help-23

[honest perch]

no

[open wedge]

oh

$\LaTeX$

[flat frigate]

1 divided by 0 equals Infinity

[sinful token]

what

[sinful token]

yes but the teacher still needs us to be be rigorous

[open wedge]

uhm, if you wanted to write your papers on pdf

[honest perch]

cool?

[sinful token]

we can close the channel to let it to someone else

[honest perch]

there are many open channels

[sinful token]

thanks yall

[honest perch]

this is not a reason to close the channel

[sinful token]

no ithnik i got my answer

[honest perch]

alright

ur welcome

[sinful token]

:))

[ebon igloo]

so now you can write well in mathematics?

[sinful token]

No but i got some advice

[safe radish]

@sinful token Has your question been resolved?

[tardy mango]

hlep channels are for specific questions

#study-discussion

.close

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[frosty musk]

In spherical space simulation 4d+, given two points on the sphere and the center which is zeros origin and a direction, like this half is forward and that is backward, what function is the travel, sine or cosine? Also, should I represent in rotation or Euclidean points?

[frosty musk]

As far as I can tell it is like a direction and an amount for travel, no spiral, so?

Welcome. X E.

[delicate shore]

What does "spherical space simulation 4d+" mean

[frosty musk]

Like on a thing where all valid points are a single distance from an origin and a sphere is 2d, like 3d Euclidean, 4d of the space, so?

[plucky elk]

that's usually just called 4d sphere

[frosty musk]

Well, the + signifies a 100d sphere is valid. X E.

[empty gyro]

Sine

Because sine starts at 0

[plucky elk]

you can just say "hypersphere in any dimensions" then

or "n-sphere" for short https://en.wikipedia.org/wiki/N-sphere

and add on the requirement "n > 3" later

[fervent valley]

How do you understand the question to be able to answer so confidently? 😯 What does "what function is the travel" mean?

[empty gyro]

I think they're trying to slerp from A to B

[primal bone]

Feels like it

Though without any of the terminology needed to actually convey it

[fervent valley]

I see

[frosty musk]

I would think cosine because like we start travel as 1 and go around to 0 then back to -1 then 0. X E.

[empty gyro]

https://en.wikipedia.org/wiki/Slerp

[empty gyro]

The convention is 0 to 1

[frosty musk]

Almost does not matter though, if I can get a rotation for one and a rotation for another and which side of forward or backward it is, I can get a rotations per time constant, then apply that, right?

In fact, just going by rotations would be better for travel and representations, less space and easy. X E.

I do not even need a Euclidean equivalent or the function of this. X E.

Sorry I made this and thanks for talking, it can work like rotations per distance right?

Like if I start at (0,0,0) in rotations with forward being like (1,0,0), (1,1,1) being forward I can say that the travel is an amount per arc length and then just need arc length. X E.

Then again, arc length, need to find that. X E.

Basic linear algebra. X E.

.close

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[novel kelp]

.

[polar shore]

I havent made any progress whatsoever and have no idea where to start (euclidean geometry)

[safe radish]

@polar shore Has your question been resolved?

[manic radish]

hint: are all the perpendicular bisectors concurrent?

[safe radish]

@polar shore Has your question been resolved?

[safe radish]

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[cold garnet]

Suppose for sake of contradiction that 0<x<1 and 1/(x(1-x)) < 4 By 0<x<1, x and 1-x must both be positive. Thus we can multiply both sides by x(1-x) without changing the sign of the inequality. 1<4x -4x^2
4x^2 - 4x + 1 < 0
(2x-1)^2 < 0
This gives us a negative square which is a contradiction since squares are always non negative. Thus by contradiction, we have proven that if 0<x<1 then 1/(x(1-x)) >= 4

Could someone help verify that this proof is rigorous?

[midnight cypress]

I believe it's rigorous

However you can probably rephrase that last statement

"Since the square of a real number can't be negative blah and so on"

[cold garnet]

yeah that sounds better

thank you. Do you see other structural problems in the writing?

[midnight cypress]

Not really no

it's good

[glacial cairn]

Why the proof by contradiction

[cold garnet]

@glacial cairn It's more straightforward imo

[midnight cypress]

No 😭

[glacial cairn]

...

[cold garnet]

lol

[glacial cairn]

A direct proof is more straightforward

[midnight cypress]

proof by contradiction is the book example of an indirect proof

[cold garnet]

I mean yes

personally I found this problem easier to approach this way

could you provide the direct proof rq?

[midnight cypress]

you can try yourself

you can get it pretty easily just try to sit with it for a while

[cold garnet]

okay fair, thanks

[severe pond]

i mean you literally have it

you just added the whole contradiction structure to it for no reason

[cold garnet]

my reasoning was

I couldn't manipulate the >= into a strict <

idk I'm new so

[midnight cypress]

you can prove a >=

by just proving that > is true

and = exists

[cold garnet]

true

[severe pond]

why would you need to

all you need is x > 0 and 1 - x > 0 to get the same inequality you had just reversed

(2x - 1)^2 >= 0 holds by the trivial inequality

[cold garnet]

I see, yeah that is simpler lol

well thanks guys I won't disappoint you next time 😉

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[fringe dock]

I need help with part b

b(i)

the markscheme is really confusing me

[plucky elk]

do you know the max of cos(stuff)

[fringe dock]

i do

it occurs when cos is 1

[plucky elk]

so max(H) = max(3 + 2 sqrt(5) * cos(stuff)) = ?

[fringe dock]

as in i dont get what happens with sin not being there

[fringe dock]

like this part

i just randomly see a square root

[plucky elk]

if you did part a), it's the same work

[fringe dock]

oh yeahh

i didnt notice that

at all

ok wait

[safe radish]

@fringe dock Has your question been resolved?

[safe radish]

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[raw pivot]

hey! i have tried to solve this and i just cant get it lol, so far i have gotten to, the rectangle perimiter is 10x-6 so therefor the triangle perimiter is also 10x-6

[plucky elk]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[raw pivot]

um ok

so basically

i have no idea what the next step is

[plucky elk]

You're given side AB so find another equation for the perimeter of the triangle

[raw pivot]

it could be split into 2 right triangles!

[plucky elk]

no

the perimeter of the triangle is the sum of the three sides

[plucky elk]

read the question again and find the length of side AB

[raw pivot]

but arnt we tryina find the area of the triangle

the perimiter though would be 8x+8

[plucky elk]

right now combine that with the information you already found to solve for x

and then yea there are multiple formulas you can use to find the area of an isoscoles triangle

[raw pivot]

yea so 8x+8 = 10x-6

2x=14 so

x = 7

oh so

midline is

625-49

so 576

so 24

so 24x7 is awnser

so 168

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[raw pivot]

tysm @plucky elk btw

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[vagrant mango]

I was thinking about what the average ratio of a right triangles hypotenuse to the sum of its other two sides would be, considering all possible right triangles.

Does anyone know a resource that goes over this that I could read up on?

My starting unbased hypothesis is 0.707.

Whenever I try to approach this from a limit perspective, I have issues with infinitesimal ratios combining with ones approaching infinity and I just get average ratio = 1

Is there a way to find the median of an infinite set? That doesn't make sense to me.

If I work this from a different perspective, the ratio would be c/(a+b). Heres an image of what I worked on for this idea

[vagrant mango]

So if there was a known average ratio for that final piece for all values of x, that would also be the answer I'm looking for

Which makes me think this is unbounded and not a thing at all

Could root mean squared this I suppose

But I think that comes out to equal 1

Which doesn't make sense big picture

[queen ingot]

i think you need to decide how you're sampling right triangles

by sampling theta uniformly from 0 to 2pi in this diagram?

there might be some other ways that lead to a different answer

[vagrant mango]

that makes sense to me as bounds

i mean it would be cool to have unlimited capacity to know the average for every possible right triangle that could ever exist but thats not realistic

[safe radish]

@vagrant mango Has your question been resolved?

[safe radish]

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[stark hull]

I need help with qualitive analysis of an Ordinary differential equation

[stark hull]

I honestly have no idea what it is but if someone can explain me I would be glad

[marsh walrus]

what kind of questions are you supposed to answer

[stark hull]

I need to analyse and write it on my report

it is this and I need to also get a nondimensional formula

whatever that is

[marsh walrus]

that parts less fun lol

i mean what kind of qualitative analysis

[stark hull]

I dont know I asked my prof and all he did was mumble some words inside of his mouth

what are even the types

[marsh walrus]

its usually just a matter of extent

maybe most basic would be like

look at the asymptotics

so describe what happens to this as t->infty

[stark hull]

It reaches terminal velocity.

mb

reaches an impossible velocity

[marsh walrus]

i dont think thats true

[stark hull]

It should

[marsh walrus]

oh wait yea

because we have something like $v(t) = C + e^{-t}$ or so

[flat frigate]

jan Niku

[stark hull]

I mean it is normal this is a really half verison of drag formula

[marsh walrus]

you can also nondim it before you do this

but the nondim kind of sucks

the other basic thing i would do is do a quick analysis of parameters you can change

so c, g probably are considered constant

m is a parameter

[stark hull]

c is a parameter*

[marsh walrus]

you should describe qualitatively what happens as m -> 0 and m-> infty

[stark hull]

we change the atmoshpere

[marsh walrus]

oh

well then, yea

[stark hull]

I mean I also have the starting point as a parameter

[marsh walrus]

this part is usually e asy

youll just notice what happens as you make the parameters large or small

and note that it agrees with what you expect physically

describe why making the atmosphere extremely thin causes velocity to increase crazy fast

[stark hull]

I am already down to reading more of this cr##

that is called fluid dynamics

[marsh walrus]

dont get too side tracked

[stark hull]

to propose a higher reality I also have to quote it

[marsh walrus]

do you think you got the qualitative part understood?

[stark hull]

I mean yeah

[marsh walrus]

okay

[stark hull]

I will change the parameters and see what happens

[marsh walrus]

and describe behavior regimes

then to nondim you usually want to start by looking at units

you rememember we have $v(t) = C + \frac{e^{(\text{stuff})t}}{c}$

[stark hull]

yeah

[marsh walrus]

so we want to pick a grouping such that C has units velocity, like it should

and stuff * t should be unitless

[flat frigate]

jan Niku

[stark hull]

uh how do we get there?

[marsh walrus]

we basically group stuff out so that each term has the same units

[marsh walrus]

it will happen more or less like you expect

like lets look at your first term

[stark hull]

ok

[marsh walrus]

$\frac{mg}{c}$ is a good 'grouping'

[flat frigate]

jan Niku

[marsh walrus]

can you make sure it has units of velocity

[stark hull]

how do I do that

I know nothing about nondiming as he did not teach us that

nor I understood any of the stuff I found on my books

like I did not clearly understand what you wanted actually

[marsh walrus]

The idea is that you scale quantities

and when you do it in a nice way, things cancel

your new equation wont have the same exact numerical behavior, but will retain the qualitative behavior

sorry im flipping frantically through my notes trying to jog my memory of how i made this make sense to me

[stark hull]

no no problem

I am also trying to scrape through whatever I have in hand

[marsh walrus]

what i remember making sense is that, we can write it like this

instead of velocity, just, v(t)

we split it into two pieces

lets say one, V(t), it changes in magnitude with v(t)

but, it has no units

then, we have another piece, call it vT

it is a constant, it has units velocity

so v(t) = V(t) vT

[stark hull]

oh ok now I understand better

[marsh walrus]

so we can make some choices, like say for $v_T$, it would make sense to call this $\frac{mg}{c} =: v_T$

[flat frigate]

jan Niku

[marsh walrus]

maybe you call this terminal velocity

[stark hull]

yeah I guess

[marsh walrus]

maybe i can scribble

[stark hull]

it would help

I am arguing with myself I am a growing shizoism

[marsh walrus]

schizoism?

[stark hull]

dont ask

[marsh walrus]

sorry something threw me for a loop

lemme share where I am

I got confused because I didn't do unit analysis first

Can you read the writing okay?

@stark hull

Well maybe you will come back

I got thrown by the 1/mg

but then i realized the units of the magic number there divided by c have to be velocity

its a force

[stark hull]

no no

I am here

I am also trying to

write

this all down

and to argue with a real person(hopefully)

[marsh walrus]

Well, now that you see it does the process make sense

we do that magical thing where we split it into two pieces

a constant with the right units, and a unitless piece with the 'right' (scaled) magnitude

because the units of all the terms will agree if the ODE makes sense, you get a bunch of cancellation once you start dividing stuff through

[stark hull]

btw I can my writings are far worse

my numbers look like original arabic numerals

[marsh walrus]

maybe we wanna go one step further and define that number away

but i personally dont understand the role that force is playing

enough to interpret what the ratio of that force to the force of gravitation means

thats the drag term?

[stark hull]

I guess so

nothing here is physics this is pure math

all the formulas are wrong

[marsh walrus]

it still has a qualitative meaning

[stark hull]

yeah

I mean

[marsh walrus]

so like

v = mg/c

this seems like

well this also seems like it has drag accounted

[stark hull]

yeah

[marsh walrus]

idk we dont have to get fancy

[stark hull]

I mean you dont account for the area or the density of the fluid

but we will act like drag coefficient has it all

[marsh walrus]

just wait til you take other math phys lol

youll see how crazy it gets when you dont make simplifying assumptions and shit

idk if youre hating on the assumptions

[stark hull]

My dad used to work on turbines and I saw thermo fluid dynamics as a child

I want to kill myself now that I remember I want to also work on Turbines

[marsh walrus]

so

[stark hull]

but nah I will probably live through

[marsh walrus]

qualitative part you get

[stark hull]

yeah

[marsh walrus]

you got the nondim now?

[stark hull]

yeah

[marsh walrus]

if you get good at the nondim

you usually want to do that first

so the process is

unit analysis

non dim

[stark hull]

I mean it ends up as v=mg/c

[marsh walrus]

then qualitative

because the nondim will be prettier and retain the same qualitative behavior

you have like $V(\tau) = 1 - Ae^{-\tau}$

[flat frigate]

jan Niku

[marsh walrus]

the only real analysis to make here i think is the magnitude of A?

which is the ratio of your magic number to mg

[stark hull]

it

[marsh walrus]

do you feel like we have hit all the parts that seem kind of familiar

do you remember doing any other things

[stark hull]

I mean I also have another scribbing from someone else

but like

very vauge

not this

this

[marsh walrus]

yea maybe solving it is in order

[stark hull]

like this is the solution

it has the mg/c = v but I dont think that has anything to do with nondim

[marsh walrus]

no, theyre solving for terminal velocity

[stark hull]

ah yeah

[marsh walrus]

well then you went above and beyond i guess

[stark hull]

yeah

[marsh walrus]

but i would call what you did just basic vanilla qualitative analysis

[stark hull]

I am gonna stick to that till I require further

[marsh walrus]

and you didnt give me any guidelines so i dont feel bad

[stark hull]

I mean you did a fantastic job tbh

I did not belived myself I would learn how to nondim

[marsh walrus]

it can be harder

in systems

or in PDE

but mostly you just need to remember

split into two parts

one constant with units

one unitless with magnitude

anyways good luck

[stark hull]

thanks

have a good day

.close

[safe radish]

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[indigo thunder]

2/3 nth degree super super root of 4 =?
2/3 nth degree super root of 4 =?
2/3 nth degree root of 4 =?

[indigo thunder]

I think 65536
16
8

Correct?

Or nah

[round egret]

what is a super root?

[indigo thunder]

Super root is inverse to tetration

And super super root is inverse to pentation

Specifically the root inverses

[round egret]

ah okay, got it

by 2/3 nth degree root of 4, you mean $4^{\frac{2}{3}}$

[flat frigate]

MxRgD

[round egret]

?

[indigo thunder]

Yup

And 2/3 nth sroot of 4 would would be 4^^(2/3)

And 2/3 n ssroot of 4 would be 4^^^(2/3)

[delicate shore]

how is fractional tetration defined

[indigo thunder]

Went into that ealier just need the answers

Im checkikg my work

😇

Woohoo

🥳

Theres a general root calculator on the interent that i can use to plug in 2/3 n root of 4

And result is 8

Just need to check my predictions for 2/3 n sroot of 4 and 2/3 n ssroot of 4

Its saying that 8 is right because 8^(2/3) = 4

[zealous ingot]

there's no universal definition really

[delicate shore]

blud I went on the wiki page for tetration and the section "extension to complex numbers" cites a paper by "Vincent Vey" who is an "independent researcher"

the paper is also very clearly AI-generated

[zealous ingot]

that's crazy

[delicate shore]

https://en.wikipedia.org/wiki/Tetration#Complex_heights

this is the cited paper (slop)

[zealous ingot]

how did bro use schrodingers equation

must be a genius

[indigo thunder]

@help

I cant figure out how to use desmos

[delicate shore]

Schröder ≠ Schrödinger

[delicate shore]

how come

[indigo thunder]

Im relly bad with it

(Im stupid)

[zealous ingot]

yikes need to be wearing specs

[indigo thunder]

Can someone plug it in? Pwease 🥺

[delicate shore]

I don't think desmos understands tetration

Also hyperoperations in general are rather poorly understood

There is not even an agreed-upon definition for fractional tetration so I doubt your question has an answer

[indigo thunder]

Wdym

I think i have the right method

[zealous ingot]

the concept used in your question itself has no clean definition

[indigo thunder]

I think ive discovered how to do it then

I know how fractional iterations work

There is a calculator that allows the inputs for normal roots for n degree to be a fraction

And if you plug in 2/3 n root 4 = 8

Which matches my predictions

[round egret]

If you want to post this amazing discovery, try vixra

[indigo thunder]

I mean deadass

[zealous ingot]

that's for exponentiation

[indigo thunder]

Did i actually stunble across something here

[zealous ingot]

what are you thinking of for pentation and tetration

[round egret]

I don't think so, unless you somehow defined fractional tetration clearly

I guess there isn't a unique way to interpolate the values of tetration at integer "height" (those number of exponents in the "tower" you'd see.) So you could pretty much define it to be anything

[indigo thunder]

Well we take 1 full iteration and split into 2 terms D where(0+) or (1×)D (n-1 degree hyperoperation)itself B-1 amount of times operation for non inverses

Equels A

Where A is 1 full iteration

And this new value D

Is a half ieration of A

Because we split it into 2

We can split it into 3 so B would be 3 times and we would get a third iteration

[zealous ingot]

like 4^^^(3/2) = 1/2 (4) + 1/2 (4^^4) is it?

[indigo thunder]

Then iterate that new value D into the main string as many times as the numerator

So my method here is

We need 2/3 root of 4

[zealous ingot]

whatever logic you're trying to apply, try applying to a few integers

goodluck

[indigo thunder]

So 4÷D÷D=1

That means D=2

Because 4÷2÷2=1

Or D= (-2)

Because 4÷(-2)÷(-2)=1

So that means 1×D×D×D=A and A÷D÷D÷D=1

So we take 2

And cube it

And that gets us 8

This method works for negitive iteration and fractional iteration for addition subtraction multiplication division exponentation roots logarithms

And im assuming it works for super roots and super super roots because it works for tetration and exponents too

And gives the right answer

All the same logic

All the same pattern

The only thing is the operation were 1×d×d=a and a÷d÷d=1 is it flips for inverses twice (so it remains the same)

[delicate shore]

[indigo thunder]

Are you guys theres no answer for 2/3 n sroot 4

I think answer is 16

[delicate shore]

I think you should write down formally how you define this fractional tetration

[indigo thunder]

And 2/3 n ssroot 4

I think 65536

[indigo thunder]

Do you think im on to something or just tweaking

Ong can get einstein in the chat

Yo where he at

Terrance tou??

[delicate shore]

Lowk I think you're just tweaking

but I can't even tell because you wrote your stuff in an incomprehensible way

[indigo thunder]

Are you sure theres no way to calculate sroots and ssroots

What about slog

Can you calculate those

Because i deffinatly figured that out

Like 100%

[delicate shore]

[indigo thunder]

I 100% figgured out how to do super logs and super super logs

Theyre easy actually

Is it possible to calculate those?

And i know logs are the inverse such that we have C and A so were looking for the iterator

B

But i can makes logs give fractional outputs too

You just just change either the value of 1 of your inputs A or C

[delicate shore]

[indigo thunder]

Is it possible to calculate super logs

Yes

No?

[delicate shore]

you can't develop a complete theory of super logs without having first having a theory of fractional (and eventually real) tetration

for instance a hypothetical super-log should satisfy slog_2(4) = 2 and slog_2(16) = 3 but it is not clear what values slog_2 should take between 4 and 16 until you have defined a continuous extension of tetration to real heights

[indigo thunder]

Okay but can you do super logs with normal intergers because i 100% figgured that out

I can explain it too

Super logs are easy

[delicate shore]

What do you mean super logs with normal integers

the intergers being the input?

how do you compute, say, slog_2(12) then

[indigo thunder]

You take 12

And repeat log 2

Iterate it to the results each time to get to exactly 1

Then when you get exactly 1 you end

[delicate shore]

what do you mean by repeat log 2

[indigo thunder]

So 12

[delicate shore]

Are you referring to this

[indigo thunder]

Hold on i need a calculator

Log2(12)=1.079181246

[delicate shore]

mhm

[indigo thunder]

Okay that one is to harc for me i need the answer before hadn but if i had a seprete algorithm i could unpack this super logarithm recursively

I can give an example like

[delicate shore]

[indigo thunder]

Slog 65536

[delicate shore]

Ok so this is the iterated logarithm then

[indigo thunder]

Alright my calculator is tweaking

Log2 of 8 = 3

Yo

My calculator is giving me somthing else

[delicate shore]

[indigo thunder]

Yo is log 2 of 8 =3

Because 2^3=8

Right

Why is my calculator bugging

Okay now im using a internet one

Idk whats going on

Log2 of (65536) =16 log2 (16) = 4 log2 (4) = 2 log2 (2) = 1

Count how many times we iterated log 2

We iterated 4 times

Therefore slog 2 of (65536) is 4

[delicate shore]

Ok but not every number will land you at 1

in fact the only numbers that do so are the tetrations 2, 2^2, 2^2^2, 2^2^2^2 and so on

[indigo thunder]

Thats where fractional iterations come in

[delicate shore]

What does "fractional iteration" mean

[indigo thunder]

I already went through that im trying to check my work

If my predictions are correct

Then

2/3 n root 4 = 8
2/3 n sroot 4 = 16
2/3 n ssroot 4 = 65536

What is 65536^^^(2/3)

[delicate shore]

when did you go through it

[indigo thunder]

Earlier today

I closed it

[delicate shore]

[indigo thunder]

8^(2/3)=?

What is it

?

@help

[indigo thunder]

Bruh

[delicate shore]

What are you trying to get at with this

8^(2/3) is 4

[indigo thunder]

Can i get help

[delicate shore]

but like

[indigo thunder]

You just keep saying that emoji

Its like your trolling me

[delicate shore]

if you don't define your "fractional iteration" in a clear way no helper can help you

it is not a well known or standard notion

[indigo thunder]

Bro i did that like hours ago and it took forever 😭 i just need the answers to 2/3 n sroot 4 and 2/3 n ssroot 4

Im trying to check my work

Please

[delicate shore]

where did you do it

like at least link the place where you defined your ssroot

[indigo thunder]

I know that there are algorithms to find ssroots you can find mathematical stack exchange

Im just too dumb to plug the inputs in 🤣🤣🤣

I need help

I can find a root calculator that works with fractions for the nth degree

[delicate shore]

can you link them

[indigo thunder]

Thats a fractional iteration for sure

Yes i can

Hold on

https://math.stackexchange.com/questions/3314712/calculating-the-nth-super-root-when-n-is-greater-than-2

https://www.calculator.net/root-calculator.html

The root calc says im correct for the root

But i cant find a sroot calc or a ssroot calc

Yo?

[safe radish]

@indigo thunder Has your question been resolved?

[solar hazel]

[indigo thunder]

Alright guys its getting kinda late where im at im gonna have to give up soon here

[safe radish]

@indigo thunder Has your question been resolved?

[indigo thunder]

What is 4^^^1.5

[burnt notch]

What does ^^^ mean?

[indigo thunder]

Pentation

[burnt notch]

Okay, I'm out

[safe radish]

@indigo thunder Has your question been resolved?

[primal bone]

Okay, I'm working atm so I can't really explain this in great detail, suffice it to say that beyond exponentiation there's no standardised way to extend the hyperpower beyond natural numbers

[safe radish]

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[lament jungle]

how do i do this?

[lament jungle]

is this channel fine?

[round egret]

yeah this channel is fine

[lament jungle]

this is way above my standard but its for a little extra credit so i thought i'd give it a shot

doesnt the pi/2 cancel out?

[round egret]

I won't lie, this integral looks rough. I guess do a u substiution of u = sqrt(1+tan(x))

[round egret]

they wouldn't cancel out no

[lament jungle]

i only understand integrals for simple numbers so yeah idk what that is

[round egret]

u substiution?

[lament jungle]

It wouldn't be like this?

[round egret]

you can't do that, that's not how integrals work

[lament jungle]

oh my bad.

[round egret]

do you know fundamental theorem of calculus?

[lament jungle]

no im just starting calculus basics

i told you that this is way above my standard

[round egret]

yeah I don't think you can attempt this question at all

[lament jungle]

i know derivatives only

my teacher said something like king's property as a hint. what's that??

[round egret]

we're working with integrals here and since you only know derivatives, you won't have the tools to solve it

[lament jungle]

early opportunity to try to learn integrals i guess

[round egret]

yeah but this is like a difficult integral to solve that isn't what you usually learn first 😭

[lament jungle]

😭 who knows i might turn out to be a prodigy related to integrals

[round egret]

something like $\int x^2 dx$ would be easier and what you do first and then continue developing

[lament jungle]

what should i start with

[flat frigate]

MxRgD

[lament jungle]

oh also, i wanted to know how to use this bot as well

[round egret]

I'd first learn what an integral is

and how it relates to derivatives

[round egret]

it's LaTeX

[lament jungle]

ohh ok ty ill start with that

[round egret]

it's a way of representing mathematics

[lament jungle]

yeah but like how do you get the bot to show your text

[round egret]

do $$ and whatever you want to put in it

like for example

$x^2$

[flat frigate]

MxRgD

[lament jungle]

ohh

[round egret]

although some require functions using the \

so like

$sin(x)$ is different to $\sin(x)$

[flat frigate]

MxRgD

[round egret]

notice the first one is in italics?

[lament jungle]

yes

[round egret]

yeah generally it's bad practice

[lament jungle]

is there a way to see all the syntax

or do you just have to memorize it

[round egret]

You can google it, though when you've done a lot, you tend to have them memorised

[lament jungle]

thank you very much

now do i need to close this channel?

[round egret]

yeah

[lament jungle]

is it just .close?

.close

[safe radish]

Channel closed

Closed by @lament jungle

Use .reopen if this was a mistake.

[lament jungle]

so is it like i cant use this anymore or what? i joined 10 mins ago so i have no idea

[frozen marlin]

if you want to post a new q new channel; to revisit this one, use .reopen

[lament jungle]

ok thanks

[safe radish]

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[spring pumice]

i need help

[round egret]

what do you need help with?

[spring pumice]

its a linear equation question

[round egret]

sure, post a screenshot of the question

[spring pumice]

The digits of a two-digit number differ by 3. If digits are interchanged and the resulting number is added to the original number, we get 121. Find the original number

this looks easy but im not getting it

[delicate shore]

Let the digits be mn, then...

[spring pumice]

thenn

[zealous ingot]

"The digits of a two-digit number differ by 3" first equation would come from here

[spring pumice]

yes

[zealous ingot]

can you tell me what eqn you'd get?

[spring pumice]

i think 10(x)- 1(x-3)

[round egret]

why minus?

[spring pumice]

cuz they differ by 3 so it can either be plus or minus

[round egret]

oh right I was thinking of one digit being x, so second will be x + 3

other way round works

[spring pumice]

okay we can work w that too

[round egret]

would be easier since you don't have to do deal with minuses

[spring pumice]

yes

[round egret]

so we would get 10x + (x + 3) as our 2 digit number

[spring pumice]

mhm

[round egret]

now it says interchanged 2 digit number as well

[spring pumice]

yes

[round egret]

what would that equation be?

[spring pumice]

uhh

10(x+3) + 10x?

[round egret]

close

[spring pumice]

10(x+3)+x

[round egret]

yes that's correct

so now we have an equation for the interchanged and the original number

and we know adding them together gives you 121

[spring pumice]

so 10x + (X+3) + (x+3) + x =121?

[round egret]

you forgot the 10 in one of the (x+3) so it would be 10(x+3)

[spring pumice]

oh yes

and then we just solve it and we'll get the answer?

[round egret]

yep

[spring pumice]

okay thank youu

[round egret]

and you can check it yourself

to see if it works

[spring pumice]

yes

[safe radish]

@spring pumice Has your question been resolved?

[safe radish]

Channel closed

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[safe radish]

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[strong nexus]

When proving that we can construct the sum of two angles using straightedge and compass, we assume we have line segments [a, b], [a, c] and [a', b'], [a', c']. Then WLOG we assume a = a' = 0 and b = b' = 1. But why can we do this? Think about two completely disconnected pairs of segments, shifting everything to the origin seems to be non-trivial..?

[strong nexus]

Placing the ruler at the right slope and shifting isn't allowed

[peak estuary]

well we can assume that one of them is already at the origin

just by calling one point the origin

and then we only need to draw a parallel of the second line through the origin

[strong nexus]

Well, we need to rotate + shift the second pair of segments

[peak estuary]

oh wait I was thinking of something else. whoops

[arctic terrace]

Is this group for uni students.

[strong nexus]

I guess it's formulated a little weirdly, it would be better if it was just exp(ialpha) and exp(ibeta) in the first place rather than these two pairs of segments - if we just want addition of angles that should be it. But if we start with these two pairs, it's a little messy

[brave wolf]

cant you just make parallels through origin, if thats what you mean?

That doesnt change the angles

and is more or less trivial

[strong nexus]

How? We could have something like this

We want to get that 'V' at the top rotated correctly and placed at the origin (that's where the two segments of the second pair meet)

[brave wolf]

wait is the compass collapsing or non-collapsing?

[peak estuary]

when this says a=a'=0 and b=b'=1 what exactly does it mean by that

[strong nexus]

We consider a set M in C with {0, 1} c M. Then M_0 := M, M_1 := SC(M_0) and so on where SC is the original set + all the points we get with the three construction rules which are (1) intersecting two lines (2) intersecting line + circle and (3) intersecting two circles

this all is about showing M_infty has certain stuff in it

[peak estuary]

cause if it means that you can just rotate [a',b'] onto [a,b] then thats basically just assuming the result, no?

ok so 0,1 are complex numbers. good

[strong nexus]

It can transfer distances

So non-collapsing

[brave wolf]

You can just move the angle with a non-collapsing compass, but then why not move it straight on [a', b']

[strong nexus]

How can you move the angle?

[strong nexus]

Oh

[brave wolf]

draw the red arc centered at the intersection

then blue arc centered at one red-segment intersection, passing through the other red-segment intersection

[strong nexus]

That's great, so this is it. It doesn't even care about rotation so we have rotation + shift in one step

[brave wolf]

yeah

[strong nexus]

Thanks!

Oh and one little thing:

[brave wolf]

generally, when you're shifting stuff you just need to measure and transport the relevant distances. You got it easy with non-collapsing compass

[strong nexus]

We "renamed" the point where the two segments of the second pair meet as 0, and b as 1. Implicitly, this is also a rotation + shift, right? The same way

[brave wolf]

this is just putting a cartesian coordinate system in your euclidean plane, which doesnt necessarily have any coordinates. It doesnt really shift anything, it just labels stuff

[strong nexus]

Hm, we are in M_infty here, a subset of C. That does seem like coordinates

a, a', b, b' are in M_infty c C

[brave wolf]

Sure ig. The point is that if you can do your construction in the 0-1 setup, you can do it in any setup, because you can just temporarily change your coordinate system. But I wouldnt call that shifting, relative to everything else, the objects stay exactly where they were. Your angles are still positioned the same relative to each other

[strong nexus]

Alright, thanks a lot!

.close

[safe radish]

Channel closed

Closed by @strong nexus

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[safe radish]

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[frank widget]

guys,is there a nice way to solve this 😭

[frank widget]

i know that there is the case making method but that is just bad

pls 🙏 i have been suffering for like 2 hours now

[brave wolf]

we could probably interpret it combinatorically

still kinda painful though

[frank widget]

yeah ik

like make groups and check

but

the problem is

this question is meant to be done in 3 mins to 4 mins

am trying to find a method

that helps me do it in that time 😭

i had already done combi method

shi is painful

[brave wolf]

<@&268886789983436800> [Resolved]

[frank widget]

man sybau

holy godspeed mod fr 🙏

my teacher did this when i was absent

and this was 7 months ago

so i dont remember what to do

[lone void]

ohhh i kinda remember a similar question

let me try to find it

[frank widget]

🙏

pls speed i need this

[lone void]

how long did combi method take u btw

[frank widget]

10 minutes to check all

would take me a lot longer generally tho

[lone void]

check all? theres technically only 2 options

[frank widget]

yeah ik but there has to be a better way, no one is spending 10- 15 mins in exam to solve ts when u have 3 mins per question 😭

[lone void]

hmm yea the question i had was different

i think combi is the only way

i think T(5) is possible

is it multiple correct or single correct

oh multiple

[frank widget]

yeah multiple

can differentiating play a role?

[lone void]

its product

not sum

[frank widget]

oh dang

[lone void]

u want to take log or smth?

[frank widget]

i mean that will still give a bad answer

[lone void]

yeah

[toxic stratus]

T(5) can be done fairly quickly

T(8) is just ass

[lone void]

yes

[frank widget]

fr 😭

[toxic stratus]

for T(5), notice that you don't care about any terms above x^5

so the only terms you care about are (1 + x)(1 + x^2)^2(1 + x^3)^3(1 + x^4)^4(1 + x^5)^5

[lone void]

u can eliminate options tho T(8) will be well above 33

[toxic stratus]

or expanding a little, (1 + x)(1 + 2x^2 + x^4)(1 + 3x^3)(1 + 4x^4)(1 + 5x^5)

then you can just count the ways to add to 5

[lone void]

yeah they have done combi they just want to know if theres a better method

[frank widget]

hmmmm that does make sense, but still

[toxic stratus]

idk what you mean by "combi" method because it takes like maybe 1 or 2 minutes to calculate T(5) at most

[frank widget]

aight imma just ask my teacher both that

imma send yall 1 more doubt

[lone void]

"then you can just count the ways to add to 5" this is combi no

[frank widget]

this

[toxic stratus]

yeah sure

but like

you gotta do it intelligently

(1 + x)(1 + 2x^2 + x^4)(1 + 3x^3)(1 + 4x^4)(1 + 5x^5) are the only terms you care about

theres 4 ways to get x^5

[frank widget]

i aint got intelligence 😭

[toxic stratus]

its like

,calc 5 + 4 + 1 + 6

[flat frigate]

Result:

16

[toxic stratus]

T(5) = 16

for T(8) you're looking at the whole (1 + x)(1 + 2x^2 + x^4)(1 + 3x^3 + 3x^6)(1 + 4x^4 + 6x^8)(1 + 5x^5)(1 + 6x^6)(1 + 7x^7)(1 + 8x^8)

which gets a little bit ass but still doable

just case bash on the highest power

[frank widget]

hmm

[frank widget]

guys pls summon yourself here and help a fellow peasant

[mortal sandal]

Are there multiple right answers?

Oml

"The which is/are true"

I'm guessing the is/are is supposed to indicate that

[frank widget]

yeah

[frank widget]

well leave that one as of now i need help with this

[mortal sandal]

I got T(8)=83 💀

[frank widget]

correct

[mortal sandal]

nah what is this random symbol spam 💀

[frank widget]

bro idk i have been trying to somehow get some recursion

nothing is working

[lone void]

what i did was group the n terms and r terms and distribute that outer bracket among them

[frank widget]

like how?

[lone void]

n(n+1) - (2r^2+3r-1)

and then split

i got that n bracket simplified to ((n+1)C(r+1))((n+r)C(r+1) into r!

2nd bracket i wasnt able to simplify tho

2r^2+3r-1 dosent have good factors

[frank widget]

yeah that is a problem

[lone void]

kinda ahh question bro

[frank widget]

is there any way to for a recursion in first bracket?

cuz if we can pull that of maybe we endup with a good equation

and some form of nCr in right bracket

cuz this is not giving any hope 😭

dang man

[lone void]

honestly bro

since they are asking S4 and S6

just put the values

[frank widget]

am NOT doing that bro 🙏

no one is getting that in 3 mins

[lone void]

then skip 💀

noone is figuring out how to actually do it in 3 mins

nah but i dont know how to do it properly

i think u have to split 2r^2+3r-1 into something again

[frank widget]

will try by tmro

till then rip maths 🙏

thx for helping 😄

.close

[safe radish]

Channel closed

Closed by @frank widget

Use .reopen if this was a mistake.

[dusky totem]

No way this is JEE main

[frank widget]

adv bro

[dusky totem]

You don’t have 3 mins then

U can take 5min

[frank widget]

yeah ik but not like we solving that in 5 mins in exam 🙏

[dusky totem]

Multiple correct is just… ugh

[frank widget]

fr

[lone void]

bro manifest motivation

and put the values

[frank widget]

no 🗿

[safe radish]

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[wicked elm]

Can somebody please explain to me at a fundemental level how an equation can be mapped onto a polar cordinate system, I'm stuck imagining it on a cartesian

[marsh walrus]

mapped?

how do you mean

[wicked elm]

yeah pretty much

I havent really had a teacher to teach me polar graphs so I just watched a few videos because while I live in the states my school does aice math which just never covered polar graphs

so it leaves a pretty big gap in my understanding

[marsh walrus]

well you know functions are just like machines

[wicked elm]

yep

[marsh walrus]

we like to think of them as pictures

[stoic dune]

https://tutorial.math.lamar.edu/classes/calcii/polarcoordinates.aspx

[marsh walrus]

but more generally they dont even have to be spatial

[wicked elm]

can you define that

"dont have to be spatial"

[marsh walrus]

really anything that takes an element in its domain to some unique element in the range is a function

[wicked elm]

okay

[marsh walrus]

well our domain doesnt have to be numbers is how i mean

[wicked elm]

syfm

[marsh walrus]

all that we require is that if something is in the domain, its mapped to a unique element

[wicked elm]

ok so what you're saying is I need to stop trying to understand it in an ordinary cartesian xy format

and rather see it as a machine or for what it is

[marsh walrus]

im just trying to lean into a more general understanding

which helps resolve your specific concern

[marsh walrus]

maybe you see how here we could have like cylindrical functions

or spherical functions

all kinds of shapes

[wicked elm]

I dont really know that many functions yet like cylindrical etc

[marsh walrus]

i just mean say we have some function defined like inside a cylinder

you give me a point and ill tell you the temperature there

does that help at all?

[wicked elm]

a little bit, you did lose me on temperature

[marsh walrus]

or do you have like more specific worries about polar

[wicked elm]

I'm just trying to get it at a more fundemental level

Ive seen on ap preview tests how you only need to know the very basics that will let you know what you're looking at

like if its sin and its coefficient is a positive its opening up

but I didnt really understand past that

[marsh walrus]

maybe a video would be helpful

[wicked elm]

do you rec any?

recommend

[marsh walrus]

or you can play around like desmos has polar

i like this one guy

1s

[wicked elm]

I like that ali the dazzling guy but he doesnt cover everything unfortunately

[marsh walrus]

https://www.youtube.com/playlist?list=PLX2gX-ftPVXXWNn8FQ8DfZ0N0YVJCxe-p

theyre short vids so skip around the playlist

he covers converting and graphing

[wicked elm]

and how can I play around with this in desmos

[marsh walrus]

and he graphs a ton of functions

[marsh walrus]

if you open settings you can pick polar

id show but im on mobile

itll look weird

[wicked elm]

alright thanks

[marsh walrus]

the circle lookin one @wicked elm

[wicked elm]

got it

[proper socket]

@knotty valve

[wicked elm]

.close

[safe radish]

Channel closed

Closed by @wicked elm

Use .reopen if this was a mistake.

[wicked elm]

@marsh walrus got it now, thanks

[safe radish]

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Remember:
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[solemn helm]

I know this is a math server but can anyone help me with this biology graph?

[tardy mango]

As per #old-network, try something like https://discord.gg/QwSdffq

the math server is for math questions

[solemn helm]

Ok ty

[tardy mango]

.close

[safe radish]

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[verbal briar]

what mistake did i make

[winter whale]

what are you trying to do?

[verbal briar]

get the final answer as 16

but im pretty sure i got all the steps right

[winter whale]

you can write ln(7) - ln(3) as ln (7/3)

[verbal briar]

still getting the same

[winter whale]

well if you use a calculator on the second last step you dont get 16

no idea how they got it

can you show the original question?

[safe radish]

@verbal briar Has your question been resolved?

[verbal briar]

its horrendously hard, im sure u agree

[main mural]

whose writing is red

[main mural]

their answer makes no sense

,w 5/16 ln(14) - 5/16 ln(16-14) -5/16 ln(3)

[flat frigate]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Feb 21, 2026, 8:11 PM

[main mural]

bruh

anyway it's what you got

[winter whale]

they thought we wouldn't notice..

[main mural]

unless i am missing something this must be a mistake on their side

[winter whale]

yeah their calculations dont lead up to 16 at all

[main mural]

what does Way 2 show?

[verbal briar]

my friends

[soft crag]

wait you multiply by 60 because it wants the answer in minutes

[main mural]

ahhhh

[verbal briar]

he was stuck asw so he gave up half way

[main mural]

yeah that makes sense

[winter whale]

ohhh

[main mural]

ffs

absolutely horrendous writing

and no units too

[winter whale]

tbf every solution manual does this

[main mural]

criminal

[verbal briar]

whos 😭

[winter whale]

quite irritating

[verbal briar]

its so vague

[main mural]

solution manuals

[verbal briar]

yh true

[main mural]

but yeah you forgot to multiply by 60 in the end

[verbal briar]

if u guys took a while to clock it, imagine me

[verbal briar]

got it now

tyyyy

[main mural]

you've got good eyes lol

[soft crag]

haha tbh i should've realised when the correct answer was 60x the answer above

[safe radish]

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Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timber anchor]

how do i prove (-1)x = -x where x belongs to the set of real numbers

i mean what is there to prove here??

[zealous ingot]

lol abstract algebra eh

[timber anchor]

huh?

[zealous ingot]

basically, proving that is equivalent to proving a + (-1)a = 0 as additive inverses are unique

apply distributivity

so (1+(-1))a

and yeah

[timber anchor]

okay,so then how do i prove that a(-1) is -a

[zealous ingot]

well, -a is the additive inverse of a by definition yeah

so proving that (-1)a is the additive inverse of a

is equivalent to proving that (-1)a = a, since additive inverses are unique in a ring

[timber anchor]

aoh i get it now

so by definition x has an additive inverse which is -x ,
and i prove that (-1)x is the additive inverse and hence it is equal to -x is that right?

[silk osprey]

You're actually really smart

[zealous ingot]

right

[zealous ingot]

I've js seen this question before 🙏

bit of a horror for me too when I saw books asking these

[silk osprey]

im not asking for anything dude

[zealous ingot]

I js can't take a compliment 😭 thanks tho

[silk osprey]

ahh i got u

say less

[silk osprey]

why bud

[timber anchor]

aight thanks for the help

[zealous ingot]

being silly

[timber anchor]

.close

[safe radish]

Channel closed

Closed by @timber anchor

Use .reopen if this was a mistake.

[silk osprey]

bet tho its up to u anyway

What are you studying?if u don't mind me asking are you in university or still in school?
@zealous ingot

[silk osprey]

btw why ur name is donkey ?

[zealous ingot]

hs senior

[silk osprey]

ahh good to know

[zealous ingot]

donkeys are too cool 😂

come dms or discussion if u wanna talk or smn tho. Better not chit chat here

[silk osprey]

ahh i got u cheers for being respectful

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pastel vault]

if each of these red points can be 0 or 1 how would i find the amount of combinations without duplicates that can be mirrored or rotated

[stoic dune]

Can we get an example? I'm not positive what's happening here

[pastel vault]

so if each point can represent a 1 or a 0 i wanna know how many different combinations like this for example i could have

but only counting one combination for combinations that could be the same if rotated or mirrored