help-0

[cloud nacelle]

and many other things

google it

[twin nimbus]

!occupied @granite iron

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[daring echo]

That's not what I meant. I meant how do I use that equality to find something

[cloud nacelle]

oh

arc length = R times theta

where theta is in radians

[granite iron]

I don’t get where the R(y+1) goes into 2R 2/(y+1)

[cloud nacelle]

create a new channel

and ask

#❓how-to-get-help

[twin nimbus]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[daring echo]

I still don't understand. If I divide arc length by circumference and I have the area of a circle then I can find area of sector by cross multiply and divide?

Is that what it means?

[cloud nacelle]

yes

[daring echo]

That's weird. I'm gonna need to find a proof to believe that

[cloud nacelle]

just input the formula and check it out yourself

[daring echo]

Yeah but how did they figure that out

Who figured all this stuff out

[cloud nacelle]

haha

[daring echo]

Too much mathing for me lol

[cloud nacelle]

math is fascinating isnt it?

💀

[lone heart]

@daring echo Has your question been resolved?

[lone heart]

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[lethal salmon]

Does this seem right

I found 5 btw

Any idea where to start on number 6 i havent done this yet

[buoyant saddle]

so sine is negative in which quadrants

note that sine is related to the y value

on the unit circle

[lethal salmon]

2?

Wait 3

[buoyant saddle]

no Q2 is above the x axis

yes Q3 is one

where else

[lethal salmon]

4

[buoyant saddle]

mhm because they’re below the x axis

meaning y is negative

so you’re working in those two quadrants but they only want between pi and 3pi/2

which is what quadrant

[lethal salmon]

3

Well

Its in between

[buoyant saddle]

yes it’s Q3

so draw your triangle in Q3

the vertical distance is down 7

the hypotenuse is 25

so just find the adjacent side

then tan is easy

and cos is easy

make sure you know

if cos is positive or negative in Q3

same with tan

what do you think

[lethal salmon]

I assume negative

[buoyant saddle]

cos corresponds to the x coordinate

yes negative because it’s to the left

now what would tan be

tan is the ratio of sin/cos

if sine is negative

and cos is also negative

what’s tan

[lethal salmon]

Negative

[buoyant saddle]

what’s negative/negative

[lethal salmon]

Oh positive

[buoyant saddle]

mhm

[lethal salmon]

Is it a right triangle too

[buoyant saddle]

yes

[lethal salmon]

I assume yea

[buoyant saddle]

it says in a right triangle

[lethal salmon]

Ah right lol

[buoyant saddle]

so what’s cos

what’s the adjacent side

[lethal salmon]

24

[buoyant saddle]

mhm

so cos=

and tan=

[lethal salmon]

Cos = -24/25 right

[buoyant saddle]

mhm

[lethal salmon]

And then yea tan is positive

[buoyant saddle]

and the ratio is

[lethal salmon]

7/24

[buoyant saddle]

yup

[lethal salmon]

Awesome thank you man

[buoyant saddle]

youre welcome

have a nice night

[lethal salmon]

Except i think i might need sone help on 7 lol

[buoyant saddle]

oh sure

no problem

8 too?

[lethal salmon]

Nah 8 should be fine

9 though i got confused by angle of declination

[buoyant saddle]

ok so you know how negative angles work

[lethal salmon]

I mean depression

[buoyant saddle]

yea

[lethal salmon]

Somewhat

[buoyant saddle]

that’s just the top angle

[buoyant saddle]

so

negative angles go clockwise

from zero

instead of ccw

[lethal salmon]

Yes

[buoyant saddle]

so you’re going down 60 degrees

is sine negative or positive

[lethal salmon]

Negative

[buoyant saddle]

which quadrant is it in

mhm

[lethal salmon]

4

[buoyant saddle]

so what’s sin(60) normally

[lethal salmon]

.87

[buoyant saddle]

30-60-90 triangles

it’s sqrt(3)/2

[lethal salmon]

Oh

[buoyant saddle]

hold

[lethal salmon]

Oh thats with sin included?

[buoyant saddle]

so for the unit circle the x is just 1

[buoyant saddle]

wdym

[lethal salmon]

Well i never really knew why 30 60 90’s worked the way they did

[buoyant saddle]

sine is just the ratio of opposite to hypotenuse

yea 30-60-90 is a special right triangle

angle opposite the 30 degrees is half the hypotenuse

[lethal salmon]

Ah

[buoyant saddle]

and you can confirm this all works with pythagorean theorem if you want

[lethal salmon]

I believe u lol

[buoyant saddle]

1^2+(sqrt(3))^2=4

it works

anyways

you should memorize the ratios

for sin cos and tan

for both the 45-45-90 and 30-60-90 right triangles

[lethal salmon]

Yea i learned them like a year ago and forgot

Well not the ratios for those ig

[buoyant saddle]

no worries

ok anyways

so sin(60)=sqrt(3)/2

so sin(-60)=..

[lethal salmon]

Sqrt-3/2?

[buoyant saddle]

yes

well

[lethal salmon]

Or -sqrt3/2

[buoyant saddle]

write it like

yea

-sqrt(3)/2

now

are you familiar with radians

[lethal salmon]

Yea i just convert them all to degrees

[buoyant saddle]

do you know how many degrees are in pi/6

[lethal salmon]

No but ik how to convert

[buoyant saddle]

30

pi/4 is 45

and pi/3 is 60

so these are the only ones you should know off memory really

and obviously pi=180

[lethal salmon]

Alright

[buoyant saddle]

so

5pi/6

[lethal salmon]

Yes i know this one at least lol

Its 150 degrees

[buoyant saddle]

is pi/6 less than pi correct

mhm

yes

so the reference angle will be what

[lethal salmon]

210?

[buoyant saddle]

no

30

[lethal salmon]

Wait

I always mess this up

[buoyant saddle]

if you drew the triangle

you’re 30 degrees from the x axis

[lethal salmon]

It’s supposed to be 180 i always do 360

[buoyant saddle]

it’s always how many degrees you are from the x axis

[lethal salmon]

Yea

I keep messing that up lol

[buoyant saddle]

so

is sine positive or negative

and what about cos

and with those tell me what tan will be

positive or negative

[lethal salmon]

Positive

[buoyant saddle]

sine is positive yes

because it’s up or above x axis

[lethal salmon]

Cos isnt?

[buoyant saddle]

yes it’s negative

because it’s to the left

[lethal salmon]

Ohh

Yea nvm

[buoyant saddle]

so if sine is positive

[lethal salmon]

Tan will be negative

[buoyant saddle]

and cosine is negative

what’s tan

yea

and i’m guessing you don’t know tan of 30

but

look at the triangle

opposite of 30 is 1 adjacent to 30 is sqrt 3 and hypotenuse is 2

so what’s tan(30)

[lethal salmon]

Wait so is 30 the degree im basing my opposite and adjacent on?

[buoyant saddle]

yes

because it’s the reference angle

[lethal salmon]

Yea im lost

What are the lengths of each side?

Im confused because someone said the hypotenuse was always 1 earlier

Is the hypotenuse 150?

[buoyant saddle]

that’s for the unit circle

and it is

i’m talking about

for the ratio

of the 30-60-90

triangle

the hypotenuse is twice the side opposite the 30 degree angle

and the side adjacent to the 30 degree angle is the sqrt(3) times the opposite side of the 30 degree angle

[lethal salmon]

Yea i dont understand what you’re trying to explain sorry

How do i find the opposite side length?

[buoyant saddle]

well the ratio is always the same

that’s why they use x

so if you were to find tan(30)

you’d do x/(sqrt(3)(x))

which simplifies to 1/sqrt(3)

[lethal salmon]

Does it end up as -sqrt3/2 again

Nvm

[buoyant saddle]

no

[lethal salmon]

I just have no clue where you got this answer

[buoyant saddle]

tan(30) is just 1/sqrt(3)

but

tan(150) is -1/sqrt(3)

[buoyant saddle]

from the special right triangle

these are just ratios you have to know

[lethal salmon]

I just realized this sections not going to be on our test

Figures since we havent talked about this ahh i just wasted so much of your time im sorry lol

[buoyant saddle]

nah you’re fine

no worries

[lethal salmon]

Ok well i just did 8 im probably right on that

9 ig ill do i do need help with one more thing if you dont mind

[buoyant saddle]

yea sure

[lethal salmon]

Just one since the rest i can figure out after one

Im just confused about graphing it

Why is it in pi and not normal units but on a graph in desmos its normal units

Well unless i highlight the units

[buoyant saddle]

wdym

[lethal salmon]

Nvm hold up

[buoyant saddle]

the x axis is in terms of pi

because

the multiples of pi are the important points for trig functions

like we don’t say sin(1 radian)

it’s always ya know 0,pi/6,pi/3,pi/4,pi/2,pi etc

[lethal salmon]

The period on this should be 2pi right

[buoyant saddle]

yes

[lethal salmon]

Yea

[buoyant saddle]

because the frequency is 1

[lethal salmon]

And the amplitude is just 1 or

[buoyant saddle]

when you go to those special points where it’s a maximum or minimum etc those are just always going to be points that are in terms of pi for trig graphs desmos does this on purpose

[buoyant saddle]

yea

[lethal salmon]

Ah ok

Yea sorry im basically learning waves for the first time again rn lol

Horizontal shift is 2

Vertical is 0 right

[buoyant saddle]

no

vertical shift

is 2

horizontal shift is zero

horizontal shifts have to do with the input

vertical shifts are some added number

[lethal salmon]

But the line of horizontal is moving up by 2?

[buoyant saddle]

yes so it’s a vertical shift

horizontal shift is left/right

like if you had

sin(x-pi/2)

that would shift the graph to the right pi/2 units

[lethal salmon]

Oh wait

It’s different for these graphs

Hmm weird idk why im thinking of it this way

[buoyant saddle]

like if you had

[lethal salmon]

I got it now

[buoyant saddle]

(x^2)+4

it’s going up 4

the whole graph

[lethal salmon]

Yea

[buoyant saddle]

but

(x+4)^2

moves it to the left 4

[lethal salmon]

Ah yea ok

So for b

Would it be amplitude of 2 period of 2pi horizontal shift of pi and vertical shift of 0?

[buoyant saddle]

so amplitude of 2

shift of pi to the right

no a stretch is essentially the amplitude atleast a vertical shift a horizontal stretch would be a factor attached to the x

meaning the frequency

[lethal salmon]

Yea sorry my bad

[buoyant saddle]

for trig

[buoyant saddle]

yea no vertical shift

and P=2pi

that’s good

[lethal salmon]

Awesome

Alr ill try the last one and then leave lol

So amplitude -3 period 2pi horizontal shift -pi/2 and vertical shift 1

[buoyant saddle]

amplitude is never negative

[lethal salmon]

Oh

[buoyant saddle]

the negative tells you it’s reflected over the x axis

amplitude is how high/low it goes

[lethal salmon]

Ah ok

[buoyant saddle]

so it’s just 3

[lethal salmon]

Oh ok

[buoyant saddle]

and then yea P=2pi vertical shift is 1 and horizontal shift is pi/2 to the left

or -pi/2 to the right is you like

[lethal salmon]

Awesome

Thanks so much man

Really appreciate it

[buoyant saddle]

you’re welcome

have a nice night

[lethal salmon]

You too

[lone heart]

@lethal salmon Has your question been resolved?

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[hazy kettle]

any explaination pls

[lone heart]

Please don't occupy multiple help channels.

[edgy walrus]

we can start by lining up the books
since they are distinct, we can first calculute how many ways there are to order 5 distinct books

what would that be?

[hazy kettle]

2^5?

[edgy walrus]

not quite

so imagine we have 5 spots in a line up like so
A B C D E

[hazy kettle]

oki oki

5P5=5!?

[edgy walrus]

now we want to order the 5 books
so we have 5 options for position A and once weve used one of the books, we have 4 options for position B, ...

[edgy walrus]

yes 5!

now that we have figure out the different ways we can line up the 5 books, we must figure out how to split between 2 bookshelves

have you learned the stars and bars method?

i can explain if you havent

[hazy kettle]

oki oki,
i think it's gonna be
5,0
4,1
2,3
and opposite?

[hazy kettle]

no I havn't learnt that..

[edgy walrus]

wellll... you can technically do it like that since the numbers are small
but theres a really nice and neat formula that is much cooler and quite simple once you get the hang of it

[hazy kettle]

pls teach me that...

[edgy walrus]

so lets arrange our 5 books and image placing 1 divider
this divider will determine which books go into shelf 1 and which books go into shelf 2

for example:
A B | C D E -> (2, 3)
A B C D | E -> (4, 1)
| A B C D E -> (0, 5)

does this make sense so far?

[hazy kettle]

yes it does

[edgy walrus]

i just added their "equivalences" using the method you previously mentioned

[hazy kettle]

can I ask a question here?

[edgy walrus]

okay perfect
so now the question is, how many places/options do i have to place the divider?

[edgy walrus]

ofc

[hazy kettle]

why have we found this before it doesn't relate to finding arrangements of books....

[edgy walrus]

ahh i see your question
the reason we first find the number of ways to arrange the 5 distinct books is because the second part of our method doesnt account for distinctness
i think it will become more clearly once we explain the full solution

[hazy kettle]

oki oki understood let's go ahead and understand complete solution

[edgy walrus]

i believe this is where we paused

[hazy kettle]

yes

[hazy kettle]

it's 3*2

[edgy walrus]

yes, we have 6 options to place the divider
| A | B | C | D | E |
as we can see by the 6 bars i have used

but we only want to divide into 2 bookshelves so we only want to use 1 divider
so the answer would be 6 choose 1

because we have 6 places to divide the books but we only want to divide once

so our complete answer would be:
ordering the 5 distinct books which is 5! = 120 and then with that "lineup" of books, we insert a divider/bar to split into the 2 bookshelves which we have 6 options for
so the answer is 120*6 = 720

a good problem to practice with would be:
In how many ways can 10 distinct books be arranged in 4 bookshelves?

i can give you a solution if you would like

[hazy kettle]

okii but it's a permutation ques 😅 (I believe as we haven't learnt combination yet till this lecture )

[edgy walrus]

yes, which is why we started by ordering the books like you said: 5P5=5! =120

ahh i see

then perhaps your professor gave you small numbers so that you could just use the method you suggested earlier:
5,0
4,1
2,3
and opposite

[hazy kettle]

yesss

[edgy walrus]

but this would be very messy with large numbers, in which case the formula would help a lot 😄

[hazy kettle]

ohh now I understood why wee had 6 divider here

[hazy kettle]

it might sound silly but why are we multiplying 6 into 5!?

I mean is this a formula?

[hazy kettle]

got it got it,
it's bcz we have 6 ways to put divider
and when we have options to choose from we use rule of multiplication

[edgy walrus]

yes this kind of like a rule
the multiplication rule for independent events

the most basic example of this would be like if you have 3 different shirts and 2 pairs of pants, how many outfits can you make?
3 x 2 = 6
this is because what shirt your wear does not depend on what pants you wear and vice versa (the 2 events are independent)

[edgy walrus]

yes exactly

[hazy kettle]

so,
ordering the 10 distinct books which is 10! = 36,28,800
we only want to divide into 4 bookshelves so we only want to use 2 dividers
which is 6 chose 2 = 15

so the answer is 36,28,800*15 = 5,44,32,000
correct?

[edgy walrus]

very very close
since we want to divide into 4 books shelves, we will need 3 dividers
the number of dividers is always one less than the number of "groups" you need to form (in this case the groups we are forming are the bookshelves)
like in the last example, since we had 2 bookshelves, we only needed 1 divider
it might help to draw out a scenario: a | b c | d e | f g h i j -> (1, 2, 2, 5)

so for the very end, it should be 6 choose 3 = 20
so the correct answer would be 10! * 20 = 72,576,000

[hazy kettle]

amazing, this was a nice practice question 😄

Thank you so much for explaining. I reallly appreciate your help!! <333

[edgy walrus]

ofc!! no problem! 😄

[hazy kettle]

.close

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[alpine sable]

(x^2+3x+1)(x^2+3x-3) pls solve this

[alpine sable]

can someone help ps

pls*

[molten pivot]

what is there to solve?

do you want to expand it? or factor it further? find the zeros?

[alpine sable]

like just simplify it

[left isle]

take x^2 and multiply it by each term in the second set of parentheses, then take 3x and do the same, and then the same for 1

add them all together

and dont forget about your negatives when multiplying

[alpine sable]

i don't need x^4 can i get my next step in ()^2??

can i send u the next step my teacher got??

[left isle]

send it here sure

[alpine sable]

just explain to me how did it happen

[left isle]

that would probably be helpful

[alpine sable]

(x^2+3x)^2-2(x^2+3x)-3

@left isle

nvm i got it

.close

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

I'm wondering if in my proof it is okay to assume that |a|=k is finite

I don't know how else you'd do this

[keen plinth]

the order of a is the cardinality of the subgroup generated by a

[vapid shuttle]

yes and couldn't this cardinality be non-finite?

that's my doubt

[keen plinth]

potentially

<f(a)> = f(<a>)

[vapid shuttle]

What is this relevant to? I'm not seeing it. And if it potentially is non-finite, then is my proof false?

[keen plinth]

injectivity of f means f is a bijection between <a> and f(<a>)

[keen plinth]

you'd have to account for k being non finite

[vapid shuttle]

why are you considering cyclic subgroups though?

[keen plinth]

because |a| = |<a>|

[vapid shuttle]

but couldn't |<a>| also be non-finite?

[keen plinth]

|<a>| is a cardinal number

when <a> is not finite then |<a>| = N

and f being a bijection says they have the same cardinality anyway

[vapid shuttle]

I guess I'm not following what the point of that is, sorry, could you elaborate?

[keen plinth]

ord(a)
= card(<a>)
= card(f(<a>))
= card(<f(a)>)
= ord(f(a))

which line doesn't make sense

[vapid shuttle]

a few, but I guess just the first line to start
|a|=card(<a>)

[keen plinth]

that is what i said up here

[vapid shuttle]

well, I see why that is true in the finite case

[keen plinth]

you can prove this statement separately

ideally you'd take it as a definition though

[vapid shuttle]

Okay for the infinite case, I'm not even sure how to make sense of |a|, because isn't |a| the smallest positive integer k where a^k=e

and if such a k does not exist

then

idek what to say about it?

[keen plinth]

it's the cardinality of the subgroup generated by a in that case

but when the subgroup is not finite

it can only be countable

[vapid shuttle]

Okay

So, taking for granted that |a|=card(<a>)

this is equal to card(f<a>) because f is a injective so both f<a> and <a> have the same amount of elements even if infinitely many

[keen plinth]

yes

this is a purely set theoretical argument

[wraith stratus]

this is just the definition of the order of an element btw

[vapid shuttle]

and why is that equal to card(<f(a)>)?

[keen plinth]

f(<a>) = <f(a)> is something you should verify

[vapid shuttle]

in the non-finite case I'm not familiar with if we defined it that way (or at all), but I will have to check and for now just accepting that

[keen plinth]

this is because f is a homomorphism

[wraith stratus]

👍

[vapid shuttle]

f(<a>) = the set of f(a^i) for i in Z
= the set of f(a)f(a)...f(a) i times =(f(a))^i because f is a homomorphism
= <f(a)> since i ranges for all of Z

[keen plinth]

yeah

true even if f isn't injective

[vapid shuttle]

objective

[keen plinth]

autocorrect

smh my smh

[vapid shuttle]

and then card(<f(a)>)=ord(f(a))

is

card(<f(a)>)=|f(a)|

so again just the definition ?

[keen plinth]

yeah

[wraith stratus]

yep

[vapid shuttle]

So we have this

for the non-finite case

that proves

|a|=|f(a)|

[keen plinth]

that proof works for all cases

your proof only works for the finite case

[vapid shuttle]

But I liked my proof :(

[keen plinth]

it's correct though

[vapid shuttle]

I'll just append this for the non-finite case I think

[vapid shuttle]

Thank you

[wraith stratus]

in your proof

aren't you only getting j >= k

[keen plinth]

last part

[vapid shuttle]

|f(a)| <= k
suppose for contradiction that |f(a)|<k
and in particular is equal to j<k

then when we contradict this

we get

equality

[wraith stratus]

oh yes carry on

[vapid shuttle]

verily

@keen plinth this is all I have for the non-finite case... odd because I don't think we defined it as you did ever

[keen plinth]

yeah that's a poor definition

[vapid shuttle]

I think I should email my ma'am about this

to be safe

[keen plinth]

well if the order is infinite

you can certainly still say that |a| = card(<a>)

[wraith stratus]

you can show that if a has infinite order, then all the a^k's are distinct which makes {a^k | k integer} countable

[keen plinth]

because <a> will be infinite too

[wraith stratus]

is how the two are equivalent

[keen plinth]

but ideally you want that infinity to be specifically countable infinity

[vapid shuttle]

We did this

[keen plinth]

[wraith stratus]

yep so map the set {a^k | k integer} to the integers by a^k -> k

and it's a bijection

[keen plinth]

that's integer

[vapid shuttle]

indubitably

that's integer

[wraith stratus]

you can just use your theorem 7.8 also to show |f(a)| = |a| for infinite case

if you don't wanna do the cardinality arugment

[vapid shuttle]

isn't it still the cardinality argument

[keen plinth]

all f(a^k) = f(a)^k distinct

infinitely many of them

[wraith stratus]

i suppose it is still a cardinality argument

[vapid shuttle]

unfortunately I feel more lost now

I don't know what emotions/thoughts KekPogFridge is supposed to illicit in me

[keen plinth]

definitely one of them

[wraith stratus]

$kekw \cap pog \cap fridge$ obviously

[keen plinth]

[ocean seal]

chebyshev's infinite pee norm

[vapid shuttle]

so

[wraith stratus]

[vapid shuttle]

by theorem 7.8, why is |a|=card(<a>)

card(<a>) = |N| I buy

[wraith stratus]

use this mapping

then integers is same cardinality as N

[vapid shuttle]

doesn't that just show card(<a>)=|N|

[keen plinth]

yeah

[vapid shuttle]

but that's not showing |a| = card(<a>)

[keen plinth]

|a| is infinite

[wraith stratus]

oh i see

[keen plinth]

so is card(<a>)

[vapid shuttle]

it could be uncountable?

[keen plinth]

what more could you want

oh

cannot

[vapid shuttle]

why cannot

[keen plinth]

this shows it

[vapid shuttle]

😭

Sir

[keen plinth]

<a> contains all the powers of a

[wraith stratus]

pleas use the map a^k -> k

[keen plinth]

so there are only countably many

[wraith stratus]

!!!

it's a bijection

[vapid shuttle]

yes

I understand that card(<a>)=|N| by that bijection

I just wasn't sure why |a| = card(<a>)

[wraith stratus]

no i was answering this part

[vapid shuttle]

I was referring to |a|

not card(<a>)

why can |a| not be uncountable?

[wraith stratus]

i understand why you're confused about the |a| = card(<a>) since you're not defining it with order of the subgroup

let me think how to answer

[keen plinth]

it doesn't make sense for |a| to be uncountable

since you didn't even define it to be a cardinal

you just distinguish the two cases of finite and infinite

[vapid shuttle]

we didn't even define it to be anything

[keen plinth]

okay use the definition you posted from the book

[vapid shuttle]

[keen plinth]

least k>0 such that a^k = e

[vapid shuttle]

doesn't exist

[keen plinth]

and infinite if no such k exist

[vapid shuttle]

so hence |a| does not exist

[keen plinth]

no, you say it's infinite

[vapid shuttle]

so hence |f(a)|=|a| is non-sensical

[keen plinth]

when it's infinite, there are only countably many powers of a

[vapid shuttle]

why

[keen plinth]

well

just count them

a^k, one for every k in Z

[vapid shuttle]

ok |a| is infinite and there are only countably many powers of a

[wraith stratus]

and infinite if no such k exists
depends on how you want to define it of course, but it is commonly defined like this , just not for your particular course

[vapid shuttle]

don't see why => |a| is countable

[keen plinth]

well you define it to be the cardinality of Z in that case

or, you just say it's infinite

[vapid shuttle]

but importantly I define it

we're not like deriving this from something vague in the text

right?

[keen plinth]

no this is how everyone should define it

your text will probably just say that two elements have the same order if they're both infinite

and consider the infinities to be the same

this is something that should be proven though

[wraith stratus]

^ this is likely

[keen plinth]

since your text doesn't distinguish between infinities

so you just regard it as the one infinity

[vapid shuttle]

[keen plinth]

one doesn't have to talk about cardinalities to talk about infinities

you can just have some object called infinity

[vapid shuttle]

No

https://tenor.com/view/no-gif-6533142189269812111

[keen plinth]

the infinity at the end of the real line isn't a cardinal

it's just some additional point that you add onto the real line

[vapid shuttle]

https://tenor.com/view/that-doesnt-exist-rerez-thats-not-real-theres-no-such-thing-gif-20893693

Okay

I will think of how to juggle with this existential threat to my algebra homework tomorrow.

thank you for your help

[keen plinth]

[vapid shuttle]

, ti austinu

[keen plinth]

if it helps think less about infinities

[ocean seal]

The current time for austinu is 02:46 AM (PDT) on Fri, 12/04/2024.

[keen plinth]

finite stuff is better for your sanity

[vapid shuttle]

well I was very happy to prove it for just the finite case

then you had to come in here talking about

daowdnaudwa

ok

.close

[lone heart]

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[keen plinth]

[wraith stratus]

@vapid shuttle

here is a sanity check

from my old algebra textbook

[keen plinth]

you can tell its old from their usage of \varepsilon

[wraith stratus]

hungerford :^)

[keen plinth]

,, \mathrm k \mathrel\varepsilon \mathbf Z

[lone heart]

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[alpine sable]

GUYS PLEASE HELP WITH THE INTEGRALLL

[lethal belfry]

hint :- $x^2-25+1=x^2-24$

[ocean seal]

ƒ(Why am. I here)=I don't know

[alpine sable]

?

[livid sage]

standard process for integrating rational functions with higher degree numerator:

use polynomial division to rewrite the integrand

[alpine sable]

Umm

[last minnow]

Just use long division

[alpine sable]

Ok i got what i gotta do but i cant do it 0_0

[last minnow]

just type long division integral

[alpine sable]

Whats that

[last minnow]

if you had typed it to google instead of asking me you would've got the answer

[livid sage]

what do you mean by this?

[alpine sable]

Ok ill re arrange the polynomial athe the top and then divide?

[livid sage]

no you just divide

[alpine sable]

Oh

I think i got what you mean

So;

[last minnow]

https://youtu.be/aHkpVpPiBLo?si=cB-dvrOoynDQAwMW

just watch this one

[alpine sable]

Ok ill watch

But why does this not work

.close

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[potent cobalt]

Hi, I want to find α(alpha)

[hushed ether]

in terms of s and a?

[wraith stratus]

[ ] is floor function ?

[potent cobalt]

it's a ()

[past musk]

is there some context

[potent cobalt]

s & mu is constant

[past musk]

hmm

i would just use alpha = 2 arcsin(sqrt(s/2a))

[potent cobalt]

thanks

[past musk]

I mean idk

is alpha supposed to be an acute angle

[potent cobalt]

idk s is a semi-perimeter a is a semi-major axis

[past musk]

bleh ellipses

in that case i have no idea

[potent cobalt]

i only need alpha so i can solve for a just need alpha

but i'm not good at trig

.close

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[lusty nimbus]

Hey

[lusty nimbus]

I need help with 5 and 6

[naive valley]

!status

[lone heart]

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

[lusty nimbus]

1

@naive valley

[lone eagle]

Idk if it's a thing that changes a lot based on what country your learn in, but the f function seems poorly defined in question 5a

[naive valley]

seems ok

[lone eagle]

Do you know how to evaluate with x = 1 in question 5?

[lusty nimbus]

Yes I guess

[naive valley]

well do you know for example what the pair (2,3) means?

[lusty nimbus]

X,y?

Vertex?

[naive valley]

well in particular it means f(2) = 3

[lusty nimbus]

I'm confused

[naive valley]

each pair is giving you an x value (the first number) and the corresponding f(x) value (the second number)

[lusty nimbus]

Alr

F(x)

Is basically y

[naive valley]

yes

[lusty nimbus]

So it's x and y

Right?

[naive valley]

yes, x and the corresponding y

[lusty nimbus]

Alr

Ima ask my teacher for now

Ok I made a chart

@naive valley

[naive valley]

yo

[lusty nimbus]

Yo

I made a graph

Now what?

Domain and range

There a number of the reals

[naive valley]

for 5(a)?

domain is the set of all possible x values

so what would that be?

[lusty nimbus]

Wdym?

Nvm I figured it out

[lone heart]

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[odd hound]

Topic: Loci & construction (mostly construction in this case)

Without using a protractor, how would I construct a 30 degree angle??
(Ps. In the image above I used a protractor which wouldn't be correct as I didn't use any construction lines)

[gray isle]

consider special triangles

[odd hound]

wdym?

[gray isle]

do you know how to construct an equilateral triangle?

[odd hound]

yes

[gray isle]

and do you know how to bisect an angle using a compass

[odd hound]

yes

[gray isle]

apply those two together

[odd hound]

ohh i see tyyy

.close

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[hexed garden]

I have a group of ML models with different average scores on some data. I have highlighted the model with the best score. How do I calculate if this model performance is "statistically significant" compared to the other models?

[hexed garden]

Okay, So i have little understanding on what to do here but I can discuss what i know.

I think I have to define a null hypothesis saying H0 NeuralNetwork h5 performance is not statistically significant. Then a H1 where some p value that occurs < 0.05 says NeuralNetwork h5 performance is statistically significant.

I am just missing the process of getting this p value

[lone heart]

@hexed garden Has your question been resolved?

[twin nimbus]

So the null hypothesis can be "all of these models do not differ in predictive power" while the alternative hypothesis can be "the best model is better at predicting based on the loss metric". What you do is find the mean and stdev of all of the models, then set up a normal distribution and find the difference between the sample mean and best model. Finally, you can use a z-score to find the p-value. You'll be looking for a value of around 2 stdev more than the mean to reach a p-value of 0.05. However, you have more than 20 models here, so even if you hit the 1/20 chance I wouldn't be too sure it's significant.

[hexed garden]

Thanks this is clear I am going to try to do this. If I have any further questions are you cool with leaving this chat open?

[twin nimbus]

I can't promise I'll be available, but if I am then I'm happy to help

[hexed garden]

Why does having 21 models mean I should not be too sure its significant?

[twin nimbus]

Your proposed p-value for significance is 0.05, right?

This value represents the probability that this outlier would have occurred by chance rather than for a good reason.

So that's a 1/20 chance

The fact that you have 21 models means you should expect to hit that 1/20 chance 1 - (19/20)^21 times

Which is about 66% of the time

I would suggest instead using a p-value of 0.0025 or so

Which will give you about a 5% chance to hit the lucky shot with 21 different models

[hexed garden]

gotcha gotcha thanks

[chrome merlin]

I need help with a question

can I ask the question here?

[hexed garden]

check this https://discord.com/channels/268882317391429632/488120190538743810

[chrome merlin]

I did and there isn't any channel showing up for me under it

[hexed garden]

@twin nimbus I have calculated the z score of my highest performing model as -0.6865. is the p value im after the area under the standard normal curve to the left of this z position?

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[alpine sable]

So let’s say I take k = 2 and n = 2

I have

My set will be {1,2,3,4}

The first thing I need to do is how many combinations can I make

1,2 3,4 1,3 2,4 1,4 2,3

Thats 6

Now I need to take these 6 elements and put them into groupings of 2, ie: {(a,b), (b,c)}

This is where I am stuck

I would say it’s 6 choose 2

But idk

Lmk if my approach is ok

[opaque narwhal]

I need help

[karmic saddle]

#❓how-to-get-help and don't just say you need help. Say the question

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

Here is my potential solution

Order is immaterial

Gotta devivide

By n!

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[timber cobalt]

Quest: find all values of a for which there are exactly 2 solutions.

[mystic swallow]

not necessarily abs x

[timber cobalt]

Here I decided that we get the solutions when the top part is 0 so we can remove the denominator.
I noticed the top part is a parabola and decided to find all values of a for which the discriminant is bigger than 0

Another parabola which gives us a values this time

Then I apply the |x|!=sqrt(a) to this range and get this answer. The area slightly left to 0 is where a is negative so I didn't exclude it

However the answer I was given is this and I'm told it is supposed to use circles in a way. However I don't see where I made any mistakes. Could you help please?

[timber cobalt]

?

[vapid drift]

to this part everything is correct

but the way you applied |x|!=sqrt(a) is just nonsensical

and to be easier for us il rewrite |x|!=sqrt(a) as x!=+-sqrt(a)

the way you should do it is, first: write x in terms of a (use quadratic fomula for x) and apply what you get to x!=+-sqrt(a)

[timber cobalt]

👀

[vapid drift]

(write expression you got in the first step instead of x)

and solve it

then you get all a's that cannot be in your final answer

[timber cobalt]

Can I ping you when I'm done with it?

[vapid drift]

yeah

[timber cobalt]

How do I solve this tho?

I applied this formula to that but am not sure it's the best way

[lone heart]

@timber cobalt Has your question been resolved?

[timber cobalt]

@vapid drift I solved it successfully! I still can't understand why I couldn't apply the restriction the first way tho

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[fervent ferry]

i have this problem

[fervent ferry]

and in the video he says f(t) is the integral of sq(t)

and he takes the integral and there's a constant term

so to find the constant term, he finds the a_0 term of the fourier series of f(t) I think, but i don't understand why he uses 1/2pi in front of the integral.

isn't that part always 1/pi ?

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[long ice]

Hey guys i have an easy one i think.

[long ice]

A spring with an unstretched length of 5 in expands from a length of 2 in to a length of 4 in. The work done on the spring is _________ in·lb .

so ive done that math and i get 4k but the answer key says -4k and im just not understanding how its negative

[lone heart]

@long ice Has your question been resolved?

[long ice]

nope

[hushed locust]

is the spring compressed or stretched? Did it get closer to or farther than its natural length?

[long ice]

closer

it starts compressed and ends slighty less compressed

[lone heart]

@long ice Has your question been resolved?

[harsh swallow]

is the question the work done by the spring or something else?

cuz you used work done on the spring not by

[long ice]

work done on the spring

that is how the question is worded

[harsh swallow]

afaik the work done by the spring is calculated using

$\int_{a}^{b} F_{s} dx = \int_{-3}^{-1} -k x dx = 4k$

[ocean seal]

Katharine

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[unkempt notch]

help!

[silent stratus]

!status

[lone heart]

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

[unkempt notch]

1

[silent stratus]

Firstly try and simplify the inside

[unkempt notch]

you cant

oh

you can

tan is sinx/cosx

[silent stratus]

Yh

[unkempt notch]

cos^2x(sinx)/cosx

[silent stratus]

Yep

[unkempt notch]

[silent stratus]

Cancellation

[unkempt notch]

cosx(sinx)

[silent stratus]

Not cos(sinx)

Do you mean cosxsinx

Yh kk

[unkempt notch]

yes

now what do i do

[silent stratus]

Now it's just easy product lu

[jagged cobalt]

or simplify it more

[unkempt notch]

how

[jagged cobalt]

double angle formula for sin(2x)=2sin(x)cos(x)

sin(x)cos(x)=1/2 sin(2x)

[silent stratus]

That works too

[unkempt notch]

yay

oh no

how do you find that then

😢

HELP!

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@unkempt notch Has your question been resolved?

[lone heart]

@unkempt notch Has your question been resolved?

[unkempt notch]

help

HELP!

PLEASE

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@unkempt notch Has your question been resolved?

[unkempt notch]

<@&286206848099549185>

[vast tapir]

where are you stuck at?

try to simplify the inside

then use the double angle formula

finally plug in the values

[thorn heron]

use the chain rule method

are use the multiplication rule

[unkempt notch]

how

sin(x)cos(x)=1/2 sin(2x)

but i don’t know how to find the derivative of that

i give up

😢💔

.close

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[runic prism]

I guess this form might easier for you to understand.
https://www.youtube.com/watch?v=17X5g9QArTc

[unkempt notch]

thank you!

[runic prism]

np! hope you enjoy study math.

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[alpine sable]

Any mistake?

Original figure:

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[mossy swift]

why can you model it as one trial (poisson) with mean 763301

.close

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[fallen estuary]

Can someone please help me with question 1 a

[fallen estuary]

i'm having trouble representing it as an equation

the stuff i've written is incorrect

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[fallen estuary]

<@&286206848099549185>

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@fallen estuary Has your question been resolved?

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@fallen estuary Has your question been resolved?

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[clear tulip]

where did this come from

video said they minused the "two equations"

not sure what it was tho

[crimson field]

What do you get if you subtract the top minus the bottom equation

[clear tulip]

a-ar^(k+1)

[crimson field]

= Sk - rSk

So solve for Sk and you get that

[clear tulip]

sorry, not really clicking yet, wdym by solve for Sk?

[barren portal]

here you have S_k and rS_k

say, what you get when you subtract them

[clear tulip]

a-ar^(k+1)

[barren portal]

ie, $S_k-rS_k= a-ar^{k+1}$ right?

[ocean seal]

.doc

[barren portal]

where S_k is the sum of first (k+1) terms.

[barren portal]

solve for S_k

[clear tulip]

just an additional question while i solve it, whats the thinking behind getting rSk?

[barren portal]

r S_k is just the sum of first (k+1) terms multiplied with r

you can always factor r from it

[clear tulip]

why do that though? is it like SOP for geometric series?

[barren portal]

SOP?

[crimson field]

It lets you do what you’re about to do

[clear tulip]

*standard operating procedure

[crimson field]

Kind of hard to see until you see it lol

[barren portal]

that’s the beauty behind geometric series,

you already have what you need

$\sum_{k=0}^{k} ar^k - r\sum_{k=0}^{k} ar^k= a-a^{k+1}$

[ocean seal]

.doc

[barren portal]

maybe doing this way shows what’s happening

[clear tulip]

hmmm

would it be proper to say that a series is the length of a line

no nvm

if the derivative is rate of change/"slope in a sense", then integral is area under a curve. what is the equivalent of series in a graph, or is it not possible to graph

perhaps that would make it easier for me to understand

[barren portal]

series in a graph?

you can represent an integral using a series

[clear tulip]

is it possible to visualise a series?

[clear tulip]

oh

how?

[barren portal]

i’ll try to answer this first

are you familiar with convergence of a sequence ?

[clear tulip]

yes, i just studied it earlier, however only the basics i would assume

[barren portal]

Say sequence {1/n}, it converges to 0 right?

[clear tulip]

yes, because of the limit

[barren portal]

visually you can think as the terms of 1/n in a number line getting arbitrary close to 0

something like this, not drawn upto marking

effectively it moves, (1,1/2,1/3,1/4,….)

[clear tulip]

as it approaches infinity, it tends closer to 0

[barren portal]

exactly

now, we use the notion of sequence to define convergence of a series

let me define the term a ‘partial sum’

[clear tulip]

hmmm so sequence is to series as derivative is to integral in a sense, not completely understanding though

[barren portal]

forget about integrals and derivatives for now

i’m just giving how to visualise series

[clear tulip]

would that be the S_k

[barren portal]

when i say S_10, i mean the sum of first 10 terms of the series

just add up the first 10 terms alr?

s_100, would be add up the first 100 terms

so, you get a sequence

(s1, s2, s3, s4, …, s101, …)

each term ‘in this sequence’ is really sum of series upto the number k

[barren portal]

visualise this as a sequence

where each term of the sequence is sum of the series upto some number

[clear tulip]

sequence is more on individual numbers, like numbers in a line. series is the sum of all those numbers?

[barren portal]

exactly

so, when we consider sum upto k terms we get individual numbers

and we want to know, where this limit of terms goes to

[clear tulip]

just to get it straight, series will always be exponentially greater than sequence, considering n,k e of N?

[barren portal]

i don’t get it sorry

[clear tulip]

what im seeing rn(althought may be wrong), a series will always be greater than a sequence because it is basically the sum of all the terms within the sequence?

[clear tulip]

wdym by limit of terms

[barren portal]

the partial sum of a corresponding series looks like (s1,s2,s3,s4,…) right?

s_100 is like sum upto 100 terms of the series

if this converges, it means if you add upto infinity it still converges

essentially, we say a series converges if it’s sequence of partial sum converges

[clear tulip]

sequence of partial sums

woah woah woah okay i kinda get it

[barren portal]

suppose you have sequence of partial sums like this (3,5,6,7.1,7,7,7,7,7,7…)

this mean, add upto be 100000th terms in your series you get 7 as the result

[clear tulip]

so, we are basically treating the series as a sort of sequence through partial sums, so like
given an=1/n
S1=1
S2=1+1/2
Sk=1+1/2+1/3+.......+1/k

[barren portal]

exactly

This S_k is really your sequence of partial sum

sum of the series upto kth term

if this converges, so does the series

[clear tulip]

oh okay so in my example, the one with Sk-rSk, we did that to find the sequence?

[barren portal]

sequence of partial sum

yes

[clear tulip]

would that be the same for most series, using partial sums so we can express the series as a sequence of partial sums?

[barren portal]

convergence of any series

this is how we define series convergence

there should be some formalising behind saying a series convergence right?

this is that formalism

$\sum_{k=0}^ {\infty} a_k$ converges if and only if $\sum_{k=0}^ {k} a_k$ converges

[clear tulip]

that is a constant series?

only convergent when a=0, otherwise it would diverge?

[ocean seal]

.doc

[barren portal]

I really mean a_k

$\sum_{k=0}^ {\infty} a_k= a_0+a_1 + a_2 + a_3 + \ldots +$