Mathematics - 2018-08-12 (page 2 of 2)

Rudi_Birnbaum

16:00

@ÍgjøgnumMeg Its usually part of the whole "Schaltgruppe"

what type have you got?

ÍgjøgnumMeg

Shimano Sora R3000

but the hanger fits the bike specifically

Rudi_Birnbaum

oh, that could be a problem. For such special cases I always bother the guys from the university work-shop :-)

ÍgjøgnumMeg

hahah nice, I know exactly what piece i need to get but apparently nobody in the UK sells them :(

Rudi_Birnbaum

Is shipping that expansive?

ÍgjøgnumMeg

I'm not sure, but all the other hangers cost like 12€ and for some reason this one is like 30€

hahaha

Bikes being so expensive!

Rudi_Birnbaum

16:06

@ÍgjøgnumMeg Yeah thats not so nice ...

ÍgjøgnumMeg

@Rudi there is one in Scotland that sells these guys apparently hahaha

but very cheap so I think it's okay

Rudi_Birnbaum

@ÍgjøgnumMeg So go for it!

@ÍgjøgnumMeg btw. what was your connection with Vorarlberg again?

ÍgjøgnumMeg

@Rudi I just lived there with my ehemaligen girlfriend

Rudi_Birnbaum

@ÍgjøgnumMeg Ah OK! Good way to learn languages it seems (vie girlfriends)

ÍgjøgnumMeg

@Rudi this is also how I learned Swedish hahaha

Rudi_Birnbaum

16:11

@ÍgjøgnumMeg yes it read that

ÍgjøgnumMeg

haha, I meet women and then absorb their language until we break up

apparently

Rudi_Birnbaum

Jag kan taler också lite svensk

ÍgjøgnumMeg

Ah okej bra då :D

Ted Shifrin

sounds oddly illicit, @ÍgjøgnumMeg

Rudi_Birnbaum

@ÍgjøgnumMeg lol

ÍgjøgnumMeg

16:12

@Ted you would think, I don't do it on purpose.... ;)

Rudi_Birnbaum

@TedShifrin In Europe stuff with girl friends is more relaxed ;-)

(legally seen)

Ted Shifrin

LOL, ohhh, that explains it, Rudi ;P

Rudi_Birnbaum

You still have the poor Austrian caught haven't you?

ÍgjøgnumMeg

Is that aimed at me? hahaha

Ted Shifrin

if that's addressed to me, I have no idea what you're talking about ... if it's not ...

ÍgjøgnumMeg

16:16

I'm not some kinda spaceship with language-based tractor beams

Rudi_Birnbaum

@TedShifrin yes it was, it was in the news, some 18 y old Austrian visited his US girl friend and did stuff what you do, and it turned out she was just 15 so he got arrested

Ted Shifrin

Oh, "you" was highly ambiguous.

ÍgjøgnumMeg

hahaha

Ted Shifrin

I don't follow news like this. There's far more alarming stuff to worry about.

Rudi_Birnbaum

@TedShifrin Over here everything would be completely legal

Well not in Austria apparently ...

ÍgjøgnumMeg

16:17

15 would not be legal in the UK tbf

Rudi_Birnbaum

Well if she agrees and age difference is not more then 4 y it seems to be OK. And on mutual agreement they can start with 12 y over here ...

(child pregnancy btw. is a complete non-issue in most of the EU except the UK)

ÍgjøgnumMeg

I mean, kids/teenagers will do what they want (it certainly happened in my school) but it becomes weird when you have someone who is older and someone who is younger

Rudi_Birnbaum

sure.

Ted Shifrin

since this is a math chatroom, I feel obliged to say it's highly unlikely that every couple (of whatever genders) will be exactly the same age, @ÍgjøgnumMeg :)

ÍgjøgnumMeg

@Ted typical pedantry! lol

Rudi_Birnbaum

16:21

For that its the 4 y max. difference I suppose.

fscd: Our maths teacher once asked us to estimate how many humans have lived before on earth

Fawad

@ted dy I need you!! How to find $11^{13}+13^{11}$ is divisible by...?

Rudi_Birnbaum

and his first approximation was about how many as live now.

Ted Shifrin

@Fawad: Divisible by what?

Rudi_Birnbaum

That was about exponential functions

Fawad

Options were like 121 , 157 etc

Ted Shifrin

16:24

oh, ugh

Well, $121$ you can rule out immediately. Why?

Fawad

Because we have 13

Rudi_Birnbaum

@Fawad: its by three

Fawad

121 is 11^2

Ted Shifrin

And, mod 11, this sum is definitely not 0.

Rudi_Birnbaum

11^13+13^11
36314872537968
./3
12104957512656.00000000000000000000000000000000000000000000000000000\
00000000000

ÍgjøgnumMeg

16:25

lol wtf

Ted Shifrin

Well, using a high-powered calculator is not by the rules, I'm sure.

Rudi_Birnbaum

and once more

Fawad

@Rudi_Birnbaum 36314872537968/157=231304920624

Rudi_Birnbaum

so ite divisible by 9

Ted Shifrin

But, mod 3, $11^{13}+13^{11} \equiv (-1)^{13}+1^{11} = 0$.

Rudi_Birnbaum

16:27

and again so its divisible by $3^3 \cdot 157$

and again

Ted Shifrin

Well, I'm done here.

Fawad

@TedShifrin can we use binomial theorem?

we don’t have modulus concept

Rudi_Birnbaum

$11^{13} + 13^{11} = 2^4 \cdot 3^4 \cdot 157 \cdot 178476019 $

Fawad

@Rudi_Birnbaum lol we don’t have calculators allowed

Rudi_Birnbaum

You don't need them calculators for division by 2 and 3 given you can calc the number by hand

Ted Shifrin

16:47

@Fawad: I don't see where that gets you ... BUT $$11^{13}+13^{11} = 11^{11}+13^{11} + (11^{13}-11^{11}).$$

By the binomial theorem, $$11^{11}+13{11} = 24^{11} - \sum_{k=1}^{10} \binom{11}k 11^k 13^{11-k},$$

and the second term is divisible by $11$.

So our friend is $24^{11}$ plus something divisible by $11$. But I don't see how such things are helpful.

ÍgjøgnumMeg

$11^{13} + 13^{11} \equiv 24 \bmod 11 \cdot 13$

wow nice mathjax

ergh

dunno why that helps either

Ted Shifrin

rats, I forgot to exponentiate my $13^{11}$. :(

ÍgjøgnumMeg

but that's just an observation

Secret

I have no idea how one does a mod on a sum since mod don't distribute

Ted Shifrin

what do you mean, @Secret?

Perturbative

16:55

Hey everyone!

Ted Shifrin

hi @Perturb

Perturbative

Hey @TedShifrin :)

Ted Shifrin

@Secret: Mod respects sums and products. That's why it's so powerful.

Secret

hmm, I must have misremembered

Perturbative

You know I used to make fun of Einstein summation convention before I knew any sort of diff geom, but now that I'm actually working with tensors it's a godsend

Ted Shifrin

16:57

I still write the sum symbols, Perturb. But generally checking the balance of upper/lower is good for making sure things are well-defined. And with hermitian geometry, you get conjugated indices, too ... :)

Secret

Let $11^{13} + 13^{11} = y$
Then $1+13^{11} (11^{13})^{-1} = y (11^{13})^{-1}$

Ted Shifrin

What does that mean, @Secret?

Secret

and... this is not helping

Ted Shifrin

When does $(11^{13})^{-1}$ make sense?

Secret

I am trying to divide away something to see if I can use mod somehow

Perturbative

16:58

Yeah, my lecturer really stressed to make sure our indices are right and stuff @Ted

Ted Shifrin

You can't leave the world of integers, unless you're working mod something relatively prime to $11$ and then the multiplicative inverse will make sense mod whatever.

Secret

$11$ has inverse in mod 13, right?

Ted Shifrin

But doing that multiplication didn't help. Just work with the original, mod $13$.

Secret

Does (a+b) mod n = a mod n + b mod n in general?

Ted Shifrin

Yes, as I said earlier. Analogous thing for product, too.

Perturbative

17:00

I'll be back in an hour see y'all soon

Secret

ok hmm...

$(11^{13} + 13^{11}) \mod 13 = 11^{13} \mod 13 + 0 \mod 13 = 11^{13} \mod 13$

Ted Shifrin

And $11\equiv -2\pmod{13}$, so ...

$11^{13} \equiv -(2^{13})\pmod{13}$.

And baby Fermat tells us $2^{13}\equiv 2\pmod{13}$.

Secret

so the remainder of this number divide by 13 is 2

Ted Shifrin

Yeah, but that doesn't help us tell the factors of the number.

ÍgjøgnumMeg

no it's $11$

Ted Shifrin

17:04

Oh, no, it's $-2$.

ÍgjøgnumMeg

Right

but i mean you can just hit it with little Fermat from the start because you already have $11^{13} \equiv 11 \bmod 13$

(just to butt in)

Ted Shifrin

Yes, of course.

Secret

hmm... if we do the same trick with mod 11, then we should get two ways to break down $11^{13}+13^{11}$ i.e.:

$11^{13}+13^{11} = q_1 13 + 11$
$11^{13}+13^{11} = q_2 11 + 2$

this can be rearranged to get a relation between $q_1,q_2$

now our target is $11^{13}+13^{11} = q_0d$

hmm... still too many unknowns...

We need a 4th equation

will a mod 143 get us anywhere?

Daminark

17:44

Hey everyone

Tobias Kildetoft

@Daminark Hi

Daminark

How's it going?

Tobias Kildetoft

One of the Danish super marked chains has had a campaign where you collected stickers to get discounts on some selected board games. So now we have expanded our collection with some classics we were missing

Eric Silva

sup nerds

Secret

[Random] Multiplication by 11

Let abcd be digits of a 4-digit number. Observe the following:

$abcd \cdot 11 = abcd0+abcd$

Now observe:

$abcd \cdot 11^2 = (abcd0 + abcd) \cdot 11 = abcd0 \cdot 11 + abcd \cdot 11 = (abcd0)0+abcd0 + abcd0+abcd = abcd00 + 2\cdot abcd0 + abcd$

Continue by induction we found the following:

Let $x$ be an integer. Then the action of $11$ on $x$ is given by:

$11\cdot x = x \cdot 11 = x0+x$

and more generally:

$$x \cdot 11^n = \sum_{k=0}^n \binom{n}{k}x\underbrace{00\cdots 0}_{\text{k times}}$$

$11 \cdot 11 = (*0 + *)(*0 + *) = (*0)(*0 + *) + *(*0 + *) = *00 + *0 + *0 + * = *00 + 2\cdot *0 + *$

Thus it checks out

so presumably...

Let $xy$ be a 2-digit integer, then we have:

$abcd \cdot xy = (x \cdot abcd)0 + y \cdot abcd$

and $abcd \cdot (xy)^2 = ((x \cdot abcd)0) \cdot xy + (y \cdot abcd) \cdot xy = (x\cdot (x \cdot abcd)0)0 + y \cdot (x \cdot abcd)0 + (x \cdot (y \cdot abcd))0 + y \cdot (y \cdot abcd)$

Now since $\cdot$ both commute and associates we have:

$abcd \cdot (xy)^2 = (x\cdot x) \cdot abcd00 + (x \cdot y) \cdot abcd0 + (x \cdot y) \cdot abcd0 + (y \cdot y) \cdot abcd$

$ = x^2 \cdot abcd00 + 2 \cdot x\cdot y \cdot abcd0 + y^2 \cdot abcd$

This means by induction:

$$x \cdot ab = \sum_{k=0}^n \binom{n}{k} a^{n-k} \cdot b^k \cdot x\underbrace{00\cdots 0}_{\text{k times}}$$

and therefore:

Daminark

18:17

@TobiasKildetoft sorry I was out for a bit, but yeah any particularly fun ones?

@Eric yo

Secret

$ab = (a\cdot *)0 +b \cdot *$

and more generally for an n digit number $u$ we have:

Tobias Kildetoft

@Daminark Ticket to Ride Europe, Carcassonne and Pandemic

Secret

$$u = \sum_{k=0}^n (u_k \cdot *)\underbrace{00\cdots 0}_{\text{k times}}$$

where $u_k$ is the (k+1)th digit of $u$

Eric Silva

@Daminark help im in probability hell

Secret

And finally, for any two integers $u,v$ their product $u \cdot v$ is given by

$$u \cdot v = \sum_{k=0}^n \left(u_k \cdot \sum_{\ell = 0}^m (v_{\ell} \cdot *)\underbrace{00\cdots 0}_{\text{$\ell$ times}})\right)\underbrace{00\cdots 0}_{\text{k times}} = \sum_{k=0}^n \left(\sum_{\ell = 0}^m (u_k \cdot v_{\ell} \cdot *)\underbrace{00\cdots 0}_{\text{$\ell$ times}}\right)\underbrace{00\cdots 0}_{\text{k times}}$$

And that, is what happens when you are multiplying two integers: a shear followed by a left shift

Daminark

18:27

Ah nice

@Eric just a few more weeks and you can return to geometry, just gotta hold out

Secret

Thus factorisation is a hard business even for one big integer into two is because when solving for $u,v$ such that $w = u \cdot v$ we are actually trying to solve $mn+2$ variables

namely, all digits of $u,v$ and the length of $u,v$ itself

O and I made a typo

I forgot the multinomial coefficients

$$u \cdot v = \sum_{k=0}^n \binom{n}{\{k_i\}}\left(\sum_{\ell = 0}^m \binom{n}{\{\ell_j\}} (u_k \cdot v_{\ell} \cdot *)\underbrace{00\cdots 0}_{\text{$\ell$ times}}\right)\underbrace{00\cdots 0}_{\text{k times}}$$

hmm....

$11^{13} + 13^{11} = \sum_{k=0}^{13} 1\underbrace{00\cdots 0}_{\text{k times}} + \sum_{k=0}^{11} 3^k \cdot 1\underbrace{00\cdots 0}_{\text{k times}}$

Eric Silva

hopefully lol

Secret

typo

$11^{13} + 13^{11} = \sum_{k=0}^{13}\binom{13}{k} 1\underbrace{00\cdots 0}_{\text{k times}} + \sum_{k=0}^{11}\binom{11}{k} 3^k \cdot 1\underbrace{00\cdots 0}_{\text{k times}}$

Tobias Kildetoft

Is anyone at all actually paying any attention to the multiplication rambling going on?

Secret

18:44

And the largest factor n of this number is such that number mod n = 0. Now the problem is that there are at least $24!$ ways to group the terms together in order to get a sum of 0 mod n s thus clearly this is not useful for this problem

However...

Now that this multiplication business is figured out, this can be easily extended to the binary case which is ultimately what I am interested in

Because it might give us a way to visualise $e^e$

G. Ünther

This time the bifurcation is much more visible, got a good set of parameters: youtu.be/AME2A2DhSxg

Secret

can't wait to see what the period point diagrams look like

G. Ünther

at 0:24 the chaos starts

This is very intersting as there are four points one might land with a random starting point

and if the point is in one of those cycles it can't escape

Secret

$11_2 = 1 \cdot 2 + 1$
$11_2^2 = (1 \cdot 2 + 1)(1 \cdot 2 + 1) = (1 \cdot 2^2) + 2 \cdot (1 \cdot 2) + 1 = 1\cdot 2^2 + (1 \cdot 2) \cdot (1 \cdot 2) + 1 = 2 \cdot 1 \cdot 2^2 + 1 = 1 \cdot 2^3 +1 = 1001_2$

$11_1^2 = 100_2 + 10_2 \cdot 10_2 + 1_2 = 100_2 +100_2 +1_2 = 1000_2 +1_2$

$11_2^2 = (10_2 + 1) (10_2 + 1) = 100_2 + 10_2 + 10_2 + 1 = 1001_2$

hmm... how can I confine that nonlocal effect...

or maybe they should not matter, since they bubble out from the product thus canbe dealt with separately...

besides if $u_k \cdot v_{\ell} = 0$ for some $k,\ell$, those terms are dead anyway...

Daminark

19:21

@Secret that less message goes way offscreen

Secret

ok nvm, that does not help because we need to compute it from the right end which takes forever

$$(\pi \cdot e)_2 = \sum_{k < \omega} \binom{n}{\{k_i\}}_2\left(\sum_{\ell < \omega} \binom{n}{\{\ell_j\}}_2 (\pi_k \cdot e_{\ell} \cdot *)\underbrace{00\cdots 0}_{\text{$\ell$ times}}\right)\underbrace{00\cdots 0}_{\text{k times}}$$

This is uncomputable thus it does not work

The only way I can think of is to take advantage of the normality of $\pi,e$ thus that will give us a pdf on when $\pi_k\cdot e_{\ell} = 1$ for every $k,\ell < \omega$

After that, convert the multinomial into binary and multiply it to this pdf, that will give us the distributions of ones in the inner bracket, and then we can compute the outer sum in that way to get the pdf of each term

and from that we should be able to extract the probability that the terms form an arithmetic sequence, hence whether $\pi e$ is rational

Secret

19:40

so we have something like:

$Pr (1 \subset \pi e | \pi_k = 1) = 2^{-1} \cdot 2^{-1} = 2^{-2}$

$Pr (10,11,01,00 \subset \pi e | \pi_k = 1) = 2^{-1} \cdot 2^{-2} = 2^{-3}$

$Pr (000,001,010,011,100,101,110,111 \subset \pi e | \pi_k = 1) = 2^{-1} \cdot 2^{-3} = 2^{-4}$

Kummer's theorem

In mathematics, Kummer's theorem for binomial coefficients gives the p-adic valuation of a binomial coefficient, i.e., the exponent of the highest power of a prime number p dividing this binomial coefficient. The theorem is named after Ernst Kummer, who proved it in the paper Kummer (1852). == Statement == Kummer's theorem states that for given integers n ≥ m ≥ 0 and a prime number p, the p-adic valuation ν p ( ( ...

G. Ünther

How is the Lyapunov Exponent calculated if the starting point determines if the iteration is chaotic? New problems with math.stackexchange.com/questions/2872160/…

Secret

7

Q: p-adic valuation for multinomial coefficients

Kummer's formula https://en.wikipedia.org/w/index.php?title=Kummer%27s_theorem&oldid=745783657 says that $$ \text{ord}_p \binom{n}{k} $$ is the number of carries required when adding the base-$p$ expansions of $k$ and $n-k$. Is there a similar formula for the $p$-adic valuation of a multinomial...

this can be useful to expand the multinomials in binary...

$$\lim_{n \to \omega}\operatorname{ord}_2 \binom{n}{k_1,\ldots,k_r}=\lim_{n \to \omega}\sum_{i < \omega}S(k_i)-S(n).$$

Now since $k_i < \omega$, it follows $k_i < n$

Thus the summand is negative in general and thus the sum is negative and unbounded

Thus $\lim_{n \to \omega}\operatorname{ord}_2 \binom{n}{k_1,\ldots,k_r}=0$

1-n+2-n+3-n+... = 1+2+3 - 3n

$k_i(k_i+1)/2 - k_i n = \frac{k_i^2}{2} - k_i n$

which blows up to minus infinity

Thus in general the multinomials will be divisible by any powers of 2

which translates to the probability of finding any binary strings of the form 10000.... is uniform

A. Hendry

20:14

Does anyone know if the torus knot is a general helix?

Secret

So blowing all of this up, we have that the probability of finding 1,10,01,00,11,100,... in $\pi \cdot e$ is corresponding to that of a normal number, thus if $\pi,e$ is normal, then $\pi \cdot e$ is normal. Hence $\pi \cdot e$ has to be irrational, and moreover, transcendental

QED (well not quite because sloppiness, but whatever will fix that later...)

Sil

You know what would be cool? If there was a page on SE or chat, that would be connected to every comment/question/answer/voting event on the site, you would just watch animations of these poping out all the time, SE traffic in real time! Just saying...

Oskar Tegby

Cool idea.

I can really recommend the 3Blue1Brown podcast called Ben, Ben, and Blue.

https://www.benbenandblue.com

Sil

Random discussions podcast? Sounds good.

Oskar Tegby

It's not entirely random. It's related to math and education.

Nebulae

21:06

Does the site not have 'new questions' tab anymore? It says interesting, featured, hot, but not new?

Sil

You must click to Questions on left nav bar

ÍgjøgnumMeg

@Nebulae you need to click "unanswered" and go on the "newest" tab

Nebulae

Oh, thanks @ÍgjøgnumMeg

Sil

You know, those days you feel people ignore you

Nebulae

@Sil Oh lool.

I didn't see you replied.

Sil

21:12

@Nebulae It's okay, I am just kidding :)

Nebulae

But actually that works better because I can see newest answered questions too, so thanks.

I know you were kidding :)

Cosinux

Hello

I need a little help

What is this shape?

Oskar Tegby

Convex

Really cool tool right here: earth.nullschool.net

No. Sorry. What do you mean by what it is?

Cosinux

It's convex, but what is it? It's not an ellipse for example..

Does it even have a name? I'd like to now it's properties..

Sil

Looks like two curves to me, not much like a single commonly used shape

@OskarTegby Nice site btw, it even maps aurora!

Oskar Tegby

21:27

Yeah! I know. I've been playing around with it for an hour now.

Cosinux

It's two straight 3D horizontal lines in a curvilinear perspective. Each line has two vanishing points.

I am trying to figure out how to calculate these 2D "curves" by knowing their 3D vectors

Sil

Maybe something from this might help en.wikipedia.org/wiki/Curvilinear_perspective ?

it has part about projection and its math

Anyway if you dont find what you are looking for, feel free to ask question on site

Cosinux

There are only calculations to map 3D coordinates to the perspective. I've already done that..

I'd like to find some equation for the curve

Mary Star

Does someone of you have an idea about my question about showing the uniqueness: math.stackexchange.com/questions/2880721/… ?

Cosinux

So I can draw it and use it in my program. The curve on the image was created manually by moving two points horizontally

Sil

21:36

Well if you can draw it, it seems you have equation of it available, or not?

A. Hendry

@Cosinux Looks like two catenoids or an oscillating string

Cosinux

I'm wondering, what kind of equation would be appropriate for this curve

@Sil it was drawn manually

A. Hendry

@cosinux check out catenoids

@cosinux sorry, I meant catenary. Check out catenaries on wikipedia

Everyone, what is the right way to ask new questions in this chat?

Cosinux

Thank you. That might be it. I'm already reading the article

A. Hendry

21:52

For a section of a torus knot having a fixed arclength, the equations of which are given by $$(R+r\cos(pt))cos(qt)\mathbf i + (R+r\cos(pt))sin(qt)\mathbf j + r\sin(pt)\mathbf k$$ is there a way to prove that as R approaches infinity the equations turn into the circular helix equations?

Ted Shifrin

22:47

@A.Hendry: That doesn't seem possible.

Leland Reardon

23:17

That shape looks like a separatrix

Separatrix (mathematics)

In mathematics, a separatrix is the boundary separating two modes of behaviour in a differential equation. == Example == Consider the differential equation describing the motion of a simple pendulum: d 2 θ d t 2 + g ...

@cosinux

jaslibra

I placed 200 bounty on this question for those interested: math.stackexchange.com/questions/2766802/…

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$Mathematics$ Mathematics