help-33

[marsh citrus]

@obtuse sky Has your question been resolved?

[knotty trellis]

okay so whats the question?

[marsh citrus]

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[cunning oak]

how do i prove pascal's theorem using menelaus?

[sweet sparrow]

try to prove that (GZ/GX)(IX/IY)(HY/HZ)=1

@cunning oak

use Melenaus repeatedly in triangle XYZ and the lines that cut through it

[marsh citrus]

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[marsh citrus]

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[limpid zealot]

need help in likelihood ratio

[limpid zealot]

why there are two different formula here

i thought likelihood is produkt of the funktions under null hypothese divided byprodukt of the funktions under alternativ hypothese

[leaden monolith]

They probably mean the same thing

[marsh citrus]

@limpid zealot Has your question been resolved?

[limpid zealot]

How?

Is first one is when n 0?

[leaden monolith]

I presume somewhere in your text it will define what these symbols means

They are probably defined to be the same thing

[limpid zealot]

Seems like i skipped the part.

But isnt likelihood is product of the functions on all xi s

[iron marlin]

Thats why check the definition

[limpid zealot]

.close.

.close

[marsh citrus]

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[past thicket]

Hi!

What are you stuck on?

[boreal rock]

Where to start?

[frank rose]

start with smaller cases

smaller observations

like three matches

instead of '5'

and observe patterms

patterns*

[still temple]

what even is the question

theyre not asking it properly

[boreal rock]

A game ends when a player wins 5 matches. There are two players and a match can only end in win or loss. In how many ways can the game end?

[marsh citrus]

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[still temple]

Hi. I am trying to work part b

[still temple]

. What I got are these: 20/36 = 5/9 - at least 2 of the scores are greater than 4, 18/36 = 1/2) for the sum of the two scores being odd and 6/36 = 1/6 for two scores being equal

I was getting assistance earlier but I didn't complete it

I have a question for the solution

I was told this and I agree -->since the dice are independent, the probability of both being <= 4 is (2/3)x(2/3) is 4/9 ....2/3 because 1,2,3,4 are 4 options out of 6... so 4/6 simpliefied is 2/3

Then I believe I should subtract it from 1

My question is, why can't I just do 2/6 * 2/6 --> Seeing you only can get a 5 or a 6?

[small berry]

It's a little hard to follow what you write out

[still temple]

This part or everything?

[small berry]

First part

Independence is an iff statement: P(A) * P(B) = P(A and B). You can just evaluate both sides.

[small berry]

Which two events are you considering to possibly be independent here? M and W?

[still temple]

F and M

I didn't realize I didn't mention that, sorry

[small berry]

What's P(F)?

[still temple]

18/36 = 1/2

P(M) = 20/36 = 5/9

I get to multiply it and I get 5/18

but how do I know the intersection?

P(F n M)

Do I have to count it?

[small berry]

I would count it, I don't see any shortcut here. Of the 20 outcomes of M, how many are also F

[still temple]

Ah, I didn't count it

[small berry]

I mean there's some amount of hand-wavey symmetry here, but I'm not sure if it can be easily shown rigorously (and tbh trying to do so would probably take more time than couting 20 events)

[still temple]

Yeahh

(1,6), (2,5), (3,6), (4,5), (5,6)

I see why I wasn't getting it

[small berry]

And (6,1), (5,2), ... right?

[still temple]

Oh, so even though the numbers repeat I should still count them

like 1,6 and 6,1

[small berry]

Yeah, don't you count them in the 20 events where at least one of the dice are over 4?

[still temple]

Yeahh

because I realize that the numerator is 5 but I forgot that the answer was cancelled for it to be 5

5/18

let me continue

[small berry]

yeah, 10/36 is what reduced to that

[still temple]

(1,6), (2,5), (3,6), (4,5), (5,6), (5,4), (6,3), (5,2), (6,1)

(6,5)

[small berry]

Yup those are the 10 events where one die is over 4 and where the sum is odd

Of the 36 possible outcomes of two die rolls, P(F and M) = 10/36 = 5/18

[still temple]

Yayyy!!

I have one more question

[still temple]

This part

I don't understand the easier method to get the20/36 without counting

P(M)

at least one of the two scores is greater than 4

I think I got the 1 - (4/6 * 4/6)

but what if I was focusing on the 2 outcomes for it to be more than 4

Why can't it be 2/3 * 2/3

sORRY

2/6 * 2/6

[small berry]

The outcome of two dice rolls are independent, we can agree on that right? One die being above/below 4 has no bearing on whether or not the second die is above 4

[still temple]

Yes, independent

[small berry]

Our "winning" event is both being above 4. Let D1 be the probability that the first die is above 4 and D2 be the probability that the second is above 4. P(D1 or D2) = -P(D1 and D2) + P(D1) + P(D2) = -(1/3)^2 + 1/3 + 1/3 = 5/9

Recall inclusion/exclusion

[still temple]

Ok, let me read through

[small berry]

This is the same thing, but you find the complement of winning (losing) and subtract 1 - P(losing). Feel free to do this as an exercise to confirm you get the same result if you want.

[still temple]

Alrighty, sure

I have to run now. I should be back in a few, hopefully because I have more questions to do. I got to go and come back

Thank you so much

.close

[marsh citrus]

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[small berry]

np

[marsh citrus]

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[strong dagger]

are they using the fact that the cyclic group generated by a is the smaller subgroup containing a in this proof to say <a^k> is a subset of <a^d> and the converse?

[strong dagger]

For example if a^d = (a^k)^t then since <a^d> is the smallest subgroup containing a^d and <a^k> also contains d, that means <a^k> contains <a^d>

Im just wondering if im reasoning about this right because they dont really explicitly say why its a subset of the the other unles im missing something

[tight furnace]

Sure, that works

you can kind of have 2 definitions for the subgroup generated by S, one of them is this, and the other is ${s_1s_2\ldots *s_n|s_1,\ldots,s_n \in S}$

[elfin berry]

Dreyuk

[tight furnace]

(for cyclic groups this can be made a bit simpler lol)

[marsh citrus]

@strong dagger Has your question been resolved?

[marsh citrus]

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[strong dagger]

Having trouble understand this second part

[strong dagger]

What tells us |a^d| = n/d? How do we know that this is the smallest possible intger such that a^d = e?

also I think theres a typo in the last sentence, do I ignore the expression before the last equal sign

oh wait

so they say that implies the |a^d| <= n/d

so I assume the next sentence shows that |a^d| >= n/d but I dont see it yet so ill try taking another look

ok I see

because the order of a is n

and i is less than n/d

so theres no ways to multiply n/d by i such that n/d = n

[marsh citrus]

@strong dagger Has your question been resolved?

[marsh citrus]

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[elder crag]

i dont really understand the solution given for b, ii
(the cards are selected with replacement)

[elder crag]

Why are we doing 4 * 4 * (1/4)^3

[stoic saddle]

i think one of the fours is extraneous

they meant only 4 * (1/4)^3

this in fact lines up with the stated answer of 1/16

[elder crag]

oh okay lol i was so confused

[stoic saddle]

typo, really

[elder crag]

and 4 * 1/4^3 is just the probability for SSD * 4

since they are all equally likely

[stoic saddle]

yes

[elder crag]

okay!, in this case the fact that its a geometric distribution is useless correct?

like for this specific question atleast

this fact is just useful for the other questions (A and B i)

thanks ann (again :p) :}

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[marsh citrus]

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[stoic saddle]

X is not even involved here

[elder crag]

ah yea, brain is kind of turned off after a full day of doing this i guess lol

[marsh citrus]

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[zealous plume]

how do i do this

[trim owl]

What have u tried

[zealous plume]

tried using synthetic but saw that there's 2 diff variables so dk what to do next

[marsh citrus]

@zealous plume Has your question been resolved?

[frank rose]

!showwork

[marsh citrus]

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

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[shy mica]

I don't know how to start

[trim owl]

Do u know the formulas of S.I and C.I

[shy mica]

yes

PRT/100

P(1 + r)^t

C.I.

[trim owl]

Alr now express them in ratio form

Also note that the r/100 represents percentage
That is if u r given r= 2% ,in place of r u will need to place 2

And in case of C.I ,if r is 2 %
Then in place of r u need to write 0.02

So it's better to express the r of S.I as r itself

$\frac{P(1+r)^n}{Prn}=\frac{169}{144}$

[elfin berry]

Victimizer

[trim owl]

Amd the question states that the value of P is equal

So u can cancel both out

[shy mica]

okay

and its for 2 years so n = 2

?

it will become sq. root. to other side?

[trim owl]

Yes

[trim owl]

Cross multiply first

[trim owl]

There is a mistake

Compound interest is P(1+r)^n -P

So it should be
$\frac{P(1+r)^n-P}{Prn}=\frac{169}{144}$

[elfin berry]

Victimizer

[trim owl]

And then
$\frac{P((1+r)^{2}-1)}{Pr2}=\frac{169}{144}$

[elfin berry]

Victimizer

[trim owl]

$\frac{((1+r)^{2}-1)}{2r}=\frac{169}{144}$

[elfin berry]

Victimizer

[trim owl]

$\frac{(1+2r+r^{2}-1)}{2r}=\frac{169}{144}$

[elfin berry]

Victimizer

[trim owl]

$\frac{2r+r^{2}}{2r}=\frac{169}{144}$

[elfin berry]

Victimizer

[trim owl]

$\frac{r(2+r)}{2r}=\frac{169}{144}$

[elfin berry]

Victimizer

[trim owl]

$\frac{(2+r)}{2}=\frac{169}{144}$

[elfin berry]

Victimizer

[trim owl]

U think u can solve from here? @shy mica

[shy mica]

it becomes little lengthy

[trim owl]

But solvable

[shy mica]

.close

[marsh citrus]

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[marsh citrus]

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[void brook]

can someone please explain how mutual exclusion does not work here? since P1 has already set the lock variable as 1 (even before pre-emption), how is P2 able to enter the critical section?

[elfin cairn]

it seems to me like just P1 is paused just before updating the lock and after checking if the lock is valid

so P2 can also enter the lock in the small amount of time the lock is still not locked

[void brook]

p1 gets preempted after changing the value of lock variable from 0 to 1, before entering the critical section

[elfin cairn]

I see a meanwhile

[void brook]

no, wait, it enters the critical section and then gets preempted

but it never sets the value back to 0, no?

[elfin cairn]

but its preempted before updating the lock to 1

thats what I think they meant

what context is this paragraph?

[void brook]

eh, how is it entering the critical section though, then?

[void brook]

lock variable in process synchronization

[copper raven]

that 'meanwhile' is doing a lot of work here

[elfin cairn]

yes

[copper raven]

it just sounds poorly written

[elfin cairn]

I don't see any other interpertation that makes sense

[copper raven]

yes indeed

[elfin cairn]

other then that the locking and validating the lock is not done atomically

[void brook]

i do not quite follow

[elfin cairn]

to prevent this, we would want to be able to check if the lock is unlocked and lock it at the same time

[void brook]

okay, thank you.

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[marsh citrus]

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[elfin cairn]

this is usually done by the CPU btw

just some instruction

[void brook]

.reopen

[marsh citrus]

✅

[void brook]

i think what i do not get is

say p1 wants to get executed

at beginning, the lock is set to 0

so it does not enter the while loop and then sets the lock to 1

and enters the critical section

but then, it gets preempted

now, p2 wants to start its execution

the paragraph says that it will be able to enter the critical section too because the lock is set to 0, but, that is not true? because p1 never updated the value?

[elfin cairn]

like I said

I think they meant that p2 is trying to enter the critical section just before p1 updates the lock to 1

and that aquiring the lock is not atomic

and it should be atomic

[ruby mulch]

P1: While (lock != 0)
P2: While (lock != 0)
P2: Lock = 1
P1: Lock = 1
P2: //Critical Section
P1: //Critical Section

Whoops

[void brook]

well, that makes sense

[void brook]

this probably sounds stupid but, is that not done before a process enters the critical section? how is that relevant here?

[elfin cairn]

I meant the same as Xwtek

!occupied

[marsh citrus]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[void brook]

i see. so if i understand this correctly, the main problem with this algorithm is that both p1 and p2 can see the value of lock variable (0) at the same time and hence both can enter the critical section simultaneously?

[ruby mulch]

Yup.

[elfin cairn]

yes

[void brook]

okay, thank you so much!

.close

[marsh citrus]

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[glad elbow]

Confused on if I can just equal it to x-1/x-1 or how to split it into a piece wise function

[late geode]

|x| = x
for positive x

[glad elbow]

Yeah, and both left and right hand limits are positive

So can I just use positive c

X*

[late geode]

yeh

[glad elbow]

Ok but then I don’t rly know what to do

Since I have 0/0

[late geode]

well you can just cancel

the x-1

[glad elbow]

Uh

Ok I guess

That seems weird tho

[late geode]

weird how?

[glad elbow]

Well if you plug in 1

Ugh math is just weird sometimes

[late geode]

but you don't really care about the value at x=1

you're more concerned about what happens around that

[glad elbow]

I see, so the value doesn’t exist

But the limit does

For showing my work, is this fine

[late geode]

cancel the whole x-1 ,
it looks like your cancellation is just on the x

[glad elbow]

Ok

Thanks

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[opal raptor]

I'm having a bit of trouble trying to figure out what boxes to check off! If someone could help guide me through it that would be great.

[opal raptor]

The question for it

[slate rampart]

What have you tried?

If anything

[opal raptor]

Well, I didn't try anything yet because the question was a bit complex for me to understand

[slate rampart]

Did your teacher teach inverse functions yet?

[opal raptor]

Yes she did

I have notws

Notes

[slate rampart]

Alright

Ok so look at what takes you from f^-1(x) to x

What rule do you apply?

[opal raptor]

We switch x and y right

[slate rampart]

Huh

No

Ignore that for now

[opal raptor]

Okay

[slate rampart]

What is the simple rule in the table

Just look at it and see the pattern

Between f^-1(x) and x

[opal raptor]

Oh wait

Okay so from 0.25

Inverse Operations

[slate rampart]

Actually

This is probably more efficient

Since we already have a table from x to f(x)

If you wanted to find it without the x->f(x) table it would require using the inverse

[opal raptor]

Ohh okay

Let me try something

[marsh citrus]

@opal raptor Has your question been resolved?

[marsh citrus]

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[noble widget]

i need help understanding how to do this

[trim owl]

Can u find the line y=-1 first

[noble widget]

yeah

i think

[trim owl]

Mark it up

[noble widget]

what

[trim owl]

The line y=-1

[noble widget]

im doing this on paper

i dont have it printed

and im in computer

wait

[trim owl]

This is y=-1,do we agree

[noble widget]

yse

[trim owl]

When a point is reflected over a line , the distance between its coordinate and the line & the distance between its reflection point and the line is equal

[noble widget]

ah

i understand it now

[trim owl]

Blue point is B'

Green is C'

Purple is A'

[noble widget]

ok

.close

[marsh citrus]

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[celest stream]

Pardon me, does anyone here do statistic math?

[celest stream]

I am learning the Understand principles of Chebyshev's Theorem

Basically I dont understand the theorem at all

(please ping me when someone is available)

[wary lichen]

I can just give you the general concept, im not that versed on statistics

[celest stream]

Um yes please yeah

What exactly is it?

And if so do you know someone that is versed on statistics

[wary lichen]

Ill assume you know what the normal distribution is

If so, you probably can remember that its usually taught that ~68% of the data is within a standard deviation of the mean.

~95% for 2 deviations
~99.3% for 3 deviations

Well, the Theorem states a similar concept but for any dataset. Specifically, it sets a lower bound for the data found based on the amount of "standard deviations" you distance yourself from the mean.

[marsh citrus]

@celest stream Has your question been resolved?

[celest stream]

Very interesting

tysm

[marsh citrus]

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[tropic berry]

guys srry for ask easy math problems but what is the result from this 6 + 6/7

[twilit prairie]

rewrite 6 with the denominator being 7 first

then simply add the numerators and your done

[tropic berry]

alr

48/7?

if u wana see all the problem there is it

[last locust]

Yooo can someone help me ?

[azure ivy]

.close

[marsh citrus]

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[still temple]

need help

[elfin stone]

yeah

[still temple]

2. im having trouble understanding how and why our components turned into equations like that, even though they are vector components

it just doesnt make sense to me how all of sudden those components because equations to graph a line

[harsh scroll]

Direction vector $\vec{v}$ is parallel to $\overrightarrow{PP_{1}}$ but there's nothing to ensure they're equal but we do know we can find a number t that when we do scalar product we get $t\vec{v}=\overrightarrow{PP_{1}}$

[elfin berry]

Alexis_Fx

[still temple]

right i agree

so you are saying a constant t will scale the direction vector to make it equal to our line

but i just dont understand why and how that magically turns into equations

that graph our line

i need help understanding how the vector components went from vector components to equations of a line

[harsh scroll]

v<a,b,x>

[still temple]

yea

thats a direction vector with direction numbers

[harsh scroll]

Sorry

Igtg

[still temple]

uts okay

its okay

[still temple]

@elfin stone

[quiet anvil]

Please don't ping individual helpers unless you have their explicit permission to.

[still temple]

ok

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[tardy narwhal]

Create a two-column proof.

[tardy narwhal]

<@&286206848099549185>

[harsh scroll]

!15m

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[harsh scroll]

Chill

[tardy narwhal]

sorry

[zenith pasture]

i think that's just the definition of trisecting an angle

[harsh scroll]

What's the definition of trisect that you've learnt

[tardy narwhal]

"2 points (or segments, rays, or lines) that divide a segment into 3 congruent segments bisect the segment. The 2 points at which the segment is divided are called the trisection points."

[tardy narwhal]

to reword the question its saying given angle aeb which is congruent to angle bec which is congruent to angle ced, prove that rays eb and ec trisect angle aed using a two-column proof

[still temple]

whats cis(x)

[zenith pasture]

it's a way of writing complex numbers

[viral cedar]

thats sooo trivial bruh

[still temple]

how is it pronounces

pronounce*

bruh

[zenith pasture]

s - i - s

i think

[still temple]

as in sister

interesting

[viral cedar]

ok pls move to chill, you two

[tardy narwhal]

its that "great part of geometry" according to my teacher

[viral cedar]

wdym

[tardy narwhal]

if we aren't spamming two column proofs, we are spamming paragraph proofs

[tardy narwhal]

shes obsessed with two-column proofs

[viral cedar]

ok lol

its a great type of proof fr

[indigo nest]

So at what point are we planning to start helping

[indigo nest]

Okay genuinely, what even is there to prove here

[still temple]

if this channel is not occupied anymore lemme know

[tardy narwhal]

idk 😭

[indigo nest]

Just get a different channel

[tardy narwhal]

ok its fine i will just ask in class tomorrow

you can close it

[indigo nest]

!done

[marsh citrus]

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

[still temple]

say less

[indigo nest]

?

[still temple]

as in i have the channel now

[tardy narwhal]

i hate geo 💔

[indigo nest]

.close please

[still temple]

facts

[tardy narwhal]

wait how do i close it again

[still temple]

type .close

[tardy narwhal]

.close

[marsh citrus]

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[still temple]

👍

[marsh citrus]

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[finite fjord]

There are 4 marbles A,B,C,D,we have to arrange it in a order such that A and B shd not be beside each other,ik the answer is 12 and it is 4! - 3!
4! is the total combination,what is 3!\

[elfin cairn]

4! - 3! = 18

[finite fjord]

yea u are right

so how to solve this

[zenith pasture]

do you know complementry counting

[finite fjord]

ik combinotorics,no

[elfin cairn]

How many options are there for ordering the marbles without the restriction?

[zenith pasture]

basicly its where you take the total outcomes and subtract the ones that aren't the right ones

[finite fjord]

4x3x2x1

[elfin cairn]

yes

[finite fjord]

ic

[elfin cairn]

Now how much did you overcount?

[finite fjord]

how do i find it?

[elfin cairn]

hmm

[finite fjord]

please explain,i will get the idea of solving these type of questions

[elfin cairn]

yeah I am tired give me one sec

[finite fjord]

ok

[elfin cairn]

there is a better way

[finite fjord]

which is

[elfin cairn]

since you want A,B to be together we can think of them as one marble

and then order the C,D and AB

[blissful olive]

3! is the cases where A,B are together

[elfin cairn]

and then order AB internally

[finite fjord]

is 4! -3! right answer tho,isnt it 12

[finite fjord]

so we get 3!?

[elfin cairn]

oh its should not be beside each other

[elfin cairn]

3! * 2!

[blissful olive]

its 18

[finite fjord]

why so

[blissful olive]

You would need to arrange A,B too

A,B and B,A

That is the 2!

So 4!-3!*2!=12

[solar crown]

There is another way to do this idk if you guys wanna listen

[finite fjord]

so we have two ways to arrange one is AB,and the other is BA,for each case there is 6 wrong,so it 6x2=12, 24 -12 = 12

thanks

[finite fjord]

sure

[solar crown]

Ok we know without restrictions is 4! Right

[finite fjord]

ye

[solar crown]

Our teacher told us the method to use P

So mine is

4! - (3P2 * 2!)

Mb

[finite fjord]

did not get it

what is Method P?

[solar crown]

Here

[solar crown]

Permutations

Mb

[finite fjord]

isnt it the same thing?

4! - 3!*2!

[blissful olive]

He's talking about P(n,r)

[solar crown]

3P2 is not the same as 3!

[finite fjord]

ohh,ncr

Here we call it as ncr,mb

[blissful olive]

No, not C(n,r) that's a different thing...

[solar crown]

I use nPr and nCr

[finite fjord]

what does nPr do?

[blissful olive]

The relation between C(n,r) and P(n,r) is:
C(n,r) * r!= P(n,r)

[blissful olive]

Number of premutations

[finite fjord]

ic thanks all

.close

[marsh citrus]

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[finite fjord]

.reopen

[marsh citrus]

✅

[finite fjord]

.close

[marsh citrus]

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[marsh citrus]

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[glossy yoke]

for calc 3, no clue where to start

[vestal sandal]

well do you know how to find an orthogonal vector to w?

[glossy yoke]

cross product?

[vestal sandal]

you can either find some orthogonal vector and then solve a system, or easier would be to find the projection of v onto w

do you know how to find the projection of v onto w?

[glossy yoke]

no

[balmy shadow]

hmmm are you aware of dot products

[glossy yoke]

yes

[balmy shadow]

we use those to find projections

[glossy yoke]

what is a projection

[vestal sandal]

a projection is something youre familiar with, you just havent heard of the technical term probably

in this art work, Vx is the projection of V on the x-axis and Vy is the projection of V on the y-axis

the thought process of using projections is basically this: you find the projection of V onto w, which is by definition parallel to W, and then the difference between V and V(proj.w) is the perpendicular part

[glossy yoke]

okay so how do i find the projection

[vestal sandal]

the projection of vector A onto vector B is just a⋅b/b⋅b *b

so in your case, you can just do (v⋅w/w⋅w)*w

and youll find your projection

[glossy yoke]

ok

and then

V - (v . w/w . w)*w is the perpendicular one

[vestal sandal]

yep

[glossy yoke]

ok swag

hmmm

that equation isnt in my textbook for some reasonp

[marsh citrus]

@glossy yoke Has your question been resolved?

[glossy yoke]

.close

[marsh citrus]

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[marsh citrus]

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[scarlet ingot]

Trying to prove a dual basis defined $\beta^={\alpha^1,...,\alpha^n}$ is a basis of $V^$ over $\mathbb{F}$, but I'm stuck on span. I've got that $\forall\phi\in V^*$ we need some $c_1, ..., c_n\in\mathbb{F}$ so that $\phi(v)=\sum_{i=1}^{n}c_i\cdot\alpha^i(v)$ and that $\alpha^i(v)=d_i$ for any $v=\sum_{i=1}^n d_i e_i\in V$, but after/apart from that im stuck. any guidance wld be nice

[elfin berry]

Syrenate

[scarlet ingot]

oh idk if any of this is standard but $\alpha^i$ is defined by $\alpha^i(e_j)=\delta_j^i$ where $\beta={e_1,...,e_n}$ is a basis of $V$.

[elfin berry]

Syrenate

[humble nebula]

could you send the original question?

[humble nebula]

thanks

[humble nebula]

to obtain the various $c_i$, evaluate $\phi$ at each $e_i$

[elfin berry]

soup_norm

[scarlet ingot]

why?

[humble nebula]

for now, try it and see what happens

[scarlet ingot]

assuming $\beta^*$ is a basis?

[elfin berry]

Syrenate

[scarlet ingot]

dont see how else I can

[humble nebula]

wait let me think of a better hint

[humble nebula]

show that for all $v \in V$,
[ \phi(v) = \sum_{i = 1}^n \phi(e_i) \cdot \alpha^i(v) ]

[elfin berry]

soup_norm

[humble nebula]

and this will suffice

[scarlet ingot]

i can see how that suffices because $\phi(e_i)$ is well defined in $\mathbb{F}$ but i really cant see how youd get to that form

[humble nebula]

You can express v in terms of the basis of V

[elfin berry]

Syrenate

[scarlet ingot]

sure, $v=\sum_{i=1}^n c_i e_i$, gives that $\alpha^i(v)=\sum_{j=1}^n c_j\cdot\delta_j^i = c_i$

[elfin berry]

Syrenate

[humble nebula]

Yes

(I’m going to not elaborate until you ask because you might be able to figure it out from here)

[scarlet ingot]

i already had that result before i came here could u pls elaborate

[humble nebula]

Ok

[humble nebula]

Consider: $\phi(v) = \phi\left(\sum_{i=1}^n c_i e_i\right)$, then expand by linearity and use the equation you sent

[elfin berry]

soup_norm

[scarlet ingot]

ohhh damn

ok i got u

thanks sm

first abstract algebra class

https://tenor.com/view/meme-bcs-better-call-saul-tv-watching-gif-9734752830431950225

but ty

[humble nebula]

linear and abstract?

[scarlet ingot]

i mean its using linear algebra

[humble nebula]

oh

[scarlet ingot]

itd better be anyway

better not be studying this for no reason

back to it!

.close

[marsh citrus]

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[marsh citrus]

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[odd rivet]

help me pls what is cosa ?

[odd rivet]

@everyone

[past thicket]

that doesn’t work bud

Also, even if it did

you wait fifteen minutes before pinging helpers

[odd rivet]

okay sorry

I am new at this server

[stoic saddle]

goes to server with 300k+ members
tries to PING EVERYONE

[past thicket]

i understand

[stoic saddle]

anyway have you tried working out some other angles here in terms of alpha

[odd rivet]

I tried

[marble gazelle]

haha

[odd rivet]

ı found [bd] to 2√6

but it doesnt work

also

cos a = 5/7 but how

[late geode]

there's no real need to determine unmarked lengths

did you determine any other angles in terms of alpha as Ann asked?

[stoic saddle]

what's angle DAC
and then what's angle DAB

work them out in this order

[marsh citrus]

@odd rivet Has your question been resolved?

[astral bough]

guys

just use 1/AD^2=1/AB^2+1/AC^2

[marsh citrus]

@odd rivet Has your question been resolved?

[marsh citrus]

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[wheat egret]

Hi can someone please tell me how I can solve this? I can't get past trying to simplify the expression 😭
*more context, I'm using u-substitution to solve it, letting u=tan(x)

[fringe nacelle]

well.first

you xan simplify the numerator, right

i mean

denominator

[wheat egret]

Yeah I got around to making (sec^2.. ) to 1

[fringe nacelle]

okay cool

[wheat egret]

i tried to make tan(2x) to become tan(x) as well using the trig identity

[fringe nacelle]

what i like doing is looking at a table of trig identities

lets do that

ohay you did

https://proofwiki.org/wiki/Trigonometric_Identities for reference though

tan(2t)=(2 tan t)/(1-tan²t)

what happened when you tried that

[wheat egret]

the equation got super big unfortunately 😭

[fringe nacelle]

also the way the formula is framed is dumb

what if i wrote F(t)+5 + C

its not unique

anyway

try it but writing "t" instead if of "tan t" to make it less messy

[wheat egret]

right

[fringe nacelle]

so we get

hold on this

needs a second

[wheat egret]

take ur time dw

[fringe nacelle]

$$\int \frac{-t + \frac{2t}{1-t^2}}{-\frac{2t}{1-t^2}t-1} , \mathrm dx$$

[elfin berry]

gfauxpas

[fringe nacelle]

okay this isn't so bad

how do we simplify this, any ideas?

multiply top and bottom by ...

[wheat egret]

take out t on top

[fringe nacelle]

yes sure

$$\int \frac{-1 + \frac{2}{1-t^2}}{-\frac{2}{1-t^2}t-1} , t ,\mathrm dx$$

[elfin berry]

gfauxpas

[wheat egret]

hmmm but then I can't seem to cancel the terms after doing so since the signs are different 😭

[fringe nacelle]

when you have a nested rational function like

[storm plume]

Why not simplify the numerator and denominator by giving common denominators to both terms on top and bottom and then cancelling the similar denominators

[fringe nacelle]

thats what i was trying to say, what Artemis was saying, he typed it out better and faster than I

[wheat egret]

OH wait

I think i get it

let me try it again

[fringe nacelle]

dont start over, you did what you did

multiply top and bottom by something clever

[wheat egret]

right

[fringe nacelle]

which is? take a look at numerator and denominator seperately

there's something you can multiply numerator and denominator by simultaneously

to simplify

[storm plume]

(-1 +t^2+2/1-t^2)/(-2-1+t^2/1-t^2)

[wheat egret]

yeah like this

[fringe nacelle]

PEMDAS my friend

[storm plume]

you'll get this

then cancel 1-t^2

[fringe nacelle]

take a look

[wheat egret]

I think at one point i got the same thing

[fringe nacelle]

what happens if you multiply top and bottom by 1-t^2

try it that way

[wheat egret]

alright

[storm plume]

to get 1+t^2/-3+t^2

I believe it to be correct

try it out

[fringe nacelle]

PEMDAS

[storm plume]

BODMAS*

[wheat egret]

(tan (x) +tan^3 (x))/(-tan^2(x)-1)

[lyric zodiac]

yeah now just take tan(x) common

in the numerator

[wheat egret]

yes i got it now XD

I realised where I went wrong initially

I didn't notice that I cancelled the terms and continued which resulted in a fat expression

thanks guys :D

[fringe nacelle]

btw

the correct way to make an intergral a function

is to set one bound, usually the lower bound, as a constant, and the upper bound as a variable

I already stated the problem with saying "let C = 0"

thats not your problem

im criticising the person who wrote the problem

[wheat egret]

yeah I get that

[storm plume]

=>(-1 +t^2+2/1-t^2)/(-2-1+t^2/1-t^2)
=>(1+t^2/1-t^2)/(-3+t^2/1-t^2)
=>(1+t^2)/(-3+t^2)
=>-(1+t^2/2+1+tan^2)
=>-(sec^2(x)/2+sec^2(x))

Guys I have my phy exam tmrw and Im stressed

[wheat egret]

Goodluck man I got calculus class 😭🙏

also last question

[wheat egret]

final answer 0?

[fringe nacelle]

you still have to integrate!

[wheat egret]

yeah after integration

so I got -tan(x) after simplifying allat

and then it is -ln|sec(x)| after integration

input 2pi

which is 0

since ln(1) = 0

right?

[fringe nacelle]

you wrote sec^2(x)/2 + sec^2(x) whcih is 3/2 sec^2(x) which has antiderivative 3/2 tan(x) so I suspect you didnt write parentheses correctly

[marsh citrus]

@wheat egret Has your question been resolved?

[wheat egret]

where did sec^2(x)/2 + sec^2(x) come from?

[fringe nacelle]

oh art3mis said that sorry, I thought you said it

my bad!

[wheat egret]

it's alright haha

so uh is it 0 or?...

[marsh citrus]

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[marsh citrus]

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[swift roost]

@thorny skiff Warum hast du mich von deinem Server verbannt?

[quiet anvil]

.close

[marsh citrus]

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[quiet anvil]

@swift roost off topic

Please take this to DMs.

[marsh citrus]

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[swift roost]

I apologize for the incident.

[calm harbor]

.close

[marsh citrus]

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[winter aspen]

How do i do this small angle approximation?

[jagged relic]

What's the approximation? cos(theta) = 1-theta ?

[robust igloo]

How small is theta? theta = -2747347234734? Or smaller?

[vapid adder]

i have also seen cos(θ)=1-θ^2/2

[winter aspen]

cos(theta) = 1 - 1/2(theta)squared

yeah hunk

[vapid adder]

just substitute that in to the left hand side

[jagged relic]

Right so replace cos(theta) by that, and cos²(theta) by (that)²

[zenith lantern]

and ignore any cubic/quartic terms

[winter aspen]

is there any @zenith lantern ?

ive got the answer but i don't get the third row where the 1/2 at the end turns into a 1/4 and there is also a -(theta) added into that same bracket at the end

[jagged relic]

There's a square sign

They just expanded

[winter aspen]

where dies the extra theta squared come from tho in the end bracket?

[jagged relic]

.

[winter aspen]

wdym

[winter aspen]

yeah i know why it becomes 1/4(theta)^4 now but whre does the other theta come from?

[jagged relic]

(a-b)² = ?

[winter aspen]

a²-2bc+b²

[jagged relic]

-2bc ??

[winter aspen]

(a-b)(a-b) right?

[jagged relic]

Yeah, where does that c come from?

[winter aspen]

oh yeah my baad

2ab

a²-2ab+b²

[jagged relic]

Right, so can you do that when a = 1 and b = 1/2 theta² ?

[winter aspen]

1-(theta)²+1/4(theta)^4

THANKS

[jagged relic]

You did it yourself

[winter aspen]

much appreciated!

[marsh citrus]

@winter aspen Has your question been resolved?

[marsh citrus]

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[buoyant jetty]

need help with this limit

[past thicket]

I’m assuming sen(𝑥) is the spanish translation of sin(𝑥)?

[fringe nacelle]

start by replacing y+1 with a new variable

note that y+1->0 is interchangeable with y->-1

[buoyant jetty]

i did a little progress

but i am not even sure if this thing is correct or no

[fringe nacelle]

you need to consider theta->0

if you use polar

[buoyant jetty]

theta is arbitrary

[fringe nacelle]

it is not

[buoyant jetty]

not necessarily theta -> 0

[fringe nacelle]

oh

insee

yeah okay

you allowed to use sin x ~ x for x near 0?

that deals with sin(w³), though not sin(1/v)

[buoyant jetty]

yes but use it as an upper bound for |sin(x)| < |x|

[fringe nacelle]

assuming the limit is 0

if not, mine is better

guess it doesnt matter

just how i would do it but youre not obligatdd to follow my style

[buoyant jetty]

can you elaborate?

what would be the upper bound

I forgot to put an absolute value to the first upper bound

my bad

[fringe nacelle]

its not bounding, its asymptotic equivalence

if lim x->0 f(x)/g(x)=1, then

you can replace f with g in a limit expression approaching 0

but your way is fine

hmm what if you took andolute values and replaced r->0 with 1/n ->0 as n-> +infinity

[buoyant jetty]

jajaja

maybeee

dude I was trying to use sandwich thoerem

[fringe nacelle]

idk if it would work but maybe try it

[fringe nacelle]

works at infinity

[buoyant jetty]

thats why the bounding

I am trying to prove this goes to 0

[buoyant jetty]

yeah

[fringe nacelle]

show me what it becomes if we change r to n

yes it works @buoyant jetty

but

in retrospect

so does the current form

[buoyant jetty]

im so fucked

exam is in a week or so dude

I dont want to say it but I am so fucked

[fringe nacelle]

but all your work is correct

you just didnt finish

use triangle inequality on |r sin t - 1|

<=|sin t - 1| (for |r| < 1 )

[marsh citrus]

@buoyant jetty Has your question been resolved?

[fringe nacelle]

squeeze theorem now

[buoyant jetty]

,align &\lim_{(x,y) \to (0,-1)} \frac{|y| \cdot \sin((y+1)^3) \cdot \sin\left(\frac{1}{x}\right)}{2x^2 + (y+1)^2} \ &\lim_{(x,y) \to (0,-1)} \left|\frac{|y| \cdot \sin((y+1)^3) \cdot \sin\left(\frac{1}{x}\right)}{2x^2 + (y+1)^2}\right| \ &\lim_{(x,y) \to (0,-1)} \frac{|y| \cdot |\sin((y+1)^3)| \cdot \left|\sin\left(\frac{1}{x}\right)\right|}{|2x^2 + (y+1)^2|} \ &\leq \lim_{(v,w) \to (0,0)} \frac{|w| \cdot |\sin(w^3)| \cdot \left|\sin\left(\frac{1}{v}\right)\right|}{|v^2 + w^2|} \ &= \lim_{r \to 0} \frac{|r\sin(\theta)| \cdot |\sin(r^3 \sin^3(\theta))| \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right|}{|r^2|}

[fringe nacelle]

f

[fringe nacelle]

ni dont know why you redid your old work though

it was correct before, wasnt it?

[buoyant jetty]

yes, but I erased it and had to re do it

It was like, in a crisis moment

[fringe nacelle]

:0

anyway, squeeze theorem time

[buoyant jetty]

,align &\lim_{(x,y) \to (0,-1)} \frac{|y| \cdot \sin((y+1)^3) \cdot \sin\left(\frac{1}{x}\right)}{2x^2 + (y+1)^2} \ &\lim_{(x,y) \to (0,-1)} \left|\frac{|y| \cdot \sin((y+1)^3) \cdot \sin\left(\frac{1}{x}\right)}{2x^2 + (y+1)^2}\right| \ &\lim_{(x,y) \to (0,-1)} \frac{|y| \cdot |\sin((y+1)^3)| \cdot \left|\sin\left(\frac{1}{x}\right)\right|}{|2x^2 + (y+1)^2|} \ &\leq \lim_{(v,w) \to (0,0)} \frac{|w| \cdot |\sin(w^3)| \cdot \left|\sin\left(\frac{1}{v}\right)\right|}{|v^2 + w^2|} \ &= \lim_{r \to 0} \frac{|r\sin(\theta)| \cdot |\sin(r^3 \sin^3(\theta))| \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right|}{|r^2|} \ &= \lim_{r \to 0} \frac{|r| \cdot |\sin(\theta)| \cdot |\sin(r^3 \sin^3(\theta))| \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right|}{|r|^2} \ &= \lim_{r \to 0} \frac{|\sin(\theta)| \cdot |\sin(r^3 \sin^3(\theta))| \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right|}{|r|} \ &= |\sin(\theta)| \cdot \lim_{r \to 0} \frac{|\sin(r^3\sin^3(\theta))|}{|r|^3 \cdot |\sin^3(\theta)|} \cdot |r|^2 \cdot |\sin^3(\theta)| \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right| \ &= |\sin(\theta)|^4 \cdot \lim_{r \to 0} |r|^2 \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right|

[fringe nacelle]

why did you write the whole thing in latex for

i mean, go ahead, why not

but we could read it

already

[elfin berry]

Renato

[buoyant jetty]

sorry

its just that the picture was kinda blurry because discord compresses images

[fringe nacelle]

it doesnt bother me, it just seems like a waste of time for you

[buoyant jetty]

https://tenor.com/view/nothing-to-do-nothing-but-time-i-have-time-nothing-else-to-do-i-got-nothing-gif-12334945

[buoyant jetty]

alright, now what?

[fringe nacelle]

for any function f

-1 <= sin(f) <= 1

[buoyant jetty]

do you agree with my work so far?

[fringe nacelle]

i checked the first version and it was right, i dont feel like checking this version lmao

[buoyant jetty]

wait but

how to squeeze theorem this?

[fringe nacelle]

agree?

[buoyant jetty]

ye, but we have |sin|

[fringe nacelle]

0 le |sin(f)| le 1

[buoyant jetty]

|sin| <= 1

[fringe nacelle]

I was editing

[buoyant jetty]

yeah

[fringe nacelle]

now multiply all three terms by anything > 0

[buoyant jetty]

anything like what

[fringe nacelle]

like |sin(t)|^4

[buoyant jetty]

dude but hear me out

[fringe nacelle]

👂

[buoyant jetty]

,align &\lim_{(x,y) \to (0,-1)} \frac{|y| \cdot \sin((y+1)^3) \cdot \sin\left(\frac{1}{x}\right)}{2x^2 + (y+1)^2} \ &\lim_{(x,y) \to (0,-1)} \left|\frac{|y| \cdot \sin((y+1)^3) \cdot \sin\left(\frac{1}{x}\right)}{2x^2 + (y+1)^2}\right| \ &\lim_{(x,y) \to (0,-1)} \frac{|y| \cdot |\sin((y+1)^3)| \cdot \left|\sin\left(\frac{1}{x}\right)\right|}{|2x^2 + (y+1)^2|} \ &\leq \lim_{(v,w) \to (0,0)} \frac{|w| \cdot |\sin(w^3)| \cdot \left|\sin\left(\frac{1}{v}\right)\right|}{|v^2 + w^2|} \ &= \lim_{r \to 0} \frac{|r\sin(\theta)| \cdot |\sin(r^3 \sin^3(\theta))| \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right|}{|r^2|} \ &= \lim_{r \to 0} \frac{|r| \cdot |\sin(\theta)| \cdot |\sin(r^3 \sin^3(\theta))| \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right|}{|r|^2} \ &= \lim_{r \to 0} \frac{|\sin(\theta)| \cdot |\sin(r^3 \sin^3(\theta))| \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right|}{|r|} \ &= |\sin(\theta)| \cdot \lim_{r \to 0} \frac{|\sin(r^3\sin^3(\theta))|}{|r|^3 \cdot |\sin^3(\theta)|} \cdot |r|^2 \cdot |\sin^3(\theta)| \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right| \ &= |\sin(\theta)|^4 \cdot \lim_{r \to 0} |r|^2 \cdot \left|\sin\left(\frac{1}{r\cos(\theta)}\right)\right| \ &\leq \lim_{r \to 0} |r|^2 \ &\lim_{r \to 0} -r^2 \leq \lim_{(x,y) \to (0,-1)} \frac{|y| \cdot \sin((y+1)^3) \cdot \sin\left(\frac{1}{x}\right)}{2x^2 + (y+1)^2} \leq \lim_{r \to 0} r^2

since both |sin(something)| <= 1

we bound it from above with this

[elfin berry]

Renato

[buoyant jetty]

this I mean

since both |sin(something)| <= 1

[fringe nacelle]

yeah sure

in fact if I were the professor I'd allow you to just say

<= lim_{r to 0} |r|^2 = 0

[buoyant jetty]

yeah

you think this proof is valid?

[marsh citrus]

@buoyant jetty Has your question been resolved?

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[marsh citrus]

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[untold spruce]

So they say that the Lebesgue-Stieltjes measure is the completion of the Borel measure mu_F but then how do they know that it must be that infinum thing they give? It seems there is a jump as isn’t it possible for the completion of the borel measure to be something else?

[main idol]

you can also try #advanced-analysis

[untold spruce]

Like someone told me it follows from defn of pre measure

Ok

[marsh citrus]

@untold spruce Has your question been resolved?

[marsh citrus]

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[novel juniper]

This would just be the number of ways to partition 8 elements into groups of 2, right

[elfin cairn]

yes

[novel juniper]

It's not 8!?

oops

I suppose it's $\binom{8}{2,2,2,2}$

[elfin berry]

wai

[lucid zenith]

yes

[novel juniper]

cool. Thanks!

Kind of unsure of part 1, do I have to brute force it?

[pallid tapir]

in proving that H is a subgroup?

[pallid tapir]

do multinomials take into account that the ordering of the cycles themselves don't matter?

[devout mauve]

yes

just like binomials

[pallid tapir]

well, choosing 1 & 2, then 3 & 4, ... should hopefully be the same as choosing 3 & 4, then 1 & 2, ...

[regal coral]

oh

[pallid tapir]

the binomials only need to care about the order of the things they're collecting in a singular bag, but multinomials only consider the ordering of the objects in each bag, not the bags themselves, right?

[devout mauve]

binomials take care of two bags

the objects that you choose and those that you dont

they are also multinomials

[pallid tapir]

yes, but in (4 choose 2), choosing 1 and 2 is different from choosing 3 and 4

[devout mauve]

oh wait

ah fuck

[regal coral]

yeah divide by 4! right?

[pallid tapir]

that's what I think, yeah

mod out permutations of the bags themselves, not just the ordering of the objects in the bags

[lucid zenith]

wait yeah fuck

but then why don't you do that when there's 2

[regal coral]

two 2-cycles?

so A_4?

[pallid tapir]

you still do that (dividing by 2!) when there's only two 2-cycles to consider

[regal coral]

yup there are 3 in A_4

[novel juniper]

sorry, I had gone to use the toilet

[novel juniper]

hmm?

[pallid tapir]

how many unique products of two disjoint 2-cycles can you make in S_n? choosing two for the first cycle, then choosing another two for the second cycle, yields (n choose 2)(n-2 choose 2) = (n choose 2,2). but wait, because you have commutativity between disjoint 2-cycles, so divide this count by 2

[novel juniper]

ah

[regal coral]

(12)(34) = (34)(12)

[novel juniper]

right, I did think of dividing by that, but it felt sus

[lucid zenith]

but you don't divide by 2! when you're just splitting 4 items into two groups of two

or do you

i think i should stop talking and not confuse wai

[regal coral]

depends if the groups are distinguishable

or... depends if the order matters

[novel juniper]

2-cycles are commutative

and their own inverses too

[regal coral]

disjoint 2-cycles are commutative

[novel juniper]

yea,my bad

Okay, I'll do this now

The 2nd part ( finding the number of elements) is simple

it's 3!( As where 1 and 3 go are fixed)

[pallid tapir]

in binomials, you don't because you only ever cared about the objects you're choosing, with no implications on the fate of the unchosen

[lucid zenith]

fate of the unchosen sounds ominous

[novel juniper]

Sounds like an indie video game lol

[pallid tapir]

it's the title of a post-capitalist horror game, next to night of the consumer

[novel juniper]

The first part would just be computations?[ proving H is closed?)

nvm

[pallid tapir]

closure under inverses and products, or the one-step subgroup test

[regal coral]

my book had 2 different theorems for proving a subset of a group is a subgroup

[novel juniper]

The question is how would I prove closure here

inverses is easy enough

[pallid tapir]

take two permutations that fix 1 and 3; does their composition also fix 1 and 3?

[novel juniper]

So I have to compute , no abstract proof is possible

[regal coral]

alpha(beta(1)) = alpha(1) = 1

[pallid tapir]

well

I mean, it's really highly unnecessary, but the collection of permutations that have two fixed points in 1 and 3 might be the kernel of a homomorphism S_n -> S_{n-2}

[lucid zenith]

😭

[pallid tapir]

but then you have to address the shuffling of the entries in the canonical choice of set that your permutations permute lol

[novel juniper]

I havn't dones this yet 😭

[pallid tapir]

ok I thought so, but that is at least another way! but the thing Axe said is like, a two-liner

[novel juniper]

Got it

thanks

So my issue here is there's no assurance that an element of $H$ has to have even order is there

[elfin berry]

wai

[pallid tapir]

if this is your first time reading the question, you might be better off thinking about it for a bit before someone else steals the experience lol

[novel juniper]

It is my first time yes

[pallid tapir]

I had to stop myself from giving a hint because the answer I had excited me

[novel juniper]

okay, I'll think about it for a bit , I just wanted to finish this as I have an exam on this in like 40 hours lol

[regal coral]

can't we at least give a small hint?

[pallid tapir]

with permission

[novel juniper]

I'll think

It's late anyway 😭

,ti

[elfin berry]

The current time for math_rocks is 12:39 AM (IST) on Thu, 18/09/2025.

[novel juniper]

I think i should just eeep

.close

[marsh citrus]

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[marsh citrus]

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[scarlet wagon]

Two points are given, each a finite distance apart. Both points are the centers of circles with radii equal to the distance to the other point. The two resulting circles have two intersection points, which are the centers of new circles, with radii equal to the distances to the centers of the initial circles. The process is then repeated indefinitely. The resulting construction consists of the intersection points of pairs of circles, which in turn are formed by this point, and a radius equal to the distance from this point to some other point besides itself. The process is repeated for all combinations of points, so that each point is the center of a circle, and each distance between pairs of points is the radius of some circle.
Determine:

1. What construction is obtained from circles and points.
2. Define it analytically for further analysis.
3. Analyze this construction.

Help with 2 pls

[scarlet wagon]

I did a little sketch

so you can get an idea of what it is

[burnt ivy]

Wait am I in the wrong channel

[scarlet wagon]

the construction is quite simple, but how do I actually define it?

please help...

it is similar to a fractal since any two points generate the same construction, but turned and resized