help-39

[pastel totem]

but that equals to just x right

[feral leaf]

Not yet

[pastel totem]

oh okok

[feral leaf]

So that means, h(x) can be rewritten as $x^{\frac{1}{3}} +1$

[jolly parrot]

CaptainNova22

[feral leaf]

Do you agree?

[pastel totem]

OH YES

[feral leaf]

Then when you plug in g(x)

You have h(g(x)) = $((x-1)^3)^{\frac{1}{3}} +1$

[jolly parrot]

CaptainNova22

[feral leaf]

Do you understand that?

[pastel totem]

i understand why its written that way but i dont understand what is next

like yes i understand it but i dont understand what to do with it

[feral leaf]

So you now, back to the table, you apply the rule $(a^x)^y = a^{xy}$

[jolly parrot]

CaptainNova22

[feral leaf]

Meaning that h(g(x)) = $(x-1)^{3 \cdot {\frac{1}{3}}} +1$

[pastel totem]

okok i see

[jolly parrot]

CaptainNova22

[feral leaf]

Then you can simplify

What's 3 * 1/3?

[pastel totem]

wait wait

its 1

[feral leaf]

Yes

[pastel totem]

so then you just have x -1 + 1

which cancels to x

[feral leaf]

Yes

So h(g(x)) = x

[pastel totem]

so every time theres a sqrt like that ill do fractional exponent

instead of just needing to distribute

[feral leaf]

You didn't need to distribute in the first place, you just needed to apply what you know with roots and exponents and simplify

[pastel totem]

right right

[lilac ocean]

is the winter math homework hitting rn 😂

seems like you've been here all day without break

[pastel totem]

who me..😅

[lilac ocean]

yeah hahaha

[pastel totem]

im preparing for college i haven’t ever been good with math loll

so im trying to quickly learn everything to college standard at least

[feral leaf]

So when it was asking to do h(g(x)), you started it off fine by plugging g(x) into h(x), it was just then applying the mathematical concepts to simplify it

[pastel totem]

okok yes i see, thank youu

this helped alottt ty:)

.close

[pearl pond]

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[lilac ocean]

fair, best of luck!

if you keep working hard you'll get there in no time :)

[pastel totem]

hehe thank youuu!!!

[pearl pond]

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[royal galleon]

for part by I can find that it is <= (b-a)^2 by saying that int_a^b f(x)dx <= f(b)(b-a) but I cannot figure out how to get the 1/2 there

[naive zinc]

<=(x-a)dx x from a to b

(=|int (f(x)-f(a))dx|<=int |f(x)-f(a)|dx<=int (x-a) dx)

[royal galleon]

what is this supposed to be? why is there an extra x at the end?

[naive zinc]

What?

You prefer in Tex?

[royal galleon]

yes please

[naive zinc]

$=| \int_{a}^{b}(f(x)-f(a))dx| \leq \int_{a}^{b} |f(x)-f(a)|dx \leq \int_{a}^{b}(x-a)dx=\frac{(b-a)^{2}}{2}$

[jolly parrot]

Cogwheels of the mind

[royal galleon]

two question. First how did we even go from $|\int_a^b f(x)dx - (b-a)f(a)|$ to what you have. And second how can $\int_a^b (x-a)dx = \frac{(b-a)^2}{2}$ would it not be $\frac{(b-a)^2}{2} - ba -a^2$

[jolly parrot]

BigBen

[naive zinc]

|integral of g|<=integral of |g|

plug in g(x)=f(x)-f(a)

Second

[royal galleon]

but if you make that sub you forget (b-a)

[naive zinc]

$\int_{a}^{b} (x-a)dx= \int_{a}^{b}(x-a)d(x-a)= \int_{0}^{b-a}ydy=\frac{(b-a)^{2}}{2}$

[jolly parrot]

Cogwheels of the mind

[naive zinc]

I forgot nothing

[royal galleon]

I don't see where our b-a goes in the sub of g(x) = f(x)-f(a)

[naive zinc]

I don’t see how bound becomes an issue

I am out

Oh I left out your first question

$|\int_{a}^{b}f(x)dx-(b-a)f(a)|= |\int_{a}^{b}f(x)dx-\int_{a}^{b}f(a)dx|=|\int_{a}^{b}(f(x)-f(a))dx| \leq \int_{a}^{b} |f(x)-f(a)|dx \leq \int_{a}^{b}(x-a)dx=\frac{(b-a)^{2}}{2}$

[jolly parrot]

Cogwheels of the mind

[naive zinc]

I thought it’s obvious to you

[royal galleon]

I missed that

[royal galleon]

how are we allowed to do this with the differential? what is the motivation? Also I don't see where we have y = x

[naive zinc]

y=x-a

[royal galleon]

why?

[naive zinc]

You never use substitute?

okay

[royal galleon]

ok then we have dy/dx = x and then we wouldl need xdx

[naive zinc]

$\int_{a}^{b}(x-a)dx=\frac{x^{2}}{2}-ax |_{a}^{b}=(\frac{b^{2}}{2}-ab)-(\frac{a^{2}}{2}-a^{2})= \frac{(b-a)^{2}}{2}$

[royal galleon]

I was thinking about it in terms of the expansion of an interval property

[jolly parrot]

Cogwheels of the mind

[naive zinc]

Though I prefer this

I really don’t understand how it becomes an issue. Maybe get used to substitute

[royal galleon]

for your sub part how did you get dx = d(x-a)

[naive zinc]

df(x)=f’(x)dx

f(x)=x-a

f’(x)=1

[royal galleon]

ok I see thx

.solved

[pearl pond]

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[pearl pond]

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[frank goblet]

Chat is this real

[carmine schooner]

spam alert

[frank goblet]

xD

[pearl pond]

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[fading ledge]

@stable dune

[brave sluice]

it's a corollary of lagrange's theorem

[fading ledge]

I see yes

How do I prove it?

[brave sluice]

what's your definition of the order of an element?

[stable dune]

Oh I lowkey thought they wouldn't even be able to use lagrange

The proof would essentially be proving Lagrange anyways I suppose

[brave sluice]

oh, you can't use it?

[waxen agate]

yeah

consider proof by cosets perhaps

[fading ledge]

order of a element mean if we increase power of it we get identity of that group

The least power

[cursive wraith]

The least positive power yes

So

What can you say about the subgroup generated by that element

[fading ledge]

<x>

[cursive wraith]

What is in it? How many elements?

[fading ledge]

we will get some elements?

number of disjoint elements?

[carmine schooner]

how many elements are in <x> ?

[cursive wraith]

Well yes that's essentially the question

When we asked how many elements are in a set

[fading ledge]

The order of this element

We will get that number of elements?

[cursive wraith]

Yes

[carmine schooner]

yes, the order of <x> is the order of x

[cursive wraith]

|<x>| = o(x)

[fading ledge]

I see

[cursive wraith]

So

[fading ledge]

but we increase power one side

And we are making subgroup?

[cursive wraith]

What does Lagrange say

[brave sluice]

<x> is a subgroup

[cursive wraith]

We are looking at the set of all possible powers of x

[fading ledge]

i meant subgroup is like we need identity and a element?

[carmine schooner]

(can you use lagrange for this question?)

[cursive wraith]

It makes a subgroup

[fading ledge]

I guess no

[cursive wraith]

Really?

[fading ledge]

we will try to see...via example)

[fading ledge]

They didn't mention it actually

[carmine schooner]

well by definition <x> contains the identity 1

[fading ledge]

any simple examples which clarify more

Yes it has

[carmine schooner]

since the elements are powers of x then <x> is closed under multiplication

[stable dune]

<x> is just all integer powers of x

[cursive wraith]

The inverse of x^n is x^(-n)

Still a power of x

[round heart]

@fading ledge given a group G and an element x of G, <x>:= {e,x,x^2,...,x^{n-1}} where e is the identity element and n is the order of x

[brave sluice]

probably your book already proved that <x> is a subgroup

[cursive wraith]

And the product of two powers of x is still a power of x

[fading ledge]

Yeah that's true

I got it

[round heart]

now if you want prove that this is in fact a subgroup of G, or just take it for granted if you feel like its obvious to you

[carmine schooner]

so we know <x> is a subgroup from all this above

[fading ledge]

Yeah

It is

[carmine schooner]

what can we say about the cosets of <x>

[round heart]

anyways, since you cant use lagrange then you can prove it if you want

[fading ledge]

Left cosets?

[carmine schooner]

yeah

[round heart]

and then the result you are looking for will follow

[fading ledge]

okay so left cosets we will have like all the elements which multiply by x left side

[brave sluice]

i would say you can use lagrange if it's stated in the book prior to this question

[carmine schooner]

likely a question like this comes after proving lagrange anyway

[fading ledge]

I see

[round heart]

hmmmm i am not sure what exactly you mean by this but if i understood you correctly then not quite

can you write what that means using some symbols?

[carmine schooner]

but what you are looking to prove is that ||the cosets partition the group and are of the same cardinality, so the order of the subgroup divides the order of the group exactly by the number of cosets it has in the group||

[fading ledge]

aH: where a is element of group G

And H is subgroup?

[round heart]

right so what are the cosets of <x> in this case?

[fading ledge]

Number of elements in group G?

[round heart]

no i meant to use this notation

[cursive wraith]

I think at this point we're into the proof of lagrange theorem

Are you sure Lagrange theorem isn't before this in your book

[brave sluice]

you should check

[fading ledge]

Ohh wait

[brave sluice]

assuming you're using a textbook

[stable dune]

Lagrange is immediate from properties of cosets anyways

Cosets have same order + partition G

[fading ledge]

I see

That's true

So what does it prove?

[carmine schooner]

for a subgroup H < G, there are [G : H] cosets. the cosets partition G and have order |H|. so |G| = [G : H] * |H|.

[stable dune]

Well if <x> has an order which divides G by Lagrange

And the order of x is the order of <x>

Then what can we say

[round heart]

side note: you could generalize more and say that given a group G and 2 subgroups H and K such that K\subset H, its true that (G:K)=(G:H)(H:K)

[carmine schooner]

<@&268886789983436800> crypto spam :(

[pearl pond]

@fading ledge Has your question been resolved?

[pearl pond]

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[gloomy berry]

Hi

[gloomy berry]

Let f be a twice continuously differentiable real-valued function on an open interval containing 0 such that f(0) = 0 and f′(0) = 1. Suppose f satisfies the equation
f(x) = integral from 0 to x of e^(x − t) · (f(t))^2 dt for all x in its domain.
Prove that f is uniquely determined, find an explicit expression for f(x), and determine the maximal interval on which this solution exists.

Can anyone solve it with proper solution

<@&286206848099549185>

[surreal siren]

I’ll try

[atomic ermine]

me too

[left steppe]

what have you tried

[gloomy berry]

Yes

[left steppe]

?

[gloomy berry]

Got maximum interval

(-infinite,0)

f(x)=?

No

@surreal siren

Are you Bhartiya?

[surreal siren]

Jä

[left steppe]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[surreal siren]

Ya

[gloomy berry]

How to find f(x)

[gloomy berry]

Me also

[surreal siren]

Cool

[gloomy berry]

Operate our language

[left steppe]

try using leibniz's rule

[gloomy berry]

Hindi

Ok

[surreal siren]

Yes

[gloomy berry]

Bhai yeh f(x) kaise ayega

Interval to agaya

[left steppe]

!noanswer

[pearl pond]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[surreal siren]

Leibritz theorem use karna padega

[gloomy berry]

Acha ruk karta hu

Ho gayyaa

[surreal siren]

Kya aya expression

[gloomy berry]

f(x) = e^x/1-e^x

Solveddd

Sahi hai kya

Bata na brother

[surreal siren]

Check karta hu

[gloomy berry]

Differentiate the equation using Leibniz rule to get f′(x) = f(x) + (f(x))²
2. Solve the ODE to get f(x) = e^x / (1 − e^x), valid for x < 0

[surreal siren]

F(0) = 0 nahi aaraha hai

Undefined hoga

Lim lagaoge toh infi ayega

[gloomy berry]

Oh sry

I used C also

f(x) = 0?

[surreal siren]

Const Func ka slope 1 nahi aasakta

F’(0)=1 Diya hai

[gloomy berry]

Ha

Answer hai 0

[surreal siren]

Kese?

F(x)=0

?

Uska slope zero hoga

Phir f’(0)=0 ayega

[gloomy berry]

Differentiate karle equation using Leibniz rule to get f′(x) = f(x) + (f(x))²
Use the initial condition f(0) = 0; the only solution to f′ = f + f² with f(0) = 0 is f(x) ≡ 0.

[surreal siren]

Leibniz check karo

[gloomy berry]

Writing op

[gloomy berry]

Sahi hai

[surreal siren]

Isme leibniz se f(0)=0 nahi aana chahiye

[gloomy berry]

It is the only sol

Unique like you

[surreal siren]

But possible nahi hai kyuki for constant function slope always zero ayega

Toh f’(0)=1 satisfy nahi ho raha hai

[gloomy berry]

Hai possible

At x = 0, the integral is zero, so f(0) = 0
Differentiate the equation using Leibniz rule to get f′(x) = f(x) + (f(x))²
At x = 0, f′(0) = 0 because f(0) = 0
The only solution to f′ = f + f² with f(0) = 0 is f(x) = 0 for all x

Yeh dekh bro

[surreal siren]

But question mein f’(0)=1 Diya hai

[surreal siren]

Aap idhar f’(0)=0 le rahe ho

[gloomy berry]

Then there is no sol

Because f'(0)= 1 and f'(0) = 0 simultaneously

No solution

Not defined

Also

Maximum interval

@surreal siren

[surreal siren]

Ha no solution hi hoga I think

[gloomy berry]

Bhai add karlo bhai jaisa

Batao

[surreal siren]

Karliya

[gloomy berry]

Padhenge aur topp

Topppperrr?

College me ho kya

Ya MIT

[surreal siren]

10th mein hu

Odisa state board

[gloomy berry]

Bruh yeh college ka question hai

[surreal siren]

Isiliye ek aur bar kisi se puch lo

I think aur help mil jayegi

Scam

Gtfo

[gloomy berry]

Apko kese pata

Me also

9th

👽

[surreal siren]

Nice

Enjoy karo life

Sone chala

[gloomy berry]

Kyu

[surreal siren]

Par pls

Question

[gloomy berry]

Odisha

[surreal siren]

Ek aur bar

[gloomy berry]

Jai Jagannath

[surreal siren]

Kisi aur se puch lo

[surreal siren]

Yes

[gloomy berry]

Jai Jagannath Ji

Me Rajasthan

North

[surreal siren]

Jai rajputana

[gloomy berry]

CBSE board

[surreal siren]

Hm

[gloomy berry]

Yes

[surreal siren]

Goodnight brother

[gloomy berry]

Kyu

[surreal siren]

Neend aarahi

[gloomy berry]

1 question karlo

[surreal siren]

Abhi nahi

Baadme baat karta hu aapse

[gloomy berry]

sin²x + tan²x=?

[surreal siren]

Iska kya karu?

[gloomy berry]

Kya hota hai

[surreal siren]

I think already kaafi simplified hai

Limit lagi hai kya?

[gloomy berry]

Nahi

[surreal siren]

X->0?

Phir?

[gloomy berry]

trigonometry hai

No lim

Just triangle

[surreal siren]

Ha par koi options?

[gloomy berry]

No waves

Yes

[surreal siren]

(Sin^2(2x)+sin^2x)/(1-sin^2x)

Ye hi hoga

[gloomy berry]

sin²x + tan²x = sin²x+sec²x-1=sin²x-sin²x-cos²x+sec²x

[surreal siren]

Ek simplification

[gloomy berry]

sec²x-cos²x

Answer

[surreal siren]

Ha wo bhi hai

Wo bhi keh sakte hai

[gloomy berry]

= (secx+cosx)(secx -Cosx)

[balmy olive]

.

[surreal siren]

Options par depend karta hai

[gloomy berry]

Yes

True

[surreal siren]

.!closed

.closed

[gloomy berry]

.Gn

[surreal siren]

Yes

[gloomy berry]

Gnnn

Bhai

[surreal siren]

Gn

[gloomy berry]

.Gn.

[surreal siren]

!closed

[gloomy berry]

Good Night Brother

. close

[surreal siren]

Good night brother

[gloomy berry]

.close

[pearl pond]

Channel closed

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[gloomy berry]

Yes

Bye

Gn

.close

[pearl pond]

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[lilac ocean]

ive never seen a problem like it

[plush bramble]

<@&268886789983436800>

[pearl pond]

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[arctic siren]

Do not spam the help channels

[cobalt hinge]

guh guh guh

[pearl pond]

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[real arrow]

im just wondering why this only works for non integer values of p?

and what would we do for integer values?? would there be a new defined function and what would the GS look like

[lean star]

holy periodhater he/him

[real arrow]

yes i am

i also read feminist literature and drink matcha instead of water

[lean star]

are u also 6ft2

[real arrow]

6'3*

(im 5'10)

[lean star]

i hope some kind lady helps this saint

[real arrow]

no no ladies have too much on their plate already

i wouldnt dare bother them

jokes aside tho any help would be appreciated

[pearl pond]

@real arrow Has your question been resolved?

[lean star]

i might be trolling

but

do jp and j-p have to be linearly independent

i plugged in 2 and -2 for p in 4.37 and got the same thing

so maybe thats why

[real arrow]

yes your right

im not sure about the GS

for integers

im pretty sure jp would be in it still because the recurrence relation is still fine

j-p would break the recurrence

i dont know what the other part of the GS would be

would i just use this to find the other one

i think it will give it to me

if i use jp

[lean star]

i checked p=1 and got Jp = - J-p
so it might be that J-p = (-1)^p Jp for p non negative integers so theyre not linearly independent

[real arrow]

yea

i found out that its not linearly independent for integers

so i was wondering what the actual GS would be

im pretty sure it would be jp(x) + some other function (x)

jp(x) would still be present but obviously j-p(x) would be something else for integers

[pearl pond]

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[real arrow]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[real arrow]

so im thinking that we use this to find the other part of the GS

[lean star]

yayyy

[real arrow]

❤️

.close

[pearl pond]

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[lean star]

maybe u can also do reduction of order stuff like y_2 = u(x)y_1(x)

maybe thats how theyre getting that fmla

[real arrow]

yea i probably could as well

oh wait the picture i sent is from reduction of order xd

[lean star]

omg fire

[pearl pond]

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[tacit jolt]

[proper nova]

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[crystal dew]

do you have a question?

[snow sail]

https://cdn.discordapp.com/attachments/1336498829473874022/1393412810092052561/attachment.gif

[tacit jolt]

Yah wait i have one

@crystal dew

Can you help me

[crystal dew]

you have accurately described my response to your question

[tacit jolt]

[crystal dew]

so do you have a proper question or shall I close this

and ofc, please do not troll in help channels.

[tacit jolt]

Noi i am high schooler

[crystal dew]

in that case, see you around

.close

[pearl pond]

Channel closed

Closed by @crystal dew

Use .reopen if this was a mistake.

[tacit jolt]

Why there is my channel

I didn't open it

[crystal dew]

because you typed something in an open channel

[tacit jolt]

Oh

[crystal dew]

now, off to #discussion if you want to talk further

[tacit jolt]

[pearl pond]

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[shut elm]

Observe how the numbers change in this sequence. 
Using the convention of starting with $a_0 = 1$, we have $a_n = a_{n-1} + 1$ for $n \in \{1, 3, 6, 10, \dots\}$ (and $a_n = a_{n-1}$ otherwise), the well-known series $S(i) = \sum_{i=1}^ki$. 
In other words, the given sequence is constructed by writing $k+1$ until $n = S(k)$.

So we write $a_{\ceil{S(k)}} = k + 1$.
We need to solve $n = S(k)$ for $k$ to determine a formula for $a_n$.
\begin{align*}
n &= \sum_{i=1}^ki \\ 
n &= \frac{k(k-1)}{2} \\
2n &= k(k-1) \\
k^2 - k - 2n &= 0 \\
k &= \frac{1 \pm \sqrt{1 + 16n}}{2} \\
k &\approx \frac{1}{2} + 2\sqrt{n}
\end{align*}
using only the positive root.

[jolly parrot]

Coolempire2026

[shut elm]

Was working to construct a proof for this by solving backwards but

My formula is slightly off

Oh

I see what happened

Observe how the numbers change in this sequence. 
Using the convention of starting with $a_0 = 1$, we have $a_n = a_{n-1} + 1$ for $n \in \{1, 3, 6, 10, \dots\}$ (and $a_n = a_{n-1}$ otherwise), the well-known series $S(i) = \sum_{i=1}^ki$. 
In other words, the given sequence is constructed by writing $k+1$ until $n = S(k)$.

So we write $a_{\ceil{S(k)}} = k + 1$.
We need to solve $n = S(k)$ for $k$ to determine a formula for $a_n$.
\begin{align*}
n &= \sum_{i=1}^ki \\ 
n &= \frac{k(k-1)}{2} \\
2n &= k(k-1) \\
k^2 - k - 2n &= 0 \\
k &= \frac{1 \pm \sqrt{1 + 8n}}{2} \\
k &\approx \frac{1}{2} + \sqrt{2n}
\end{align*}
using only the positive root.

[jolly parrot]

Coolempire2026

[shut elm]

Applied the quadratic formula wrong

Okay now my question is how to turn this from observations into a proof

In reality n = ceil(S(k)) so we actually have

$0 \leq k^2 - k - 2n \leq 2$

[jolly parrot]

Coolempire2026

[shut elm]

Any help would be appreciated!

[shut elm]

Slight error - should be writting k* until n = S(k) [and starting the next number, k+1, exactly when n = S(k)

With this in mind, the observations become

Ah I just realized that ceil(S(k)) doesn't even make sense since it's an integer always 😂

It should be

[round heart]

yea i was going to tell you that

[shut elm]

S(k-1) <= n < S(k)

[round heart]

but also i am not really sure i understand what you are saying here tbh

it might be me being stupid rn

[shut elm]

Did you read the image with the sequence

[round heart]

yea

for some reason i cant comprehend what is being said before the n=\sum...

[shut elm]

You mean the sentence that has the incorrect notation? or the whole paragraph

[round heart]

starting from "the well known series" up to and excluding the beginning of the calculations

[shut elm]

I just followed the textbook's way of writing

'constructed by' and then the procedure

Observe how the numbers change in this sequence. 
Using the convention of starting with $a_0 = 1$, we have $a_n = a_{n-1} + 1$ for $n \in \{1, 3, 6, 10, \dots\}$ (and $a_n = a_{n-1}$ otherwise), the well-known series $S(i) = \sum_{i=1}^ki$. 
In other words, the given sequence is constructed by writing $k$ until $n = S(k)$.

So we write $a_{S(k-1) \leq n < S(k)} = k$.
We need to solve $n = S(k-1)$ and $n < S(k)$ for $k$ to determine a formula for $a_n$.

[jolly parrot]

Coolempire2026

[shut elm]

So it would be interpreted as

a0 = 1

S(1) = 1
S(2) = 3

So for n in [1,3) we write a_n = 2

S(2) = 3
S(3) = 6
So for n in [3,6) we write a_n = 3

S(3) = 6
S(4) = 10
So for n in [6,10) we write a_n = 4

We can see how this gives us the original sequence

1 entry (0) includes 1, 2 entries (1 and 2) include 2, 3 entries (3, 4, 5) include 3, 4 entries (6, 7, 8 and 9) include 4

And so on

[round heart]

hmmm i see what you are saying

,w solve (x-1)(x-2)=2n

[jolly parrot]

Results provided by WolframAlpha

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

Requested by pirateking0723

[shut elm]

Hmm what did they do with the b^2 term that turned into a 9

Ah

8(n-1)

ofc

Observe how the numbers change in this sequence. 
Using the convention of starting with $a_0 = 1$, we have $a_n = a_{n-1} + 1$ for $n \in \{1, 3, 6, 10, \dots\}$ (and $a_n = a_{n-1}$ otherwise), the well-known series $S(i) = \sum_{i=1}^ki$. 
In other words, the given sequence is constructed by writing $k$ until $n = S(k)$.

So we write $a_{S(k-1) \leq n < S(k)} = k$.
We need to solve $n = S(k-1)$ and $n < S(k)$ for $k$ to determine a formula for $a_n$.
\begin{align*}
n &\geq \sum_{i=1}^(k-1)i \\ 
n &\geq \frac{(k-1)(k-2)}{2} \\
2n &\geq k^2 - 3k + 2 \\
k^2 - 3k - 2n + 2 &\leq 0 \\
0 \leq k &\leq \frac{3 \pm \sqrt{9 + 8(n-1)}}{2} \\
0 \leq k &\leq \frac{3 \pm \sqrt{1 + 8n}}{2} \\
0 \leq k &\approx \frac{3}{2} + \sqrt{2n}
\end{align*}
using only the positive root.

[jolly parrot]

Coolempire2026

[cinder flower]

coolemplud

[round heart]

no

[shut elm]

And then comes the S(k) root

[shut elm]

Hiiiii

I guess I'll leave it in the unsimplified form because the S(k) gives

[cinder flower]

i feel like i did this problem a long time ago

[shut elm]

Observe how the numbers change in this sequence. 
Using the convention of starting with $a_0 = 1$, we have $a_n = a_{n-1} + 1$ for $n \in \{1, 3, 6, 10, \dots\}$ (and $a_n = a_{n-1}$ otherwise), the well-known series $S(i) = \sum_{i=1}^ki$. 
In other words, the given sequence is constructed by writing $k$ until $n = S(k)$.

So we write $a_{S(k-1) \leq n < S(k)} = k$.
We need to solve $n = S(k-1)$ and $n < S(k)$ for $k$ to determine a formula for $a_n$.
\begin{align*}
n &\geq \sum_{i=1}^(k-1)i \\ 
n &\geq \frac{(k-1)(k-2)}{2} \\
2n &\geq k^2 - 3k + 2 \\
k^2 - 3k - 2n + 2 &\leq 0 \\
0 \leq k &\leq \frac{3 \pm \sqrt{9 + 8(n-1)}}{2} \\
0 \leq k &\leq \frac{3 + \sqrt{1 + 8n}}{2}
\end{align*}
and
\begin{align*}
n &< \sum_{i=1}^ki \\ 
n &< \frac{k(k-1)}{2} \\
2n &< k(k-1) \\
k^2 - k - 2n &> 0 \\
k &> \frac{1 \pm \sqrt{1 + 8n}}{2} \\
k &> \frac{1 + \sqrt{1 + 8n}}{2}
\end{align*}
using only the positive roots.

[jolly parrot]

Coolempire2026

[shut elm]

What did I do wrong 👀

[round heart]

nvm its just me hallucinating at 1:20 AM

[shut elm]

Ah wait I've solved it!

That was it

[round heart]

yea

[shut elm]

Wait no I made a contraduction

!

No I didn't

!

[round heart]

yea its fine

[shut elm]

Observe how the numbers change in this sequence. 
Using the convention of starting with $a_0 = 1$, we have $a_n = a_{n-1} + 1$ for $n \in \{1, 3, 6, 10, \dots\}$ (and $a_n = a_{n-1}$ otherwise), the well-known series $S(i) = \sum_{i=1}^ki$. 
In other words, the given sequence is constructed by writing $k$ until $n = S(k)$.

So we write $a_{S(k-1) \leq n < S(k)} = k$.
We need to solve $n = S(k-1)$ and $n < S(k)$ for $k$ to determine a formula for $a_n$.
\begin{align*}
n &\geq \sum_{i=1}^(k-1)i \\ 
n &\geq \frac{(k-1)(k-2)}{2} \\
2n &\geq k^2 - 3k + 2 \\
k^2 - 3k - 2n + 2 &\leq 0 \\
0 \leq k &\leq \frac{3 \pm \sqrt{9 + 8(n-1)}}{2} \\
0 \leq k &\leq \frac{3 + \sqrt{1 + 8n}}{2}
\end{align*}
and
\begin{align*}
n &< \sum_{i=1}^ki \\ 
n &< \frac{k(k-1)}{2} \\
2n &< k(k-1) \\
k^2 - k - 2n &> 0 \\
k &> \frac{1 \pm \sqrt{1 + 8n}}{2} \\
k &> \frac{1 + \sqrt{1 + 8n}}{2}
\end{align*}
using only the positive roots.

Thus we have 
\begin{equation*}
\frac{1 + \sqrt{1 + 8n}}{2} < k \leq \frac{3 + \sqrt{1 + 8n}}{2} = \frac{1 + \sqrt{1 + 8n}}{2}
\end{equation*}
That is, that $k = \floor{\frac{1 + \sqrt{1 + 8n}}{2}} \approx \floor{\frac{1}{2} + \sqrt{2n}}$.

I don't have an equation environment?

wtf

No I put begin at the end

[jolly parrot]

Coolempire2026

[shut elm]

So now I have that $a_n \approx \floor{\frac{1}{2} + \sqrt{2n}}$

[jolly parrot]

Coolempire2026

[shut elm]

Showing that they are equal is beyond me thoug

So now that's my new question 🙂

And then how to formalize this compendium of observations

[feral sedge]

this is so cursed

you've observed that it's sufficient to find where each sequence increases by 1, right

[shut elm]

Where each sequence?

Like the entries yeah

[feral sedge]

a_n and floor(1/2 + sqrt(2n))

[round heart]

something here must be wrong in the inequalities

probably not the calculation itself but what you were calculating to begin with

[feral sedge]

like you know a_n last has a value of k when n=k(k+1)/2

then it switches to k+1

[shut elm]

Oh right it's k(k+1) not k(k-1)

But yes sure

[round heart]

because $\frac{1+\sqrt{1+8n}}2>\frac 12+\sqrt{2n}\geq\floor{\frac 12+\sqrt{2n}}$ so reaching $k>\frac{1+\sqrt{1+8n}}2$ is definitely wrong

[feral sedge]

for the kth triangular number yeah

[jolly parrot]

ali yassine

[feral sedge]

so you just need to show that plugging k(k+1)/2 into the sqrt thing gets you k, but plugging in k(k+1)/2 + 1 gets you k+1

[shut elm]

Shoudl be -1 + sqrt(1 + 8n)

Since it's k+1 not k-1

[round heart]

ohhh i see, in that case it might work

tbh i am not really able to focus rn so i cant really be very helpful 🥀

sorry vro

[shut elm]

I'll try this

This is a lot of work for me to do now

I will have to do it later/tommorow

.close

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

[pearl pond]

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[sharp smelt]

So I want to show that $\sum_{i=1}^{\infty} \frac{1}{i}$ diverges using cauchy

Proof: Let the series converge. Then for $\varepsilon= \frac{1}{2}, \exists N \in \N : n,m≥ N \implies \abs{\sum_{i=1}^{m} \frac{1}{i} - \sum_{i=1}^{n} \frac{1}{i}}< \frac{1}{2}$.

Chose $m=2N;n=N$. We then have $\sum_{i=N}^{2N} \frac{1}{i} ≥ N \cdot \frac{1}{2N} = \frac{1}{2}$. This is a contradiction to our assumption that $\sum_{i=1}^{\infty} \frac{1}{i}$ converges.

Thus $\sum_{i=1}^{\infty} \frac{1}{i}$ diverges

[sharp smelt]

oops

[jolly parrot]

wai

[plush bramble]

You kinda avoid using cauchy anywhere

sum u=N to 2N of 1/i > 1/2 is independent of Cauchy

[sharp smelt]

well, my idea was I show the difference exceeds 1/2 for large enough N

[sharp smelt]

that's basically the difference in cuachy no

[plush bramble]

Cauchy is more than just about the specific choice of m=2N and n=N

Once you fix m=2N, it's not really using Cauchy anymore

Using Cauchy would be more like any arbitrary m and n greater than some N

[sharp smelt]

I'm confused

isn't this correct for a proof by contradiction using cuahcy

[plush bramble]

Your proof is like 90% contradiction and 10% Cauchy

The question is phrased that you shouldn't be contradicting convergence, but contradicting the sequence is Cauchy

[sharp smelt]

could you show me how that'd work please

[plush bramble]

https://math.stackexchange.com/questions/2132993/prove-that-a-cauchy-sequence-is-convergent#2133001
Just substitute your actual sequence

[sharp smelt]

cool

thanks!

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

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[pearl pond]

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[timber schooner]

Hi

[timber schooner]

This is for f(x) = x^2 -2x +2

I completed the square

(x-1)^2 + 1 greater than or equal to 1

but should the final answer be f(x) is greater than or equal to 1

rather than 0

because the minimum value is 1

[heavy onyx]

the minimum x value is 1

the minimum occurs at (1,0)

[graceful leaf]

what is the question

[heavy onyx]

at x=1 for example, f(x) = 0

[timber schooner]

Why is it f(x) is greater than 0

[heavy onyx]

if anything, the answer should be f(x)≥0

[timber schooner]

should it be f(x) is greater than equal to 1

[heavy onyx]

yes

no

0

[timber schooner]

I thought the minimum value is 1

[heavy onyx]

like i told u

plug in x=1

[timber schooner]

so why is the minimum value 1

[heavy onyx]

it isnt

[timber schooner]

look

this is in the answer sheet

[heavy onyx]

they should say the minimum value f can attain is at 0 and it happens when x=1

[timber schooner]

no but how

when x = 1

it’s 0^2 +1 which is 1

[heavy onyx]

okay i realize now

[heavy onyx]

the quadratic is (x-1)^2 + 1

yes?

i thought (incorrectly so) the quadratic in question is (x-1)^2

u r correct, f(x)≥1 is the answer

[timber schooner]

yeah

so why does the answer sheet say

it’s greater than 0

[graceful leaf]

because its also true

[heavy onyx]

answer sheet is wrong

i mean it is true

but like

im sure the teacher desired u to do f(x)≥1

[pearl pond]

@timber schooner Has your question been resolved?

[timber schooner]

Tanks

[pearl pond]

Channel closed

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[light wasp]

When milk freezes, its volume increases by one-fifth.
When the milk melts, by what fraction does its volume decrease?

[pearl pond]

Please don't occupy multiple help channels.

[light wasp]

Ok @pearl pond

[rigid mist]

close this channel and stick to #help-27 please

[toxic lichen]

.close duplicate channel -- go to #help-27

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

[light wasp]

.close

[pearl pond]

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[inland laurel]

On solving inequations i get x = 2 and -4

After that what to do

[quasi estuary]

There should be a rangle probably cuz we need to chose 3 points

[inland laurel]

For x=2 the y is ±√10 and for x = -4 the y is ± 2

Like this we have 6 distinct points

[quasi estuary]

y is a integer tho

[inland laurel]

Then that value of y is rejected

±√10

[quasi estuary]

so we just have 2 points ??

[quasi estuary]

4 tho

[inland laurel]

What to do next now?

[quasi estuary]

2 points cant form a triangle

u solved it wrong I think

Make an ellipse

[tardy reef]

there are a ton of points

the inequality has an overlap of regions

idk where you got x=-4 coz thats not even in that parabola

[inland laurel]

I can reject -4

[quasi estuary]

It'll be an ellipse th

[tardy reef]

make a graph or a rough sketch and see what are the actual regions of intersection

[quasi estuary]

*tho

[tardy reef]

one is an ellipse, the other is parabola

[quasi estuary]

ye

[tardy reef]

and X is set of the intersection of the two regions

[inland laurel]

I'll do it later i should go now

[pearl pond]

@inland laurel Has your question been resolved?

[pearl pond]

Channel closed

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[pearl pond]

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[zenith mural]

how 10080 i think im losing it

[turbid summit]

So if we want to group the vowels, we treat it as 1 unit

[zenith mural]

I GOT IT

i got it

im sorry

lmao thanks

.close

[pearl pond]

Channel closed

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[uneven cypress]

. That was a quickie

[zenith mural]

. yeah thanks adhi

[oak quiver]

you dont need to use a . whena channel is closed

[oak quiver]

only after this msg appears

[pearl pond]

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[eager jewel]

rxa-(b+c)xa=0
(r-(b+c))xa = 0

r-(b+c) = lambda(a)

[toxic lichen]

brackets

(r-(b+c)) × a = 0

yes there we go

[eager jewel]

this is all ive done till now

[toxic lichen]

well, λa has to be orthogonal to b-c.

so maybe it's time to calculate $\vec{a} \cdot (\vec{b} - \vec{c})$, for a start.

[jolly parrot]

Ann

[eager jewel]

oh but why? r.(b-c) = 0 and r=lambda(a) + (b+c) means lambda a orthogonal to b-c?

[toxic lichen]

oh hold on my bad

misread the question

r itself is orthogonal to b-c sorry

not λa

[pearl pond]

@eager jewel Has your question been resolved?

[ancient knoll]

r-(b+c) is parrallel to a

$$(\vec{r} - (\vec{b} + \vec{c})) \times \vec{a} = \vec{0}$$

oh the cross

[jolly parrot]

Robo_17

[eager jewel]

yes..

[eager jewel]

oh ok..do u know how to proceed after this?

[toxic lichen]

.

have you done this

[eager jewel]

oh no not yet

1 min

its -593

but what to do w this

[toxic lichen]

$r \cdot (b-c) = \lambda a \cdot (b-c) + (b+c) \cdot (b-c)$

[jolly parrot]

Ann

[toxic lichen]

cbf to put \vec everywhere but you hopefully get my point

[eager jewel]

ohh ok thats = 0 from there lambda

[toxic lichen]

yes

[eager jewel]

alr alr got it ty

.close

[pearl pond]

Channel closed

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[ancient knoll]

assume r-(b+c) to be equal to lambda times a

and r ans

as

rx + ry +rz

[toxic lichen]

too late + too vague + overcomp

are "L + ratio" jokes still funny

[pearl pond]

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[pastel totem]

im lost on what to do next, i - 3 both sides and was left with |2x-5| >1

[proper nova]

|2x - 5| > 1

leads to 2x - 5 < -1 or 2x - 5 > 1

[pastel totem]

yes

wait

how

[proper nova]

you don't know how why i got the -1?

[proper nova]

what's the definition of absolute values

$\abs{A} = \begin{cases}
A \text{ if } A \geq 0\
-A \text{ if } A < 0
\end{cases}$

[jolly parrot]

1 divided by 0 equals Infinity

[pastel totem]

im confused how u got the -1 and why its < 1

[proper nova]

hold up

[proper nova]

is this the correct definition

[pastel totem]

since its -5, it then switches to < ?

[proper nova]

no no no no no

you got |2x - 5| > 1

so split into 2 cases: 2x - 5 >= 0 and 2x - 5 < 0

[pastel totem]

okok so i splitt niow

[proper nova]

when 2x - 5 >= 0 then |2x - 5| = 2x - 5

and |2x - 5| > 1 so 2x - 5 > 1

[pastel totem]

okay i didnt realize we split alr okay hold up

[pastel totem]

wait why 0's

[proper nova]

since the definition splits into 2 parts, it's best to split our absolute value into 2 cases

[proper nova]

what does the definition say

[pastel totem]

yesyes i understand that part

[proper nova]

you split it like that

[pastel totem]

but its > 1 not < 1

[proper nova]

the definition of absolute values

[pastel totem]

i see the definiton but its not lining up

[proper nova]

it will make sense later

[pastel totem]

??

wdym later

im confused now mane

i cantr learn if i dont know what im doing now

[proper nova]

when you split into cases like what im doing here

don't really care about what the question says now

[pastel totem]

yes but u used 0 not 1 howd you get 0

[proper nova]

we split by definition

[pastel totem]

im still lost sir, the definiton doesnt line up

[proper nova]

ma'am, the question says |2x - 5| > 1

not 2x - 5 > 1

[pastel totem]

oh wait

yes

is it zero bewcause 1 is positiv?e

[proper nova]

🙏

|A| >= 0 for any A

anyways

[pastel totem]

wait wait

i get it now but i havent needed to do that any other times

how do ik when?

if its = to 1?

[proper nova]

the |2x - 5| = 1?

or 2x - 5 = 1

[pastel totem]

nono

okay pre split

when i got 0

[proper nova]

the splitting is just me proving to you

[pastel totem]

you knew to get 0 bc by definition, but im lost because i didnt need to do it

do i go by definition because its 1?

[proper nova]

you want some shortcuts?

[pastel totem]

nono im just trying to understand when to and not to intergrate it

[proper nova]

|A| > c => A < -c or A > c

|A| < c => -c < A < c

these shortcuts can be proved by splitting into cases

which is what i did

[pastel totem]

thats not what im askigg ut telling me ant splitting right?

okok

wait

i am just explainging badly

[proper nova]

now if you want >= or <=, apply the same definition but instead of >, it's >= and instead of <, it's <=

[pastel totem]

so when i get to |2x-5| >1 i was lost, you told me to go by definiton, did you go by definion because it was > 1 or because you always go by ddefinition at that part

[proper nova]

that was because im trying to prove to you

im sorry for the confusion

[pastel totem]

wait i understand that the next part i split, i just was confused if i missed a part inbetween or not

but when you said next you split

you had 0’s and you go by definition. did you go by definition because its 1?

or because you always automatically follow definition before splitting

[proper nova]

again, the splitting is because im trying to explain why this works

but if it's multiple choice like this, you can try to use shortcuts like these

[pastel totem]

but your supposed to split regardless

so are using shortcuts?

[proper nova]

you can use shortcuts like what i said above

[pastel totem]

okok i see why im confused

can you show me normal way

i dont wanna learn shortcuts if i dont have a basic understanding of it in general

[proper nova]

split into 2 cases: $2x - 5 \geq 0$ and $2x - 5 < 0$

[jolly parrot]

1 divided by 0 equals Infinity

[pastel totem]

bro😭

[proper nova]

that's the normal way

😭

[pastel totem]

im stil confused abt the 0’s

u never answered

my question abt ot

[proper nova]

you can actually split into anything

for example: $2x - 5 \geq 1$

[jolly parrot]

1 divided by 0 equals Infinity

[pastel totem]

do i use 0’s because its 1 or because you automatically change then into 0

[proper nova]

and $2x - 5 < 1$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

but if you use $2x - 5 < 1$, it turns really confusing

[jolly parrot]

1 divided by 0 equals Infinity

[rough forge]

When we say positive, then something is greater than 0, else smaller

[proper nova]

so i use $2x - 5 \geq 0$ and $2x - 5 < 0$, split by definition is an easier approach

[jolly parrot]

1 divided by 0 equals Infinity

[pastel totem]

so its 0’s because you just wanted to use them???

[pastel totem]

yes ik

that doesn’t answer my question tho😭

[proper nova]

yeah + it's easier to handle than splitting into other numbers

[pastel totem]

but what abt the 1

how r you able to just get rid of ut

[proper nova]

.

[proper nova]

that's when the definition of absolute values come in

if $2x - 5 \geq 0$ then $\abs{2x - 5}$ is just $2x - 5$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

boom, first inequality $2x - 5 > 1$ which you can solve it yourself

[jolly parrot]

1 divided by 0 equals Infinity

[pastel totem]

is that basically not what it just was😭

[rough forge]

You should tell her that you subtracted 3 on both sides

after resolving the absolute values

[pastel totem]

how is it easier now when all you did was change < to >

[pastel totem]

ik that part but that is skipping parts

[proper nova]

no it isn't, because this one is $2x - 5 > 1$ while the question asks for $\abs{2x - 5} > 1$

[jolly parrot]

1 divided by 0 equals Infinity

[rough forge]

Skipping parts?

[proper nova]

the shortcut

[pastel totem]

why did i need to do all that 0 stuff when all u did was take away the absolute value

[proper nova]

this is what she meant

[proper nova]

?

[proper nova]

hold on, we also have 1 case left

[rough forge]

They aren't really shortcuts,,,

[rough forge]

Sorry for interrupting, I saw you had a help channel, so I thought you'd be kinda stressed out

[proper nova]

im fine with that

[pastel totem]

so i have |2x-5| > 1 and i split into 2x-5>1 and 2x-5<-1 right

then just -5 both sides?

[proper nova]

ill let you take over the channel

[rough forge]

the and is wrong

you get 2x-5>1 or 2x-5<-1

[pastel totem]

yeah

okay

but the rest is right?

also how do ik if its and or or

[rough forge]

You'd add 5 on both sides, and ig solve the inequalities

[pastel totem]

BRUH

[proper nova]

lord

🙏

[pastel totem]

DOBT SAY LORD

O LIT ASKED U LIKE SO MANY TIEMS

[rough forge]

Under the constraint 2x-5 >= 0 which means x >= 5/2 you try to solve 2x-5 > 1, see which solutions satisfy the constraint

[pastel totem]

okok wait ik what to do

how do ik if its and or or

[proper nova]

<@&268886789983436800> user id: 423444344037703680

[bronze heath]

🤦

[proper nova]

2x - 5 > 1 and 2x - 5 < -1 seems impossible

because literally no number is > 1 and < -1

so we use or in this case

[rough forge]

|x| > c means x > c or x < -c where as |x| < c means -c < x < c which means -c < x and x < c

[proper nova]

best way to know if it's an or or an and is to visualize on a number line

[pastel totem]

ohhh okok

in college do i rlly need to know that tho

pls say no

[rough forge]

in college?

[pastel totem]

yes

[rough forge]

especially in college

yes

[pastel totem]

do i rlly need to lnow and or pr

aUghh

Bro

[bronze heath]

Yes

Unless u dropout

[rough forge]

you probably will encounter many other logical operators as well

[pastel totem]

yeah whatever that means…

[proper nova]

||and or not nor xor xnor nand||

basically all of them

[rough forge]

nand

[pastel totem]

well ty thooo

.close

[pearl pond]

Channel closed

Closed by @pastel totem

Use .reopen if this was a mistake.

[rough forge]

[pearl pond]

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.

[ebon skiff]

hey! combinatorics problem here.

[ebon skiff]

Let me put it in google translate

Determine the number of ways to distribute 65 students taking advanced mathematics into three courses, two of which consist of 22 students and one of which consists of 21 students.

[rough forge]

@cinder flower

[ebon skiff]

Oh no

[cinder flower]

hello

[ebon skiff]

Hi

[frank violet]

<@&268886789983436800>

-# wtf

[ebon skiff]

So im thinking

That

[unborn abyss]

way worse today than ever i've seen

[ebon skiff]

We have 3 cases.

21, 22, 22
22, 21, 22
22, 22, 21

[cinder flower]

wtf

[ebon skiff]

Layla wtf is happening

Lmao

[cinder flower]

did that guy get banned or do i need to ping mods still

[ebon skiff]

Hm

Idk

[cinder flower]

hm i don’t think that’s the idea here

[ebon skiff]

,rccw

[ebon skiff]

So basically i would add those 3 up

And thats it

[cinder flower]

or maybe it got lost in translation. but i would have expected the course student numbers to already be set

so like, this particular one has 21, this one has 22, this one has 22

not just any one of them can have 21 and the others need to be 22

[ebon skiff]

It just says

[cinder flower]

but anyway it only affects the answer by multiplying by 3 so we can do it either way

[ebon skiff]

2 of the courses have 22, and one has 2q

21

So im guessing the 21 course can be either of the 3

[cinder flower]

it’s not totally clear but ok

[ebon skiff]

If it said:
The first one has 22, the second one has 22, and the third one has 21

Then

I'd have left out the *3

[cinder flower]

this is almost right

[cinder flower]

ok, we can do it your way that’s fine

[ebon skiff]

Ok

[cinder flower]

you probably know better than me, understanding the original text

[ebon skiff]

Well maybe im wrong

I mean

I trust your intuition

[lilac quartz]

viel glück

[cinder flower]

actually with your way this looks right

[ebon skiff]

Danke 👍

[cinder flower]

and actually the difference in the two interpretations is different than a factor of 3

[ebon skiff]

Right

Wait

You mean "different by a factor of 3"

[cinder flower]

with your way you are right

i meant with my proposed interpretation, the answer would not just be 3x more

[ebon skiff]

Hm

What would be your way

That the courses are fixed

?

The order

[cinder flower]

yea

[ebon skiff]

22,22,21

Or what

[cinder flower]

sure

i don’t know if that’s the correct interpretation though

[ebon skiff]

Kk

So why did you say my solution was almost correct?

Oh you said its correct

I saw now

[cinder flower]

if the courses are distinguishable and any two of them can have 22 students, your answer is correct

i cannot confirm 100% that is the intended interpretation

[ebon skiff]

okok

[ebon skiff]

well, this interpretation makes the most sense to me

so i'll take it

thank you layla

[cinder flower]

i do not feel very helpful but no problemo

[ebon skiff]

bro stop stalking the channel

[ebon skiff]

[cinder flower]

i guess it would be actually since the courses are (i presume) distinguishable

[ebon skiff]

well

[hollow cosmos]

halo

[ebon skiff]

here we go...

[cinder flower]

you are so pretty

[ebon skiff]

fr

and that origami

magnificent

Wait why is 43 choose 22 = 43 choose 21

But 65 choose 22 != 65 choose 21

huh where'd my messages go 😭

nah bro this has to be a simulation or sum

[dry lagoon]

chose 21 in 65
chose 22 in 44
and its just it

[ebon skiff]

but i could also choose 22 in 65 and 22 in 43

[cinder flower]

say you have 43 blank marbles. you will paint 22 of them black and the rest white. how many ways to do that? you could choose 22 out 43 marbles to be black. that’s the same same as choosing 21 marbles to be white

[dry lagoon]

yeah and its the same result

[ebon skiff]

omg

[hollow cosmos]

any discussion room in this server guys?

[ebon skiff]

yes

[ebon skiff]

but why is this the case?

oh

yeah i see

cuz 22 + 21 is not 65

[cinder flower]

yea

[ebon skiff]

nice

[cinder flower]

there’s a saying that goes something like