Mathematics - 2017-03-17 (page 3 of 5)

Meow Mix

20:04

@BalarkaSen do pythagorean triples arise in this at all?

Balarka Sen

@MeowMix Not in my solution, but try it out!

I think there are many ways to do it, of which I only know 2

Meow Mix

my pythagorean theorem solution only works for multiples of 8

so i'm missing 7/8 of the naturals

:P

well, that's certainly not the way to go

Balarka Sen

Oh really? How did you do it

Meow Mix

Oh wait

Does the center have to have natural radius / center?

Balarka Sen

Note that I wanted points to be inside the circle and not on the circle

Meow Mix

20:06

OH

LOL

I read as "on". Sorry!

Balarka Sen

"On" is still an interesting question but I haven't solved it.

Astyx

I doubt "On" is true

Balarka Sen

It is.

Astyx

Once again, my intuition sucks

Meow Mix

Actually

All you'd need to do to is find a circle containing $n-1$ points, such that when you expand it so that it contains another point, there's a radius such that it will only contain that point

Balarka Sen

20:10

Correct. You have to find appropriate centers and radii to do that.

Meow Mix

That is, a circle containing $n-1$ points such that there is a unique point closest to the center

Balarka Sen

Not enough, but yes.

Meow Mix

Anyways

So assume there is a circle containing $n-1$ points

If there is a unique point, QED

If there isn't, smidge it only a tiny bit to the left, right, up, or down

And re-expand the circle so that it contains some unique point

I suppose I should put this in formal terms

Balarka Sen

Sounds like an idea I initially had :) I couldn't make it work, but let's see if you can

Meow Mix

Oh, I think I see how to do this

Let me bring out geogebra

So, suppose we have $k$ points which are equidistant from our circle's center (and outside of our current circle, and are all the minimal distance away)

Consider each pair of points

Let's call these points $B,C$ WLOG

and let's call the center $A$

The set of points equidistant to $B$ and $C$ is a line through $A$

$A$ cannot lie on this line, since then we wouldn't have a unique point

Construct all lines for each pair of points, and you have a bunch of lines through $A$ which $A$ can't intersect

Balarka Sen

20:18

Very astute.

Meow Mix

Since there must only be finitely many lines, there MUST be some point in the circle not contained in any of these lines

In fact, we could get arbitrarily close to $A$ if we wanted

Just take a line that's not in the set of "no-go lines", and approach $A$ along it.

And we're guaranteed a unique point

Balarka Sen

All good ideas. Can you fix the center? Your current solution-sketch moves the center around.

Meow Mix

The reason I think that I'm allowed to move it around is because I could make it arbitrarily close to $A$

So, if there were any points very close to the boundary of the circle (that might be excluded), we can find some point very close to $A$ that will account for it.

Astyx

Yeah, that has to do with the disk being compact I think

Balarka Sen

I think you're allowed to move it too. But I'm asking if you can modify your proof so that it works for a fixed center.

@Astyx Nah.

I mean, sure, but it's the countability of Z^2 which is more important.

Meow Mix

20:22

Oh, like there exists a center where there is always a unique closest point?

Balarka Sen

@MeowMix Err, unique closest point from the center doesn't guarantee there would be a circle containing for every $n$ exactly $n$ lattice points

What do you mean?

Meow Mix

I mean that for any circle, that there is a unique closest point

Balarka Sen

Oh. Yes.

Meow Mix

So that when you expand the circle to contain that point, there is also another unique closest point

Balarka Sen

Correct. There is such a fixed center.

Akiva Weinberger

20:24

In other words, no two points on the lattice have the same distance from Balarka's point

Meow Mix

Hm... I'm thinking

Astyx

You've already solved it in some sense

With what you stated earlier

Balarka Sen

it's very close to the set of ideas you're hovering about

Meow Mix

Yeah, take any two points on the lattice

the set of points equidistant from those two points must be a line

however, there are countably many lines

so that can't cover $\Bbb R^2$

Balarka Sen

yeah man. great job

Meow Mix

20:25

so there exists a point where no line is intersecting it.

Now, I'm wondering. What point is that? :P

Balarka Sen

An explicit example of such a point, btw, is (sqrt(2), sqrt(3))

oops

Meow Mix

Oh, happy St. Patrick's day to those who celebrate it.

My research class teacher brought in Lucky Charms today...

But I didn't eat them. Too much sugar.

Hi @Mike

Mike Miller

hi

Meow Mix

Today I had a benchmark in geometry, and so for the last question, I was finished and bored, so I wrote a linear algebra proof of it as a joke.

Balarka Sen

King's son is lying dead, none to care for
King's palace is grey with dust, none to clean it up
Thousand lilies lying on the ground, none to pick them up

(It's a - sort of not very good - translation of a riddle I like a lot)

Akiva Weinberger

20:36

So apparently someone needs a positive integer solution to $\frac x{y+z}+\frac y{z+x}+\frac z{x+y}=4$

and also apparently the solutions are all really large

Meow Mix

did you try making it a giant big fraction?

Akiva Weinberger

No

and I kind of don't want to

Meow Mix

ill take the liberty of doing so and then doing nothing

$\frac{x(x+z)(x+y) + y(y+x)(y+x) + z(z+x)(z+y)}{(x+y)(x+z)(y+z)}$

be right back.

Alessandro Codenotti

good evening chat

how's everyone?

Meow Mix

(i followed through on my promise to do nothing)

@AlessandroCodenotti still disappointed

Alessandro Codenotti

20:40

By what?

Meow Mix

loss of data and code

Alessandro Codenotti

How did it happen?

Meow Mix

Overwriting a partition with windows 10

Balarka Sen

@AkivaWeinberger Maybe one can do a greedy expansion on 7/(x+y+z) to estimate about how large x, y and z needs to be

Meow Mix

That is, the partition had stuff on it, and it was overwritten with the windows 10 data, the deed wasn't done by windows 10

Alessandro Codenotti

20:42

Ah, I see

I started learning standard ML, I'm doing a functional programming course in uni, pretty interesting stuff

Meow Mix

have you done assembly yet

also ML = markup?

Alessandro Codenotti

Not in school, but yes

it's the name of the language, I din't know that ML stands for

Meow Mix

oh

I thought you meant general markup

like XML, HTML, YAML

Alessandro Codenotti

Nope, nothing of this sort

Meow Mix

anyways, for which processors?

Alessandro Codenotti

20:47

Only intel stuff, x86 and higher, up to x64, with all the SSE and AVX instruction sets

Meow Mix

oh

i don't like x86 that much

the instruction set i mean

Balarka Sen

i'm bad at combinatorics

should work on that for exams

Alessandro Codenotti

@MeowMix it has been expanded a lot though

I mean compared to the 386 or what was the original instruction set

When will you have exams?@Balarka

Balarka Sen

monday

Akiva Weinberger

Courtesy of Wolfram Alpha:$$\frac{3 x y z + (x + y + z)^3 - 2 (x + y + z) (x y + x z + y z)}{-x y z + (x + y + z) (x y + x z + y z)}$$

is the fraction above in terms of symmetric polynomials

Meow Mix

20:52

@BalarkaSen exams for what?

Balarka Sen

school

Meow Mix

yeah, but you have combinatorics in school?

Akiva Weinberger

or $(3s_3+{s_1}^2-2s_1s_2)/(-s_3+s_1s_2)$

Balarka Sen

a thing or two in high school, yes

Meow Mix

interesting..

i'm going to be taking algebra 2 next year

Balarka Sen

20:53

(i suck at it)

Meow Mix

sighs

Alessandro Codenotti

What kind of combinatorics?

Because I can't imagine you having troubles with the combinatorics done in high schools here

Balarka Sen

basic counting problems, permutations/combinations, binomial formula

It confuses the hell out of me

Meow Mix

you're not alone

Daminark

Lol, you should totally take a combinatorics class when you get the chance though

Balarka Sen

20:58

@Daminark Indeed, I plan to fill in a lot of what I don't know in my undergrad, even though I am constantly told to just jump into a PhD or something.

Daminark

Woo!

Meow Mix

@Balarka Good idea.

Alessandro Codenotti

There's no undergrad course in higher cathegory theory :P

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$Mathematics$ Mathematics