help-39

[bitter herald]

alternatively

,align
\vb P(C_2) &= \vb P(C_2\where C_1)\vb P(C_1) +\vb P(C_2\where \comp{C_1})\vb P(\comp{C_1}) \
&= \vb P(C_2\cap C_1) + \vb P(C_2\cap\comp{C_1})

[bitter herald]

i dont get your point

[green aurora]

here

if i put this into $P(A|B)=\frac{P(A\cap B)}{P(B)}$

[jolly parrot]

Goofy Joe

[bitter herald]

i mean you are not doing that though

[green aurora]

ok,sorry

[bitter herald]

like, yes, it would be redundant if that was what you are doing

[bitter herald]

the point of what i said was to have this correspondence

[green aurora]

but there is no dependence here, the fact of removing or not removing that particular card does not change the probability of the second event?

the result is as if he were saying that $P(A|B)=P(A)$

[jolly parrot]

Goofy Joe

[green aurora]

so indipendence

[bitter herald]

There is dependence, though. The outcome of the first card physically alters the deck for the second draw

its like having only two cards, lets say an ace and a king

[green aurora]

but

[bitter herald]

if you take one out, you are guaranteed to have the other be the one besides it

the probability changed

[green aurora]

doesnt P(A|B)=P(A)->indipendence?

[bitter herald]

yes

and thats not the case here

[west sail]

how did you end up with that in the first place?

[green aurora]

P(A)=13/52=1/4

the solution is 1/4

[bitter herald]

yeah thats right but

im still puzzled by what your confusion is, sorry

[green aurora]

the confusion is why they are independent

[bitter herald]

they are not

who said they were

[green aurora]

me

because i can use P(A|B)=P(A)

but its strange

[west sail]

let A = first card is club, A' = first card is not a club, and B = second card is a club
the solution is a union between P(B | A) and P(B | A'), not just a singular condition!

[green aurora]

yes

[green berry]

hi

[thorn agate]

ig he is confused on the fact P(B | A) and P(B | A') results in P(B)

[green berry]

If you define the events as

R: a generic card has been removed (i.e. a card was removed, without knowing which one)

B: the revealed card is a club

[green aurora]

I'm getting confused myself

[green berry]

they are independent, but only in a trivial sense

P(R) = 1

[green aurora]

they are not independent

I'll try to solve it

[bitter herald]

i feel like you are mixing up marginal probability and conditional probability

[green berry]

$P(B | R) = \frac{P(R | B) P(B)}{P(R)}$

[jolly parrot]

alee

[green berry]

P(R) = 1

P(R | B) = 1

[green aurora]

I can't calculate the probabilities

i got P(A|B)=12/25

😢

[west sail]

in my humble opinion, you should first figure out what probabilities to calculate before starting work on calculating them.

[green aurora]

i consider A="probability that the second card is a club" B="probability that the first card removed is a club"

[west sail]

and so how did 12/25 come about?

[green aurora]

I did it all wrong

but if I use my events writing P(B|A) doesn't make sense right?

[bitter herald]

you'd want P(A|B) rather, yes

[west sail]

okay, let's think of this conceptually.
suppose I am randomly given one of two tests, A and B, each with a probability P(A) and P(B) respectively.
now, given either test, the chances of me failing the test is P(F | A) and P(F | B) respectively.

can you come up with an expression to find the overall chances that I fail the test?

once you've understood this, you can reapply the concept in your problem.

[green aurora]

idk how to calculate the intersection

[west sail]

actually let me revise the question a little bit.

revised the conceptual question.

[green aurora]

$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{?}{\frac{13}{51}}$

[jolly parrot]

Goofy Joe

[green aurora]

im stucked here

[west sail]

there are no intersections required here, theoretically speaking.

[green aurora]

how do I do it?

[west sail]

again, we should consider these two different first-card events separately.

[green aurora]

yes

[west sail]

let's start by considering the event where the first card removed is a club.
knowing this, what are the odds of getting a club next?

[green aurora]

this is only the first part

12/51

[west sail]

good.
now, consider the event where the first card removed is not a club.
what are the odds of the next one being a club?

[green aurora]

13/51

[west sail]

ok, pause here.

these two values you have found are P(B | A) and P(B | A') respectively.

[green aurora]

?

[west sail]

and why would that be confusing? the odds of drawing a club as the second card depends on whether you removed a club as the first card or not.

as you've found out by our little experiment here.

this is why everyone has been telling you that you already know what the two conditional probabilities are.

there is no need for an intersection calculation here because you already got the conditional probabilities by direct counting.

the only thing remaining now is to weight each outcome by P(A) and P(A') respectively.

[green aurora]

?

[west sail]

if you don't know why we are doing this, you should consider this again.

[west sail]

correct. you forgot to weight the outcomes.

perhaps I was not clear in that statement - I apologize. you did consider the first card in your conditional probabilities, but you forgot to weight those conditional probabilities by the probability of the conditions themselves.

that was what I meant by that statement.

[green aurora]

so

its basically P(C)=P(B|A)P(A)+P(B|A')P(A')

[west sail]

that is precisely the law of total probability that Lex mentioned, and yes, that is exactly right.

[green aurora]

because it would be that {second heart card drawn, first heart card drawn} intersected with {second heart card drawn, first non-heart card drawn}

$P(C)=P(B\cap A)\cup P(B\cap A')$

no union in the middle

[jolly parrot]

Goofy Joe

[west sail]

I'm going to leave this to someone else to confirm.

[green aurora]

ok anyway thanks

<@&286206848099549185>

1/4 finally comes out

[green aurora]

I think what I wrote doesn't make sense because we're not talking about sets anymore

[bitter herald]

It doesn't lol

[green aurora]

ok

thanks all

.close

[pearl pond]

Channel closed

Closed by @green aurora

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[steep saddle]

what part

[fervent siren]

b

and the rest

[fervent siren]

horimiya is peak

[bitter herald]

Try anything?

[fervent siren]

i was thinking

and i dont know where to start

[steep saddle]

what is a residual

[fervent siren]

the space between two points on a graph no?

[bitter herald]

Residual = actual y - predicted y

[fervent siren]

yeah

y - yhat

[steep saddle]

i guess you could say that

so what is the answer to part b

[fervent siren]

11 thousand?

just 30000-41,256.10?

[bitter herald]

Those are the xs

Not the ys

[fervent siren]

oh

uhh

79.04-79.43

[bitter herald]

Where did u get 79.04 from

[fervent siren]

the top bit

the table

[bitter herald]

no ignore the table

[fervent siren]

oh ok

[bitter herald]

For this part

[fervent siren]

?

[bitter herald]

Like

They already gave you your actual y in the question: 79.43

[fervent siren]

yep

[bitter herald]

Your job is to figure out what the predicted y is

Which is to run jt through your model

[fervent siren]

and how do i do that?

[steep saddle]

u have an equation

solve for the other variable

[bitter herald]

Use the thingy u got in part a

[fervent siren]

oki doki

i got

x=3.47

wait i forgot ln

oops

now i got e^3.47

which is 32.1

[bitter herald]

where is e popping up from

Like

[fervent siren]

Ln

log base e

[bitter herald]

You have [
L(x) = 56.67 +6.566\ln(x)
] right? You need to compute $L(41.2561)$

[fervent siren]

what

oh

i put

79.43=56.67+6.565ln(x)

[bitter herald]

no

Its x = 41.2561

[fervent siren]

i see where i went wrong

yeah

i got 327.516 now

[bitter herald]

Can you show what you did to get that

[fervent siren]

56.67+6.565(41.2561)

[bitter herald]

U just like

Obliterated that ln

[fervent siren]

i keep forgetting.

i got 81.0905

[bitter herald]

Yea gold

Good

[bitter herald]

Do this now

[fervent siren]

which is 79.43- that

-1.6605

[bitter herald]

Yep

Das it

[fervent siren]

YAY

[bitter herald]

My energy running low so maybe wait on somebody else to help u with the rest

[fervent siren]

can i @ them

because this is some panic studying

[bitter herald]

who's 'them'

[fervent siren]

helpers

[bitter herald]

Ye

Go for it

[fluid rose]

Hello

[fervent siren]

i need help with C

is this just sub in 43500 into the x value?

[fervent siren]

hello?

<@&286206848099549185>

[dull steeple]

[torn wave]

Ping after every 15 mins...

Pls

[fervent siren]

sorry i remember it being 10

ap precalc

.close

[pearl pond]

Channel closed

Closed by @fervent siren

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[woven isle]

Aren't a and c the same

[steep saddle]

,rotateccw

[steep saddle]

A would imply C but C does not imply A

it's asking what is ALWAYS true

[woven isle]

So wouldn't A be a better answer

Now I got to this

But Im not able to prove that X=Y starting from XY=YX

But I feel like that doesn't matter to the question

[steep saddle]

what does that tell you then

if you can get XY = YX but not X = Y from the equation

let's revisit

XY = YX does not imply X = Y

X = Y implies XY = YX

but you can't get X = Y

what is always true

[woven isle]

Idk i was discussing this question and my peers keep saying that c is more general than a

So I thought if I could get C=A from C and the other isnt true

Then i would shut them up

[steep saddle]

okay but you can't get X = Y from the equation

XY = YX does not imply X = Y

[woven isle]

Even tho that are matrices?

[steep saddle]

the reverse is true but you can't use it

yeah?

consider the case
1 0
0 0
for X and
0 0
0 1
for Y

what is XY and YX? is X = Y?

[woven isle]

Okay so why cant I say that A and C are both right and the question is the wrong kne

[woven isle]

No

[steep saddle]

because A is not always true

simple as that

like i said

you got to XY = YX from the equation

which is sound

however you cannot use that statement to claim X = Y

it simply isnt always true

[woven isle]

Okay I got this part

So C is the answer

[steep saddle]

yeah

[woven isle]

Ah fuck

[steep saddle]

A can be true but not always

[woven isle]

I was so persistent 😅

When can XY =YX in matrices but Y doesn't equal X

[steep saddle]

i gave an example

[woven isle]

I thought those were numbers

Oh okay

I see

[steep saddle]

yeah i'm not about to type latex on a phone

[woven isle]

Makes sense

Well thanks for your help

.close

[pearl pond]

Channel closed

Closed by @woven isle

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dense jasper]

<@&268886789983436800>

[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shell plover]

so i'm currently trying to solve an induction proof involving summation, but i'm just a bit confused as to how we got 4k + 3 in the sequence

2(2k) + 1 = 4k +1
2i + 1 = 4k + 3
2(2(k+1)) + 1 = 4k + 5

however, i would just like clarification as to what value i might have for us to get 4k + 3, or if there is a particular rule regarding sequences and/or summation that i am not aware of (i'm not very used to working with sigma notation)

[brave sluice]

that term is from i = 2k+1

[shell plover]

thank you so much !!!

[brave sluice]

you're welcome

[shell plover]

.close

[pearl pond]

Channel closed

Closed by @shell plover

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dawn sundial]

hello does anyone know how to solve this
I’ve been trying to solve this problem the way my teacher taught us but I keep getting a different answer from the actual answer (2nd image)

[dawn sundial]

this is how I tried solving it but I keep getting stuck on simplifying the two fractions together

[fierce aspen]

can you see any terms u can simplify?

[hoary birch]

just multiply the second fraction by * y/y

and now you have a common demoninator to work with.

[fierce aspen]

works

[hoary birch]

What exactly is the point of this exercise?

[proper nova]

Golden rule: find da common denominator

[hoary birch]

also why did you just only multiply 2x/2x to first term?

[dawn sundial]

cuz that’s how my teacher taught us how to dit

do it*

[proper nova]

I don't see the point ngl, and the work is wrong too

[dawn sundial]

this is one of her notes I’m only following her instructions

[proper nova]

I think your teacher wants you to find the common denominator

[proper nova]

Your teacher found the common denominator which is $t^2(t + 1)$

[jolly parrot]

1 divided by 0 equals Infinity

[hoary birch]

Don’t worry about working in each pieces

just solve the fraction and find a common denominator.

[proper nova]

In this case, the common denominator $2x(2xy) = 4x^2y$ which works, but if you kept multiplying denominators for the common denominator, it will be harder to simplify

[jolly parrot]

1 divided by 0 equals Infinity

[dawn sundial]

mmmmm I think I get it now

[proper nova]

Now tell me, how do you solve your problem?

Tell us your steps first

[dawn sundial]

I multiplied -6x over 2x by y in the numerator and denominator

[proper nova]

Mhm

[proper nova]

Okay, so this step is alright

Now you're trying to subtract the 2 fractions

[proper nova]

But why is there a -1 in the numerator?

[dawn sundial]

that’s to flip the -6xy and 3

for some reason she does it in some problems and doesn’t it others

[proper nova]

That's already wrong if you multiplied -1

[dawn sundial]

lemme find the note for it

[proper nova]

You gotta understand why she multiplied by -1

[dawn sundial]

I don’t really understand why she does😭

[proper nova]

Does what?

$b - a = -1(a - b)$

[jolly parrot]

1 divided by 0 equals Infinity

[dawn sundial]

like idky she flips the numbers sometimes and doesn’t like in here, the answer to 7 was flipped but 8 wasn’t flipped

[proper nova]

Wdym?

[dawn sundial]

nvm 😭

[proper nova]

I don't understand why this ws even exists 😭😭😭

[proper nova]

I mostly flip if i get a case like that

[dawn sundial]

that’s what I mean bc in number 8 it was written like b-a and so was number 7 but she only flipped number 7

basically I’m just trying to say I’m confused on when I’m supposed to know when to flip the numbers

.close

[pearl pond]

Channel closed

Closed by @dawn sundial

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crude mist]

Show that a polyhedron with 3 triangular faces meeting at each vertex must in fact have a total of 4 faces

[pearl pond]

@crude mist Has your question been resolved?

[pearl pond]

@crude mist Has your question been resolved?

[crude mist]

can someone help me please

[tardy reef]

You can use Euler characteristic formula for polyhedra

[crude mist]

Each face is a triangle, so each face has 3 edges but each edge is shared by 2 faces. This means that while counting edges by taking help of faces, we tend to count every edge twice which makes the eq
3F=2E?

is this correct?

[tardy reef]

yea

similarly, you can also relate the vertices to the other quantities

[crude mist]

and Each face has 3 vertices but each vertex is shared by 3 faces. So we again tend to count every vertices thrice?
3F=3V?

[tardy reef]

yep

[crude mist]

NICE

ty

.close

[pearl pond]

Channel closed

Closed by @crude mist

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[brittle harbor]

Find the differential equation of all circles which pass through the origin and whose centres are on the x-axis.
(x-h)^2 + (y-k)^2 = r^2
The circle passes through (0,0) and (h,k) = (h,0)
So:
(x-h)^2 + (y^2) = h^2
Differentianting 2 times with respect to x gives;
1 + yy'' + (y')^2 = 0
Is my process correct?

[pearl pond]

@brittle harbor Has your question been resolved?

[midnight haven]

yh

looks good

unless they want it ot be of specific order

[pearl pond]

@brittle harbor Has your question been resolved?

Channel closed

Closed by @brittle harbor

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cosmic fox]

i dont understand how a 3d shape is created

[cosmic fox]

is it something liek this?

but then what would be the depth of it

[unborn abyss]

they're like this; imagine a square built on the pink line coming out to bonk you in the head

[cosmic fox]

ahh i see

thanks

.close

[pearl pond]

Channel closed

Closed by @cosmic fox

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sleek sandal]

hi

[sleek sandal]

i feel like i only got the answer right because I can see the answer choices.

how would i come up with the domain if I wasn't given the answer choices

[rough stream]

Anything under a square root must be ≥ 0

So we have:
x + y ≥ 0
x - y ≥ 0

[sleek sandal]

im here so far

[sleek sandal]

yo

wait i think i got it

upsetting it's not intuitive tho

.solved

[pearl pond]

Channel closed

Closed by @sleek sandal

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[iron basin]

im not sure how the sin(sin(sin...)) n times is supposed to be handled

[prisma kernel]

(just wanted to ask, is this JEE Advanced limits?)

[iron basin]

sameer bansal yes

[cinder flower]

asking as if no was a possibility

[ebon skiff]

Haha

Is ts uni or high school

[prisma kernel]

ug entrance exam

[ebon skiff]

Ug?

[prisma kernel]

undergraduate

[cinder flower]

for high schoolers, to demonstrate to indian universities that they are very smart

[autumn fossil]

Have u handled the rest already?

[iron basin]

yeah most adv questions look performative asl

[cinder flower]

i cannot imagine giving ts to students

[toxic lichen]

neither can i

looks like an extremely ass limit

like one would need to get a handle on the behavior of iterating sin(x) on small inputs

[prisma kernel]

can't even expand the sin, you get even more sin

multiverse of sine

[somber tinsel]

u write it as a sum?

[frank violet]

Have you tried that one trick y=sin(y)

[somber tinsel]

the sin thing

[iron basin]

i mean im guessing the largest term (base) with n as an exponent are only considered around x = pi/4

[cinder flower]

we need to use trick 119.II.f in JEE handbook volume 14 for this

ok i'm done shitting on JEE i promise

[prisma kernel]

yeah but y = sin(y) though, like Fionna suggested

[frank violet]

It's okay, the exam deserve that hatred 🥀

[iron basin]

well yeah but im not sure if the n-times composition is supposed to mean anything

ill try and come back though

[prisma kernel]

anyways it'd always be between -1 and 1

[autumn fossil]

I'd probably start by simplifying everything u can

Guess what the function behaves like, divide by it, do lim of quotient = quotient of limits and simplify

...

You can probably do similar stuff with the numerator now

you can split it to 2 limits

[iron basin]

... does it just simplify to f(x) = csc(x) at x near pi/4

actually why not graph it 1 min

[autumn fossil]

oh i probably fucked up with the lim in denom

it's not 1, yeah

i guessed wrong

[prisma kernel]

wait what, you can put values in denominator separately?

[autumn fossil]

That expression is equal to the original expression (i just divide numerator and denominator by the same thing). The point was to find a function which behaves just like the thing in the denominator, but is simpler, and then apply limit of quotient = quotient of limits in hope that the limit in denominator will simplify to 1 (since i chose the function to behave just like the og denominato)

but it didnt quite work here, because i chose a wrong function

i wish we oculd just swap the limits

[autumn fossil]

this can still help in solving the RHS limit though

$\frac{3^{n}+\sin\left(x\right)\left(\sqrt{2}\cos\left(x\right)+2\right)^{n}}{3^{n}\left(1+\frac{\sqrt{2}}{2}\right)}$

For x > pi/4, this'll tend to 1/ (1 + sqrt2 / 2)

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

because the cos will be slightly lower than sqrt(2)/2 and so that thing ()^n will be slightly lower than 3^n

$\frac{3^{n}}{3^{n}\left(1+\frac{\sqrt{2}}{2}\right)}+\frac{\sin\left(x\right)\left(\sqrt{2}\cos\left(x\right)+2\right)^{n}}{3^{n}\left(1+\frac{\sqrt{2}}{2}\right)}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

so if we split it like this, the first part goes to 1/(1+sqrt2/2)

and the second part goes to 0 for x > pi/4

[prisma kernel]

strange question, but like we did with denominator, why not just sub numerator x with pi/4 too?

[autumn fossil]

it was kinda faulty reasoning from me, we first take lim of n and then lim of x

[autumn fossil]

in fact, if we just substituted cos(x) = sqrt2 / 2 here, the limit would be diiferent (since there would be 3^n in the numerator)

while when x > pi/4, cosx is slightly lower, making the second fraction go to 0

and when we take lim from right side as x -> pi/4, x is > pi/4 so the limit will be just 1 / (1 + sqrt2 / 2)

so the RHS lim simplifies to lim of $\frac{3^{n}\sin\left(\sin\left(...\sin x\right)\right)+\left(\sqrt{2}\cos\left(x\right)+2\right)^{n}+2^{n}\cos\left(x\right)}{3^{n}}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

and this is pretty simple already

$\sin\left(\sin\left(...\sin x\right)\right)+\frac{\left(\sqrt{2}\cos\left(x\right)+2\right)^{n}}{3^{n}}+\frac{2^{n}\cos\left(x\right)}{3^{n}}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

the first thing -> 0 (we dont need to worry about the rate, luckily)

[prisma kernel]

the other two terms work, but what about sin(sin(sin.....(sinx)))))

[autumn fossil]

it goes to 0

[prisma kernel]

oh wait yeah, it converges right?

[autumn fossil]

yeah

intiutively, i can draw a pic

[iron basin]

ohh sqrt(2)cosx + 2 approaches 3

[autumn fossil]

rigorously, I'd do induction

[autumn fossil]

careful there

[iron basin]

i kept taking it as 3.28 or sth

might be tipsy

[autumn fossil]

the issue is that we are first taking the limit of n and then limit of x

[frank violet]

I'm still skeptical about this, isn't that changing the limit order

[prisma kernel]

same

[autumn fossil]

no, that's the correct limit order, no?

first n, then x

i just evaluated that limit pointwise for all x > pi/4

it's literally x > pi/4, not x -> pi/4

[frank violet]

Oh right

Hang on

[prisma kernel]

I might be slow but I still don't get why you evaluated x for pi/4 in the denominator

before n that is

[frank violet]

Oh yeah

[autumn fossil]

i didnt i belivee

[autumn fossil]

though i probably overcomplicated the limit by doing this monstrosity

[frank violet]

So x doesn't even have to approach pi/4

Just has to be below it

[prisma kernel]

@autumn fossil I might be asking a lot but could you explain this to me in dms (I want to have the ability to review it later and not clog up this channel), could you, I'm actually interested in this?

[autumn fossil]

Sure, I'll try to make my work less bloated, because looking at it retrospectively now, I did a lot of useless stuff

[prisma kernel]

Alright thanks man🙏

[eager jewel]

what is this bruh 🫠 this aint jee level

[frank violet]

Can we just divide both num and denom by (sqrt(2)cos(x)+2)^n for the LH limit

[eager jewel]

dosent sameer bansal have a solutions book

[round heart]

[iron basin]

ok but uh im not sure where the division by 1+sqrt(2)/2 came in handy here

[frank violet]

I don't think it has anything to do with the limit lol

[iron basin]

wait does it now

[eager jewel]

yea

[autumn fossil]

yeah, its better to divide just by 3^n

that constant was useless, it cancels out later

i originally introduced it because i thought it'd make the lim in denominator go to 1, but it instead goes to that very constant

[pearl pond]

@iron basin Has your question been resolved?

[autumn fossil]

Alright, so i cooked up sth more elegant in dms

[prisma kernel]

indeed🙏

it's so elegant even I can explain it

[autumn fossil]

Alright, go ahead

[prisma kernel]

😭

okay so

first of all, we evaluate the n limit

and we separate the limits into denominator limit and numerator limit

we evaluate the denominator limit first

(we divide both sides by 3^n to start with)

[autumn fossil]

here's the pic

[prisma kernel]

yes thank you

now in the denominator, the 3^n just becomes 1

and for the other function, we generalise the power

now for x>pi/4 (and less than pi/2 to prevent any issues), we see that cos is always < 1/sqrt(2), and hence the numerator is always smaller than 3

therefore when that fraction is raised to n (infinity), we get 0, hence the denominator just becomes 1 + 0, i.e. 1

and the entire function becomes the numerator only

[autumn fossil]

denominator limit:

[prisma kernel]

thanks man🙏

||in the numerator sin(sin(sin..........(sinx))))))) just converges to 0; for the second term we do the same thing (generalise the power) and see that it's always 0 for x>pi/4, and for the third term (2/3)^n converges to 0||

[autumn fossil]

numerator limit:

[prisma kernel]

||hence we get 0 + 0 + 0, which is just 0||

🤌 🤌

now for x<pi/4, instead of 3^n, this time we divide by (sqrt(2)cosx + 2)^n on both sides, to get the same thing on our terms

[prisma kernel]

so there, the denominator becomes 0 (again it converges to 0 as the denominator is always >3) + sinx, or simply sinx

[autumn fossil]

denominator limit with x < pi/4:

[safe mason]

Man what is this math I feel stupid looking into this

[prisma kernel]

||and the numerator can be simply solved, for the first term, it just converges to 0 (both for the sin function and the (3/(sqrt(2) cosx + 2))^n, the second term becomes 0, and the third term converges to 0 too||

||hence simply 0 + 1/sinx + 0, or just cosecx||

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[prisma kernel]

||which at pi/4 becomes sqrt(2)||

[autumn fossil]

numerator limit

[safe mason]

Can i ask you something

[prisma kernel]

sure

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[safe mason]

I’m impressed like how do you solve all of that

[safe mason]

Isn’t it like very hard?

[prisma kernel]

all thanks to @autumn fossil 🙏, he explained all of this to me and let me explain what I understood, again thanks a ton man

[autumn fossil]

It's not really difficult, lots of symbols doesnt generally mean difficult. It is somewhat tedious and requires some insight, but it's still a relatively simple thing

[prisma kernel]

yeah like the pattern seems kinda intuitive to me now, once I see how it works

[safe mason]

I see your perspective

Well to be honest I wish to get to your level at math guys

I don’t think I could solve one of these at all lol

[prisma kernel]

neither me man

[autumn fossil]

You'll get there

but ig this can be closed now @safe mason ?

you accidentally opened a channel

[safe mason]

Oh mb😭

[warped violet]

Should write all this in LaTeX

[autumn fossil]

my latex preamble is kinda fucked, so it'd make it unreadable

[warped violet]

😭

[safe mason]

uh..how to close the channel 🥀

[autumn fossil]

.close

[safe mason]

.close

[pearl pond]

Channel closed

Closed by @safe mason

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[winter otter]

what does "such that" actually mean?

[winter otter]

for example, when negating ∀y€Y ∃x€X | y=f(x)

what do we do

like is it ∀y€Y (∃x€X | y=f(x)) or (∀y€Y ∃x€X) | y=f(x)

can it just be replaced by an implication -> ?

[grim fractal]

basically means "satisfying..."

[winter otter]

so A such that B is saying that when A is true B is true?

but not vice versa?

[grim fractal]

Well "such that" usually follows "there exists"

it basically tells you what exists

in this case, there exists x in X such that y=f(x)

otherwise if you just have for all y in Y, there exists x in X

that doesn't tell you anything meaningful

[grim fractal]

are you trying to negate the original statement

[winter otter]

yes

but to do that i first need to translate it into ∨s ∧s and ¬s

[grim fractal]

I don't think you need to do this

but you can if you want Ig

[glass meadow]

You can think of "such that" as a non-commutative "and"

[winter otter]

so (A such that B) = (A and B) != (B and A)

that doesn't feel quite right

[grim fractal]

well "B != B" is a little problematic

[glass meadow]

Kinda depends what you mean by A and B

[winter otter]

ok scrap the translatig part, how do i negate the original statement

what's the general rule

[grim fractal]

so we have ∀y€Y ∃x€X | y=f(x)

a general rule is for all and exist sort of "swap" when you negate but it's important to understand why

[glass meadow]

Negate "for all, (...)" into "there exists, (not ...)"

[winter otter]

∀ is a bunch of ∧s and ∃ is a bunch of ∨s

[glass meadow]

Negate "there exists, (...)" into "for all, (not ...)"

It's literally just common sense: the negation of something being true in all cases is that thing being false in at least one case

And the negation of something being true for at least one case is that thing being false in all cases

[winter otter]

well common sense can only help me when i understand what the formula is saying

ideally i'd want to not think about that

[grim fractal]

I think it's better to not think about formulas right now

and just understand logically what's happening

then try to formalize

[glass meadow]

You can just read it in plain english

For all y in Y, P(y)

That P(y) is "there exists x in X such that y = f(x)"

So the negation is that there is at least one y that does not satisfy P(y)

In other words, there exists y in Y, not P(y)

(the comma is an implicit "such that")

[winter otter]

i don't know how to extrapolate that for larger propositions though

[glass meadow]

You just take it step by step

You had $\forall y \in Y, \exists x \in X, y=f(x)$

[jolly parrot]

Nel

[glass meadow]

The first part is $\forall y \in Y$, so take the $\exists x \in X, y=f(x)$ and just call it $P(y)$ (since it depends on the y you just defined)

[jolly parrot]

Nel

[glass meadow]

Then you have $\forall y \in Y, P(y)$, and you can negate that

[jolly parrot]

Nel

[winter otter]

so ∀y€Y (∃x€X (y=f(x)))

[glass meadow]

Yes

[winter otter]

¬(∀y€Y (∃x€X (y=f(x))))
∃y€Y ¬(∃x€X (y=f(x)))
∃y€Y (∀x € X ¬(y=f(x)))
∃y€Y (∀x € X (y!=f(x)))

[glass meadow]

Exactly

[winter otter]

so i basically want to invert all the quantifiers and negate the statement

at least when it's in this form with all the quantifiers before and the statement at the end

[glass meadow]

The original statement is basically "f: X -> Y is surjective (it can reach any y with the appropriate x)"

The negation is "f is not surjective, there exists some value y it cannot reach no matter what x you try"

[winter otter]

thank you very much

.close

[pearl pond]

Channel closed

Closed by @winter otter

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tacit bough]

How do I solve for x here? $\frac{3}{7} = \frac{-2-4}{-4-x1}$ I don't want the answer, just the thought process

[jolly parrot]

Vortac

[west sapphire]

what does x1 mean here?

[tacit bough]

something multiplied by 1

[west sapphire]

why not just write x?

[tacit bough]

although I first read it as xsubscript 1

[tulip ore]

thats awfully suspicious

[tacit bough]

the problem set shows it as x1

[west sapphire]

hmm

[tulip ore]

no one puts a 1 to the right of an x unless they mean something about it

[west sapphire]

ok well assuming it just means x.. try cross multiplying by the denominators

[tacit bough]

If I do that and solve for x I get 10

[grim fractal]

that's correct

[tacit bough]

but then wouldn't that give me $\frac{-3}{7}$?

[jolly parrot]

Vortac

[grim fractal]

no

it's (-6)/(-14)

[tacit bough]

ohh

I was confused because $-\frac{3}{7} = \frac{-3}{7} = \frac{3}{-7} != \frac{-3}{-7}$

[jolly parrot]

Vortac

[tulip ore]

type \ne for not equals

you can count odd or even - signs

you can view the - on the left as a "negation" operator that swaps the sign, and the / sign as something that lets negations get applied one after another

so -3/-7 = --3/7 = 3/7

[pearl pond]

@tacit bough Has your question been resolved?

Channel closed

Closed by @tacit bough

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[glossy creek]

can someone help with this

[glossy creek]

i get the limits are gonna be 1 and 0

drawed it out too

but idk why im not supposed to integrate x-x^2

[prisma kernel]

well find the region bounded by y = x^2, find the volume, and subtract the region bounded by x (if you haven't already done that)

no wait

[glossy creek]

is this gonn be the region

[prisma kernel]

even I'm confused, I think that it's that though

[glossy creek]

mmm

i treid it using 1 and 0 as limits with the func inside as x-x^2

ans was wrong

[rotund nymph]

,,\int_0^1x-x^2\dd x

[jolly parrot]

bored amogi

[glossy creek]

why isnt this region alos part of the area tho

[rotund nymph]

is that it?

[prisma kernel]

what I think you should do is find the volume after rotating x^2 from y = 0 to 1, then do the same for x, and subtract the volume from x^2 by the volume of x

[glossy creek]

yeah

exactly ehat i did

[prisma kernel]

yeah that's the one I'm confused about too

[rotund nymph]

are you using shell or washer

[glossy creek]

im trying washer

the donut shape right

[rotund nymph]

k then

washer about a vertical line will be wrt y

the bigger radius is sqrt y
the smaller radius is y

so try

[glossy creek]

idk i understood this

but my ans was wrong

[rotund nymph]

,,\pi\int_0^1(\sqrt y)^2-y^2 \dd y

[rotund nymph]

wdym you understood?

[glossy creek]

like this is what i did before

but after putting in the limits my asn wasnt right

[rotund nymph]

uh

do you know the formula for the washer method?

[jolly parrot]

bored amogi

[rotund nymph]

oops it should be that

[glossy creek]

this is the one we use

[rotund nymph]

oh thats exactly what you got

hmm

oh right

so your radius should be from the axis of revolution to the curve

[glossy creek]

K

That would be from x=-2 to the curve

[rotund nymph]

yes

[glossy creek]

Oh

Yeah

Gotta find y in terms of x before putting in an x value

Mb

[rotund nymph]

which would be (square root y + 2)^2 and (y + 2)^2

[rotund nymph]

yes that too

[glossy creek]

cant it be (underrot (y) -y)^2

and then integral of that form 1 to 0

[rotund nymph]

tell me how you got that formula please

[glossy creek]

oh crap

teh formula

[rotund nymph]

@glossy creek the formula says youre squaring then subtracting not the other way around

[glossy creek]

yeah i got that

[rotund nymph]

but also thats not a correct radius either

ok are you working on it then?

[glossy creek]

yeah thnx

[rotund nymph]

k

[pearl pond]

@glossy creek Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[naive hare]

whats my error here

[midnight haven]

error is in this step

$\phi + \frac{200\phi}{n}5 = 10000$

$\implies \phi( 1+ \frac{5*200}{n}) =10000$

[jolly parrot]

Hi

[pearl pond]

@naive hare Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[flint burrow]

this is a problem in my game theory class - can anyone help for part b, i got a pretty compelling answer for a being 5 moves (the same solution is also availabel online), but b seems to be an entirely new extension to a problem made solely by my instructor.
You are locked in a room and have to solve a puzzle to get out. Here is the description: there is
a small cylindrical symmetric table with four large identical cupholders, each wide enough to
fit your hand inside. Each cupholder contains a coin which you can easily feel with your hands,
and you can detect whether the coin is heads up or heads down. This puzzle is happening in
a dark room where you cannot see inside the cups, but only feel with your hands. If all four
coins have the same orientation at any point (all flipped heads up or all flipped heads down) then you have solved the puzzle and you may leave the room. The rules of the puzzle are as
follows:

[flint burrow]

A move in solving the puzzle consists of putting each of your two hands simultaneously into two
of the cupholders (either adjacent to each other, or along a diagonal), checking the orientation
of the coins in the cupholders, and choosing to flip either 0, 1, or 2 of the coins in their
respective cupholders, and immediately removing both hands. After removing your hands, the
table starts spinning around its axis. It spins such that the coins cannot reverse orientation
by themselves – they can only be reversed during one of your moves. When it stops, you
cannot know what coins were accessed in the previous move because it’s too dark to see and
you don’t know how many rotations, including fractional rotations, the table made.
(a) What is the minimum number of moves to guarantee your escape?
(b) Let’s replace the number of cupholders and hands to eight large identical cupholders set
up analogously, and with a friend, you are allowed to each simultaneously put two hands
into a total of four cups. You are allowed to communicate the orientations of the coins
in your cupholders, and change 0, 1, 2, 3, or 4 of the coins’ orientations. What is the
minimum number of moves to guarantee your escape in this new situation?

this is an extremely challenging problem

[autumn fossil]

I dont have a full solution, but here is at least an insight:
You can't surely access all 4 cups during the game, you can surely acess at least 3 though, by doing diagonal and adjacent choice.

So in what I think is the worst case, you'll have to flip the 3 coins you acessed to the same orientation, and then flip them back to the other orientation, in case the secret 4th coin wasnt the same

[pearl pond]

@flint burrow Has your question been resolved?

[flint burrow]

yeah thats what i did for a

b is the extremely hard one

[midnight haven]

i feel i have seen this somewhere

[pearl pond]

@flint burrow Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[unborn charm]

Let C ⊂ [0, 1] be the Cantor set. Prove that there exists a function g : [0, 1] → [0, 1]
such that it is continuous at every point in [0, 1] \ C and discontinuous at every point
in C.

[west sapphire]

any thoughts so far?

[unborn charm]

i want to show that if $x_0 \in C^c$ then $\forall \epsilon > 0, \exists \delta > 0$ such that $g(B_{\delta}(x)) \subseteq B_{\epsilon}(g(x_0)$

ouch

[jolly parrot]

ginny

[west sapphire]

tbh i would start simple (literally), does the characteristic function of C satisfy the requirements, and if not, can you modify it somehow to make it so?

[unborn charm]

we haven't covered characteristic function

for continuity we have the epsilon delta definition, equivalenetly this open ball definition

[west sapphire]

characteristic function of C just means 1 on C and 0 on C^c

[unborn charm]

oh lol

[west sapphire]

some people call it the indicator function

[unborn charm]

fair, that is a good start it seems

okay so let's say it is that

g is that for now

[unborn charm]

I still need this, but for a known g rather than an arbitrary one

[west sapphire]

right

the easier part is checking whether it's continuous on C^c

[unborn charm]

which again is this

so for an epsilon <= 1 we have B_{\epsilon}(g(x_0)) = {1}

so we have to find a delta > 0 such that all x in (x_0-delta, x_0+delta) get mapped to 1

[west sapphire]

do you know some basic facts about C?

[unborn charm]

supposing that doesn't happen, we have an open interval in C

which is false

[west sapphire]

like is it open, closed, neither

[unborn charm]

yeah I see that

[unborn charm]

has no open intervals, so neither I would assume

[west sapphire]

hmm

what about C^c

think about how you construct C

[unborn charm]

should be open since it is a union of open sets

[west sapphire]

yep good

C^c is open so C is closed

[unborn charm]

ah yeah

[west sapphire]

the fact that C^c is open should make your argument easy for continuity on C^c

[unborn charm]

ah that for every x_0 we will have some delta such that (x-delta,x+delta) is also in C^c

delta > 0 ofc

that does it yeah

[west sapphire]

yep

great

now the slightly harder part

checking whether g is discontinuous at every point of C

[unborn charm]

hmm

[west sapphire]

this will be useful:

the fact that C contains no open intervals

[unborn charm]

right

yeah I got it

thanks very much

[west sapphire]

nice, what was your finding?

g is discontinuous on C?

[unborn charm]

I have defined g to be 0 on C

[west sapphire]

should be "take epsilon < 1" on line 3, otherwise your ball will include 1 as well

(minor detail)

[unborn charm]

I have $\leq$

?

[jolly parrot]

ginny

[west sapphire]

which would include the possibility of epsilon = 1, no?

oh wait, your B_epsilon is open right

so that's fine

[unborn charm]

yes

[west sapphire]

ok no worries then

[unborn charm]

thanks!

my instructor gave the proof considering for every y_0 in C a sequence y_n in C^c such that |y_n-y_0| < 1/n

[west sapphire]

yep that works too

[unborn charm]

hmm

but that's not something I would be able to come up with lol

[west sapphire]

either way you're using the fact that there are points of C^c arbitrarily close to any given point of C

(because C contains no open intervals)

[unborn charm]

hmm

makes sense

[west sapphire]

oops said it backwards, fixed now

[unborn charm]

well then, see you around, thanks again

.close

[pearl pond]

Channel closed

Closed by @unborn charm

Use .reopen if this was a mistake.

[west sapphire]

yw

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vestal thistle]

Hello, could someone check if this proof looks good please?
⁨⁨```latex
\begin{Theorem}
Assume $n \in \bZ$. If $n^2$ is even, then $n$ is even.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose $n$ is odd.
Then, $n = 2k + 1$ for some integer $k$.
So, $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$.
Since $k$ is an integer, so is $2k^2 + 2k$.
Then, $n^2 = 2l + 1$, where $l = 2k^2 + 2k$.
Thus, by the definition of odd integers, $n^2$ is odd.
Therefore, by the contrapositive, if $n^2$ is even, then $n$ is even.
\end{proof}

[autumn fossil]

n^2 = 2l?

[pulsar flax]

should be n^2 = 2l +1

[autumn fossil]

other than that, it looks fine

u're just missing that +1

[vestal thistle]

Typo 😬

[jolly parrot]

Mor Bras

[vestal thistle]

Thank you!

.close

[pearl pond]

Channel closed

Closed by @vestal thistle

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vestal thistle]

Hello, could someone check if this proof looks good please?
⁨⁨```latex
\begin{Theorem}
Assume $n \in \bZ$. If $4 \nmid n^2$, then $n$ is odd.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose $n$ is even.
Then, $n = 2k$ for some integer $k$.
So, $n^2 = (2k)^2 = 4k^2$.
Since $k$ is an integer, so is $k^2$.
Thus, by the definition of divisibility, $4 \mid n^2$.
Therefore, by the contrapositive, if $4 \nmid n^2$, then $n$ is odd.
\end{proof}

[ashen ivy]

i recommend $\not\mid$ or $\nmid$ for a little more clarity

[jolly parrot]

ηασιβ ♥

[ashen ivy]

actually don't use $\not\mid$, whoops

[brave sluice]

Looks good 👍

[jolly parrot]

ηασιβ ♥

[ashen ivy]

the proof looks good - clever use of contrapositive

[brave sluice]

Yeah \nmid is good

[vestal thistle]

Yes, I didn't know ⁨⁨\nmid⁩⁩, thanks!

[jolly parrot]

Mor Bras

[ashen ivy]

fwiw detexify helps a lot with symbol lookup, especially more obscure symbols

[cinder flower]

there is always “4 doesn’t divide n^2” or “n^2 isn’t a multiple of 4” too

[vestal thistle]

There's ⁨\nshortmid⁩

[brave sluice]

\neg 4\mid n^2
(Just kidding)

[viscid shale]

I think no pre-set space actually fits correctly to make this work, lmao.

,, \not\hspace{3pt}\mid

[ashen ivy]

$n_2 \not\in [0]_4 \subseteq \mathbb{Z}/4\mathbb{Z}$

[jolly parrot]

ηασιβ ♥

[ashen ivy]

how's that for symbol spam

[vestal thistle]

Thank you all for you comments!

.close

[pearl pond]

Channel closed

Closed by @vestal thistle

Use .reopen if this was a mistake.

[eager jewel]

😈

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vestal thistle]

Hello, could someone check if this proof looks good please?
⁨```latex
\begin{Theorem}
Assume $n \in \bZ$. If $3 \nmid (n^2 - 1)$, then $3 \mid n$.
\end{Theorem}

\begin{proof}
We use the contrapositive.
Suppose $3 \nmid n$.
Written differently, $3 \nmid (n - 0)$.
This means, by the definition of congruency, that $n \not\equiv 0 \pmod{3}$.
So, either $n \equiv 1 \pmod{3}$ or $n \equiv 2 \pmod{3}$.

\begin{case}
Suppose $n \equiv 1 \pmod{3}$.
Then, by the definition of congruency, $3 \mid (n - 1)$.
So, by the definition of divisibility, $n = 3k + 1$ for some integer $k$.
And then, $n^2 - 1 = (3k + 1)^2 - 1 = 9k^2 + 6k + 1 - 1 = 3(k^2 + 2k)$.
Since $k$ is an integer, so is $k^2 + 2k$.
Thus, by the definition of divisibility, $3 \mid (n^2 - 1)$.
\end{case}

\begin{case}
Now, suppose $n \equiv 2 \pmod{3}$.
Then, by the definition of congruency, $3 \mid (n - 2)$.
So, by the definition of divisibility, $n = 3k + 2$ for some integer $k$.
And then, $n^2 - 1 = (3k + 2)^2 - 1 = 9k^2 + 12k + 4 - 1 = 3(k^2 + 4k + 1)$.
Since $k$ is an integer, so is $k^2 + 4k + 1$.
Thus, by the definition of divisibility, $3 \mid (n^2 - 1)$.
\end{case}

In both cases, we have that if $3 \nmid n$, then $3 \mid (n^2 - 1)$.
Therefore, by the contrapositive, if $3 \nmid (n^2 - 1)$, then $3 \mid n$.
\end{proof}

[jolly parrot]

Mor Bras

[pearl pond]

@vestal thistle Has your question been resolved?

[vestal thistle]

Well spotted 👌
Here's the fix:

⁨```latex
\begin{proof}
We use the contrapositive.
Suppose $3 \nmid n$.
Written differently, $3 \nmid (n - 0)$.
This means, by the definition of congruency, that $n \not\equiv 0 \pmod{3}$.
So, either $n \equiv 1 \pmod{3}$ or $n \equiv 2 \pmod{3}$.

\begin{case}
Suppose $n \equiv 1 \pmod{3}$.
Then, by the definition of congruency, $3 \mid (n - 1)$.
So, by the definition of divisibility, $n = 3k + 1$ for some integer $k$.
And then, $n^2 - 1 = (3k + 1)^2 - 1 = 9k^2 + 6k + 1 - 1 = 3(3k^2 + 2k)$.
Since $k$ is an integer, so is $3k^2 + 2k$.
Thus, by the definition of divisibility, $3 \mid (n^2 - 1)$.
\end{case}

\begin{case}
Now, suppose $n \equiv 2 \pmod{3}$.
Then, by the definition of congruency, $3 \mid (n - 2)$.
So, by the definition of divisibility, $n = 3k + 2$ for some integer $k$.
And then, $n^2 - 1 = (3k + 2)^2 - 1 = 9k^2 + 12k + 4 - 1 = 3(3k^2 + 4k + 1)$.
Since $k$ is an integer, so is $3k^2 + 4k + 1$.
Thus, by the definition of divisibility, $3 \mid (n^2 - 1)$.
\end{case}

In both cases, we have that if $3 \nmid n$, then $3 \mid (n^2 - 1)$.
Therefore, by the contrapositive, if $3 \nmid (n^2 - 1)$, then $3 \mid n$.
\end{proof}

[jolly parrot]

Mor Bras

[pearl pond]

@vestal thistle Has your question been resolved?

[vestal thistle]

Does this proof looks good?

[round heart]

looks good to me

in fact you can say something more general: let n be a positive integer and let a_1,...,a_n be n consecutive integers, then n divides exactly one of the numbers a_k

[vestal thistle]

I'll look up that, thanks!

[round heart]

np!

[vestal thistle]

.close

[pearl pond]

Channel closed

Closed by @vestal thistle

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tardy reef]

theres no intuition there, thats how work is quantified

ppl experimentally found that this dot product quantity seems useful and decided to call it work

[unborn abyss]

if u sit on a box while someone pulls it you're not helping

[south ice]

Ehh

Idk about that tbh there’s always gotta be an explanation

[tardy reef]

as for why the dot product thing is of some consequence, you need to see that some force affects the motion of a body in the direction of application, and as such, the component of motion thats along the force gets affected

[south ice]

Oh yeah im asking about the dot product lol

You thought I was asking why work is defined the way it is?

[tardy reef]

yea

[south ice]

Oh yeah no I was asking about the dot product

[south ice]

I didn’t understand

So work is force * distance right

[tardy reef]

no...

its dot product

[south ice]

Yeah

That’s not multiplying tho

Right

[tardy reef]

this is a superspecific case when the displacement is parallel to the direction of force

[south ice]

But work is generally defined via multiplication

[tardy reef]

maybe thats the case back in middleschool where extremely simple cases were handled

[south ice]

What’s the intuition of why the scalar produced by the dot product is the work

[unborn abyss]

dot product gives you the component of the force in the direction of motion

if the force and the motion are perpendicular the dot product is 0

if they're parallel it's just the regular product

[south ice]

Sorry but can you please elaborate

It’s not clicking tbh

[unborn abyss]

ummm what do you know about the dot product

[south ice]

Great question

All i know is it tells you the angle between 2 vectors

Or how the angle behaves (acute obtuse, flat)

[unborn abyss]

hmmm ok

[south ice]

But also I have no idea what the resultant scalar is

So yeah

[unborn abyss]

what you're describing is that $\vec a \cdot \vec b = |a|\cdot|b|\cos\theta$

[jolly parrot]

hayliänus austrǎlis

[south ice]

Yeah

[unborn abyss]

okay

and if theta is π/2 then that dot product will be 0

[south ice]

Yes

[unborn abyss]

if theta = 0 then that dot product is just |a| * |b|

[south ice]

Ye

[unborn abyss]

so we want to take force, times displacement, times the amount that your force is actually helping (cos theta)

and that turns out to be exactly the dot product

[south ice]

Why are you multiplying three things

I thought it was W = F · D

[unborn abyss]

it's force dot displacement

but like ok

if i told you i was helping to move a box and was exerting a force of 100N on it and it moved 3m

how much work did i do?

[south ice]

300

[unborn abyss]

300 what

[south ice]

N

[unborn abyss]

what

what is M

[south ice]

Newtons

[unborn abyss]

that's a unit of force not work

[south ice]

Typo

Oh

[unborn abyss]

how did you get 300

[south ice]

Multiplied distance and force

Distance is 3 meters and force is 100N

[unborn abyss]

okay, so how did we multiply 3m and 100N and get 300N

[south ice]

Just regular multiplication

3 · 100

[unborn abyss]

that poor lonely m

forgotten

unloved

abandoned

you should get 300 Nm aka 300J (joules)

[south ice]

Oh

[unborn abyss]

but regardless here's something i haven't told you

[south ice]

Yes

[unborn abyss]

i was sitting on the box while my friend moved it

[south ice]

Okay

So you just added more weight

However much that was

[unborn abyss]

yes, and i didn't help move it in any way

[south ice]

Ye?

[unborn abyss]

so i didn't do any work on the box

[south ice]

Yes

[unborn abyss]

but you said i did 300J worth of work on it

[south ice]

Wait not you

The one who moved it

[unborn abyss]

what?

no i did a force and it moved

[south ice]

Your friend moved it

Right

So your friend used work of 300J

[unborn abyss]

why's that

how do you know how much force my friend was using

btw my friend is a ford f150 truck with a trailer hitch

[south ice]

Oh

Uh okay I see what you’re asking

[unborn abyss]

it matters what direction the force is being applied

we care specifically about the component of the force in the direction that the box is moving

[south ice]

Okay now I’m completely lost 🌝

[unborn abyss]

which is |F| cos theta

[south ice]

What?

[unborn abyss]

i am applying a gravitational force to the box

which is pushing it down

but the box is moving left

so the component of my force in the direction of left is 0

[south ice]

Wait wait before this

What are we looking for here

In this picture

What do we want to find

[unborn abyss]

the amount of work that i perform on the box as it moves 10m

or did i say 3m

yeah 3m

[south ice]

Okay

So yeah then how do we find that

[unborn abyss]

we take the dot product of force and distance

[south ice]

What’s force and what’s distance here ?

[unborn abyss]

well... the force is the force of gravity, and the displacement is 3m

as i mentioned i'm exerting 100N on the box

it's just due to gravity so it's 100N straight down

and the box is moving 3m to the left

[south ice]

100N that is gravity and your weight combined ?

[unborn abyss]

wdym gravity and my weight

that is the force that i'm exerting due to gravity

also known as my weight

[south ice]

Oh okay

[unborn abyss]

so how much work am i doing on the box as it moves?

[south ice]

The force vector produced by your weight dot displacement vector

[unborn abyss]

uhhh right

what is that amount?

[south ice]

300J

[unborn abyss]

what are those vectors?

[south ice]

Yeah so

Displacement vector is <-3, 0> and your force vector is <0, -100>

Is it negative 100 or just 100

[unborn abyss]

we can say -100, that seems fine

[south ice]

Okay

Yea

[unborn abyss]

so what do you get if you take the dot product of those?

[south ice]

Bro I get 0

[unborn abyss]

correct

that's expected though right?

[south ice]

Idk 😂

[unborn abyss]

since i wasn't helping with this operation i was just sitting still and looking pretty

[south ice]

Well yeah you weren’t putting any work

[unborn abyss]

let's try one more thing first

[south ice]

Okay

[unborn abyss]

you are annoyed and get out and help

because of the angle you have to push kind of into the ground

so how much work do you perform on the box as it moves 3m to the left?

[south ice]

My force vector dot distance vector ?

[unborn abyss]

yup

[south ice]

Force is mass times acceleration tho right

How is that accounted for by just taking my force vector