Mathematics - 2018-02-18 (page 6 of 7)

ÍgjøgnumMeg

19:00

The website says I need to send my Hochschulzugangsberechtigung though, which makes no sense to me and is gonna be a real pain in the ass :(

MatheinBoulomenos

You have to prove that you have an equivalent of the Abitur, yes

It's like the grades don't matter but you have to have it

ÍgjøgnumMeg

That's so STUPID

hahaha

Eric Silva

that is a really long word

MatheinBoulomenos

Hochschulzugangsberechtigung is a nice word. You have to give them your Hochschulzugangsberechtigungsbescheinigung to prove that you have a Hochschulzugangsberechtigung

ÍgjøgnumMeg

hahaha

Still I'm really angry, I need to pay like 100 pounds to have some stupid useless document sent to me because the English exam boards are assholes

MatheinBoulomenos

19:03

I've you've lost your Hochschulzugangsberechtigungsbescheinigung you have to make a Hochschulzugangsberechtigungsbescheinungswiederausstellungsantrag

ÍgjøgnumMeg

lol

MatheinBoulomenos

German is fun

That sucks, my high school gives me new copies of my documents for free

ÍgjøgnumMeg

That's because you live in a not-backwards sh*thole

lol

that kinda sounds like I'm saying you live in a sh*thole

but that's not what I meant of course

MatheinBoulomenos

I thought you meant to put the "not" in some other place

ÍgjøgnumMeg

Not-backwards not-shithole

oops

MatheinBoulomenos

19:06

lol

ÍgjøgnumMeg

I broke my own censorship rule

Daminark

Double negative?

MatheinBoulomenos

not really, if you negate each part separately

ÍgjøgnumMeg

Geldgierige Wichsteufel

Eric Silva

double the negative for emphasis

MatheinBoulomenos

19:08

Lol, your German swearing is really authentic

ÍgjøgnumMeg

sarkastisch gemeint oder wie

hahaha

MatheinBoulomenos

No, I was serious

ÍgjøgnumMeg

ah right

Semiclassical

most complicated german words I know are

ÍgjøgnumMeg

Actually I have a really horribly strong farmer accent

Semiclassical

19:10

bremstrallung and zitterwebegung (no idea if i'm spelling those right)

MatheinBoulomenos

bremsstrahlung

Semiclassical

nope, bremsstrahlung and zitterbewegung

right

ÍgjøgnumMeg

Zitterbewegung?

lol

okay

haha

Perturbative

So if $(X, \Gamma)$ is a cell complex and $\Phi : D \to X$ is a characteristic map for some open cell $e \in \Gamma$, how can I show $\Phi[D] = \operatorname{Cl_X}(e)$

ÍgjøgnumMeg

Gadoladolälla is a funny word people from "my" region always teach people

MatheinBoulomenos

19:11

Hochschulzugangsberechtigungsbescheinungswiederausstellungsantrag is not really a complicated word

2

Semiclassical

and those I'm only aware of because physics

ÍgjøgnumMeg

Meaning something like.. "the hook on the outside of the window of your parents' bedroom that you use to hold the shutters open"

0celo7

lol

Eric Silva

don't u dudes just smash words

wordsmash

ÍgjøgnumMeg

wordsmaschifikationshammer

Semiclassical

19:13

A word should not be a sentence.

Eric Silva

@Semiclassical what's the difference

Semiclassical

The headache I get from trying to read one versus the other.

Eric Silva

pssshhhhhhh

the harder it is to read the better tbh

ÍgjøgnumMeg

lol once you know all the subwords of a word it really becomes very easy to read the long words

Semiclassical

seems like it'd make it hard to look up words

Eric Silva

19:17

while we're at it we should remove all punctuation

MatheinBoulomenos

you don't have to look up the whole word when you can break it into parts correctly

ÍgjøgnumMeg

^

Eric Silva

in fact fuck separate words lets just have one big fuckin word

spanning pages and pages

Daminark

Kant debunked language in Groundworks already

MatheinBoulomenos

There was a law in Germany that was officially called Rindfleischettiketierungsüberwachungsaufgabenübertragungsgesetz

ÍgjøgnumMeg

19:19

ahahaha

0celo7

My German is rusty, reading stuff like that is hard

Eric Silva

Kant debunked Kant years ago

ÍgjøgnumMeg

I was a Küchenbauteilettiketierer in Austria for a while

etikettierung* wow

double-t in the wrong place

MatheinBoulomenos

But honestly, 100 pounds as a Hochschulzugangsberechtigungsbescheinungswiederausstellungsantragsgebühr is way too much

ÍgjøgnumMeg

hahahaha

Eine Hochschulzugangsberechtigungsbescheinigungswiederausstellungsantragsgebührübertr‌eibung ist das

Akiva Weinberger

19:26

@EricSilva A bunch of languages don't use spaces

ÍgjøgnumMeg

It's very common in Germanic languages!

Akiva Weinberger

Chinese and Japanese don't use spaces, I don't know if there are any other examples actually

~~Korean~~

Eric Silva

english is just a poser germanic language

for nerds

ÍgjøgnumMeg

hahaha

Akiva Weinberger

> Vietnamese uses spaces between syllables instead of between words (except some few recent loanwords)

Huh weird

The anti-German

Eric Silva

19:30

whoa

i didnt know that

i should ask my roommate about that

ÍgjøgnumMeg

Dziewięćsetdziewięćdziesiątdziewięćmiliardówdziewięćsetdziewięćdziesiątdziewięćm‌ilionówdziewięćsetdziewięćdziesiątdziewięćtysięcydziewięćsetdziewięćdziesięciodzi‌ewięcioletniego

Polish wins?

Akiva Weinberger

Tiếng Thái

Tiếng Thái (ภาษาไทย, chuyển tự: phasa thai, đọc là Pha-xả Thai) là ngôn ngữ chính thức của Thái Lan và là tiếng mẹ đẻ của người Thái, dân tộc chiếm đa số ở Thái Lan. Tiếng Thái là một thành viên của nhóm ngôn ngữ Thái của ngữ hệ Tai-Kadai. Các ngôn ngữ trong hệ Tai-Kadai được cho là có nguồn gốc từ vùng miền Nam Trung Quốc ngày nay và nhiều nhà ngôn ngữ học đã đưa ra những bằng chứng về mối liên hệ với các ngữ hệ Nam Á, Nam Đảo, hoặc Hán-Tạng. Đây là một ngôn ngữ có thanh điệu (tonal) và phân tích (analytic). Sự phối hợp thanh điệu, quy tắc chính tả phức tạp, tạo liên hệ (có thể là liên tưởng?…

Too many diacritics ^

(Vietnamese)

Eric Silva

wow the latin alphabet is not appropriate for the phonology of these languages

Akiva Weinberger

I wonder how close Thai phonology is to Chinese. Pinyin seems to work reasonably well

It's not intuitive for people who speak other Latin-using languages, but it's internally consistent

Eric Silva

my SO is apparently very very confused by pinyin

Akiva Weinberger

19:34

Pinyin "er" is pronounced more like "ar" and weird things like that

but it's always pronounced the same

Eric Silva

she can never tell what anything is saying unless its written in like traditional or whatever it is

ÍgjøgnumMeg

zherrrrrrr

na'errrrr

Akiva Weinberger

huārr

Beijing is apparently pirates

ÍgjøgnumMeg

hahaha

Hey cool fact

Manolis Lyviakis

can i use derivatives to recursive sequences?

ÍgjøgnumMeg

19:35

You know the weird "arrrr" accent that pirates have?

Akiva Weinberger

Cornwall innit?

@ÍgjøgnumMeg

The one guy from that one film was from that one place

ÍgjøgnumMeg

@AkivaWeinberger Yeahhhh, I live like.. just on the border

A lot of people in my city speak like that

hahaha

Akiva Weinberger

Is Bristol there? My British geography isn't that great.

ÍgjøgnumMeg

Bristol is about 1 and a half hour drive away from here

but bristolians have a similar accent

Manolis Lyviakis

i want to prove that $x_{n+1}=x_n - \frac{(x_n)^2 -a}{2x_n} $

is monotone

can i take derivative?

use*

im trying with induction

but i must know what $(x_n)^2 -a$ sign is

since $x_n \geq 0$

Semiclassical

19:42

What I might do is write the $n-1$ case, $x_n=x_{n-1}-\frac{(x_{n-1})^2-a}{2x_{n-1}}$

and then compute the difference of those two equations i.e. $x_{n+1}-x_n=$ stuff

Manolis Lyviakis

doesnt lead somewher

its the same as $x_{n+1} - x_n=- \frac{(x_n)^2-a}{2x_n} $

only with n-1 instead :p

suppose $x_0=1 $

mercio

maybe you should look at $x_{n+1}^2-a$

Manolis Lyviakis

thats what i wrote

Semiclassical

@ManolisLyviakis Not what I meant.

Manolis Lyviakis

i need to know the sign

mercio

19:55

no you wrote $x_n^2-a$

Manolis Lyviakis

ohh

Semiclassical

First, we can write the initial equation $x_{n+1}=x_n-\left(\frac12 x_n+\frac{a}{2x_n}\right)=\frac{x_n}{2}+\frac{a}{2x_n}$

Manolis Lyviakis

ok

maybe i should take 2 cases

Semiclassical

so then $$x_{n+1}-x_n = \left(\frac{x_n}{2}+\frac{a}{2x_n}\right)-\left(\frac{x_{n-1}}{2}+\frac{a}{2x_{n‌-1}}\right)=\frac12(x_{n}-x_{n-1})+\frac{a}{2}\left(\frac{1}{x_n}-\frac{1}{x_{n-1‌}}\right)$$

Manolis Lyviakis

a>=1 and 0<=a<=1

Semiclassical

19:58

which further simplifies to $$x_{n+1}-x_n = \frac{1}{2}(x_n-x_{n-1})\left[1+\frac{a}{x_n x_{n-1}}\right]$$

Which feels pretty suggestive to me

Manolis Lyviakis

suggestive?

0celo7

$t\ge 0, p\ge 1\implies (1+t)^p\le 1+p\max\{1,2^{p-2}\}t+\max\{1,2^{p-2}\}t^p$...hmm

convexity?

mercio

I mean, $x_{n+1}^2-a$ simplifies to $((x_n^2-a)/2x_n)^2$

Semiclassical

@mercio oh, nice

Main thing that I can see from what I wrote is that if all $x_n$ are positive, then $x_{n+1}-x_n$ has the same sign as $x_{n}-x_{n-1}$

Which is exactly what you'd want for monotonicity

MatheinBoulomenos

@ÍgjøgnumMeg da zählen nur die Noten vom Bachelorstudium

Manolis Lyviakis

20:01

xn are positive

Semiclassical

Yeah, that looks to be easily proven.

$a>0$ here?

Manolis Lyviakis

yeap

ÍgjøgnumMeg

@MatheinBoulomenos Okey aber trotzdem benötigt man das Abiturzeugnis lol

Manolis Lyviakis

i still dont get why the same signs

Semiclassical

I guess it'd have to be, since this looks to be Newton's method with $f(x)=x^2-a$

MatheinBoulomenos

20:02

@ÍgjøgnumMeg ja. Du darfst nicht studieren wenn du kein Abitur hast

Semiclassical

4 mins ago, by Semiclassical

which further simplifies to $$x_{n+1}-x_n = \frac{1}{2}(x_n-x_{n-1})\left[1+\frac{a}{x_n x_{n-1}}\right]$$

ÍgjøgnumMeg

@MatheinBoulomenos Das ist mir schon klar, nur blöd dass man das noch beim masterstudium benötigt lol

Semiclassical

take a look at that second factor $1+a/(x_n x_{n-1})$

0celo7

@MatheinBoulomenos Wei\st du was ein Ami brauchen w\"urde?

Manolis Lyviakis

thats exactly what it is @Semiclassical

0celo7

20:03

SAT/ACT?

Manolis Lyviakis

thats what im trying to look xD

Semiclassical

If $a>0$ and $x_n>0$, what can you say about the sign of $1+a/(x_n x_{n-1})$

@ManolisLyviakis yeah, in which case $a<0$ would be pretty nonsensical

Manolis Lyviakis

oh x_n are all positive right

MatheinBoulomenos

@0celo7 Ich glaube ein Highschoolzeugnis würde als äquivalent zum Abitur zählen

Semiclassical

Right. So that factor is definitely >0 as well.

Manolis Lyviakis

20:04

so whatever the first factor does so does the left hand side

Semiclassical

Right. They both have to have the same sign

Manolis Lyviakis

so that means it is monotone

pretty awkward proof maybe im not used to them

did cercio came up with something else?

Semiclassical

It's a technique I like to try in these cases.

6 mins ago, by mercio

I mean, $x_{n+1}^2-a$ simplifies to $((x_n^2-a)/2x_n)^2$

Manolis Lyviakis

is that something?

0celo7

god, the worst inequalities are those with numbers

Semiclassical

20:06

Well, it means that $x_{n+1}^2-a$ is definitely not negative , since it equals the square of something else

Manolis Lyviakis

yeah

0celo7

like, say at least if you used convexity or whatever

ÍgjøgnumMeg

Ich glaub ich hab eh kein Abitur gekriegt um ehrlich zu sein, meine A-Level Noten waren scheiße, hab eigentlich nur so 2 Fächer bestanden und dann bin ich einfach nach Österreich gezogen, jetzt hab ich gute Noten im Bachelorstudium aber kann wie's aussieht wahrscheinlich nicht in Deutschland studieren weil ich "technisch" gesagt keine Hochschulzugangsberechtigung hab lol

Semiclassical

@0celo7 yeah, not saying anything seems sloppy

"By convexity on the interval X, we have inequality Y."

MatheinBoulomenos

@ÍgjøgnumMeg Was? Echt? Das klingt seltsam, aber ich kenne mich nicht so genau aus

0celo7

20:08

@Semiclassical I need $(1+t)^q\le 1+\mu t+\nu t^q$

I believe it, I think

Semiclassical

Seems plausible

But it's a bit of a weird bound.

Manolis Lyviakis

@mercio how dd you find that?

Semiclassical

@ManolisLyviakis I think you can show it by plugging the recursion for $x_{n+1}$ into there and simplifying sufficiently

0celo7

@Semiclassical it's amazing how much of PDE/measure theory is reduced to convexity of functions on R

Semiclassical

@0celo7 hmm

mercio

20:09

I looked at $x_{n+1}-x_n$ and used the recurrence relation until its sign became obvious.

ÍgjøgnumMeg

@MatheinBoulomenos Ja also da braucht man nur so 3 Fächer, ich hab Deutsch, Mathe und Erweiterte Mathe genommen, hab so gar keine Arbeit gemacht und hab das Buch nicht gelesen wo wir zur prüfung eigentlich lesen müssen hätten, also hab ich einen zweier in Deutsch geschrieben, einen zweier in Mathe und sonst hab ich nichts gekriegt hahaha

Semiclassical

@mercio I like the result, though I'm guessing the algebra is a tad tedious

@0celo7 convexity is great when it works

MatheinBoulomenos

@ÍgjøgnumMeg ah okay, das ist hier anders hier im Abitur

mercio

well not really, things cancel out pretty neatly

Semiclassical

ah, nice

mercio

20:11

$x_{n+1}= x_n - (x_n^2-a)/2x_n$

0celo7

@Semiclassical well $x\mapsto x^q$ is indeed convex when $q>1$

so maybe I need to use convexity there

Semiclassical

Oh sure

mercio

so $x_{n+1}^2-a= x_n^2 - (x_n^2-a) + ((x_n^2-a)/2x_n)^2-a$

0celo7

@Semiclassical that gives me $(1+t)^q\ge t^q+1$

welp

Semiclassical

@mercio yeah, that's not too bad

@0celo7 yeah, that's where I'm confused

the linear bound seems more like a tangent line statement

ÍgjøgnumMeg

20:13

@MatheinBoulomenos Ja das weiß ich schon, in Österreich ist es glaub ich ähnlich bei der Matura, finds trotzdem blöd dass man das zeugnis braucht fürs masterstudium wenn ich eh beim bachelorstudium gute noten kriege, aber okey is wurst, zahl ich halt 100 pfund für so an stück

hoiiiii

Semiclassical

whereas the $t^q$ part seems like a secant line statement

ÍgjøgnumMeg

Rant over

Manolis Lyviakis

@mercio yeah that what i did just a sec before

so since that is positive

0celo7

@Semiclassical well you expect the function to be $\sim t^q$ so really I would expect $(1+t)^q\le \text{big constant}+t^q$ to work

Semiclassical

sure

0celo7

20:14

maybe make that $2t^q$ or something

idk

Manolis Lyviakis

then $x_{n+2}-x_{n+1}=- \frac{(x_{n+1})^2-a}{2x_{n+1}} $ is negative

Semiclassical

If $1\leq p \leq 2$, this becomes $(1+t)^p \leq 1+p t+t^p$

Manolis Lyviakis

does that make the sequence decreasing for all n>=1?

Semiclassical

If $p>2$, it's $(1+t)^p\leq 1+2^{p-2}t+2^{p-2}t^p$

Manolis Lyviakis

@Semiclassical which way looks more legit?

0celo7

20:16

@Semiclassical How about this:$$\left(\frac{1+t}{2}\right)^q\le \frac 12 1^q+\frac 12 t^q$$

Semiclassical

@ManolisLyviakis Both work.

0celo7

using convexity

Semiclassical

hmm, but that'd be $(1+t)^q\leq 2^{q-1}+2^{q-1}t^q$

0celo7

Oh crap, do I need that to be 1 out front...

Semiclassical

yeah

I think there's a thought there, though

0celo7

20:17

well I mean for the purposes of the proof

I do, unfortunately.

Semiclassical

the linear lower bound would be $(1+t)^q\geq 1+q t$

nah, my line of thought goes nowhere

0celo7

that's just it lying above the tangent line at 0, right?

Semiclassical

Right. Which isn't that useful here.

0celo7

does a Taylor expansion work?

this function is analytic, isn't it?

Semiclassical

Dunno.

it seems reasonable

actually, no, that won't help

a taylor expansion near zero $t=0$ would give $q(q-1)$ coefficients

that doesn't seem like an improvement in this context.

0celo7

20:22

someone claims $(1+t)^q\le 1+q(q-1)t+(q-1)^{q-1}t^q$

which seems so random

Semiclassical

$q>1$?

0celo7

yeah

Semiclassical

that one could be doing second order Taylor stuff, since it has that $q(q-1)$ coefficient

but I feel like we're missing a trick here.

0celo7

wolframalpha.com/input/?i=Taylor+expand+(1%2Bt)%5Eq+in+t

Semiclassical

hmm

the inequality is equivalent to $$\frac{(1+t)^q-1}{t}\leq \mu+\nu t^{q-1}$$

Right?

0celo7

20:25

yeah

Semiclassical

That seems really suggestive

0celo7

My tangent line calculation gives $(1+t)^q\le 1+q(1+t)^{q-1}t$

induction?

$q$ isn't an integer

Semiclassical

shrug

yeah

If I write the LHS above as $\frac{1}{t}(f(t)-f(0))$ with $f(t)=(1+t)^q$

then the RHS is of the form $\mu+\nu f'(t-1)$

Ahah, I just noticed something while plotting

when $q>2$, that difference quotient is concave up. When $1<q<2$, it's concave down

($q=2$ makes it $(2t+t^2)/t=t+2$ i.e. linear)

So for $1<q<2$ I think it amounts to noting that the difference quotient is concave down and therefore as a function of $t$ it's bounded above by the tangent line at $t-1$

For $q>2$, I think one instead does a secant line bound of some kind. Maybe the line connecting $(0,0)$ to $(2t,f_{diff}(2t))$?

@0celo7 okay, here's an attempt. Let $g(t)=\frac{(1+t)^q-1}{t}=qt+\frac12 q(q-1)^2 t^2+\text{higher order terms}$

...nope

oops, expansion should be $g(t)=q+\frac12 q(q-1)t+\frac16 q(q-1)(q-2)t^2+\text{higher order terms}$

ok. note that the coefficient of $t^2$ is positive for $q>2$ and negative for $1<q<2$. I think this holds more generally for $g''(t)$ with $t\geq 0$, and I'll assume it from here. So $g(t)$ should be concave up for $q>2$ and concave down for $1<q<2$

0celo7

20:42

Maybe one has to compare the derivatives.

Semiclassical

possibly

0celo7

Use a Gronwall type argument

they're equal at $0$, and maybe one gets some info from the derivatives

so one has to lie above the other

idk

Semiclassical

yeah I dunno

0celo7

yeah not very helpful...

Semiclassical

I do think it can be done by looking at $g(t)$ for $1<q<2$ vs. $q>2$ and doing appropriate convexity arguments

But I'm not confident about the details.

0celo7

20:47

the things you find on google lol

this guy is giving an axiomatic theory of inequalities

Semiclassical

jfc

0celo7

I would call this a reverse Bernoulli inequality

Akiva Weinberger

Which is more important, the invention of writing or the invention of paper?

0celo7

this inequality.

Xander Henderson

@AkivaWeinberger The invention of writing. We have written documents that predate paper, and no one uses paper anymore, anyway.

0celo7

20:51

@Semiclassical I got it for $q>2$, I think

Xander Henderson

Writing itself is clearly longer lasting and of greater overall impact.

0celo7

There's some tomfoolery going on when the derivative becomes concave

because the growth is barely superlinear

Akiva Weinberger

I wonder why clay tablets didn't stay popular. Wouldn't they be cheap? Especially since paper was expensive for a while

Or is that completely wrong

0celo7

@Semiclassical By using the first derivative characterization of convexity, I got $(1+t)^q\le 1+q(1+t)^{q-1}t$,

Akiva Weinberger

You can write the whole Bible in a bathtub's worth of clay, it's not the most convenient but it's not impossible

0celo7

20:53

Now assuming $q-1\ge 0$, we have $(1+t)^{q-1}\le 2^{q-2}+2^{q-2}t^{q-1}$ using midpoint convexity.

Xander Henderson

clay tablets are not terribly portable, which is a problem

but paper bibles are a relatively recent invention

the Torah is written on sheepskin (not paper)

0celo7

So $(1+t)^q\le 1+q 2^{q-2}t+ q2^{q-2}t^q$

Akiva Weinberger

Isn't it cowskin?

Xander Henderson

not generally, no

though any kosher animal is probably okay

Semiclassical

@0celo7 nice

0celo7

20:55

@Semiclassical that's neither of the bounds i find in the literature

Xander Henderson

but the jews were shepherds, not cattle ranchers

so most Torahs are on sheep

0celo7

And it leaves $1<q<2$ open

Semiclassical

hmm

yeah

I may have an argument there, but I'm not convinced yet

0celo7

of course I mean $q-1\ge 1$ above.

Semiclassical

sure

0celo7

20:57

@Semiclassical Ah, when $q-1\le 1$, the derivative is concave, hence subadditive.

So $(1+t)^{q-1}\le 1+t^{q-1}$

Semiclassical

I always forget. concave in that sense = concave up or concave down in the calc 1 sense?

Xander Henderson

up, no?

0celo7

concave means concave down in the calc 1 sense

Akiva Weinberger

> Calfskin is usually used — about 65 animals are used to make a complete torah.

Xander Henderson

concave down is convex!

0celo7

20:58

convex means $x^2$

Akiva Weinberger

So the one I read from is probably calfskin

though I could ask my Rabbi when I next see him

Semiclassical

convex = concave up

0celo7

yes

Semiclassical

concave = concave down

0celo7

x2

Semiclassical

20:59

poor language = everywhere

0celo7

concave functions are subadditive, convex functions are superadditive

Transcript for

$Mathematics$ Mathematics