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

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

and I have way more pigeons that needed

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

@TedShifrin oups yes sorry

but is my idea correct?

12:27 AM

No.

grrrr

can we have same element in the set?

Oh, maybe.

I don't know if you have 48 possible sums

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

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

12:28 AM

no they want all possible sums of subsets

there are 63 non empty ones

2^6 -1

Subsets of what set?

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

I honestly do not understand the question.

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

Yes, Demonark. I think so.

12:30 AM

well I was not clear

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

So we fix 6 numbers between 1 and 14.

each element in S is less than or equal to 14

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

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

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

No, it doesn't have to be.

12:33 AM

you can have all the set or 1 element subset

Sure.

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

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

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

ohhhhhhh =p true ><

12:35 AM

Then consider all possible sums of subsets of them.

I have no idea why 14 is relephant.

well from what I understood to find upperbound

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

well daminark corrected my lower bound

1<= S_A <= 69

I don't think this is helpful.

hmm

12:38 AM

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

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

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

How horribly can I abuse notation

I have faith in you, DogAteMy.

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

12:41 AM

wait: we can't add power sets.

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

That is horrendous.

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

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

But I'm only guessing.

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

12:43 AM

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 $[14]$

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

Akiva you doing my problem ? =p

This might help cut down the upper bound

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

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

12:45 AM

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

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

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

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

That was my point, yeah, Demonark.

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

12:47 AM

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

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

my lower bound should be in the fixed set

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?

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

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

12:48 AM

(I know it's only five things)

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

Oh wait yeah yeah you're good then @Ted

Don't look at the solution

or if you do don't tell us

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

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

12:49 AM

I like it

I wont do it yet

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

waiiiiiiiit

:D

subset of S

does not include whole S right?

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

@KasmirKhaan Yes it does

grrrrrrr

12:50 AM

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

nah

It's necessarily greater than every other sum

because if the whole set is not included

then we have 62 subsets

and only 5 elements at max in each subset

Kasmir, think about what Ted said

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

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

12:51 AM

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

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

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

Oh that's 60, right?

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

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

yes akiva

that was my idea

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

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

12:53 AM

Oh I mean you guys basically have it at this point

62 pigeons and 60 pegonholes sounds right to me =p

1 < S_A < 60

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

(Ex: all numbers from 1 to 61)

what do you mean ?

60 numbers from 1 to 60 right? ><

Yeah but 60-1=59

The difference between the max and the min there is 59

@TedShifrin What's this new world order?

12:55 AM

LOL ... You tell me :P

@AkivaWeinberger that does not matter

Oh, wait, I think I've confused myself

=p

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?

Yes

12:56 AM

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)

Well, why don't you each talk separately?

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

But then it isn't chaos! :(

Lol jk

Not good for this guy <---

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

Yepp the key idea is excluding the whole set =p

daminark what is your idea? :)

12:59 AM

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

@Daminark …plus 1

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

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

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

stay tooned for the next problem =p

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

1:01 AM

Ah, right

Have fun, drive/walk/boat safely

Um, thanks :)

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

Whatever mode of transport floats your goat

Night, @Kasmir.

1:02 AM

Speaking of, goat safely

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

Kek

DogAteMy: When do you disappear into intensive Spanish?!

I think stuff starts Monday

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

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

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 $f(x)$?

1:06 AM

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

Wait are $a$ and $b$ functions?

No and yes.

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

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/…

Oh wait a second I'm illiterate

Sorry

1:09 AM

I think this is way underdetermined

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 $f$.

1:43 AM

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

1:58 AM

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

2:12 AM

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?

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

that's the fundamental theorem of calculus basically

C what? (I mean what is that)?

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

ah ok

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

yeah i mean some arbitrary antiderivative

2:22 AM

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?

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

The span of $v_{\lambda}$

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

sure

ok

i feel weird abt it

cause 0 is the only point where this is wonky

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 $tv_{\lambda}$

@Kirill The derivative isn't even defined at zero

2:30 AM

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

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

so what do you mean

I meant it mod 1

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

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

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

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

2:41 AM

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

what did i miss

another nb question, nothing more

:)

also amin talking about basic dynamics stuff

i'll badger him about that laterz

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

2:43 AM

probably cause it's so much new terminology

@EricS I know what the Riemann curvature tensor is now

really?

aw yiss

wowowow

@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

2:46 AM

idk dude talk to your people

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

it's troubling

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

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