Mathematics - 2016-10-23 (page 1 of 5)

12:00 AM

@TedShifrin They're pretty useful when all you want are partials.

In multivariable analysis we make a big deal about how being differentiable is so much stronger than having all directional derivatives.

For example, I can give you a function all of whose directional derivatives are 0 at 0, and yet it's not even continuous at 0.

Axoren

@TedShifrin You mean having all of them but they're not equal, right?

Oh

That feels like cheating.

Ted Shifrin

But partials are useless unless you know the function is actually differentiable in a stronger sense. You won't get linearization.

No, it's not cheating. It just shows you that in more than one variable you need the right definition to get reasonable results.

DHMO

What do you guys think about this: in chem.SE, we forbid the use of $\LaTeX$ in question titles because $\LaTeX$ messes up the searching.

Axoren

A function like that which is just a bunch of rays starting at $0$ with random heights $h_\alpha$ and pointing outward at angle $\theta$ for each angle theta.

syzygy

12:03 AM

@DHMO sounds reasonable

Ted Shifrin

No, @Axoren, if they have random heights you won't have directional derivatives.

Axoren

Since all the rays are flat, approaching $\vec 0$ from those directions would be gradient $0$

DHMO

Do you guys think math.SE should be subject to similar ban?

Ted Shifrin

You have to have them anchored at $0$. But still won't be continuous.

Axoren

Oh, way, nevermind.

That would be $(0, h_\alpha)$ not $\vec 0$

Ted Shifrin

12:04 AM

If you have time to kill, you can browse through some of my lectures on YouTube and see lots of discussion of these things, @Axoren.

syzygy

@DHMO while I don't feel very strongly about this, I think it would make sense. I mean surely you can give a descriptive titles without using a lot of LaTeX

Axoren

I might work in the living room tonight and chuck you on the TV.

Ted Shifrin

Agh.

DHMO

@syzygy but the situation is different in math.SE; sometimes your question is just a specific integral and the question itself would be more informative if put in the title

Ted Shifrin

The titles should be clear enough for you to guess which ones :)

Axoren

12:05 AM

Are you saying Agh because you can't tell which ones?

Or because you don't like being a TV star? :P

Ted Shifrin

No, I said Agh with the thought of TV ...

syzygy

well that's a good point, but these questions are usually not the ones that at least i would look up :D

Axoren

Generally, I only use TV with guests.

Ted Shifrin

@DHMO: In general, I think titles should be more descriptive and people should not put the question in the title. It's particularly bad, though, when the question in the title and the question in the body disagree.

Axoren

I have one in my room, but it's opposite my PC so I can't even watch it while I work

DHMO

12:07 AM

@TedShifrin what if the question is just "find the following integral"?

Ted Shifrin

Still use some imagination in the title and don't just put the integral there ... imho.

syzygy

you can describe the integrand

Ted Shifrin

Maybe something like "I tried substitution and integration by parts, but ... to no avail." :P

syzygy

perhaps it is a rational function or something more strange

DHMO

but there is no title more informative than the actual integral

syzygy

12:08 AM

that's true

Ted Shifrin

The title is not meant to be the question. That's clearly in guidelines.

But a title like "Do this integral for me" would not be ideal.

DHMO

so we should reform most of the questions XD

Ted Shifrin

I think you'll find that most don't have the statement of the problem/question literally in them.

syzygy

there should be thread collecting all these integrals

would be easier

Axoren

Indexing integral threads by their coefficients?

Ted Shifrin

12:10 AM

It would be good, in general, if things got grouped so that most question-askers would at least search before asking their question. Soooo many questions appear numerous times.

syzygy

in any case it is not my business since i never look at these threads. this is exactly what i mean @TedShifrin

so much repetition

no i mean by the form of the integrand axoren

Ted Shifrin

Agreed.

Axoren

Google's getting really good at anticipating google searches depending on if you share your browser info with Google.

syzygy

i figure most integrals must be special cases of reasonably general integrals

Axoren

Maybe it could anticipate homework questions and link to duplicates before they become duplicates.

syzygy

12:11 AM

at least the easier ones (which make about 95%)

Ted Shifrin

Well, pedagogically speaking, most question-askers can't deal with the ultimate generality.

Axoren

@syzygy You're not wrong. That is very likely the case.

syzygy

that is true

Axoren

At least, the ones we expect calc students to ask about

syzygy

but perhaps it might make sense to them when you tell them put a=2, b=10, etc.

Axoren

12:12 AM

I can't expect many people wanting the integral of $e^{e^{x^e}}$ or anything silly.

Actually, now that I mention it, what is that integral? I don't think it has a closed form.

Ted Shifrin

I was about to say Good luck with that.

syzygy

i think that would be closed as trolling :D

Ted Shifrin

Remember that even $e^{x^2}$ has no elementary antiderivative.

DogAteMy!!!!

Akiva Weinberger

Dr. Shifrin!!!

!

Axoren

Akiva's name used to be DogAteMy?

Ted Shifrin

12:14 AM

LOL, deleting the Dr. is OK too :)

DogAteMyHomework :)

Akiva Weinberger

It used to be columbus8myhw

(Incidentally, my dog's name is not Columbus)

Ted Shifrin

No, that was the name of your turtle who 8 your hw

Danu

Herpaderp

time to go to bed

Axoren

Night, Danu.

Danu

Curvature is up next

Then finally Chern classes (!!)

Ted Shifrin

12:16 AM

Night, @Danu.

Axoren

@TedShifrin or @syzygy Do you guys happen to know Fractional Calculus?

Danu

Bye, erryone

Ted Shifrin

Not I, Axoren.

Akiva Weinberger

Here's a nice puzzle; suppose you have a square pyramid and a tetrahedron.

The triangles of both solids are all equilateral, and the triangles of the square pyramid are the same size as those of the tetrahedron.

Glue the two solids together along a triangular face of each.

How many faces does the resulting solid have?

I actually have to go now, sorry

Axoren

@AkivaWeinberger 6?

DHMO

12:20 AM

@AkivaWeinberger originally you had 5+4 = 9 faces; you eliminated 2 faces; so you are left with 7 faces

Axoren

@DHMO You didn't take into consideration the angles.

DHMO

@Axoren but the triangular faces are congruent to each other

user340082710

Is there a way to request that a question be moved to a different stackexchange site?

Axoren

If I'm picturing it in my head properly, two pyramid faces adjacent to the glued tetrahedron are coplanar with the other faces of the tetrahedron

So two pairs of faces are actually turned into a single face each

DHMO

@user340082710 flag it for migration

Akiva Weinberger

12:22 AM

@DHMO Nope! Not seven

Axoren

I think I did the math wrong.

5.

DHMO

@AkivaWeinberger what does "along" mean?

Akiva Weinberger

@Axoren Yup. Why?

Axoren

My reasoning that I gave DHMO

Two of the tetrahedron faces become coplanar with adjacent pyramid faces.

Akiva Weinberger

Why are they coplanar, though

Axoren

12:23 AM

I dunno. That's what it looks like in my head.

DHMO

what??

you had 9 faces

how do you eliminate 4 faces by gluing them together?

Akiva Weinberger

Along means, put lots of glue on one triangle of each solid and put them together

DHMO

exactly

Axoren

So, you lose two faces because they're now internal to the solid.

Akiva Weinberger

So the two triangles with glue end up inside the shape, eliminating 2 of the 7…

DHMO

12:25 AM

@Axoren how?!

Akiva Weinberger

and then — the hard-to-visualize part — two pairs of triangles join together to form a single parallelogram each!

DHMO

how?

Axoren

@DHMO, draw an X on each of your palms.

Akiva Weinberger

How on the coplanar bit I think s/he means

Axoren

Then, clamp your hands together. You had 2 Xs showing, now you don't have any Xs showing.

OH

I dunno. I just put the pieces together in my head and the faces were in the same direction.

Those are easy objects to visualize, so I just did that

I guess you could find an angular reason for it.

But that's nasty. I don't know the angle between triangular faces of a pyramid.

Akiva Weinberger

12:28 AM

Hm. Name the faces. The square pyramid has a base on the bottom, a Front triangle, a Left, a Back, and a Right

Glue onto the R face

Right face

So the tetrahedron has a face that disappears inside, a thing connected to the front face (F'), a thing connected to the back face (B'), and a thing connected to the base

We need to show F and F' are coplanar, as well as B and B'

Axoren

I think I see why DHMO's confused and it's reminded me of a puzzle, as well.

Akiva Weinberger

Consider just the four faces F, F', B', and B, and ignore the rest of the solid.

Axoren

A right triangle has a base of 4 and a height of 3. What is the area of such a triangle?

DHMO

@Axoren 6

Axoren

@DHMO No such triangle exists.

Akiva Weinberger

12:31 AM

We can flatted those faces down on the ground to get four triangles, each connected to the next, getting a sort of part-of-a-hexagon-thing.

DHMO

@Axoren why?

@AkivaWeinberger still not convinced that they are coplanar

Akiva Weinberger

Now, if we pick it up to put the faces back on the solid, we only fold along the edge dividing the F' and B' faces

That's a hard argument to get through text, so I might want to share a different one

Axoren

@DHMO Because for a triangle to be a right triangle, it's base functions as the diameter of a circle on which it's right-angle vertex is on the edge of a circle with radius $\frac {base} 2$.

Because the base is the diameter.

DHMO

@Axoren ????

Axoren

Consider now that you had that vertex any lower, could that angle be 90 degrees?

DHMO

12:33 AM

you are saying that the hypotenuse is the base?

Axoren

Oh, my mistake. I had assumed that.

In the case where both legs are height and base respectively, such a triangle exists.

Lay it flat on the hypotenuse and no such triangle exists.

DHMO

@Axoren sure

Akiva Weinberger

Axoren

GAH! Akiva what is that monster.

DHMO

@Axoren the glued thing

Akiva Weinberger

12:36 AM

Heres a bird's-eye view. Black means z=0, White means z=1

Ignoring the bottom-right vertex, that's our square-pyramid

Axoren

Yeah, that looks right

DHMO

alright

Akiva Weinberger

Ignoring the black vértices on the top and the left, that's our tetrahedron

Axoren

It's weird in dots and lines.

Akiva Weinberger

The two edges that are partially drawn don't really exist because the faces on them are coplanar, I claim

They're in parallelograms, because opposite sides are clearly planar

(X and Y coordinates are all integers by the way, so that all edges are the same length)

48 secs ago, by Akiva Weinberger

They're in parallelograms, because opposite sides are clearly planar

^^QED

DHMO

12:39 AM

@AkivaWeinberger alright, I'm convinced

Akiva Weinberger

Sides means edges there

Axoren

This makes me want to get some wooden block geometric shapes.

Just to play around with

Akiva Weinberger

A confusing bit is seeing why the tetrahedron is a tetrahedron. It's essentially just standing on an edge.

Axoren

What happens when you do it with a pentagonal pyramid and a square pyramid?

Akiva Weinberger

OH GOD NO

Axoren

12:41 AM

Obviously, not with equilateral triangles.

Akiva Weinberger

You can do equilateral triangles. Why not?

Axoren

GL Drawing that, Akiva

Akiva Weinberger

Just not hexagonal pyramids and beyond

Axoren

Can a pentagonal pyramid have equilateral triangles?

I didn't think about it enough to be convinced.

Akiva Weinberger

because a hexagonal pyramid with equilateral triangles would be flat

But pentagonal should be fine

Axoren

12:43 AM

I'll take your word for it.

Akiva Weinberger

Wiki has a pic en.m.wikipedia.org/wiki/Pentagonal_pyramid

Axoren

Pentagonal pyramids aren't a simple enough object for me to think of without a visual aid.

Oh, it's a slice of an icosahedron.

Funky.

Akiva Weinberger

Ah, yes ^^

Axoren

Definitely convinced.

Akiva Weinberger

A similar dot diagram thing shows that a tetrahedron with side-length $\sqrt2$ has volume $1/3$ by the way

Linear algebra also does it probably

And an octahedron with the same side-length has volume $4/3$

Axoren

1:01 AM

This is just creepy. Thought I'd post this here since we were talking about signal stuff earlier: youtube.com/watch?v=FsVSZpoUdSU

A recurrent neural network with long-short-term memory model trained on 10 minutes of anime girls talking.

Simple Art

1:15 AM

@DHMO At that residue stuff, you did it right, and then you would have $$(1-e^{2\pi iz})\int_0^\infty\frac{v^z}{(1+v)^2}dv=-2\pi ize^{\pi iz}$$

Mussulini

is it okay to digress a bit on your answers?

DHMO

@SimpleArt oh, and then the complex definition of sine

Simple Art

Yup

DHMO

$\sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}$ right

Simple Art

We originally had $\Gamma(1+x)\Gamma(1-x)$ where $\Gamma(1+x)=\int_0^\infty s^xe^{-s}ds$

Multiplying the two integrals and using $u=s+t$ and $v=s/t$, you would get the above integral

which becomes all that

:D

Thus, we have

$$\Gamma(1+z)\Gamma(1-z)=\frac{\pi z}{\sin(\pi z)}$$

DHMO

1:20 AM

What is $\Gamma(1-z)$?

Mussulini

I hate the offset on the gamma function :(

DHMO

@SimpleArt Can I use your profile pic to derive that $\displaystyle \sum_{n\mathop=1}^\infty n = \left[ \sum_{n\mathop=1}^\infty 1 \right]^2$?

Simple Art

@DHMO $$\Gamma(1-z)=\int_0^\infty t^{-z}e^{-t}dt$$

DHMO

@SimpleArt I see

Simple Art

@DHMO XD Go right ahead, but it's supposed to be alternating ;)

DHMO

1:23 AM

@SimpleArt You know, since the left hand side equals $\dfrac1{12}$...

Simple Art

@DHMO No, the alternating version. $\mu(-1)$ dirichlet eta function

Specifically, we have $\mu(0)^2=\mu(-1)$

DHMO

@SimpleArt I know, I've seen your profile

I'm talking about my version

Simple Art

Ofc not, its supposed to be a joke XD

DHMO

@SimpleArt ??

Simple Art

Zeta regularizing $\sum_{n=1}^\infty n$ gives $-1/12$, and the square of something can't be negative....?

DHMO

1:25 AM

@SimpleArt alright

Simple Art

lol

DHMO

@SimpleArt what is the value of $\mu(0)$?

Simple Art

Fun times

$1/2$

DHMO

and $\mu(-1)$?

Simple Art

$1/4$

DHMO

1:26 AM

how to prove it?

Simple Art

Consider the binomial expansions of $(1+1)^{-1}$ and $(1+1)^{-2}$

I guess that's one sorts of regularizing. :/

Note that the original definition of $\mu(x)$ is defined for $\Re(x)>0$

So $\mu(0)$ is a limit point.

DHMO

Are you saying that $\displaystyle \lim_{x\to0^+} \mu(x) = \dfrac12$?

Simple Art

Yeah.

Akiva Weinberger

Isn't it $\eta(x)$, not $\mu(x)$?

2

Being the eta function

DHMO

lol

Simple Art

1:29 AM

Huh, I forget tbh

XD fail

@AkivaWeinberger GG

DHMO

@SimpleArt does $\zeta(0)$ have a value?

I thought that is $\zeta(-1)$?

Simple Art

Oops

Akiva Weinberger

I think a summation method (something that gives values to divergent series) is called "stable" if it gives the same answer to $0+a+b+c+\dotsb$ as $a+b+c+d+\dotsb$

Simple Art

$\zeta(0)=-1/2$

Akiva Weinberger

and whatever method is used to derive $1+2+\dots=-1/12$ is not stable

Simple Art

1:31 AM

You saw nothing

@AkivaWeinberger Nah, analytic continuation

DHMO

@SimpleArt why?

@AkivaWeinberger why?

Simple Art

Well, we can use the following:

$$\zeta(x)=\frac1{1-2^{1-x}}\eta(x)$$

And take $\lim_{x\to0^+}$

This is found by using the original definitions of $\zeta(x)-\eta(x)$ for $\Re(x)>1$, noting absolute convergence.

DHMO

alright

Simple Art

And then analytic continuation

Lol, why are you guys asking me, don't you guys already know these things?

DHMO

why would I?

Simple Art

1:34 AM

Good point

@AkivaWeinberger Nice pic. I like its geometry

DHMO

but then, $\dfrac14 = \eta(-1) = 1-2+3-4+\cdots = (1-2)+(3-4)+\cdots = (-1)+(-1)+\cdots = -\zeta(0) = -\dfrac12$?

Akiva Weinberger

Well, it can't both be stable and linear ($\sum a_n+\sum b_n=\sum (a_n+b_n)$) because you could use it to derive $0=1$

DHMO

@AkivaWeinberger what is not stable?

Simple Art

@DHMO And that's why your not allowed to do that. The moment you did grouping, RIP

DHMO

how do you use it to derive $0=1$?

Simple Art

1:35 AM

@DHMO Wait

DHMO

@SimpleArt I thought only rearrangement is not allowed

Mussulini

I'm triggered

DHMO

@Mussulini by what?

@Mussulini why?

Simple Art

@DHMO Grouping is also disallowed

DHMO

@Mussulini when?

@Mussulini by whom?

@Mussulini how?

Mussulini

1:35 AM

by you mostly

Simple Art

lol

Akiva Weinberger

5 mins ago, by Akiva Weinberger

I think a summation method (something that gives values to divergent series) is called "stable" if it gives the same answer to $0+a+b+c+\dotsb$ as $a+b+c+d+\dotsb$

@Mussulini where?

LMAO

RIP convergence

1:36 AM

4 mins ago, by Akiva Weinberger

and whatever method is used to derive $1+2+\dots=-1/12$ is not stable

DHMO

@AkivaWeinberger I mean, which sum is not stable?

Simple Art

@AkivaWeinberger Ding Ding

DHMO

@AkivaWeinberger I mean, how do you derive $0=1$ from that?

@SimpleArt :o

Simple Art

@DHMO Take the following:

$1+1+1+\dots=\zeta(0)=-1/2$

Akiva Weinberger

Take $x=1+2+3+\dotsb$. By stability, $x=0+1+2+\dotsb$. Subtract the second from the first; $0=1+1+1+\dotsb$. With me so far?

Simple Art

1:37 AM

Add $1$ to the beginning.

$1+1+1+\dots=\zeta(0)?=-1/2$

Akiva Weinberger

By stability, $0=0+1+1+\dotsb$. Subtracting one from the other, we get $0=1$.

Simple Art

But if we add $1$ to both sides of the original, then we get $-1/2=1/2$

Akiva Weinberger

(I'm assuming our summation method gives the right answer to convergent series, so that $1+0+0+\dotsb=1$. Otherwise it's kind of a rubbish summation method.)

Simple Art

Similar contradiction.

@AkivaWeinberger Like that rubbish description XD

Akiva Weinberger

Without that condition, we could make our summation method just send everything to zero.

Simple Art

1:39 AM

Can you Riemann sum contour integrals?

Or can we not because $\mathbb C$ is not well ordered set?

Mussulini

you can Riemann sum anything

DHMO

@SimpleArt why not? contour integrals can be converted to normal integrals...

Akiva Weinberger

To be formal, a summation method is a function whose domain is (some subset of) infinite sequences of numbers and whose codomain is numbers.

Simple Art

It just seems... weird

Akiva Weinberger

Which ideally does the right thing to convergent stuff.

DHMO

1:40 AM

@SimpleArt for example?

Simple Art

nothing, just ignore me lol

DHMO

@AkivaWeinberger so no summation method of divergent series is stable?

Simple Art

@AkivaWeinberger Why would it ever not give the right thing to convergent stuff?

@DHMO No... it depends on the series

stability lies more on the series itself

Akiva Weinberger

@SimpleArt If we define it to just send every sequence of numbers to zero

It would be a rubbish summation method

Simple Art

XD

DHMO

1:41 AM

@SimpleArt is there any stable divergent series?

Akiva Weinberger

No, "stable" applies to the summation method

Simple Art

Hm, oh well

I'm no expert in math

Akiva Weinberger

But no stable, linear summation method can sum $1+2+\dotsb$.

Simple Art

:P

Akiva Weinberger

Because it would prove $0=1$.

There are stable linear summation methods that sum $1-2+3-4+\dotsb$, though.

Simple Art

1:42 AM

mhm

Akiva Weinberger

And they always sum it to $1/4$ I believe (exercise)

DHMO

Let $x=1-2+3-4+\cdots$.

Simple Art

Lmao

DHMO

By stability, $x=0+1-2+3-4+\cdots$

By linearity, $2x=1-1+1-1+\cdots$

By stability, $2x=0+1-1+1-1+\cdots$

By linearity, $4x=1$

Therefore, $x=\dfrac14$ (QED)

Simple Art

$$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$

@DHMO Isn't that on the wikipedia?

By Euler I think?

DHMO

1:45 AM

@SimpleArt I didn't look at Wikipedia

Akiva Weinberger

Here we go: en.wikipedia.org/wiki/…

Mussulini

@SimpleArt I think it is by Ramanujan

Simple Art

@Mussulini Yeah, ur right

Akiva Weinberger

Harder exercise is to show that you can't get a contradiction doing that, which you would do by explicitly constructing a summation method that can sum that series

Alternatively, @DHMO, show that a linear stable series that sums $1^2-2^2+3^2-\dotsb$ would sum it to $0$.

Which you can do along the same lines.

DHMO

$x=1-4+\cdots$

$x=0+1-4+\cdots$

$2x=1-3+5-7+\cdots$

$2x=0+1-3+\cdots$

$4x=1-2+2-2+\cdots$

$4x=0+1-2+\cdots$

$8x=1-1=0$

Simple Art

1:48 AM

@AkivaWeinberger Say, is there a summation method that uses analytic functions to solve classes of series in a similar form by moving the problem to convergent series?

DHMO

$x=0$

Akiva Weinberger

Probably? The wiki page I linked to probably has information related to that

Simple Art

@DHMO Prove for me that $1^{2k}-2^{2k}+3^{2k}-\dots=0\forall k\in\mathbb N$

@AkivaWeinberger I don't recall it being there, but I'll look it over

@DHMO Hint: probably induction

@AkivaWeinberger en.wikipedia.org/wiki/Divergent_series#Analytic_continuation

Nvm, found it

Akiva Weinberger

I, for one, have no idea how to solve that

DHMO

$x=1^21^{2k}-2^22^{2k}+\cdots$

Simple Art

1:50 AM

@AkivaWeinberger Explosive binomial expansions hehe

Oh god XD

@DHMO tell me how that goes XD

rofl

Akiva Weinberger

But, assuming this accurately reflects the values of the eta (alternating zeta) function, and using the formula relating the eta and zeta functions, this would give us the trivial zeroes of the zeta function

DHMO

$2x = 1^21^{2k} - (2^2 2^{2k} - 1^2 1^{2k}) + \cdots$

Simple Art

@AkivaWeinberger ofc

Akiva Weinberger

You can't type a^b^c, you need a^{b^c}

Simple Art

T_T What's next....?

DHMO

1:52 AM

binomial expansion

$(n+1)^{2k+2} - n^{2k+2} = \displaystyle \sum_{r=0}^{2k+1} \binom{2k+2}r n^r$

then I'm lost

Akiva Weinberger

Maybe try $\eta(-4)=1^4-2^4+\dotsb$ by hand first

Forever Mozart

2:22 AM

try

Jessy Cat

0

Q: Having trouble understanding the proof that $D_{n}$ is nonabelian for $n \geq 3$

I saw a proof that $D_{n}$ is nonabelian for $n \geq 3$ that went the following way: Let $a,b$ are arbitrary elements of $D_{n}$. Suppose to the contrary that $ab = ba$. Then, we must have that $ab = ba \, \implies \, abb^{-1} = bab^{-1} \, \implies \, a = bab^{-1} = a^{-1}$. Therefore, $a^{2...

Adeek

2:53 AM

0

Q: Showing that $l_{\infty}$ is complete proof (verification)

Consider $l_{\infty} = \{f : \mathbb{N} \rightarrow \mathbb{K} : sup_{n \in \mathbb{N}} |f(x_i)| < \infty\}$. Suppose that $\{f_i\} \subset l_{\infty}$ is a cauchy sequence. This mean there exists $N \in \mathbb{N}$ such that whenever $s,m \geq N$ we have $sup_{n \in \mathbb{N}} |f_s(n) - f_m(n)| <...

functional-analysis analysis proof-verification proof-writing

If someone would like to check it

user340082710

3:35 AM

Is there a name for a permutation that consists of exactly one cycle?

It isn't quite a cyclic-permutation, since I want to exclude self-loops in my permutation

so I guess this would be a cyclic-permutation that is also a derrangement?

Axoren

3:57 AM

Are three points enough to uniquely determine a parabola?

More specifically, a quadratic function of one variable?

Brody

4:13 AM

@Axoren dunno, perhaps we should look at a system of equations?

@Axoren on another hand, you may refer to this math.stackexchange.com/questions/1415370/…

user116211

4:41 AM

Has anyone seen the proof of a function having an isotone extension iff the function maps order-bounded sets to order-bounded sets?

user116211

Hmm, I am reading Kelley; there he shows the sufficiency of the theorem.

user116211

But some of his statements seem to be vague to me.

user116211

Here is the theorem:

user116211

> Let $f$ be an isotone function on a subset $X_0$ of a chain $X$ to an order-complete chain $Y$. Then $f$ has an isotone extension whose domain is $X$ if and only if $f$ carries order-bounded sets to order-bounded sets. (More precisely stated, the condition is that, if $A$ is a subset of $X_0$ which is order-bounded in $X,$ then $f[A]$ is order-bounded in $Y .$)

user116211

4:49 AM

In the proof, he first takes $A\subseteq X_0$ which has a lower bound in $X;$ then he writes apparently from nowhere that $f[A]$ has also lower bound.

user116211

How did he conclude that? I'm not getting that.

Balarka Sen

5:43 AM

@Axoren No. Take $y = x^2$ and $x+1 = (y-1)^2$.

Two distinct parabolas may even agree at 4 points: take the first and $x+2 = (y-2)^2$.

Prince M

6:03 AM

If Y is some subset of (X,d), possibly empty, is it possible for the compliment of the boundary of Y to be empty?

Ive been trying to think up a pathological example all night but cant

Cause if it were the case then it would mean that the boundary was all of X

which i think would give a contradicts my intuition but I'm hesitant b/c there could be some strange metric or strange choice of X that would let this work?

Mussulini

8:30 AM

$$\pi=6\displaystyle\sum_{h=0}^\infty{-1/2 \choose h}\frac{(-1)^h}{2^{2h+1}(2h+1)}$$

what a nice identity

is there a way to make it converge faster?

Mussulini

8:41 AM

wait, I forgot a h! in the denominator

Luigi M

9:32 AM

Can someone give me some references for the computations and characterisations of the Edge Homomorphism in the Atiyah-Hirzebruch Spectral Sequence? The only result I found is in Davis & Kirk, Lecture notes in Algebraic Topology at page 246, but it's not proven. I'm mostly interested in the horizontal one

Mahmoud

9:54 AM

Hello.

Fargle

Greetings!

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$Mathematics$ Mathematics