general discussion

[open whale]

$z_{n(k+1)}=p^n(z_{nk})$

now, the important part

if c solves zn

[vivid gulch]

yoavmal

[open whale]

then it should solve z2n due to cyclicism

and so

it solves pn and pkn

meaning really

no it doesn't help it's cyclicism

i want to show it divides it wit hthe same amount of roots

uhh

suppose zn has a root r with multiplicity m

$z_{(k+1)n}=p^n(z_{kn})$

[vivid gulch]

yoavmal

[open whale]

and lets suppose it has the same multiplicity for zkn

then does it have it for pn(zkn)

oo wait this isn't even about iterative sequences

just about polynomial composition

if we have $p(z)=(z-r)^mh$

[vivid gulch]

yoavmal

[open whale]

and also r isn't a root of h

then what about p(p(z))

$p(p(z))=((z-r)^mh-r)^mh$

[vivid gulch]

yoavmal

[open whale]

$=h\sum_{j=0}^m\binom{m}{j}\qty((z-r)^mh)^j(-r)^{m-j}$

[vivid gulch]

yoavmal

[open whale]

$=h(-r)^m+h\sum_{j=1}^m\binom{m}{j}\qty((z-r)^mh)^j(-r)^{m-j}$

[vivid gulch]

yoavmal

[open whale]

$=h(-r)^m+h^2(z-r)^m\sum_{j=1}^m\binom{m}{j}\qty((z-r)^mh)^{j-1}(-r)^{m-j}$

[vivid gulch]

yoavmal

[open whale]

$=h\qty((-r)^m+h(z-r)^m\sum_{j=1}^m\binom{m}{j}\qty((z-r)^mh)^{j-1}(-r)^{m-j})$

[vivid gulch]

yoavmal

[open whale]

ah, the opposite is what we want

$=h\qty(h(z-r)^m-r\sum_{j=0}^{m-1}\binom{m}{j}\qty((z-r)^mh)^j(-r)^{m-j-1})$

[vivid gulch]

yoavmal

[open whale]

uhh

i'm stuck, i'll get back to that later

oh it doesn't have to say anything lol

it is iterative

[hollow ice]

hmm

[dusty swallow]

U donut

Give full latex file else I won’t understand in 100 years

Bro is always throwing snippets around the whole server in different intervals

[open whale]

Well what I said is that my corrected conjecture is that if z(n) has r as a root m times, then z(nk) has r, m times as well

[candid cedar]

Yes, so the axioms of the real numbers have to be rechosen

[lethal wigeon]

$z_x = x$

[vivid gulch]

Miguel

[lethal wigeon]

Lol

[open whale]

Lol

My favourite automorphism

[placid bough]

What’s it like to do an reu

[open whale]

ok, i'm going to make it now

the proof in full

@dusty swallow i have it

let $p(z)=a_0+\sum_{j=m}^da_jz^j$ where $a_j$ are polynomials in $c$, and also let $m$ be the lowest nonzero power in the polynomial.\
we define the sequence $z_n$ such that $z_0=0$ and also $z_{n+1}=p(z_n)$.\
then, for all $n,k\geq1$, there exists some polynomial $h$ so that $z_{nk}=h\cdot z_n^m+z_k$

[sour sinew]

wait then what was even the point of defining p

or is this not the problem statement this is the solution

[vivid gulch]

yoavmal

[open whale]

proof:\
we will first show that for any $n,k>1$, there exists a polynomial $h$ so that $z_{n+k}=hz_n^m+z_k$.\
for $k=1$, we get $z_{n+1}=p(z_n)=a_0+\sum_{j=m}^da_jz_n^j=z_0+z_n^m\sum_{j=m}^da_jz_n^{j-m}=hz_n^m+z_1$\
now, assume that there exists $h$ satisfying it for $k$, meaning that $z_{n+k}=h\cdot z_n^m+z_k$.\
then, we get $z_{n+k+1}=p(z_{n+k})=p(hz_n^m+z_k)=a_0+\sum_{j=m}^da_j(hz_n^m+z_k)^j=a_0+\sum_{j=m}^da_j\sum_{v=0}^j\binom{j}{v}(hz_n^m)^v(z_k)^{j-v}=a_0+\sum_{j=m}^da_j\qty(z_k^j+\sum_{v=1}^j\binom{j}{v}(hz_n^m)^v(z_k)^{j-v})=a_0+\sum_{j=m}^da_jz_k^j+\sum_{j=m}^da_j\sum_{v=1}^j\binom{j}{v}(hz_n^m)^v(z_k)^{j-v}=z_{k+1}+\sum_{j=m}^da_j\sum_{v=1}^j\binom{j}{v}(hz_n^m)^v(z_k)^{j-v}=z_{k+1}+z_n^m\sum_{j=m}^da_j\sum_{v=1}^j\binom{j}{v}h^v(z_n^m)^{v-m}(z_k)^{j-v}=hz_n^m+z_{k+1}$

[vivid gulch]

yoavmal

[open whale]

you will have to handle the cropping lmao

[open whale]

now we'll show with induction that there exists $h$ so that $z_{nk}=hz_n^m+z_n$.\
for $k=2$ we obtain $z_{2n}=hz_n^m+z_n$.\
now, we assume there exists $h$ so that $z_{nk}=hz_n^m+z_n$, and now obtain $z_{n(k+1)}=z_{nk+n}=g(hz_{n}^m+z_n)^m+z_n=g(z_n(hz_{n}^{m-1}+1))^m+z_n=gz_n^m(hz_{n}^{m-1}+1)^m+z_n=hz_n^m+z_n$

[vivid gulch]

yoavmal

[open whale]

@dusty swallow repeated ping since it's the whole thing

@ashen agate too

i have the generalised thing

[ashen agate]

That is so unbearably ugly

[open whale]

i know

i don't want to get it cropped out by the ends though

[ashen agate]

You can do it without getting cropped out

Using align* environment

And newlining at proper places

But don't edit it now

Because it will mess up the order

[open whale]

i did

anyways this is just pure algebra extracting the things i want out to the form i want

[ashen agate]

ok, needless to say I will read it later

[ashen agate]

No you didn't, you just opened a $

[lethal wigeon]

how to get grilfriend 😿

[vague dagger]

This is not a maths topic dear @lethal wigeon

[lethal wigeon]

mb

how to get a grilfriend that likes math 😢

[vague dagger]

That's better

[lethal wigeon]

https://tenor.com/view/spank-bank-mochi-bunny-gif-24384259

[ancient helm]

Use my go to math pickup line
“Hey girl are you the integral of tan(x) • sec(x) cuz you’re sec C”

[vague dagger]

I like how you checked it first in #bot-cmd-latex

[ancient helm]

I was 99% sure but still

It’s like using a calculator for a simple multiplication equation

[lethal wigeon]

My doing my first order logic exam:
Is 5x3 =15 or 12??

[ancient helm]

Base 13:

[sour sinew]

4*2 is 8 no matter what base

even base 8 or less

8 still means 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

even if it's not really a symbol in the base

[ancient helm]

But 5 • 3 is 12 in base 13

[lethal wigeon]

ehhhhhh

ah yeah

xd

true

Btw, did you guys ever read "A mathematician's lament"?

Pretty good essay

I even based my english final project on it

[lethal wigeon]

I don't think that works, does it? 8 is just the representation of a quantity []

Oh you're right

8 is 1+1+1+1+1+1+1+1 regardless of the base

[sour sinew]

agreed

[open whale]

i did consider it*

i forgot to type the rest

but i ended up thinking "not worth it"

[hollow ice]

math is objective
all 1 billion schools of thought related to formal logic:

[sour sinew]

we put those people in a set and then we make statements about the complement

[hollow ice]

"The validity of LEM is so obvious"

"LEM is clearly false"

"The Continuum hypothesis is true"

"The Continuum hypothesis is false by mathematical folklore"

[sour sinew]

as far as im concerned if the domain of discourse does not contain anybody who doubts LEM, it's obviously true

[hollow ice]

haha

yes

brilliant

I love LEM

[solar quarry]

X

[sour sinew]

y

[solar quarry]

I am just pressing X

[sour sinew]

oh to pay your respects?

[solar quarry]

What

[sour sinew]

press x to pay your respects

it rhymes

[solar quarry]

https://i.kym-cdn.com/entries/icons/original/000/023/021/e02e5ffb5f980cd8262cf7f0ae00a4a9_press-x-to-doubt-memes-memesuper-la-noire-doubt-meme_419-238.png

[sour sinew]

you doubt that magma loves lem?

[solar quarry]

That or LEM itself

[sour sinew]

so you also doubt lem

well rip

[sour sinew]

guess this fails

[solar quarry]

Either or

Which means one of two is true, not all the two

[sour sinew]

could also be all the two

[solar quarry]

Yep

[lethal wigeon]

I don't get this

Can't it get proven or disproven?

[solar quarry]

No, it cannot be either proven or disproven.

It is proved that (it=) continuum hypothesis cannot be proven nor disproven.

[lethal wigeon]

Great, no subjectivity there

[hollow ice]

This is just wrong. Consider the theory with axioms $\text{CH is true}$. Denote this axiom $\Gamma$. Then $\Gamma \vdash \text{CH is true}$. Thus it is proved that it is provable that CH is true.

[vivid gulch]

Magma (N,^)

[solar quarry]

Wh, why wouldn’t you assume ZF here

[hollow ice]

I was just saying that the continuum hypothesis is independent of ZF, not all theories

[solar quarry]

Ugh, I mean there are not many axioms which is equivalent to continuum hypothesis.

[hollow ice]

Any number of extensions of ZF can prove the CH

Sure, maybe not axiom of choice

[solar quarry]

What "any number of extensions"

[hollow ice]

Star axiom

Which apparently in 2021 is a result of MM

[solar quarry]

Yea, I guess there's one

Anything else?

[hollow ice]

Maybe not indirect ones that are obvious, but certainly

I mean, assuming the generalized continuum hypothesis as an axiom is certainly viable

It was proven in ZF and ZFC that CH was unprovable

unprovable

Not false

Sorry I just woke up here

And so

It’s not universally true

[solar quarry]

Oh yes, what I meant was it is unprovable (in most contexts)

Ofc if you assume CH, it is true. If you assume something opposite, it is false

(Also sorry for my tone, I was overly annoyed when I got pinged)

[lethal wigeon]

That feels objective to me

As objective as any other axiom

[hollow ice]

Yes

[hollow ice]

Your tone was fine

don't apologize

[open whale]

@lethal wigeon check this out

[open whale]

Typo

let $p(z)=a_0+\sum_{j=m}^da_jz^j$ where $a_j$ are polynomials in $c$, and also let $m$ be the lowest nonzero power in the polynomial.\
we define the sequence $z_n$ such that $z_0=0$ and also $z_{n+1}=p(z_n)$.\
then, for all $n,k\geq1$, there exists some polynomial $h$ so that $z_{nk}=h\cdot z_n^m+z_n$

[vivid gulch]

yoavmal

[dusty swallow]

Just ping: everyone at this point

Imma read ur rant later today

[open whale]

Just read that image

Proof is just sheer induction

And algebraic sum manipulations

[lethal wigeon]

[lethal wigeon]

isn't h just a constant in terms of z?

sorry

[open whale]

yes

p(z) is just the tool to generate the next polynomial in the sequence

[open whale]

@sour sinew

[sour sinew]

hi

[open whale]

Want to help me show that there is an environment around each root in a polynomial in the Mandelbrot set that converges to a cycle - the same cycle length?

[sour sinew]

around each c that eventually goes to 0, we have a neighbourhood of constant cycle length?

[open whale]

Technically infinite cycle length

Since they never enter a perfect cycle

But they converge to a cycle

[sour sinew]

ah

they all converge to some cycles with the same length

but then
that sounds unbelievable

[open whale]

Whyso?

[sour sinew]

also i now see why the server pfp is mandelbrot set cuz the server owner worships mandelbrot set

[open whale]

Yes lol

[open whale]

So why not?

[sour sinew]

idk, it sounds like a strong condition for every point in a dense set to have a neighbourhood in which all points are (1) almost cycling (2) with the same period

presumably n where z_n is the first z(c) = 0?

maybe its only dense on countably infinitely many places other than the real line, idk

owo

seeming like the area around each zero doesnt even need to be really tiny

these are so cool

for all z with -0.5 < Re(z) < 0.2, -0.5 < Im(z) < 0.5, they seem to spiral in towards a single fixed point

maybe there are no zeroes of any z in this area except for c = 0

is the zeroes just the centres of the circles

and the environments you say are just the circle

s

[open whale]

Ye

[open whale]

Well they aren't precisely circles

[sour sinew]

whaa

[open whale]

The only two perfect shapes I know are the big cardioid

We can prove it indeed is a cardioid

And the little circle to its left

Which is, too, a perfect circle

So uhh

Lets see how to prove the cardioid

So we have the proper convergence

I.e.

It satisfies

$z_{n+1}=z_n^2+c$

[vivid gulch]

yoavmal

[open whale]

Well, actually, since we want convergence

We'd want

$z_n+\varepsilon=z_n^2+c$

[vivid gulch]

yoavmal

[open whale]

And since this is differentiable we can simply remove the epsilon

[sour sinew]

wat

epsilon is definitely nonconstant

[open whale]

Well we want to find an n such that the difference < epsilon

$\abs{z_n^2+c-z_n}<\varepsilon$

[vivid gulch]

yoavmal

[open whale]

Technically speaking

[sour sinew]

some n for which for all N > n the above holds

[open whale]

Ye, but effectively

Since p is continuous

We can infact demand them to be equal

If it converges to the value, the value also satisfies it

So we can actually demand

$z=z^2+c$

[vivid gulch]

yoavmal

[open whale]

Now, here's the odd thing

There's some magic done here I'm not 100% sure of's formalism

We solve for z using the quadratic formula

[sour sinew]

the magic of the quadratic formula

[open whale]

That's not the magic I don't understand

Oh, I understand it now

$z=\frac{1\pm\sqrt{1-4c}}{2}$

[vivid gulch]

yoavmal

[open whale]

Now, how do we tell which is it?

Ah

c=0 converges to 0

Still not enough, how do we prove it's - for all?

I suppose we can plug to the original equation and see which holds?

[sour sinew]

is one of the roots sometimes out of the 2 radius circle

[open whale]

Probably

No idea though

[sour sinew]

time to rig desmos up to compute complex square roots

[open whale]

I know it should be minus

But I want to prove it

[sour sinew]

well, square root is going to be a problem if you loop around the origin

it appears that if c is not very close to 1/4, one of the roots is inside the cardioid and one outside

[open whale]

And you can infact loop

The in would be the minus

Can we maybe prove it's universal?

Like, if it's - for one it's - for all?

[sour sinew]

nvm you would have to loop around 1/4

so its fine

i guess that does imply its - for all

[open whale]

How?

[open whale]

Oooooooo

Very nice

[sour sinew]

ummmm because (?? my knowledge about this topic is limited to 1 video) the thing inside the root needs to go around

[open whale]

Ah wait right

It's continuous

Nice

So, that's one done

So we have

$z=\frac{1-\sqrt{1-4c}}{2}$

[vivid gulch]

yoavmal

[open whale]

Now, we have another important rule

The derivative of the function needs to be less than 1 in magnitude, for the thing to converge

Otherwise it's repelling

How do we prove that first?

So lets suppose we have some iterative function f

Let me just watch 3b1b's video lil

Lol*

[sour sinew]

does that hold for complex

[open whale]

Yes, magnitude is the key here

Back

Ok so

[sour sinew]

i think i messed it up

[open whale]

Messed what up?

[sour sinew]

the figuring out where the derivative is > 1

[sour sinew]

i looked at the derivative of this 💀

[open whale]

Oh lol

No we look at the derivative of z²+c

Ok, I have it

So this works properly, it's just a proof on Julia sets from which we can deduce it on here

So basically

For a given c, we have a fixed point z

Which we have found the value of

Now, the fixed point is attractive, meaning points around it converge to it (which is what we're interested in, that precise infitesimal convergence)

If the derivative is less than 1

Which is rather obvious generally speaking

[sour sinew]

can you produce the cardioid though

[open whale]

Yes, now it's way simpler

We have that if 2|z|<1

Then it works

Which means that we can set z=re^iθ

Where 0<r<1/2

And get uhh

$re^{i\theta}=\frac{1-\sqrt{1-4c}}{2}$

[vivid gulch]

yoavmal

[open whale]

Uhhh

I suppose we can solve for c now

Though we didn't need the quadratic for that

$c=re^{i\theta}-r^2e^{2i\theta}$

[vivid gulch]

yoavmal

[open whale]

$c=re^{i\theta}(1-re^{i\theta})$

[vivid gulch]

yoavmal

[open whale]

Which corresponds to all 'cardioids' filling the area up to the boundary using r=1/2

So that proves it

now

we have 2 things we want

first off

we want to somehow do this for all convergences in the polynomial

secondly

we want to do this for a general polynomial

[sour sinew]

c = 1/2e^itheta(1 - 1/2e^itheta) is a cardioid?

[open whale]

ye

[vague dagger]

Cool bean

[open whale]

So what we want to replicate is this

For some n iterations

I.e. $p^n(z)$

[vivid gulch]

yoavmal

[open whale]

So we differentiate once

And what do we get?

A very large and hard to solve polynomial

How do we go on?

We want all regions in which the derivative in magnitude is less than 1

So we have fixed points and attractive values

Ahhh

No, actually, nope

Well, nvm, I do have an idea

We can use the roots to say

$z_{n+k}=z_n$

[vivid gulch]

yoavmal

[open whale]

And then for regions around the roots, if the function is attractive

Then there is a small region satisfying this

[solar quarry]

What is this?

[open whale]

General discussion

On maths

[solar quarry]

No, z_n+k = z_n

[open whale]

Right now, I am trying to prove the connection between the roots, and the bulbs in the Mandelbrot set

@solar quarry do you want the full explanation?

[solar quarry]

Ohh

Interesting subject, but it's okay you do not have to explain further

[sour sinew]

(from #1130945077099905104)

mr. atf has n^2 rectangular prisms of dimensions 1x1x1, 1x1x2, 1x1x3... 1x1xn^2

and he puts the 1^2 faces in a nxn grid

then what's the expected surface area

so

first we find the total surface area of all the prisms

$1 + 4(1) + 1 + 1 + 4(2) + 1 + 1 + 4(3) + 1 \cdots = 2n^2 + 4(1 + 2 + 3... + n^2)$

[vivid gulch]

cute rizzly bear (won't eat you)

[sour sinew]

$ = 2n^2 + 4\frac{n^2(n^2 + 1)}{2}$

ummm where is my latex

[orchid lodge]

He got bored

$ = 2n^2 + 4\frac{n^2(n^2 - 1)}{2}$

[sour sinew]

anyway, = 4n^2 + 2n^4

[orchid lodge]

$frac{1}{1}-1$

[sour sinew]

you need the \frac

we will continue

[orchid lodge]

Yes I just did it to test it XD

Sure

[sour sinew]

now to find the expected area between 2 adjacent prisms that is covered

[orchid lodge]

I've derived it until there but I don't know how to derive the «expected» number of things

[sour sinew]

there are n^2 - 1 ways it can be 1, n^2 - 2 ways it can be 2, etc.

so we have the average is

$\frac{\sum_{k=1}^{n^2 - 1} k(n^2 - k)}{\sum_{k=1}^{n^2-1}k}$

[vivid gulch]

cute rizzly bear (won't eat you)

[sour sinew]

hecka ugly

[orchid lodge]

XD

[sour sinew]

im just going to do the rest on 1 message

[orchid lodge]

Sure

[orchid lodge]

Shouldn't it be 2(n²-1) since we can swap the towers?

[sour sinew]

then you would be dividing by 2k as well

[orchid lodge]

Right

[sour sinew]

$\text{there are exactly n(n-1)(2) adjacencies on the combined shape,}\
\text{each loses twice the expected touching area }\
4n^2 + 2n^4 - \frac{\sum_{k=1}^{n^2 - 1} k(n^2 - k)}{\sum_{k=1}^{n^2-1}k}(2)(n)(n-1)(2)\
= 4n^2 + 2n^4 - \frac{\sum_{k=1}^{n^2 - 1} kn^2 - \sum_{k=1}^{n^2 - 1}k^2)}{\sum_{k=1}^{n^2-1}k}(2)(n)(n-1)(2)\
= 4n^2 + 2n^4 - \frac{n^2\frac{(n^2-1)(n^2)}{2} - \frac{(n^2 - 1)(n^2)(2n^2-1)}{6}}{\frac{(n^2-1)(n^2)}{2}}(2)(n)(n-1)(2)\
= 4n^2 + 2n^4 - (n^2 - \frac{(2n^2-1)}{3})(2)(n)(n-1)(2)\
= 4n^2 + 2n^4 - (\frac{n^2 + 1}{3})(4)(n^2-n)\
\text{etc you can do the rest}$

[vivid gulch]

cute rizzly bear (won't eat you)

[orchid lodge]

Got it

Simpler than I thought

[agile roost]

your brackets are fucking everywhere

line 2

[open whale]

why would brackets do it to everywhere

anyways

short tip to help

[sour sinew]

omg, one stray bracket, oh no

[open whale]

add \qty infront of (

[vivid gulch]

PDF File (apex predator)

[sour sinew]

i know how to do \left and \right

i just do not feel like it

as it is pointless

[open whale]

\qty( helps

$\qty(\frac{a}{b})$

does it for you

[vivid gulch]

yoavmal

[sour sinew]

nice

[orchid lodge]

That's a nice trick

[open whale]

makes it 300% more legible

[agile roost]

unnecessary

[open whale]

$\qty{\frac{a}{b}}$

[vivid gulch]

yoavmal

[agile roost]

if you can waste time, waste it

[open whale]

go ahead

walk in a circle for the rest of your life

[agile roost]

do you not have a house? if you do then you already are

[open whale]

i live with my parents

[agile roost]

ok, so you're walking in a circle for the rest of your life more of less

[open whale]

actually i am sitting in place

[agile roost]

size of the circles varies

[open whale]

based

[orchid lodge]

Radius=0

[agile roost]

maybe he never returns to the same point

in which case it's just a line

until he overlaps and it's some twisted lemniscate

[vivid gulch]

PDF File (apex predator)

[orchid lodge]

Maybe it's a circle in a non Euclidean manifold

[agile roost]

curved spaces 🥴

[orchid lodge]

You're living in one right now

[agile roost]

actually

nevermind

icba

I have done too much GR for a man's entire life

[orchid lodge]

Your momma so fat

Her Riemann Curvature tensor invariant approaches infinity

[open whale]

me when sqrt(z)

[agile roost]

fuck outta here with your Riemann surface ASS BULLSHIT!!

[open whale]

we need to make that gif but with a mathematician

[orchid lodge]

Literally

Was looking for the same meme

[agile roost]

wrong one

propaganda

we want the truth

[agile roost]

this is the truth

[agile roost]

this is heresy

[orchid lodge]

Wow

Enlightened

[open whale]

i chose a terrible meme intentionally

[open whale]

delighted, actually

[agile roost]

Evil Gödel says what the people want to hear

a force to be reckoned with

[open whale]

we're not bringing newton to discussion

[agile roost]

GODDAMN IT MAN

I mean, relatively speaking

[orchid lodge]

Relativity?

[agile roost]

generally, even

[orchid lodge]

More like

Idk

[agile roost]

most beautiful formula? Kerr-Newman metric

[open whale]

lets break the discussion to things we can quantitate

[agile roost]

your dick is a single quanta

[sour sinew]

the most beautiful formula is $\forall x \in \mathbb{C}, x = 0$

[vivid gulch]

cute rizzly bear (won't eat you)

[open whale]

wow why is the world so pixelated all of the sudden

[agile roost]

liar, your world is clearer than ever

[open whale]

oh this is actually simple to prove

[orchid lodge]

Formula of deceivement, prohibiting and the enlightened youth from making tendentious science

We feed these formulas

Into the youth

[open whale]

we know $\sin(x)=0$ for all $x$

[orchid lodge]

Making them believe it's real like sheep

[vivid gulch]

yoavmal

[orchid lodge]

Dumb asses

[open whale]

and since sin(x) is not a constant function

then its range is the entire complex plain

[agile roost]

Kerr-Newman is NOT fed to kids, Kerr metric maybe

Schwarzchild metric even more

[open whale]

now, we can use a second identity

that $\sin(x)=x$

[agile roost]

even the Reissner-Nordstrom metric

[vivid gulch]

yoavmal

[agile roost]

but not Kerr-Newman

[open whale]

to get the simple result that $x=0$ for all $x\in\bC$

[vivid gulch]

yoavmal

[agile roost]

proof by induction actually

x=x obviously
assume x=0
0=0 obviously
x=x
x=0 so x=0

[orchid lodge]

Take this

https://ravens-2-progressive-matrices-clinical-edition.netlify.app/

What will you guys get

[agile roost]

advertising??? banned

[orchid lodge]

Damn it

[open whale]

well you missed a step

1=0 by the law of big numbers

now you can make steps properly

[orchid lodge]

cryptic symmetries of Lie groups artfully intertwine with the intricate fibrations of fiber bundles

unveiling the enigmatic nature of Riemannian curvature and the profound implications of noncommutative algebras

[open whale]

those are lies

[orchid lodge]

That's what the government wants you to think

[open whale]

lie groups

[orchid lodge]

Hairy ball??

[agile roost]

looks like a sham but I tried it

it is a sham

153±4 😴

[dusty swallow]

Ur a sham

Tf even is a sham

,w define sham

[vivid gulch]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by dappppy#0

[dusty swallow]

Sounds like u

[agile roost]

actually sounds exactly like you

riddle master

[dusty swallow]

Rizzle me this

x=w

[agile roost]

shut the fuck up

[dusty swallow]

Determine why ur dad left

[dusty swallow]

U donut

[agile roost]

unlike yours he's still around

I feel for you though

must be tough

[orchid lodge]

It's legit but slightly inflated

[agile roost]

inflated? it's squished

[dusty swallow]

The cope shows what pain u have been through

My best wishes

[agile roost]

lmao, your sheathed cope

[orchid lodge]

I've got plenty different professional IQ tests illegally

[agile roost]

illegal iq test lmao

[orchid lodge]

Illegal to redistribute

[agile roost]

ah

[orchid lodge]

SBV, WAIS-IV etc

[dusty swallow]

HIV, AIDS

[orchid lodge]

Try his one @agile roost https://brght.org/

[agile roost]

MRSA!

[agile roost]

"the most accurate iq test" bruh

[orchid lodge]

Normalized on participant data

[dusty swallow]

20 minutes

[orchid lodge]

Has a fine g-loading

[agile roost]

😕

can't have many participants seeing as I've never heard of it

[orchid lodge]

There are plenty takers

Just try it out bro

It's fun

I don't take these seriously

[agile roost]

I don't consider IQ tests as fun, they're just trivial shapes

and actual tests aren't really just shapes

[orchid lodge]

Mate I know that, I can send you a pdf of what actual one looks like

These just measure your PCI

[agile roost]

I've done an actual one lmao

[orchid lodge]

Really? Which one

[agile roost]

Cattell III B

illegally

[dusty swallow]

He has an IQ of a potato approx 20 +/- 30

[orchid lodge]

Kinda niche

[agile roost]

and Stanford Binet legally

[orchid lodge]

Ohhh

Now that's legit

What you got?

[agile roost]

both are recognised by mensa

160

[orchid lodge]

Insane

What age did you take it?

[agile roost]

magma got 160 on wais or whatever

[agile roost]

16

[orchid lodge]

There is no available proctoring for such tests since I live in a non anglophone country

[agile roost]

ah

[orchid lodge]

Never got myself tested

XD

I just test others

[agile roost]

lmao

I checked my answers on the first test you sent and I got 100% (as I should hope lmao) and it maxes out at 153

[orchid lodge]

Yup that's the ceiling

Take the second one

Wonder what you'll get

[agile roost]

i don't find them fun, unfortunately

[orchid lodge]

Duh

[agile roost]

take this sonic boom diagram

[orchid lodge]

Hot

[agile roost]

"he turned into me!!"

[orchid lodge]

This scenery looks familiar

[agile roost]

fr

[agile roost]

I should add that this was SB5, not the old one with SD 16

[orchid lodge]

That makes it even more impressive

I hate being a non native speaker

Literally all my scores will be deflated for the verbal sections

I still do pretty well on most verbal tests

I got something 700+ for the 1980 SAT

https://rentry.co/ud2nt

[agile roost]

tbf the SAT is just American English which is absolute gobshite

but

you do type quite well for a nonnative speaker

[sour sinew]

you can probably obtain a copy in which the math problems are translated to british english

[orchid lodge]

Bear in mind how I specified it's 1980's SAT specifically for that

Check out the article I sent

It's not much of an tendentious statement to call these bygone forms virtually IQ tests

[agile roost]

... the math problems

bro what

[orchid lodge]

That's why they reformed the SAT

[sour sinew]

what are they supposed to be

[sour sinew]

wth

lemme just

study for 244 hours

[agile roost]

they pretty much are just IQ tests

or nowadays just how fast can you do maths in your head

[sour sinew]

student affluence test

[agile roost]

which can be trained

[orchid lodge]

They reformed it so that people who practice yet limited by IQ could just achieve better

[orchid lodge]

Exactly, it's just a matter of practice for the Modern SAT

[sour sinew]

so it's the how much you practiced? test cuz in that case i got 0%

[agile roost]

funny

I looked at an SAT question once

does that count?

[sour sinew]

100% counts

[agile roost]

(prior to sitting the exam)

[sour sinew]

practically cheating

[agile roost]

NOOOOOO

[orchid lodge]

Pretty much

Your execution trial is by

The next day

[agile roost]

fuck

it's getting late therefore I mustache

[sour sinew]

you mustache

[agile roost]

I mustache

[orchid lodge]

Thus mustache

[agile roost]

I'll just hope you guys got that one, bye

[orchid lodge]

I'd hope for you to explain it some time

[agile roost]

I mustache is pronounced similarly to I must dash

[orchid lodge]

Depends on the accent

I read it as must ach XD

[agile roost]

kind of uniquely south west English

sometimes stimulates a double take if you slip it into an otherwise normal conversation

already bye fr now

[orchid lodge]

Bye

https://pdfhost.io/v/F3fb0u6uV_SAT_1980pdf.pdf

If anyone would like to take the older SAT

[placid bough]

Who voluntarily takes the sat lol

And isn’t it 3-4 hours

[lethal wigeon]

Is that university entrance exams?

It doesn't look too hard ngl (at least part 3 and 5, I skipped 2 and 4 because we'll I'm lazy to read)

Part 1 I think I could do good enough not being native English

This one has confused me tho

Is the belt tied to the circles or is it like moving thanks to the opposite movement of the circles

Idk if runs over means the second thing here

Anyway

The type of exam is at least more interesting than the ones we do here to get to uni, but there are a lot of mental math which seems pointless ngl

[placid bough]

The difficulty lies more in the time constraint than the problems

[sour sinew]

idk about you but when i roll a wheel, it doesn't push the ground in the opposite direction

so the belt is moving with the wheels

[lethal wigeon]

The ones in my supermarket work in the opposite direction

[candid cedar]

Can you send video

[lethal wigeon]

Yeah whenever I get to my city back, I'm in vacation right now

Remind me the third week of August

[sour sinew]

i am highly confused how your wheels break the laws of physics

[lethal wigeon]

Maybe it was just me confusing a wheel with something else

But when you touch like the extreme of the supermarket thing, you can feel like there is something rotating there in the opposite direction

[halcyon heart]

what part confuses you?

I'd say the answer is 2

[lethal wigeon]

I was saying that some wheels like the one at my supermarket move the opposite to the belt movement

[orchid lodge]

Just did it an got 780 with 10 minutes spare for the M section

[last briar]

this should be renamed to generalesqueuest

[sour sinew]

im interested in taking geometry problems and finding their dual, is there any information on how to deal with circles

you can apparently get the dual of a curve by finding its tangents and using the dual of those

but idk

my intuition says the dual of a circle should be a circle

then the question is what circle is centred on a line

[solar quarry]

What dual are you talking about?

[sour sinew]

well, the way points and lines can be exchanged
2 lines share a point, 2 points share a line, som sets of 3 lines coincide, some sets of 3 points are collinear
theorems about lines and points work the same if you switch them in this way

[solar quarry]

Ah that one

Sounds like projective geometry

[sour sinew]

yep

oh, i guess circles are not preserved by projective transformations

RIP ig

[solar quarry]

Indeed.

That's why you had problems

[sleek pasture]

.

[hollow ice]

wtf don’t take this out of context

[lethal wigeon]

Wow, that's insane.

[lethal wigeon]

Hmm

[ornate thunder]

how understandable is this notation?

[sour sinew]

i'd say it's pretty understandable

idk what S1 is

[ornate thunder]

S is the successor function

S(1) would be 2

[sour sinew]

for all x4,x5 either
(x4+2)(x5+2) != (x2+2)(x3+2) or x4+x5 = x1+x1
(x4+2)(x5+2) = (x2+2)(x3+2) implies x4+x5 = x1+x1

[ornate thunder]

yeah, !a or b is the same as a implies b

fun fact: unless i made an error (which is possible) the thing notated should be equivalent to a well known problem

[sour sinew]

x4x5 = (x2+2)(x3+2) implies x4 + x5 = 2x1 + 4

how is this possible

oh these are integers aren't they

i assumed it was over R

[ornate thunder]

clarification: the set being quantified over is N

[sour sinew]

positive?

[ornate thunder]

nonnegative integers

[sour sinew]

for all nonnegative integers x, there are x2,x3 with average x, such that ab = (x2+2)(x3+2) implies that a,b have average x+2

is this goldbach conjecture

[ornate thunder]

bingo

[sour sinew]

it seems like x2+2 and x3+2 should be primes ya

[oak cliff]

yes i am sure there is a proof somewhere

[sour sinew]

i don't think so

i think it converges for sufficiently negative x

π is not a liouville number, it can only be approximated so well

it's just that wolfram is extremely terrible at correctly determining the fact that it doesn't know whether this converges or not

[undone cove]

hi

[orchid quail]

hellos