serious-discussion

[south dawn]

can lambert help us

[magic python]

Wait what is a formula then for it

Using non elementary functions

[south dawn]

for quintic?

tberes one

theres

its like the fountain of youth tho

[magic python]

Like obviously you could define a function and just say it's output is the roots of the polynomial

[south dawn]

u gotta be juan ponce de leon

[magic python]

But what is a well defined way that isn't approximation?

[south dawn]

million dollar question right there

[magic python]

Sadly not

[south dawn]

with things like the lambert W function and the erf its been studied and it can be graphed in the complex plane

[neat lintel]

What’s the integral of

1 over ln(x) dx?

It looks simple tbh

[worldly horizon]

non elemntary i think

[south dawn]

can we get rid of the absolute value

pretty please

[neat lintel]

yeah
I found that out and shit felt like I was standing on the edge of it hah

[worldly horizon]

even if you do it’s non elementary

you’d need the function li(𝑥)

[neat lintel]

Which is larger

e^pi or pi^e

[south dawn]

i found something interesting

[neat lintel]

by intuition only

[neat lintel]

Sure

[south dawn]

https://en.wikipedia.org/wiki/Bring_radical
@worldly horizon @magic python

[worldly horizon]

[magic python]

We really just defining whatever now, huh

[vast wraith]

there are many, here is one:

[south dawn]

just sent the wiki link for the bring radical mentioned at the bottom

for those who are curious

[magic python]

Wow uhhh so a multivalued series?

[vast wraith]

no, that is just one of the solutions

[magic python]

Awww 3:

Sad

[south dawn]

if u have one of the solutions could it not be reduced to a quartic with the one root factored out?

either long division or plain factoring

[magic python]

Yes ig

[left patio]

how it feels seeing Riemann get glazed just for plugging complex numbers into the oiler zeta function.

[tidal acorn]

Hello

https://tenor.com/view/nice-gif-13099583323393004498

Wait it didn't embed

can someone post it embedded pls

:(

[left patio]

mathematics dc can’t afford embeds

too expensive for our unemployed asses

[tidal acorn]

fine then just go to the link to see the gif or whatever

[gritty heath]

Its not as simple as just plugging it into Euler's series formula, because that doesn't always work, Riemann did the analytic continuation to do this

[left patio]

let me slander fraudmann

goatler solos

[hallow plume]

https://tenor.com/view/epic-embed-fail-epic-embed-fail-embed-fail-gif-9875677717620262384

[tidal acorn]

go on. post the embedded gif then.

[hallow plume]

why lol

[worldly horizon]

It’s a Rickroll

[terse bay]

https://tenor.com/view/no-image-perms-no-image-perks-no-image-gif-12889793216262706171

[zinc tendon]

https://tenor.com/view/epic-embed-fail-ryan-gosling-cereal-embed-failure-laugh-at-this-user-gif-20627924

[patent phoenix]

I'm thinking of learning Lean before my Discrete class next term. have any of yall learned it?

and how helpful is it

[surreal sapphire]

what kind of benefit are you hoping for

i dont see how knowing lean would be particularly useful

[patent phoenix]

proofs

discrete is an inherently proof-heavy/logical class

was wondering if lean would help w this

[surreal sapphire]

not really

[zealous bronze]

i mean you could use it, but its not going to directly help you in any way

for an intro discrete math class

if you want to learn it independent of that then yeah do it

[patent phoenix]

I was thinking if I have exposure or practice of learning how to write theorems and proofs in lean, that skillset could benefit me in discrete

i realize now that this isnt a property of just discrete tho lol

ig i could learn it just to have it

[zealous bronze]

kind of, but the way of writing proofs in a proof assistant is different

[surreal sapphire]

humans don't write proofs like you do in lean

its almost always significant extra work

[patent phoenix]

hmm

maybe i can absolutely flex on my teachers

(im joking)

[turbid sedge]

go work bih

[rose ridge]

Unc McDonald's is calling...

[turbid sedge]

NAW

[twin meadow]

Hi CHATTT

how's everything going

[neat lintel]

It’s dead

[worldly horizon]

hello!

[worldly horizon]

good. Life goes on.

hbu?

[viral oracle]

[viral oracle]

[neat lintel]

Weird ahh stickers

[twin meadow]

yep same

[viral oracle]

lol

[worldly horizon]

hello hanako!

[small dust]

[viral oracle]

[small dust]

[viral oracle]

[small dust]

[dusty lark]

Best way to unbias yourself is by randomly generating opinions.

[old oak]

Depends on the distribution

[neat lintel]

for example this is stupid imo

lol

[silent junco]

nope

not allowed here

no? youre just not allowed advertise.

nobody is allowed to

recruitment is also not allowed

[charred thunder]

Huh, whats happening here

[magic python]

"Looking for people who—" Yeah advertising.

[charred thunder]

The or is wild 💀

[silent junco]

take a day to read the server rules

we can move on now, thanks

[zealous garden]

GA cult days are numbered

The set of statements I have left to make about GA is in bijection with a proper subset of itself

[twin meadow]

Hi chat

everyone slept

its raining outside

AND LIGHT WENT OUT TOI

ANYTHING WORSE CAN HAPPEN NOW 😭

[cerulean kite]

hi theree

i really want some tips on how to make practicing math a habit every day even when i dont have it in school i still want to become really good at aptitude and solving stuff fast and accurately so some tips would be nice (im in middle school btw)

[cerulean oracle]

I guess you have to practice a lot in order to become good at solving stuff

[fathom tendon]

Random question - has anyone been able to sign into mathprograms in the last hour

I've been unable to sign in and I'm not receiving login tokens to anywhere in my email inbox

[harsh lion]

Know urself

[fathom tendon]

What

[cerulean kite]

yeah ik but like how do I make it a routine every day yk

[cerulean oracle]

Practice every day. I'm not sure what else to say tbh

[vale coral]

maybe find a way to make math fun? I'm not sure how with with middle school math but once you do you'll naturally get better at it and you won't need a strict routine for it (since it's something you do/think for fun). Maybe try to understand it better, find some applications or talk to people about it. All those might be fun

[astral axle]

I have an Instagram profile called nicemath_ where I make math videos using Manim. Anyone who can help me out there, I’d really appreciate it

[frozen patio]

guys lets be real: 1+1=3

[tulip flume]

@frozen patio Please elaborate that.

[frozen patio]

ok

[sand orbit]

2 = 0

[desert rapids]

Dumb question
$$ \left(1 + \gamma M_1^2 \right) \sqrt{M_2^4 + \frac{2 M_2^2}{\gamma - 1}} = \left(1 + \gamma M_2^2 \right) \sqrt{M_1^4 + \frac{2 M_1^2}{\gamma - 1}} $$
Is there any good way to solve for $M_2$ without doing the huge expansion?

[silent junco]

M_1 = M_2 works

[fathom swallow]

supersonic

[desert rapids]

You aren't wrong

I figured it out

Solve[(1 + g  M1^2) Sqrt[
    M2^4 + 2  M2^2/(g - 1)] == (1 + g  M2^2) Sqrt[
    M1^4 + 2 M1^2/(g - 1)], M2]

Mathematica for the win

[still granite]

Is calc ab low for a senior in hs?

[torpid bay]

For people interested in math, then I generally hear calc ab or bc done in senior year, or maybe calc 3 if your school offers it

Some specific schools have programs where you can take a lot more mid to late stage college stuff while in high school, but those are the exception. Alternatively, you can take a summer course or 2 at a college while still in hs if you meet the requirements. But in general for most people I hear them doing calc in college

[verbal quest]

no, not at all if you look at the entire HS population
or even just the 4 year college bound
there are just a bunch of accelerated progress types in mathcord

[past heron]

No.of soln of sinx+cosx=1 in interval [0,2pie]
Ans will be 2 or 3?..
Help me out

[silent junco]

square it

$\sin(x)\cos(x) = 0$

[fathom swallow]

råys

[past heron]

No

I m just confused that either should I take 9pie/4 or not..like after solving the equation it will come to sin(x+pie/4) = 1/root2

And if we equate sinx= 1/root2 in interval 0 to 2pie..then the no of soln will be pie/4,3pie/4 and 9pie/4

But some said u shouldn't take 9pie/4..

Means it will lie on the same line as pie/4

If we look at the quadrant

[silent junco]

the equation has solutions at 0 and pi/2

and 2pi

[past heron]

Not at 2pie?

Accha

Ok ok

[silent junco]

bangali?

[past heron]

Nah..indian (hindu)

[silent junco]

ah i see

[past heron]

Btw how u did it

[silent junco]

squared it, then removed extraneous solutions

the candidate solutions are $\frac{k\pi}{2}$ but some of them are due to -1 getting squared to 1

[fathom swallow]

råys

[silent junco]

so you just plug them into the original equation to see if they work

[past heron]

Yaa..ok ok I did it by harmoinc additional theorem..it's easy

Thanks though

[silent junco]

np

[zealous summit]

Guys can someone join my Microsoft teams school meeting please, it’s in Bulgarian though.

[inland hornet]

bro left

[charred heath]

Random question, but is anyone good at height comparisons here?

[long matrix]

tall ask

[torpid bay]

@foggy meadow did you find the x^x thing or nah?

[dry coral]

Me and my teacher were actually trying to integrate x^x the other day, we couldn't figure it out, but it was cool to see what problems came up along the way

[quasi raven]

I thought $sin^2(x)+cos^2(x) = 1$

[silent junco]

yes

[fathom swallow]

Jake

[silent junco]

$(\sin(x) + \cos(x))^2 = 1 + 2\sin(x)\cos(x)=1$ in this case

[fathom swallow]

råys

[quasi raven]

From $(a+b)^2=a^2+2ab+b^2$

[fathom swallow]

Jake

[silent junco]

indeed

[quasi raven]

I love that formula and that saved my time

[neat lintel]

Give us the cubic one

[quasi raven]

Which one sum of cubes or difference of cubes

[neat lintel]

Both

[foggy meadow]

Na.

[rose ridge]

Natrium

[daring totem]

Guys I have a question. How can someone study Einstein's special theory of relativity without even knowing real analysis or even basic calculus

One guy I know claims that

[latent edge]

[snow ice]

I sob with you

[daring totem]

Bro what happened @snow ice

[snow ice]

nothing

[sharp portal]

Sorry, to intrude but does anyone have book recommendation or resources for self studying physics?

[snow ice]

#old-network #book-recommendations

[viral oracle]

[hexed tangle]

[viral oracle]

[tropic pivot]

https://tenor.com/view/simpsons-ned-okily-dokily-gif-4289162988260002514

[smoky turret]

do u know alzebra and calculas?

[fiery stump]

no idk alzebra

[rich quail]

Marisa

[loud tinsel]

yo

[shell locust]

It is also from the pythagorean theorem

[verbal steppe]

hii

[cyan karma]

Cloudflare saying they have detected unusual traffic from my network when accessing an article though it's just me alone at home and it's the first time I open a researchgate article today looooool

[tall escarp]

how long does it take to learn the DI method?

[magic python]

Learning what it is is simple iirc, but learning when to use it may take longer

[tall escarp]

yea i was referring to the latter

im in hs so i gotta use it for IBP that includes cyclical integration

[magic python]

You might be able to learn it in a month or so (Probably an exaggeration, it's been a while since I've messed with it)

[wraith magnet]

someone is definitely stealing your wifi and using it to do research

[tall escarp]

woahhh thats a long time

what makes it take so long?

[magic python]

Probably could in a week tbh

It all depends on how much exercises you do

Practice is really good if you want to be able to spot when you use it

You could probably learn the basics in a day or two tho

[tall escarp]

its applicable everywhere where IBP can be done right

[magic python]

DI is IBP

It's just formatted easier to understand

iirc

Idk you could probably do it in a week, maybe even less

Considering you seem to know calc decently enough already

iirc you can find lots of videos explaining it on YouTube, most of the time would probably be getting used to noticing when to use it, that comes from practice

[daring totem]

Guys reply

[idle tiger]

I think Brian Greene is absolutely amazing at explaining relativity: https://www.youtube.com/watch?v=XFV2feKDK9E

But if you don't have a good grasp of the math it'll be hard to derive the Lorentz equations and understand it precisely

[placid shadow]

a*b=c, can you subtract b from both sides?

[cyan karma]

Ok fine I'll let them if they do research

[wraith magnet]

!xy

bru

[placid shadow]

What?

[wraith magnet]

what's the original question?

[placid shadow]

There isn't one. This is a conceptual question, in general.

[wraith magnet]

sure you can. whether or not it helps you is a different thing though.

[placid shadow]

I tried subtracting cos(55) here but it gave me an incorrect result.

Why?

[wraith magnet]

open a help channel.

[placid shadow]

no ty

[rose lance]

You don’t need to subtract, just multiply ED with cos(55) to get ≈0.573ED and then to find ED you just divide the value of 0.573 by 4 to get an estimated answer of 6.98. Hope this helps

[loud herald]

Can someone help me with 24=4 x p ??

[rose lance]

Sry I may be a little late

[placid shadow]

!help

[quasi jetty]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[bronze rock]

if the whole thing about HCP is that any complex conjugated amplitude based on direction is changed but only reflected as a mirror image then why do they remain pointing the same direction but are still not topologically equivalent? if they are mirroed then surely they are?

cant they be deformed to match regardless if there is a different start

[pastel needle]

Drink vending machines intentionally drop it forcefully so that your drink explodes and you have to get a new one

[daring totem]

No bro I'm not asking how can I do that, I'm asking how's that even possible for anyone? One of my friends claims that.

[idle tiger]

To strictly answer your question, you don't need to know real analysis or even basic calculus to derive much of "special relativity"

However, you need to know some algebra for sure (for example to calculate the lorentz transformation), and some geometry for example the Pythagorean theorem for light clock calculations

[drifting flower]

Hey guys, I want to get into college to become an apprentice. Thing is I struggle a lot with math and have a learning disability. A lot of these apprenticeship trades require lots of math or little math. Could anyone give me some advice?

[gentle birch]

take your time with the concepts and practice a lot

this might seem as generic advice but it's what works for a lot of people
idk much about any learning disabilities though so sorry about that, but what's tried and tested is worth a shot

[drifting flower]

Okay, would you think getting a tutor would help? I mean they'd have to be patient with me and I learn differently I mean everyone does, and I don't know much about my learning disability I'm diagnosed with it and I know one of the stronger things I struggle with is math so it kinda brings down my hopes for making the decision to go to college, I didn't do bad in school I averaged 70s, my dad says he would pay for it but I'd have to be really committed so there's a lot of pressure.

[hallow glacier]

hi

[gentle birch]

Getting a tutor does help, yeah
I've never had the need to get one but I know people who did and it helped them improve their grades
Still, I'd first try and teach myself, there are various sources
the best ones being khan academy, and then of course textbooks

if that doesn't work out feel free to get a tutor

[drifting flower]

Okay thank you

[neat lintel]

twenty-four minus four equals twenty, therefore, P=20

[drifting flower]

@gentle birch what's the best way/method to learn division? I can't do division whatsoever

[gentle birch]

go through worked examples slowly and then practice a lot

could also watch a youtube video

[drifting flower]

True

[gentle birch]

start from basics like

40/10

20/2

then go to stuff like 528/13

remember that division is just repeated subtraction

[drifting flower]

I don't understand

[gentle birch]

40/10 = 4
how many times can you subtract 10 from 40 until you can't anymore

so you have

40 - 10 = 30, that's 1

then 30 - 10 = 20, that's 2

20 - 10 = 10, that's 3

10 - 10 = 0, that's 4

can you do 0 - 10? nope

so we have 40/10 = 4 with remainder 0

[gentle birch]

remainder's just the number left at the end, 0 in this case

[drifting flower]

So for example

[drifting flower]

20÷2 can be divided 10 times with the remainder as 0

But 20÷2 what would it equal

[gentle birch]

yeah

10

[drifting flower]

Okay so basically

[gentle birch]

can be subtracted* not divided

[drifting flower]

Ohhh okok

So any division problem you subtract it till you can't and the number of times it can be subtracted till 0 is the answer?

Like the answer is whatever

[gentle birch]

that's a slow way to do it, there are better ways to do it

but that's one way to do it

that's basically the idea of what division actually is

[drifting flower]

What are the other ways I've got my notebook open right now

But I need it to be easy

[gentle birch]

you can look up long division on youtube right now

[drifting flower]

Okay thank you

[zealous garden]

not quite, but it answers a question related to repeated subtraction

if you think about it, modulus is repeated subtraction

5 % 3 = 2

15 % 2 = 1

repeatedly subtract until doing so again would yield a negative number

division tells you how many times you could repeat that subtraction, but is itself not the repeated subtraction

interestingly, you could also look at mod as the canonical projection of the quotient by the ideal generated by the second argument

so div is divide the number, mod is divide the space

I think that's cool

[cold needle]

?

What if i put in a negative number in the left argument of %

[blazing raft]

then the earth combusts and we all die unfortunately

[cold needle]

ohh ok i see

I will avoid that then

[blazing raft]

thank you

[cold needle]

👍

[agile fiber]

you subtract the second argument a negative amount of times

[tawny cosmos]

6 7

[random tangle]

does anyone know of a method to solving a bernoulli question faster?

[tulip wren]

just master your current method

[true fox]

Anyone here a expert at pre calc? Need some help

[agile fiber]

can you please not spam the same message in three channels

[round abyss]

dude just write the help channel post

check #precalculus probably. you get faster help from asking any questions you have directly

[magic python]

<@&268886789983436800> spamming too

[silent junco]

how did they even spam that i already timed them out

[magic python]

They sent it again so idk

[silent junco]

oh wait this is discussion 2

[magic python]

I just say it's spam if it's copy pasted to more than one channel

[silent junco]

nvm

[silent junco]

no it was definitely spam you were right to ping

[magic python]

Kk

[silent junco]

i just thought they sent it after i timed them out

[magic python]

Ah kk

[mint patio]

me trying to look for engineering stuff:

google:

maybe this belonged in chill sorry lol

[ripe dirge]

I am definitely buckling under pressure

Funny how I chose this major by my own volition and I am struggling with it

Definitely has to do with time management

[jagged forge]

math major?

[ripe dirge]

Yes

[gritty heath]

My friend describes this as "picking an academically challenging degree and then going 'wow this is academically challenging'" haha

Its hard for many of us, dont worry too much about it, you'll get through

[rapid scaffold]

patty cake patty cake with no hands

[ripe dirge]

I hope I do

My first year was a disaster, thanks to numerous health issues

[crisp kindle]

I hope this channel is kinder to mee

[latent leaf]

Sup

[half skiff]

sky

[proper umbra]

can any one say some importent properties of hyperbola

[crystal cave]

how to get motivated quick

quick tell me the secret

wrong chat

[ripe dirge]

Simple
Existential dread and anxiety about the future

[inland raft]

you either need to feel the pressure of an upcoming exam

or have someone threaten you

[verbal spindle]

spite

[normal torrent]

can anyone help me with creating linear equations

[barren anchor]

sup

1. motivation: be anxious
2. creating arbitrary linear equations is incredibly easy, you've probably been doing it forever (y = mx + b bruh)

[austere scaffold]

how's diff eq. vs multivariable calculus? i've heard different things, like it's easier but more tedious. i know it varies greatly br institution, but im curious about others' experiences

[ripe dirge]

Multivariable calculus is similarly tedious

definitely as tedious as Differential Equations

[turbid sedge]

Hi fiscussy 2

[sharp zenith]

^ i agree they are equal

[magic quiver]

taking intro to proof-writing before diff eq was a mistake

i can no longer compute anything lol

[rare summit]

I know it's quite a while later, but I ended up getting a travel whiteboard that I take with me to lecture each day. It's very useful for communicating concepts to friends, and I can draw pictures that are on the board so I can later replicate them to put in my notes. I'm still a little sad that I can't quite work on the same surface as my peers, but this is good enough most of the time

[coral sand]

0+0=1

[frosty zephyr]

Anyone in university

In Malaysia?

[pastel blade]

Huh

[viral oracle]

welcome to the server! a fellow countryman?

[last rose]

[rocky shuttle]

on an elliptically fibered K3 surface, there are families of Ricci-flat Calabi--Yau metrics whose Gromov--Hausdorff limit is the base S^2 with a specific McLean metric... away from the 24 singular fibers the K3 looks like tiny flat 2-tori collapsing over S^2

[last rose]

thank you this is a really cool maths fact

[quartz compass]

does anyone here like flowers?

[narrow compass]

dead chat

[neat lintel]

Yes!

[turbid sedge]

https://giphy.com/gifs/alert-gem-f5nQLULe1SlQkXrVjr

[worldly horizon]

[vast wraith]

guh this seems like a nice article except this typesetting, i hate when they squish things like that

[frosty zephyr]

Hii

U study in malaysia?

[viral oracle]

technically I'm not studying, but I am a Malaysian, yes

[frosty zephyr]

Ohh

So ur in school?

[viral oracle]

no, I'm a teacher

oops, sorry, didn't see your msg

[pure hollow]

orz

pro

[viral oracle]

NO

[pure hollow]

Idk being a teacher is pretty pro thing to do

[viral oracle]

am nub

[pure hollow]

Lies

[viral oracle]

no lies

smh

[pure hollow]

you are pro , you're able to progress very fast

[viral oracle]

I've been stuck in chp1 of abbott for four months :|

[pure hollow]

Try 3 sections in a year, you're fast

[viral oracle]

am still bad!

[pure hollow]

Nope

[true zinc]

me here for 16 months, haven't tried to go further in a while, but you're doing better than we are

[viral oracle]

guarantee you I'm not if you peek into our threads.

[velvet crypt]

Math teacher?

[viral oracle]

music teacher first and foremost, STEM teacher on the side

[velvet crypt]

Cool

[limber thunder]

<@&268886789983436800>

[radiant sparrow]

grrrrrr

[viral oracle]

[radiant sparrow]

[pure hollow]

I'm in the thread and yes you're good

[rocky shuttle]

a teacher in school?

[vernal gorge]

guys i have to pick between lie groups and matrix operations next semester which should i take

[latent edge]

Are you trolling?

[torpid bay]

take neither, do geometric calculus instead

[velvet dagger]

I'm guessing "matrix operations" isn't basic linear algebra but rather something more advanced? Like numerical analysis?

[viral oracle]

not really

[agile sundial]

I feel like lie theory is analogous to Fourier theory in the sense that you pick up most of the bits you’d be interested in while in other classes.

[foggy meadow]

Pick linear algebraic forms.

[turbid sedge]

Hii

[white lichen]

hi

[rustic mountain]

Hi

[rich quail]

Hi

[unkempt crest]

hi

[bronze pelican]

I don't like that

Seems like a lazy way to get around the page limit

[viral oracle]

https://tenor.com/view/cat-rabbit-cat-rabbit-cat-and-rabbit-cat-sleep-gif-5780726925060253864

[viral oracle]

btw what made you think I was a good teacher lol

[rustic mountain]

i see people spam that

[viral oracle]

just for the record, no

[rich quail]

BNUYY AND KITTY

[viral oracle]

YES

https://tenor.com/view/holaroberto3-gif-15249034458188762381

[zealous garden]

I can't believe no one pinged me for this banger

[inland hornet]

my bad

@zealous garden

[true zinc]

Considering the username, I think so because they're certifiably "clashing" random subjects together

[zealous garden]

Thanks!

[small dust]

yes you are

[stone ferry]

You have geern role, that's how it works

[viral oracle]

no I bad, special case

[stone ferry]

I see. I'll report this to the council of green, the algorithm must be perfected.

[fresh comet]

[junior turtle]

no one did it because everyone knows that you will be pinged automatically whenever geometric algebra is mentioned

[latent edge]

thought you were talking about number fields

[junior turtle]

i see, this explains why you appeared. it seems that you got pinged

[digital maple]

Hello?

[true zinc]

don't ask to ask, ask your question directly please

[digital maple]

@fresh comet can we dm and I tell you about bad discord user who sometimes join to troll server?

[dapper garden]

I'm your fan, ur creator of k theory,respond my dm

[main plaza]

how is this formatting allowed lol

polyhedra though 👀

[magic python]

:3

[noble tinsel]

:3

[simple reef]

no

[torpid bay]

I wanted to ask you about sequences but ig its a little too late

[latent edge]

[torpid bay]

oh u here

allright, so I'll define a a sequence A in two parts: 1 is the initial conditions and 2 is the iterated part

Let 'a' be a sequence with terms $a_0,a_1...a_k$

[fathom swallow]

Yeatte

[torpid bay]

and let A be the sequence where A_n = a_n for $0\leq n \leq k$

[fathom swallow]

Yeatte
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[torpid bay]

now let any term greater than k be defined as such

or rather, have the bound

$A_{m+1}\leq A_m + f(\prod_{n=1}^{m}{(1-\frac{1}{A_n})^{-1}})$

[fathom swallow]

Yeatte

[latent edge]

hmm

[torpid bay]

I mean for this to essentially mimic primes and to get the lowest expected prime number, but I realized i didn't know enough about this the thing i set up, so I was wondering a few things about this sorta thing

a few things i noticed was

suppose I have A_0 = B_0, and A_1 > B_1

and suppose the two f's are the same for each, and certain conditions liks a>b implies f(a) > f(b)

then the entire f term for the A sequence is less than the entire f term for the B sequence,

so in other words

A_1 > B_1 implies A_2 - A_1 < B_2 - B_1

in a sense balancing it out, maybe? my naiive guess was that any two sequences defined in this manner will have the same growth rate

or rather, $\lim_{x\to\infty}{\frac{A_x}{B_x}} = 1 \forall A,B$

[fathom swallow]

Yeatte

[latent edge]

no i literally have no idea what you are trying to do

[torpid bay]

oh i think i missed a condition, in the thing above i had A_m+1 <= A_m stuff, but in the growth rate conjecture i want it to be A_m+1 = A_m + f stuff

anyway, ill go get an example of what im thinking of

the reason, is that when i made this to make a fake prime sequence, it seemed to have the same growth rate as the actual prime sequence

so i wanted to use this to bound the real primes

[latent edge]

you seem to be interested in these things a lot huh

why not read Apostol's analytic number theory?

eeveekawaii~1

[torpid bay]

oo ill write dat down

dis?

[latent edge]

yes

[torpid bay]

https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/blms/6.3.337

huh

turns out oeis does have that fake prime sequence in it

and i got this link

[latent edge]

many such cases

[torpid bay]

rip i guess i wasnt the first to attempt this method

[latent edge]

i mean it's fine

[torpid bay]

it means they prob have results on it

49 bucks to get a pdf? wot

ig ye my only questions were about that sequence thing

well, there is one, but i def dont have the prerequisites to understand the answer to it

[graceful quiver]

Hi wanna chat dm?

[latent edge]

hmm

[foggy meadow]

Analytic Number Theory is what exactly?

[torpid bay]

dunno

im beating up bugs rn

[foggy meadow]

hell divers?

[torpid bay]

silksong

[shadow palm]

What'd @neat lintel do now?

[ornate glade]

MATHEMATICS!!

[willow herald]

Why

[old oak]

they know what they did

[real pagoda]

yo guys

[eager cedar]

anyone applying to sumac

[neat lintel]

Whatis this harassment

[ripe dirge]

Hmm

seems I will have to drop the idea of passing Real Analysis this semester

lemme save my other papers then

[wraith rose]

Hey

[unborn patrol]

Hey! I am fresher. Can anyone give me guidance for maths for learning Machine learning

[lunar adder]

Alguém br!?

[inland raft]

https://www.youtube.com/watch?v=S8enYup_FaY

for those interested

[lime fractal]

😭

[main plaza]

did you add the wise man

[civic canopy]

Boi 67 so tuff

[noble tinsel]

are we deadass

[civic canopy]

Boi

[noble tinsel]

https://tenor.com/view/unoriginqll-gif-16933207306122549340

[civic canopy]

I swear i used to have a boi so tuff jif

I have this though

https://tenor.com/view/troll-face-meme-troll-face-scary-troll-face-phonk-those-who-know-those-who-snow-gif-8838879525100589615

[shadow palm]

<@&268886789983436800> wise man is back

[hard oasis]

[agile fiber]

this has been addressed.

[hard oasis]

You may like these two as well

The Iwaniec--Kowalski is like a bible, and the Bateman--Diamond is lesser known but very cool

Also there's always Keith Conrad's lecture notes on the many topics of ANT

[tall pond]

https://x.com/i/status/2016622367343030495

Alex Kontorovich (@AlexKontorovich)

My lecture last week on “Teaching Real Analysis as a Video Game” (in the Lean Together conference) is now posted: https://t.co/NS0GHi20dj

X · Jan 28, 2026, 9:20 PM

[latent edge]

So many number theory propaganda

Alex Kontorovich and matplotlib recommending ANT

[deft dust]

Oh wow those covers are so good

Really pretty

[brave flare]

whos recommendation holds more weight

[bronze pelican]

ends the talk with "I actually think we should stop doing that [natural language proofs]" is an insane conclusion to the experiment

[royal tangle]

"Moreover I am of the opinion that Carthage should be destroyed"

[noble tinsel]

carthago delenda est 🗣️

[vivid halo]

broke: we should formalize all mathematics with a computer because god forbid we ever make a mistake
woke: we should skip proving half of anything so that we can catch up to the star trek era math that physics people are doing

I'm tired of waiting we need to accellerate

[foggy meadow]

@torpid bay So, silly math we did does cool things with path integrations. Also, Sum and Products (SP) seem impossible to escape from for path based stuff.

[torpid bay]

hm

would taking different paths for a path integral yeild different inverses? and so a multivalued inverse can be done that way ig

[foggy meadow]

Basically for calculus over C the input function must be g(x)^{2} for it to be the identity, and the function g(x) must be \int^{\infty}_{\infty} g(z)^{2} dz < \infty. Which looks a lot like some properties of physics.

[torpid bay]

ah, schwartz function iirc

[foggy meadow]

I think someone (sharp) here was nerding out about those the other day. But not really.

[foggy meadow]

I used it to integrate the path function actually. The inverse path would just swap stuff.

Like it'd be pretty simple to do I think.

I'm just not sure if one needs to fractional integrate the path or like derivate the path at all. More so this is just interesting rather than useful feeling to me.

[torpid bay]

[foggy meadow]

Besides just giving intuition for basic lemmas in quantum physics. I don't feel like it offers much, but it's neat to be able to do.

[foggy meadow]

What's new with your math stuff?

[torpid bay]

well

I wanted an upper bound for primes

and i based it off of factorization

heere's the basic idea

if i have two consecutive numbers, then at least one of them is not divisible by 2

therefore

either 2+1 or 2+2 is the next prime

however, what if i have the primes 2 and 3? then wahts the chance a number is divisible by neither of them?

$(1-\frac{1}{2})(1-\frac{1}{3})$

[fathom swallow]

Yeatte

[torpid bay]

which is just (1/2)(2/3) = 1/3

so i take x * (1/3)= 1 where x is to be the number of consecutive numbers to check at most

which if actually calculated, checks out

so x = 3

therefore 3 + 1, 3+2, or 3+3 is prime

and i continue

(1-1/2)(1-1/3)(1-1/5) = (1/3)(4/5), so x = 15/4 = 3.75, but i round x up to 4

so 5 +1,5+2,5+3,5+4 at least one is prime

and we continue

what if we take the worst case scenario where the next predicted prime is that p + x value?

then instead when we start at 2, we get a certain sqeuence

2,4,7,11,15,19...

if I am to write this out using notation it is

$p_{k+1} \leq p_k + \ceil{\prod_{n=1}^{k}{(1-\frac{1}{p_n})^{-1}}}$

[fathom swallow]

Yeatte

[torpid bay]

but in this case we just set it equal for A_k+1 = A_k + ceil(stuff)

so the question is, how bad of an approximation is it?

if we label this sequence of fake primes as A_k

then from desmos testing

$\lim_{k\to\infty}{\frac{A_k}{p_k}} = 1$

[fathom swallow]

Yeatte

[torpid bay]

so its actually pretty good

[foggy meadow]

why is 16 in set p?

[torpid bay]

uh

wait

thats wierd

[foggy meadow]

P U {16}

[torpid bay]

16,49

seems to be squares

[foggy meadow]

How did you calculate P? I'm assuming it means primes, btw.

[torpid bay]

essentailly, compare number n against primes via modulo

if no factors, then add to list

[foggy meadow]

https://en.wikipedia.org/wiki/List_of_prime_numbers ✅

[torpid bay]

ugh

lets just make 16 prime or smthn

huh

for some reason it listed 16 and 49 in it as well

[foggy meadow]

Na, It's like 2 minuets and some regex in note pad to make

https://www.desmos.com/calculator/qw6p2ezfu3

Just take that then come back.

[torpid bay]

I wanted an efficient algorithm that spits out primes at will :\

[foggy meadow]

🤷‍♂️

Gotta make sure prime times prime isn't like (x*x = p or p mod x)\

[torpid bay]

oh crap

for the prime numbers, i didnt actually reset the prime i was using against n

wait, why would that affect this then?

[foggy meadow]

Idk what is 7*7 mod 7?

[torpid bay]

0

[foggy meadow]

Yeah so, the off chance a number is p*p*... may not be caught by the seive?

[torpid bay]

7^k mod 7 = 0 for positive integer k, so it should be fine i think

for some reason, it just isn't checking 16/2

like

[foggy meadow]

Do you reset n = 3?

Then it'll never mod 2?

[torpid bay]

its just not checking 2 for any of them

wtf

then why is 49 failing as well

[vivid halo]

[torpid bay]

oh

did i hallucinate 49

its not here anymore

[foggy meadow]

16 is.

and 12?

[torpid bay]

12 is divisible by 3 so it gets checked out

but then why not 8?

[foggy meadow]

What would you rather talk about nG?

[torpid bay]

wait

8 gets checked by 2

[torpid bay]

ye, but the checker value starts at 2

[foggy meadow]

???

[torpid bay]

3 is the initial number wher i ask "check dis" and p[0] should be the first number it uses in the modulo against it

[foggy meadow]

just add in the if statement if( n mod 2 == 1 && ...) and call it a day?

[torpid bay]

but for 8 it actaully does check 2, but for 9 it doesnt

wait

hold on

this might be stupid

no nvm i dont get it

yeah its just not running that first prime p[0] sometimes

wait

hold on

[gritty heath]

int p[] = new int[5000];

void setup() {
p[0] = 2;
p[1] = 3;
int ptr = 2;
for (int n = 5; ptr < 5000; n += 2) {
for (int k=1; k < ptr; k++) {
if (n % p[k] == 0) {
p[ptr] = n;
ptr++;
break;
}
}
}
}

[torpid bay]

yeah

that'll work as well

maybe its because i'm trying to change b while its in it's own for iteration?

[gritty heath]

maybe

that makes it hard to reason at least, i try to not modify the looping variable

[torpid bay]

oh hold on

i see what's going on

aha

yeah it is

so what;s happening is if its not prime, then i set b = 0, but at the end of the for statement, it increases by 1, so i end up cheking all primes greater than 2

and skipping it

[gritty heath]

another way is to recall that the 5000th prime is larger than 5000(ln 5000 + ln ln 5000 - 1) ~ 48926
make an array of size 48230 for safety, and do a sieve of eratosthenes

[torpid bay]

ok yeah that fixed it

b = -1 now

fun little problem lel

im glad 49 turned out to not exist

[foggy meadow]

So how does your prime thing look now in desmos?

[torpid bay]

pretty much the same

[gritty heath]

what is your S and P

[torpid bay]

S = pseudo primes via a certain recurrence relation, P = true primes

[young dagger]

Guys, if anyone is from the quant sector, can they please dm me? I need some help 🙂

[foggy meadow]

#❓how-to-get-help

[foggy meadow]

What's a true prime?

[torpid bay]

2,3,5 etc

[young dagger]

tbh i asked for dms cause i thought this is for pure math

[young dagger]

same as a prime

[foggy meadow]

Okay what's a fake prime?

[young dagger]

1

That was settled much before, but true prime is still used to refer to the set of prime numbers such that x > 1

[torpid bay]

one defined by the recurrence relation: $S_{k+1} = S_k + \ceil{\prod_{n=1}^{k}{(1-\frac{1}{p_n})^{-1}}}$

[fathom swallow]

Yeatte

[torpid bay]

tho this does lead to the generalization of

$A_{k+1} \leq A_k + f({\prod_{n=1}^{k}{(1-\frac{1}{p_n})^{-1}})$

[fathom swallow]

Yeatte
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[young dagger]

Is this a true prime though?

Isn't this a formula that builds a sequence using the usual prime numbers to study their overall distribution?

[torpid bay]

im pretty sure a true prime would satisfy this

the less than or equal one

[young dagger]

Can you tell me the problem? I am lost

[torpid bay]

yep

take 2 consecustive numbers, at least one of them is coprime to 2

at least and exactly one of them to be precise

if i take 3 consecutive numbers, then 2 of them are coprime to 2

but then one of those 2 is coprime to 2

so if i want a number coprime to 2 and 3, then i need just 3 consecutive numbers and itll be in there

so i take 3, and so the next prime after 3 is either 3+1,3+2, or 3+3

going back, the next prime after 2 is either 2+1 or 2+2

[young dagger]

-# Might be a miscommunication, but I meant the what the question that you were trying to solve was, but please, if im wrong, go on.

[torpid bay]

oh, im just trying to iteratively find an upper bound for primes

using iteration

[young dagger]

your logic has a flaw though

[torpid bay]

there is an assumption ye

[young dagger]

Among any
𝑘
consecutive numbers, some will be coprime to a given set of primes, but being coprime to small primes does not guarantee being prime, only that the number avoids those factors.

[torpid bay]

im not checking any random k numbers tho

just the k numbers right above the primes we already have

so no room for any extra possible primes betwene the known primes and the possible set of k numbers we're checking

[young dagger]

it should still fail if im correct

please wait

\textbf{Claim.}
Let $p_k$ be the $k$-th prime. There is no guarantee that the interval
[
p_k+1,, p_k+2,, \dots,, p_k+k
]
contains a prime.

\textbf{Proof.}
Fix $k \ge 1$. Choose $k$ distinct primes
$q_1, q_2, \dots, q_k$ such that $q_i > p_k$ for all $i$.

We know that by the Chinese Remainder Theorem, there exists an integer $N$ such that
[
N \equiv -i \pmod{q_i} \quad \text{for } i=1,2,\dots,k.
]

Then for each $i$,
[
N+i \equiv 0 \pmod{q_i},
]
so $N+i$ is composite.

Thus the $k$ consecutive integers
[
N+1, N+2, \dots, N+k
]
are all composite.

Let $p_k$ be the largest prime less than $N$. Then the interval
[
{p_k+1, \dots, p_k+k}
]
contains no prime.
\qed

[fathom swallow]

diddy_lovesme

[young dagger]

@torpid bay

[torpid bay]

the interval isn't based on k

above the p_k'th prime

[young dagger]

oh...

Can you please give me the formal question again then?

Sorry but english isnt my mother language so I have dificulties

I usually do math in my language so theres a boundary

[torpid bay]

if I have the first k primes, what is the most number of consectuvies numbers above p_k that one has to check to get a new prime? naiively a basic upper bound would be 2^k, tho instead i use actual factorization to cut that down

[young dagger]

from what i understand, your intuition is each known prime removes residue classes, so after finitely many steps something must survive

[torpid bay]

(1-1/2)(1-1/3)(1-1/5)...(1-1/p_k) * x => 1, smallest integer x is the new upper bound of number of consecutive numbers to check above p_k, being the numbers p_k + 1, p_k + 2.. p_k + x

[young dagger]

Am I correct?

[torpid bay]

i havent heard of a residue class

but if it is what it sounds like, then yes

[young dagger]

A residue class (or congruence class) is a set of integers that all share the same remainder when divided by a specific positive integer (n), known as the modulus

[fathom swallow]

diddy_lovesme

[young dagger]

so is it still a yes?

[torpid bay]

removing the residue classes of 0 with respect to each prime modulo yeah

[young dagger]

Oh then I guess I see where you went wrong

please wait

The error was:

\textbf{Error.}
The argument incorrectly replaces the statement
[
\text{expected number of survivors } \ge 1'' \] with \[ \text{a prime must exist.''}
]
The product only controls divisibility by primes $\le p_k$, not
compositeness by larger primes.

[fathom swallow]

diddy_lovesme

[young dagger]

Formally,

[torpid bay]

any larger prime would be a new prime

being the new one added to the list

[young dagger]

well that statement in itself is true

but it still doesnt account for your logical gap

[torpid bay]

no?

[young dagger]

\begin{theorem}
Let $p_k$ be the $k$-th prime.
The fact that every prime divisor larger than $p_k$ is a new prime does not
imply that an integer coprime to all primes $\le p_k$ is itself prime.
\end{theorem}

\begin{proof}
Suppose $n$ satisfies
[
\gcd(n, p_1 p_2 \cdots p_k) = 1.
]
Then no prime $p \le p_k$ divides $n$. Consequently, every prime divisor of
$n$ is strictly larger than $p_k$, and hence is a new prime.

However, this does not imply that $n$ has only one prime divisor.
Indeed, let
[
n = p_{k+1} \cdot p_{k+2}.
]
Then $n$ is composite, yet all of its prime divisors exceed $p_k$.
Thus $n$ is coprime to all primes $\le p_k$ but is not prime.

Therefore, the implication
[
\gcd(n, p_1 p_2 \cdots p_k)=1 ;\Longrightarrow; n \text{ is prime}
]
is false.
\end{proof}

[fathom swallow]

diddy_lovesme
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[foggy meadow]

Thanks for also talking about your prime function, btw.

It's really neat!

[torpid bay]

I'm not saying that an number coprime to all p: 2 <= p <= p_k is necessarily a prime, what im saying is that within the interval p_k... p_k + x, if there was a number corpime to p_k etc, then it would either a: be a new prime, or b: be a composite that is divisible by a new prime that would still be in that interval from p_k to p_k + x

and this is with the condition that our p_1 ... p_k list is entirely complete up to the point of p_k at that point

[young dagger]

\begin{theorem}
Let $p_k$ be the $k$-th prime and let $x \ge 1$.
The existence of an integer
[
n \in (p_k,; p_k+x]
\quad\text{with}\quad
\gcd(n, p_1p_2\cdots p_k)=1
]
does \emph{not} imply that there exists a new prime in the interval
$(p_k,; p_k+x]$.
\end{theorem}

\begin{proof}
Assume, for contradiction, that the existence of such an $n$ forces a new
prime $q$ with
[
p_k < q \le p_k+x.
]

Choose a prime $Q$ such that
[
Q > p_k + x.
]
(Such a prime exists by Euclid’s theorem.)

Let
[
n := Q^2.
]

Then the following hold:
\begin{enumerate}
\item $n$ is composite.
\item Every prime divisor of $n$ equals $Q$, so all prime divisors of $n$
satisfy $Q > p_k$.
\item Hence
[
\gcd(n, p_1p_2\cdots p_k)=1.
]
\end{enumerate}

Thus $n$ is coprime to all primes $\le p_k$ and is a valid “survivor” of the
small-prime sieve.

However, $n$ has no prime divisor $q$ with
[
p_k < q \le p_k+x,
]
since its only prime divisor is $Q > p_k+x$.

Therefore, the existence of such an $n$ does not force the existence of a new
prime in the interval $(p_k,; p_k+x]$, contradicting the assumption.

Hence the claim is false.
\end{proof}

[fathom swallow]

diddy_lovesme
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[young dagger]

I dont know mate I might be heavily misunderstanding

Sorry if I am wrong in this regard

please, do consider better help than me if I dont seem to be understanding what you mean

[torpid bay]

if Q > p_k + x, then its entirely outside the interval we care about

[young dagger]

https://tenor.com/view/discord-gif-11683950601656251288

[young dagger]

holy shit i never seem to understand questions

Wait ill really rack my brain

[torpid bay]

oh x is a specific number btw

not just any x => 1

[young dagger]

Ohhhhh

OHHHH

That makes much more sense now

😭

sorry for not understanding your question

[torpid bay]

nah its fine

specifically i chose x large enough to make sure that 100% there be a comprime within the interval right above p_k

[young dagger]

can you give me your full statements will all restraints ill need to consider? I am very bad at keeping things in mind. I need it all laid out for me, if you dont mind

[torpid bay]

all irght

[young dagger]

and the method (formula) you use now

[torpid bay]

Given the set of primes ${2,3...p_k}$(complete set from 2 to $p_k$, no gaps), there will always exist at least one prime within the interval of consecutive numbers ${p_k+1...p_k+x}$ that is coprime to all prime in that set, and thus is either a new prime, or is a composite divisible by a new prime larger than $p_k$ and is also thus in the interval. $x \coloneq \ceil{\prod_{n=1}^{k}{(1-\frac{1}{p_n})^{-1}}$

[fathom swallow]

Yeatte
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[young dagger]

Given the set of primes
[
{2, 3, \dots, p_k} \quad (\text{the complete set of primes from } 2 \text{ to } p_k),
]
there will always exist at least one number in the interval
[
{p_k+1, p_k+2, \dots, p_k+x}
]
that is coprime to all primes in that set. Such a number is either:
\begin{enumerate}
\item a new prime, or
\item a composite divisible by a prime larger than (p_k), which must also lie within the interval.
\end{enumerate}
Here,
[
x \coloneqq \Bigg\lceil \prod_{n=1}^{k} \Big(1 - \frac{1}{p_n}\Big)^{-1} \Bigg\rceil.
]

[fathom swallow]

diddy_lovesme

[young dagger]

Any constraints

?

[torpid bay]

unless im forgetting anything, that's all

[young dagger]

okkkk

[torpid bay]

should give the minimal x that still guarantees a coprime within interval

[young dagger]

so you want to prove this claim rgiht?

[torpid bay]

yep

[young dagger]

hmm alr

Ill work on it too

wait kay

[torpid bay]

[young dagger]

can you check it for the set {2,3,5,7}?

[torpid bay]

then x = 4, and so we check 7+1, 7+2, 7+3, 7+4

[young dagger]

see?

[torpid bay]

7+4 = 11

[young dagger]

oh there exists at least one?

[torpid bay]

yep

[young dagger]

Im retarted

[tribal talon]

Integral from 0 to pi over 2 to ( x * cos (x) - sin (x) ) / (x^2 + sin^2 (x) )

[torpid bay]

23 is also interesting

first gap of 6 and works a ok

quite trivially, the x will increase without bound

that 8 to 11 interval is probably the closest it has to failing

[young dagger]

I have a proof i guess but it is very hand waby

wavy*

[torpid bay]

ooo 113 to 127 gap is also quite close

tho actually, there does become this problem that i thought i might into

where it actuall fails

so yeah 113 to 127 fils

and im pretty sure i know why

[young dagger]

I conjecture the statement is likely TRUE, but a rigorous proof is beyond the scope of elementary number theory

\begin{theorem}
Let
[
S_k = {p_1, p_2, \dots, p_k}
]
be the first $k$ primes, and define
[
x = \left\lceil \prod_{i=1}^{k} \left(1 - \frac{1}{p_i}\right)^{-1} \right\rceil.
]
Then there exists at least one integer
[
n \in [p_k+1, p_k+x]
]
such that
[
\gcd(n, p_1 p_2 \cdots p_k) = 1.
]
\end{theorem}

\begin{proof}
Let
[
I = [p_k+1, p_k+x]
]
and for each $i = 1, \dots, k$ define
[
A_i = { n \in I : p_i \mid n }.
]
We want to show
[
|I \setminus \bigcup_{i=1}^k A_i| \geq 1.
]

By the inclusion-exclusion principle:
[
|I \setminus \bigcup_{i=1}^k A_i| = x - \left| \bigcup_{i=1}^k A_i \right|.
]

Using the Möbius function $\mu$:
[
|I \setminus \bigcup_{i=1}^k A_i| = \sum_{d \mid N} \mu(d)
\left( \left\lfloor \frac{p_k+x}{d} \right\rfloor - \left\lfloor \frac{p_k}{d} \right\rfloor \right),
]
where $N = \prod_{i=1}^k p_i$.

Write
[
\left\lfloor \frac{p_k+x}{d} \right\rfloor - \left\lfloor \frac{p_k}{d} \right\rfloor = \frac{x}{d} + \epsilon_d, \quad |\epsilon_d| \le 1.
]

Thus
[
|I \setminus \bigcup_{i=1}^k A_i| = x \sum_{d \mid N} \frac{\mu(d)}{d} + \sum_{d \mid N} \mu(d) \epsilon_d.
]

Observe that
[
\sum_{d \mid N} \frac{\mu(d)}{d} = \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right) = \frac{1}{x - \epsilon_x}, \quad 0 \le \epsilon_x < 1.
]

Also,
[
\left| \sum_{d \mid N} \mu(d) \epsilon_d \right| \le \sum_{d \mid N} |\mu(d)| = 2^k.
]

Therefore,
[
|I \setminus \bigcup_{i=1}^k A_i| = \frac{x}{x - \epsilon_x} + O(2^k).
]

Since the error term $O(2^k)$ dominates for large $k$, this bound is insufficient to guarantee $|I \setminus \bigcup_{i=1}^k A_i| \ge 1$ rigorously.

Nonetheless, for small $k$, the main term exceeds 1 and the bound $x$ ensures that at least one integer in $I$ is coprime to all $p_i$.
\end{proof}

[fathom swallow]

diddy_lovesme
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[young dagger]

before i go to eat, can you run a test for k = 30?

[torpid bay]

i think it fails for 113

[young dagger]

def sieve(limit):
    is_prime = [True] * (limit + 1)
    is_prime[0] = is_prime[1] = False
    for i in range(2, int(math.sqrt(limit)) + 1):
        if is_prime[i]:
            for j in range(i*i, limit + 1, i):
                is_prime[j] = False
    return [i for i in range(2, limit + 1) if is_prime[i]]

primes = sieve(200)[:30]  # First 30 primes
product = 1.0
for p in primes:
    product *= p / (p - 1)

x = math.ceil(product)
print(f"x = {x}")

# Check interval [114, 122]
for n in range(114, 123):
    factors = []
    temp = n
    for p in primes:
        if temp % p == 0:
            factors.append(p)
            while temp % p == 0:
                temp //= p
    print(f"{n} has factors in S_30: {factors}")

it failed this but idk if i did smth bad in code

[torpid bay]

the x for 113 is about 8.7, but the gap from 113 to 127 is 14

and i think the propblem is multiplying the fractions together

[frail juniper]

Hey guys

I have a question regarding cantor’s theorem

If a set led to a set of all sets e.g. the set necessarily N turned to a self-referential set of all sets, would that mean the set necessarily N is contradictory not only that it’s improperly formatted?

e.g., a necessary disjunction of all insufficient explanations Z where Z in itself is an explanation and is found in itself and every other set L is found in Z and is an insufficient explanation.

[young dagger]

yhh

[torpid bay]

hm

yeah it must be the fractions

then the true x must be between $\ceil{\prod_{n=1}^{k}{(1-\frac{1}{p_n})^{-1}}}$ and $\prod_{n=1}^{k}{\ceil{(1-\frac{1}{p_n})^{-1}}}$

[fathom swallow]

Yeatte

[torpid bay]

the latter just equals 2^k

the question is, whats the minimal number of primes where i need to switch the inside to ceil and keep the rest having the ceil on the outside

actually, why does it fail?

[stiff fable]

halo

[torpid bay]

hm

suppose i take x consectutive numbers

[young dagger]

all check later kay

I have my own homework now

but a very fun problem man

[torpid bay]

then floor(x/2) is minimal number of numbers divisible by 2, same for floor(x/p) for prime p, the thing is, if i do floor(floor(x/2)/3) then im that gets the true value for number of minimal primes in that interval, but floor(x/6) does not

[young dagger]

been a long time since i had to use python for proving

[torpid bay]

yeah, x was a little lowballed because i didnt account for that

oop

a better example

yeah, this fracion error occurs much more in higher numbers

the true x should be higher because these are lower

mm

well, onthe other hand

kinda feels like p_k <= p_(k+1) <= 2p_k is in sight

eh

[sonic orbit]

YWNBAW

[young dagger]

\documentclass{article}
\usepackage{amsmath, amssymb}

\begin{document}

\section*{Backpropagation Mathematical Model}

\subsection*{1. Forward Pass}

For a network with $L$ layers:

[
a^{[0]} = x
]

[
z^{[l]} = W^{[l]} a^{[l-1]} + b^{[l]}, \quad l = 1, 2, ..., L
]

[
a^{[l]} = \sigma^{[l]}(z^{[l]}), \quad l = 1, 2, ..., L
]

where
\begin{itemize}
\item $W^{[l]}$ is the weight matrix of layer $l$
\item $b^{[l]}$ is the bias vector of layer $l$
\item $\sigma^{[l]}$ is the activation function
\end{itemize}

The final output is:
[
\hat{y} = a^{[L]}
]

\subsection*{2. Loss Function}

Let the loss function be $L(\hat{y}, y)$, e.g.:
[
L = \frac{1}{2} (\hat{y} - y)^2 \quad \text{(MSE)}
]
or
[
L = -\sum y \log(\hat{y}) \quad \text{(cross-entropy)}
]

\subsection*{3. Backward Pass – Error Terms}

Define the error term $\delta^{[l]}$:

[
\delta^{[l]} = \frac{\partial L}{\partial z^{[l]}} = \frac{\partial L}{\partial a^{[l]}} \odot \sigma'^{[l]}(z^{[l]})
]

\textbf{Output layer:}
[
\delta^{[L]} = \nabla_{a^{[L]}} L \odot \sigma'^{[L]}(z^{[L]})
]

\textbf{Hidden layers ($l = L-1, ..., 1$):}
[
\delta^{[l]} = (W^{[l+1]})^T \delta^{[l+1]} \odot \sigma'^{[l]}(z^{[l]})
]

\subsection*{4. Gradients of Weights and Biases}

[
\frac{\partial L}{\partial W^{[l]}} = \delta^{[l]} (a^{[l-1]})^T
]

[
\frac{\partial L}{\partial b^{[l]}} = \delta^{[l]}
]

\subsection*{5. Weight Update (Gradient Descent)}

[
W^{[l]} \gets W^{[l]} - \eta \frac{\partial L}{\partial W^{[l]}}
]

[
b^{[l]} \gets b^{[l]} - \eta \frac{\partial L}{\partial b^{[l]}}
]

where $\eta$ is the learning rate.

\end{document}

[fathom swallow]

diddy_lovesme
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[young dagger]

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\section*{Why We Take the Derivative of the Sigmoid Function}

The sigmoid function is defined as:
[
\sigma(x) = \frac{1}{1 + e^{-x}}
]

During backpropagation, we want to update the weights of a neural network to reduce the loss. To do this, we need to know how the loss changes with respect to each weight. Using the chain rule, for a weight $w$ connected to a neuron, we have:

[
\frac{\partial L}{\partial w} = \frac{\partial L}{\partial y_{\text{hat}}} \cdot \frac{\partial y_{\text{hat}}}{\partial a} \cdot \frac{\partial a}{\partial z} \cdot \frac{\partial z}{\partial w}
]

Here:
\begin{itemize}
\item $z = w \cdot x + b$ is the input to the neuron,
\item $a = \sigma(z)$ is the output (activation) of the neuron,
\item $y_{\text{hat}}$ is the output of the network,
\item $L$ is the loss function.
\end{itemize}

The term $\frac{\partial a}{\partial z}$ is the derivative of the sigmoid function:
[
\sigma'(z) = \sigma(z) (1 - \sigma(z))
]

\subsection*{Intuition}

\begin{itemize}
\item The derivative measures \textbf{how sensitive the neuron's output is to changes in its input}.
\item If the output $a$ is near 0 or 1, the derivative is small, meaning the neuron is \textbf{already confident}, so its weight should change less.
\item If $a$ is around 0.5, the derivative is larger, meaning small changes in input affect the output more, so the weight should adjust more.
\item Multiplying the error by $\sigma'(z)$ correctly scales the weight update for each neuron.
\end{itemize}

\end{document}

[fathom swallow]

diddy_lovesme
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[humble jacinth]

@molten grove is it ok for me to dm you? (to talk about the thing in your bio)

[eager knot]

Is there slow mode here

Omg

No soowmofe

Yay

Slow mode

[molten grove]

Sure

[last rose]

what the blud

[old oak]

There should be

[mint patio]

Mannn

be feather, do idiot things that'll hurt your career

get stressed when you potentially have to face the consequences of said idiot things

Professor I want to do research with is ASKING ME FOR MY TRANSCRIPT and it DOES NOT LOOK THE BEST 😐

He likes me but still

😔

[eager knot]

A

[eager knot]

Noo

[gleaming kite]

heyyy does anybody by chacne have studydrive premium?

[keen berry]

ahh don't be a spoilsport

we all need to blurt out thinking

especially in mathematical situations where instant resolve is needed

slowmode just delays things

and to be honest only profound mental deficiency would cause someone to spam here

[old oak]

precisely, gives people a chance to think longer before hitting "send"

[old oak]

I can't think of any tactful reply to this

[keen berry]

haha fair

[keen berry]

i agree

[old oak]

I'm mostly

in favour of slow mode

because

a lot of people

type

like this

[haughty lion]

And then someone sends a wall of text halway through and following the convo becomes impossible

[molten grove]

[errant zenith]

likely spammer

[molten grove]

Read. My. Status.

I’m trying to get it fixed

[errant zenith]

ik

[molten grove]

Then why say it

[errant zenith]

idk its just funny and unusual

also it says spammer not scammer

and ur status says scammer

[mental vector]

How to use foundations channel

Please

[young elbow]

is 1/infinity in an rng still possible

[old oak]

What does "possible" mean here? And what does "in an rng" mean here?

[dire sandal]

well 1 doesnt exist in a rng

[vast wraith]

well rings are also rngs but not every rng is a ring

[dire sandal]

honestly when someone says rng i automatically assume they specifically dont want 1 in there

tho ur still right ofc

[tame egret]

<@&268886789983436800> is this allowed

[wild lantern]

Nope

[tame egret]

[torpid bay]

indeed, PDA in a starbucks is not allowed

[vast bloom]

[tranquil kayak]

Why would they do this…

[latent edge]

what about 67?

I know the quadratic field with discriminant d=-67 is an example of a non euclidean ring of integer

[main plaza]

rizz sigma skibidi ohio

[latent edge]

go to sleep

[young dagger]

i feel personally attacked

oh shi 😭

[fervent kite]

This app keeps sending me side quests instead of the main story.

[normal orchid]

most game have more side quest than main story ¯_(ツ)_/¯

[eager cedar]

GUYS HAS ANYONE QUALIFIED RMO?

[keen berry]

<@&268886789983436800> was my postgrad refused or did i accidentally select undergrad

[bright condor]

Send your queries to modmail

[keen berry]

okay

im new, where is modmail

[hard oasis]

@polar panther

Dm that bot

[zealous birch]

Hi

I'm preparing for ioqm

Finally found someone

[ripe dirge]

Hmm

I wonder how my fellow math lovers cope with crippling depression

I am unable to focus

[rose ridge]

See psychiatrist

[ripe dirge]

hmm

can't afford it

I guess I will coast through it somehow

[tranquil kayak]

Smh

[sharp mulch]

Take a break

[ripe dirge]

You are right

I need a break

I am tired

[azure rune]

the number 2

secret ,

One marriage had generated the world around us

from who they made natural numbers

the number 2 for couples .male and female in humans ,plants ,animals .

[charred wigeon]

Yoo

[reef dew]

Yoyoyoyo

[shadow palm]

@azure rune @charred wigeon
Hi, new nerds! welcome to #discussion, home of general/non-mathematical discussion. We also have #math-discussion for more mathematically-oriented discussion

[azure rune]

[pastel blade]

[coarse sierra]

Hi

[foggy meadow]

@vivid halo Before the chat got server political. I assume that Artin Stacks add new properties that Orbifolds don't have, due to their continuity?

[vivid halo]

they allow for much more general quotients among other things

like here is something funny this allows: a quotient X/G should have dimension dim(X/G)=dim(X)-dim(G), and this dimension will be negative as soon as dim(G)>dim(X)

finite groups have dimension 0 so no negative dimensional examples arise for Deligne-Mumford stacks, but plenty of more general Artin stacks are like this

maybe the most basic example is BG=pt/G

[foggy meadow]

Dimension like literal dimension or?

[vivid halo]

any good definition of dimension will have to be like this yes

it's either that or you just give up being able to define dimension for stacks which is the worse alternative

[foggy meadow]

R^{-1} sounds wild.

[vivid halo]

it's not quite like that though

for stacks individual points sort of remember their "internal symmetries" and if this internal group of symmetries has positive dimension then this should modify the overall dimension accordingly

when dim(G)<dim(X) this is a very classical thing that people are comfortable with

[foggy meadow]

How do I get -1 for example?

Just like R^{2}/So(3)?

[vivid halo]

already taking pt/R is simple enough

[foggy meadow]

what's pt?

[vivid halo]

point

[foggy meadow]

Okay keeep going?

[vivid halo]

a nice example which is endlessly useful in practice is to consider a quotient like C/C* where C* acts by the obvious multiplication action on C

the resulting quotient has two points: one is the quotient C*/C*=pt which is a classical point with no internal symmetries, one is the quotient {0}/C*=pt/C* which is a stacky point with internal symmetries C*

this stack is incredibly useful when you consider vector bundles over it

a vector bundle over the quotient C*/C*=pt is just a vector space of course

a vector bundle over the quotient {0}/C*=pt/C* turns out to be a graded vector space V=\oplus_{n\in Z}V_n

a vector bundle over the quotient C*/C*=pt turns out to be a filtered vector space (...\subseteq F^{-1}V\subseteq F^0V\subseteq F^1V\subseteq...)

[split basin]

how to see this intuitively?

[vivid halo]

now everything in terms of filtered and graded vector spaces can be interpreted in terms of geometry over this quotient stack

[foggy meadow]

pt/C* calculus type thing, or what?

[vivid halo]

the general explanation for this is Cartier duality which is a bit overkill for this example but in those terms it is because BS^1 is Cartier dual to Z

[split basin]

is there a more elementary approach here?

idk algebra

[vivid halo]

yeah hold on lol

I mean here the observation is just that a vector bundle on BC*=pt/C* is the same thing as a C-linear representation of C*

such a representation V splits into a direct sum of eigenspaces V_n for this C*-action where V_n is the subspace upon which z in C* acts by multiplication by z^n

if you have a filtered vector space then both the underlying vector space and the associated graded vector space are recovered by just pulling this vector bundle back to either of these two points

[split basin]

how can i see this without knowing about classifying spaces?