Mathematics - 2021-10-26 (page 2 of 3)

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5:00 PM

Koro Nice !

i'm joking a little bit. the moore of the moore method was a huge racist who did not want black people in his class. this may have been incidental to what people call the moore method - roughly, providing students with definitions and theorems to be proved and not presenting them with a textbook or outside information that would help along the way.

i don't know how you would even implement it in the internet era.

that would be one hell of an honor system.

monoidaltransform

Let $e_1,.....,e_n$ be a basis for $V$ and $T:V\rightarrow V$ a linear map. What can I do to $e_1,.....,e_n$ for them to swap the columns of the matrix representation of $T$?

no moore please

I know some classes that have honor systems like that, incidentally, so I don't think it'd be unprecedented

(but the pressures students are under don't broadly favor academic honesty - up to the university to provide higher pressure to discourage)

monoidal, someone was asking something equivalent to this, i want to say, a few months ago. maybe they will weigh in. think about re-ordering basis vectors. maybe worth thinking about what happens if you re-order on one side (domain or codomain) but perhaps not the other.

monoidaltransform

5:06 PM

swapping $e_i$ and $e_j$? @leslietownes swaps the i and j columns?

one critique of the moore method is that it basically allows the instructor to mentally go on vacation. an advocate would say, that's not the true moore method. but hard to tell between a good professor using the method with students who aren't putting in the work, vs. a bad professor using the method with students who are. which may have been the point.

but yeah, enough about some dead racist.

monoidaltransform

no that swaps rows

Akiva Weinberger

Is it obvious somehow that $\dfrac{\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}} = 1+\sqrt{2}$

Mmm, been sitting on a lot of math reading lately... gotta pick that up again, soon.

Akiva Weinberger

I tried to simplify $\sqrt{2+\sqrt{2}}$ and ended up with $\sqrt{1+\sqrt{\frac{1}{2}}}+\sqrt{1-\sqrt{\frac{1}{2}}}$

which would've been fantastic in an alternate universe where 1/2 is a square

5:13 PM

monoidal: are you familiar with how row operations are implemented as matrix multiplication? roughly speaking that corresponds to changing bases for the codomain. for example, swapping rows i and j corresponds to swapping the roles of your ith and jth basis vectors for the codomain. because e.g. if T(some domain basis element) = c e_1 + d e_2 that is going to be represented in the matrix as a column (c, d), and if i want that to be (d,c) i should swap the order of e_1 and e_2.

monoidal: if you want to fiddle with columns, consider what happens if you fiddle with the order of the basis for the domain while leaving the basis for the codomain fixed (even if the domain and codomain are the same space, change one basis but not the other).

@AkivaWeinberger yes. rationalisation?

Akiva Weinberger

I wonder if I do stuff like mod 47 (where 2=7^2) if that gives me anything useful

$\sqrt{2+7}\equiv\sqrt{1+1/7}+\sqrt{1-1/7}$ mod 47

$3\cong\sqrt{28}+\sqrt{21}\pmod{47}$

Well I can't say I knew that before

Well I mean it's actually possible 28 and 21 aren't quadratic residues mod 47. But in some field extension where they are squares...

They are, they're $13^2$ and $16^2$

Er, $3\cong-13+16$ so close enough

@AkivaWeinberger Isn't this algebraic - $\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}=\left(1+\sqrt{2}\right)^2$? Or are you looking for a different intuition?

Akiva Weinberger

and $42\cong\sqrt{2-7}\cong13+16$

Yeah nah I guess that's good enough

Also means $\sqrt{2+\sqrt{2}}\left(\sqrt{2}-1\right) = \sqrt{2-\sqrt{2}}$

which simplifies to $\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}=\sqrt{4+2\sqrt{2}}$, which feels nonobvious

This is reminding me of Klein 4-group Galois theory.

@AkivaWeinberger Do we know a priori that these square roots exist?

Akiva Weinberger

5:27 PM

Don't think so. But I checked and they do

You have to be careful selecting which roots, though (every square number has two square roots of course)

Ah, you verified. Never mind.

Akiva Weinberger

In any case,$$\sqrt{a+t\sqrt{b}} = \sqrt{\frac{a}{2}+\frac{\sqrt{a^{2}-t^{2}b}}{2}}+\sqrt{\frac{a}{2}-\frac{\sqrt{a^{2}-t^{2}b}}{2}}$$which is useful if it turns out the norm of $a+t\sqrt b$ is a square

because if $a^2-t^2b$ is a square then you get rid of nested roots

but otherwise it becomes a mess

Is it true that for arbitrary free $\Bbb Z_2$-space $(X,\nu)$, if $X = A\cup B$ for some $\nu$-invariant subspaces $A$ and $B$, then $ind_{\Bbb Z_2}(X)\leq ind_{\Bbb Z_2}(A)+ind_{\Bbb Z_2}(B)+1$? Because my professor wonder if it's true for arbitrary topological space (unable to prove).

runs and hides from general topological spaces

5:43 PM

Hey... I have a question. I am trying to prove that every graph (undirected) with m edges is O(√m)-colourable. I am going via proof by contradiction method but am confused.

6:41 PM

I'm so proud of myself

I almost posted an undeletion request today on meta

what a waste of time that would have been :)

Interaction on meta is probably universally a waste of time.

agreed

7:14 PM

Question is to find the worst case time complexity of a brute-force algorithm for scheduling the talks by examining all possible subsets of talks.

Suppose we have $n$ talks that we want to schedule using greedy algorithm. There is a theorem which says that $n$ elements could give $2^n$ subsets. So, we have a combination of $2^n$ talks. Then we an compare $n\times(n-1)\times2^n$ pairs of talks and thus the worst-case complexity is $\mathcal{O}(n\times(n-1)\times2^n)$.

Each pair of talks in a subset have to be compared against each other to check for overlapping. There are $n$ talks and thu…

Why we have to multiple by $2^n$ please? I understand why there is $n(n-1)$ above, which is there to tell how many ordered and distinct pairs we have, but not sure why we multiple by $2^n$ :/

Xander Henderson

7:30 PM

Well, I just spend the last 10 minutes of a precalculus class talking about the Sierpinski triangle...

(I'll note that my lecture took an unexpected turn in the last 10 minutes; I had not planned to talk about fractal geometry).

@XanderHenderson this is a good thing, though!

Akiva Weinberger

The imperial system is very mathematical. For example, a mile is e^(π√73/3) feet, up to a few microns.

@XanderHenderson How did it come up?

Xander Henderson

Before getting to the Sierpinski gasket, we were talking about Minecraft.

Oh that's where you went wrong. ;)

Akiva Weinberger

Wait now I'm very interested in the flow of this conversation

Xander Henderson

7:38 PM

@AkivaWeinberger The lecture was on scaling relations. "Length is proportional to length" (e.g. the circumference of a circle scales like the the radius), "area is proportional to length squared" (the surface area of a cube scales like the square of side length), and "volume is proportional to length cubed".

Akiva Weinberger

Ah now I see how that leads to Minecraft

Xander Henderson

For the last one, I suggested that any "sufficiently nice" set in three dimensions can be approximated by cubes. So if you understand how cubes scale, you understand how any three dimensional object scales.

Akiva Weinberger

Fill a volume with cube pixels (voxels); the error should be small as the cubes get smaller

Yes

Xander Henderson

e.g. Minecraft, which is made of voxels.

@AkivaWeinberger exaclty.

Akiva Weinberger

Fun fact: given that a double tetrahedron has 8 times the volume of a normal one, you can use "cut and rearrange" logic to find the volume of a regular tetrahedron

because a double tetrahedron minus four unit tetrahedra, plus two unit tetrahedra in different spots is a parallelohedron

parallelopiped

7:42 PM

Question: So we know that, in general $\|Tx\|\leq\|T\|\|x\|$, but is there a simple (by some opinions) $g(\|T\|,\|x\|)$ such that $\|Tx\|\geq g(\|T\|,\|x\|)$ for any operator $T$ between normed linear spaces?

Minecraft doesn't really use the full potential of voxels

See this one:

New Voxel Engine Reveal - Crystal Islands Experiment

Akiva Weinberger

which can be cut and rearranged into a box (rect. prism) the same way a parallelogram can be turned into a rectangle

@BalarkaSen 'Cause you can't subdivide?

Voxels are indeed wild

Akiva Weinberger

I think, given one of the goals is that anyone should be able to build anything, a minimum voxel size was a good technical choice

It's just "How big are the voxels relative to the player?"

Akiva Weinberger

7:43 PM

Imagine trying to build a virtual house and spending hours on inch-size details

Sounds like how I play Factorio

@AkivaWeinberger It doesn't use voxel rendering techniques

Though, a lot of the look of the game comes down to shaders more than geometry

Akiva Weinberger

@XanderHenderson Dimensional analysis also helps here

'cause like ft^2 (12 in/ft)^2 = 144 in^2

though it's not a proof

Wait, I implicitly assumed that measuring in inches is the same as scaling up by 12 and measuring in feet

Well, that's reasonable I suppose

@BalarkaSen What voxel rendering techniques?

Like shaders? There are lots of third-party shaders you can download for Minecraft

I mean as far as I know it uses polygons to display the voxels; it just saves the maps in terms of voxels.

Akiva Weinberger

7:48 PM

By the way, do you think $\sqrt{4+2\sqrt2}-\sqrt{2+\sqrt2}-\sqrt{2-\sqrt2}$ is positive or negative

Xander Henderson

@AkivaWeinberger Yeah, we do a lot of dimensional analysis in there, too.

One method I know of which generates virtual geometry is "marching cubes," but that's more a matter of "figuring out the shape of the world" as opposed to "how do you represent the shape of the world?"

Akiva Weinberger

I'm guessing the conversation moved to the Menger sponge because it's a cool voxel-y shape?

and then the Sierpińsky gasket 'cause similar

Wait did you say you talked about the Menger sponge? I could be misremembering

By the way, $\inf\emptyset>\sup\emptyset$

which doesn't feel too horrible if you're used to the empty set being weird

but it does mean that if you type

Math.min > Math.max

into JavaScript it will return True

Xander Henderson

@AkivaWeinberger No, it was more "Hey, look... a $d$-dimensional object is defined by the property that $m = s^d$, where $m$ denotes the 'measure' (length, area, volume, whatever) of that object and $s$ represents some kind of scaling factor".

In high dimensions simplices are superior to cubes

$n+1$ more efficient than $2^n$

Akiva Weinberger

7:54 PM

(I like to imagine it as, $\inf$ sends a laser from $-\infty$ upwards and it stops when it first hits an element of the set. Same with $\inf$ but going down. But with the empty set the lasers are never stopped and just pass through each other)

Xander Henderson

@BalarkaSen It depends on what you are trying to do.

Akiva Weinberger

@XanderHenderson Ahhh

"Here's a thing whose measure is tripled when you double it! But 3 isn't a power of 2!"

Xander Henderson

@AkivaWeinberger Yup. That was fun.

Of course, I have one poor student in that class who experiences a lot of anxiety, and all she wanted to know is "Will this be on the test?"

:'(

Akiva Weinberger

Aw

Can't blame 'em, though

I mean, you are giving them tests, and they will impact their future, so they kinda have to worry about them

@XanderHenderson A fun thing to do is to take the middle-fourths Cantor set and square it (Cartesian product with itself)

Middle-fourths Cantor dust

Has dimension 1 because when you scale it by 4 it becomes 4 times larger

Has the same measure as the sidelength of its convex hull

Hausdorff measure is a pretty strange creature.

Akiva Weinberger

8:02 PM

I have one meter of dust

I used to have a square root meter but I squared it

In fact if $C_4$ is the middle fourths cantor set then I'm pretty sure $2C_4+C_4=[0,3]$ almost injectively

An interesting fact about Brownian motions in R^d for d >= 2 is that it has Hausdorff dimension 2. Compare with the fact that the traces of two iid Brownian motions in 3D intersect almost surely.

Akiva Weinberger

where $+$ is the Minkowski sum $A+B:=\{a+b|a\in A,b\in B\}$

A variant of the more well-know fact that $C_3+C_3=[0,2]$ (not almost surjectively though)

which you can ask me for a hint for if you don't see the proof

($C_3 ={}$middle thirds Cantor set)

@BalarkaSen Oh interesting

For any $d\ge2$??

Yeah

Akiva Weinberger

How do you show that

It's easy to show $\leq 2$.

$\geq 2$ requires capacity stuff

Akiva Weinberger

8:08 PM

Oh wait misread

Yeah I mean HD

Akiva Weinberger

Thought you meant the R^1 and R^2 cases

Sorry, I realized, yeah. Although BM in 1-d has Hausdorff dim $3/2$

Xander Henderson

@AkivaWeinberger Yup. I've done that.

Akiva Weinberger

@BalarkaSen wait isn't it an interval

Oh its graph?

8:11 PM

Yes, I was about to correct. Graph, yes.

Akiva Weinberger

Sorry I thought you meant its image

But dim graph = dim im in dim >= 2

Akiva Weinberger

Interesting

Not too bad, because graph is moving it up in 1 dimension. You can use Fubini

Akiva Weinberger

Hm. Exercise: find a homeomorphism between $(C_4)^2$ and $C_5$

Since the middle-whatever Cantor sets are all homeomorphic, this means $C^2\simeq C$

8:16 PM

...homeomorphism?

Yeah

Exactly

Akiva Weinberger

@BalarkaSen …Yeah?

I mean you can cook up a homeo C^2 = C without being explicit

It's general nonsense

Akiva Weinberger

Wait I don't mean $C_5$

uhhh

But since you put numbers I suppose you want an explicit one

Akiva Weinberger

Yeah crap I need new numbers because what I wrote (while true) isn't pretty and not what I meant

OK, new definition. $C_n$ is the set of all numbers whose base-$n+1$ representation only uses the digits $0,1,\dots,n-1$

so $C_2$ is half the usual Cantor set

basically you break the interval into $n$ pieces each time

Now biject $(C_2)^2$ with $C_4$

Yeah OK now that I write it out it's pretty obvious

so ignore me I guess

8:23 PM

Few days ago I learnt an example of a pair of spaces $X, Y$ with covering dimension of $X, Y$ both $2$, but covering dimension of $X \times Y$ is 3

shrieks

Maybe upper shriek.

Funny you say that

Akiva Weinberger

wait uh what

also gonna have to remind myself all the different types of dimensions

Oh OK so if I cover it with open sets and squeeze 'em real tight the best I can do is have an $n+1$-way overlap

Aye

Akiva Weinberger

So, uh

How on earth?

8:30 PM

It's related to $\Bbb Z/2 \otimes \Bbb Z/3 = 0$

Now lower shrieks as well!

Exactly lol

we should re-axiomatize topology so that can't happen

Akiva Weinberger

@BalarkaSen Yeah that doesn't help me

Something something fill them with circles that wrap 2 and 3 times around themselves something

No idea

Yeah thats it

That was Pontryagin's idea

Akiva Weinberger

8:33 PM

but like how do I turn that into a space

Take a surface, triangulate it very fine, then fill it in like you said by replacing the disk given by the interior of the triangle with your new disks wrapped twice (or thrice). Repeat.

You get two spaces $X$ and $Y$

by doing it mod 2 vs doing it mod 3

Akiva Weinberger

Do I have to triangulate them "very fine" if I'm repeating infinitely?

nah that was just for good measure

Akiva Weinberger

Oh I'm sure whatever measure this space has is very bad

lol

Akiva Weinberger

8:37 PM

I assume these don't embed into any Euclidean space

Maybe an infinite-dimensional one

i think finite covering dimensional metric spaces admit Euclidean embeddings

Akiva Weinberger

("Euler's sum is very geometric, it's saying the infinite-dimensional cube with sidelengths 1, 1/2, 1/3, ..., has diagonal pi/sqrt6")

Wait hold on I've just figured out how to square the circle

Step 1, Construct that with straightedge and compass

these embed in $\Bbb R^4$ apparently

Akiva Weinberger

Oh wow

you're forgetting Step 0: Construct the straightedge and the compass

Akiva Weinberger

8:41 PM

Ah, yes, the greatest precision engineering challenge of the 19th century, constructing a straight line

user image

oh lol

user image

https://en.wikipedia.org/wiki/Peaucellier–Lipkin_linkage

By the way the identity $\sqrt{4+2\sqrt2}-\sqrt{2+\sqrt2}=\sqrt{2-\sqrt2}$ feels like sloppy subtraction @BalarkaSen

Like someone thought $\sqrt a-\sqrt b=\sqrt{a-b}$ and then also messed up a sign

$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{2}+\sqrt{6}}{2}$$

Xander Henderson

8:56 PM

Hrm... I just discovered something: there are two light switches in my office, and two banks of lights. Now, if it were me, I would wire things so that switch A turns on bank A, and switch B turns on bank B.

But that isn't how it works. :\

Switch A turns on bank A. So far so good.

But switch B turns on bank B, as well as half of the lights in bank A.

WHY!?!

Whose idea was that?!

What if I only want bank B to be on?!

ARG!

Akiva Weinberger

How is the overlap handled? OR gate?

(eg on if either switch is on)

Xander Henderson

If both switches are on, both banks of lights are on.

Akiva Weinberger

Weird

Xander Henderson

It makes no sense.

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