help-27

[lunar harbor]

i feel a bit dumb asking but here 1/2 only applied to the 10 and not the rest of the RHS?

everything is being multiplied together though
then when they div both sides by 5 how come x and sin93 are unaffected?
you’re essentially cancelling a factor of 5 from the right hand side. Neither x nor sin(93) is involved in this factor of 5, and hence they’re unaffected.

[pseudo basin]

1/2 * (xyz) is not the same as x/2 * y/2 * z/2. multiplication does not distribute over itself

[torn jackal]

what does it mean when its all being multiplied together? i see that but i thought it is still seperate from each other so what i was thinking is
(10xsin93)/2 = 70

[torn jackal]

ah

[pseudo basin]

yes sure. and if you look at that as a fraction, you can simplify it via 10/2=5

[torn jackal]

so itd be
5*sin93x = 70

then when you bring 5 over it divides instead of subtract bcz multiply yeah

[pseudo basin]

you do not "bring it over" at all. you should just never use that wording.

but yes you divide both sides by 5

to undo the multiplication

[torn jackal]

i think i got it confused bcz of that wording

i thought the 5 was seperate and positive instead of being altogether

alr i see now thanks

[devout snow]

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[restive river]

In wavy curve method for single inequality it makes sense to include '+' and '-' as at 1 ; x-1 would be 0 , greater than it +ve less than it -ve.

[restive river]

but for this i dont understand how it makes sense to include '+' and '-' for an inequality with more than 1 modulus as it dosent seem to represent anything.

[devout snow]

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[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[restive river]

we take values in between the two points marked and see whether the function gives a positive or negative answer, then we put +/- accordingly

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[hollow minnow]

how is this not symentric

(a,b) can be divisible by (b,a)

if it cannot be symentric then how come its transitive?

[peak needle]

(3,6) is in R, but not (6,3)

this disproves symmetricity

in general, it need not be true that if a divides b, then b divides a

[hollow minnow]

then how is it transitive?

[peak needle]

because if a divides b and b divides c then a divides c

[hollow minnow]

then a b c can be used for symentric

oh wait

your right

[peak needle]

these are simple facts that follow from the definition of divisibility 😄 you can sit and prove them fairly easily

[hollow minnow]

its only reflexive

[peak needle]

symmetricity and transitivity have nothing to do with each other!

[hollow minnow]

i got it now

thanks

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[peak needle]

this is not true, though 😄 it is also transitive

its just not symmetric

[hollow minnow]

huh? in my book it shows its only reflexive

oh yeah it shows transitive too

but wait hold on

.reopen

[devout snow]

✅

[hollow minnow]

how is transitive but not symentric?

[peak needle]

i just told you!!

[peak needle]

.....

[hollow minnow]

okai i get it but

trasitive means (a,b) (b,c) (c,a)

symentric means (a,b) (b,a)

[peak needle]

okay...

[hollow minnow]

so it says y divisible by x

[peak needle]

this is not correct, btw

[hollow minnow]

what

why?

[peak needle]

transitive means: if (a,b) and (b,c) in R, then (a,c) in R

[hollow minnow]

yeah same thing

[peak needle]

symmetric means: if (a,b) in R then (b,a) in R

[hollow minnow]

yeah exactly the same thing

[peak needle]

no, its is not the same, our sentences are different

[hollow minnow]

alr but in transitive 4 is y and 2 is x (they are divisible)

then what else divisible

see it says y is divisble by x

4 divisible by 2

6 is divisible by 2

but 4 and 6 is not divisible

[peak needle]

okay, so what?

(2,4) and (2,6) are in R but (4,6) is not in R

this does not disprove that the relation is transitive

[hollow minnow]

so its not transitive

[peak needle]

you need to find me three numbers a b and c in A such that a divides b and b divides c but a does not divide c

in order to disprove that R is transitive

do you understand?

[hollow minnow]

OHHH IGET

I GOIT IT

OHHH ALR ALR

THANKS

[peak needle]

😄 glad to help

[hollow minnow]

thanks a lot

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[long trout]

What is the number of non isomorphic directed simple graphs with v vertices and e edges? I found out there is this formula on math stack exchange to this problem but they didn’t happen to explain how it works

[pseudo basin]

direct = directed?

can you link the formula

[long trout]

Ah yes mb

[long trout]

The formula is in the bottom answer of the math stackexchange link: https://math.stackexchange.com/questions/354062/how-many-nonisomorphic-directed-simple-graphs-are-there-with-n-vertices-when

I enumerated it out for smaller number of vertices and it seems to be correct

[devout snow]

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[devout snow]

@long trout Has your question been resolved?

[devout snow]

@long trout Has your question been resolved?

[devout snow]

@long trout Has your question been resolved?

[long trout]

I’ll leave this open for now, but I might not be available because I have to sleep now. I’ll read later if someone offers to help me

[kind tiger]

ive been on and off thinking about it

i assume this, somehow, relates to ants

my thought regarding the formula is that perhaps you want to thinking functions E -> V because of the v^e term

then the / e! is because the "order" of the edges doesnt matter

and the extra v ... idk, get to that bridge later haha

[long trout]

Lol

What I really wish happened was that the person who provided an answer like that put an explanation behind the formula. Even someone else asked how that formula works

[kind tiger]

yh fr

let E = {1, ..., e},V = {1, .., v}, X = {(i, j) in V^2 | i < j}
f : E -> X defines a directed graph, there are (v(v+1)/2)^e such functions
we can also define h : E -> {-1, 1} of which there are 2^e functions. this gives our edges direction
so in all we have (v(v+1))^e directed graphs with e edges and v vertices

(im just riffing this may be totally useless)

ok alternatively lets count injective functions directly

( v(v+1)/2 choose e ), and that gives us the /e! to ignore order

perhaps stirling gives us something good enough

ah e and e 😭

seems pretty gross

[restive river]

<@&268886789983436800>

[devout snow]

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[worthy herald]

im not even sure where to start with this

[wind mason]

Do you know the definition of the derivative?

$f’(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$

[woven radish]

;(

[worthy herald]

not sure, my class has been moving so fast

[wind mason]

Okay, well, I just gave it to you

So think about what this’ll mean

[worthy herald]

ive seen that before

[wind mason]

Okay

[worthy herald]

it seems like my teacher is just saying things but not explaining them too well

this isnt making much sense to me

[wind mason]

Try to derive it, then

Or watch a YT video

[worthy herald]

would it help if i multiply (x-3) to both sides

so f(x)=6x-22

[gritty vapor]

no, instead look at how the equation defining the derivative matches the problem statement

[worthy herald]

would that be an acceptable way to solve?

[gritty vapor]

looks right

[worthy herald]

is this what you were talking about?

[gritty vapor]

it's better than what i was talking about

[worthy herald]

this way is probably easier?

[gritty vapor]

not easier, but more honest and general

i would've pattern-matched a = 3, f(a) = -4, f'(a) = 6 by comparing the equations, but that doesn't work if the problem isn't in this convenient form

[wind mason]

That’s not legal lmfao

[worthy herald]

its right tho

so if it wasnt equal to 6 it wouldnt have worked?

[wind mason]

But bro could have just concluded from definition of derivative

[gritty vapor]

if it was any other number than 6, it'd still work

what matters is the arrangement of f(x), x - 3, etc

[wind mason]

No

[wind mason]

What you did was illegal but you got lucky

[worthy herald]

mb

[wind mason]

“Right answer, wrong method”

[worthy herald]

i did have 3,-4

about the middle of the paper on the left

[wind mason]

What you’re supposed to do is notice that the point will be (3, -4) and slope is 6

Then you can write in point-slope form

[worthy herald]

whats point slope form

i forgot

need refresher

why is this right

[wind mason]

Your x is feet, y is dollars

y/x=dollars/feet

[wind mason]

$y-y_1=m(x-x_1)$

[woven radish]

;(

[worthy herald]

oh

why is it 1100 dollars for 1 foot

[wind mason]

Think about estimation using the tangent line

[worthy herald]

i think im stoopid, what is that supposed to mean

[wind mason]

It ain’t that bad

Just use the noggin

[worthy herald]

ive only had 2 classes for my calc class

i have 11 classes left till the end of calc 1

[wind mason]

💀

[gritty vapor]

"an additional foot" means you change x by 1 foot from x1

[worthy herald]

how do i find the derivative of a graph

[gritty vapor]

derivative is slope. how steep is the graph at each point? which way does it slope?

[worthy herald]

isnt it like the steeper the graph is the more extreem is it but if it were to flatten at the top would the derivitive be 0?

like the rate of change?

[gritty vapor]

exactly

[worthy herald]

so it would be that one

[gritty vapor]

no it would not

wait

maybe

[worthy herald]

it took it

said its right

[gritty vapor]

yes it is, good job

[worthy herald]

with the same logic it is this

[gritty vapor]

the graph part looks right, but not the checkbox question

[worthy herald]

am i missing one

[gritty vapor]

you're missing a couple and wrongly checked a couple

think: what does s mean, in common terms? what about s'?

[worthy herald]

im not to good at select all that apply

s is position and s' is acceleration

[gritty vapor]

s' is not acceleration

[worthy herald]

rate of change

[gritty vapor]

but as one word, in context of the car?

[worthy herald]

velocity?

speed

[gritty vapor]

velocity it is

check each statement again

[gritty vapor]

what did you think to get to your answers for D to F

[worthy herald]

it would be f because it would be moving if it is positive?

if it were to slow sown wouldnt it go into the negatives?

[gritty vapor]

it can be slowing down while having a positive velocity = s'

[worthy herald]

so its not like rate of change of velocity?

im confused

oh t is time and y is distance

and s'(t) is just when they are moving?

so this graph cant go negative?

[devout snow]

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[dull thunder]

I know lagrange but this one says |f^(n) (x)| <= n/n+1 so I'm kinda confused on how I should solve

[devout snow]

@dull thunder Has your question been resolved?

[supple knot]

Do you know the formula for the error term

[dull thunder]

yea

[supple knot]

Yea so bound the right term using the inequality on the derivative

[dull thunder]

$\frac{f^{4}\left(x\right)\left(x-1\right)^{4}}{4!}$

[woven radish]

ruby

[dull thunder]

would it be like this?

idk how im gonna use the given condition

[supple knot]

Put absolute values and bound this from above using

a <= |a|

So not so much "putting absolute value signs" but rather using the simple inequality above

[dull thunder]

$f^{4}\left(x\right)\le\frac{4}{4+1}=\frac{4}{5}$

[woven radish]

ruby

[dull thunder]

this is for f right?

$R\ \le\frac{4}{\frac{5}{4!}}\cdot1^{4}$

[woven radish]

ruby

[dull thunder]

so R is like this?

hence its B after simplified?

[supple knot]

Yes

You should understand why (x-0)^4 <= 1^4

And why n=3

[dull thunder]

isnt n=3 cuz its third degree and 1^4 cuz it told to use 1 and (x-0) cuz its f about x=0

[supple knot]

Half right. The point x in the remainder is "between" two other points in this case 0 and 1. So 0 < x < 1.

[dull thunder]

its like the interval for max right

[supple knot]

Correct

[dull thunder]

ok

thanks for helping 🙂

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[eternal fable]

this is the question that i'm struggling with, this is what i have so far, i just want to see if someone can double check that my answers are correct or work for the question.

[pseudo basin]

yeah seems ok

[eternal fable]

Are the steps clear enough do you think for submission? Or should I add comments saying exactly what I did?

[pseudo basin]

i think it's fine

[eternal fable]

Ok ok, tysm 🙏🙏

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[hollow sun]

can anyone help me with this question? (11)

[hollow sun]

I’m not sure where to start

[lament kraken]

hello

Firstly you know that the arc measure of TV is 12x and that angle TUV is 3y

So what did that tell you

[hollow sun]

uhh that there’s two different variables I’m working with? not too sure

[lament kraken]

hmm ok

In the figure right

What does the arc measure of TV mean

like what angle does it correlate to

[hollow sun]

TUV right?

[lament kraken]

Yes

So 12x = 3y right

[hollow sun]

ohhh is the inscribed angle equal to the arc?

[lament kraken]

yes

[lament kraken]

you got it yourself

[hollow sun]

so is there 9y piece of information relevant here at all or is it just there

[lament kraken]

wait I'm coming to that

[hollow sun]

kk

[lament kraken]

but for now you understand that 12x = 3y so y = 4x right

[hollow sun]

yup

[lament kraken]

okay good

Now what angle does the measure of arc UV relate to

[hollow sun]

uhhh I can’t see any one specifically I just see the semi ish circle with the 9y

[lament kraken]

ok

Connect TV and try again

[hollow sun]

3y?

[lament kraken]

wat

i meant which angle

Like

[hollow sun]

idek😭

[lament kraken]

oh fuk wait

shit i got it wrong

@hollow sun sry forget that 12x = 3y

sorry

[hollow sun]

kk it’s good

[lament kraken]

connect SV

S is the center

So the measure of TV is angle TSV right

[hollow sun]

yup I see that

I connect it

[lament kraken]

Ok

So angle at center = 2 x angle at circumference right

essentially TSV = 2* SUV

can you find TSV now?

[hollow sun]

central angle is 2 times the corresponding arc?

[lament kraken]

yes

This

[hollow sun]

so would I do to get TSV is just divide the 12x by 2? So 6

6x for TSV

[lament kraken]

no

TSV is defined as 12x

because the measure of TV is 12x

What we want to do is find TUV in terms kf x

And since TUV is angle at circumference

It is equal to half of TSV

Which is...?

[hollow sun]

6?

[lament kraken]

yes

so now we have 6x = 3y right

[hollow sun]

Yes

[lament kraken]

good

so y = 2x

[hollow sun]

Yes

[lament kraken]

Now if we connect SV

what is the measure of arc UV in terms of angle

like which angle is it equal to on the diagram

[hollow sun]

9y?

[lament kraken]

on the diagram

like arc means angle respective to center

in this case what is that angle

[hollow sun]

I don’t see a number for it but whichever one with S so like VSU or USV

[lament kraken]

Yes thats correct

So we have USV is 9y right

[hollow sun]

Yes

[lament kraken]

Nice

So now we have TSV and USV

Their sum is 180

[hollow sun]

because they’re linear pairs?

[lament kraken]

Yes

[lament kraken]

You can see it right

[hollow sun]

Yes

[lament kraken]

So 9y + 6y = 180

Whats y

[hollow sun]

12 if I divide 180 by 15y

[lament kraken]

correct

Now TUV is 3y

which is?

[hollow sun]

Huh

[lament kraken]

if y = 12 whats 3y

because TUV is 3y in the diagram

[hollow sun]

yeah are u asking me to combine both the y together?

OH

replace it

36

[lament kraken]

yes

that's our answer

yay

the 12x was useless

[hollow sun]

yup in the formative said it was right✅tyyyyy!

[lament kraken]

np

[hollow sun]

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[south lagoon]

why does the d^2 not get differentiated into 2d

[south lagoon]

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[fossil locust]

the principle is $\frac{d}{dx} (d(x))^2 = 0 \implies 2 d(x) d'(x) = 0$

since you know the distance is not $0$, differentiating $d(x)^2$ and setting to $0$ is the exact same as solving $d'(x) = 0$

[woven radish]

south

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[eager stone]

Im confused on why we are suppose to take f(x,y,z)=x^2+y^2+z^2

also pls ping me if u answer

[fossil locust]

$\sqrt{(x - 0)^2 + (y - 0)^2 + (z - 0)^2}$ first

[woven radish]

south

[fossil locust]

if you want to minimise sqrt(d(x)) then it suffices to minimise d(x)

it literally says "square of the distance from origin (0, 0, 0)" also on your picture

[eager stone]

i guessed this from that text yea but we can just toss the root?

[fossil locust]

yes

[eager stone]

ohh okok i get it

ty

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[south lagoon]

when d gets substituted into d^3, the cube power becomes 3/2, why?

[sand quarry]

What does "d gets substituted into d^3" mean?

[sand quarry]

What is d

[south lagoon]

d is the distance here

its a pythagorean theorem formula

[pine patio]

What are we trying to find?

[south lagoon]

full question:

full answer manual if that helps too

[sand quarry]

The reason why is because [
d = \3{(h-4)^2 + x^2}
]
In this case by pyth theorem

[south lagoon]

OH

its just multiplying 1/2 by 3

[pine patio]

Exponent law

[south lagoon]

right

[sand quarry]

So,
[
d^3 = \9{\3{(h-4)^2 + x^2}}^3
]

[sand quarry]

Yes

[pine patio]

[south lagoon]

alright I see now, thank you. i missed that square root

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[scenic pawn]

hello, i'm stuck with this question :

We consider a matrix A ∈ M3(R), which we will not attempt to calculate, verifying the following properties.

Give the eigenvalues ​​(and for each of them a basis of the associated eigenspace) of A,
and deduce that A is diagonalizable over C.

it's clear that the first eigenvalue is 1 and the second is (-2+i), how do i find the third ?

[devout snow]

@scenic pawn Has your question been resolved?

[devout snow]

@scenic pawn Has your question been resolved?

[lunar harbor]

Insufficient information to determine the third eigenvalue

Example 1 with a third eigenvalue of 0

Example 2 with a third (repeated) eigenvalue of 1

[scenic pawn]

so the third is just 0 ?

[fossil locust]

no, Wolfram seems to be telling you that there are only two eigenvalues

1 and -2 + i

distinct eigenvalues imply that the matrix is diagonalisabl

the converse is however not true

[scenic pawn]

but A ∈ M3(C), if there's only two, that means one is a double

[fossil locust]

yeah so 1 is the double eigenvalue

[fossil locust]

oh wait that was an example sorry

I misread

[fossil locust]

the third could be anything

[scenic pawn]

so i just take x as the third ?

but without the third eigenvalue, i can't get the associated eigenspace

[lunar harbor]

congrats - you have a bad question or missing context

[scenic pawn]

here is the full exercice ( in french):

[lilac moat]

oh thats a very different question

[scenic pawn]

1 Give the eigenvalues ​​(and for each of them a basis of the associated eigenspace) of A,
and deduce that A is diagonalizable over C.

2 solve the differential system Y'=AY

[lilac moat]

you miswrote the first line

[lunar harbor]

[scenic pawn]

?

👀

i don't get it

[lilac moat]

not what you wrote here

[scenic pawn]

yes mb it's A ∈ M3(R) not M3(C)

but does it change smthing ?

[lilac moat]

yes very much

[scenic pawn]

can i deduce the second and the third eigenvalue with the 2nd poperties ? or do i need something else ?

[scenic pawn]

and btw what is this tool ? it seems to be useful

[fossil locust]

https://www.wolframalpha.com/

[scenic pawn]

thx

[lunar harbor]

Real entries -> characteristic polynomial has real coefficients -> ||what does this mean about your non-real roots?||

[scenic pawn]

it means the characteristic polynomial isn't split on R

[lunar harbor]

I have no idea what “split” means

[scenic pawn]

i will search the right word

does the third eigenvalue is 2-i ?

and the characteristic polynomial have at least 2 complex solution ?

[sullen island]

yes indeed

sometimes I can't read shit

-2-i should be your other eigenvalue

complex conjugate of -2+i

[scenic pawn]

i get it

thanks for helping and sry for my misswrote

.close

[devout snow]

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[sullen island]

well you still need to find the eigenspace for that other eigenvalue

gg if you already did somehow

[devout snow]

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[rain coral]

Can someone explain

[rain coral]

The flipping part

Of d

To find a

W

I don’t quite understand

I did 1- it’s wrong

But if u negate, it it’s correct

[devout snow]

@rain coral Has your question been resolved?

[devout snow]

@rain coral Has your question been resolved?

[devout snow]

@rain coral Has your question been resolved?

[devout snow]

@rain coral Has your question been resolved?

[charred zodiac]

Hello. Consider P(G > w) = P(Z > z1),
and P(B < w) = P(Z < z2)

For both probabilities to be the same, then z1 = -z2 because of normal distribution's pdf symmetry

That's why they negate it

[devout snow]

@rain coral Has your question been resolved?

[green bone]

My calculus exam is coming up and I forgot some important stuff

#help-27

My calculus exam is coming up and I forgot how to solve half the questions, can you help me with some of them

[somber heron]

Which parts in calculus

[devout snow]

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[lilac moat]

ooh that's a classic

do you know what |X|<=|Y| means if one of them is infinite?

well, i mean, it means the same thing if they're finite too @cunning kettle

[devout snow]

@cunning kettle Has your question been resolved?

[cunning kettle]

oh sorry the ping didnt go through

uhm

well |Y| isnt like "defined" if Y is infinite right

but we take it as like

the cardianity of an infinite set is greater then one of a finite set

at least from my understanding

i also know theres like layers

countable uncountable etc

countable if there exists a bijection w the natural numbers

@lilac moat

[lilac moat]

hi

[lilac moat]

we define the inequality even if we dont define the symbol on its own

[cunning kettle]

yeah ok

[lilac moat]

|X|<=|Y| means there is an injection from X to Y

similarly

[cunning kettle]

oh yeah

[lilac moat]

|Y|>=|X| means there is a surjection from Y to X

which is a cooler way to say the theorem bc then you see the notation doing its thing

[cunning kettle]

yeah those two makes sense like logically

[lilac moat]

so any idea which one to do first? one direction is easier and doesnt need choice

[cunning kettle]

why do we need to do this one

[lilac moat]

nbd if you choose the harder one first

[cunning kettle]

i mean we can do both sure

uh

[lilac moat]

wdym do this one? its the premise in the exercise

the exercise says to prove if and only if

[cunning kettle]

the question only asks to prove iif surjection tho

yeah

[lilac moat]

thats both ways

|X|<=|Y| iff |Y|>=|X|

[cunning kettle]

so prove if surjection exists then |X|<=|Y|, and then prove if |X|<=|Y| then surjection exists right?

[lilac moat]

yes

im just saying it in a different notation

[cunning kettle]

oh i see i see

mb

ok

uh

"surjection exists then |X|<=|Y|"

feel like this one would be easier no?

doesnt rlly matter tho

[lilac moat]

yeah you have to do both anyway

but only one way needs choice

[cunning kettle]

do we know that

[lilac moat]

so there is a surjection s:Y to X

[cunning kettle]

the imf of a function

the size of the image of a function is less then or equal to the size of the domain

am i mixing it up

[lilac moat]

you're right but "size" gets weird for infinities so this isnt so helpful

[cunning kettle]

ok im skipping steps sorry

oh true

[lilac moat]

youre not wrong

[cunning kettle]

ok

so

surjection exists

so the image of the function = X

[lilac moat]

meaning s(X)=?

noo surjection is s:Y to X

oh

i misread you! my fault

[cunning kettle]

ok

[lilac moat]

s(Y)=X

we need to go from X to Y

[cunning kettle]

wdym by s(X)? sorry im not familiar w that notation

[lilac moat]

s(X)={y in Y|y=s(x) for some x in X}

the image of s under X the image of X under s (I got it backwards at firstt)

[cunning kettle]

oh wait

im stupid

u named the function s

in my head i named it f

[lilac moat]

yeah s for surjection

[cunning kettle]

ok yeah

yeah so s(X)=Y

nope other way

lmao

s(Y)=X

[lilac moat]

suppose s were a bijection, how would you find an injection from X to Y?

[cunning kettle]

inverse?

[lilac moat]

yeah what does that inverse look like

might help to draw a picture depends on how you think

[cunning kettle]

how do we know that the inverse exists tho

[lilac moat]

we don't, I'm just helping you get the proof off the ground

[cunning kettle]

oh ok

well the inverse would be s^(-1)(X)=Y

[lilac moat]

btw which version of the choice axiom are you using, there are 3

[cunning kettle]

this is the one my teacher gave me

[lilac moat]

sure that's valid

I meant to quote the picture

[lilac moat]

sure that's valid

[lilac moat]

yeah and what about elementwise?

[cunning kettle]

hm

wdym

it would be

[lilac moat]

s^-1(x)=s^-1({x}) is what im trying to get you to say lol

see the difference between those two things?

$s^{-1}(x)=s^{-1}({x})$

[woven radish]

gfauxpas

[cunning kettle]

why would those be equal though

[lilac moat]

the left hand side is an inverse function, the right hand side is a set

[lilac moat]

because I said, "if s were bijective..."

[cunning kettle]

wait

[lilac moat]

it's a theorem that a function is bijective iff invertible

[cunning kettle]

how does s^-1({x}) make sense

how does the function take in a set

as its input

[lilac moat]

the notation means

$s^{-1}({x})={y \in Y: s(y)=x}$

[cunning kettle]

oh ok

[woven radish]

gfauxpas

[cunning kettle]

ok yeah that makes sense

[lilac moat]

so then s^{-1} would be a mapping, and in particular it would be injective

because it's invertible

and so its inverse is invertible too

[cunning kettle]

yes

[lilac moat]

problem is we dont have that s is injective, only surjective

[cunning kettle]

i see how the logic would carry over with surjective now

tho

[lilac moat]

o ya?

[cunning kettle]

bc the image of s is the codomain of s, so the domain of s^-1 is the image of s which therefore is the codomain of s
so the inverse of s is defined for every element in X

and by definition of a function (two inputs cannot map to a single output), in s, the surjection, every element of X in the codomain is mapped by a distinct element of Y
so in the inverse s^-1, an element of X maps to a distinct element of Y, which is the defintion of an injection

smth like this?

[lilac moat]

you're clsoe

problem is

s^{-1}({x}) is not guaranteed to be a singleton for all x in X

in fact, there may very well be infinitely many x in X such that s^{-1}({x}) has infinitely many elements of Y

you are right that a function has to have one output for every input

so if there is some x in X such that s^{-1}({x}) contains more than one element

you need some way to ... well,

choose an element

if you know what I mean

[cunning kettle]

lmao

wait ok

so

we have the inverse s^-1

[lilac moat]

which is NOT GUARANTEED TO BE A FUNCTION

but IS guaranteed to be a non-empty set s^{-1}({x}) for every x in X, because s is a surjection

In other words:

[cunning kettle]

😭 let me think about it

[lilac moat]

okay!

[cunning kettle]

my brain is so fried

[lilac moat]

you're close!

the other direction of the proof is much easier btw

[cunning kettle]

i think s^{-1}({x}) is throwing me off ive never used it

[lilac moat]

Here's another way to say it that might make it easier to think about

[cunning kettle]

is s^{-1}({x}) just the inverse where the domain of the inverse is the area of the codomain of s that is defined by s(x)

[lilac moat]

know about partitions and equivalence relations?

[cunning kettle]

yeah

[lilac moat]

s^{-1}({x}) means "the set of all element y in Y satisfying s(y)=x"

[lilac moat]

alright cool, so, think about X=s(Y)

and partition it into equivalence classes of:

two elements a and b are equal if s(a)=s(b)

[cunning kettle]

ok

[lilac moat]

wait did I say it backwards hold on

[cunning kettle]

okok

ok

i think im getting ti

so

s, our surjection

maps Y to X

[lilac moat]

I was backwawrds

s:X to Y is a surjection

[cunning kettle]

its a surjection so we know that every element in X is connected

[lilac moat]

which means X can be partitioned in a way that

[cunning kettle]

but theres more Y elements then X elements

[lilac moat]

a~b if s(a)=s(b)

[cunning kettle]

so when we take the inverse

a certain element in X

can map to multiple elements in Y

so we use the axiom of choice

just to pick one?

[lilac moat]

yeah basically!

you just need to say

what the collection of nonempty sets is

[cunning kettle]

and the nonempty sets is the set of elements in Y that map to a certain element x in X

which is the notation u used

ok ok

it clicked

[lilac moat]

YEAH!

https://tenor.com/view/confetti-style-gif-19616552

${ s^{-1}({x}) | x \in X}$

[woven radish]

gfauxpas

[cunning kettle]

can i write this down formally before we move on ._. i wanna write it down before it unlicks

[lilac moat]

of course

[cunning kettle]

thank u so much

[lilac moat]

i have to go actually but the other direction is much easier

[cunning kettle]

oh ok

[lilac moat]

and doesnt need choice

good luck

[cunning kettle]

thank u

[lilac moat]

🙌

[cunning kettle]

❤️

.close

[devout snow]

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[floral cradle]

How do I integrate tan^4x 😭😭😭

[lost laurel]

$\int tan^2(x) \cdot tan^2(x)dx = \int tan^2(x) (sec^2(x)-1) dx$

can you do it from here

[hollow ice]

$tan^2x$ is $sec^2x - 1$ not $sec^2x + 1$

[woven radish]

Bacter10Fr4g

[lost laurel]

oops

yes

😔

[woven radish]

What a wonderful world !

[lost laurel]

Do it from here

[floral cradle]

If I do u=tanx I'll get du=sec^2xdx. But what then? I have a (1-sec^2x)

[lost laurel]

see this?

can you simplify this further

[floral cradle]

I'm not sure

[lost laurel]

Notice this is $\int tan^2(x) sec^2(x) - sec^2(x) dx$

[woven radish]

What a wonderful world !

[hollow ice]

Do you know your derivatives? Like those of sec or tan or sec^2 or you know, in general the terms involved in this integral? If you dont, calculate those, and see if that helps

[violet wind]

Hint: ||split up the integral w/ the -||

[floral cradle]

I got here

I'm very lost, not sure if I should have u subbed before this or if I should just integrate each of these separately.

[gray coyote]

cant u just solve this by taking tanx = u for the first integral and second is tanx anyways

[floral cradle]

oh you're right, I just realized that

[devout snow]

@floral cradle Has your question been resolved?

[devout snow]

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[devout snow]

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[solid marlin]

What this means?

[pseudo basin]

$20 \uparrow^9 6$

[woven radish]

Ann

[pseudo basin]

this?

i would guess it's Knuth up-arrow notation.

[solid osprey]

https://en.wikipedia.org/wiki/Knuth's_up-arrow_notation this no?

[solid marlin]

Yes I think

[solid marlin]

Is it useful? I found the number related to an fps for a game and I'm not sure what it's equal to

[topaz axle]

so it's 20 six times, with ↑8 in between

[pseudo basin]

????

can you show the context in which you found it

[solid marlin]

OK one moment

[pseudo basin]

it's a fuckoff massive number, what's it doing in a FPS

[solid marlin]

Yes so I made this level for a game and apparently it is that fps to play

Something like that

[solid osprey]

gd impossible levels ahh

[lament kraken]

fr

[pseudo basin]

???

[solid osprey]

afaik its supposed to be like the theoretical minimum fps to beat a level in a platformer

[pseudo basin]

someone decided to mess around

frames per second????????

[solid osprey]

note that its 31st place

[solid osprey]

yeah

[pseudo basin]

yeah they're messing with you

[solid marlin]

The mechanism I made basically like counted your clicks and it was like layered so when you click once it moves a target block

[solid osprey]

which list is that

[solid marlin]

Wtpll

[pseudo basin]

how did you get this up-arrow notation as the fps

[solid marlin]

I didn't know it was a big number though like absurd big

[pseudo basin]

where did it come from

[solid marlin]

That's what the list moderator put it as

This is the list explanation

[solid osprey]

wtf is this top 1 💔

[solid marlin]

Oh yeah I saw that

[lament kraken]

what

What game is that

[solid osprey]

gd

[solid marlin]

Geometry dash

[lament kraken]

Actually?

no shot which website

last hope was named correctly 💀💀💀

[solid marlin]

It's a Google doc

Tpll

Can I link it here?

[pseudo basin]

hm

googologically big numbers...

[solid marlin]

Idk what that is

https://docs.google.com/document/d/1GvNWez3Vh3Q6AQXLshYOmA3sXG-9JbV6Zf7ZzRpba0o

Yeah it's that list

Idk that it was that big tho I just saw the mechanism in other levels and made my own like it

Here's my mechanism

[wicked rover]

back when i played gd ten years ago we didnt have half of this newfangled tech. those were the days

[solid osprey]

i remember i could only play the lite and clutterfunk was just added in lite

[solid marlin]

Better angle

I'm pretty sure it does something like this

[solid osprey]

iad is still a thing?

[wicked rover]

i joined when electrodynamix was new. i remember when hexagon force came out and everyone went nuts over slopes and duals

[solid marlin]

It's an old video but I think it's still around

[devout snow]

@solid marlin Has your question been resolved?

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[wind mason]

Sure

I cannot LaTeX

One second

Pls work

fuck

OAWDHOA

close but thank you

Like you can assign $\frac{p}{q}=\frac{a}{\frac{b+\frac{c}{d}}}\ldots}$

[woven radish]

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

[wind mason]

aight im out

[devout snow]

@light hazel Has your question been resolved?

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[cursive bolt]

Hello! First time here but I am working on Algebra 2 right now, and we are doing divison and multiplication of rational expressions. I am not understanding any of it and am hoping for a step by step breakdown of the solving process (Explain like I'm five)

[lilac crescent]

you are 5 years old ?

[cursive bolt]

Explain like I'm five I mean lol

[lilac crescent]

lol

send the exercice you are struggling to understand

and we can help

[cursive bolt]

Okay!

[pseudo basin]

im also gonna throw this probing question out:

are you comfortable with multiplying and dividing fractions (specifically, ones that have just numbers, and not algebraic expressions, on the top and bottom)?

[cursive bolt]

Yes

I can do basic fractions, it's just when you start adding letters in that it turns into a mess

[pseudo basin]

that so?

well, again, show us the question(s) you're struggling with.

[cursive bolt]

Trying to upload it, apoligies

[lilac crescent]

so we will start with the first one ?

[cursive bolt]

Yes please

[lilac crescent]

!status

[devout snow]

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

[cursive bolt]

1

[pseudo basin]

ok let's see then

[lilac crescent]

you know that dividing is multiplying by the inverse ?

[pseudo basin]

... ok yeah rajel you can handle this ig

[cursive bolt]

Mhm

So we flip it and multiply correct?

[lilac crescent]

yep

,, \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b} \cdot \frac{d}{c}

[woven radish]

<rajel />

[cursive bolt]

Okay, so 21x^3 over 5z^2?

[lilac crescent]

why 5z^2 ?

[cursive bolt]

z^4

My bad

[lilac crescent]

yep , correct

[cursive bolt]

Okay, so from there, what's the next step

[lilac crescent]

you cant simplify it no more

[cursive bolt]

So the answer is just that?

[lilac crescent]

yeah

[cursive bolt]

Okay, that makes more sense

So for all the division you simply flip the second one and then multiply?

Then simplify if possible?

[lilac crescent]

yep

,, \frac{\frac{a}{b}}{c}=?

[woven radish]

<rajel />

[cursive bolt]

And multiplication you just skip the flipping step?

[lilac crescent]

you would also use this one

[cursive bolt]

Okay

[lilac crescent]

what is the value ?

[cursive bolt]

Of the equation in the image?

[lilac crescent]

yep

[cursive bolt]

A/B x C? Sorry for all the questions I'm trying to wrap my head around this

[lilac crescent]

good

it should be clear now , try the other questions

[cursive bolt]

Okay

[lilac crescent]

you'd also simplify long expressions to a multiplication of 2 expressions

[cursive bolt]

Could I get an example of that?

[lilac crescent]

i.e if you know the quadratic formula

[cursive bolt]

Oh okay

[lilac crescent]

in the 5th question , how can you rewrite the first expression

,, y^2-10y+9

[woven radish]

<rajel />

[cursive bolt]

It looks like it is already in that form, at least the first numerator is

I think I'm missing something

[lilac crescent]

you know the quadratic formula ?

[cursive bolt]

Yes

[lilac crescent]

try it here

[cursive bolt]

Ax^2 + Bx + C

[lilac crescent]

\begin{align*}
&P(x)=ax^2+bx+c \
&\alpha,\beta \text{ are zeros of the polynomial } P \ \
&\rightarrow P(x) = a(x-\alpha)(x-\beta)
\end{align*}

@cursive bolt

[cursive bolt]

Okay

[lilac crescent]

this is how you simplify a polynomial

[woven radish]

<rajel />

[cursive bolt]

So solving for alpha and beta?

[lilac crescent]

yep

,, x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

[woven radish]

<rajel />

[cursive bolt]

Oh okay

Checking my equation for # 2

X^3 + 4X^2 - 9x - 36 over X^2 + X - 12

.resolved

[lilac crescent]

!done

[devout snow]

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

[cursive bolt]

.close

[devout snow]

Channel closed

Closed by @cursive bolt

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[cursive bolt]

Thank you!

[devout snow]

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• Show your work, and if possible, explain where you are stuck.
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[granite bough]

how to solve this?

[granite bough]

i changed y for y=rsintheta

and i changed dx to dr and dy to dtheta

and i havw drawn the circle and filled in the area i am supposed to find

the area above y=-x but below y=x is a fourth of the circle

and half of the lower first quadrant and half of the upper fourth quadrant

the jacobian for polar coordinates is equal to r

so i have the double integral of omega r^2sin(theta)drd(theta)

im having a hard time finding the limits or whatever they are called for integrals

i need 4

domain and bound maybe they are called idk

0 < r <= 1

for r

idk for theta

is it -pi/4 <= theta <= pi/4

the only reason im guessing that is because y=x and y=-x

and i just know from before they will cross the circumferance of the circle at sqrt2/2 and -sqrt2/2

[faint gorge]

yes

[granite bough]

when doing the double integral after finding the domain and bounds

do i include the constant after the first integral is done

actually nvm

they are definite

no constant

oops

[devout snow]

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[lone marsh]

how do i do this?

[lone marsh]

<@&286206848099549185>

[brittle badge]

!15min

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[lone marsh]

??

OH

Sorry I didn't notice

(why did i think 10 minutes was 15 minutes)

My bad sorry

.....

[devout snow]

@lone marsh Has your question been resolved?

[halcyon cargo]

what have you done so far

[lone marsh]

nothing

[halcyon cargo]

did you calculate p(-t-1)?

try that first

[lone marsh]

ok

one sec doing

actually wait

[halcyon cargo]

[lone marsh]

got it

thanks

.close

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[halcyon cargo]

very nice

[gray coyote]

sry but how did 4th step become the 5th step

[halcyon cargo]

for example, derivative of (x+1)(x+2)(x+3) is (x+1)(x+2)(x+3)*[1/(x+1) + 1/(x+2) + 1/(x+3)]

[gray coyote]

oh so its a standard result then

[halcyon cargo]

so the thing inside is just derivative of (x+1)(x+2)...(x+2020)

[gray coyote]

yeah ok thanks

[halcyon cargo]

guess you could say that

just product rule tho

[valid iron]

Reverse chain rule pretty much

[devout snow]

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[tribal root]

Question is: Triangle ABC is in a circle with the mid point M. Length of AC is the same length as the circles radius. The angle BAC = 40 degrees.

[tribal root]

I’m stuck on what do to do figure out C angle

[radiant dune]

One side is given to be equal to the radius, can you tell what the other 2 are

[tribal root]

i figured that is the circles radius

Those other lines

Not sure on the one with no angels in it

[radiant dune]

angles opposite to equal sides

Are also equal

[tribal root]

So C is 60 + 60

?

[radiant dune]

No

Im talking about the triangle made by 2 radii and AC

[tribal root]

Yeah

Thats why each corner has 60 degrees

We need to find angel V

Forgot to say

[radiant dune]

Cool, now do you know the relation between angle subtended by an arc at the centre and at the circumference

[tribal root]

You mean that M/2 is C?

[radiant dune]

Because angle AOC, as you said is 60

[radiant dune]

What is M

And no im not even talking about C yet

[tribal root]

Mid point

[tribal root]

Oh alright

[radiant dune]

You mean the centre

Angle ABC = 1/2 AOC

Where O is the centre

[tribal root]

O is where?

Where M is

[radiant dune]

Yes

O is the centre of the circle

Or M as youre calling it

[tribal root]

So V is = O/2

[radiant dune]

And v is just equal to ABC

[radiant dune]

Not good nomenclature but yes

[tribal root]

So v is 30?

[radiant dune]

Yes

[tribal root]

And the rest of C is 110

[radiant dune]

The angle ACB is 110 yes

[tribal root]

Alright thank you

.close

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[worthy herald]

im confused

[worthy herald]

.clo0se

.close

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[barren quiver]

how do you find the area of a sector of a circle

[lament kraken]

Are the radius and angle given?

[barren quiver]

only the angle

which is 64