help-17

[lilac sluice]

so 42+10(1-(7/10)^8)

over 1-7/10

[elder scaffold]

$42+\frac{10(1-{\frac{7}{10}}^8)}{1-\frac{7}{10}}$

[lilac sluice]

ohhhhh

[twin meteor]

Galaxy

[lilac sluice]

i never would’ve gotten that wtf

[elder scaffold]

like the 10 is in geometric progression since thats what is decreasing each week

then u just add 42 since its initially 42 cm tall

[lilac sluice]

but why isn’t it 10 instead of the 42 on the outside

and vice versa

since T1 would be 42

[elder scaffold]

t1 is week 1

so it should be 52

[lilac sluice]

so then why wouldn’t it be 52 instead of the 10

basically 10 on the outside and 52 next to the bracket

[elder scaffold]

because the 42 isnt going to change, its base height is always 42 cm tall

[lilac sluice]

ohhh

[elder scaffold]

whats actually changing is the amount it grows by each week

[lilac sluice]

so hypothetically if we used T2 instead of T1, we would put 7 instead of 10?

[elder scaffold]

yeah

[lilac sluice]

ahhhh okay

i’m cooked then

[elder scaffold]

your fine, you've got a whole year to learn this stuff

itll be easy by the end of it

[lilac sluice]

nah it’s just cus i got a 20min quiz today

for dr du

and it’s based on last weeks content

which is this

[elder scaffold]

do the results actually matter?

[lilac sluice]

kind of

i wanna promote to A2 next term

rn i’m in A3

[elder scaffold]

oh theres tiers to his tutoring?

[lilac sluice]

yea

A1, A2, A3, B

[elder scaffold]

sounds odd, whats the difference?

[lilac sluice]

basically all ‘A’ tier classes learn extension content

but A1 learns stuff that’s out of the syllabus too

so advanced things

most students in A1 plan on doing ext 2 maths

so it’s very competitive

A2 learns advanced stuff that’s within the syllabus

and we learn stuff that’s a little less advanced to A2

but all A classes cover the very basics of what u have to know for that topic

B is like advanced/standard maths i believe

if they do learn ext 1 its incredibly basic

like easier than this

[elder scaffold]

thats weird

[lilac sluice]

it’s alright

and to promote u need a termly average of 90% + top 5% in the cohort

[elder scaffold]

are you yr 12 rn?

[lilac sluice]

year 10

they teach in advancr

[elder scaffold]

ah so u learn content early then

[lilac sluice]

yep

some of which i can actually apply to my school exams which is helpful

like trig

[elder scaffold]

yeah trig is in everything

[lilac sluice]

yes

[elder scaffold]

good thing to be useful at for hsc

[lilac sluice]

mhm

i’m pretty good at trig

speaking of trig can u help me with this one question

[elder scaffold]

sure

[lilac sluice]

c ii

[elder scaffold]

construct a triangle it seems

[lilac sluice]

wdym

oh like right angled triangle?

[elder scaffold]

yeah

[lilac sluice]

ok

[elder scaffold]

arccot and arctan are just angles

[lilac sluice]

so the angle is x

?

[elder scaffold]

nah thats one of the lengths

take cot of both sides

[lilac sluice]

i don’t understand

[elder scaffold]

youll see that $x=cot(\frac{\pi}{2}-\arctan{x})$

[twin meteor]

Galaxy

[lilac sluice]

true

but then what

[elder scaffold]

probably start somewhere else

let arctanx = theta

so tan(theta)=x

[lilac sluice]

yes

[elder scaffold]

now construct a triangle

which satisfies tan theta = x

[lilac sluice]

does this work?

[elder scaffold]

yeahg

[lilac sluice]

then what do i do

[elder scaffold]

then notice that the other angle is pi/2-theta

[lilac sluice]

we take tan of that?

or cot of it

[elder scaffold]

cot

so u should get $\cot(\frac{\pi}{2}-\theta)=x$

[twin meteor]

Galaxy

[lilac sluice]

yes

so pi/2 - theta = arcccotx

[elder scaffold]

yep

[lilac sluice]

ah

okay

what’s the range of y=arccos(e^x)

[elder scaffold]

sorry mb i went to get a drink

[lilac sluice]

allgs

[elder scaffold]

first determine the domain

[lilac sluice]

x must be less than or equal to 0

[elder scaffold]

right

so at x=0 we get y=0 right

[lilac sluice]

yes

[elder scaffold]

and as x approaches -infinity e^x would approach 0 correct

[lilac sluice]

uhhhh

how

is e^x just a log graph

[elder scaffold]

so $e^{-\infty}=\frac{1}{e^{\infty}}$ by indice laws

[twin meteor]

Galaxy

[lilac sluice]

yes

[elder scaffold]

and the denominator becomes a infinitely large

[lilac sluice]

yes

[elder scaffold]

which makes the number really small

[lilac sluice]

yes

[elder scaffold]

practically 0

[lilac sluice]

mhm

[elder scaffold]

so then it becomes arccos(0)

which is pi/2

[lilac sluice]

so 0 is less than or equal to arccos(e^x) which is less than or equal pi/2

wait

it can’t equal pi/2

right

[elder scaffold]

yeah thats true

[lilac sluice]

and it’s bc x approaches infinity but never reaches it

so it never reaches pi/2

[elder scaffold]

yep

[lilac sluice]

ah alright

[lilac sluice]

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[azure oyster]

$\lim_{n \to \infty} (2^n+e^n)^{1/n}$

[twin meteor]

Falco

[peak matrix]

I would try finding the limit of logarithm of that

[hard atlas]

can also squeeze

[azure oyster]

$\lim_{n \to \infty} log(2^n+e^n)^{1/n}$? I'm not sure what you mean

[twin meteor]

Falco

[cobalt oxide]

take e^ln

$\lim_{n \to \infty} e^{ln(2^n+e^n)^{1/n}}$

[twin meteor]

cps

[azure oyster]

I remember something about this but I'm not sure if I remember correctly. Would this be $\lim_{n \to \infty} e^{\frac{1}{n}{ln(2^n+e^n)}}$ and then you can insert the 1/n somewhere inside? It's been a while since I did logs

[twin meteor]

Falco

[cobalt oxide]

log(a)^n is nlog(a)

thats right

[azure oyster]

nice, thanks

so what is the next step? I have no clue how to proceed

[cobalt oxide]

input the limit and check where it tends to

think of it as

$e^{g(x)*ln[f(x)]} where f(x) = 2^{n} + e^{n}, g(x)=1/n$

[azure oyster]

its e^0?

[twin meteor]

cps

[azure oyster]

welp, its $e^{0 \times ln(\infty \times \infty)}$?

[twin meteor]

Falco

[peak matrix]

Sorry, i had to go for a while

anyway now you can factor out 2^n

from 2^n + e^n

or maybe e^n

yeah, e^n is better

Wait I just realized that whole the log thing might not be needed

Yeah

you can just factor out e^n

that makes things so much simpler

[cobalt oxide]

wait yeah

the 1/n bring out just the e

[azure oyster]

you mean $\lim_{n \to \infty} (e^n(\frac{2^n}{e^n}+1)^{\frac{1}{n}}$?

[twin meteor]

Falco

[peak matrix]

Yes, exactly

now apply few exponent rules

[cobalt oxide]

wouldnt it be only e outside?

[peak matrix]

i think they just forgot to add parens

[cobalt oxide]

ahh

[azure oyster]

damn, thanks guys

[peak matrix]

$\lim_{n \to \infty} \left(e^{n}\left(\frac{2^{n}}{e^{n}}+1\right)\right)^{\frac{1}{n}}$

[twin meteor]

MæthIsAlwaysRight

[azure oyster]

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[vast shale]

Hi @everyone: does anyone here know Hoover Tung (Nikon/ExploringSolitude)?

[vast shale]

Or Bokeh_123?

[flat whale]

[vast shale]

<@&268886789983436800> .

[flat whale]

[vast shale]

<@&286206848099549185> .

@drowsy igloo .

[flat whale]

uhhh bot probably

[vast shale]

@regal parrot .

[inland sail]

Ping a lil more and you’ll get help probably

[mild flower]

im banning

[limber falcon]

yeah looks like a bot

[mild flower]

oh

[flat whale]

help = ban in this context

[limber falcon]

lemme help you to the door

[flat whale]

[mild flower]

only in this context!! modpings should not be a gamble

[devout kindle]

lmao

[limber falcon]

welp

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[lilac plaza]

Hello

[lilac plaza]

Does a) linearly independent because there are no free variables in the REF and is not a spanning set of R^3 because there are only two standard basis?

And is d) linearly dependent because it has free variables in the fourth column and is a spanning set of R^3?

[tawny sorrel]

yes

[dull bear]

Also any set of strictly more than 3 vectors in R^3 is immediately linearly dependent

[lilac plaza]

Make sense :)))

Thank you

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[silk rampart]

chaewon doing math now

[lilac plaza]

Chaewon also doing electricity and magnetism & vector calc & linear circuits & boolean algebra & numerical methods 😭

[silk rampart]

i'm also doing whatever chaewon is doing 😭
lol nice to meet a fearnot

[lilac plaza]

Yayyyy

Are you also in comp eng?

[silk rampart]

No, I'm in HS

[lilac plaza]

Wait what

[silk rampart]

uh i dont think we should talk in help channels

[lilac plaza]

How

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[spring gate]

going over some basic differentiation

[spring gate]

wondering if this looks correct?

its basic chainrule so it should be fine right?

[cobalt ocean]

looks good to me

[spring gate]

:3

ty for the confirmation

also this should become (1.75*sin(2.143x))/2.143?

[cobalt ocean]

sure

[spring gate]

ok good, i always get mixed up when dealing with this on how your ment to write it

hows about this? im feeling very uncertain about integration and its iffy

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[undone burrow]

Not sure how to tackle this one

I thought itll be smthng to do with the identity C(n+2, 3) = C(n+1,2) + C(n+1,3)

[hushed pewter]

induction maybe?

[undone burrow]

I was hoping to get a

more combinatorial proof

[hushed pewter]

ah I presume this is for a combinatorics class then?

[undone burrow]

Not really self learning

i jus reckon itll help my understanding

kinda struggle with counting and all ig

[hushed pewter]

I see I see. Well you can try #combinatorial-structures maybe. I don't know much about the subject matter myself

[undone burrow]

I dont think i can use it but ill look into that

thanks

[hushed pewter]

yeah. Sorry, I am not a good counter myself. It's why I do analysis

[undone burrow]

fair enough 😭

counting really gets confusing

Btw I am allowed to use an ug role even if im not

[hushed pewter]

we sound like idiots out of context

[undone burrow]

ug myself right?

[undone burrow]

fr counting is so trivial, whats hard about 1, 2, 3 and so on

[hushed pewter]

ug?

[undone burrow]

undergraduate role

[hushed pewter]

ah

I mean, if you are studying undegratuate mathematics, I don't see why not

[undone burrow]

fair enough

great thanks

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[desert glen]

Would it still be correct if I leave the + C in there or should I move it outside or does it not matter

[silk osprey]

2C is another constant

[desert glen]

so it doesnt matter whether i move it in or outside?

[silk osprey]

you might want to distinguish the two if you do though

like by writing C_1 and C_2

[desert glen]

alright i see

ty

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[ionic marlin]

can I just make the argument that

Let M = A1A2
Let C = A3A4

det(MC) = det(M) * det(C)
= det(A1 * A2) * det(A3 * A4)
= det(A1) * det(A2) * det(A3) * det(A4)

[sly sierra]

yep this is fine

[ionic marlin]

thanks

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[verbal blaze]

I'm supposed to use triple integrals to calculate the volume of a solid, but I got the question wrong and I"m not sure what's wrong with my approach

[tulip cedar]

Write six triple integrals 😭

[verbal blaze]

the teacher marked these all as wrong but I don't know why

apparently all but the outside integrals are wrong

[verbal blaze]

tbf they're pretty simple triple integrals

[tulip cedar]

every outer integral is incorrect?

[verbal blaze]

yeah

and I got part a correct

[verbal blaze]

my approach was just to use the linearity of the solid to set it up as just a bunch of triangles

apparently that doesn't work tho?

[tulip cedar]

It should, isn't necessarily hard to picture the sweeping when fixing each variable

[verbal blaze]

that's kinda what I was thinking

<@&286206848099549185>

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[tidal hull]

can someone help me with this

I did 220 km/h [east] for the first question a)

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[tidal hull]

<@&286206848099549185>

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[brittle wasp]

The same method for this is used for part b

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[lone star]

is there a faster way besides substituting them one by one

im lazy

[sly sierra]

row reduction, if you know how to do that

[lone star]

how

[sly sierra]

do you know what row reduction is?

[lone star]

maybe but idk the names

so how do i do row reductiion

[kind light]

have you learned mateicies?

[lone star]

no

[kind light]

im p aure row reduction uses matricies

[lone star]

wait is it the
|a b| =dx |e f| = dy
|e f| |c d|

thats all i learnt tho

[kind light]

uhh no clue

[lone star]

oh its fine then

i used chatgpt

ty anyways

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[patent crescent]

can someone help with 9 and part b of 10?

[patent crescent]

,rotate

[marsh charm]

whats that little circle mean

f(g(x))?

for g(f(x)) u literally did the right thing

its suppose to be sqrt(4-x)^2

[fringe smelt]

composite function, which i think u did correctly?

but part b of q10 has some mistakes

[hexed kestrel]

9a is correct?

Ok idk

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[patent crescent]

yes

[patent crescent]

it is????

I’m getting so confused idk what to do w/ that

[patent crescent]

oh so I can’t simplify it?

[hexed kestrel]

No

Also for 10b

The range is -infinity to 4

[patent crescent]

4 is brackets right

ok i lo2k3yly have no idea what im doing wrong😭

@hexed kestrel for 9a would i just ave (1/x-3)+2?

cause wouldn't that form also work???

idk

[hexed kestrel]

Yes, but idk how does that form the same with this

[patent crescent]

?

sorry what do you mean

[hexed kestrel]

idk

[patent crescent]

😭🙏

9b is right right

[patent crescent]

what part the domain or the g(f(x))

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[hardy shadow]

Can someone explains how the top part becomes 18x?

[proper minnow]

The 6x from bottom goes to top

That's how division works

[hardy shadow]

Is that just like a general rule?

[proper minnow]

Yeah?

X/y ÷ a/b is just x/y*b/a

[hardy shadow]

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[prisma rain]

hallo

[prisma rain]

f(x) = (3x + 1)/2
g(x) = x/2

Consider all possible compositions: f(x), g(x), f(f(x)), f(g(x)), g(f(x)), f(f(f(g(f(g(x)))))), etc. i want to prove they all result in different functions

what kind of math do i need to do this? i know this forms a transformation semigroup but idk if it's necessary to involve that

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@prisma rain Has your question been resolved?

[verbal blaze]

maybe approach this in terms of function properties?

I'm a high schooler so I don't actually know what I'm talking about

but you could maybe look at how they transform the slope and y intercepts of the functions

like okay instead of functions we think of a vector of these two properties
and then the functions become
f(x,y) = <x/2,y/2>
g(x,y) = <3x/2, 3y/2 + 1/2>

sorry I'm not even helping atp I'm just trying to work this out

[verbal blaze]

okay so then you can identify the amount of f(x) and g(x)s were used to compose a function

just by factorization

[verbal blaze]

oh wait and then the two starting vectors would be
for f(x) as the root, <3/2, 1/2>
and for g(x) it'd be <1/2, 0>

[proper minnow]

Just solve all of them

[verbal blaze]

of the slope

[proper minnow]

And show they yeild different result

[verbal blaze]

in the case where you only have one g(x), you can identify its position pretty easily

[proper minnow]

Do you know what you yapping about ?

[verbal blaze]

not at all

[proper minnow]

Ok stop the yapping

[verbal blaze]

just by factorizing the y intercept

[proper minnow]

The only way to prove it ,is just solving it and showing it

There is no other proofs to prove them

[proper minnow]

This is just so wrong

[verbal blaze]

why

[proper minnow]

Why?

[verbal blaze]

yeah

bcs if you have like
f(f(f(f(f(g(f(f(f(x))))))))

it'll have a y intercept of like

[proper minnow]

Show you can get count of f(x) , g(x) by factorising mathematically

[verbal blaze]

1/128

[verbal blaze]

this part is easy actually

so

[proper minnow]

Do it

[verbal blaze]

the slope of the function is gonna be determined by which one of these functions gets applied

[verbal blaze]

this is the transformation they do to the function

[proper minnow]

😭

[verbal blaze]

as a vector of <slope, y intercept>

since one multiples it by 3/2, and the other doesn't

[proper minnow]

Prove by solving a set of function fr

And you will see the issue

[verbal blaze]

the amount of g(x)'s is gonna be determined by the amount of factors of three in the numerator

the amount of f(x)'s is gonna be determind by subtracting that from the number of 2's in the numerator

qed

[proper minnow]

Ngl idk tf you talking about

Show examples

[verbal blaze]

I don't want to

that sounds really annoying to do

[proper minnow]

🥺

[verbal blaze]

for like, more than a few repetitions of this

[proper minnow]

Are you trolling ? Or something

[verbal blaze]

I can't figure out the induction step here

[proper minnow]

How do you even manage to convert a 1 variable into 2 variable ?

[proper minnow]

Because it's not possible

Slope doesn't work that way

[verbal blaze]

what do you mean?

[proper minnow]

F(something ) will have the slope of f(x) but that doesn't mean

F((g(x) will have a combination of two slope

It will just have slope of f(x)

Or it will have different slope

[verbal blaze]

if f(x) = x/3 and g(x) = x/2, f(g(x)) will have the slope of their product

this is true for all linear functions

[proper minnow]

F(x) = 4x-3 and g(x) = 3x-9

Try now

[proper minnow]

• what you will get is x/6

Its not slope

[fringe scarab]

composition, not multiplication

[verbal blaze]

I'm trying to figure out whether or not two g(x)'s can be located

[proper minnow]

Oh yeah mb

[fringe scarab]

idk what kind of induction u can do

[verbal blaze]

honestly it looks like collatz

but that might just b a coincidence

or they're trying to do their own investigation

[fringe scarab]

wait Is it

[verbal blaze]

it would be funny but I don't think it is

[fringe scarab]

it is

lol

[verbal blaze]

wait is it actually

[gentle sleet]

yea it is
f is a 2 step collatz I think

[verbal blaze]

man
I thought I was onto something

[fringe scarab]

wait no it almost is

oh yeah what sepdron said

[verbal blaze]

how is this statement equivalent to collatz

[fringe scarab]

it's not

it's just asking if each combination of compositions is unique

i think u can cover Z adjoin 1/2 with the constant term tho

oh except the negatives

[verbal blaze]

I think this might be equivalent to collatz in some weird way

[fringe scarab]

no it's a simpler question

[proper minnow]

Just solve all of them and see

Fr

[gentle sleet]

there's infinitely many to check

[proper minnow]

Eh ? What

[proper minnow]

I am talking about op's question

[gentle sleet]

all possible composition

[fringe scarab]

u cant just look at the constant term, ff and gffff give constant terms of 1

[verbal blaze]

ah

was I onto something, then

[gentle sleet]

yea, I'm taking about OP's question

[proper minnow]

They given all the composition and the f(x) and g(x)

[fringe scarab]

idk! im more focused on answering the question they posited but go and look at it some more and see if it does connect in that way

[fringe scarab]

"etc"

[verbal blaze]

you can at least figure out how many there are of each with the slope

[proper minnow]

Oh

[verbal blaze]

that doesn't seem right

[fringe scarab]

wait

[verbal blaze]

yeah no it's not

[fringe scarab]

yeah nvm

teehee

[verbal blaze]

what I know so far is:
the slope uniquely identifies the number of both of them
if you have one f(x) you can identify exactly where it is w the constant

er I've been mixing them up

there

[fringe scarab]

yeah the numerator is 3^n where n is how many compositions of f u have

[verbal blaze]

if you have two g's can you uniquely determine their positions

[proper minnow]

And you can get number of f function using dem - the number of f

[verbal blaze]

this is where I got stuck

[proper minnow]

But again what if the order is different but number of function is same ?

Like g(f(x) and f(g(x))

[verbal blaze]

something tells me you can

[fringe scarab]

fg gives $\frac{3x}{4} + \frac{1}{2}$
gf gives $\frac{3x}{4} + \frac{1}{4}$

[twin meteor]

betterert

[verbal blaze]

yeah

#g = 1 case is trivial

[proper minnow]

You can't check the constant term like you do for slope tho?

[fringe scarab]

how do u mean

[verbal blaze]

wym

like if there's only one g

[fringe scarab]

i dont understand their question

[verbal blaze]

oh

they're just a little confused I think

I'm actually gonna get a repl real quick and run through like, the first 128 combinations or smth
just to peek around

[proper minnow]

We could check the slope easily and say it's different for each number of f(x) and g(x) as both doesn't have a common factor

But how will you get constant are different for each combination

[verbal blaze]

tryitonline my beloved

[gentle sleet]

oh, I think you can simplify the problem a little if we have k=3x+1
then f=g(k)
not sure if that helps though

[verbal blaze]

k doesn't exist as a combination of f's or g's tho

so it's kinda outside the domain of the question

[verbal blaze]

nevermind I haven't prgorammed in months I don't feel like googling syntax

[fringe scarab]

i aint have enough experience with discrete mathematics like OP alluded to to do anything like that
but i have enough to say that including k might make a proof easier

[gentle sleet]

ah I can do it

[verbal blaze]

could you print out the constant term for
f(x)
g(x)
f(f(x))
f(g(x))
g(f(x))
g(g(x))
f(f(f(x))
etc

[fringe scarab]

gfg and gf have the same constant term

[verbal blaze]

do they?

[proper minnow]

Different slope

[verbal blaze]

oh yeah

they do

[fringe scarab]

yeah

[gentle sleet]

from your thing, we don't need to check if the number of f and g in 2 sequence is different right?

[proper minnow]

Only need to see if they have the same constant with same slope

[fringe scarab]

u cant determine the problem based off of the constant term

[proper minnow]

Yeah

[verbal blaze]

could u append the functions also

[fringe scarab]

nor with the coefficient alone

[proper minnow]

Only gotta check where the order of f and g differ

[proper minnow]

We need both

[verbal blaze]

that's the word I was looking for

[proper minnow]

We already know that

Ig we could prove it always yields different results when the order is different

[fringe scarab]

that's the question asked

[verbal blaze]

the constant for n f(x)'s composed together would be
uhhh

[fringe scarab]

0

[proper minnow]

For the same number of f and g

[verbal blaze]

srry i keep mixing them

[proper minnow]

Could be just n/2

[fringe scarab]

f multiplies by 3/2 and then adds 1/2

[proper minnow]

Oh

[fringe scarab]

be a weird series

[verbal blaze]

that's so annoying to think about

[verbal blaze]

yeah

[proper minnow]

Eh probably possible to prove by induction but won't be a proper proof tho

[fringe scarab]

prove what by induction

[proper minnow]

Constants differ for different order of f and g

[fringe scarab]

what would u be inducting tho

[proper minnow]

The pattern

[fringe scarab]

how did u get 9/4

i got 5/4

[verbal blaze]

oh

oh my god im stupid sorry

lemme try again

I should have the computer do this

lemme get the first 20

[fringe scarab]

$f_n = \frac{3^nx}{2^n}$ plus the constant term

[twin meteor]

betterert

[proper minnow]

Ok so if x isn't -1/2 then the order of f and g doesn't matter

If we exchange the order of f f f f or ggggg it doesn't matter too

Because all of them will just count as a single combination

[proper minnow]

Can anyone prove -1/2 can't be formed by any operation of f and g

[fringe scarab]

-1/2 given what value of x

[proper minnow]

Any value

[fringe scarab]

idk

[proper minnow]

The different order will only be equal when the operation gets -1/2 and then next operations as different

[fringe scarab]

either way i feel like u lose generality there

if u fix x

[proper minnow]

You aren't fixing x

You just need to prove

-1/2 can't be formed by any x anywhere

But it could be formed when x = -1 and then the first operation is g(x)

So ig it's only possible to prove the whole thing

If x is greater than 0

[fringe scarab]

i dont think that would prove the question we're asking

if any combination of compositions cant reach a certain value
that doesnt mean two different combinations of compositions cant be the same value

[proper minnow]

It could tho? The different order can only ever result into same answer if it has -1/2 inbetween the values followed by different type of function

I got -1/2 by equating the f(x) and g(x)

The f(x) and g(x) can only ever replace each other if they give equal values for x

And the equal value ever happens if x= -1/2

[verbal blaze]

1: 1/2 : 0.5
2: 5/4 : 1.25
3: 19/8 : 2.375
4: 65/16 : 4.0625
5: 211/32 : 6.59375
6: 665/64 : 10.390625
7: 2059/128 : 16.0859375
8: 6305/256 : 24.62890625
9: 19171/512 : 37.443359375
10: 58025/1024 : 56.6650390625
11: 175099/2048 : 85.49755859375
12: 527345/4096 : 128.74633789062
13: 1586131/8192 : 193.61950683594
14: 4766585/16384 : 290.92926025391
15: 14316139/32768 : 436.89389038086
16: 42981185/65536 : 655.84083557129
17: 129009091/131072 : 984.26125335693
18: 387158345/262144 : 1476.8918800354
19: 1161737179/524288 : 2215.8378200531
20: 3485735825/1048576 : 3324.2567300797

[proper minnow]

By proving -1/2 can't ever happen

[verbal blaze]

constant terms acquired

upon further inspection this is kinda just garbage

xd

[fringe scarab]

u double and add half of 2^n

nice

[verbal blaze]

w decimal values added

[fringe scarab]

decimal values are useless tho

[verbal blaze]

I can probably print the prime factorization of the numeral too

wait that sounds a lot more useful lemme do that

[fringe scarab]

yeah try that

[fringe scarab]

for any two combinations of compositions, there exists some x value such that they are equal

unless they have the same coefficient

[proper minnow]

Oh nvm I got it

The constant term can never be less than 0

Yea it was a simple proof

Ok so what I did is I just assumed a inbetween term we will get as

z=yx+c

Put them inside both f(x) and g(x) and equated the constant part

The constant part becomes equal only and if only c = -1/2

Which is never possible here because the term which gives constant is +1/2 from f

[fringe scarab]

we're doing repeated composition tho

[proper minnow]

Will be same

[proper minnow]

You can't replace f by another f cuz it results in same combination

The only issue happened when f and g was replaceable by each other

Which i just proved not possible

By @verbal blaze proof of different number of f and g give different slope

And mine

[fringe scarab]

gfg and gf have the same constant value
and fg and gf have the same slope?

[proper minnow]

You can say it will be always a different term regardless of the order or number of functions

[proper minnow]

But I just showed fg and gf can't have same constant value

[fringe scarab]

but they do

[proper minnow]

And gfg and gf have different slove proven by @verbal blaze

[proper minnow]

Eh they don't ?

You showed the result

It was 1/4 and 1/2

Different constant term

[fringe scarab]

$fg = \frac{3\frac{x}{2} + 1}{2} = \frac{3x+3}{4}$

[twin meteor]

betterert

[proper minnow]

So yea that's the solution

[fringe scarab]

$gf = \frac{3x+1}{4}$

[twin meteor]

betterert

[verbal blaze]

fg and gf having the same constant disproves my previous result

which

[proper minnow]

Yes constant terms is different?

[verbal blaze]

is weird

[fringe scarab]

fg has a coefficient of 3/4
gf has a coefficient of 3/4

[proper minnow]

They don't

[proper minnow]

That's slope

The constant are different

[verbal blaze]

I mean this makes sense

[fringe scarab]

fg has a constant term of 1/2
gf has a constant term of 1/4
gfg has a constant term of 1/4

[verbal blaze]

given you can only divide it by two or increase it

[fringe scarab]

ah i fucked up my arithmetic

[verbal blaze]

this makes sense I think

[proper minnow]

Listen carefully

We got two proofs

1. When number of function differ their slopes differ (lexica proved)

2. Order of operations doesn't matter for same number of function ,constants will always be different (my proof)

[fringe scarab]

no but

what

[proper minnow]

Yeah

That's the whole sum

It was easy I just thought it was a hs school sum and didn't think too much of it

[verbal blaze]

oh wait

okay so

u know how to convert a decimal number to binary

[proper minnow]

Yeah ?

Why you need that for?

[verbal blaze]

where it's basically taking the biggest chunk out of the number you can before moving down

[proper minnow]

??? What

[fringe scarab]

can u reply to the message where she proved that a different number of functions proves the coefficient differs

[proper minnow]

You were there ?

[vast shale]

Babi is right u don’t need much more

[fringe scarab]

if i remembered it i wouldnt ask

[verbal blaze]

it's just the prime factorization of the slope

remember?

[fringe scarab]

where

[proper minnow]

The slope will be always just 3^nf(x) at top

[verbal blaze]

the amount of 2's in the denominator is the total of the functions
the amount of 3's in the numerator is the amount of f(x)'s

[fringe scarab]

oh ok

[verbal blaze]

yea

[proper minnow]

Yea

So proven

[verbal blaze]

okay wait

you proved the constants differ? where?

[proper minnow]

Up

[proper minnow]

Its mostly logical

Hear carfully

Replacing a f by f will just be same order

Same goes for g

Right ?

So only issue happens if and only if f can replace function g

Which can only happen if and only if both yield the same constants

Right ?

@verbal blaze

[verbal blaze]

right okay sure

[fringe scarab]

yeah

[verbal blaze]

how do you prove that two changes can't cancel each other out tho

[proper minnow]

Now just put y= zx+c in both function

You will get c in both function

Equate them

You get c=-1/2

Which means

[gentle sleet]

isn't your proof only for 1 swap?

[proper minnow]

The constants will be same if and only if c=-1/2

[verbal blaze]

that's what I'm saying

[proper minnow]

Its same for all swaps

[vast shale]

But u can induct it

[proper minnow]

You can never get c = -1/2

[gentle sleet]

yea it's like f(...) vs g(...)

[verbal blaze]

can you?

[proper minnow]

So the number of swaps doesn't matter

[vast shale]

Yes

[proper minnow]

Let's say I got g f g f g f

[verbal blaze]

okay

[proper minnow]

And f g f g f g

I already showed one swap will lead to different coefficient right?

You understand that right

[verbal blaze]

yeah sure

[fringe scarab]

yeah

[proper minnow]

Now I swap the first 2 function

I will get

F g gf gf

Now he second pair and so on

we will never ever get c = -1/2 anywhere

C is always will be greater than 0

[verbal blaze]

this is the part I'm struggling to get

[proper minnow]

I just equated both constant terms of the function

[fringe scarab]

how do u know gfgfg and fggfg wont have the same constant value
u know fgfg and ggfg wont, but what about gfgfg and fggfg?

[proper minnow]

They only equal od the constant term was -1/2

Which isn't possible

[verbal blaze]

bcs gf =/= fg

[proper minnow]

They won't because

^

[verbal blaze]

yeah

[proper minnow]

And replacing both f by f or g by g will be just same combination

[gentle sleet]

what about like f g g f and g f f g?

[proper minnow]

Thus problem solved

[verbal blaze]

this is what comes to mind

[proper minnow]

Same you never get -1/2

It doesn't matter the order

You never get -1/2 for constant

[verbal blaze]

can u clarify wym by never getting -1/2

that seems to be a key step in your argument but I don't quite understand what u mean

[proper minnow]

Ok so

C is just constant term

[gentle sleet]

what you've showed is f g g f ≠ g f g f
but how do you then get to g f f g?

[proper minnow]

And its always positive because c is given initially only by f(x) right?

[gentle sleet]

I understand the -1/2 thing
but if you do 2 swaps, they don't stay unequal

[proper minnow]

G(x) can't give the c

[fringe scarab]

given some function h
gh and fh cant give the same constant term
in order for them to do so, h would require a constant term of -1/2

[proper minnow]

They don't

[verbal blaze]

okay right

[proper minnow]

That's the point

We don't want them to be equal

We want them to be different

Thus problem solved

It doesn't matter whatever order you got, you can never get negative constant

[gentle sleet]

how?
bc from what I understand, you show that sequence ≠ swap(sequence)
and swap(sequence) ≠ swap(swap(sequence))
but that doesn't mean sequence ≠ swap(swap(sequence))
they could be equal

[verbal blaze]

I think for any string fgfg etc etc, if you take the last term and put it at the start the constant changes

[proper minnow]

They are because let's say I got from one swap and get a different seq and then again swap to get what you wanted

But I did two swap

By my proof , the two swap can't have equal constants

Unless c = -1/2

[verbal blaze]

would it be possible to get to g f f g from f g g f in an odd number of swaps?

[fringe scarab]

ok can we define "swap"
cuz idk if ur referring to fgfgfg -> gffgfg
or fgfgfg -> ggfgfg

[proper minnow]

If the swap was just fg gf it will still lead to different coefficient

[verbal blaze]

I think we're saying a swap is changing any term gf to fg or vice versa

[proper minnow]

The swap will only be equal if the initial seq = final seq

[fringe scarab]

well fgh != gfh

[proper minnow]

Yeah

[verbal blaze]

I don'tthink so

yeah no u cant

[proper minnow]

Try equating with two swap, you will still need the constant term to be negative

Problem solved

No more questions

Go home

[verbal blaze]

okay wait but I think for any odd string you can reach any other odd string with the same characters

using swaps

[proper minnow]

Ok so?

.close this fr

[verbal blaze]

like
fgg
gfg
ggf
and that's all of them

[fringe scarab]

if u show it's associative for all combinations of 3 characters then i think u win

[proper minnow]

?

[proper minnow]

Me?

[verbal blaze]

I just did

well

[fringe scarab]

or maybe im misapplying my group theory knowledge

[verbal blaze]

no wait I did

[proper minnow]

Dw we solved it

[verbal blaze]

just invert them

[gentle sleet]

you proved that fggf ≠ gfgf and gfgf ≠ gffg
but can't fggf = gffg still?

[proper minnow]

Don't think too much

[fringe scarab]

how did they prove fggf != gfgf
all they did was prove fh != gh

[gentle sleet]

oh what's h
I missed that

[fringe scarab]

any function

like any combination of compositions

[verbal blaze]

well fg =!= gf
so swapping them anywhere should make it unequal

[gentle sleet]

oh I thought that they proved gfh ≠ fgh

[verbal blaze]

I did too

[fringe scarab]

i was gonna ask "are the functions one to one" but they're lines so yeah

ok

[verbal blaze]

xd

uh I think gfh =!= fgh is kinda trivial

[fringe scarab]

yeah

[verbal blaze]

hgf =!= hfg also

[fringe scarab]

actually no

[verbal blaze]

why not?

[fringe scarab]

no not that

how do u know that given different sequences of compositions, like i and j
that hi != hj

given i and j are not equal

[verbal blaze]

you don't

[gentle sleet]

h is 1 to 1
so if fg ≠ gf, then hfg ≠ hgf

[verbal blaze]

that's just not true

you have to say that i and j are permutations of each other

in which case it is

[fringe scarab]

oh

[gentle sleet]

wait, wouldn't that prove OP's question?

[proper minnow]

There is also another way to look at this

Gf is always greater than fg always

More like the later you get f the bigger your constant term will be

[verbal blaze]

which is fine tho bcs if they're not permutations of each other then they're disambiguated by slope

[verbal blaze]

it's conjecture xd I don't have a proof

also it doesn't work for the fggf ? gffg case

[proper minnow]

In theory constant term of fggf should be higher than gffg

[gentle sleet]

from your first thing
they would be permutations of each other though

[proper minnow]

Someone check that rq

[fringe scarab]

u check it bruh

[verbal blaze]

wym

[proper minnow]

Where ?

[gentle sleet]

that's a different thing
but it might be something

[fringe scarab]

with a pen and paper

[proper minnow]

Eh no

Too much work

[gentle sleet]

if 2 sequence is equal, the number of their f and g must be the same

[verbal blaze]

sure

[fringe scarab]

yeah we showed that earlier

[verbal blaze]

our result was stronger than that actually

for the record

[fringe scarab]

wdym it's the contrapositive of our result

[verbal blaze]

uhh wait

oh yeah

[proper minnow]

Wat

[verbal blaze]

I'm a dumbass

it seemed cooler in my head than the way they said it tho

no wait

our result IS stronger

[proper minnow]

Why could that have any effect when talking about constant terms tho?

[verbal blaze]

bcs we didn't just get equality

[proper minnow]

Which result is weaker?

[gentle sleet]

@proper minnow I'm talking about this

[fringe scarab]

if the total number of functions composed is unequal, then the sequences are unequal
if the sequences are equal, then the total number of functions composed is equal

these are logically equivalent

[verbal blaze]

yeah my result says more than that

it was that slope uniquely determines the number of both functions

[fringe scarab]

oh yeah

[verbal blaze]

thanks to which we only have to worry about permutations

[proper minnow]

Why we tracing back to the initial proof

[proper minnow]

Yeah

[fringe scarab]

if n is number of f's and m is number of g's
then $slope = \frac{3^n}{2^{n+m}}$

[verbal blaze]

yeah that

[fringe scarab]

$slope = \frac{3^n}{2^{n+m}}$

come on bot u got it

[verbal blaze]

so this property follows

[fringe scarab]

did i fuck it up somehow?

[verbal blaze]

I mean

latex is p easy to read

[fringe scarab]

$\frac{3^n}{2^{n+m}}$

ok i give up

[verbal blaze]

anyways

okay so

[gentle sleet]

bot's broken I think

[verbal blaze]

conjecture:
for any permutation of a binary string of size n (where n is odd), swaps generate all other permutations

tru for n = 3

anyone know the math we need for this

[proper minnow]

They don't for all n tho

F f g g can't yeild f f f f by swapping

[verbal blaze]

permutations

[fringe scarab]

permutation not combination

[verbal blaze]

yeah what she said

[proper minnow]

Oh yeah mb

[gentle sleet]

I think it does
but what would be the continuation?

[verbal blaze]

we have to prove it

[gentle sleet]

I mean after that

[verbal blaze]

well

okay assuming this conjecture we've proven the original statement for all odd n

well

no

not true

false actually

the swaps are weaker than I first thought

[proper minnow]

Again why are we proving this tho?

What's the use

[verbal blaze]

I'd rather do this than my homework

[proper minnow]

I mean doing this why tho

[verbal blaze]

also I just failed my multivariable calculus final and need to blow off some stress

[fringe scarab]

in the space of functions are all h one to one ...

[proper minnow]

We already got the op sum done

[verbal blaze]

yeah

they're all lines

[fringe scarab]

no not R to R

[proper minnow]

Eh how many variables ?

[fringe scarab]

in the spac eof functions ...

[verbal blaze]

oh

uh

[proper minnow]

X y z?

[verbal blaze]

yes? yes

yes I'm pretty sure

[fringe scarab]

cuz if all h are one to one then hi != hj

u know

then we kinda win a little

[verbal blaze]

yeah

[verbal blaze]

oh true

[fringe scarab]

if ur lucky u even throw a w in there 😱

[proper minnow]

You in hs right?

[fringe scarab]

she got the undergrad math role so i dont think so

[verbal blaze]

yeah and i make questionable decisions with my enrollment

[fringe scarab]

ok nvm

[proper minnow]

Eh which country?

[verbal blaze]

I should not have signed up for that dual enrollment class

[verbal blaze]

guns

[fringe scarab]

how old r u

[proper minnow]

America?

[verbal blaze]

17

[proper minnow]

America?

[verbal blaze]

yeag

[verbal blaze]

wait actually

uh

uhhhh

no nvm

[proper minnow]

You guys got calculus? I heard it's only in college for usa

[verbal blaze]

at least we have gh =/= fh

[fringe scarab]

i mean i took 1 and 2 with dual enrollment

[verbal blaze]

im just built different
I mean
not different enough to not fail the class but still

[fringe scarab]

girl took multivar i didnt even know u could do that

[verbal blaze]

first math class I failed

boutta get ALL my applications rescinded

er admissions I mean

[proper minnow]

Don't your whole class got calculus?

[fringe scarab]

u took multivariable tho
not many hs students do that

[verbal blaze]

no just the special kids

[verbal blaze]

ig

idk I failed today bcs I don't know how to take surface integrals

[proper minnow]

Yeah usa good in that

[fringe scarab]

double and triple or what

[verbal blaze]

yeah

[proper minnow]

In India gotta take every math courses 😭

[fringe scarab]

yeah my university is teaching that neutered ass multivariable too

pissing me off

[verbal blaze]

idk r u supposed to get the jacobian of the polarized form of the surface or what

[fringe scarab]

i want line integrals and all vector calc

[verbal blaze]

?

we had line integrals on the exam

those were the easy parts

[fringe scarab]

ok

nv,

yeah see

[proper minnow]

Sat doesn't even have that , where are you gonna use it ?

[verbal blaze]

just green's theorem

or cry

lowk I did the second one because I kept getting non closed curves

[fringe scarab]

my university "multivariable calculus" class didnt do any of that

[verbal blaze]

lmao

[verbal blaze]

is this a good enough condition

[fringe scarab]

oh wait

lmao i think it is

[verbal blaze]

is it?

[proper minnow]

My School just did upto getting volume of shapes using integrals

[fringe scarab]

cuz gh and fh just turn into new h's

so gh != fh again

[verbal blaze]

god I wish it was just that

[fringe scarab]

oh wait that's only ffffffffh and gggggggh

):

[proper minnow]

• differential equations

[verbal blaze]

oh ew

I don't even know how to touch diff eqs

[verbal blaze]

not even lol

[proper minnow]

We got multiple variables integrals too

[verbal blaze]

god I wish it was just that

[proper minnow]

Its important

[proper minnow]

What else you have ?

[verbal blaze]

although

using the slope condition then yeah

[fringe scarab]

u think so?

[verbal blaze]

yeah

trivially

uhm

fh =!= gh might be the strat actually

[fringe scarab]

gh != fh
fgh != gfh

[verbal blaze]

is that enough

[fringe scarab]

what about ffgh and ggfh

[proper minnow]

Ah there is also another statment I made

If f comes at the very last the constant term will be big

[verbal blaze]

surface integrals

which are MISERABLE

they cost me the class

[proper minnow]

Isn't just getting areas of a shape ?

[fringe scarab]

that's with the fuckass limits of integration right

[verbal blaze]

and now I have to explain to my colleges why this lovely student didn't pass her very hard and impressive and cool college class

and why they should still let me apply as a math major pleasepleaseplease

[verbal blaze]

its where you take the integral along the surface

[verbal blaze]

perimeter

[proper minnow]

Calculus doesn't matter for many topics

[verbal blaze]

hm

[proper minnow]

So no big deal

[fringe scarab]

like shells in calc II?

[verbal blaze]

we're not guaranteed that

[proper minnow]

^ me when no one reads my statements

[verbal blaze]

kinda?

more like fluid flow into and out of a boundary