Mathematics - 2017-07-09 (page 1 of 7)

Kasmir Khaan

00:00

and ofc forgetting about that those odd numbers like 3 divides 27

Ted Shifrin

Which even number did you pick?

Kasmir Khaan

can I still argue that it works using pigon ?

any like 20

Ted Shifrin

Well, $20$ is divisible by $5$.

Kasmir Khaan

yes but hmm if I work in general without knowning that fact

Ted Shifrin

For DogAteMy and Demonark: Try to prove this one by induction!! It's a cool argument that requires real thinking.

No matter what you pick, it has an odd factor, @Kasmir. Done.

Kasmir Khaan

00:02

@TedShifrin if I pick the number 2 ?

Ted Shifrin

Now think about your $y$'s when you write $x=2^n y$.

Then it's divisible by $1$ (which was on your list).

Kasmir Khaan

ahh true ><

Ted Shifrin

In fact, if you put $1$ on your list, you're automatically done.

Kasmir Khaan

well the y's are 100 and I have to pick 101 so thats solved

Ted Shifrin

But you can't do a proof by trying to consider all possible lists.

Kasmir Khaan

00:03

Yes yes =p h

@TedShifrin are there really 100 numbers from S that dont divide each other?

Daminark

@TedShifrin I mean...

:P jk

Akiva Weinberger

Oh, I made a stupid mistake in my solution before

Kasmir Khaan

i mean like can we find 100 numbers from S that dont work ?

Ted Shifrin

Yes, you really need 101.

Kasmir Khaan

Very nice problem :D

Akiva Weinberger

00:06

Everything from 101 to 200 works I think @KasmirKhaan

Ted Shifrin

Try it with a smaller number. Like 4 numbers from 1 to 6. Now use only 3 numbers?

Or listen to DogAteMy :P

Akiva Weinberger

Oh sorry

Kasmir Khaan

clever idea from akiva =p

Ted Shifrin

Try to prove it by induction, DogAteMy.

I always put that as a challenge problem on my first problem set in abstract algebra.

Akiva Weinberger

So, uh, base case, if I try to choose 2 numbers from {1,2} I clearly have one dividing the other

Daminark

00:07

Inducting to prove it for all even numbers and 1+half?

Ted Shifrin

right

Akiva Weinberger

So the induction case is knowing that if I choose $n+1$ numbers from $[2n]:=\{1,2,\dots,2n\}$, one divides another

and I want to show that the same happens for $n+2$ numbers in $[2n+2]$

Ted Shifrin

Right. Actually, you guys are super-clever, so you'll get it too easily.

Akiva Weinberger

Well, I want $\{2n+1,2n+2\}$ to be a subset of my list

Daminark

I think the idea should be about where you choose the numbers. Clearly if you choose all but one of them from the first 2n, you're dead

Ted Shifrin

00:09

You do?

All or all but one ... OK ... that's two of the cases.

Akiva Weinberger

Otherwise at least $n+1$ of them end up in $[2n]\subset[2n+2]$

Daminark

Meaning as Akiva said, you're immediately reduced to the case where you look at both 2n+1 and 2n+2

Ted Shifrin

OK. 2 of the 3 cases.

Mike Miller

@Daminark When are objects of functor categories C^D compact?

Ted Shifrin

WTF?

Akiva Weinberger

00:10

^

Objects of general categories can be compact?

't the hell

Mike Miller

Just tryna read a paper. I don't know all the yoga of this fancy stuff.

Akiva Weinberger

("All the yoga"?)

Daminark

What's a compact object in a general category?

Akiva Weinberger

OK, hold on, call our list $L$

Daminark

Also @Ted I think I get it

nigel

00:13

@AkivaWeinberger just nerd talk for theory

Daminark

So you know for sure you have 2n+2 in your list

Akiva Weinberger

Then consider $(L\cap[2n])\cup\{n+1\}$

Ted Shifrin

We were assuming both $2n+1$ and $2n+2$, so yeah.

Daminark

So you don't want to choose n+1 or 2

Ted Shifrin

Yeah, you guys are too clever. But it's a cool induction proof, eh?

Akiva Weinberger

00:14

$L\cap[2n]$ has $n$ elements, so the above has $n+1$ elements

(because $n+1\notin L$)

and then it has a pair that does the thing

Ted Shifrin

I've even had math faculty ask me how to do it :P

Daminark

Oh I really like this problem

Akiva Weinberger

and if one of the pair is $n+1$, it's clearly $a|n+1$ and not the other way around

so we have $a|2n+2$

Ted Shifrin

Yuppers.

Maybe I'll give that to my AoPS students the first class.

Akiva Weinberger

I guess I could have done $(L\cap[2n])\cup\{2\}$ instead

Ted Shifrin

00:17

I still have no idea what the actual curriculum is.

Akiva Weinberger

because it's clearly $2|a$ and not the other way around (hopefully $L$ doesn't have $1$ in it!)

and oh wait never mind I'm wrong

So yeah I need to append $n+1$ instead

@TedShifrin Back up. AoPS students?

Ted Shifrin

Why should I back up?

Daminark

@TedShifrin It'll probably be trying to make sure they have arithmetic down. Maybe it'll use Serre?

Ted Shifrin

No, Demonark. Only you would be that nuts.

Daminark

Lmao

Ted Shifrin

00:21

So did you learn all the stuff you promised to during June? Differential forms? Integrals?

Daminark

I did go through forms more, though I didn't yet do the Mayer-Vietoris sequence

Ted Shifrin

didn't I send you my notes on that stuff?

Daminark

Yeah, that's what I was using

Ted Shifrin

of course, I wanted you to get better at computing with forms ... but that's a bad word for you.

Daminark

I'm slowly downgrading the profanity level of the word "compute" due to Titchmarsh

Some of the stuff we did here was fun

Ted Shifrin

00:25

you're no longer ranting about his open closed intervals?

Kasmir Khaan

Let S be a set of six positive integers whose maximum is at most 14 . show that the sums of the elements in all non empty subsets cannot all be distinct.

my idea is like this 1+2+3+4+5+6<S_A< 9+10+11+12+13+14

but that gives me 48 possible sums

Daminark

Oh I mean that's unforgivable, just that I've found some computation-type problems that I'd actually describe as fun

Kasmir Khaan

and I have way more pigeons that needed

Ted Shifrin

you need $\le$, not $<$, Kasmir.

Kasmir Khaan

@TedShifrin oups yes sorry

but is my idea correct?

Ted Shifrin

00:27

No.

Kasmir Khaan

grrrr

can we have same element in the set?

Ted Shifrin

Oh, maybe.

Daminark

I don't know if you have 48 possible sums

Kasmir Khaan

well I based my idea that we can have a set with same member

Ted Shifrin

Why do they say non-empty subsets? Don't they want you to fix six numbers and then consider sums of subsets?

Kasmir Khaan

00:28

no they want all possible sums of subsets

there are 63 non empty ones

2^6 -1

Ted Shifrin

Subsets of what set?

Kasmir Khaan

of S , where S has 6 members and maximum of each member is 14

Ted Shifrin

I honestly do not understand the question.

Daminark

I think the point is, given a 6-element subset of $[14]$, you take the set of sums of its subsets?

Ted Shifrin

Yes, Demonark. I think so.

Kasmir Khaan

00:30

well I was not clear

ok we have a set S and |S| = 6

Ted Shifrin

So we fix 6 numbers between 1 and 14.

Kasmir Khaan

each element in S is less than or equal to 14

eg S = {1,10,11,12,13,14}

Ted Shifrin

So note in your example that $1+10=11$ or $1+11=12$, etc.

Kasmir Khaan

hmm that is the question

we have to show that the sum of all possible subsets cannot be distinct

or wait I think you got it right =p @TedShifrin

but my idea it does not have to be a subset of size 2

Ted Shifrin

No, it doesn't have to be.

Kasmir Khaan

00:33

you can have all the set or 1 element subset

Ted Shifrin

Sure.

Kasmir Khaan

hmm the way I was trying was to find max and min of each sum

21 < S_A < 69

1+2+3+4+5+6 <= S_A <= 9+10+11+12+13+14

Daminark

Your issue is that you're not necessarily summing up 6 numbers

Ted Shifrin

No. No. You have to fix 6 numbers to start with.

Kasmir Khaan

ohhhhhhh =p true ><

Ted Shifrin

00:35

Then consider all possible sums of subsets of them.

I have no idea why 14 is relephant.

Kasmir Khaan

well from what I understood to find upperbound

Daminark

The smallest sum you could get would be just the smallest element, the largest sum you could get would be the sum of everything in the set

Kasmir Khaan

well daminark corrected my lower bound

1<= S_A <= 69

Ted Shifrin

I don't think this is helpful.

Kasmir Khaan

hmm

Daminark

00:38

You'd need to tighten that more than just 69, because the set of subsets of a 6 element set is just 63

Kasmir Khaan

yepp like Ted said it cant be done that way ._.

Ted Shifrin

For the $A$ that gives you 69, what is the smallest possible sum?

Akiva Weinberger

How horribly can I abuse notation

Ted Shifrin

I have faith in you, DogAteMy.

Akiva Weinberger

$\sum\mathcal P(S_6\subseteq[14])$ should have a repeated element

Ted Shifrin

00:41

wait: we can't add power sets.

Kasmir Khaan

@TedShifrin Ted we cant fix 6 numbers and only work on those

Ted Shifrin

That is horrendous.

Akiva Weinberger

Nah OK that was less abuse of notation and more pure gibberish

Ted Shifrin

Yes, that's what the problem is about, Kasmir, as far as I can tell.

But I'm only guessing.

Kasmir Khaan

hmm its good guess because 63 subsets and 68 possible sums dont work =p

Akiva Weinberger

00:43

Is it OK for me to think out loud even if my thoughts are horrible

This is like a linear map from $\{0,1\}^6$ to $\Bbb N$

where the image of each basis vector is in $[14]$

Daminark

I'm thinking that when you take the smallest number, you must force that no two of the larger numbers can differ by it

Kasmir Khaan

Akiva you doing my problem ? =p

Daminark

This might help cut down the upper bound

Akiva Weinberger

Yeah but I have no idea what I'm doing :P @KasmirKhaan

Kasmir Khaan

@AkivaWeinberger haha keep going you might find something ! =p

Daminark

00:45

Meh that doesn't work out quite the way I wanted it to

Akiva Weinberger

So we know that $\{1,2,4,8,16,32\}$ has unique sums. But that way exceeds our 14 bound.

Kasmir Khaan

@Daminark I was trying to consider the worst case , the sum will be 69

Daminark

True, but in that specific worst case, your smallest number is 9, so then yeah you're alright

Ted Shifrin

That was my point, yeah, Demonark.

Daminark

Point is, there are too many possibilities floating around to consider the best and worst case across all choices

So you need to do something else

Kasmir Khaan

00:47

I see your point now >< I was doing it the wrong way

Daminark

In particular, you need to consider the set in question and narrow things down

Kasmir Khaan

my lower bound should be in the fixed set

Ted Shifrin

Right.

And your upper bound should be the sum of the 6 numbers in the fixed set.

What's the biggest difference those two can have?

Akiva Weinberger

Does $\{2,3,4,8,14\}$ have unique sums?

Kasmir Khaan

Ermm ._. I have the solution btw I can look at it if you want

Akiva Weinberger

00:48

(I know it's only five things)

Ted Shifrin

Kasmir: Why waste our time if you're looking at solutions?

Daminark

Oh wait yeah yeah you're good then @Ted

Akiva Weinberger

Don't look at the solution

or if you do don't tell us

Kasmir Khaan

@TedShifrin I did not mean that way, maybe we got the question wrong

Akiva Weinberger

or if you tell some of us don't tell me

Daminark

00:49

I like it

Kasmir Khaan

I wont do it yet

Ted Shifrin

I think I framed it correctly up there ... ^^^^

Kasmir Khaan

waiiiiiiiit

:D

subset of S

does not include whole S right?

Akiva Weinberger

Hm maybe $14=2^4-2$ is relevant

@KasmirKhaan Yes it does

Kasmir Khaan

grrrrrrr

Akiva Weinberger

00:50

But why do we care? It can't equal any other sum anyway

Kasmir Khaan

nah

Akiva Weinberger

It's necessarily greater than every other sum

Kasmir Khaan

because if the whole set is not included

then we have 62 subsets

and only 5 elements at max in each subset

Daminark

Kasmir, think about what Ted said

Let's say you've already chosen the 6 numbers between 1 and 14

Ted Shifrin

wow, that's the first time Demonark's ever told someone to listen to Ted :D

Kasmir Khaan

00:51

I know Ted is allways right but I have to find solution bymyself for once ><

Daminark

On that particular set, what are the smallest and largest sums possible?

Akiva Weinberger

$14+13+12+11+10={}?$

Oh that's 60, right?

'Cause the average is 12 so it's like five times 12

Kasmir Khaan

I dont have to pick specific set because that wont be a proof danimark =p

yes akiva

that was my idea

Daminark

No no, I'm not telling you to choose the example

Akiva Weinberger

Our specific set is $\{a_1,a_2,a_3,a_4,a_5,a_6\}$ in ascending order :P

Daminark

00:53

Oh I mean you guys basically have it at this point

Kasmir Khaan

62 pigeons and 60 pegonholes sounds right to me =p

1 < S_A < 60

Akiva Weinberger

Well if the difference between lowest possible and the highest is 60, we actually have 61 pigeonholes

(Ex: all numbers from 1 to 61)

Kasmir Khaan

what do you mean ?

60 numbers from 1 to 60 right? ><

Akiva Weinberger

Yeah but 60-1=59

The difference between the max and the min there is 59

Daminark

@TedShifrin What's this new world order?

Ted Shifrin

00:55

LOL ... You tell me :P

Kasmir Khaan

@AkivaWeinberger that does not matter

Akiva Weinberger

Oh, wait, I think I've confused myself

Kasmir Khaan

=p

Daminark

Alright think about it like this

So, given $\{s_1,\ldots,s_6\}$, you the lowest possible sum of stuff is $s_1$, the highest is $s_1 + \ldots + s_6$, yeah?

Kasmir Khaan

Yes

Akiva Weinberger

00:56

Yeah, ignoring the sum of all the things ('cause it can't equal anything else anyway), the minimum value of a sum is $1$ and the highest value is $10+11+12+13+14$

(Daminark's doing it slightly differently than me)

Ted Shifrin

Well, why don't you each talk separately?

I mean ... finish one train of thought — then do the other.

Daminark

But then it isn't chaos! :(

Lol jk

Kasmir Khaan

Not good for this guy <---

Akiva Weinberger

So the sums (excluding the sum of all the things) are between $1$ and $60$, and there are $62$ possible sums (excluding the everything and nothing), and $62>60$, so.

OK ignore me and listen to Daminark and then un-ignore me when he's done I guess

Kasmir Khaan

Yepp the key idea is excluding the whole set =p

daminark what is your idea? :)

Daminark

00:59

I mean it is basically the same idea wrapped up differently. The number of numbers between $s_1$ and $s_1 + \ldots + s_6$ is $s_2 + \ldots + s_6$

Anyway, so yeah that's capped at 60, so we win

Akiva Weinberger

@Daminark …plus 1

But that's capped at 61, so we still win

Kasmir Khaan

I guess thats it then >< ill put a mark on this question and think about it later and do other one =p

Akiva Weinberger

(There are $b-a+1$ things between $a$ and $b$!)

Kasmir Khaan

stay tooned for the next problem =p

Ted Shifrin

Glad I helped so much. I'm heading out for the evening.

Daminark

01:01

Ah, right

Akiva Weinberger

Have fun, drive/walk/boat safely

Ted Shifrin

Um, thanks :)

Kasmir Khaan

@TedShifrin Thanks Ted for all help as allways ! have a nice night / day =p

Akiva Weinberger

Whatever mode of transport floats your goat

Ted Shifrin

Night, @Kasmir.

Akiva Weinberger

01:02

Speaking of, goat safely

Daminark

But what if Ted's gonna grow wings and fly to his destination? Huh Akiva?

Kek

Ted Shifrin

DogAteMy: When do you disappear into intensive Spanish?!

Akiva Weinberger

I think stuff starts Monday

I didn't realize you were waiting so eagerly :P

Ted Shifrin

After all this goat-floating, it did occur to me :P

user76284

How can I find $a,b,c,d$ from $f(x) = \frac{a(x)+b(x) (1 - x/r)^{-c}}{x^d}$ in terms of $f(x)$?

Ted Shifrin

01:06

Do you have a specific $f(x)$ at hand, @user76284?

Akiva Weinberger

Wait are $a$ and $b$ functions?

user76284

No and yes.

Daminark

presses f to pay respects to the law of the non-contradiction

user76284

I'm trying to get them purely in terms of $f(x)$, without knowing $f$ in advance.

I was thinking I might be able to do it through contour integrals and derivatives, but am not sure how.

The formula comes from here: en.wikipedia.org/wiki/…

Daminark

Oh wait a second I'm illiterate

Sorry

Akiva Weinberger

01:09

I think this is way underdetermined

user76284

I forgot to mention $a$ and $b$ have a radius of convergence greater than that of $f$ and $b(r) \neq 0$.

Basically I'm trying to find the asymptotic expression for the coefficients of $f$.

Daminark

01:43

Alright so, let's say we have a hyperbolic toral automorphism (just on the 2-torus)

I'm thinking that anything in the unstable manifold of 0 is a recurrent point

Daminark

01:58

Oh wait scratch that, $\mathbb{Q}^2$ should do it

Kirill

02:12

hi!

We say that $f'$ ist contiuous, if $f$ is differentiable. Can we say that every antiderivative of $g$ is differentiable, if $g$ is continuous?

Eric Silva

the antiderivative of any continuous function is $C^{1}$

that's the fundamental theorem of calculus basically

Kirill

C what? (I mean what is that)?

Eric Silva

$C^{1}$ means it's differentiable with a continuous derivative ($g$ being its derivative)

Kirill

ah ok

but you mean some function, not a set when you say "antiderivative", right?

Eric Silva

yeah i mean some arbitrary antiderivative

Kirill

02:22

At the same time, $f(x)=-1$ for $x\le0$ and $f(x)=1$ for $x>1$ is not contiunous, but partially derivated gives a continuous zero function. So, the answer on my question is "no", right?

Eric Silva

@Daminark what do you mean by "the unstable manifold of 0" bc HTAs don't move 0 around at all

Daminark

The span of $v_{\lambda}$

We call $W^u(x) = \{x + tv_{\lambda}\}$

Eric Silva

sure

ok

i feel weird abt it

cause 0 is the only point where this is wonky

Daminark

But yeah so like, rational points are eventually periodic, and thus periodic. Which ofc implies recurrence, so that's good. My concern right now is finding points which are not recurrent

Which is sneaky

Actually the span is not the same as $tv_{\lambda}$

Akiva Weinberger

@Kirill The derivative isn't even defined at zero

Daminark

02:30

So maybe the span of the eigenvector doesn't actually have a sequence converging to it

That or maybe the stable manifold?

Wait hold on that should do it

Eric Silva

$\{x + tv_{\lambda}\}$ this isn't in the torus

so what do you mean

Daminark

I meant it mod 1

If you take $w = v_{\frac{1}{\lambda}}$, then $\|Aw\| < \|w\|$

Kirill

@AkivaWeinberger so, its' derivative is $0$ without 0, or how? I can still derivate in any point besides 0, right?

Eric Silva

the domain of definition for the derivative doesn't include $0$ so there's no contradiction

so if you keep track of the domain the result is that an anti derivative of this is continuous on the set $(-\infty, 0) \cup (0, \infty)$ which is true of the function you started with @Kirill

Kirill

@EricSilva yes, I only want to make my mistake clear. So, $f'$ ist not continuous, now right? :)

Eric Silva

02:41

I mean $f'$ is continuous on $(-\infty, 0) \cup (0, \infty)$; that's everywhere it's defined.

you can't really say "it's not continuous at $0$" because it's not even defined there

so the result still applies, it just applies to a different domain

Balarka Sen

what did i miss

Kirill

another nb question, nothing more

:)

Eric Silva

also amin talking about basic dynamics stuff

Balarka Sen

i'll badger him about that laterz

Daminark

Yeah it's taking longer than it should to grapple with this

Eric Silva

02:43

probably cause it's so much new terminology

Balarka Sen

@EricS I know what the Riemann curvature tensor is now

Eric Silva

really?

Balarka Sen

aw yiss

Eric Silva

wowowow

Daminark

@Eric I've currently signed up to lecture on Wednesday, is it likely a bit too late to back down from that if I message the group or something? I still would like to do it if possible but I'm ever so slightly concerned that I'm jumping the gun

Eric Silva

02:46

idk dude talk to your people

@Balarka idt there's a "good" way to think of the big daddy curvature tensor

it's troubling

Balarka Sen

yeah. either understand it via the vague parallel transport thing, or as coding up all the sectional curvatures of the sub-2-planes, i was told

Eric Silva

i like to think of it as being the second order properties of the metric

but like none of these are things that are universally useful

Balarka Sen

mhm

interesting typo

Eric Silva

like they're all good interpretations of what curvature means but the big thing is still tooooo big

i think gromov refers to the curvature tensor as the multilinear monster or something

Balarka Sen

i'm trying to get on terms with it at a symbolic level rn

oh yeah he said nobody understand the riemann curvature tensor lol

Eric Silva

02:50

i mean on some level that's the point of riemannian geometry

if it weren't obscenely hard it wouldn't be interesting

Balarka Sen

tru

i have to run now but we cal continue this discussion later soon

Eric Silva

you should compute the taylor expansion of the metric in normal coordinates

Balarka Sen

i'll keep that in mind

Eric Silva

it's a long exercise but it gives you practice

Jasch1

How would you prove that x^(ln(y)) = y^(ln(x))

Akiva Weinberger

02:57

@Jasch1 Replace $x$ with $e^{\ln x}$

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$Mathematics$ Mathematics