Mathematics - 2018-10-18 (page 2 of 4)

Akiva Weinberger

13:24

So I was thinking about the problem of,

say two people are dividing a yard, so they're putting up a wall

but they still want to be able to walk from one yard to the other, so they leave a whole in the wall

but they also want privacy, so they each want it such that you can't see into their yard from the other.

And there's a lot of ways to do this, but it was interesting thinking of the different shapes you can put

Like this, which has a big pillar in the middle

(You can't see from the top half-plane into the bottom half-plane unless you're inside the structure in the middle, but you can still walk from one to the other)

Secret

Akiva Weinberger

or this thing

Eran

Let $X$ be a set, $\cal{E} \subset \mathbb{P} (X)$ and $\cal{M} = \sigma(\cal{E})$

Secret

I wonder what does the general solution to the yard problem look like...

What do all the possible fence configuration have in common geometrically?

Akiva Weinberger

or this thing (which is kinda like the inverse of the first one)

(And you can also divide it in half if you're OK with the assymetry)

(which would look kinda like ーコー)

Ninja hatori

13:34

any application or motivation to study quadratic forms

Secret

This might be the common feature in all these fences

a hole sanwiched between two walls

Akiva Weinberger

I kinda want to turn this into an optimization problem but I dunno what quantity I'd want to optimize

Secret

well... for -( O )-, that... might be the above hole component split into two halves

All of these configuration cab be made to equally divide the yard into congruent pieces, so that does not seemed to be an interesting to optimise

what about... the length of fence used, as that is a practical consideration

People will like to use as less resource as they can to achieve a task

Thus one possible optimisation parameter is the total length of fence used

subjected to the constraint listed in the problem

Akiva Weinberger

If you have really thick walls, you can get away with this:

(If you're playing Minecraft or something, where everything's made out of 1m x 1m x 1m cubes)

Secret

that seemed to be the most optimised perimeter possible

any turns with only make things longer

Akiva Weinberger

13:42

With infinitesimally thin walls, you can still approximate the "thick wall" solution like this:

but that seems kinda ugly to me, I dunno

Secret

Is it the most optimised? It seems longer to me

Akiva Weinberger

The most optimized one is probably the - ]-- one

(i.e. half of the "I"-shaped one I drew, which I realize is kinda hard to approximate with keyboard characters)

With infinitesimally thin walls, anyway.

I don't think this has any practical application

Like, nothing's wrong with just using doors, right?

Secret

lol yup

(and you can lock them if you want to)

But I suppose for certain garden designs, if you want to conceal something without using a door, then such geometry will be useful

There are also public toilets that are built using similar principles thus no doors needed. You see many of these throughout the world

Lozansky

Imagine a spherical shell of radius $R$, carrying a uniform surface charge $\sigma$, set spinning at $\mathbf{\omega} = \omega \hat{z}$. Now I'm interested in the magnetic dipole moment of the sphere and there are two ways that I see to approach this:

1) Write down the current density $\mathbf{K}(\mathbf{r})$ and compute $\mathbf{m} = \dfrac{1}{2} \int (\mathbf{r} \times \mathbf{K} )\text{ da}$. With $\mathbf{K} = \sigma \mathbf{v} = \sigma (\omega \times \mathbf{r}) = \sigma \omega R\hat{\phi}$ and $\mathbf{r} = R \hat{r}$ so $\mathbf{r} \times \mathbf{K} = \sigma \omega R^2 \hat{z}$ toge…

First result is wrong, second is correct

Where does approach 1) fail?

Akiva Weinberger

No idea, I know nothing about electricity

but maybe there was a spherical integral somewhere and you forgot the $rdrd\theta$ or something?

Wait, that's not spherical coordinates, that's polar coordinates

Lozansky

13:52

Surface element of a spherical shell is $R^2 \sin \theta d\theta d\phi$

Akiva Weinberger

Arright

Can't help you then

@Secret Hey, the second one I drew actually has a redundant segment: the small vertical segment on the bottom left isn't needed.

If we delete that, we have something with only five units of fencing (not counting the main wall).

Is this the most efficient, if we're only allowed to build on one side?

Secret

Will this work?

Akiva Weinberger

I guess we need to make some qualifications as to the smallest object required to pass through it

@Secret Line of sight

Secret

ah...

In that case, that separation in the middle of the second is is necessary, since the same line of sight can be drawn

Loong

That's something we need for radiation shielding design.

Akiva Weinberger

14:07

No, not that one

The vertical segment to the bottom-left of it @Secret

In any case, I think these are the most efficient, where you can use only one side or both sides respectively

Sorry for the inconsistent scale; both arcs are meant to have radius 1 (the width of the hole)

On the top one, that's a 135 degree arc, combined with a short unit-length segment

Secret

sounds legit, I wonder if we can prove there is no more efficient ones.. but then geometry is not something that is easily turned into proofs, hmm...

Akiva Weinberger

Did the first one in Desmos to look less rough

Line of sight just blocked

So that was some random stuff

ÍgjøgnumMeg

design it in such a way that the area inside the triangle is the same as the area outside and then cut off a corner

lol

Akiva Weinberger

14:23

Heh

ÍgjøgnumMeg

or, alternatively

Akiva Weinberger

Yeah, if "no-man's land" is that $1\times1$ square in the bottom-left, then it works

ÍgjøgnumMeg

hahaha

I mean

Akiva Weinberger

Dunno why you didn't just draw the diagonal of the rectangle and chop off a corner

But yeah if it's an enclosed space you can just push the hole to the side

ÍgjøgnumMeg

I'm trying to minimise the amount of space wastage caused by some kind of "line of sight blocking mechanism"

Akiva Weinberger

14:26

Also, the area inside the triangle is always equal to the area outside

Half base times height

ÍgjøgnumMeg

I mean, not if you just stick a random triangle in the middle of the yard

this is what I was getting at

cool question tho

Akiva Weinberger

14:48

imgur.com/a/E6e0GOz

Desmos-ified drawings

Fargle

16:33

@TedShifrin: apparently someone's developed a font based on Palatino that has a compatible mathmode font.

It doesn't look half bad.

Alucard

can this be? a function takes arbitary large values, but not infinity.

Fargle

@Alucard How do you mean taking "arbitrarily large values"? Certainly $f : \Bbb R \to \Bbb R$ by $f(x) = x$ takes arbitrarily large values, but is never infinite.

Semiclassical

Not really. If it takes arbitrarily large value, then as such you can find a sequence of function values which grow without bound

(And "grows without bound" is about the only way you'd ever say a function takes on infinity as a value)

Corellian

test: $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)$

$\left(\begin{array}{c} 0 \\ 1 \\ 2 \end{array}\right)$

Semiclassical

an alternative to that is \begin{pmatrix} a & b \\ c & d\end{pmatrix}

Corellian

16:42

oh special commands

thanks

Semiclassical

the drawback is that it doesn't let you right/left justify your columns

Corellian

in the 2 x 2 case the upper half of the curvy brackets (parentheses) are invisible. hmm

Semiclassical

what?

\begin{pmatrix} a & b \\ c & d\end{pmatrix}

Leaky Nun

Sandbox

Where you can play with chat features (except flagging) and ch...

2

Semiclassical

displays fine on my end

The main reason for array is when you want to do stuff like tables or piecewise functions

for doing matrices, pmatrix/bmatrix/vmatrix are sufficient

Corellian

16:47

seems to be a problem in Firefox browser

Semiclassical

weird

Corellian

Not just chatjax, anywhere it's a latex render of brackets with certain dimensions. Oh well

Semiclassical

I use chrome, so i wouldn't know

Alucard

how does a echolonform of a hypermatrix look like? (where are the zeroes?)

Rithaniel

So, question worded as follows: Let $(X,\tau)$ be a regular topological space and $F$ a closed subset of $X$. If we define an equivalence relation $R$ on $X$ via the partition $\{x|x\in F\}\bigcup (\bigcup_{y\notin F}\{y\})$ Show that $X/F := X/R$ is Hausdorff.
I think I'm missing something, because I don't know if this should necessarily be Hausdorff.

Michael.P

17:08

The problem and the solution. Can anyone rephrase the solution, because I really don't get it

mercio

which part do you not get ?

Michael.P

Specifically, in the third step, are we replacing in the expanded (1+x)^{n+m} for n+m over l ?

Semiclassical

the point is that you're looking for which terms in the left-hand side are of order L (writing it like that to disambiguate from 1)

and the only way to get terms of order L is when k+j=L

mercio

and also they are not explicitly finishing the proof by pointing out that the coefficient of x^l on the right is binom(n+m,l), and that the coefficient of x^l on the left is equal to the coefficient of x^l on the right

Michael.P

And we see that by looking at the power of x^k * x^j ?

Semiclassical

17:16

yes.

Martin Hopf

Find the next one in the sequence: 11, 23, 47, 2351, 4703,...?

Michael.P

The problem is that these powers will overlap several times
0 1 2 3 4 5 ... n+0
1 2 3 4 5 ... n+1
...
m m+1 m+2 ... n+m

mercio

when you expand the product you get sum for k=0 to n of sum for j=0 to m of binomial(n,k) * binomial(m,j) * x^k * x^j

and x^k * x^j is x^(k+j)

and so when you look at the coefficient of x^l you only look at the terms of the sum where k+j is l

Semiclassical

@MartinHopf note that the terms are increasing rapidly, probably too fast for a polynomial function to be what the asker had in mind

But do you notice anything about how each term relates to the next, in terms of their overall size?

Martin Hopf

2*11+1=23

2*23+1=47

Semiclassical

17:21

yeah, but then 47*2+1=95

but what I was going to suggest doesn't help much either. The change from the third to the fourth term is so big

Michael.P

@mercio umh, okay. There will be a special case when the equation we're proving will make sense, what about all of the 'superfluous' values of k+j that don't fit into l

Martin Hopf

50*47+1=2351

Ninja hatori

0

Q: about exact sequenceof witt group of rational

I want to understand this lemma. $|l|<p $ is clear. But why < $pn_1n_2m$ > - < $prm$ > $=$ < $pn_1n_2m$ > -<$lm$> -<$prm$> mod $L_{p-1}$ is unclear 2nd step all other steps after and before that are clear.

Semiclassical

@Michael.P Those won't fit into a particular value of l. But you're not considering just one value of l, but all possible ones

@MartinHopf sure, but that 50 is pretty arbitrary

Rithaniel

(it took me way too long to determine that "regular"referred to separation.)

Semiclassical

17:24

the most I noticed is that, when you shift all the numbers up by one, then they're all multiples of 12

11=12*1-1
23=12*2-1
47=12*4-1
2351 = 12*196-1
4703 = 12*392-1

Martin Hopf

2*2351+1=4703 that is the first result.

Semiclassical

sure, but then why is 2351=50*47+1 relevant?

mercio

I hate those kind of questions

Semiclassical

same

for instance, I hate the one that showed up here: fivethirtyeight.com/features/…

Michael.P

@Semiclassical so how do I prove it -_-

Semiclassical

17:27

9, 10, 19, 24, 31, 40, 51, 64, 79, 90, ?

@Michael.P By understanding how polynomial multiplication works. Try multiplying out (a0+a1x^1+a2 x^2+a3 x^3)(b0 +b1 x+b2 x^2) and see which terms end up being what degree

The answer to the above? It's A9, because the pattern is (N+2)^2 in hexidecimal

Martin Hopf

start instead with ord(2,x)=11 the multipicative order of x in base 2=11.

Semiclassical

if we're going to have to go to something as abstract as that, then I'm really not interested

The more arbitrary the pattern, the less I care.

Martin Hopf

thes smalles number of multiplicative order of 11 in base 2 is: 23

Semiclassical

I think the pattern may have something to do with the fact that you've got (23)51 and (47)03

but I honestly can't bring myself to care

Martin Hopf

what is the smallest number with multiplicative order of 23 in base 2 then?

Semiclassical

17:34

What?

base 2 isn't a group or a ring or anything. so i have no idea what you intend by multiplicative order

Martin Hopf

see en.wikipedia.org/wiki/Multiplicative_order

Semiclassical

That's modulo 2, not base 2

And even so that doesn't help a whit, since 11^1 is already 1 mod 2. So the multiplicative order of 11 mod 2 is 1.

Every odd integer has multiplicative order 1 mod 2.

Martin Hopf

yes, and ord_23(2)=11

Semiclassical

That's the multiplicative order of 2 mod 23.

Ninja hatori

@mercio math.stackexchange.com/questions/2960986/… have a look at it

Semiclassical

17:43

So now I guess you'd want to do ord_x(2)=23

Martin Hopf

yes, that is right.

Semiclassical

well, the multiplicative orders of 2 mod odd integers is listed here: oeis.org/…

Ted Shifrin

Hi DogAteMy, @Semiclassic

Semiclassical

hi

Fargle

Hey @Ted.

Ted Shifrin

17:45

Hi @Fargle. I saw your Palatino note.

MatheinBoulomenos

Hi @Ted

Ted Shifrin

Hi Mathein

Rithaniel

Heya Ted

Ted Shifrin

Heya Rithaniel ... Nice $\Theta$.

MatheinBoulomenos

finally taking an algebraic geometry class! there was an interesting exercise on the first sheet, but they changed it and made it easier

Ted Shifrin

17:48

LOL ... You could still do the more interesting version :)

MatheinBoulomenos

so the hard direction that one had to show was that if the Zariski topology on $k^{n+m}=k^n \times k^m$ agrees with the product topology of the Zariski topologies for some $n$ and $m$, then $k$ is finite

Ted Shifrin

Yeah, otherwise you can have varieties that are not fibers of the obvious projections.

Martin Hopf

sorry for the misstype: the smllest k such that 2^k = 1 (mod 23) is 11 so ord_23(2)=11

Ninja hatori

@LeakyNun are you there math.stackexchange.com/questions/2960986/…

Semiclassical

yeah, ignore it

looks like it's the right pattern

i.e. order_47(2)=23, order_2351(2)=47

MatheinBoulomenos

17:53

@TedShifrin I used the compactness of $k^m$ to get that the projection $k^n\times k^m \to k^n$ is closed and then constructed a subset of $k^n$ that is proper, closed and dense

Semiclassical

and ord_4703(2)=2351

so i guess the task is to find x such that ord_x(2)=4703

Fargle

@Ted I sent a sample of it to your email. I rather like it.

Semiclassical

which...yuck

Ted Shifrin

@Mathein: Assuming $m=n$ for the moment, can't you just look at the variety $y=x$?

Martin Hopf

so what do you think is next in the sequence: 11, 23, 47, 2351, 4703,...?

Ted Shifrin

17:54

@Fargle: I'd rather see a lot more math ... But that does look fine.

Fargle

I'm workin' on it!

MatheinBoulomenos

yeah, the case $m=n$ is easier, since you have the non-closed diagonal by non-Hausdorffness of $k^n$

Ted Shifrin

But you can modify that easily enough.

MatheinBoulomenos

my proof is probably too complicated :D

Semiclassical

no idea. from mathematica, i know that it's no smaller than x=500000

MatheinBoulomenos

17:56

didn't really have a good idea to show that stuff is not Zariski closed/open other than using a density argument

Ninja hatori

@ÍgjøgnumMeg any idea math.stackexchange.com/…

Semiclassical

(I'm having it compute tables of multiplicative orders of many many many integers and seeing if 4703 shows up)

Ninja hatori

@ÍgjøgnumMeg math.stackexchange.com/questions/2960986/…

Ted Shifrin

Ninja: Please don't go pinging everyone who might not be in the room.

Either ask a direct question or wait for someone to answer it on main.

Ninja hatori

ok sorry

Ted Shifrin

17:59

@Leaky: There may be a Chinglish origin of that phrase, but of course it's been around in English as long as there have been automobiles.

MatheinBoulomenos

@Ninjahatori and please don't ping me thrice in another room after I explained that I'm not knowledgeable about the question; your obstrusiveness doesn't make me want to answer it

Martin Hopf

the smallest k such that 2^4703 = 1 (mod x)

Semiclassical

smallest x, you meaen

Martin Hopf

yes,sorry

Semiclassical

and it seems really big, just based on mathematica work

Ted Shifrin

18:01

Last time I asked number theory experts, they told me that the order of $2$ in the multiplicative group mod $m$ is "unknown" — i.e., no known formula for it.

Rehi DogAteMy

Semiclassical

I'm testing up to x=4000001 right now.

Martin Hopf

in other words find the smallest factor of 2^4703-1.

Semiclassical

oh, that's easier to do in mathematica I think.

Ted Shifrin

If the number doesn't exceed capacity

Martin Hopf

the sequence so far: 11, 23, 47, 2351, 4703, (the smallest factor of 2^4703-1), ...

Semiclassical

18:02

well, mathematica can compute 2^4703-1 and print out the number

Ted Shifrin

Oh, then it should factor it

Semiclassical

and it comes up false when i test for primality

Ted Shifrin

howdy, a @Balarka

Rithaniel

Quick question: Suppose we start with $\mathbb{R}$ with the standard metric topology and define an equivalence relation $\leftrightarrow$ s.t. $x\leftrightarrow y$ iff $(x-y)\in\mathbb{Q}$. Is it fair to say that it's trivial to show that $\mathbb{R}/\leftrightarrow$ has the trivial topology (heh, trivial to show trivial), or does that claim warrant a proof?

Balarka Sen

Hi

Ted Shifrin

18:04

What do you mean by trivial topology?

Martin Hopf

the sequence so far 11, (the smallest factor of 2^11-1)=23, (the smallest factor of 2^23-1)=47, (the smallest factor of 2^47-1)=2351, (the smallest factor of 2^2351-1)=4703, (the smallest factor of 2^4703-1)=?,...

MatheinBoulomenos

Hi @Balarka

Balarka Sen

\o

Rithaniel

The topology where only the entire set and the empty set are open.

Ted Shifrin

So you're talking about the quotient topology?

MatheinBoulomenos

18:07

every prime factor of $2^{4703}-1$ is of the form $1+2n \cdot 4703$, that could speed up the search for the smallest factor by a constant factor

Semiclassical

nice

Rithaniel

Yes, the quotient topology being effectively the same as the trivial topology in this case.

Ted Shifrin

Well, you do need to work with the definition of the quotient topology. I wouldn't dismiss this as "trivial." I'd write the few sentences.

Rithaniel

Okay, will do, then. Danke.

Martin Hopf

yes, but consider 2^4703-1 a 1400 digit number ECM factorisation is fast for finding factors in the range of 10^25

your method is fast for finding small factors

MatheinBoulomenos

18:11

ah, I know nothing about ECM factorization other than the name and that it is uses elliptic curves

Semiclassical

yeah, this has quickly wandered into the realm of "beyond mathematica's range"

Ted Shifrin

I thought so, @Semiclassic.

Semiclassical

ah, this looks handy: alpertron.com.ar/ECM.HTM

Martin Hopf

yes unfortunately no factors of 2^4703 have been found. see factordb.com/index.php?query=2%5E4703-1

Semiclassical

luckily, you can input 2^4703-1 into the above calculator directly

(and anyone who asks what the next number in the sequence after this unbelievably large number is going to be laughed out of the room)

Martin Hopf

18:17

with the tool from Dario Alpern it will took about an hour to find a factor >10^25

it will be sure a very large number > 10^40

Semiclassical

i mean, the largest the factor could be is 2^4072-1 ~ 10^1415

sooooo

MatheinBoulomenos

maybe if you run Sage or Magma natively on your machine, the perfomance will be better than with an online tool (I'm no expert at this computational stuff)

Semiclassical

possibly.

Daminark

Hey there ya nerds

Martin Hopf

no the largest factor could be < 2^(4702/2)

Semiclassical

18:21

can't say this seems very appealing though, absent a better approach

Abdullah UYU

Hola

MatheinBoulomenos

hi @Daminark

Semiclassical

I was just using the naive sqrt(2^4073-1) bound, tbh

Daminark

How's it going?

MatheinBoulomenos

pretty well, thanks. Finally taking an alg geo class. And yourself?

Martin Hopf

18:23

yes sqrt(2^4073-1) is about 2^2036

Semiclassical

that still leaves you having to worry about factorization over integers up to ~700 digits tho

ah, you're right

wait

yeah, alright. (it's closer to 2^2351, amusingly)

But regardless this no longer seems like a good use of time.

Martin Hopf

and dont forget the smallest factor can be 2*k*4073+1 with k>0

you're right, nevertheless the sequence has one more element:

11, 23, 47, 2351, 4703, (the smallest factor of 2^4703-1), ...

Alessandro Codenotti

@MatheinBoulomenos I know who to ask about sheaves and stuff then :P

Semiclassical

A good question at this point would be how long it'd take to compute the next term

I suspect the answer is going to be prohibitive.

MatheinBoulomenos

@AlessandroCodenotti I do like sheaves

Alessandro Codenotti

18:31

Good, because I have a lot of confusion

Semiclassical

I have to wonder if there's the possibility of the sequence hitting the exponent of a Mersenne prime

Martin Hopf

thats a good question :)

Semiclassical

in which case it'd just stop there

Alessandro Codenotti

Ah, unrelated random coincidence: my algtop professor was a student of Tom Dieck! And you can see that in his style and approach to teaching algebraic topology

Semiclassical

I mean, I imagine the probability of this happening is small for this sequence

but I sorta doubt you'd ever be able to forbid it outright

Martin Hopf

18:34

for the mersenne primes start the sequence with 3:

3,7,127,?

MatheinBoulomenos

@AlessandroCodenotti nice!

Semiclassical

oh, mersenne primes require that p and 2^p-1 both be prime

Martin Hopf

2^3-1 is a Mersenne prime 2^7-1 is so and also 2^127-1 is a Mersenne prime

is 2^170141183460469231731687303715884105728-1 also a Mersenne prime ?

Semiclassical

given that the largest known one is 2^77,232,917-1

MatheinBoulomenos

@Alessandro there's value to both the algebraic Tom-Dieckesque and the geometric Hatcheresque approach, I only had the latter in the lectures and the former just from my own reading

Semiclassical

18:38

i think the answer is going to be "no one is even remotely close to knowing whether that's prime"

i mean, there's no particular reason to think it should be

Martin Hopf

there is no factor known of 2^170141183460469231731687303715884105728-1

Semiclassical

i don't see any reason, absent other considerations, why one should expect that particular number to be prime

Alessandro Codenotti

@MatheinBoulomenos I'm definitely glad I read singular homology from Hatcher since here we just started by defining an abstract homology theory and we will eventually build an explicit one

So that I have some intuition for what's going on

Semiclassical

it could be, but it's so beyond our computational range that speculating about it just seems futile

MatheinBoulomenos

@Alessandro yeah, I can see a point for doing an intuitive/geometric approach first

if you actually want to compute things, cellular homology is nice

Martin Hopf

18:46

yes, primes distribution is known, it is very unlikely that it will be prime

Alessandro Codenotti

So I've heard, I only learned about singular and simplicial homology

Daminark

@Mathein also finally taking an alg geo class so that's something :P

Today we talked some about primary ideals, the fact that ideals can be written as finite intersections of primary ones, and using it to show that projective varieties can be written as a finite union of irreducible ones

Alessandro Codenotti

We did stuff about presheaves and sheaves today in alggeo but I can't say I followed a lot of what was going on

Primary decomposition is cool

Daminark

In AT we did homology axiomatically, gonna have to redo the computations done in class since they were a bit of a mess

But yeah after this I think we're gonna actually construct homology

Oh also in AG we started talking about the Hilbert function

MatheinBoulomenos

19:06

@Daminark oh, so you're doing dimension theory, that's one of the more difficult parts of commutative algebra

Alessandro Codenotti

I was actually surprised by how much stuff you can prove just from the axioms

That's definitely a cool thing to see. It has some kind of foundational flavour after all

MatheinBoulomenos

yeah, true

we did singular first, then axiomatic, then cellular, than simplicial

Eric Silva

19:25

sup nerds

MatheinBoulomenos

sup @Eric

Eric Silva

how goes it

MatheinBoulomenos

pretty well, and yourself?

Daminark

Hello nerd

Eric Silva

@MatheinBoulomenos aight im kind of bored and stuck in a lecture on hodge theory

but it aint that bad cant reallly complain

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