help-36

[tribal ocean]

there was a prev part

[sly forum]

oh it is 2 parts

[tribal ocean]

so actually

you can write x y and z in terms of a third parameter, say t

since its one dimensional

but what you have is also probably fine

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[late gazelle]

Idk how you would make t the correct parameter

[tribal ocean]

parameterize the circle

of y and z

[late gazelle]

Since I don’t see any easy way to sub t = some combination of x y and z

[tribal ocean]

in terms of t

use r cos t and r sin t

plus a constant

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[late gazelle]

… now we getting I to stuff idk so I think it’s best to leave it

[tribal ocean]

bro but i think your answer is not final though

i can already see the answer

i forgot about it until now

[late gazelle]

It just needs to be expanded

And it’s correct so idk

[tribal ocean]

it is but its not an explicit expression

for x y and z

if you can solve an explicit expression you usually should

and in this case its actually not that hard

if this is calc 3 then thats what they would usually expect

[late gazelle]

Ahh

It’s not calc 3

lol

[tribal ocean]

oh lol

even still they might expect it

[late gazelle]

It’s vector calc or whatever we’re doing

[tribal ocean]

uh

is that after calc

[late gazelle]

Yea

[tribal ocean]

then its the same thing

so u should simplify

[late gazelle]

But it’s also not calc 2 or 3

Our unit is just “vectors in 3D and vector calc in 2D”

So what we have is fine for the level im at

[tribal ocean]

okay

i just want to see what the final answer looks like to double check

[late gazelle]

And I’ve gtg soonish so I don’t have time to do much more

[late gazelle]

Just expand the answer I got at the end

[tribal ocean]

will u be here in like a few mins

[late gazelle]

Uh probably not, I’ll be here later today more than likely

Ty for the help, I’ll look into parameterisation when I get the chance, since it sounds like it simplifies the answer more

[timber leaf]

Dude is doing Specialist Maths 🥀

[late gazelle]

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[radiant igloo]

They didn't, I edited the ping in.

[radiant igloo]

Oh shit

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[bronze vector]

I need help idk how to do thks

[tiny gorge]

try finding a formula?

[bronze vector]

Well ik we must use the inf geometric one

[tiny gorge]

say, total distance traveled when it hits the ground for the N'th time

[bronze vector]

first, when it says its thrown out the window from 24 feet, what is the 1st term?

[glossy zephyr]

Thats the "problem", the first term isnt part of the series itself.

Consider, for each bounce, the ball goes up and down, making the same distance twice.

Except for the first one, where its thrown directly down

[bronze vector]

this is why its confusing

[pliant smelt]

$2\sum_{i = 1}^{\infty} s_i - s_{1}$

[robust mulch]

the easiest way imo is to track the first down, the first up, then find you series based on that total distance

[soft zealot]

destiny

[bronze vector]

uhhh i dont need sigma notation for this problem i dont think

[robust mulch]

no, you dont, but if youre using a geometric sequence, you are anyway

[bronze vector]

ok so then explain what i should do

[robust mulch]

track the distance in the first box, then base your series off that

[vital crag]

How else did you learn to sum an infinite number of terms

[bronze vector]

first term over 1-ratio i think

[pliant smelt]

This kinda assumes the final bounce's covered distance tends towards 0 as i approaches infinity, does that mean extra work?

[vital crag]

Well yea that's only after you get the infinite sum using sigma notation here

[robust mulch]

what

[bronze vector]

yall are confusing me i should mind aswell ask ai or something

[robust mulch]

ok have fun

[glossy zephyr]

what do you know about geometric series?

[bronze vector]

geometric series multiplies by a number

[glossy zephyr]

Would you be able to give an example

[bronze vector]

yes

we could say a series could have a geometric series of 5(2)^n-1

[vital crag]

That would be infinity

[bronze vector]

thats a geometric equation

[vital crag]

And you wouldn't be able to apply this formula

[bronze vector]

he didnt ask for an infinite geometric series

[vital crag]

The word series implies infinite number of terms

[pliant smelt]

I think it would be helpful for him to unravel the origin of that formula

[robust mulch]

thats the plan

let him work

[bronze vector]

i asked ai and it said the first bounce is 2/3 of the first drop which is 24 so it would go up 16 and because a bounce is up and down the first bounce is 32

so, we plug it in to the formula

[pliant smelt]

[bronze vector]

32/1-2/3 +24

and so we should get 120

[glossy zephyr]

🥀

[robust mulch]

good for you

[pliant smelt]

bruh

[glossy zephyr]

Well, crash course of what a geometric series is:

[pliant smelt]

it's not even correct

[robust mulch]

yes it is

anyway can we get someone in the door instead of all jamming the way

[vital crag]

!done

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[bronze vector]

Am i wrong?

Yall aint helping just confusing me thats why i asked ai, but am k right or wrong?

So like is everyone just going to disappear

[robust mulch]

i guess so

anyway 120 is correct

[bronze vector]

Destiny said its not correct though

Idk if its the blind leading the blind or what

[surreal vault]

yes, 120 is correct. although, from backreading, i'm sort of not convinced that you got what the problem is trying to show you.
like for example if i made it so that i want the distance travelled in the ball's first three bounces instead of until it comes to rest, how would you approach that?

[bronze vector]

Well, i would probably do y= 24(1-2/3^3)/1-2/3 then i think i would times it by 2 cus of the double bounce

[surreal vault]

sorry, i dont get your notation. do you mean $\frac{24(1-(\frac{2}{3})^3)}{1-\frac{2}{3}}$?

[soft zealot]

insult

[bronze vector]

Yes

Then times 2

Cus we got the double bounce

[surreal vault]

its close. however, the 24 is wrong here. you're basically saying that the first bounce of the object will reach 24 ft, but it doesnt

for posterity, let's try your equation for just the first bounce. following your equation it would $\frac{24(1-(\frac{2}{3})^1)}{1-\frac{2}{3}}=24$, multiplying it by two like you said gives us 48. but that's wrong.

[soft zealot]

insult

[bronze vector]

So how do we find this first term?

[surreal vault]

ill throw the question back to you. how do you think you should find it?

[bronze vector]

Well i think the first term should be where it starts which is 24 but its prob just the first bounce including both directions

[surreal vault]

24 is sort of important, yes. however, you need to do something to it first.

here is the diagram above but ive annotated it a little. where do you think i got the 16ft from?

[bronze vector]

The 2/3 of 24

[surreal vault]

ok, good. now, how do you think we could get the height reached by the second bounce?

[bronze vector]

2/3 that

[surreal vault]

how about the height reached by the next bounce?

[bronze vector]

2/3

[surreal vault]

2/3 of what?

[bronze vector]

Of the previous nimber

[surreal vault]

would you agree with me if i said that for bounce $n$, the height reached is given by $16(\frac{2}{3})^n$?

[soft zealot]

insult

[surreal vault]

e.g. bounce 3 would reach a height 16*(2/3)^3 = 4.74m

[bronze vector]

Sorry im brushing my teeth gimme a moment

[bronze vector]

Would you have to account for the distance both ways down and up?

[surreal vault]

i mean yes, but for now we are only counting the height reached

[bronze vector]

Ok the height not the distance?

[surreal vault]

i mean the distance is trivial, since its just twice the height.

[bronze vector]

Ok so were just doing 16(2/3^3)?

[surreal vault]

yes, for the sum of three bounces, you could get it from $2(\frac{24(1-(\frac{2}{3})^3)}{1-\frac{2}{3}})+24$. note the two outfront since the ball travels twice the height per bounce, and the initial 24 ft drop height

[soft zealot]

insult

[surreal vault]

however, ill continue the derivation for the original problem. notice how for each of these green "bounces", the distance travelled is twice the height, once going up and once going down

thus, if you add up all of these bounces so on and so forth to infinity you'll get the distance travelled for the entire motion

[bronze vector]

This makes sense

[surreal vault]

here, ive done some of the heavy lifting in factoring.

do you follow what i did in factoring the equation?

[bronze vector]

Not really but its fine i understand ehat i need to know thanks for the help

[surreal vault]

ok. the first part is i factored out the two from each of the terms, and put it outside, the second, i factored out 16, leaving it outside. the two multiply together to give us 32.

the yellow part highlighted should look very familiar to you.

[bronze vector]

Yea it looks like it does

[surreal vault]

it should be, because it is an infinite geometric series with ratio r=2/3.

thus, if you write it out, the distance becomes (via $S_\infty = \frac{a}{1-r}$), the distance travelled through all that bouncing is $d = 32(\frac{1}{1-\frac{2}{3}})$.

[soft zealot]

insult

[surreal vault]

this should be exactly what ai has given you. then, just add 24 for the initial drop.

[bronze vector]

Yes thats what it gave me

[surreal vault]

if you have no further questions about the approach, you may close the channel with .close

[bronze vector]

Ok thanks for the help

.close

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[tropic crest]

<@&268886789983436800> here too

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[coral ingot]

Would PL be 110*?

[timber leaf]

Nah

LNP is the big arc LP

[coral ingot]

So LNP would be 305 then?

[timber leaf]

Mhm good

[coral ingot]

thank you!

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[jagged flare]

How many ways can you choose $3$ numbers from the set ${1,2,3,\dots,21}$ such that the three numbers are sides of a non-degenerate triangle?

[soft zealot]

ihave<skissue>

[jagged flare]

is there any neat way to do this

other than like "manually" counting them

[glossy zephyr]

I assume some ceiling function bs will be involved in it if so.

[jagged flare]

mm wdym by that

[glossy zephyr]

Well, for a non-degenerate triangle, you have that a+b > c for any ordering of sides you choose

Notably, 1,2,3 isnt a valid choice, and you can start counting from 4 to 21, and only considering the choices below those numbers

[serene breach]

the bay harbour mathematician?

[jagged flare]

err what

[grizzled tusk]

Is the problem assuming without repetitions?

[glossy zephyr]

My idea for it is to fix one of the lenghts, and go about your way finding valid combinations of side lengths lesser than the one you fixed.

[jagged flare]

yeah

[grizzled tusk]

Any pair of sides has a sum larger than the third side
Let a, b, c be the lengths, and wlog, a < b < c, then the only inequality that matters is a+b > c
So we can start by counting triplets st a+b <= c and subtracting them from 21 choose 3

I think you can see it as a sum over c where you choose a and b with a bound over the sum

[jagged flare]

the answer is in the thousands im p sure, so you would need to do alot of them

actually i just had an idea what if you try recursion?

[jagged flare]

hmm

[grizzled tusk]

My intuition would be in the hundreds rather than thousands, since 21C3 is not that big

[jagged flare]

oh true

oh ok i got it

thanks guys

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[fiery comet]

Find the number of positive integers (n \le 100) such that (n \cdot 2^n + 1) is divisible by 3.

[soft zealot]

Ilikecats

[fiery comet]

n×2^n+1 is odd if n is odd
it's odd in both cases
odd multiple of 3
now that is clearly
3×odd
= 3×odd= n×2^n+1
now hm what?

That's the work I have done

I can think n×2^n leaves remainder 2 when /3

[radiant igloo]

hi so

[fiery comet]

Hello

[radiant igloo]

can we replace 2^n by something equivalent when n is for eg even

[fiery comet]

When n is even it's 2^(2n+1)?

[radiant igloo]

uhm no

when n is even

what would be the remainder when 2^n is div by 3?

[fiery comet]

1??

[radiant igloo]

exactly

[fiery comet]

We take two cases?

[radiant igloo]

yea

[fiery comet]

N is even n is odd

(N+1)/3 if n is even
That's

[radiant igloo]

since we only care about the remainder, we can take n2^n+1->n+1 (for n even)

[fiery comet]

Ok understood

What about n is odd

[fiery comet]

This is for n even check cases

[radiant igloo]

for n odd, 2^n gives 2 as remainder

[fiery comet]

Yeah

[radiant igloo]

so we can replace it with 2n+1

[fiery comet]

Ohhh

I understand

I solved the problem before but can't put values 😭

Thanks for clearing that

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[autumn agate]

lemme try smth

1s

[dire glacier]

uhh is (x^2 + y^2)^2= (x^4 + x^4)?

[autumn agate]

you dont know

im squaring both sides and trying to substitute

in the 2nd eq

lets see if it makes anything easier

[hazy bay]

1

[hazy bay]

it does

[autumn agate]

uh

i think i messed smth up

oh

wait

im getting -1

huh

oh

i forgot its -y/x

oops

yep 1

tysm

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[autumn agate]

.reopen

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✅ Original question: #help-36 message

[autumn agate]

new q

how exactly do i complete the identities

because iont think chain rule is the ideal way here

[timber leaf]

Omar

[candid pulsar]

use $\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1}a - \tan^{-1}b$ to simplify the terms

[autumn agate]

uh

[candid pulsar]

<@&268886789983436800>

[autumn agate]

oh so a = 5x and b = x?

[candid pulsar]

yes

[autumn agate]

crap im dumb

how dint i think of that

umm

we ping mods for that?

<@&268886789983436800>

[autumn agate]

i got final ans as

5/1+25x^2

does it look correct

nvm its correct

ty

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[lyric obsidian]

im lost

[lyric obsidian]

Basically i did the equation for the ratio or whateber you call it

r^2=6r-9

you get r^2-6r+9=0 -> (r-3)^2=0 so r=3

but clearly we aren’t dealing with a geometric series of ratio 3 so uh yeah

Idk how to do this

[rugged merlin]

yeah so

it is a repeated root

for a repeated root $r$, the general closed-form solution is always of the form [
u_n = (A+Bn)r^n
]

[lyric obsidian]

ohhhh

thanks

lesse

$4=(A+B)3^1$\
$6=(A+2B)3^2$

[soft zealot]

;(

[lyric obsidian]

so then A+B=4/3 and A+2B=2/3 -> B=-2/3 and A=2

$u_n=\qty(2-\frac23n)3^n$

[soft zealot]

;(

[lyric obsidian]

looks about right

thanks lex

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[lyric obsidian]

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✅ Original question: #help-36 message

[lyric obsidian]

one more question, part d)

im a lil rusty with group theory so im not sure how i should approach this

i know that a bijection is possible forsure, but idk about the homomorphism part

[royal gust]

One way is to identify orders

That is, the order of each matrix needs to match the order of the permutation it maps to

[lyric obsidian]

ah

[royal gust]

Order being the power n such that aⁿ = e, sry

[lyric obsidian]

i think i had B=3, A=C=D=2, E=4?

For the orders

[royal gust]

Oh lol the other questions ask you to find the orders. Sorry I'm slow

[lyric obsidian]

nah yg

i dont have mt paper with me anyways 😭

[royal gust]

Then we also want the order of the permutations

[lyric obsidian]

yeah that i forgot

[royal gust]

Know how to read the permutations? For example a says that 1 is sent to 2, 2 is sent to 3, 3 is sent to 1

What happens if you do that permutation twice?

[lyric obsidian]

you send 2 to 3, 3 to 1 and 1 to 2

and you do it again and you get 3 to 1, 1 to 2 and 2 to 3

so we’re back at e

so the order is 3

[royal gust]

This is what happens when you do it once

[lyric obsidian]

ou

[royal gust]

What we care about is what's actually sent where

[lyric obsidian]

oh shoot

would it be like 1 to 3, 2 to 1 and 3 to 2

[royal gust]

I'll let "applying a" be represented with an arrow, and I'll use two arrows to represent applying a two times
1 -> 2 -> 3
2 -> 3 -> 1
3 -> 1 -> 2

[lyric obsidian]

ahhh

that makes sense

[royal gust]

That's b. So we have a² = b

[lyric obsidian]

so then applying 3 times we get 1 to 1, 2 to 2, 3 to 3 which will be e

k so order of a is 3

Is there a faster way to do this

1 -> 3 -> 2 -> 1
2 -> 1 -> 3 -> 2
3 -> 2 -> 1 -> 3

b is 3

1 -> 1 -> 1
2 -> 3 -> 2
3 -> 2 -> 3

[royal gust]

We can do something like:
b³ = a⁶ = e² = e

And that's the lowest power of b that goes to e, since b² = a² ≠ e

Arguments like that can make it a little faster

[lyric obsidian]

Ahhh

k, that makes sense

[royal gust]

The order of every element must divide 6, so all orders are 2 and 3 (or 1 for e)

[lyric obsidian]

c is 2

[royal gust]

But the real fast way is to see how many elements get swapped

[lyric obsidian]

i see

cause like in c 1 just maps to itself so it only relies on the permutation of 2 & 3 right

Nd since they map to each other order is 2

[royal gust]

Yeye. We call any swap a "transposition"

[lyric obsidian]

d=2

f=2

so we have a=3, b=3, c=2, d=2, f=2

[royal gust]

Well done! That's not enough to find the homomorphism but it helps a lot

Since the order of the matricies need to match the order of the permutations

[lyric obsidian]

ill brb im doin smth

k im back

[lyric obsidian]

what else would we need

[devout mist]

Can't we use the homomorphism theorem

[royal gust]

I did get ahead of myself earlier. We identified a² = b

Is there two matricies, of order 3, such that m1² = m2?

[lyric obsidian]

i think i screwed up order of E

uhh lemme recalculate

,w {{0, 1}, {1, 1}}^2

[soft zealot]

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[lyric obsidian]

what

[royal gust]

Where 2 is 0

[lyric obsidian]

yeah

hmm so its actually order 4

,w {{1, 1}, {1, 0}}^2

[soft zealot]

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[lyric obsidian]

huh

interesting

,w {{0, 1}, {1, 1}}^2

[soft zealot]

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[lyric obsidian]

ok what

💀

[royal gust]

What's up?

[lyric obsidian]

why is it returning that

[devout mist]

Didn't you just do that?

[vital crag]

what do you expect it to return

[grizzled tusk]

as sets I think

[lyric obsidian]

Fuck

Fuck im an idiot 💀

my god i need to multiply it by E again

sigh

,w {{0, 1}, {1, 1}}{{1, 1}, {1, 0}}

[royal gust]

E

[soft zealot]

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[lyric obsidian]

ok so E is order 2 thank god

wait

ughhhhhhh

i messed up somewhere

A C or D is order 3

[devout mist]

Shouldn't it be B

bc B²=E

[lyric obsidian]

gaaaaaah

jesus im selling

[lyric obsidian]

ok so B^2=E

Therefore a should correspond to B and b should correspond to E

what about the other ones

[royal gust]

That's the right logic! I haven't checked the matricies and am trusting the computation is correct lol

[lyric obsidian]

^^ this

[royal gust]

We have to distribute the other order 2 matricies to order 2 permutations

[lyric obsidian]

wait E^2=B

bruh

[royal gust]

We could do something like calculate AB and see what we get

We have free choices here lol

[lyric obsidian]

ok

,w {{0, 1}, {1, 0}}{{1, 1}, {1, 0}}

[soft zealot]

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[royal gust]

AB = D

[lyric obsidian]

im on like the mobile website scrolling is impossible 😭thank you

hmm

[royal gust]

Wait what did we say B was again? Lol

[lyric obsidian]

E^2=B i think

,w {{0, 1}, {1, 1}}^2

[soft zealot]

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[royal gust]

So B maps to b

[lyric obsidian]

yes

[royal gust]

E maps to a

I'm realizing that both are squares of the other

Of course they are lol. That's the only way you can have two order 3 elements

[lyric obsidian]

cd=a it looks like

[grizzled tusk]

I = {{1, 0}, {0, 1}}
A = {{0, 1}, {1, 0}}
B = {{1, 1}, {1, 0}}
C = {{1, 1}, {0, 1}}
D = {{1, 0}, {1, 1}}
E = {{0, 1}, {1, 1}}
now you can copy paste for quick calc

[lyric obsidian]

thx

[lyric obsidian]

AB=D but this wont match

AC gonna be what

[grizzled tusk]

,w {{0, 1}, {1, 0}}*{{1, 1}, {0, 1}}

[soft zealot]

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[lyric obsidian]

which is E

AC=E
cd=a

so A->c and C->d

whats the last one

we got B->b, E->a, A->c and C->d

D->f i thi k

and thats the isomorphism?

[grizzled tusk]

we should sum up everything to check

let's list orders of elements of G

I = {{1, 0}, {0, 1}}
A = {{0, 1}, {1, 0}}
B = {{1, 1}, {1, 0}}
C = {{1, 1}, {0, 1}}
D = {{1, 0}, {1, 1}}
E = {{0, 1}, {1, 1}}

,w {{0, 1}, {1, 1}}^3

[soft zealot]

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[grizzled tusk]

ok so B and E are of order 3

A, C, D are 2

https://cdn.discordapp.com/attachments/908076393198419988/1507450688043618504/IMG_0642.jpg?ex=6a11f25b&is=6a10a0db&hm=c308fcbe9dc548f99613c65e46d8f06edb55d60b74a7ee63566ddb2f68c7d0db&

a is of order 3, b is too

[royal gust]

B² = E and E² = B. So, we have a free choice here

[grizzled tusk]

c, d, f order 2

the cheat code at the end is to check with the group table

p and q are the element of order 3

you can place your 3 elements of order 2 and check you don't have problems

[royal gust]

I'm realizing everything is free. Any placement of an order 3 works, any placement of an order 2 works

[grizzled tusk]

yeah the placing of order 3 can only be free since it determines after a choice for order 2 too

but you must check the last choice

because pr has to be s

on the table

which doesn't matter much because nobody would write the same element twice ig

[royal gust]

Every swapping of elements is an automorphism of S3

But we can't really use that, probably need to prove by construction

[grizzled tusk]

you're making a weird statement tho

even tho you can swap the orders two

the automorphism can swap order 3 too

doesn't have to be independent

S3 has 2 generators and then the full table is generated

so you really only have to make two choices

[grizzled tusk]

like here ^

[royal gust]

Yes. Swapping two transpositions also swaps the order 3 elements

You're right, it's not all free

[grizzled tusk]

<@&268886789983436800>

I didn't read all you did before but I suppose you calculated products bc of the idea of leveraging generators

so you could have found the 6 different automorphisms of S3 by making different choices for say, B and A

[royal gust]

We started manually multiplying things and seeing what matched to what, but assigning the order 2 elements is a lot smarter

[grizzled tusk]

not necessarily because you can use a line from the table

[royal gust]

Like, I started with order 3 thinking that would be easier, but order 2 elements are not uniquely defined that way

[grizzled tusk]

choosing an element of order 3 fix the other, then you choose the image of a transposition, and there are only two products left

[grizzled tusk]

indeed but that's the nice part, you have a choice for order 2 so only two calc to make

I think your way was fine

[final saddle]

@lyric obsidian Has your question been resolved?

[lyric obsidian]

oop

yall wanna continue the discussion or nah

[grizzled tusk]

I mean depends on how much you're sure you have a good iso

we didn't see all your final choices I think?

[lyric obsidian]

its up there somewhere

[lyric obsidian]

^^

and D->f

i feel like its good

[grizzled tusk]

I'll do one quick check

I = {{1, 0}, {0, 1}}
A = {{0, 1}, {1, 0}}
B = {{1, 1}, {1, 0}}
C = {{1, 1}, {0, 1}}
D = {{1, 0}, {1, 1}}
E = {{0, 1}, {1, 1}}

,w {{1, 1}, {1, 0}} * {{0, 1}, {1, 0}}

[soft zealot]

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[grizzled tusk]

BA is C

so you should have bc = d

I think there is a problem

[lyric obsidian]

sob

i uh gotta eat lunch

ill leave it open and come back ig

[grizzled tusk]

anyway it's either C->d and D->f or D->c and C->f

so you can check which one is right by calculations (I check, you indeed made a mistake, you chose B->b, E->a, A->c, so BA = C -> bc = f, you can check my calc but seems right to me)

[chrome bramble]

guys what is 1 raised to the power of infinyti

@everyone

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[tranquil jackal]

Translation?

[drowsy epoch]

Doesn't exist

[soft zealot]

thecrumbeler2

[gentle zephyr]

wait, should be a some sort of cylinder?

let me refine the drawing

something like this

[tranquil jackal]

I would assume this is an application of Stokes theorem

[gentle zephyr]

math isn't about guessing

stop spoiling answers @subtle ginkgo

!nosols

[final saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[subtle ginkgo]

well there are two methods to solving this obviously first is just using stokes' theorem and the second is just direct parameterization of the curve

[soft zealot]

thecrumbeler2

[subtle ginkgo]

so just evaluate each component from there

now that i think about it direct parameterization might be easier

or cleaner at least

[gentle zephyr]

you are not explaining why you are using stokes at all you are jumping straight to compute

stokes is a theorem with some conditions

[tranquil jackal]

The curve your given is the boundary of an ellipse

[subtle ginkgo]

its pretty trivial in this case i feel

or not trivial more like its obvious they want you to do this

but if you want i can go through like every step of justifying

but in short the vector field $\mathbf{F}$ is continuously differentiable on $\mathbb{R}^3$, the surface $S$ is smooth, and the boundary curve $C$ is a simple, closed, smooth curve with a correctly matched orientjuation

[soft zealot]

thecrumbeler2

[subtle ginkgo]

satisfies the 4 steps

if you were to justify then you would just say since polynomials are infinitely differentiable ($\mathcal{C}^\infty$) everywhere on $\mathbb{R}^3$ and each component of the function ($P = y^2 z$, $Q = x z^2$, $R = x y^2$) is a polynomial (since the vector field is given by $\mathbf{F}(x, y, z) = (P, Q, R) = (y^2 z, x z^2, x y^2)$) cause $\mathbf{F}$ is $\mathcal{C}^1$ on the entirety of $\mathbb{R}^3$, it is continuously differentiable on any of the open neighborhood(s) containing surface $S$ and the boundary curve $C$

[soft zealot]

thecrumbeler2

[subtle ginkgo]

and thats for the differentiability of the vector field

next is smoothness and boundary of the surface S and then so on

[subtle ginkgo]

so like what are you asking here

[gentle zephyr]

this is my definition of stokes

how do you know you can apply it

F is C1 yeah

what else

[subtle ginkgo]

well there are 5 conditions laid out

[soft zealot]

thecrumbeler2

[gentle zephyr]

is our parametrization regular

?

[subtle ginkgo]

well

that is what we are trying to prove first

[soft zealot]

thecrumbeler2

[subtle ginkgo]

and as you probably know parameterization is regular if its tangent vectors are linearly independent (meaning their cross product is never zero)

[soft zealot]

thecrumbeler2

thecrumbeler2

thecrumbeler2

[gentle zephyr]

let me show you my definition of regular parametrization

[subtle ginkgo]

you can prove it many different ways

[gentle zephyr]

we need to show C is open simple and smooth

[gentle zephyr]

.

@subtle ginkgo

[soft zealot]

thecrumbeler2

thecrumbeler2

thecrumbeler2

[subtle ginkgo]

which i am assuming you can probably do

[gentle zephyr]

no

I mean yes but

let me look at a different def because this regular def sucks

[subtle ginkgo]

its not the cleanest but it works

[subtle ginkgo]

if you can do it then i suggest doing it

[gentle zephyr]

this better

@subtle ginkgo

[gentle zephyr]

are you here?

this fixes the problem we were talking about

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[rapid frost]

can someone help me with this, ive failed this same assignment 3 years in a row now. its got me in a deep depression lmao. might as well drop out cause clearly this isnt for me

[tiny gorge]

show a screenshot, no one wants to click some dodgy pdf

[zinc bluff]

yeah, could be malware

[grave trout]

!img

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@rapid frost Has your question been resolved?

[rapid frost]

my bad fo real, im sorry

[fiery comet]

Well u can let equation as cubic

Ax³+bx²+cx+d

Then plug x,y

X,y

Get some equations and solve them

But according to graph

It's not cubic

If u want according to graph

Look

In x€(-2,0)

What's the slope?

[tiny gorge]

how can a cubic give you a piecewise-linear graph?

[fiery comet]

Well i meant

If u want according to the points only

And not according to graph

U can get cubic

[tiny gorge]

ah i see

[fiery comet]

If u strictly want according to graph

U can take slopes

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[normal solar]

What is the radius of convergence R of the Taylor series?

The following is a taylor series of a function centered at 8.

[normal solar]

I am trying to find the radius of convergence of R of this series.

[bronze grove]

power of n

[normal solar]

My guess is that the radius of convergence is infinity because we have (-1/7)^n which approaches to 0 as n approaches infinity.

[bronze grove]

can u guess which test we can apply

[scarlet sequoia]

does that mean (anything in terms of n) * (-1/7)^n approaches 0?

[bronze grove]

try ratio test

we can't tackle ^n

[normal solar]

no, if I multiply it by something that grows even faster than 7^the limit will blow up to infinity

oh nvm i got it

..close

.close

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[mint forum]

Hi guys could someone help me with part b please

[mint forum]

i ended up doing this so far

i did substitution so u = root h- 9
idk if the integrals are right, so from 130 - 50

[grave trout]

I dont think you need a definite integral

[mint forum]

i think its meant to be from 200 to 50

to find out the time

[grave trout]

Still no

Once you solve the integral

You'll get a relation ship between h and t

Right?

To get rid of the +C, you know that at t=0, the water height is 200cm

[mint forum]

yea

okk wait lemme try that

so you're saying to just input 200 in h

[grave trout]

Where is the +C?

All the math teachers are crying

[mint forum]

😭 sorryy

ok so c is -20root191

then i input h = 50?

[grave trout]

Is C in the lhs or rhs?

[mint forum]

i mean does it matter cuz i integrated both? but I put it on the rhs

where t is

but is it better to put it on the lhs so then c could be positive

[grave trout]

C does matter. But if you put it on the rhs its fine then

[mint forum]

sorry not if c matters, but rather the side?

[grave trout]

Doesn't really matter as we are nearly finished with the question

[mint forum]

like it could be lhs or rhs

okayyy nice thank you so much

so then after i input 50 for h?

[grave trout]

Yes

[mint forum]

omg ur so smart

same answer

so how did you know not to put definite integral?

cuz the mark scheme said that but i guess this way is easier to know

[grave trout]

Think of it like this. If C is on the lhs, you'll get 1 value for C. If its on the rhs, you'll get another value for C. But at the end of the day both values solve the question

[mint forum]

okay fair enough, love that

[grave trout]

Not really. There are so many people smarter than me

[mint forum]

u areeee, take the compliment 🥰 , alright i got the answer thanks to you!

[grave trout]

Because there is no need to. The question asked for a value of t when h=50, indicating a relationship between h and t. A definite integral wouldnt give us that relationship

[grave trout]

Thank you ❤️

[mint forum]

fair enough the mark scheme was giving all sort of bs

[mint forum]

ok nicee ! Thank you again

.close

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[grave trout]

Who makes ts

[grave trout]

No problem

[mint forum]

🤣 edexcel exam board

in england

[grave trout]

No one can read that. It genuinely looks like a mess

But wait

Defenite integrals is possible

Like they did in the solution

Do you want me to explain how @mint forum ?

[mint forum]

Literallyy its so useless like it overstimulates me reading that shit

[mint forum]

Yes please 😭

Sorry for wasting your time

[grave trout]

Reopen the channel first

[mint forum]

okay 🤣

.open

wait wtf

idk all these commands tbh

[grave trout]

I have quite a bit of time. Your not wasting any

[mint forum]

.reopen

[final saddle]

✅ Original question: #help-36 message

[grave trout]

So basically

[mint forum]

Yay

[grave trout]

A definite integral is literally a fancy way to solve this question

Because the 2 integral were equal

[mint forum]

Okayyy

[grave trout]

If you were to integrate them both from points that both the values are equal at, then you'll get the same value

[mint forum]

Why was the 50 at the top compared to 200

[grave trout]

<@&268886789983436800>

[mint forum]

Okayy nice

[grave trout]

Think of it like this

At h=200, what is the value of t?

[mint forum]

0

[grave trout]

Yes

And at h=50, you can set the value of t as some constant T

[mint forum]

Ohhh

So it's like "bigger"

[grave trout]

So if integrate h from 200 to 50, then you can integrate t from 0 to T

[mint forum]

Idk how to work around that

But you'd still get the same answer if it was the other way around

so 200 at the top and for t , 0

[grave trout]

True

[mint forum]

Okayyy so thats fine then

[grave trout]

But you'll have to change the other one too

[mint forum]

Okk it needs to be consistent

Dude my teacher could neverrrr omg

[grave trout]

Liek if you integrate form 50 to 200, you'll integrate t from T to 0

[mint forum]

Thank you for explaining that

Yess, amazing!

Sorry I have one more question

But let me finish doing it first 😭

[grave trout]

Do say

[timber leaf]

Lowkey thought I opened a channel

[grave trout]

Similar name to you so its understandable

[mint forum]

i'm crying 🤣

hi twin

okay with part b

i've done this so far

excuse the handwriting, litch writing with a mouse

idk how to get it into that form

[timber leaf]

Helo

[grave trout]

The second part seems a bit wrong

Give a minute ill try solving it

[grave trout]

Yes your second part is wrong

You dont get $1/P +P-2$

[soft zealot]

-TimeLord-

[grave trout]

Could you maybe show your working?

[mint forum]

but then i multiplied it all by e

[grave trout]

I would highly reccomend you write this on a piece of paper

[mint forum]

i am 😭

its cuz my phone is downstairs charging i cant take a pic

[grave trout]

But ur so close

The on with the natural logs in it is correct

But how do you do the step after it

[mint forum]

okayy thank you

so then i multiplied it all by e

to get rid of the ln

then it gave (p)^-1 + (p-2)

= e^1/2sin2t + c

[grave trout]

Just wait

[mint forum]

$ (p)^-1 + (p-2)
= e^1/2sin2t + c $

[mint forum]

ugh i cant make it into ur format like this

[mint forum]

okayyy thanksss

[grave trout]

You should be getting the second line

,rccw

[grave trout]

Im still confused on how you solved it after

[mint forum]

but e cancels out ln?

why would it be e^p-2

[grave trout]

Oh fck. Assume there is ln there

$e^{ln(P-2)}e^{-ln(P)}$

[soft zealot]

-TimeLord-

[grave trout]

After this hope for you solve it

[mint forum]

okay thank you lemme try

[grave trout]

Wait. You do realise that if $f(x)+g(x)=h(x)$, then $e^{f(x)} + e^{g(x)} \neq e^{h(x)}$. And instead $e^{f(x)+g(x)}=e{h(x)}$

[soft zealot]

-TimeLord-

[grave trout]

Basically if you do e raced to the power of both sides, then you have to do it for BOTH SIDES and not each ELEMENT?

[mint forum]

yes im aware thank you, I got the answer 😭 i appreciate u so much for ur help

I shouldn't of done e so soon but i ended up getting it right thanks to uuu , again

[grave trout]

No problem

!done

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[mint forum]

.close

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[grave trout]

<@&268886789983436800>

<@&268886789983436800>

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[dire glacier]

ok

soooooo

where did

^3 go

oh im stupid

mb

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[dire glacier]

<@&268886789983436800>

opened multiple channels

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[summer yew]

stuck on this problem for some hours now can't seem to find the next step

[summer yew]

my solution:

wait I think I figured it out

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[summer yew]

.reopen

[final saddle]

✅ Original question: #help-36 message

[summer yew]

Just to check is this correct?

[real patio]

The derivation in your notes contains a fundamental error in the section where the p should be equal to one

[final saddle]

@summer yew Has your question been resolved?

[summer yew]

Where?

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[summer yew]

.reopen

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✅ Original question: #help-36 message

@summer yew Has your question been resolved?

[zenith coral]

Since you use Cauchy Schwartz then he’s implying that p=+-1

[summer yew]

yeah I've done it in my corrections

so that's it? I've cleaned everything up

[zenith coral]

Yes

When p=0 it assumes and assume a deterministic linear relation it becomes inconsistent with the rest

When p=-+1 it becomes consistent

[summer yew]

thanks

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[bold turtle]

(and if you're confused because you're thinking something like "Isn't it obvious that the inverse of an inverse is itself?", you need to rely heavily on the verbatim definition of what an inverse is in your course notes)

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[shy mica]

.reopen

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✅ Original question: #help-36 message

[shy mica]

thank you so much! i can go see my family now

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[velvet pivot]

How is it possible to have (a_n)(x_n) terms? Are (a_n)(x_n) terms just theoretical?

[glossy zephyr]

a_n is a sequence, x_n is a coordinate

[velvet pivot]

Here a_n is a coefficient, right?

and x is any variable?

[glossy zephyr]

so, say x_n from n = 1 to n = 3 is a """sequence""" (x,y,z)
and a_n from n = 1 to n = 3 is a sequence (3,7,2)

They are any and different variables

In here, the linear equation would be
3x + 7y + 2z = B

[velvet pivot]

If my understanding is correct, then that just refers to how many terms and coefficients of those terms there are?

that's what the n means?

[glossy zephyr]

It has to do with the space the elements of the solution set reside in

[velvet pivot]

Tysm! I thought that this could just happen infinitely, and that maybe for a given equation there's just an invisible a_n xn or something where they just equal 0

[glossy zephyr]

Say, in the example i gave, the solutions are vectors in R^3

[velvet pivot]

Yep

[glossy zephyr]

yea, the coefficient can be 0

[velvet pivot]

Anything that answers the solution forms a plane in this equation?

[glossy zephyr]

And if you want you can sometimes extend to n -> infty

[velvet pivot]

Infinite dimensions/terms?

[deep condor]

generally if we write ... + a_n x_n, we're stopping at an n-th term so it's going to be finite. but if we want to represent an infinite number of terms, we would write a_1 x_1 + a_2 x_2 + ... , without adding a "last term"

[velvet pivot]

As a learner of this book, do I have to completely understand all of that perfectly before moving on?

[glossy zephyr]

The concept of plane is a bit arbitrary, since in reality it depends on which dimension youre working on

[velvet pivot]

So 2x+4y=3 can be considered a plane kinda?

[glossy zephyr]

say, Ax = B is a point
Ax+By = C is a line
Ax+By+Cz = D is a plane

[velvet pivot]

that's what i'm most familiar with

[velvet pivot]

Ah ok!

What about x_infinity?

is that also a thing?

I believe I understand it now! the term after the ... in the image I showed is just the last term in the equation you're looking at?

[glossy zephyr]

Yea, at that point you usually are working from a purely symbolic standpoint relating to lin. algebra

[velvet pivot]

And the equation has to take the form of that to be considered in standard form of a linear equation?

[velvet pivot]

Thanks!

I really appreciate clearing it up!

Here x_n is also just the last term that is defined/exists?

Thanks for the help! I'm pretty sure I understand it

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[glossy zephyr]

yea

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[blissful meadow]

Did you like even read the problem?

[gentle zephyr]

yeah

[blissful meadow]

It would be nice if you attempted something.

[gentle zephyr]

well green is out of the question, so either Stokes or Gauss, but since the region is I believe in R3 we would be prolly needing gauss for this one

other than that, I was trying to understand the geometric intuition on the region S, seems like a disk intersecting with a plane

[blissful meadow]

It's a solild cylinder intersecting the graph of the function 2 + f(x,y).

[gentle zephyr]

yeah

solid cylinder

[blissful meadow]

So you can imagine taking the graph z= 2+ f(x,y) and keeping just the part that's above the disk x^2+y^2 <=1 in the xy plane.

You do have the information that f(x,y) >=0 and that f(x,y) = 1 on the boundary

[drowsy epoch]

hello

[gentle zephyr]

care to elaborate?

[blissful meadow]

That’s just what it means to intersect the graph z=2+f(x,y) and a solid cylinder

[gentle zephyr]

however we dont know what f(x,y) even is dawg

[glossy zephyr]

Id swear you can actually use green if youre crazy enough, lmao.

[blissful meadow]

But you’re given info about what it looks like on the boundary

[gentle zephyr]

the vector field is of R3

[glossy zephyr]

Ill spoil you into it, you have to use stokes

And it happens to be that you can find the intersection pretty easy by the definition they gave you of f(x,y)

[gentle zephyr]

I need more help and avoid giving out spoilers because I prefer hints

@glossy zephyr @blissful meadow @drowsy epoch

[blissful meadow]

Make a rough sketch of S.

You can use divergence

[glossy zephyr]

Whats the value of f(x,y) at the boundary of the cylinder

[gentle zephyr]

the other lad said stokes tho

@glossy zephyr @blissful meadow

what are we using gauss or stokes dawg?

[gentle zephyr]

z = 3

[blissful meadow]

The issue with Stoke's is you'd need to find G such that curl G = F.

[glossy zephyr]

So you have a surface which bound is a constant circle (because its the intersection of a plane and a cylinder) at z = 3

[blissful meadow]

Stokes takes in the curl of a vector field in the integrand. Seeing as F is given as the integrand, you need that integrand to be expressed as the curl of some vector field, no?

[blissful meadow]

The main idea is that because f(x,y) >= 0 and f(x,y) = 1 for x^2 + y^2 = 1, then you know that S touches the plane z=3 on the outside (x^2 + y^2 = 1) and it has to bulge upwards in some way within x^2 + y^2 < 1 because f(x,y) >= 0.
To use the Divergence Theorem, you would need a closed surface (like we did in the past. you need it to enclose a solid 3D region). What simple shape could you stick to it in order to close it up?

[gentle zephyr]

z = 2

[gentle zephyr]

either x^2 + y^2 = 1 or z = 2

well this shit isnt about guessing

[glossy zephyr]

Oh, mb, get what you mean, still, i think stokes turns out to be easier since closing the surface takes a few steps.

Its pretty easy to find G

[blissful meadow]

Oh okay I thought it was hard at first. May be simpler than I thought.

Divergence is just nice because ||div F = 0|| I guess

[glossy zephyr]

yea ||div(curl(G)) = 0||

[blissful meadow]

Indeed

[gentle zephyr]

i need help

[glossy zephyr]

There are two ways to do it.

You can use Divergence Theorem (Gauss) and struggle with calculating the Flux Integrals of the non-meaningful parts of the solid.

Or you can use Stokes Theorem and struggle finding the Vector Potential for F.

Id advice you go with the second, but you can do it however you like

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[vale quarry]

basically you take the x and factor it

[timber plume]

Ahhh nvm I know how to do this

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[vale quarry]

no problem happy to help

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[patent compass]

24th problem, Idk how to factorise this equation and I think that maybe graphing will help but I'm not allowed for that.

[patent compass]

Ping me if yk what to do here pls.

[brittle lagoon]

@patent compass you might want to draw out a diagram and maybe solve it by proportions

[patent compass]

proportions?

wdym

[brittle lagoon]

like with the ratios given you can find AP/PD

[patent compass]

Sry but I struggled on 24th

not 23.

[brittle lagoon]

oh mb i cant read lmao

im sorry

[patent compass]

I shouldve cropped that part out mb

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[loud sundial]

Hint: The left hand side is usually bounded between which two perfect squares?

Oh also idt factoring is super fruitful here

[patent compass]

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[boreal laurel]

How do we start such problems?

[novel kettle]

take three arbitrary elements of P(S) and check if they satisfy transitivity

your approach will be a bit different based on whether it's actually transitive or not

[novel kettle]

this is required if the relation is transitive, but otherwise you get to choose 3 specific elements and show that they don't satisfy transitivity because you just need a single counterexample

in this case the relation is actually not transitive, so you need to find three sets A B C where A ⊄ B and B ⊄ C are true but A ⊄ C is false

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[boreal laurel]

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[boreal laurel]

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✅ Original question: #help-36 message

[boreal laurel]

if I take power set as {(1),(2),(3),(12)(13)(23)(123) phi}

A-->(1,2)

B-->(23)

Clearly A is not subset of B

C-->(1,2)(1,2,3)

B Is not subset of C

But A is subset of C

[crimson zenith]

wdym phi?

[novel kettle]

empty set i assume

[crimson zenith]

probably

just checking

[novel kettle]

is C (1,2) or (1,2,3)
it can't be {(1,2),(1,2,3)} because that's not an element of the power set, that's a subset of the power set

[polar obsidian]

Hey man, are you doing an intro to math course?

[boreal laurel]

Yeah...new in math cord

If we take c-->(1,2)

B is not subset of C but A is subset of C...but they should make the subset sign more convenient because it seems proper subset@novel kettle

[novel kettle]

yeah that works

oh wait proper subset

hmm

might as well take A = (1), B = (2), C = (1, 3) to be safe

[polar obsidian]

perhaps, but shouldnt the proof be for all possible combinations? i'm sorry im doing a foundations to maths course too

soo addressing this to other experts here :D.

[novel kettle]

it's a disproof so you only need one counterexample to show that it's not transitive