help-36

[slate tiger]

which one r u stuck at @autumn agate

[patent compass]

The problem is in Z team now

[autumn agate]

^

[slate tiger]

is this Permu nd Combi

[autumn agate]

yeah

[patent compass]

Good problem

[patent compass]

That approach ain't getting us anywhere

[slate tiger]

yea nah havent revised PnC from a while

[slate tiger]

obv it wont, ydk permu nd combi

[patent compass]

Ikr

[slate tiger]

you'll study tht in 11th

Mid 11th

[patent compass]

I have studied

[slate tiger]

hm ok wht

[autumn agate]

that q is like an easier pyq

i think

[patent compass]

Later I'll tell

[slate tiger]

werent you confused w trignomentry a while back

[autumn agate]

from adv 2024

[slate tiger]

ya it i

[patent compass]

Later

[slate tiger]

orey

orey teri

.

[patent compass]

.

Back to study

[slate tiger]

alr we didnt see tht

nothing happened

[slate tiger]

👍🏼

[autumn agate]

can i do smth like this

or is it wrong

[patent compass]

Leme

Idk if it's wrong

But i got smthin wait

[dusty quarry]

!nosols again, as a reminder

[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.

[autumn agate]

man dw we both arent able to figure it out

mostly trying things lol

[dusty quarry]

hence, as a reminder. I've seen quite a lot of newcoming helpers just slap the solution and call it a day

[autumn agate]

okay i guess

[autumn agate]

umm sorry to ping but do you have any idea how i can begin with that q

[dusty quarry]

well I didn't look at it closely, but I guess I'll take a look

[autumn agate]

i appreciate it

[dusty quarry]

oh alright

I would do this by complementary counting

[autumn agate]

whats that

[dusty quarry]

assume no restriction exists, then subtract all invalid cases

[autumn agate]

ah so

set theory?

[autumn agate]

somewhat like this?

[dusty quarry]

I don't see why it would be set theory, it's a combinatorics counting technique

[dusty quarry]

incomplete

[autumn agate]

thats the start

like

[dusty quarry]

2 x 8C1 only deals with the first individual in team X. what about the other 7?

[autumn agate]

Total - n(A U B)?

[dusty quarry]

I would probably focus less on seeing these as sets

and just seeing these as events

I mean I suppose you can see them as sets, then which the final answer is exactly what you said.

[autumn agate]

2 x 8C1 will give me all the teams which can formed when S1 is a member right

[dusty quarry]

or rather, it will be using inclusion exclusion.

[dusty quarry]

all the possible team X, yes. but you want the overall number of team comps, don't you?

[patent compass]

Bro

I did it ig

N(X) =56

[autumn agate]

it was from the sets pov

[patent compass]

N(Y)=120

[autumn agate]

like i say the invalid teams of X is that and then find the invalid teams of Y

and subtract w set identities

[patent compass]

N(Z)=6

[dusty quarry]

you know what? perhaps you should finish your idea and then I'll take a look at it

and see? what did I say about nosols?

[patent compass]

Am I right?

[dusty quarry]

not only did you not even provide any explanation about where these numbers came from

[patent compass]

Bro

Let me

Tell

X= 8•7

=56

[autumn agate]

@patent compass that was the very first thing

i did

also

[dusty quarry]

then you have failed to understand the question at all

[autumn agate]

its wrong

[patent compass]

Bro

I did it diff

[patent compass]

Then u explain if u get it

[dusty quarry]

why do you think I'm stepping in to help?

but I'm not gonna dish out the full explanation right away

please let me do my thing with OP

[dusty quarry]

now OP, back to this

you may ping me once when you're done

[autumn agate]

ye 1s

okay

[patent compass]

K

[dusty quarry]

I will drop OP a hint: the final answer is a 3-digit number

[autumn agate]

this has to be the ans

[dusty quarry]

compute it and let me know

[autumn agate]

oh

ok

[dusty quarry]

but from the grouping it does look promising, even though that is not how I am looking at things

[autumn agate]

i messed up the multiplication while doing manually so used a calculator

665

[dusty quarry]

correct

[autumn agate]

oh wait

tysm

appreciate your help

[dusty quarry]

now if you wanted to know how I saw it

[autumn agate]

yeah

[dusty quarry]

I didn't use any sets

from the get-go, team Z is kinda picking up what's left after teams X and Y are assigned, so it is always 4C4 = 1

so we don't have to really consider team Z

then I know that if I count the total, I will then need to subtract the case where A is in team X, and where B is in team Y.
but if I do that, then I would have subtracted any case where BOTH A is in X and B is in Y one time too many, so I need to add any case where both A is in X and B is in Y once

[autumn agate]

i didnt understand the 2nd part of this

[dusty quarry]

so that ends up being:
(9C2 x 7C3) - (8C1 x 7C3) - (8C2 x 6C2) + (7C1 x 6C2) = 665

[dusty quarry]

specifically?

[autumn agate]

entire thing

[dusty quarry]

but if I do that, then I would have subtracted any case where BOTH A is in X and B is in Y one time too many, so I need to add any case where both A is in X and B is in Y once
so this part

[autumn agate]

yeah]

[dusty quarry]

now, the total is clear

and subtracting the two cases are also clear

[autumn agate]

yeah

[dusty quarry]

but now let's say I've subtracted the cases where A is in X

amongst those cases will be cases where B is also in Y

[autumn agate]

wait what exactly is A

is A = S1

here

[dusty quarry]

oh I used A to denote S_1

and B = S_2

[autumn agate]

ohh i see

okay

[dusty quarry]

cuz two less keystrokes

and one less shift key press

[dusty quarry]

anyway this

[autumn agate]

oh

wait

i get it

so thats what i did right

like n(A int B)

[dusty quarry]

basically what you did (which is called inclusion-exclusion)

[autumn agate]

is the 1 case which you added

[dusty quarry]

just without the set-theoretic framing

[autumn agate]

oh new term for me

[dusty quarry]

because for a problem like this having to come up with sets is a minor bit slower for me

[autumn agate]

oh i see

[dusty quarry]

implicitly I guess I'm still using sets. but the main computation is combinatorics

[autumn agate]

our sir also told us to use sets alot of times so yea

i went with sets

[dusty quarry]

then sure, use whatever you're comfy with

I will not dictate what you should use

[autumn agate]

rather than comfy its like

im used to it

[dusty quarry]

used to it = comfy with it

hopefully?

[autumn agate]

well

i guess so

[dusty quarry]

either way, you seem well-versed enough to be comfy with it anyway so that works enough for me

anything else?

[autumn agate]

nope

[dusty quarry]

aight

[autumn agate]

ill reopen this if i have any doubt

or create new 1

thanks so much

.close

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[keen hare]

did i start this correctly? this question is very confusing

[drowsy epoch]

why y2'=0y1+0y2=0

[keen hare]

i dont know, before someone told me that i should set y' = y1 and y = y2

thats all i got, i dont really understand how to do this problem at all

[drowsy epoch]

yes

though i would do y1=y and y2=y' rather be cause of the order

[drowsy epoch]

That's how you get y2'

[keen hare]

okay let me try this on my paper

[final saddle]

@keen hare Has your question been resolved?

[keen hare]

wow

this works

thank u

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[silver oar]

Anybody here familiar with Gina Wilson all things algebra ? If so please dm

[final saddle]

Please don't occupy multiple help channels.

[left trail]

You should probably go to #book-recommendations if you haven't been there yet

[covert tree]

i think the one they are referring to is some courses not book

[bold turtle]

It's a website/course, yh

[slow coral]

#help-6

here

[bold turtle]

[this is also a duplicate channel, yh]

[tulip coyote]

.close

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[tulip coyote]

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[topaz kite]

<@&268886789983436800>

[final saddle]

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[stable dew]

numiro 15

[stable dew]

what I did was convert cos into sin by sending to the other side

then found values of xsqr +x =0

is this method correct

.close

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[winter dagger]

Check the domain first.

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[honest crater]

Please help me to solve it, idk what to do

[topaz kite]

,rccw

[rustic scaffold]

hi

[honest crater]

Hello

[rustic scaffold]

so what are your thoughts for part i?

[honest crater]

Idk, put 300 in place of x?

[rustic scaffold]

you want to put something in place of x, but not 300

[honest crater]

Idk what to put

[sour tendon]

can you explain what h(x) is?

and what x represents

[honest crater]

It's height function

[sour tendon]

it is the height at any x, then when the van begins moving, what is x?

[honest crater]

0

[sour tendon]

yes! try solving (I) now

[honest crater]

Ok I am

[sour tendon]

it seems to me that x is the horizontal distance travelled by the van

and h(x) is the height along the x-direction

[honest crater]

Ok I have put x=0 and solved the function the answer is 3

Now what

[rustic scaffold]

yes but 3 what?

[sour tendon]

do you realize the answer to (i) now

[honest crater]

Hmmm....

[sour tendon]

read the last line of the question

[honest crater]

3 times 100?

[sour tendon]

where h(x) is the…

[rustic scaffold]

yup. 3 times 100 meters, specifically

[honest crater]

h(x) is height function

Nice ok

[sour tendon]

yes but (in 100m)

[honest crater]

So i have proved (i).. now how to solve second?

[honest crater]

Yes it's proved now

[sour tendon]

do you know what turning points of a graph are?

[honest crater]

No

[sour tendon]

do you know the cubic function's graph?

y=x³?

[honest crater]

Yeah ik

[sour tendon]

do you know differenciation?

[honest crater]

Yeah i do

Do I have to differentiate the function of h

[supple pawn]

Yeah

[honest crater]

h'(x)

Ohkk

[sour tendon]

at any point of a graph, where the derivative is zero, that is, at the local maximum, minimum, or the point of inflection of a graph

it is the turning point

a cubic graph has two

[rustic scaffold]

a turning point is a place where the function changes from concave up to concave down, or the other way around

[sour tendon]

do you get it?

[supple pawn]

Then solve for x?

[honest crater]

So i have to h'(0)

Right

?

Or i have to find the value of x?

[rustic scaffold]

apologies, I got mixed up with point of inflection. Ignore what I said

[sour tendon]

the ATMs exist at the turning points of the path traced by the van ( x , h(x) )
the turning points of the path are at where h'(x) = 0

(II) requires you to deduce the location of the ATMs

[honest crater]

See am i right?

[sour tendon]

I would assume you need to state the x-coordinate, and the h(x) (height) they exist at

[sour tendon]

sure

[honest crater]

Ok

And i assume in the third one i have to put 4 and 2 in place of x right?

[sour tendon]

don't assume. be certain

[honest crater]

Ohkk

[supple pawn]

Wait one of the x values has to be rejected right?

[honest crater]

I think i have to put 2 and 4 in h(x) right? Then substract them

[supple pawn]

Let me guess

[sour tendon]

there are two ATMs

[honest crater]

No there r two ATMs

[supple pawn]

Oh

[sour tendon]

yes
that would give you the height at 2, and 4

[honest crater]

Ok I'll solve

Is this correct?

[sour tendon]

it appears to be
does the book not have answers at the end?

[honest crater]

It's the question from my final examination

[sour tendon]

A past question paper? I see

[honest crater]

Yes

I left this question

And lost marks

[sour tendon]

be sure to realize the concepts this time hahah

[honest crater]

Sure haha

What about the last one ? What do i have to do?

[sour tendon]

think about it

[honest crater]

I don't understand what the question is asking about

[sour tendon]

Do you understand the first sentence?

[honest crater]

Yes difference between the heights of location of atms is more than 1 km

Ohh yes it's refering to 3 part

[sour tendon]

if it is more than 1000m then we need 1 more guard

[honest crater]

My answer is coming 800m.. which is less than 1000m so no guard needed

LoL thanks a lot dude

I feel so bad now for leaving this question

[sour tendon]

No worries. you can .close now

[honest crater]

.close

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[lament gale]

can someone help me read slope fields, I dont know how to differentiate between the bottom two

[alpine dove]

<@&286206848099549185>

[final saddle]

@lament gale Has your question been resolved?

[lament gale]

guys

😄

[grave trout]

Try integrating it. The answer is that the slope field is reversed (at least that's what I think)

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@lament gale Has your question been resolved?

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[ruby crypt]

hello, help me check answer ples

[ruby crypt]

and for this, I need help hehe

thanks in advance

[blissful meadow]

First one is good.
For the second part, what's an easy example of a function you know which is continuous everywhere and differentiable everywhere but at one point?

[ruby crypt]

f(x) = 1/x?

f(x) = lnx?

[blissful meadow]

Hmmm there's an easier one.

It's defined on all of R.

It's continuous

But there's a pesky corner

[ruby crypt]

f(x) = |x|?

[glossy zephyr]

1/x works too btw

[ruby crypt]

how is it differentiable

[blissful meadow]

Oh yeah true

[ruby crypt]

?

[blissful meadow]

Verify that it satisfies the formula you have and that the limit exists. Is it differentiable at x=0?

[glossy zephyr]

Ill try to hint at it, try to think why the definition of derivative includes f(x0) either for left or right evaluation

[ruby crypt]

|x|?

[blissful meadow]

Yes either one

[ruby crypt]

because we are finding if it's differentiable at x = x0?

[ruby crypt]

wdym either one

[blissful meadow]

--You can do 1/x as well as Laplace transform said--

[glossy zephyr]

But beyond that, what could be the problem if instead of f(x0) we use the limit of x tending to x0

[ruby crypt]

I searched on google and didnt understand, whats laplace transform

[glossy zephyr]

Which is the same as x0 + h with h -> 0

[glossy zephyr]

im laplace transform (my nick)

[ruby crypt]

lol no idea

[ruby crypt]

then f(x0) may not exist

but everything surrounding it may

[glossy zephyr]

What happens if f(x0) doesnt exist and we try to evaluate the derivative

[ruby crypt]

we can't

[glossy zephyr]

and this one does not include f(x0)

Just gets really close to it.

[ruby crypt]

OOOOO

but isn't f(x0) well defined?

[glossy zephyr]

Not necesarilly.

[ruby crypt]

it is given in the first sentence, how

[blissful meadow]

Wait actually why does 1/x work as a counter example though? The limit exists? Am I tripping?

[glossy zephyr]

In this part of the problem, we want to show why that isnt suitable as a definition for the derivative.

[glossy zephyr]

limit to the left + limit to the right = 0

so the derivative at x = 0, equal 0 too

[blissful meadow]

Shouldn't it be 1/x^2?

[glossy zephyr]

quick sec.

[ruby crypt]

I see I see

[glossy zephyr]

The two sides slope version runs into the problem of not being able to tell if the function is well defined at x0

[ruby crypt]

now that i think abt it, yea 1/x isnt well defined on 0

[ruby crypt]

got it got it

holy this is tricky

[glossy zephyr]

Btw, |x| isnt exactly a good example

[ruby crypt]

yea I dont know how to differentiate that

[glossy zephyr]

the left sided derivative, the right sided derivative and the two sided slope all exists.

but they dont match

[blissful meadow]

Yes but isn't the point to find such a function for which the limit with +h and -h exists, but for which the function isn't differentiable?

For |x| that limit exists, it's 0. But it's not differentiable at x=0.

For 1/x the limit doesn't exist. 1/x^2 works

[glossy zephyr]

the problem is that youre assuming a function being differentiable is already a thing we can conclude.

And we want to show why the two-sided slope isnt a good method to tell if thats the case.

[blissful meadow]

They're asking whether it is necessary that if that limit exists, then the function is differentiable.

1/x isn't a counter-example since that limit doesn't exist in this case.

[glossy zephyr]

The limit for the 2 sided one does exist

its 0

[uneven cloak]

hey may I ask what the intial problem statement is cuz I would like to help if I can

[ruby crypt]

part b from my pic ples

[uneven cloak]

ok

[ruby crypt]

this

[glossy zephyr]

oh no mb, wrong two sided, lmao.

[ruby crypt]

I now understand the reasoning on why it's false, but I don't get why this reasoning makes sense given that it is well-defined on x0

[blissful meadow]

The first part is essentially showing that differentiable implies two sided limit exists.
The second part is showing the converse (or that the converse doesn't hold, which it doesn't).
So you need an example where the two sided limit exists and the function isn't differentiable.

[ruby crypt]

isn't differentiable at that one specific point right

[blissful meadow]

Yes

I think any function that is not differentiable at a point but is even around that point works as a counter example.

[ruby crypt]

yea so |x| is elligible for this

i cant think of any other tho lol

[blissful meadow]

1/x^2 works and is even more striking since you can argue that 1/x^2 is not even continuous at 0.

But |x| is the easier/typical example.

[ruby crypt]

what but isn't 1/x not cont. at 0 too?

[blissful meadow]

The issue with 1/x is that the limit doesn't actually exist.

[ruby crypt]

isn't it -1/x^2?

except at 0

[blissful meadow]

If you plug it in the limit, you end up computing $\lim_{h\to 0}\frac{\frac{1}{h} - \frac{1}{-h}}{2h} = \lim_{h\to 0}\frac{1}{4h^2}$, which blows up to infinity. (Should've been 1/h^2 but same thing)

[soft zealot]

Azyrashacorki

[ruby crypt]

AHHHHHHHHHHH

I see

yea it is given that the expression exists

[blissful meadow]

Whereas with 1/x^2 they add together to 0 in the numerator.

Same with |x|

[ruby crypt]

but for |x|, do I need to split cases

where x neg and pos

nvm lemme just write first

how do we bound |x0 + h| - |x0|

I mean can I write lower bound as |x0 + h - x0|?

[blissful meadow]

What you get when you plug in $|x|$ for the function at $x_0 = 0$ is $\lim_{h\to 0} \frac{|h| - |-h|}{2h}$.

[soft zealot]

Azyrashacorki

[ruby crypt]

shoot i am tripping, its supposed to be f(x0 + h) - f(x0 - h)

mb

why u write x0 = 0

the question generalizes it to x0

[blissful meadow]

Well you want a counter example at a certain point.

[ruby crypt]

ah right lol

[blissful meadow]

The claim in a) is "differentiable at x_0 => symmetrically differentiable at x_0".
In b) you're trying to show that the converse doesn't hold.

[ruby crypt]

yes understood

[blissful meadow]

The problem essentially shows that this new limit is a weaker notion of derivative (it's called a symmetric derivative). Weaker in the sense that differentiable => symmetrically differentiable, but not the converse.
There is an analog of this I think for the second derivative which, in this case does yield a correct equivalent definition of the second derivative without having to take the derivative of the derivative.

[ruby crypt]

I see

[ruby crypt]

well after reaching this, do I find upper and lower bound then squeeze

or check 0+ and 0-

[blissful meadow]

Well the numerator should be 0 at this point

So the limit is 0

[ruby crypt]

but denominator is also 0

[blissful meadow]

But for any nonzero h, the numerator is already 0

So the expression is 0 for any h < 0 and any h > 0.
You can take the left and right sided limits if you wish but it shouldn't be a surprise that the limit of something which is always 0 (except at 0) is 0.

[ruby crypt]

yes

understood

then for the 2nd half, I just show lim when x -> 0+ and 0- is not equal and conclude its not differentiable at x = x0 righ?

x = 0

[turbid bough]

hello everyone

[blissful meadow]

Yes you may show that |x| is not differentiable like that explicitly using left/right sided limits.

[ruby crypt]

any other easier way? I would like your view on this

[blissful meadow]

No I think if you're using the definition then that's how you'd show it's not differentiable.
It depends how in depth you want to go. To me it would be sufficient to explain that you know |x| is not differentiable at x=0, but showing it directly is great.

[ruby crypt]

ok just wanna confirm once more, I should find lhs limit = -1 and rhs limit = 1 right?

[blissful meadow]

Yep, those essentially agree with the fact that the right branch is a line with slope 1 and the left branch is a line with slope -1.

[ruby crypt]

thank you so much good sir

I will come back with more problems 😛

.solved

[final saddle]

Channel closed

Closed by @ruby crypt

Use .reopen if this was a mistake.

[sonic cairn]

Task 11. (0–2)

For a performance in a certain theater, tickets were sold according to the following price list.

A total of 200 tickets were sold for this performance.

After paying the costs related to organizing the performance, amounting to 25% of the ticket sales revenue, the organizers had 4665 zł left.

Calculate the number of reduced-price tickets sold for this performance. Show your calculations.

[blissful meadow]

This channel hasn't been made available by the bot yet so you can't claim it.

[sonic cairn]

Logik

[final saddle]

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[left trail]

Is my understanding of why $g(x) \neq 0 \forall x\neq a$ . First we can see if $g(x) = 0 \forall x \in I$ it would be the limit of something undefined which is undefined. Then if it is $g(x) = 0$ for some $x\in I$ I think we would just have the same issue because as x approaches a there would be points where the limit is undefined

[soft zealot]

BigBen

[cerulean epoch]

if g is 0 at one point in I you would be able to solve this by making I smaller so that the point goes outside

essentially the condition is so that it makes sense to take a limit

there are weaker conditions where it makes sense as well but for the purposes of what the author will talk about this is all you need to know

[left trail]

im not following

[honest gust]

a limit of a function that is not defined at a point can exist

they're just called removable discontinuities

[left trail]

I was thinking of say f (x)/0

[cerulean epoch]

ok so say we want to take lim_(x → 0) f(x)/g(x) in I = [-2, 2] but g(1) = 0

[honest gust]

it can sometimes happen that if you had x approaching a where g(a) = 0

but it's like continuous and differentiable

(alongside f)

it may be true that

[cerulean epoch]

to solve this we would take a new I to be [-0.8, 0.8]

[honest gust]

the limit can exist via l'hopital's rule

[left trail]

im saying that it was 0 to begin with

[cerulean epoch]

i'm sorry but the issue you're talking about isn't related to the problem they have on hand

[left trail]

I don't see why we take a smaller interval

[cerulean epoch]

this is so f(x)/g(x) is defined in I \ {0}

for example you would actually have an issue if g was 0 at all 1/2, 1/3, 1/4, 1/5, etc.

[left trail]

so we change the interval to make the quotient defined. but what about removable discontinuites?

[cerulean epoch]

you don't care if g looks like this, f(x)/g(x) will still make sense

[left trail]

what if it was just undefined at say 0 1 and 2

[cerulean epoch]

then you would take I = [-0.8, 0.8] which excludes 1 and 2

you don't care about 0 since that is your a

[left trail]

so then my initial thought was right we need our quotient to be defined on the interval we are working on

[cerulean epoch]

it was in the right direction but hopefully what i said brought you some clarification

in practice you will not need to worry about this at all

the condition is imposed there solely to restrict the class of functions g you can talk about

[left trail]

ok so to make sure the distinciton you made it sure it could be undefined at some points but then you can change the interval

[cerulean epoch]

but this class will include all nonzero rational functions

[cerulean epoch]

yeah

[left trail]

ok thank you

.solved

[final saddle]

Channel closed

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

why is the domain x > 0 and not x =/ 0

[formal trail]

the domain of a differential equation solution should have no gaps in it

[modest sequoia]

i think it has to be one continuous interval

containing the initial x-value

[lament gale]

so anything including and bigger than 1?

[formal trail]

it can go before 1, just can't go past a discontinuity

[modest sequoia]

its still continuous between 0 and 1

you want the largest interval possible

[lament gale]

ah alright

thank you guys

.close

[final saddle]

Channel closed

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[final saddle]

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• Show your work, and if possible, explain where you are stuck.
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• Type the command .close to free the channel when you're done.
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[frigid laurel]

ignore the answer on the right it's wrong,i need to calculate the elements of matrice X if X*A=C-X

[frigid laurel]

only thing i knew how to do so far is make X*A+X=C

[final saddle]

@frigid laurel Has your question been resolved?

[north valve]

That's true, one step further,
$$ X(A+I) = C $$

[soft zealot]

wifix

[north valve]

$$ X = C (A+I)^{-1} $$
If $ A+ I $ is invertible

[soft zealot]

wifix

[north valve]

And obviously $A+I$ is invertible, since $\det(A+I) \ne 0$ .

[soft zealot]

wifix

[final saddle]

@frigid laurel Has your question been resolved?

[frigid laurel]

thank you so much man

[final saddle]

Channel closed

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[plain oracle]

out of context but how does it feel to write latex by hand

or did you copy the codes

[final saddle]

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[drowsy epoch]

[sonic crystal]

[trail mango]

[gentle zephyr]

how do you prove this theorem using the theorem from above

[final saddle]

@gentle zephyr Has your question been resolved?

[blissful meadow]

Is the sequence ${b_n-a_n}_{n\in\N}$ monotone? Is it bounded?

[soft zealot]

Azyrashacorki

[gentle zephyr]

unsure

[blissful meadow]

Then check.

[gentle zephyr]

yes because the limit of bn - an is 0

[blissful meadow]

Well that's what you're trying to prove.

[gentle zephyr]

so L = 0 and then bn - an is bounded and monotonic

[gentle zephyr]

oops

[blissful meadow]

Actually no it's assumed. Is there any conclusion to this "consequence"?

[gentle zephyr]

care to elaborate?

[blissful meadow]

The consequence you've written down is saying "If I_n are intervals such that 1) and 2) hold".

There is no conclusion.

The conclusion should be something like "then there is some x in R such that the intersection of all the I_n is {x}."

[gentle zephyr]

my bad

@blissful meadow

[blissful meadow]

Consider the sequences ${a_n}{n\in\N}$ and ${b_n}{n\in\N}$. Show that both are bounded and monotone. This will use condition 1) in how $I_n$ are defined.\
Then you can use condition 2) to show they actually converge to the same value.

[soft zealot]

Azyrashacorki

[blissful meadow]

Then you can show that this value, say x, lies in the intersection, and that it further is the only point in the intersection.

[gentle zephyr]

how?

[blissful meadow]

Is ${a_n}{n\in\N}$ bounded? If so by what?\
Is ${a_n}{n\in N}$ monotone?

[soft zealot]

Azyrashacorki

[gentle zephyr]

but i dont understand

[blissful meadow]

What don't you understand?

[gentle zephyr]

idk if its boubded or mone

[blissful meadow]

Do you understand intuitively why the general claim is true (that the intersection of intervals with those properties is just a single point)?

[gentle zephyr]

yeah

it keeps shrinking until the interval gets smaller and smaller

[blissful meadow]

Okay. Now a_n is just the sequence of lower endpoints of those intervals.

Do you think it's bounded?

[gentle zephyr]

what?

[blissful meadow]

The intervals are [a_n, b_n]

The sequence with terms a_n is the sequence of left endpoints of the intervals.

I'm asking you whether you think it's bounded or not.
If you don't know or you're unsure, try to come up with some bound you think may work, or an explanation as to why you think it's not bounded (if that's what you think).

[gentle zephyr]

wdym

[blissful meadow]

a_n is the left endpoint of the interval [a_n,b_n]

[gentle zephyr]

yes

[blissful meadow]

The sequence (a_n) is the sequence of left endpoints of the intervals [a_n, b_n].

[gentle zephyr]

wut

[gentle zephyr]

wdym?

[blissful meadow]

x_n is a banana
The sequence (x_n) is a sequence of bananas.

[gentle zephyr]

yes

[blissful meadow]

That's the same thing. Replace banana by "left endpoint of the interval [a_n, b_n]"

[gentle zephyr]

ah i see but how come xn is banana and (xn) is a sequence of banana

[blissful meadow]

Because each x_n is a banana?

[gentle zephyr]

(xn)_[x >= n]

[gentle zephyr]

In = [an, bn]

In is ONE interval

[blissful meadow]

Yes. I'm saying that you should consider the sequence (a_n) where a_n is the left endpoint of I_n for each n.

[gentle zephyr]

AH

[blissful meadow]

So now this

[gentle zephyr]

how to prove is bounded

[blissful meadow]

From how the intervals are defined, how come can a_n not just go to infinity?

[gentle zephyr]

In is closed

lim bn - an = 0

so bn - an is something

[blissful meadow]

Regardless of whether b_n - a_n goes to 0. From condition 1) alone you should be able to find a bound.

[gentle zephyr]

what

In+1 = [an+1, bn+1]

In supset In+1 => an >= an+1

maybe?

[blissful meadow]

Note that since $I_{n+1} \subseteq I_{n}$ for all $n$, in particular, $a_n \in [a_1, b_1]$ for all $n$.

[soft zealot]

Azyrashacorki

[blissful meadow]

Does that give you bounds on a_n?

[gentle zephyr]

well ordering principle?

[blissful meadow]

No

It's just by definition. The first interval contains the second interval which contains the third interval ...

[gentle zephyr]

how did you get this

[blissful meadow]

So any one of the I_n is a subset of I_1

[gentle zephyr]

sure

[gentle zephyr]

lower bound for an ?

[gentle zephyr]

why

[blissful meadow]

It should give you both a lower and an upper bound on a_n if you agree that a_n is in [a_0, b_0].

[gentle zephyr]

n in N so n >=1

[gentle zephyr]

how?

[blissful meadow]

Do you agree that a_n is in [a_1, b_1] for all n?

[gentle zephyr]

I'm not sure

[blissful meadow]

Well if [a_n, b_n] is a subset of [a_1, b_1] for all n ...

[gentle zephyr]

yes

[blissful meadow]

Then it should be clear that a_n is in [a_1, b_1] for all n

[gentle zephyr]

no the other way around

[blissful meadow]

No, ... [a_3,b_3] is a subset of [a_2, b_2] is a subset of [a_1, b_1]

[gentle zephyr]

In supset In+1 forall n in N

[blissful meadow]

Read the statement again.

[gentle zephyr]

ln = [an, bn] so [an, bn] supset ln+1

if n = 1 then

[a1, b1] supset l2

and l2 supset l3

so l1 supset ln

[blissful meadow]

Yes, which means that a_n is in [a_1, b_1] for all n.

[gentle zephyr]

sure.

[blissful meadow]

Ok. so is (a_n) bounded?

[gentle zephyr]

bounded from below

and above

[blissful meadow]

By what

[gentle zephyr]

a1 and b1

[blissful meadow]

Ok, so the sequence (a_n) is bounded.

[gentle zephyr]

both act as lower and upper bound

[blissful meadow]

Now why is it monotone?

[gentle zephyr]

we need to prove is decreasing or increasing sequence

[blissful meadow]

Yes. Do you think (a_n) is increasing or decreasing?

[gentle zephyr]

how would I know dat

[blissful meadow]

Again it comes from the fact that I_(n+1) is a subset of I_n.

How can you conclude that a_{n+1} >- a_n from that?

[gentle zephyr]

an+1 in ln

[gentle zephyr]

a1 is lesser than or equal than every other

[blissful meadow]

Well you just want to show that a_(n+1) >= a_n.

Not that a_n >= a_1

[gentle zephyr]

unsure

[blissful meadow]

You said yourself that a_(n+1) is in [a_n, b_n]

Can you conclude that a_(n+1) >= a_n?

[gentle zephyr]

how

[blissful meadow]

What does it mean for a_(n+1) to be in [a_n, b_n] in inequality notation?

[gentle zephyr]

ahh

an <= an+1 <= bn

@blissful meadow

[blissful meadow]

Yeah

So (a_n) is monotone increasing and bounded.

What can you conclude?

[gentle zephyr]

an = L

[blissful meadow]

For some L yes.

Now do the same thing with the sequence (b_n).

[gentle zephyr]

ln = [an, bn]
l1 supset l2
l2 supset l3
l1 supset ln
bn in [a1, b1]

ln supset ln+1

in particular a1 <= bn <= b1 forall n in N

so (bn) is bounded

ln supset ln+1 so bn+1 in an <= bn+1 <= bn

so (bn) is bounded

and (bn) is monotonically decreasing

@blissful meadow

[blissful meadow]

Yes, so (bn) converges to some value M.

Now, if you know (b_n - a_n) converges to 0 and both (a_n) and (b_n) converge to L and M respectively, can you say something about L and M?

[gentle zephyr]

lim bn - an = 0 <=> lim bn = lim an <=> L = M

@blissful meadow

[blissful meadow]

Yes.

So now let's call x = L = M.

Show that x is in I_n for each n.
Show that if y is in I_n for each n, then y=x.

[gentle zephyr]

how

[blissful meadow]

Start with the first one. Why must x be >= than any a_n? Why must x be <= than any b_n?

[gentle zephyr]

unsure

[blissful meadow]

a_n is an increasing sequence with limit x.

[gentle zephyr]

so

[blissful meadow]

It should be clear that a_n <= x for all n. If not, then assume that a_n > x for some n and show a contradiction.

[gentle zephyr]

if lim an = L1 and (an) is increasing and bounded how is L1 = sup(an)

[gentle zephyr]

the proof seems hard

[gentle zephyr]

this one is a little easier to read

[final saddle]

@gentle zephyr Has your question been resolved?

[covert tree]

@gentle zephyr is there translation of the original question?

[gentle zephyr]

yes of course

[covert tree]

can you translate it, is it just "prove monotone convergence theorem"?

[gentle zephyr]

no

[covert tree]

no to "translate it" or no to the latter or no to both

[gentle zephyr]

yes to the former no to the latter

[covert tree]

i see pls do translate it

[gentle zephyr]

this latex bot sucks

[covert tree]

if texbot don't work ill use the tex bot built into my brain

[gentle zephyr]

\noindent\fbox{
\begin{minipage}{\textwidth}
\vspace{0.2cm}
If $(a_n){n \in \mathbb{N}}$ is a monotonic and bounded sequence of real numbers, then there exists $\lim a_n = \ell$, or equivalently, $(a_n){n \in \mathbb{N}}$ converges.
\vspace{0.2cm}
\end{minipage}
}

\vspace{0.6cm}

\noindent \textbf{Some consequences}

\vspace{0.4cm}

\noindent \textbf{Nested Intervals Theorem}

\vspace{0.3cm}

\noindent Given a closed interval $I = [a,b]$

\begin{equation*}
I_1 = [a_1, b_1] \supseteq I_2 = [a_2, b_2] \supseteq \dots \supseteq I_n = [a_n, b_n]
\end{equation*}

\vspace{0.2cm}

\begin{center}
INTERVAL LENGTH = $b - a = \ell(I)$
\end{center}

\vspace{0.4cm}

\noindent For each $n \in \mathbb{N}$, let $I_n = [a_n, b_n]$ be a closed interval on the real line such that the following conditions hold:

\vspace{0.2cm}

\begin{itemize}
\item[1)] $I_n \supseteq I_{n+1} \quad \forall n \in \mathbb{N}$
\vspace{0.2cm}
\item[2)] $\lim_{n \to +\infty} \ell(I_n) = 0 = \lim_{n \to +\infty} (b_n - a_n)$
\end{itemize}

\vspace{0.4cm}

\noindent Then there exists a unique $x \in \mathbb{R}$ such that $\displaystyle x \in \bigcap_{n=1}^{+\infty} I_n$

\vspace{0.3cm}

\noindent $\iff x \in I_n \quad \forall n \ge 1$

[covert tree]

which one do you need help with, or both?

[gentle zephyr]

I need help with everything

[covert tree]

OK do you have least upper bound property from class?

[gentle zephyr]

what is it

[covert tree]

supremum of a bounded above set always exists if you are working with real numbers

[gentle zephyr]

yeah we can use this

[covert tree]

well ${a_i:i\in \mathbb N}$ is a set. And we can set the least upper bound to be $a$. Show that $a=\mathrm{lim}_{i\to infty} a_i$

oh texit bot is offline that's why

[covert tree]

you can try copying the text to your local latex thing

[gentle zephyr]

what does it say?

[gentle zephyr]

i am not in my home sorry

@blissful meadow

@covert tree

[gentle zephyr]

what?

[covert tree]

I put your sequence into a set

that set is a bounded above set, and by Least upper bound property it has a supremum: a.

[gentle zephyr]

thats illegal

[covert tree]

how so

[gentle zephyr]

the sequence is bounded not the set

[covert tree]

Yeah use the fact that the sequence is bounded, to show that the set I gave is bounded

the set is this

[gentle zephyr]

this proof seems non standard

[covert tree]

how so?

[gentle zephyr]

.

[covert tree]

the first step is literally the same

"define S to be the set in terms of {a_n} and let L be the supremum of S"

i think their notation is a bit weird

it should be (a_n)_n\in \mathbb N

[gentle zephyr]

then proof that the sequence converges to the sup

[gentle zephyr]

in N not in Z

[covert tree]

this is only the first step

[covert tree]

mb typo

[gentle zephyr]

care to elaborate

[gentle zephyr]

what is least upp bound prop

@covert tree

[covert tree]

supremum exists

supremum of a upper bounded set exists

[covert tree]

^^^

[gentle zephyr]

how?

[covert tree]

first try to show that the set is bounded

[covert tree]

i thought u said you can use this fact

that supremum exists for bounded above sets

[gentle zephyr]

how do you know this set is bounded...

[covert tree]

you use the fact that the sequence is bounded

what does it mean for the sequence to be bounde?

[gentle zephyr]

the sequence is bounded the set isnt

[covert tree]

what does it mean for the set to be bounded

[covert tree]

no the set is bounded. you can prove it

[gentle zephyr]

there exists a supremum

[covert tree]

use the fact that the sequence is bounded to show that the set is bounded.

[covert tree]

I think you are confusing the definition of bounded with the stronger thing that supremum exists

[gentle zephyr]

if the sequence is bounded from above then there exists some n in N such that an <= ai forall i in N

[gentle zephyr]

this is for a set

[covert tree]

yes

[gentle zephyr]

what about a sequence

@covert tree

[gentle zephyr]

yeah so its upperly bounded

there exists some upper bound

@covert tree

[covert tree]

so you can see how (a_n) being bounded above would imply {a_n} being bounded above

[gentle zephyr]

how so?

[covert tree]

give it a try, try to show it from definition I gave above

[gentle zephyr]

(an) bounded then {an} bounded

ok

[covert tree]

yeah try and show this

[gentle zephyr]

yes is proved

[covert tree]

show pls

[gentle zephyr]

suppose (an) is bounded above then there exists some upper bound for (an) so call it b such that b >= an forall n in N, then grab every an in N and put it in a set call it S = {an : forall n in N} then since there exists some b >= an forall n in N, HOWEVER IS B IN S?

[covert tree]

b is not necesarrily in S

[covert tree]

^here

in definition 1.5 I put b in R

anything in R works

[gentle zephyr]

then we done mate

@covert tree

[covert tree]

good now you see

that (a_n) being bounded immediately givea {a_n} being bounded

so now, by what you said, supremum of this set must exists

Let's go back to what I said:

we have a least upper bound (ie, supremum), a. It should be somewhat intuitive that a should be the limit

so I want you to try to show that $a$ is the limit

[soft zealot]

qwertytrewq

[covert tree]

oh bot is back

[gentle zephyr]

latex bot sucks

[gentle zephyr]

wut

[covert tree]

a is defined above in the image

[covert tree]

^^^

[gentle zephyr]

a >= an forall n in N

a <= bn forall n in N

so an <= a <= bn

[covert tree]

what's b_n?

[gentle zephyr]

sorry

a >= an forall n in N

[covert tree]

and a is the minimum possible value that satisfy this

[gentle zephyr]

is not easy or intuitive

[gentle zephyr]

how

[covert tree]

definition of supremum

[gentle zephyr]

we are trying to prove a is the supremum

we cannot assume it

@covert tree

[covert tree]

u said this^

so u r saying we can't use the least upper bound property?

[gentle zephyr]

is it for sequences or sets?

[covert tree]

for sets

sets have the least upper bound property

https://en.wikipedia.org/wiki/Least-upper-bound_property

[gentle zephyr]

ok

so {an} has a supremum

how do you prove the limit of an is equal to the sup

[covert tree]

why is why I asked if we can assume this

if we cannot the proof goes a bit differently

[gentle zephyr]

we can.

[covert tree]

and much longer

[covert tree]

try it urself: try to use proof by contradiction

if a_n does not converge to a

show that a will not be the supremum

[blissful meadow]

Note that this is a general case which isn't really necessary to continue the proof you were already doing Renato (although it's good to know). It's relatively simple to show that a_n <= x and b_n >= x without invoking the supremum property.

[gentle zephyr]

how

[blissful meadow]

Assume you have some a_N for which a_N > x and show (using the monotonicity and convergence of (a_n)), that you get a contradiction.

[gentle zephyr]

(an) is bounded from above

in particular there exists some p >= an forall n in N

[blissful meadow]

The point is that you want to show that x is an upper bound.

(of (a_n))

and then a lower bound of (b_n)

[gentle zephyr]

how

[covert tree]

whats bn and x

[blissful meadow]

You don't need to be rude. <@&268886789983436800>

[ornate token]

Please remove this message.

That is extremely disrespectful to OP.

People have rights to ask, and you do not have to swear at him and move on.

[blissful meadow]

They've magically disappeared

[gentle zephyr]

the proof we were working on, In is an interval such that In = [an, bn] , we were proving the nested interval theorem

[blissful meadow]

While OP had mentioned that they needed help with everything they were really already done with showing that (a_n) and (b_n) converge to the same limit in their problem. What was left was showing that this limit, x, was contained in each interval and was thus in the intersection.

[covert tree]

yeah can use monotone convergence for that

fairly simple afterwards

[covert tree]

I thiught everything included the monotone conv theorem

[covert tree]

like here

[blissful meadow]

No they were proving a corollary

[covert tree]

oh bruh all these time i coulda just assumed mvt? lol

[blissful meadow]

If there is such an a_N, then by monotonicity a_n > x for any n >= N. Try to conclude that the limit wouldn't be x in that case.

(a contradiction)

[gentle zephyr]

i might need more hints

[blissful meadow]

Use epsilon = a_N - x > 0 in the definition of the limit.

[gentle zephyr]

lim aN = x?

[blissful meadow]

No. The fact that $\lim_{n\to \infty} a_n = x$ means the usual definition of the limit : $\forall \epsilon > 0, \exists K \in \N$ such that $$ n \ge K \implies |a_n - x| < \epsilon$$.

I'm saying that you can take epsilon = a_N - x to get a contradiction to the fact that x is the limit.

[gentle zephyr]

contradiction like what

[soft zealot]

Azyrashacorki

[blissful meadow]

If you use epsilon = a_N - x, by assumption that a_N > x, there won't be a K like the definition requires.

The contradiction is that the limit of (a_n) is x, but the above would mean that it's not.

So there is no such a_N > x.

[gentle zephyr]

(an) is upper bounded so there exists some x in N such that x >= an forall n in N
so if epsilon = aN - x
then epsilon <= 0

@blissful meadow

[blissful meadow]

Again, the goal is to show that x, the one you've found before, is an upper bound itself.

x is the limit of (a_n) you computed.

You need to show that it's in the intersection of all the intervals, so in particular, you need to show that a_n <= x <= b_n for all n.

To show that a_n <= x, you assume NOT, that is that there is some N such that a_N > x, and you can contradict the (proven) fact that the limit of (a_n) is x using the argument I gave you above.

[gentle zephyr]

suppose there exists some N such that aN > x then ?

since aN > x and x >= an forall n in N
we get aN > an forall n in N

[blissful meadow]

You don't know that x >= a_n forall n in N. That's what you're proving.

Since a_N > x, and (a_n) is increasing, then a_n > x for any n > N.
Now use epsilon = a_N - x in the definition of the limit.

[gentle zephyr]

forall epsilon > 0 there exists some K in N such that N >= K => |aN - x| < aN - x

[blissful meadow]

For all $\epsilon > 0$, there is some $K \in \N$ such that $n \ge K \implies |a_n - x| < \epsilon.$.

[gentle zephyr]

wut

[soft zealot]

Azyrashacorki

[blissful meadow]

This is the definition of the limit.

In particular, since $a_N - x > 0$, there should be some $K \in \N$ such that $n \ge K \implies |a_n - x| < a_N - x$.

[soft zealot]

Azyrashacorki

[gentle zephyr]

x - aN < an - x < aN - x

[blissful meadow]

Is that possible if $a_n > x$ for any $n \ge N$?

[soft zealot]

Azyrashacorki

[gentle zephyr]

the absolute value complicates things

[blissful meadow]

It doesn't. Eventually you'll have a_n > x, so |a_n - x| = a_n -x >= a_N-x.

[gentle zephyr]

what if an < x

[blissful meadow]

It might be the case but eventually n >= N.

And at that point a_n >= a_N > x.

[gentle zephyr]

ƴeah since an is monotonically increasing if n >= N then an >= aN > x

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@gentle zephyr Has your question been resolved?

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[outer hemlock]

im not sure whats happening in the denominator, can someone explain

[brazen breach]

they multiplied it with a conjugate since a^n-b^n=(a-b)(a^(n-1)+a^(n-2)b+...+b^(n-1) where in this case, since the powers are 1/6 they just kinda did it with 1/6 at the bottom

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@outer hemlock Has your question been resolved?

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[mossy chasm]

not sure what to even start to manipulate here

[sturdy cypress]

i think you solve it like separately

find x from the equation

ok

[mossy chasm]

i wish

i did try to find the roots though just in case if it actually was

i dont think it is still

[final tangle]

consider multiplying both sides of the first equation by (x-1) and you should recognise something special about x

[mossy chasm]

diff of cubes?

[barren fossil]

is there anything special about the roots of x^2 + x + 1 = 0?

[final tangle]

yes

[mossy chasm]

i actually did that but i didnt know what to do after (x^3-1)/(x-1)=0

[mossy chasm]

not that i know of

[final tangle]

um how did you get that?

[scarlet sequoia]

so, diff of cubes, not sum of cubes

[final tangle]

still wrong

[scarlet sequoia]

there we go

x^3-1 = 0 then

so, x^3 = 1

[mossy chasm]

yeah ok

so uhh

do you think i would be able to figure out from here

[scarlet sequoia]

yes

how can we use that x^3 = 1

to simplify the expression you have

[mossy chasm]

factor

oh

is it -1

[mossy chasm]

oh wait

cube roots of unity

i see

[barren fossil]

yes

[mossy chasm]

so im assuming the expression is equal to -1

got it

thanks

.close

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[dusk shoal]

What makes a sine or cosine parent function positive or negative

[gritty chasm]

parent function is a strange term, so im going to imply you mean sine function and cosine function

[topaz kite]

unit circle. have you heard of it?

[radiant igloo]

-# hello this is a test message so it looks like a lot of ppl are answering

[dusk shoal]

I mean the parent function like g(x)=cosx

[thin dagger]

tuff

[honest gust]

jobs bro jobs

[sturdy cypress]

this is spam and has no merit

[gritty chasm]

this should clarify it

[topaz kite]

looks like Itsuki wasn't even needed for that to happen. alright.

[radiant igloo]

fahhh

[gritty chasm]

if further explanation required, distances to the RIGHT of the origin on the x-axis are negative.

[chilly granite]

https://tenor.com/view/matematica-gif-7322246

[dusk shoal]

I dont really get it

[chilly granite]

okay oaky sooo sososo-

[radiant igloo]

okay but like can one person take over instead of the whole of #discussion

[chilly granite]

LOOK AT THE RED LINE HERE-

[honest gust]

let's just shut up for now and let @gritty chasm do ts

[spare summit]

wish some people got that memo

[radiant igloo]

the wheel of truth spun and said annie should take over

[dusk shoal]

Doesn’t it have to do anything with the function being even or odd in a graph

[gritty chasm]

if further explanation required, distances to the RIGHT of the origin on the x-axis are negative. therefore if you consider the definition of the sine, and let's say the angle is in the second quadrant, then:\
$$\cos A = \frac{\text{Adjacent}}{{Hypotenuse}}$$
Now observe that the hypotenuse is ALWAYS taken positive in such situations(its the distance b/w two points).
the adjacent on the other hand, is the distance from the origin in the negative direction of the x-axis. so it is negative. i'll quickly pull up a figure if you don't get it. (so the cosine is negative)

[soft zealot]

Annie Maqionde

[gritty chasm]

yes, but the unit circle is a much better explanation

here you go.

its a bit small but its visible

the hypotenuse is positive,

the adjacent is on the negative x-axis therefore it is negative

@dusk shoal from the diagram here, can you tell me whether sine is positive or negative?

use the definition of the sine as opposite/hypotenuse

[normal heron]

its not visible, the text isnt readable

well (ig you dont need it)

[gritty chasm]

the text isnt really required

[normal heron]

yeah

[dusk shoal]

So basically sine = opposite/hypotenuse

But I don’t really get how to determine if either is negative or positive

[gritty chasm]

the hypotenuse is ALWAYS positive

[gritty chasm]

did you get why the adjacent is negative?

its on the negative x-axis

correct?

left of the origin

[dusk shoal]

Yeah

[gritty chasm]

now, tell me, is the OPPOSITE side, is it on the positive y-axis, or negative y-axis?

[dusk shoal]

Positive

[gritty chasm]

it can be parallel to also, you just have to tell me whether its 'above' or 'below' the x-axis

[gritty chasm]

precisely

so the value of the opposite is positive

therefore sine is?

[dusk shoal]

Positive

[gritty chasm]

precisely

do the same for tan

and tell me what you get.

[dusk shoal]

Since it’s sine = +/+ right

Basically if the adjacent is above the x axis it’s positive and below it’s negative

Hypotenuse is always positive

[gritty chasm]

the adjacent here is on the x-axis, is it on the negative x axis, or the positive x axis?

[dusk shoal]

And adjacent if it’s in the positive area of y axis then it’s + and vice versa

[dusk shoal]

eh

Positive?

I mean the value of x=0 can’t be either right?

[gritty chasm]

ok so lemme make a quick table.\

1. Hypotenuse: always positive\
2. Adjacent: On the negative x-axis(left of origin): $-$\
   On the positive x-axis(right of origin): $+$\\
3. Opposite: Above the x-axis(or on the positive y-axis): $+$\
   Below the x-axis(or on the negative y-axis): $-$

[soft zealot]

Annie Maqionde

[chilly granite]

-# not to interruput but is it okay if i add a bit of visuality to ur current explination?

[gritty chasm]

@dusk shoal use this always.

[gritty chasm]

ofc.

[dusk shoal]

Oh wait no mb it’s on the negative side of x

My bad

[chilly granite]

you can really see this in this gif here!!

https://tenor.com/view/matematica-gif-7322246

notice how sin(the red line) is potive when above the x axis and negative when its below)

[chilly granite]

and cos is pos when its to the right of the y axis and neg when its to the left)

[radiant igloo]

ASTC guys

[dusk shoal]

Okay I got it

[chilly granite]

-# yes but this is why ASTC exsists

[dusk shoal]

Sine is basically the y axis and cosine the x axis

[chilly granite]

-# in this gif yes!

[dusk shoal]

And if sine is on the upper region then it’s positive

And if cosine is on the right then it’s positive

And vice versa

Tysm

[chilly granite]

exactly!!! as annie said earlier here^!!

and tan would just be sin/cos which is why tan is POS in the 3rd quadrant!!

[radiant igloo]

!done

[final saddle]

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