is pedantic can be very contextual

I agree. I'm not saying that I am/was sloppy. I'm just saying I wasn't as interested in nitpicking when it adds no understanding.

I think a few people have caught errors of that type in my YouTube lectures.

Yes. The user Did would comment on some of my answers, prompting me to add details where I had thought them unnecessary. It cluttered up the answer to the point that it was no longer understandable, but it was more complete. I liked the understandable answer better.

I would always support that. Of course, what's "understandable" depends on the audience. But pedagogy is not always a strong point of math geeks.

I've taken a better-of-both-worlds approach; I write a hopefully understandable proof, with explanation below a lot of lines that can go deeper.

Sort of like footnote to amplify in a text. I've done something like that a few times, too.

a lot of my answers have a similar appearance because of that

@TedShifrin precisely

I sometimes have a final paragraph for "in case you wanted more detail above ..."

Yeah, I'll often add a section below with more detail after the main event.

But I'm notoriously bad at writing incomplete answers, especially when I think that my answer will be used to cheat. For example, I just wrote this, and I'm sure some people will bitch that that is a comment and not an answer.

@TedShifrin Do you see immediately how to give the volume of a spherical wedge explicitly, geometrically (instead of integrating)? Stewart invokes MVT on the way to approximating, but I don't like it.

Oh, you're talking about the approximate parallelepiped for the volume element in spherical coordinates?

You want something more rigorous analogous to the area of the sector computation in polar?

Yeah

I confess I never have bothered to look for that, but let me think a moment.

I see how to do it by integrating, though that's too much of a pain to do in class. I'd rather not just say "By geometry, the volume is roughly blah blah blah" though.

Like a sphere between two longitudes?

I just draw the box and point out that the three coordinate directions are orthogonal, so no big deal.

You're using the Jacobian picture, right?

I was going in the other order. Present cylindrical, spherical first; then unify

Ah, @robjohn has it. You need to know the surface area of a zone of a sphere.

I did that with middle school students. You can do it with similar triangles totally with no calculus.

That is how Archimedes did it :P

Interesting, though probably also too much of a tangent to do in class.

I actually have a handout I wrote for that presentation. Do you want it? Yes, too much of a waste of time for calculus. The area of a zone often is done as an exercise in calc II.

Sure. I don't know how to do it with triangles.

I always did it as an application of surface integration in calc III.

It's clear enough how to do it with calculus provided Fubini.

It took me a while to figure it out. I first did this in a math circle presentation at UGA years ago.

Shall I paste it in here or email you the pdf?

Just email me the pdf.

I probably don't have your current email.

smm2344@columbia

smike.smiller@gmail in perpetuity

Now I do.

I think I last used ucla.

Sent.

I will probably write down the actual volume of the wedge and then point out that for tiny tiny wedges you can approximate the volume by the volume of a box w/ corresp sidelengths and factor rho^2 sin(phi) in front of it. You get to use linear approximations to explain why cos(phi) - cos(phi') ~ sin(phi)(phi - phi'). And linear approximation is the idea you use to get the formula in general.

I get the $\sin\phi$ factor from $r=\rho\sin\phi$.

The $\Delta\phi$ needs no fixing.

I think your approach sounds less geometric. :P

It does, I gave up on explicit geometry. I'll give an intuitive explanation of why the formula makes sense. Note that the formula for volume of a spherical wedge between angles phi, phi' involves a cos(phi') - cos(phi)

Invoking linear approximation here is a good way to hint that's the right approach in general, IMO.

Right, that's robjohn's hint. Just not the way I would choose to do it in a calc class. Engineers should think about the little parallelepiped.

By which you mean the linear approximation given by Df(box), f being the transformation from spherical to Euclidean, yes?

I just vary $\rho$ by $\Delta\rho$, $\phi$ by $\Delta\phi$, and $\theta$ by $\Delta\theta$ and draw the almost-parallelepiped.

The $\sin\phi$ term comes in the $\Delta\theta$ length.

I think we're saying the same thing in different language

@MikeMiller glad it helped. I had to go for a bit after I made the comment, sorry.

I actually don't believe that. I'm never using the area of a zone.

I'm just taking plain ol' $\rho\Delta\phi$ for that dimension.

I'm saying what you're doing is the Jacobian computation in different language

No?

I'm never mapping a rectangular box, no.

Then I don't understand your picture, but without a blackboard between us, this is probably hopeless

In polar coordinates, I'm not, either. I'm just looking at the area of a $\Delta r\Delta\theta$ wedge.

Hmm, I can find the graphic I drew for my book and paste it in here. Hold on.

this image is useful in showing that the surface area of a sphere is equal to that of a cylinder tangent to its diameter. That should work for a sector.

Oh god.

and the equality is good between any two latitudes

and longitudes

project from the axis onto the cylinder and the area is equal

That's what I'm doing.

You've all convinced me I'm no geometer

@robjohn: Yes, I love the fact that that projection is area-preserving. I pointed that out in a recent answer. However, I don't think students know that before you assign it in a diff geo exercise :P

But the geometry proof of it, @robjohn, is what I just emailed to Mike, as Archimedes knew it.

(This is written for exceedingly curious/smart middle-schoolers or high-schoolers.)

Yeah, that image is from this answer

Gotcha @robjohn. You and I should just combine forces; we're so wasteful :D

I no longer have Illustrator to make those pictures, dammit. Oh, I mentioned Inkscape to you, but then I think you subsequently asked me what the app was when I was gone.

I used to use Intaglio for my illustrations, but now I use Mathematica since I wrote a small package to include LaTeX in my illustrations.

I never needed to do that, because when I used Mathematica I could export as .eps and then use LaTeXiT to paste into Illustrator. I'm too old to learn new tricks.

I have stopped using Adobe products (other than Reader) because of their subscription licensing.

I used to use Photoshop a lot for my astrophotos, but not any more

Yes, we already complained about this.

Anyhow, you had asked what I'd found to replace Illustrator, so check out Inkscape.

Ah, I will look at that, thanks!

So is the empty set the union of countably infinite many intervals? $\varnothing= \bigcup_{n=1}^{\infty}(0,0) = (0,0)$ no?

Sure, that works

Yes, it would seem so, if you allow empty intervals. Ugh.

@TedShifrin I wonder if it will use the photoshop plugins I have?

Or the empty union!

So I sit corrected.

@robjohn: I have no earthly idea.

I think the more important thing to recognize is that empty unions are unions, too, as are empty intersections

The usual axioms on the empty set and the total space in a topology, for instance, are really instances of the other axioms

@MikeMiller but isn't the union above not the union of countably infinite intervals? When someone sais countably infinite intervals, I take it to mean countably infinite $\textbf{different}$ intervals

Yeah but that's wrong

@TedShifrin Oh, I missed that Inkscape replaces Illustrator, not Photoshop. I never used Illustrator.

Oh, sorry.

No, thanks!

I appreciate the info

Hi all, I was wondering if the arc length formula in calc is exact or proven to be exact?

How do you professors teach analysis in rudin. His proof is bloody irritating since I always need to re-engineer the proof and it seems he works backward. Deriving the proof is like sailing a boat without compass but sometimes you get to your destination.

The first 20 pages are exhausting. But I have seen much more simpler proof than his one.

No, his proofs aren't backwards. It's just you don't know where he comes up with the stuff he starts with, but logically it's all forwards.

I do not recommend Rudin unless one has a lot of prior experience with proofs (e.g., Spivak's calculus and/or an abstract algebra course).

It sounds odd that you are recommended to have experience with abstract algebra course before Baby Rudin.

Is the set consisting of all subsets of $\mathbb{R}$ that are the union of countably infinite many intervals a sigma algebra?

I didn't say need, but Rudin is about the hardest text there is for real analysis.

Algebra has much simpler sentences with quantifiers than analysis does.

Somehow math students have this macho idea that one must do Rudin to be a real math student.

He does have good exercises, but one needs a lot of experience. (A students coming out of my rigorous multivariable math course were able to do Rudin, but my course pushed them hard.)

@orientablesurface: So the intervals need to be open or can be closed, half-closed?

@TedShifrin Yeah your YouTube video on multivariable and your book :P

they can be any @TedShifrin. They can be open, closed or half-closed

Right, @Stupidquestioninc.

What do you think the answer to your question is, @orientable?

I think the answer is no @TedShifrin

Why?

well, I don't see a reason why the complement of a union of countably many intervals to be the union of countably many intervals

What is the complement of a union?

intersection of complements

OK, so is the intersection of countably many intervals a union of countably many intervals?

no

because of the cantor set

Oh, that's a good answer.

So if you end up with a totally disconnected, uncountable set you lose.

dust

never mind

a interval has lebesgue outer measure 0 if and only if it is the empty set or a singleton

right?

yes