Mathematics - 2020-07-27

Beliod

01:49

I still don't understand the special product (x + a)^2 = (x + a) (x + a).

If you have (x + 7)^2, I don't see how you can get to the "correct" answer. I understand what to do when I see it, but I just can't understand the mathematical reason that it works.

And it seems I'm not alone. I distinctly remember everyone in my class solving it as "x^2 + 7^2." Without changing it into a different equation, where does the 14x actually come from?

Meaning, just using (x + 7)^2, where is the ability to add 7 to itself?

Mike Miller

02:21

or: if you believe (a+b)c = ac + bc (like the above, draw a rectangle to see this), then (x+7)(x+7) = x(x+7) + 7(x+7) = x^2 + 7x +7x + 49

Knight

04:19

@robjohn Hello sir! Meeting you after a long time.

feynhat

04:33

@AlessandroCodenotti Btw, I was using this version of the ping pong lemma: Let $G$ be generated by $a$ and $b$ both of which have infinite order. Suppose $G$ acts on $X$ such that there is partitioning $X = A \sqcup B$ satisfying $a\cdot B \subset A$ and $b \cdot A \subset B$, then $G$ is free with $a, b$ as the free generators.

Balarka Sen

@feynhat Ahh nice

Ted Shifrin

Oh oh, a @Balarka has arisen.

feynhat

There is more general version where you don't require $A$ and $B$ to be disjoint, infact you don't even require their union to be $X$. You just want that one of them should not be contained in the other. But, then you will have to check $a^n \cdot B \subset A$ for all $n \ne 0$ and likewise $b^n \cdot A \subset B$.

Balarka Sen

@TedShifrin Actually I haven't gotten much sleep lol

Ted Shifrin

Oh oh.

Balarka Sen

04:36

@feynhat Yeah you basically want a Cantor set type pattern out of iterating $A$ and $B$ back and forth

The various orbits of $A$ and $B$ should be like a rooted binary tree under inclusion, and the group "acts on the boundary Cantor set", which is how you know it's free

@feynhat Your ping pong proof of $\langle a^2 b, ba^2 \rangle = F_2$ looks solid yeah

Here's a way to phrase what we were doing earlier without orbifold theory. Let $X = S^n \setminus \{n \, points\}$ and $\Bbb Z_n$ acts on $X$ in the way described. $X \to (E\Bbb Z_n \times X)/\Bbb Z_n \to E\Bbb Z_n/\Bbb Z_n = B\Bbb Z_n$ has a section since the action of $\Bbb Z_n$ on $X$ has a fixed point. This gives a split exact sequence $1 \to F_{n-1} = \pi_1(X) \to \pi_1(X_{\Bbb Z_n}) \to \Bbb Z_n \to 1$

It remains to argue $\pi_1(X_{\Bbb Z_n}) = \Bbb Z_n * \Bbb Z_n$. How do we do this? Well, observe the action of $\Bbb Z_n$ on $X$ has exactly two full-stabilizer points, and on the complement of these the action is completely free.

Let's call these points $p$ and $q$. With some thought based on these comments, one sees $X_{\Bbb Z_n} = (X - \{p, q\})/\Bbb Z_n \times E\Bbb Z_n \sqcup \{p, q\} \times B\Bbb Z_n$

As a set-theoretic decomposition. Basically homotopy equivalent to $X/\Bbb Z_n$ but at the images of $p$ and $q$ you have two copies of $B\Bbb Z_n$ stuck. Siefert van Kampen now says $\pi_1(X_{\Bbb Z_n})$ picks up these two fundamental groups at $p$ and $q$, so $\Bbb Z_n * \Bbb Z_n$

This is morally the same orbifold SvKT computation

Now you have to tell me why $\pi_1^{orb}(M//G) \cong \pi_1(M_G)$ in general

$X = S^2 \setminus \{n\, points\}$ I meant

Knight

04:56

@robjohn Sir, why this extension doesn't work on Mac?

Balarka Sen

It's a charming observation that $F_{n-1} \rtimes \Bbb Z_n \cong \Bbb Z_n * \Bbb Z_n$. I don't know why I never noticed this, especially in light of the fact that $\Bbb Z_2 * \Bbb Z_2$ is the infinite dihedral group

Ted Shifrin

How does that look dihedral?

Balarka Sen

For $n = 2$ you get $\Bbb Z \rtimes \Bbb Z_2$ which is $\langle r, s | s^2 = 1, srs = r^{-1} \rangle$, so dihedral but $r$ has infinite order

Ted Shifrin

Ohhh, duh, $F_1 = \Bbb Z$. Got it. Thanks.

Balarka Sen

yup

Ted Shifrin

04:59

I totally don't intuit this.

Balarka Sen

It's pretty wicked, yeah

You can think of that group as the dihedral symmetry group of the apeirogon if you want to be fancy

Which is like a horocircle but regular polygonal

Ted Shifrin

So how do I see this as $\Bbb Z_2*\Bbb Z_2$?

Balarka Sen

It should basically be because you're generated by two reflections. Think of the apeirogon as $\Bbb Z$, and all distance-preserving transformations of $\Bbb Z$ generates the infinite dihedral group.

Consider reflection along $0$ (so $x \mapsto -x$) and reflection along $1$ (so $x \mapsto -(x-1)+1$)

These two freely generates the whole dihedral group, so $\Bbb Z_2 * \Bbb Z_2$

Ted Shifrin

Oh, I see. So we could use two different reflections in the finite case, rather than reflection and rotation.

I've never pondered that.

Balarka Sen

Yeah, it's the same thing as that

Ted Shifrin

05:03

Yup, that's the insight I was lacking.

That would have made a good exercise for my algebra course/book. Oh well.

Balarka Sen

Haha yeah I am pretty proud to have discovered this in class when my group theory professor asked us to come up with a concrete example of a free group and everyone was giving some memorized matrix example

Ted Shifrin

pats a @Balarka on the back

Balarka Sen

Hm, I think the exercise was: If $a$ and $b$ are finite-order in a group, is $ab$ of finite order?

Ted Shifrin

Ah, that's a standard exercise, I guess.

Balarka Sen

Yeah, you do something with matrices

Bleh I never remember that

Ted Shifrin

05:06

Well, or isometries of $\Bbb R$ would do.

Balarka Sen

So I gave this

Yeah

When I google apeirogon a novel by Colum McCann comes up??

Oh actually seems like something I'm supposed to read

Ah but Aljazeera has bad things to say about it so I won't

robjohn

05:50

@Knight I know nothing about that extension.

Knight

06:07

@robjohn Does your bookmark script work on gmail chats?

robjohn

@Knight I have not tried a gmail chat, but it works on most web pages with MathJax content that I have tried. Have you tried it on a gmail chat?

Knight

@robjohn Yes, it’s not working on hangouts and Gmails (at least not on Mac).

Someone told me that a simple change in script will do the job.

I went to gmail.com, sent a latex code to one of my contacts and then clciked on your bookmark. But it didn't get render.

robjohn

Depending on how the editable text fields are handled, the bookmark might not be able to access/recognize the text to be rendered. It may need tweaking for a different environment.

The pages on SE use AJAX, and that is what "start ChatJax" and "render MathJax" use.

Knight

06:35

@robjohn Will it be a hard and long time taking task?

I have few friends who know intermediated level of JavaScript, if you give some guidance then I will convey it to them and they might be able to change your chatjax script so that it would work on gmail chats.

And one more point to be noted here is that when I click on the bookmark on the hangout page, the hangout windows gets away on its own unlike this chat window where clicking on bookmark retains the window.

Mats Granvik

08:58

Where can I find a proof that the limiting ratio for the Fibonacci numbers exists?

proofwiki.org/wiki/Ratio_of_Consecutive_Fibonacci_Numbers

Astyx

09:17

It folllows from it being a recursive sequence

Therefore it writes $\alpha x^n + \beta y^n$

So if you compute the ratio and take the limit, only the dominant term survives (let's say that $|x|>|y|$) which gives $x$

Mats Granvik

But why are you allowed to say the sequence satisfies $\alpha x^n + \beta y^n$?

Leaky Nun

you can either use matrix or generating function

Mats Granvik

I am reading this now: proofwiki.org/wiki/Euler-Binet_Formula/Proof_4

Astyx

09:35

You can show that sequences that satisfy a given recursive equation form a vector space, and you can show the dimension of this vector space is the order of the equation you're looking at (in our case $u_{n+2} = u_{n+1}+u_n$ means it's dimension 2)

To find a basis you can try to look for geometric series

ie $x$ such that $x^{n+2}=x^{n+1}+x^n$, or simply put $x^2=x+1$ (assuming $x\ne0$, which is trivial)

this is a simple quadratic and you find two solutions $x={1\pm \sqrt{5}\over 2}$

Mats Granvik

Not that I understand everything you say, but does that proof require solving a quadratic equation?

Astyx

that means that the sequence you're looking for is of the form $u_n =\alpha {1+\sqrt 5\over2} +\beta {1-\sqrt 5\over2}$, and you can find $\alpha$ and $\beta$ by looking at initial values

yes

You could take a look at this

Constant-recursive sequence

In mathematics, a constant-recursive sequence or C-finite sequence is a sequence satisfying a linear recurrence with constant coefficients. == Definition == An order-d homogeneous linear recurrence with constant coefficients is an equation of the form s ( n ) = c 1 s ( n − 1 ) + c 2 s ( n − 2 ) +...

Feel free to ask if you have any questions

Mats Granvik

@Astyx It is the algebra part that worries me. The limiting ratio that gives an approximation to the Riemann zeta zeros probably requires solving a (truncated) polynomial(?) that is of much higher degree than the quadratic equation.

Astyx

I'm not sure where you're going with this

Mats Granvik

This: https://mathoverflow.net/q/364186/25104
and this: https://math.stackexchange.com/q/3735496/8530

Taylor polynomial approximations.

Astyx

09:52

So what is it exactly you want to know ?

Mats Granvik

I want to know if the limit giving zeta zeros, exists.

Astyx

That a whole different question to the Fibonacci one

And I don't really know

Mats Granvik

But both use limiting ratios.

These are the coefficients of the exponential polynomial we are solving with the limiting ratios from a more complicated recursive sequence:
$$\left\{\zeta (z),\zeta '(z),\zeta ''(z),\zeta ^{(3)}(z),\zeta ^{(4)}(z),\zeta ^{(5)}(z),\zeta ^{(6)}(z),\zeta ^{(7)}(z),\zeta ^{(8)}(z),\zeta ^{(9)}(z),\zeta ^{(10)}(z),0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0\right\}$$

The zeros at the end are there on purpose.

But in the case of the exponential polynomial the limiting ratio is (n-1)*a(n-1)/a(n) instead of a(n-1)/a(n).

Abram Ivanov

hey, how much reputation does one need so that when he joins a new SE network, a free +100 rep is awarded?

anyone?

robjohn

10:31

@AbramIvanov If you are an experienced Stack Exchange network user with 200 or more reputation on at least one site, you will receive a starting +100 reputation bonus to get you past basic new user restrictions.

from this page

user123456789

Anyone online >

?

I have a doubt:
We have a pair of linear equations
(a1)x + (b1)y + c1 = 0 say equation-1
(a2)x + (b2)y + c2 = 0 say equation-2

If we multiply equation-1 by b2, equation-2 by b1 and subtract them we get:

x = (b1c2 - b2c1) / (a1b2 - a2b1)
Similarly we can get
y = (c1a2 - c2a1) / (a1b2 - a2b1)

I get stuck when (a1b2 - a2b1 = 0) i.e. a1/a2 = b1/b2
When this is the case 'x' becomes '(b1c2 - b2c1) / 0' therefore should be undefined.

The problem is, when a1/a2 = b1/b2 = c1/c2 and (a1b2 - a2b1 = 0)

robjohn

@user123456789 what do you expect when the lines are parallel?

user123456789

No solutions then

robjohn

@user123456789 or infinite solutions

in the 0/0 case

user123456789

10:47

Okay! so ?

Could you please explain what am I missing

Thorgott

13:20

why is there no textbook called "A course in coarse geometry"

Edward Evans

13:50

Winging my way through Lubin-Tate theory so I don't embarrass myself tomorrow

the dream

uhoh

2

Q: Roughly how many kinds of closed or periodic orbits are there in the circular restricted three-body problem?

Question: Roughly how many kinds of closed or periodic orbits are there in the circular restricted three-body problem? Has anyone made an attempt to enumerate and classify/categorize them? Notes: If there is a distinction between closed and periodic required to answer the question, you can choose...

I've added a bounty

Mats Granvik

Mathematica Program 2 here: https://pastebin.com/1cgqhBdj gives Riemann zeta zeros.
Mathematica Program 2 here: https://pastebin.com/mAZCuXta gives Golden ratio minus 1.
The Mathematica progam is the essentially the same except for the input polynomial.

Both programs use limiting ratios to compute Riemann zeta zeros or the Golden Ratio minus 1.

But I was told earlier, that it is not known if the limit exists.

So in principle the Riemann hypothesis should be equivalent to showing that the limit exists and that the real part of it converges to 1/2 regardless of starting point in the critical strip. Numerically I have found that the imaginary part of the starting value must be greater than some number between 9 and 10, and the real part can be arbitrarily chosen at least in the interval 0 to 1.

@Astyx

Lucas Henrique

14:25

hey chat. what's the intuition of a series not being commutatively convergent? since all the terms eventually appear in the partial sums

Thorgott

you mean that all the rearrangements don't necessarily converge to the same limit?

Mike Miller

14:49

@Thorgott I once thought about giving an intro lecture to course geometry where the gimmick was that course geometry is about what's left after your vision gets fuzzy where I drank continuously through the lecture

decided against

Thorgott

lmao

Lucas Henrique

@Thorgott yup

Konformist Liberal

15:42

I am pretty sure that this question makes no sense, \textbf{but}:
Do you have any examples of "natural things" that exhibit a homotopy in a "plain" way?
As an example, I imagine a [line segment thingy] getting collapsed to a [point thingy] or vice versa

Mike Miller

I don't know what that means, sorry

Maybe you're talking about how X x I deformation retracts onto X, because I deformation retracts onto {0}

All of the standard facts (cones are contractible, mapping cylinders deformation retract onto the codomain) follow from this

that I def retracts onto {0}

Konformist Liberal

@MikeMiller I would consider an example on the lines of deformation retraction from X x I to X as also plain. But, as a bad follow-up to a bad question, I try to imagine something that visually, mentally feels correct. For (a pretty bad) example, if I asked about a natural thing which exhibit a sphere in a plain way, you can say orange that would be nice, apple is pretty good, banana is in some sense correct but not really what I am trying to feel.

Mike Miller

Yeah, I'm sorry, but I think I can't help until there's a math question.

Konformist Liberal

15:57

:) It's fine, maybe someone else would brainstorm/say something that can be interesting, just wanted to ask out

An elastic object (I imagine something like a string that bounce back or some water that takes back it's natural form), at least for me, is a good picture for homeomorphism but I am specifically looking for something not homeomorphic

Thorgott

16:36

@Lucas if you have a series that converges conditionally, but not absolutely, it will contain negative and positive terms and the positive terms on their own sum to +\infty whereas the negative terms on their own sum to -\infty, so what you are rearranging is the indeterminate expression \infty-\infty

like, if you want a rearrangement to converge to a given limit L, take enough positive terms till they sum to over L, then subtract some till you're below, add some more till you're above again, rinse and repeat

and this is possible precisely because the positive and negative terms on their own each diverge

William Sun

Let $X$ be the spectrum of $\Bbb Z$. Do we have $C(X, \Bbb R)$ isomorphic to $Z$ as a ring?

Lucas Henrique

@Thorgott well, that makes sense.

that's the riemann rearrangement theorem, isn't it?

@WilliamSun what do you mean by $C(X, Y)?$

heya @Ted!

Thorgott

yup, and what I just said is basically the proof when you make it precise

@William $C(X,\mathbb{R})$ is uncountable, so no

Lucas Henrique

16:52

@Thorgott the set of functions from X to R, I suppose?

Mike Miller

continuous

Thorgott

yeah, continuous maps from X to R

Mike Miller

all continuous functions are constants, so the ring you get is R

Mats Granvik

17:10

I apologize for interrupting but this is such a pleasant interview to listen to: youtube.com/watch?v=IsCUlZahQMc

Ted Shifrin

18:10

@LucasHenrique You mean absolutely convergent. If a series is conditionally convergent, then the positive terms alone have divergent sum, and the same is true for the negative terms. Therefore, you can add up positive terms to get close to any number you want (say $100$) — just go past — then add negative terms to just get under $100$, and continue in this fashion. With such a rearrangement, you'll converge to $100$ because the terms are eventually going to $0$.

The partial sums of this rearrangement look nothing like the partial sums of the original ordered series.

Mats Granvik

@TedShifrin What did you think of my limiting ratios for the Riemann zeta zeros?

Ted Shifrin

18:25

Sorry, Mats, this is not at all my thing.

Mats Granvik

In case it is somebody else's thing:

The first program in the image also uses limiting ratios, I forgot to say that.

And these are the programs: oeis.org/draft/A002410/a002410.txt

Lucas Henrique