Mathematics - 2017-04-03 (page 1 of 3)

In other words, if given a list of numbers, there is a number "k" in this, such that every number before it is ordered descendant and after it ascendant

I can only come up with checking for each number, but there has to be a better way

Alucard

Ok, let's do this

I crash my fathers car, and you serve him, k thanks

arctic tern

@Alucard Who are you talking to?

I have noticed you saying random things to no one for some time now.

@Maks You mean given a list $a_1,\cdots,a_N$, find if there is a $k$ such that $a_i>a_{i+1}$ when $i<k$ and $a_i<a_{i+1}$ when $i>k$?

Maks

Yep, exactly that

Ted Shifrin

2:52 AM

Hi @Maks, tern

arctic tern

you could do some kind of binary search on consecutive terms I guess

hi

Maks

Oh no, wait
$ a_i < a_(i+1)$ when $i < k$ and $a_i > a_(i+1)$ when $i > k$

First ascedant, then descendant

like a peak, you go up, find k and go down

If there is a k

Hi @TedShifrin ! How is it going ?

Ted Shifrin

Doing OK. thanks. Haven't seen you in ages.

Maks

Yeah its been a while

I dont have many math courses this year

Only numerical analysis

Did you miss me ?

Ted Shifrin

Of course :)

Maks

2:58 AM

Hahah thanks

Would you like to help me ?

Like in the old times :D

Ted Shifrin

I have no idea about what you're talking about ...

Maks

@arctictern That :D but inverted

I need to create an algorithm

arctic tern

like I said, binary search on consecutive pairs

Maks

To find if there is a $a_k$ in a list of numbers $a_1,..,a_n$, such that $a_i < a_(i+1)$ if $i < k$ and $a_i > a_(i+1)$ when $i > k$

That's the best way ?

There isnt a quicker way ?

arctic tern

binary search should be pretty quick no?

I just improved your "check every k=1,2,..." algorithm exponentially

Maks

3:02 AM

Yeah it is quite fast, but is it the fastest ?

arctic tern

I'd wager it is, more or less

Maks

I know, I'm not complaining

Oh, I'll try that, thanks !

King KONG

hi

can someone answer my question math.stackexchange.com/questions/2215435/…

Alucard

3:41 AM

@arctictern most of the time i talk to myself, walked now alot and actually screwed what i have set in my mind.

arctic tern

for the most part, a chatroom is a place to talk to others, not to oneself in front of others.

Semiclassical

Right up to the point where I start rambling on something and subsequently realize that no one is listening anymore. :/

arctic tern

there is room for talking to oneself in ways which leave room for others jumping in depending on others' interest and ability to contribute, of course. but this exception does not apply to streams of unexplained non sequitors lacking all context.

Mike Miller

Hi @JoshKeneda.

Adeek

hi @arctictern in H do you define the inner product as <x,y> = Re(x\bar{y}) ?

arctic tern

3:52 AM

yes

well, that's a formula for it

personally, I would define <x,y> as the one corresponding to the quadratic form (which has a formula via polarization), and then define the Re() operator with respect to the inner product

Adeek

okay awesome. I just wanted to verify something I am calculating the following. conjugation by elements $\eta \in S^{3}$ gives us an element of $SO_4$

arctic tern

indeed $S^3\times S^3\to{\rm SO}(4)$, where $(u,v)$ corresponds to the map $f(x)=uxv^{-1}$, is a double cover

Adeek

yeah

arctic tern

conjugating by $\eta\in S^3$ preserves the real axis, so restricts to a double cover $S^3\to{\rm SO}(3)$ (now $\Bbb R^3$ being the space of imaginary quaternions)

Adeek

hm I see

@arctictern you would define it as $<u,v> := 1/2(u\bar{v} + \bar{v}u)$ ?

given that $u,v \in H$ ?

arctic tern

3:58 AM

With a normed division algebra, we're already assuming there is some kind of quadratic form corresponding to a (necessarily unique) inner product. So the inner product is in some sense in-built. The standard way to go between quadratic forms Q(-) and symmetric bilinear forms B(-,-) is Q(x):=B(x,x) and B(x,y)=[Q(x+y)-Q(x)-Q(y)]/2 when the characteristic is not 2.

But it so happens that $\langle u,v\rangle=(u\overline{v}+v\overline{u})/2$ is true

(slightly different from what you wrote)

Adeek

what would be the quadratic form associated to the hamiltonian ?

I want to see how that would give that $<u,v>$ is given as above

ramanujan_dirac

@adeek

Adeek

@ramanujan_dirac hi

ramanujan_dirac

the quadratic form is just the norm i guess? $Q(x,x)= \overline{x}x$. Substituting this in the formula that interpolates between Q and B, u get ur inner product B, which we discussed before.

Adeek

yeah I see that makes sense

okay cool yeah I agree with everything now @ramanujan_dirac I have seen this stuff as well in my quadratic form class in the start

ramanujan_dirac

4:05 AM

@arc

arctic tern

@ramanujan_dirac personally I'd define $\bar{x}$ using the ${\rm Re}()$ and ${\rm Im}()$ operators, and define those operators using the inner product

but yes $Q(x)=x\bar{x}$ is true

Adeek

okay cool

ramanujan_dirac

or in other words if $h= a + bj$, then $Q(x) = \sqrt{a \bar{a} + b \bar{b}}$

arctic tern

my preferences are geared towards the classification of normed real division algebras, in which the inner product is in-built but things like conjugation and real parts are not.

ramanujan_dirac

*$Q(h)$

arctic tern

4:07 AM

@ramanujan_dirac that is also true, but kind of ugly

Adeek

I see

arctic tern

@ramanujan_dirac you can edit your own messages within a 2 minute window

hover over your own message, click the drop-down menu button on the left, click edit

ramanujan_dirac

Thanks!

arctic tern

one thing I proved the other day is that if $X\in M_2(\Bbb O)$ (with $\Bbb O$ the octonions) then $X^3$ is well-defined if and only if $X\in M_2(H)$ for some quaternionic subalgebra $H\subset\Bbb O$. hopefully generalizes to $M_n(\Bbb O)$.

Adeek

oh that is awesome. Btw is there some result about connection of division algebras and homotopy groups of $O_n$ because as we go to $O_{\infty}$ there seem connection between number of repetition some pattern and number of division algebras @arctictern

arctic tern

4:14 AM

yes

this is explained in baez's The Octonions

Adeek

oh

very cool

arctic tern

Baez explains that the division algebras yield elements of the corresponding homotopy groups (k=1,2,4,8), although he doesn't prove they are generators, that nothing happens for other k=3,5,6,7, or Bott periodicity.

I can prove the mod 8 periodicity for (morita equivalence classes of) clifford algebras, but not of homotopy groups of O(inf).

Adeek

oh

arctic tern

it's also related to the parallelizability of spheres (i.e. the trivializations of spheres' tangent bundles)

for $S^1$, if you draw tangent lines to it at every point, you get a kind of "crop circle." one can "unpress" the stalks, essentially rotating them into a third dimension, in order to get a cylinder $S^1\times\Bbb R$. the question is if this can be done to any sphere $S^n$. turns out to only be possible for $n=0,1,3,7$ (division algebras)

division algebras are also inherently linked to the classification of projective spaces, jordan algebras, and simple lie algebras

Adeek

why ? I am interested to know more about this

arctic tern

4:22 AM

Well, given the unit sphere in a division algebra, you can trivialize it by applying left or right multiplication maps to each tangent space to move them all to the tangent space of the identity. In other words, for example for $S^3\subset\Bbb H$, if $TS^3=\{(u,v):u\in S^3, v\perp u\}$, we have $TS^3\to S^3\times \Bbb R^3$ given by $(u,v)\mapsto (u,vu^{-1})$.

(In fact I just used that in my most recent answer to prove the space of all oriented 2D planes of 4D space itself forms $S^2\times S^2$.)

As for why nothing else works in any other dimension, that doesn't have an obvious explanation, but it's a corollary to the generalization known as Adam's Theorem or the Hopf Invariant One Theorem.

I tried to prove it myself with more elementary topology, but to no avail. One thing I noticed is that any trivialization of $S^{n-1}$ is equivalent to a subgroupoid of the action groupoid $S^{n-1}/\!/{\rm SO}(n)$ with trivial automorphism groups.

(in the hopes of using groupoid theory)

Adeek

cool

@arctictern are you currently working on this ?

arctic tern

daydreaming and scribbling about it

Adeek

awesome

i am currently working reading UCLA paper by paul balmer generalizing spec of a ring

to triangulated tensor categories

arctic tern

dunno what triangulated means

Adeek

@arctictern So you have an abelian categories and triangulated categories is defined as follows a triangle is if you have the following sequence of given by the following data : an additive automorphism from category D to D given as $T : D \rightarrow D$

then a triangle is a sequence of morphism $U \rightarrow V \rightarrow W \rightarrow T(U)$

This is part of definition and it there need to be class of distinguished triangles satisfying certain conditions

Daminark

4:52 AM

Hey there everyone!

Hello !

hi

How's it going?

good just about to head to sleep

nights everyone cya @Daminark

Daminark

Alright, catch you around @Adeek!

Dattier

6:28 AM

hi, a little problem, it can be solve in one line (compréhensible d'un éléve de prépa) : math.stackexchange.com/users/408734/dattier?tab=questions , who can find the line ? Maybe @LeGrandDODOM lol

if @LeGrandDODOM doesn't find, he must change his name for another more adapted, I propose lepetitdodom

lol

And he must change his class A+ for the class Z-, lol

hello, @s.harp do you know what it hapens here : math.stackexchange.com/questions/2213224/…

Balarka Sen

7:31 AM

@Daminark rick rolling is frustrating and not humorous :P

Daminark

I mean I can find a number of people who will strongly contest that

skill patrol

I'm one :P

Balarka Sen

it's a bit trolly, but not funny

Daminark

Well you can't really say that something "is not funny", like there's no objective standard

If people are amused, then as far as they are concerned, it's funny

skill patrol

Rick Ashley's music is a joke.

Joke = funny

Balarka Sen

7:38 AM

@skill Now that you parse it that way...

skill patrol

:-)

Balarka Sen

"April is the cruellest month, breeding / Lilacs out of the dead land, mixing / Memory and desire, stirring / Dull roots with spring rain."

starts with april 1st

skill patrol

nvm

rickrolling :P

skill patrol

7:53 AM

Practical joke

A practical joke is a mischievous trick played on someone, generally causing the victim to experience embarrassment, perplexity, confusion or discomfort. A person who performs a practical joke is called a "practical joker". Other terms for practical jokes include prank, gag, jape, or shenanigan. Practical jokes differ from confidence tricks or hoaxes in that the victim finds out, or is let in on the joke, rather than being talked into handing over money or other valuables. Practical jokes are generally lighthearted and without lasting impact; their purpose is to make the victim feel humbled or...

In Western culture, April Fools' Day is a day traditionally dedicated to conducting practical jokes.[4]

Alessandro Codenotti

8:14 AM

I have a sequence of continuous functions $f_n:\Bbb R\to\Bbb R$ and it converges pointwise on some set $X$, how bad can $X$ be?

s.harp

9:22 AM

For a sesquilinear form, does positivity imply hermiticity?

ie $(x,x)≥0$ for all $x$ implies $(x,y)=\overline{(y,x)}$

Well, that was way to simple, $(x+iy,x+iy)=(x,x)+(y,y)+i(x,y)-i(y,x)≥0$ so the real part of $(x,y)-(y,x)$ must be zero and $(x+y,x+y)=(x,x)+(y,y)+(x,y)+(y,x)≥0$ so the imaginary part of $(x,y)+(y,x)$ must be zero

Dattier

I propose a challenge, in cryptographie : math.stackexchange.com/questions/2215753/…, in analysis : math.stackexchange.com/questions/2212738/…, in convexity : math.stackexchange.com/questions/2213224/… ,

for show : It isn't because a solution is short and easy to understand that it is easy to find

for to show : It isn't because a solution is short and easy to understand that it is easy to find

SteamyRoot

I think everyone knows that sometimes short and easy solutions aren't easy to find...

Also, one of those questions is closed

And the most recent one states "This question have an answer of less ten lines with a level of license begining (L1-L2).". I find the "number of lines" to be an awful metric, because it depends heavily from author to author how much is written on a single line; and I have no idea what level "license beginning (L1-L2)" is.

Vrouvrou

9:40 AM

Hello, i have $F$ is a continuous function such that $\lim_{|t|\rightarrow+\infty}\dfrac{F(t)}{|t|^m}=+\infty$

how we deduce that $\forall M>0, \exists C(M)>0; F(t)\geq M|t|^m-C(M)$

SteamyRoot

Use the definition of a limit?

Vrouvrou

it is $\forall M>0, \exists B>0, |t|>B\Rightarrow F(t)> M |t|^m$

how to find the constant $C(M)$ ?

SteamyRoot

Well, how many "candidates" for $C(M)$ do you have?

Wait, nevermind.

Vrouvrou

i don't know

i see if $|t|\leq B$ then $F$ is continuous on a bounded then $|F(t)|\leq C$

SteamyRoot

You know that $F(t)$ is large enough outside the interval $[-B,B]$

And inside $[-B,B]$, it's a continuous function on a compact set

Vrouvrou

9:50 AM

yes i understand thank you

SteamyRoot

So you can use the extreme value theorem

$F(t) \geq N$ for some $N$ when $|t| \leq B$

Vrouvrou

ohh , i just use $F(t)>M |t|^m$

and $F(t)\geq-C$

then $F(t)\geq M |t|^m-C$

SteamyRoot

Not necessarily!

Vrouvrou

how ?

SteamyRoot

The $M|t|^m$ can ruin things inside the interval $[-B,B]$, I think.

You can subtract another term to fix that, though.

(I mean, suppose that $F(t) = -C$ for some $t \neq 0 \in [-B,B]$. Then $F(t) < M|t|^m - C$

Vrouvrou

10:05 AM

i don't understand

SteamyRoot

You have two inequalities: $F(t) \geq -C$ and $F(t) \geq M|t|^m$

You can't just claim that $F(t) \geq M|t|^m - C$

Vrouvrou

why ?

please

SteamyRoot

Because if $x \geq 1$ and $x \geq 2$, you can't conclude that $x \geq 3$ ? (take $x = 2$ for example...)

Vrouvrou

i sum the two nd devide by 2

SteamyRoot

then you have to divide the right-hand side by $2$ too

Vrouvrou

10:11 AM

yes

SteamyRoot

So you get $M/2$ and $-C/2$

Vrouvrou

yes

SteamyRoot

You want to show that $F(t) \geq M|t|^m - C(M)$, not $F(t) \geq \frac{M}{2}|t|^m - C(M)$

Vrouvrou

they say for any M>0 there exists

so as when we use \varepsilon

SteamyRoot

So $M$ is already fixed.

Secret

12:40 PM

[Position dependent pseudoinverse]
Let $S$ be a semigroup. Suppose we have the following axioms:

$\forall x \in S, axb=x$

$xx=x^2$

$y=x^5$

$\forall y, cyd=y$

Now compute:

Secret

12:59 PM

$y=cyd=cx^5d=caxbaxbaxbaxbaxbd=c(axb)^5d$

$=c^5(axb)^5d^5$

$(caxbd)^5=(caxbd)(caxbd)(caxbd)(caxbd)(caxbd)=(cxd)^5\neq c^5(axb)^5d^5$

Or more explicitly:
$c^5x^5d^5\neq (cxd)^5$

I think I need some more advanced tools. Just ramming a bunch of elements together permutatively and see what pops out is too inefficient

Yes, it seems I do need the green's relations to characterise this further

nbro

1:26 PM

hey dudes!

what's up!? :D

Akiva Weinberger

Heh

BAYMAX

so theorems can be false also ?

@Akiva

Akiva Weinberger

Technically they're not really "theorems", I guess

BAYMAX

"Mathematical theorems you had no idea existed, cause they’re false"

Akiva Weinberger

It's a jokey title

SteamyRoot

1:39 PM

Oh wow

That's a really neat one.

BAYMAX

Ohh Ok.

nbro

how can a paradoxical proof be really a proof?

SteamyRoot

Should go in my list of fake proofs that all integers are equal :O

Akiva Weinberger

You have a list?

nbro

I mean, how can a paradox be considered a proof?

Akiva Weinberger

1:40 PM

Well, clearly, it has 100 entries, because it has an integer number of entries and all integers are equal :P

SteamyRoot

Haha :P

nbro

the proof to the halting problem is simply a version of the liar's paradox

wtf

SteamyRoot

It's a list me and 2 friends made while we were Master students

nbro

and decidability theory is based on a faked proof

lol

SteamyRoot

Partially because we loved to mess around during boring classes, partially because we shared the belief that these fake proofs were the best way of convincing people that certain pitfalls really are dangerous

Akiva Weinberger

1:42 PM

@nbro The proof of Gödel's incompleteness theorem is also inspired by the liar's paradox.

Harry Evans

Hello, good morning! Anyone up for some good questions on finite fields with cyclic extensions?

nbro

@AkivaWeinberger Everything which is self-referential can be somehow paradoxical

I don't think it's a valid proof

I bet we can prove a lot more things by using self-reference-like proofs

Harry Evans

Say, you have $F$, an extension of $\mathbb {Z}_2$, such that $|F| = 2^3 = 8$. What are the three automorphisms of $F$ fixing $\mathbb {Z}_2$?

Anonymous

Can someone please help me with this: math.stackexchange.com/questions/2215878/…

Anonymous

1

Q: Which is the correct method for estimating error in $f$?

In the formula [called the Lens Equation] if I have to find maximum permissible error in $f$ (from the graph) which is the correct method? $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$ Method 1: $${df}=f^2\left(\frac{dv}{v^2}+\frac{du}{u^2}\right)$$ Method 2: $${f}=\frac{uv}{u+v}$$ $$\ln(f)=\ln...

Harry Evans

1:58 PM

Hi blue

$f$ depends on both $u$ and $v$, right?

Semiclassical

@blue A minor observation: In method 2, it should be $f=uv/(u-v)$.

Harry Evans

Thus,

$f = \dfrac {uv} {u-v}$

And therefore,

$df =[\dfrac {uv} {u-v}]_u + [\dfrac {uv} {u-v}]_v$

where you're getting the partial derivative of $f$ with respect to $u$ and $v$

I'm missing something, silly me hahaha

$df = [\dfrac {uv}{u-v}]_u du + [\dfrac {uv}{u-v}]_v dv$

Semiclassical

@HarryEvans I don't think this is intended as differentials as such, but rather in the sense of error analysis. For instance, if you do $z=x/y$ then $dz/z=dx/x-dy/y$. If you know that the experimental $x$ and $y$ differ from the true values by $1\%$, then you could have as much as $2\%$ error.

(You could also find that the errors effectively cancel out, e.g. 2.02/1.01=2. But in a typical experiment you don't know that, so one assumes the worst case scenario for error analysis.)

Akiva Weinberger

@nbro All that means is that you don't understand the proof well enough. (I'm actually a little hazy on the details of the halting problem thing, but Gödel's proof actually uses a very clever construction to make the "self-referential" thing work.)

Semiclassical

@AkivaWeinberger Godel's argument = Cantor's diagonalization argument applied to a version of the Liar paradox, if memory serves.

nbro

2:13 PM

@AkivaWeinberger I understood the proof of the halting problem, it's a simple proof by contradiction

but it's based on the liar paradox

Semiclassical

Modified in such a way as to not be a paradox, though.

nbro

at this point is like saying that proof by contradiction do not really work

*proves

Akiva Weinberger

@Semiclassical Huh, the version I read didn't look like Cantor diagonalization

Semiclassical

I don't actually remember the details either, though.

Akiva Weinberger

@nbro **proofs

nbro

2:14 PM

the thing is, the contradicting fact contradicts the assumption directly

Semiclassical

Well, the version you saw may not have been Godel's original argument?

Akiva Weinberger

…Isn't that how proofs by contradiction are supposed to work?

@Semiclassical Probably

nbro

@AkivaWeinberger the contradiction, as far as I know, doesn't have to be a direct contradiction of the assumption, but some contradiction in the reasoning

Semiclassical

Details here: cs.cornell.edu/courses/cs4860/2009sp/lec-23.pdf

Relevant bit: "Godel’s important modification to [Cantor's] argument was the insight that diagonalization on computable functions is computable, provided we use a Godel-numbering of computable functions."

Balarka Sen

I have heard of a Cantor diagonalization proof but I didn't read it except pop-math stuff about it "list an infinite collection of PA truths you can prove by writing them as an infinite word of 0's and 1's. write the diagonal and replace 1's by 0's and vice versa. That's a new truth not in your list"

No idea how that formalizes

The proof I know is by Tarski schemas.

Semiclassical

2:20 PM

That's discussed in the link, though probably not in such a pop-math way.

I'm admittedly not interested in reading through it myself, though.

Akiva Weinberger

@BalarkaSen I vaguely remember discussing something about this with you a few months ago

I did, but I didn't give nearly enough details

Dec 28 '16 at 23:42, by Akiva Weinberger

@BalarkaSen In any case, this all has to do with Gödel's construction of a sentence that says "This sentence is unprovable". $g$ is "The statement with Gödel number $x$ is unprovable", and $f$ is "Replace all instances of the free variable in the formula with Gödel number $x$ with the number $x$."

I think the idea was to consider $h=f\circ g\circ f$, so that $g\circ h=g\circ (f\circ g\circ f)=(g\circ f)\circ(g\circ f)=f\circ(g\circ f)=h$

Balarka Sen

Ah, I remember now.

Akiva Weinberger

But now I'm confused as to how exactly that gives us the desired sentence

Anonymous

@Semiclassical Actually it doesn't matter, we need to use u and v with signs....both can be considered positive in this case

Anonymous

u is the object distance which will be negative

Anonymous

2:34 PM

and v should be positive

Anonymous

So if we take the magnitudes

Anonymous

then the formula still holds

Anonymous

(I should have mentioned that in the question though :-P)

Anonymous

It seems more of a physics question than a maths question :/ I'm not sure it will get answers on this site

Anonymous

@HarryEvans I took the logarithm...

Anonymous

2:36 PM

And then differentiated

Anonymous

I took only the magnitudes to remove the confusion arising due to sign conventions

Anonymous

@AlessandroCodenotti LOL :-P I assume that refers to Terence

Anonymous

"though" problem..spelling error there :-D

SBM

Um, I'm looking for a faster way to solve sums like

$$\int \dfrac{1}{2 \cos 2x + 1} \: \mathrm{d} x$$

I know that the answer to this is fairly simple

$$\DeclareMathOperator{artanh}{artanh} \dfrac{1}{\sqrt{3}} \artanh \left(\dfrac{\tan x}{\sqrt{3}}\right)$$

plus the constant.

Alessandro Codenotti

@blue what's the correct spelling? I'm not english

SteamyRoot

2:43 PM

tough

Anonymous

"tough"

Anonymous

@SBM Put $u=tan(x)$

SteamyRoot

though is short for although

SBM

Ok

Anonymous

It should take only 1 minute or so

Anonymous

2:46 PM

After using u=tan(x)

Anonymous

I don't know any shorter method

Arrow

3:09 PM

Hello,
This is probably silly, but why does $S\cap (N+P)=0$ imply $N\cap (S+P)=0$ in the proof below?

Akiva Weinberger

@BalarkaSen Here's something that should have a simple proof, but I'm not actually sure how to do it elegantly.

Consider a plane in $\Bbb R^3$, along with a sphere tangent to it. The complement of that has three components: a half-space, a ball, and a half-space-minus-a-ball.

Consider the closure of that last one. It clearly deformation retracts onto its boundary, but is there an easy way to show it?

Balarka Sen

@AkivaWeinberger What do you mean by a simple proof? You just contract everything from infinity to wrap around the sphere component in the boundary of the closure, yes? That should be the deformation retract.

Akiva Weinberger

Something like that. But it's not very rigorous.

Balarka Sen

I don't know how to write it down, if you're looking for that.

Nah, sure it is :)

Akiva Weinberger

I guess you'd probably use something about how that's a closed half-space with a circle on its boundary plane quotiented to a point

@BalarkaSen If you can't write it down, it's not rigorous :P

Balarka Sen

3:23 PM

Sure. Pictures are rigorous. I can't write down half the pictures I say.

@AkivaWeinberger Yeah, I guess.

Akiva Weinberger

That is an odd sentence.

Alessandro Codenotti

@AkivaWeinberger can't you do a linear retraction even?

Balarka Sen

@Akiva Project everything deformation retract the half-$\Bbb R^3$ until you reach the equator of the sphere (so you're retracting half space to $\Bbb R^2$ with the unit disk removed and the upper unit hemisphere attached to the boundary).

The rest needs to be something non-linear.

Alessandro Codenotti

Or at least a retraction such that the segment between $t$ and $r(t)$ doesn't leave the space

Balarka Sen

(also answer to @Alessandro: you can do it linear 'till it reaches the equator of the sphere, not afterwards)

Alessandro Codenotti

3:29 PM

Ah, right, you took out a whole sphere, not half of it

Balarka Sen

So you want to - what - deformation retract the half-space (interior of the parabolid removed) to $xy$-plane with a paraboloid touching the origin in the $xy$-plane?

I seriously don't know how to write that down.

Akiva Weinberger

You're fitting a paraboloid in between the plane and the sphere?

SteamyRoot

Can't you like, first project downward onto a surface that looks like a ball covered by a sheet, and then tuck in the sheet underneath the ball?

Balarka Sen

@SteamyRoot Akiva wants to write that.

But yes.

SteamyRoot

I'm guessing that in this case it's enough to write the deformation retract for a two-dimensional section, and then expand it by rotational symmetry...

Balarka Sen

3:32 PM

@Akiva No, I mean you just retract the bit of $\Bbb R^3$ in between $z = 0$ and $z = 1/2$ to the lower hemisphere that fits between them.

Because for the upper bit it's projection, like I, Alessandro and Steamy said.

Akiva Weinberger

Right, I see

I feel like this should be a special case of some theorem

Balarka Sen

So upto homeomorphism I guess you want to retract $D^2 \times I$ to the cone $C(D^2)$ inside it.

That should be super-obvious.

Akiva Weinberger

@BalarkaSen How do you see that?

Remember that it's being deformation retracted onto the plane $\cup$ the sphere, not just the sphere

Balarka Sen

That's what I meant, I just don't want to repeat the plane everytime I say it. Pinch $D^2 \times \epsilon$ to the circle around the cone at level $z = \epsilon$ for some small $\epsilon > 0$.

Then push it down.

Semiclassical

hi chat

@blue The errors can be considered as positive/negative, but you need to treat u/v consistently when it comes to signs.

Akiva Weinberger

3:44 PM

I don't get what you're saying but whatever

Semiclassical

In general, I think it's best to do all error analysis with definite signs at first, and only at the very end take magnitudes.

Balarka Sen

It's the same tucking idea. Anyway, whatever, I don't care enough to write this down. Don't see the point in doing this.

@Akiva Actually you don't even need that. def ret $D^2 \times I$ to $C(D^2)$, just by projecting. In doing so you do get the plane, because you retract $\Bbb R^2 - D^2$ to $\Bbb R^2 - pt$ where $D^2$ is the unit disk in the boundary plane.

What you're doing is you're projecting outside the solid cylinder in which the lower hemisphere is inscribed in. So you retract to $(\Bbb R^2 - D^2) \cup (D^2 \times I - L)$ where $L$ is the solid lower hemisphere.

Then retract $D^2 \times I - L \to \partial L$ by projecting each $D^2 \times t - L_t$ to $\partial L_t$ where $L_t$ is the (disk) level set of the solid lower hemisphere at $z = t$.

In this process you also retract $\Bbb R^2- D^2$ to $\Bbb R^2 - 0$ where $0$ is the vertex of $L$. This finishes everything.

Mary Star

Hello!! Is there an other way to solve 2^n=2000 without using logarithms?

Hello @SteamyRoot !! Do you have an idea?

Adeek

Is it true that if we have linear transformation over $R^n$ then we have $T(R \times R \times R)$ m-times is equal to linear transformation with R m-times ?

Semiclassical

4:00 PM

@MaryStar Given that the answer is literally $n=\log_2 2000$, no.

If you want to estimate that result, there are ways. (For instance, 2^11=2048 so therefore n is a bit less than 11. A linear approximation to log yields the more precise estimate of 11-0.024.)

(ignore that 'more precise estimate', it's wrong.)

Balarka Sen

Hi @Ted

Ted Shifrin

Hi Balarka

Semiclassical

Problems like this make me cranky, if only because of how lazy they are: math.stackexchange.com/q/2216179/137524

You have a hint. Use it.

Ted Shifrin

@Semiclassic: Did you find any of the missing pages back in the main office?

Mary Star

@Semiclassical Ok, thanks!!

Semiclassical

4:12 PM

Not yet. Haven't gone over there yet since our TA meeting in that building isn't till 2:30.

Was hoping to catch my advisor on research stuff before that, but haven't seen him yet.

Balarka Sen

@AkivaWeinberger Ok... I guess your idea was sound too. $\Bbb H^3$ deformation retracts to $\Bbb R^2$ fixing a circle $S^1$ on the boundary plane; so $\Bbb H^3 / S^1$ deformation retracts to $\Bbb R^2/S^1$. Sounds like $\Bbb H^3/S^1$ is precisely the component of the complement you talked about.

(I think; by universal theorem of quotient maps)

Mary Star

4:30 PM

We have an equilateral triangle with edge a. Inside there is an other equilateral triangle where the vertices are on the middle of the first triagle. This method is repeated.

I want to find the area of the nth triangle.

The area of the first triangle is $\frac{a^2\sqrt{3}}{4}$.

So the next area is $\frac{(\frac{a}{2})^2\sqrt{3}}{4}$ ?

So, is the area of the nth triangle $\frac{(\frac{a}{2^{n-1}})^2\sqrt{3}}{4}$ ?

Daminark

Hey everyone!

Mary Star

Hello @TedShifrin !! Do you have an idea bout my question?

Semiclassical

@MaryStar It'll be a proof by induction. Suppose you've got a equilateral triangle of area $A_n$. Show that $A_{n+1}=cA_n$ where $c$ is an appropriate scaling factor.

(you've pretty much got $c$ already.)

Mary Star

It is true that the side of the new triangle is the half of the previous one, or not? @Semiclassical

Semiclassical

Should be, yeah. To be nerdy, after the first step you've got a triforce: