Mathematics - 2017-02-12 (page 2 of 4)

Ted Shifrin

06:00

It's better to note that oriented volume and determinant have the same properties with regard to row (or column) operations.

Daminark

Likely over the spring break or summer I'm gonna need to go and get 4 theorems down

Ted Shifrin

Night for now!

Daminark

Inverse function theorem, implicit function theorem, change of variables

And this one thing that my physics TA called partitions of unity, which is apparently why we ask that manifolds are second countable

Good night!

Ted Shifrin

Partitions of unity are in my book, too, of course :)

Night !

Akiva Weinberger

I'm annoyed that there doesn't seem to exist an IPA for /b͜ʙ/ with a single period in the trill. (It sounds like two 'b's in a row, and it's one of my favorite consonants)

Jasper Loy

06:03

When you have partitions of unity, it is no longer unity.

@AkivaWeinberger What language is that?

Akiva Weinberger

It's not a language

IPA is the international phonetic alphabet, which theoretically has a symbol for every possible sound the mouth could make

(More specifically, it tries to have symbols for every sound used in a language somewhere)

Jasper Loy

I mean what language does the sound exist in?

Akiva Weinberger

I don't know of any

It's fun to say, though

Jasper Loy

@Adeek Better to sleep 8 hours a day. If you only need 4 hours a day, it means you are going to die soon.

Akiva Weinberger

@JasperLoy I'm pretty sure Balarka has clinical insomnia

I don't think it's been formally diagnosed but it's become fairly obvious

On the other hand, I'm currently awake at 1:10am, so maybe I'm a hypocrite

Harnoor Lal

06:09

I actually have a really inspiring story

I was diagnosed with insomnia 3 years ago

Jasper Loy

There are people who don't need to sleep. There are even those who don't need to eat. This is not a joke.

Harnoor Lal

And all the pills the doctors tried to give me although they worked made me feel like shit

And then I found math

Jasper Loy

A certain man in India has not eaten for years. Many are sceptical but I believe this case is real.

Harnoor Lal

I literally just find some problem and solve it and generalize it and see where it leads for about 2 hours every night

And it helps me sleep

better than any medicine

Jasper Loy

He has been studied by the Indian government and there is no reason for them to lie, because they don't get any money out of it.

Akiva Weinberger

06:11

That's really nice

@HarnoorLal Do you know any group theory?

Harnoor Lal

I've dabbled in it a bit

Jasper Loy

@HarnoorLal Good for you.

Akiva Weinberger

Do you know about the free group with two letters? Like, the set of things like $a$, $b$, $aba^{-1}$, $aba^2b^{-3}$, etc.?

Harnoor Lal

I am not too familiar with it actually

Vrouvrou

hello, if $||\alpha||^2 =\langle \alpha,\alpha \rangle$, where $\alpha is a function depending on $t$ what can be $\frac{d}{dt} ||\alpha||$

Harnoor Lal

06:13

Care to elaborate?

Akiva Weinberger

So, let's consider all "words" you can make with these four letters: $a$, $b$, $a^{-1}$, and $b^{-1}$

Harnoor Lal

Ok

Akiva Weinberger

So that would be things like $a$, $b$, $aaab$, $baab$, $a^{-1}aa^{-1}$, etc.

and the empty word: $~$. (Which we'll call $e$, for "empty" I think, so that we have something to look at when thinking about it.)

Harnoor Lal

I understand

Akiva Weinberger

So our group operation is "multiplication", by which I mean putting the two words next to each other

so $a$ times $b$ is $ab$, and $ab$ times $b^{-1}a^{-1}$ is $abb^{-1}a^{-1}$

and $e$ times $b$ is just $b$ (since $e$ is supposed to be the empty word)

In fact, $e$ times anything is itself.

$e$ times $aba$ is $aba$.

Vrouvrou

06:16

can someone help me

Harnoor Lal

Interesting

Akiva Weinberger

One more thing we need to do:

We need to set $aa^{-1}$ equal to $e$

and also $a^{-1}a$

and $bb^{-1}$ and $b^{-1}b$.

So, for example:

$abb^{-1}a^{-1}$ would equal $a(bb^{-1})a^{-1}=$aea^{-1}$=aa^{-1}=e$

Am I explaining that well?

@HarnoorLal

Harnoor Lal

Yes

Akiva Weinberger

$a^{-1}$ is pronounced "$a$ inverse", and $b^{-1}$ is pronounced "$b$ inverse"

An inverse is something you can multiply something by to get $e$.

Can you guess what the inverse of $ab$ would be?

$ab$ times what is $e$

Harnoor Lal

Hmm

Akiva Weinberger

06:20

Can I tell you?

Harnoor Lal

Would it be $b^{-1}a^{-1}$?

Akiva Weinberger

Yep!

Harnoor Lal

Yay!

Akiva Weinberger

Because $(ab)(b^{-1}a^{-1})$ and $(b^{-1}a^{-1})(ab)$ are both $e$.

(All inverses are two-sided:

it doesn't matter if you multiply them on the left or on the right.)

Harnoor Lal

But it isn't commutative, correct?

Akiva Weinberger

06:21

Right. $ab\ne ba$.

Harnoor Lal

Ok

Akiva Weinberger

So, we can write that like this: $(ab)^{-1}=b^{-1}a^{-1}$ (the inverse of $ab$ is $b^{-1}a^{-1}$).

Harnoor Lal

Oh wow

That's cool

Akiva Weinberger

There are also shorthands we can use: $a^3$, for example, means $aaa$, and $a^{-3}$ means $a^{-1}a^{-1}a^{-1}$

You can check that the laws of exponents all work

Well, with the exception that $(ab)^2\ne a^2b^2$

$(ab)^2=(ab)(ab)=abab$, since it's not commutative

So, that's the free group on two letters. Three letters and more works the same way.

Call the free group on two letters $F_2$, and the free group on three letters $F_3$.

A cool thing is that we can find a copy of $F_3$ inside $F_2$.

Harnoor Lal

Really?

Akiva Weinberger

06:25

Let $X$ be $aa$, $Y$ be $ab$, and $Z$ be $ba$

Harnoor Lal

Mhmm

Akiva Weinberger

(You have LaTeX on, by the way, right?)

Harnoor Lal

Yes

Akiva Weinberger

Close $\{X,Y,Z\}$ under multiplication and inverses — meaning, consider everything we can get from the set $\{X,Y,Z\}$ by taking multiplication and inverses of elements

So, this would include $X$, $Y^{-1}$, $ZX^{-1}Z^{-1}X$, etc.

Harnoor Lal

And $e$ still represent the empty number?

Akiva Weinberger

06:28

Yeah

Harnoor Lal

Ok

Akiva Weinberger

This gives us a subset of $F_2$ (remember that $X=aa$, $Y=ab$, and $Z=ba$). It's not equal to it, though — there are things in $F_2$ that aren't in this new set.

Like, $a$ isn't in this new set. You can't get $a$ just from taking $X=aa$, $Y=ab$, and $Z=ba$, and their inverses, and multiplying them together.

Harnoor Lal

Oh wow

Akiva Weinberger

An easy way to see this is that, in $X$, $Y$, $Z$, their inverses, and everything you get from multiplying them, we have an even total number of $a$s

Harnoor Lal

And this applies for any $F_n$ and $F_{n-1}$

?

Akiva Weinberger

06:30

So, the cool thing is that this subset is isomorphic ("the same as") the free group on three elements.

So this is a copy of $F_3$ inside of $F_2$.

@HarnoorLal Yeah, you can find copies of $F_n$ inside of $F_{n-1}$ for all $n\ge3$

You can't find a copy of $F_2$ inside of $F_1=\{\dots,a^{-2},a^{-1},e,a,a^2\}$, though, since $F_1$ is commutative and $F_2$ isn't

Harnoor Lal

Of course

Akiva Weinberger

@HarnoorLal So, this subset is called the subset generated by $X$, $Y$, and $Z$. It's written like this: $\langle X,Y,Z\rangle=\langle aa,ab,ba\rangle$

Here's a question: Is $bb$ in this subgroup?

That is, is $bb\in\langle aa,ab,ba\rangle$?

Can you write $bb$ in terms of $aa$, $ab$, $ba$, their inverses, and multiplication?

Harnoor Lal

I believe so

Akiva Weinberger

Want me to tell you?

Harnoor Lal

$ba * a^{-1}b$

Would that make it?

Akiva Weinberger

06:35

How do we know $a^{-1}b$ is in the group, though?

Harnoor Lal

Ohh

Akiva Weinberger

We're just allowed $aa$, $ab$, $ba$, $(aa)^{-1}$, $(ab)^{-1}$, and $(ba)^{-1}$

(and multiplication)

Harnoor Lal

I suspect that it isn't but can't find any logical reason why

Akiva Weinberger

$bb=(ba)\,(aa)^{-1}\,(ab)$

(since $(aa)^{-1}=a^{-1}a^{-1}$)

So, it is in it.

Harnoor Lal

Huh

Semiclassical

06:39

Is there a good way to tell which elements aren't in the subgroup?

Akiva Weinberger

This means, by the way, that if we define $X=aa$, $Y=ab$, $Z=ba$, and $W=bb$, we can't just say that $\langle X,Y,Z,W\rangle$ is the same as the free group on four letters.

@Semiclassical Even total number of $a$s, even total number of $b$s

@HarnoorLal This is because we just showed that $W=ZX^{-1}Y$,

and that's not allowed in a free group.

Semiclassical

Eh? I thought you just showed that $bb$ was in the subgroup

Harnoor Lal

It's strange because I keep wanting to foil it but then I remember that this is a completely new field of mathematics for me.

Haha

Akiva Weinberger

@Semiclassical The next comment was a reply to Harnoor, not you

$bb$ is in the subgroup, since there's an even total number of $b$s (two)

Semiclassical

I think you misread what I said. I asked which ones aren't in the subgroup.

Akiva Weinberger

06:41

Oh. I did.

Whoops.

The ones with an odd number of $a$s or $b$s, then.

Semiclassical

Isn't ab in it, though?

Akiva Weinberger

Ohh, whoops

Correction, then: The elements of the subgroup are those with an even number of $a$s and $b$s total

So, both odd or both even

Semiclassical

Hmm, okay.

I can buy that.

Akiva Weinberger

Alternatively, it's the kernel of the map $F_2\mapsto\Bbb Z_2$ with $a\mapsto 1$ and $b\mapsto 1$ @Semiclassical

which shows that it's a normal subgroup

Semiclassical

Hmm, okay.

What does $F_2/F_3$ look like? I suspect it's simple.

Akiva Weinberger

06:43

$\Bbb Z_2$, then, by the map I just gave

Semiclassical

Ah, of course.

Akiva Weinberger

@HarnoorLal Here's another example: $\langle aaa,aab,aba,abb,{}$$baa,bab,bba,bbb\rangle$ is not the same as the free group on eight letters, since you can prove that that group is equal to $\langle aaa,aab,aba,baa\rangle$.

On the other hand, it is the same as the free group on four letters; it can be shown that $\langle aaa,aab,aba,baa\rangle$ has no "nontrivial relations" (nothing like the $W=ZY^{-1}X$ that our last example had), so it's free.

@HarnoorLal You can even find a copy of $F_\infty$ (free group on infinitely many letters) inside $F_2$.

The subgroup generated by elements of the form $a^nb^n$ — that is, the subgroup $\langle\dots a^{-2}b^{-2},a^{-1}b^{-1},e,ab,a^2b^2,\dots\rangle$ — works.

(That's the set of all elements where there's the same number of $a$s as $b$s.)

Another is the subgroup generated by $a^nba^{-n}$, the set of all elements with "zero total $a$s"

(counting $a^{-1}$ as "negative one $a$s")

Those are both copies of $F_\infty$ sitting inside $F_2$.

Haha

2

Holy crap, is Tuesday Valentine's Day?

Daminark

07:03

As it turns out!

Balarka Sen

07:33

@AkivaWeinberger Alternatively F_2/F_3 is deck transformation group of the double cover of wedge of two circles by wedge of three.

That's Z/2

Akiva Weinberger

Yup.

Hatcher was the only reason I knew you could put $F_m$ in $F_n$ ($m>n$) in the first place.

Balarka Sen

also, yes, Tuesday is that cursed day

Akiva Weinberger

I'm not quite sure how to respond to that

@BalarkaSen Here's a question. Consider $F_3=F(\{a,b,c\})$, and the three maps $f_a$ defined by $\begin{matrix}a\mapsto1\\b\mapsto b\\c\mapsto c\end{matrix}$, $f_b$ defined by $\begin{matrix}a\mapsto a\\b\mapsto1\\c\mapsto c\end{matrix}$, and $f_c$ defined by $\begin{matrix}a\mapsto a\\b\mapsto b\\c\mapsto1\end{matrix}$.

I have no idea what's the best way to write that.

Balarka Sen

@AkivaWeinberger I only realized that after knowing higher genus surface groups fit inside lower genus ones

Akiva Weinberger

But $f_a$ gets rid of $a$, etc.

Balarka Sen

07:38

Gotcha

Akiva Weinberger

$F_3$ is the fundamental group of the wedge of three circles, right?

What cover of that corresponds to the intersection of the kernels of those maps?

I don't know.

Balarka Sen

hm

Akiva Weinberger

So, like, $(aba^{-1}b^{-1})c(aba^{-1}b^{-1})^{-1}c^{-1}$ is in it

$aba^{-1}b^{-1}cbab^{-1}a^{-1}c^{-1}$

'cause if you remove any of $a$, $b$, or $c$, it vanishes

Balarka Sen

Ah

Akiva Weinberger

On the other hand, $aba^{-1}b^{-1}$ isn't in it

since removing $c$ doesn't make it vanish

Balarka Sen

07:41

Right, right.

Akiva Weinberger

I want to know what rank this group ($\ker f_a\cap\ker f_b\cap\ker f_c$) has

like, if it's $F_3$ or $F_\infty$ or what

but I have no idea how you'd approach that

And the algebraic topology approach with covering spaces doesn't work if I have no idea what the covering space should be.

Balarka Sen

If you work with $F_2 = F(a, b)$, is $\ker f_a \cap \ker f_b$ generated by $a^nb^{m}a^{-n}b^{-m}$?

Akiva Weinberger

Yeah, I'm pretty sure.

Oh, right, so it's probably going to be $F_\infty$.

Balarka Sen

Ya

Akiva Weinberger

But there's a different measure you could use, which is

how many elements of it do you need such that the normal subgroup of $F_3$ generated by those elements is $\bigcap\ker f$.

So, like, in the $F_2$ case, you just need $aba^{-1}b^{-1}$ now

since the smallest normal subgroup of $F_2$ containing that is all commutators. (You can do $a(aba^{-1}b^{-1})a^{-1}$ now, which you wouldn't be able to do if you were just doing the generated subgroup rather than the generated normal subgroup.)

Balarka Sen

07:47

That does make sense.

Akiva Weinberger

In other words, if I wanted to get the quotient group $F_3/\bigcap\ker f$, I could write it as $\langle a,b,c\mid r_1,r_2,\dots,r_n\rangle$ with a bunch of relators $r_n$

and the question is how many relators do I need.

Balarka Sen

Right, that sounds much harder.

Akiva Weinberger

(And you can see that, in the $F_2$ case, just $aba^{-1}b^{-1}$ suffices, since the quotient is $\langle a,b\mid aba^{-1}b^{-1}\rangle$.)

Balarka Sen

I suppose it's not just $[[a, b], c] = 1$?

Akiva Weinberger

But you also want $[[a,c],b]=1$ and stuff

and I don't know if that's independent or not

Balarka Sen

07:50

Good point.

Akiva Weinberger

I'd conjecture you just need $[[a,b],c]$, $[[b,c],a]$, and $[[c,a],b]$, but I have no idea how I'd prove that they generate the entire subgroup.

How can I show that everything that vanishes under the removal of any letter can be written in terms of those, inverses, multiplication, and conjugates?

And how could I show that it can't be done with fewer?

But now I think I want to go to bed.

Balarka Sen

:P

Well, that's a good question and I don't know.

user400188

08:03

is anyone here willing to check if I function I have thought up is injective or not?

wait im not supposed to ask to ask. Here it is anyway

I have constructed (i think) a very weird injective function that only has 1 element in its domain and co-domain but accepts multiple inputs at once.

Imagine a function which has domain {1} and co-domain {1}. It is injective because it is one to one.
Now Imagine that the functions domain is the truth value of an expression in propositional logic as opposed to a number.

We may write a domain of {$A\lor\bar A$} and co-domain {1} for this case.

Next we can imagine writing {$A\lor\bar A=$B\lor\bar B} and co-domain {1}

user400188

08:32

I've gone ahead and asked it on SE now because no one responded in chat the two times I have brought it up.

JeSuis

08:55

@BalarkaSen hi, when we take the set $\{x\in \Bbb{R}: x^2\in \Bbb{Q}\}$, is it countable ?

Daminark

It is

So given any positive rational number, we map it to its square root

This is an injection from $\mathbb{Q}_+$ to your set

Now, if we consider the positive numbers in your set, map them to their squares, and map the negative numbers to their negative squares, you get an injection from your set into $\mathbb{Q}$

Or actually replace the first one by the inclusion map

But yeah, by Schroeder-Bernstein you have a bijection

And $\mathbb{Q}$ is countable

Sophie

can someone explain me why $\int\frac{v^{10}}{\sqrt {v^{12}+1}}dv$ is "clearly hypergeometric"?

JeSuis

@Daminark I don't understand your argument, for exemple $\sqrt{2}$ is an element of my set

Daminark

09:12

OK I'll try to explain more coherently :P

Call your set $A$

Now, $\mathbb{Q} \subset A$, right?

Thus, we can find an injection $f:\mathbb{Q}\to A$ by inclusion

JeSuis

lol of course

Daminark

Conversely, take $x\in A\cap \mathbb{R}_+$

And let $f(x) = x^2$

Two different positive numbers have different squares, so this is an injection

JeSuis

ah square

Daminark

Now, if $x\in A\cap \mathbb{R}_-$, let $f(x) = -x^2$

And have $f(0) = 0$

JeSuis

lol I was thinking you are saying that maps to $\sqrt$...

Daminark

09:14

This gives us an injection from $A$ to $\mathbb{Q}$

So by Schroeder-Bernstein, there is a bijection between them

Actually wait a second maybe that function was the bijection

I think we hit every rational number by doing this

So yeah we've constructed a bijection between $\mathbb{Q}$ and $A$

JeSuis

yeah

Daminark

So by countability of $\mathbb{Q}$, we have $A$ is as well

Lol from first quarter analysis I'm used to doing the double injection thing so I wasn't actually paying any attention to surjectivity even though it was there

JeSuis

If we have a countable set $A,$ I was trying to find a function such that $f^{-1}(A)$ is not countable

Daminark

Careful with notation

$f^{-1}(A)$ feels like you're saying the preimage of $A$ under a function $f$, not necessarily "f inverse"

JeSuis

yeah the set of preimage

Daminark

09:18

Oh

Take the floor function

JeSuis

if $f$ is a bijection then $f^{-1}(A)$ is countable

Daminark

Maps any real number to floor

JeSuis

ah yeah

Daminark

The preimage of $\mathbb{N}$ is all of $\mathbb{R}$

JeSuis

nice

Daminark

09:20

(I'm assuming countable means infinite countable, otherwise just map any uncountable set to a single element in your countable set)

JeSuis

infinite countable yes, do you know the complement countable topology ?

Daminark

A set is open if its complement is countably infinite?

Or if it's the whole set ofc

Wait a second

That isn't a topology

JeSuis

@Daminark yeah with the empty set

Daminark

See, a singleton is clearly an open set, right?

JeSuis

right

Daminark

09:23

But if you have a topological space where every singleton is open, you're dealing with the discrete topology

Because every set becomes the union of singletons

Alessandro Codenotti

@Daminark not if your set is uncountable

Daminark

We don't require countable unions only

JeSuis

I am working with $\Bbb{R}$..

Daminark

A topological space is closed under all unions period

Alessandro Codenotti

The cocountable topology agrees with the discrete on countable sets, just like the cofinite agrees with the discrete on finite sets

Daminark

09:25

Oh right

Sorry I'm being stupid

It's 3AM here

Yeah you're fine

JeSuis

3AM, need to sleep :p

Daminark

I should do that soon

Well, any last questions before I head out?

JeSuis

I would like to find connectes sets, If a set $A$ has more than two points and countable ("infinite+finite") then $A$ is not connected, and if $A$ is the complement of countable set the equality $A\cap U\cap V=\emptyset$ is imposibble

so for connected set we get singletons ?

Alessandro Codenotti

I agree with your first message but not with the conclusions you drew

Is the whole space connected?

JeSuis

arf the whole space, i forgot to think about it

it is

even "worth" every open set intersects non trivially

Balarka Sen

09:34

@AkivaWeinberger I don't buy this anymore. What about like $ababa^{-2}b^{-2}$?

Alessandro Codenotti

Right so if every pair of open sets intersects nontrivially you can't write the whole space as a disjoint union of 2 of them. Whay about uncountable subsets?

Balarka Sen

I don't think you can express this in terms of words of the form $[a^n, b^m]$

Alessandro Codenotti

Hi @Balarka

Balarka Sen

Hi @Alessandro

@JeSuis As probably explained to you already, it is countable. In fact the set of all algebraic numbers is also a countable subset of $\Bbb R$

Daminark

Alright, I guess this is being taken care of, so I now bid you all adieu. Hibye @Balarka

Alessandro Codenotti

09:36

Good night @Daminark

JeSuis

Good night

@BalarkaSen ah right, union of zeros! nice

Balarka Sen

Hi/bye @Daminark

user400188

@Patrick Stevens you have given me a very good answer; however I am still unsure why $A\lor\lnot A$ is not literally its truth value. As far as I know; the meaning of an expression is propositional logic is taken to be its truth value.

Balarka Sen

@JeSuis The point is every algebraic number comes from some integral polynomial. And there are countably many integral polynomials.

JeSuis

integral polynomial: ? ; I was thinking to write this as union of countable sets, and such sets are the zeros of polynomial.

Balarka Sen

09:40

Zeroes of polynomials with integral coefficients. With arbitrary coefficients that family would not be countable.

user400188

!

does the comma in the representation of a set i.e. {1,2} represent a logical AND ?

Alessandro Codenotti

See if you can find a characterization of the convergent sequences in $\Bbb R$ with the cocountable topology when you go back to the question you asked before @JeSuis

Leila Hatami

Hi, I will be very pleased if you response to this question. Thank so much

8

Q: Counter intuitive examples in probability.

I want to teach a short course in probability and I'm looking for some Counter intuitive examples in it. Results that seems to be obvious to be false but they are not or vice versa... I already found somethings. For example these two videos: Penney's game How to Win a Guessing Game In additi...

Balarka Sen

10:00

@Alessandro Here's something cool. Suppose you look at the sphere $S^n$; think of it as the unit sphere in $\Bbb R^{n+1}$ if you like. Then projection to the vertical axis is a function $f : S^n \to \Bbb R$ with two critical points (where $df = 0$) on the north and the south pole respectively. Agreed?

Danu

Congratulations on your birthday, @TedShifrin!

Alessandro Codenotti

Yes, where's the trick? :P

Balarka Sen

@Alessandro Nowhere, I wasn't finished. Suppose $M$ is a compact $n$-manifold with a function $f : M \to \Bbb R$ having exactly two critical points (what must those critical points be?). Theorem: $M$ is homeomorphic to $S^n$.

JeSuis

@AlessandroCodenotti constants sequences ?

Alessandro Codenotti

10:22

That's a very interesting fact! I can see why this works for stuff like a cube with slightly smoothed edges and intuitively fails for a torus

yes @JeSuis so $T_1$+"all converging sequences have a unique limit" does not imply $T_2$

JeSuis

ah! cool

user400188

are elements of a set put together with a logical AND $\land$ or a logical OR $\lor$ ?

Alessandro Codenotti

And I still disagree with the fact that uncountable sets are disconnected in the cocountable topology

user400188

@AlessandroCodenotti was that in response to me or someone else?

Alessandro Codenotti

to @JeSuis

user400188

10:29

I saw sets and disconnected and got exited.

Balarka Sen

@AlessandroCodenotti Right. If you put a torus upside down and look at it's "height" than that's a function with 4 critical points.

In fact that's true for any sufficiently nice function (the word is Morse) I think

You know what's really trippy? $M$ admitting such a function $f$ need not be diffeomorphic to $S^n$ ($n$-sphere with smooth structure obtained from realizing it as the unit sphere in $\Bbb R^{n+1}$).

These were the first examples of smooth manifolds homeomorphic, but not diffeomorphic to each other, constructed by Milnor in his celebrated paper. First known example (of spheres) comes at dimension $7$.

Alessandro Codenotti

ok that's really weird. I don't know any example of manifolds being homeomorphic but not diffeomorphic actually

ah, I see, that's very weird indeed. And I thought point-set topology was the one with the weird counterexamples!

Secret

10:47

O wow...
I can only wonder how many of my questions get silently deleted by community or others...

Luckily, I already got that one answered by someone in my uni

user400188

Why would that get deleted?

Secret

No idea, except it is no longer found in my question list

user400188

very odd. Are you sure you didn't delete it yourself? (either by accident or when you got it answered at uni?)

Balarka Sen

@AlessandroCodenotti Any smooth 1, 2, or 3-manifold homeomorphic to each other must also be diffeomorphic, so you're not expected to know any counterexample. The big open question is whether there's a smooth manifold homeomorphic, but not diffeomorphic to S4

Secret

I don't delete questions that got answered.

In general, I rarely delete my questions because it is not necessary

Alessandro Codenotti

10:51

is the answer known in dimension 5 or 6?

user400188

In that case I am very surprised it disappeared. Even more so that there was no notification on it.

Secret

Deleted questions don't give notifications, thus the user just have to check often

I suspect one possibility it is deleted by the community bot because it is a 1 year tumbleweed

(no updown votes and no answers)

Balarka Sen

@Alessandro Yes, there are no exotic spheres in dimension 5 or 6.

user400188

ah that makes sense. I've noticed if you go back far enough (that is the very last page) into the unanswered question section the questions are surprisingly young.

Secret

It does means, however that I might have to kinda backup my questions if I want to check which questions is still unresolved as it is often difficult to remember questions that had been asked especially when they are not related

user400188

11:02

Us new people don't stand a chance at answering logic questions when @Mauro ALLEGRANZA is the fastest gun in the west.

Secret

O and btw, the true background of that PDE is I saw that in a dream of mine, possibly because during that time I have been playing with semilinear PDEs and navier stokes equations.

Of course, when I ask my prof. about it (especially because he does not knew me much and thus does not know about my weird personality traits) I cannot say directly I saw that in a dream. This is because people don't usually take dreams seriously. However I cannot lie either, thus I just said the next most accurate thing: I conceived it when I read about those lecture notes (which is technically true, since drea…

user400188

Mathematics is more about finding interesting things than it is about finding applications to those things.

Usually when we find an application for something it is only an application "In Theory" anyway.

Secret

True, but PDEs are more of the applied maths side of things. Unless they have unusually behaving solutions, they often only caught attention of the math community at large if it arises from a known problem

I and a few users here are leaning towards more on the pure maths side of things, that is, we like to find interesting, pathological and patterns in the mathematical objects and try to understand them, before we start thinking about applications

user400188

I only do math for recreation; so I really have no need to find anything with an application. Besides giving me a new way to think.

Out of curiosity; when you dreamed up that PDE, did you visualize what it looked like or did you just come up with the equation?

Little Rookie

Hi all, i have a question regarding the big O notation.

From the Taylor series, ln(1+x) = x - (x^2)/2 + (x^3)/3 - ...

Why is it that ln(1+1/x) is O(1/x) ?

Secret

11:16

Snipplet: ...The teacher was seen busy writing down these equations (along with 5 other students who are in charged of some acoustic detectors known as "Cylinders" that are obtaining data for some background sound levels in order to be prepared for the sports event tomorrow') on the white board during the assembly where a broadcast informed the strict dress code of tomorrow's' sports event...

So basically in the dream, these equations are use in some kind of acoustic modelling

(But a quick reality check that no sound waves can act like this unless the "Cylinders" have some very unusual structure)

user400188

Wow thats weird. I don't think I've ever heard of someone coming up with an equation (except in pop culture) like that.

Secret

My dreams are known by may SE users to be weird. One of them comments as follows:

Point is, dreams are well known to be incoherent illogical mess, but my dreams are often coherent, real life like and consistent

that is one reason I can sometimes extract insights from them and discuss with my peers and professors

some of my MSE questions are actually inspired from my dreams, btw

user400188

Seems a very convenient way to dream.

Or maybe your just physic.

I usually only ask MSE questions when I stuff up a definition and I want to know what I broke.

Secret

I have questions of that origin too, such as some of my algebra questions

5 months ago I was working on some recreational mathematics involving division by zero. Background reading on semigroups and my investigation result in some of the algebra MSE questions

user400188

Actually: just a while ago I discovered the reason why x=y=z is different from x=y $\therefore$ y=z.

In most cases they are the same; but when the sets the terms are made of differ; the equals will differ from the therefore; because one ensures that they are the same in every case while the other AND's the two statements together. This creates a smaller set where the equal signs will hold.

Secret

11:30

You also have a difference for the "if then" case from the "and" case. For the former, if x=/=y, then nothing can really be concluded about y and z due to the logical structure of "if then"

whereas for the "and" case, x,y,z must all equal

user400188

indeed

I have made that mistake a few times and I often see other people making it by mistake too.

As a matter of fact, I think I will remember your example now because its easier to understand.

Secret

Another way to distinguish between them is that x=y=z is shorthand for "x=y and x=z". One can then easily produce the truth tables to compare between them

user400188

actually never mind I read the first part as x=y and y=z instead of x=z. My bad

user21820

@user400188 @Secret: Actually, it's not very right to use "therefore" if you just mean "implies". "P, therefore Q" conveys that "P and Q" and the reason for Q can be seen from the truth of P.

user400188

My definition for therefore was: $A\therefore B=A\land (A\rightarrow B)$. Which is equivalent to $A\land B$

user21820

11:44

That's more or less what "therefore" conveys.

The problem is not there.

The problem is that "x = y = z" does not convey that at all.

user400188

Taking it to mean: "We assume the first part to be true, and we also say that if the first is true then the second

user21820

Yes.

user400188

Yes x=y=z would not convey that. We were talking about the distinction between them

I just often see people using x=y=z like they would x=y $\therefore$ y=z. Which is a common mistake.

user21820

@Secret It seems @Secret didn't get that point, because his comment here suggests that there is a logical difference.

user400188

Which comment? It could be the one I misread myself. Where 1 letter difference gives it a completely new meaning

user21820

11:48

The linked one.

@user400188 <-- I fully agree with your characterization here. If it is easy to prove x=y and y=z (or previously proven) then you can and should write x=y=z to convey the subsequent deduction that x=z. But if you want to conclude "y=z" then you probably want "y=x=z" or the reverse. Otherwise there's no point to chain the equalities.

user400188

Well its nice to see someone agrees with me; but to be honest I am struggling to find the logical difference in what secret said to what I was talking about. Although I haven't worked through it on paper yet.

user21820

As far as I can tell, @Secret was assuming your latter statement meant "x=y implies y=z", rather than what you actually were talking about. Under that assumption, he/she correctly said that if x!=y then "x=y=z" is false but "x=y implies y=z" is true.

So it's just that you two seem to be talking about different things, which is why I commented on it. =)

Secret

@user21820 Yes, you are correct, I did not know that "therefore" is not the same as "implies"

user400188

Oh. I just assumed you meant the two way implication in = when you referred to the if then structure.

Secret

I am actually using $P \to Q$ when I interpreted "therefore"

user21820

11:57

@Secret: Well yea English is like that haha..

@LittleRookie <-- To answer your question here, none of those statements are precise enough, and that's the problem. ln(1+x) ∈ x - x^2/2 + O(x^3) as x -> 0. The asymptotic behaviour of x is crucial otherwise the statement is meaningless. Now we can see that as x -> ∞ we have 1/x -> 0 and hence ln(1+1/x) ∈ 1/x - O(1/x^2). Makes sense now?

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