help-23

[gentle oxide]

yes

$$z = 2\cos(\theta) \pm 2i\sin(\theta)$$

[cerulean cloud]

our root here is

[flat frigate]

W*

[cerulean cloud]

2

what does that say then

x=1 and y=1??

[gentle oxide]

does this remind you of something

[cerulean cloud]

yeah 😭

but i still dont get

how to get y and x

[gentle oxide]

what are equations for x, y in the complex plane?

or well

what is the form of a complex number $z$ in rectangular?

[flat frigate]

W*

[gentle oxide]

it is $z = x + iy$, is it not?

[flat frigate]

W*

[cerulean cloud]

yes

it is

[gentle oxide]

so

[cerulean cloud]

in its rectangular form

that's

[gentle oxide]

$$z = x + iy$$
$$z = 2\cos(\theta) + i(2\sin(\theta))$$

[flat frigate]

W*

[gentle oxide]

any resemblance :3

[cerulean cloud]

√(x²+y²)= 2

[gentle oxide]

this is fine sure

then it is true that $x^2 + y^2 = 4$?

[flat frigate]

W*

[gentle oxide]

now what does this imply 🙂

[cerulean cloud]

right

hold on

[cerulean cloud]

x² is 2 and y² is also 2??

or it can be

other things??

[gentle oxide]

have you never seen the equation $x^2 + y^2 = r^2$?

[flat frigate]

W*

[gentle oxide]

,w graph x^2 + y^2 = 4

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by von_neumann_algebra

[cerulean cloud]

im not exactly familiar with these yet

no

[gentle oxide]

hm

that's odd

[cerulean cloud]

i still got me

questions

[gentle oxide]

from here

you should deduce $x = 2\cos(\theta)$ and $y = \pm 2\sin(\theta)$

[flat frigate]

W*

[gentle oxide]

in other words, $x^2 + y^2 = 4$

[cerulean cloud]

oh yeah that's true

[flat frigate]

W*

[gentle oxide]

is this satisfactory?

[gentle oxide]

this is the rectangular version of $z = 2\cos(\theta) \pm 2i\sin(\theta)$

[flat frigate]

W*

[cerulean cloud]

that's

it???

[gentle oxide]

yes

[cerulean cloud]

it's js x+yi? 😭

shouldn't we like find x and y from x²+y²=4

[gentle oxide]

no

x^2 + y^2 = 4 describe all complex solutions

[cerulean cloud]

oh rlly

you're saying it x²+y²=4 no matter what?

[gentle oxide]

yes

in the complex plane

[cerulean cloud]

do u have any sources on it i can read

[gentle oxide]

hmm not particularly

pauls online notes are good

[cerulean cloud]

ok thank you ❤️‍🩹

i highly appreciate it

ok actually

i still have 2 more questions 😭

would z be equal to 2e^(π/2)i?

wait no it should be

2e^(θi)

±θ

[gentle oxide]

yes

[cerulean cloud]

so does its rectangular form

change if we put θ as pi/3

[ionic wren]

Why?

[cerulean cloud]

idk

im asking

[ionic wren]

Do you like it better with me following you or without?

[gentle oxide]

i gtg now

[cerulean cloud]

Yeah works better

[ionic wren]

Is that a yes?

[cerulean cloud]

yes

[ionic wren]

You're not hurting me on purpose?

[solar hazel]

[median sand]

That is a confusing question.

[cerulean cloud]

no 😭

[cerulean cloud]

which one

[median sand]

Missan's reply.

[cerulean cloud]

ok anywho

[gentle oxide]

@cerulean cloud just an fyi, $z = 2e^{(i\theta)}$ defines a circle in the complex plane of radius $2$

[ionic wren]

Ok

[flat frigate]

W*

[gentle oxide]

this should be common knowledge if you are working like this with complex numbers, in conjunction with $x^2 + y^2 = r^2$ forming a graph of a circle centered at $(0, 0)$ with radius $r$

[flat frigate]

W*

[cerulean cloud]

ok thank u

u can close this now

[outer pollen]

you're the OP. only you can close it.

[cerulean cloud]

ohh my bad

!close

.close

[safe radish]

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Closed by @cerulean cloud

Use .reopen if this was a mistake.

[solar hazel]

well not quite true, green names can

but yes you should just close it yourself if you are done

[safe radish]

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[covert talon]

Im confused

How do i do this if theres no numbers given

is it like rearranging with the

Fraction surd thing

[still stone]

you don't have to assign individual numbers, but you can do it with variables

let's just say

AE = x

so EF = 2x

now find the sides of the right angled triangle AJM and you can use whatever you get from obtaining the tangent

[covert talon]

o

Oh ok thank u

.close

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[copper wraith]

Integration of 1/(cos(x-a)(cos(x-b))

[nimble wyvern]

multiply and divide by sin(a-b)

write sin(a-b) in the numerator as sin((x-b)-(x-a)) while taking out the sin(a-b) in the denominator outside the integral

the rest is just applying sin(A+B) and we're done

[median sand]

Perhaps help the OP step by step, not showing all the works.

[copper wraith]

I got the hint thanks

Is there any other method?

[nimble wyvern]

yes but this is the cleanest one

[nimble wyvern]

good idea but this is one of the integrals that will just trap beginners unless guided

[median sand]

Could try substitution using tangent function.

[median sand]

You are basically telling all the steps anyways...

Spoiling the fun.

[nimble wyvern]

🙏

[copper wraith]

Surely a=\b

[nimble wyvern]

yea

[safe radish]

@copper wraith Has your question been resolved?

[copper wraith]

But if at the start a=b

1/cos(x-a)^2

@nimble wyvern

[nimble wyvern]

then it just becomes sec^2(x-a)

[copper wraith]

tan(x-a)

[nimble wyvern]

yep

[copper wraith]

Yeah

Why am I getting different answers?

i meant one side we get 0 and other side tan(x-a)

[nimble wyvern]

???

[nimble wyvern]

show your working

[copper wraith]

the above integration where

I don't know who deleted my post

A=b we get 0

[nimble wyvern]

uh oh 1/sin(a-b) is multiplied with the log term not added

rest is great 👍

actually if you observe carefully, if we put a=b in the final result, we get an indeterminate (0/0) form

[copper wraith]

I see

Thanks

Tell me one more thing how did you get idea of multiply by 1/sin(a-b)

@nimble wyvern

[tawdry plover]

it might be a backwards way of explaining it, but the problematic issue at hand is the fact that you have mutliple cos terms in the denominator. the only way to produce such terms is if you use angle addition rules of either sin or cos

now the brute force way you can do is perhaps through the weierstrauss substitutions (tan half angle), but that may produce some annoying algebra

i invite you to try both cos(b- a) and sin(b - a) to see why we take the sin option over the cos option

you will see in general that a lot of integration tricks rely on adding/subtracting 0s (i.e. a - a) or multiplying by 1 with a top and bottom term

[safe radish]

@copper wraith Has your question been resolved?

[safe radish]

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[valid jacinth]

ik my part b is incomplete i just found the angle not the area

but

why is my approach to part c wrong?

all my coordinates r iincorrect

im new to vectors so idk

[vague phoenix]

You don't do the cross product for b?

[valid jacinth]

wdym

which vector are you referring to as b

[vague phoenix]

b

[valid jacinth]

oh do u mean part b

[vague phoenix]

Ah, are you using the formula $\frac{1}{2} |\vec{u}| |\vec{v}| \sin{\theta}$?

[flat frigate]

❤ Neferia ❤

[valid jacinth]

yes

its just incomplete but i understand part b

im just lost on c

[fathom jewel]

c seems right from the idea (your sketch)

[vague phoenix]

Yeah, if D = (x,y,z) then you can have combinations like ACBD, ABDC or ABCD

[valid jacinth]

yes, the solutions do something using the Origin tho

which is why im confused

[vague phoenix]

But lookin' good so far

[valid jacinth]

i dont understand why my approach gives a completely different answer

because my sketch makes sense to me

[median sand]

In addition to Nefer, for each cases, suppose ABCD, AB = DC, then B-A = C-D.

Similarly to 2 other cases, you can quickly work out the coordinates (x,y,z) of D

[vague phoenix]

Fair point

[fathom jewel]

You need a reference point (like the origin) to make sense of when you add vectors where you will end up.

For example in order to get D1 you'd need OA+AB+AC

[fathom jewel]

Your sketch is correct, your computations are wrong

[valid jacinth]

ohhh

im having the same problem with another q

[median sand]

Ask away!

[valid jacinth]

because i havent been using the origin as like a reference poiint

im not too sure how tp

one second ill dhow the q

part b finding the centre

[valid jacinth]

pink circle sorry the black dashed one is a mistake

[median sand]

So I supposed you got A done.

[valid jacinth]

yes

my answers to A were correct

.rotate

[median sand]

Have you found the midpoint?

,rcw

[median sand]

You got the coordinates of P and R already, right?

[valid jacinth]

i got 1/2(-1,4,-11) but its incorrect bc i didnt use the origin

[valid jacinth]

yes yes

[median sand]

That's the center.

[valid jacinth]

yup i got that but its wrong because i didnt use the origin

i dont undertsand how to use the origin with it

i understand the rest

[median sand]

If coordinates $P (x_1,y_1,z_1)$ and $R(x_2,y_2,z_2)$, then center is: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$

[flat frigate]

Mercury.

[valid jacinth]

yes that makes sense i used that approach

[vague phoenix]

Yes, what did you get?

[valid jacinth]

but the solutions say i have to use the origin as a reference poiint which i dotn get

[valid jacinth]

what i wrote at the bottom

thats the coordinates for the centre

[median sand]

The hypotenuse is the side opposite the right angle right? which is the segment PR.

[fathom jewel]

You include it implictly since you can write PQ+QR=(OQ-OP)+(OR-OQ)

[valid jacinth]

i dont understand what implicitly is sorry

[valid jacinth]

yes

[median sand]

The center of the circle is the midpoint of the hypotenuse PR, right?

[fathom jewel]

implicit means indirectly or hidden in your notation

[fathom jewel]

The only issue is when you found your direction another example:

AC+AB then that still needs some "anchor point" and so naturally you'd choose OA (working with the origin is easiest)

[valid jacinth]

yes i get that

[median sand]

Then the center is the formula i just pointed out

[valid jacinth]

i completely get why its the midpoint the centre of the circle

im just trying to understand the origin anchor poiint part

i see why an ‘anchor point’ is needed im just confused on how to use it

[median sand]

Anchor point?

[valid jacinth]

^^

[median sand]

Maybe anchor point is the connection between origin and the point?

[fathom jewel]

You are just saying from point A I will go along first AB then AC to get to D2

because AB+AC is just a vector in space, a simple direction and without applying to some point "meaningless"

[valid jacinth]

yes thats what i did but the O bit is ocnfusing me

yes

ohh

OH

IT JUST CLICKED THAT MAKES SENSE

THANK U

[median sand]

Haha.

[safe radish]

@valid jacinth Has your question been resolved?

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[fathom jewel]

.Are you sure this is correct?

[valid jacinth]

my answer?

[fathom jewel]

the C yes

I suppose you'd want to find the coordinates of C, right?

[safe radish]

@valid jacinth Has your question been resolved?

[valid jacinth]

yes, its wrong bc i didnt use the origin

[fathom jewel]

Hmm

Well PR/2 is only displacement, like you realized by now

So you'd need to add OP for example to get to OC

[fathom jewel]

When trying to find the coordinates of a point which is just a location vector then it helps to make a sketch to find out what you need

PR/2 would just represent the green vector but as said before wouldnt mean much since vectors are translation invariant

[safe radish]

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[crisp remnant]

I would do the 3rd question just tell me what is the 1st and 2nd

[agile scaffold]

do you know how to do them, or do you just want the answers?

well, because

!noans

[safe radish]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[crisp remnant]

I don't know a thing

[karmic merlin]

Which means you don't know what it means by square root spiral?

[crisp remnant]

I was absent

[karmic merlin]

Oh no I was just asking yes or no I'm not judging

[agile scaffold]

a square root spiral is constructed by multiple repeated right angle triangles

[crisp remnant]

can you send a image?

[agile scaffold]

you start out by using an isoceles right angle triangle of base 1 and height 1, the hypoteneuse would be sqrt2

you then use sqrt2 as the new base, draw a perpendicular line on the edge with side length 1, and draw another hypoteneuse

[karmic merlin]

Would showing them the image counts as answer cause honestly it would be somewhat hard to visualize what that means

[normal moss]

You can literally Google "Square Root Spiral"

[crisp remnant]

and continue?

[agile scaffold]

im typing...

by pythagorean theorem, this new hypoteneuse would have a length of sqrt 3

[crisp remnant]

okay what about 2nd?

[agile scaffold]

you then use that as a base, draw a perpendicular line on the edge with side length 1, and draw another hypoteneuse, so on

[crisp remnant]

thanks

[agile scaffold]

im not really sure how to help you on this like

you represent the number system in terms of a flow chart

[crisp remnant]

wasn't able to understand in google so I asked here

[agile scaffold]

perhaps something like real numbers are made of rational numbers and irrational numbers so you draw an arrow from real numbers to the other two and mark the arrow with like "consists of"

or something

the usual representation is with sets

i.e.

[crisp remnant]

okay

[karmic merlin]

Ye like that but in flow chart form

[crisp remnant]

ok thanks both of you

.close

[safe radish]

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[safe radish]

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[south wyvern]

hi

[south wyvern]

im new to algebra

[golden nebula]

Do you have a specific question?

[south wyvern]

yup

how can i work out formulas

[terse sand]

Could you provide with an example?

[south wyvern]

here

wait

how many types are there

like (2x3)+36= and 64+362+22+

:[ hello?

bruh

.close

[safe radish]

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[azure harbor]

specify types

wdym by types

[south wyvern]

like multiplying in bracket and etc

[median sand]

Specific questions please?

[south wyvern]

.reopened

[azure harbor]

its .reopen

[south wyvern]

here again

[loud wren]

skul

[azure harbor]

uhh please type .reopen

[loud wren]

.reopen

[safe radish]

✅ Original question: #help-23 message

[south wyvern]

.reopen

[azure harbor]

sigh i was blacklisted from helpful

[south wyvern]

bruh can smth or someone answer me

[main mural]

bro is unhelpful :(

[south wyvern]

k bruh

.close

[safe radish]

Channel closed

Closed by @south wyvern

Use .reopen if this was a mistake.

[azure harbor]

if you browse meta discussy lance lance mentioned that mod tickets can blacklist you from helpful

...

[main mural]

that's crazy lmao

[azure harbor]

oh yeah i have 6 mutes

[azure harbor]

sorry your question was too vague

can you please specify exactly what you are struggling with

[south wyvern]

formulas

[azure harbor]

which one

[south wyvern]

like here is a example

(2x6)+13=

idk if this was a formula so pls dont timeout me

[azure harbor]

uhh this is simple arithmetic

use pemdas

[south wyvern]

oh

how to use it

like multiplying in brackets

[south wyvern]

thx for the name

now what to do

[azure harbor]

do the thing in brackets first

also i suggest you to type .reopen

[south wyvern]

.reopen

[safe radish]

✅ Original question: #help-23 message

[south wyvern]

ok

2x6 is 12

12+13 is 25

[dense furnace]

could you screenshot the problem if you could, i’m a little confused by what exactly you’re asking

[azure harbor]

good.

[south wyvern]

nah i already learned it

and im on my pc

k im done thx

.close

[safe radish]

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[obsidian hound]

yup you got it. just practice some order of operations equations and it should stick.

[vague phoenix]

Cooked.

I got plenty of mod logs too but still green...

[azure harbor]

i think u got green before any mutes

[vague phoenix]

Don't have mutes but accidentally say some slurs

[glacial meadow]

Wild message to read upon opening a help channel

[azure harbor]

average closed help channel convo between helpers

[safe radish]

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[brazen nimbus]

hello can any1 help with the 2.2

[brazen nimbus]

so

[dry ether]

Can anyone help with task 1-2-3?

[brazen nimbus]

i already made a tree diagram

[nimble wyvern]

!occupied

[safe radish]

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).

[median sand]

This channel is occupied, sorry. Please send in other channels.

[nimble wyvern]

-# sniped

[median sand]

Ah, I didn't see.

Do you want to take over, I can step back.

[nimble wyvern]

no

[brazen nimbus]

oh mymy god

myfault i couldnt read

.close

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[median sand]

<@&268886789983436800>

Scam bots.

[safe radish]

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[edgy crest]

how am i even supposed to start i have to figure out what x y and z is

[queen saddle]

that blue thing is a rhombus, with an acute angle equal 60. So x=30.

[edgy crest]

wait what

how do u know that

[queen saddle]

almost the same way I recognize my relative's faces 🙂

I just know )

[edgy crest]

is there like a theorem or formula?

[queen saddle]

ok, first prove that the blue obtuse angle at M is 120

[agile scaffold]

boop

[edgy crest]

how do u do that

well alr

but how do u know its its an equatential triangle or wtv its called

i do maths in sweish

swedish

[queen saddle]

because they are two radii

[edgy crest]

ik that but unless they gave us the info then that triangle to the left should be an isosceles

[agile scaffold]

well, how the diagram is presented makes me inclined to believe that the blue shape is a rhombus

[queen saddle]

I suppose all blue segments are equal?

[edgy crest]

they didnt give me that info

[queen saddle]

then there is no unique answer to your questions

[edgy crest]

wait a rombus all sides are equal?

[queen saddle]

yep

[edgy crest]

aha then yes ur correct

they say its a rombus but i didnt know that was important

well alright its 120 and x is 30 i think its easy from here on

hol on imma try to do it really fast

[queen saddle]

You might have skipped some more info, since with your diagram there is no unique answer for y and z

I can only vaguely guess that x=z as they are both red

though they are not drawn equal

[edgy crest]

nah i got it right x=30 z=15 and y=75

[queen saddle]

oh, so black sides of the triangle mean it is isosceles

[edgy crest]

yes

dang i skipped that too mb

thx alot

might come back later

[agile scaffold]

usually everything in a question is put there for a reason!

[edgy crest]

yea but i forgot to mention those things

since theyre in swedish and it would be a pain for yall to translate that

[safe radish]

@edgy crest Has your question been resolved?

[edgy crest]

anyone available?

[errant ravine]

!status

[safe radish]

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

[errant ravine]

also what does the q want?

[edgy crest]

2

[edgy crest]

wym

[errant ravine]

what is the questions asking?

[errant ravine]

okay what have u done so far?

[edgy crest]

prove that d is equal that last bit

[errant ravine]

-# sorry i am confused what is "last bit"?

[edgy crest]

d= square root of 20 -2

[errant ravine]

OKAY! so bassicly we need to cauluate D and prove its equal to that!

okay soso what have u done so far?

[edgy crest]

yes

uhhh i know the raidus is 2 lol

[errant ravine]

OKAY OAKY

sooo lemme show u something cool!

[edgy crest]

alr

[errant ravine]

what do u think of this line?

[edgy crest]

it has a length of two

[errant ravine]

hmm how about this?

what shape is abc?

[errant ravine]

true!

[errant ravine]

do you see it?

hej?

-# @edgy crest if u need a hint lmk!

[edgy crest]

mb i didnt get any notifications

[errant ravine]

all good!! so what shape is abc?

[edgy crest]

aha ye

triangle

[errant ravine]

YESS

SO QUESTION DO YOU KNOW HOW WE COULD POSSIBLE GET BC

[edgy crest]

pythagoras

[errant ravine]

YESS! and once we get BC we can do what? to get D

[edgy crest]

omg it was that easy

[errant ravine]

YEAH!!

see you figured it out!!

[edgy crest]

alt thanks alot

alr

[errant ravine]

np! glad i could help!!

-# also dont forget to do .close when ur done!!

[edgy crest]

alr

thx

.close

[safe radish]

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[safe radish]

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[soft plover]

How can I visualize the total derivative?

[soft plover]

the partial derivative dy/dx1 is easy, you just treat x2, x3, etc. as constants, and 'nudge' x3, and the resulting 'nudge factor' in y is the derivative

[crude bear]

What is total derivative

@soft plover

[summer coral]

are you talking about the derivative of some function z = f(x(t), y(t))?

if so, $\frac{\dd z}{\dd t} = \frac{\partial f}{\partial x} \frac{\dd x}{\dd t} + \frac{\partial f}{\partial y} \frac{\dd y}{\dd t}$

[flat frigate]

Krish

[soft plover]

https://en.wikipedia.org/wiki/Total_derivative

[soft plover]

it's y = f(x1, x2, x3, ...)

[crude bear]

So the total derivative is the gradient

The total derivative is the linear map that aproximates f most closely around a point

[soft plover]

is a gradient not a vector?

[crude bear]

Yes

Linear maps from R^n -> R are dot producs

$<\grad f, v>$

[flat frigate]

tales

[soft plover]

with the 'offsetting' handled how?

[crude bear]

I don't know what do you mean but we have

$f(a+h) \approx f(a) + \langle \grad f(a), h\rangle$

[flat frigate]

tales

[soft plover]

ahh, ok

thanks!

[crude bear]

There is thsi drawing (in more dimensions)

[soft plover]

wait, so I know the gradient of f(x1, x2, x3) at every point.
how do I get f'(a, b, c), the total derivative?

[crude bear]

Can you tell me what in this page you call total derivative

[broken pewter]

Might be worth mentioning at some point thinking of a visual for math just doesn't work

Or at least doesn't help

[crude bear]

Yeah, it's hard to draw functions R^n -> R

R^n -> R^m is easier

[soft plover]

R^2 is sufficient

[soft plover]

wait mb, this is what I want

from https://en.wikipedia.org/wiki/Differential_of_a_function#Differentials_in_several_variables

[crude bear]

so you have f(u(x), v(x))

[soft plover]

yeah

[crude bear]

So (u(x), v(x)) is a curve on the place. The total derivative is the derivative of f aong that curve

My man @summer coral probably remembers calc 3 more

[safe radish]

@soft plover Has your question been resolved?

[safe radish]

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[hoary knot]

How’d you do a

[open wedge]

let's see

notice that you can't just divide $\sin x$ freely

[flat frigate]

1 divided by 0 equals Infinity

[hoary knot]

Haven’t u js done it there tho

[open wedge]

well can you divide by 0?

[hoary knot]

No

[open wedge]

that's the problem

$\sin x$ could be 0

[flat frigate]

1 divided by 0 equals Infinity

[hoary knot]

Ah

Should I change cos into smt else

[open wedge]

uhhh idk

i don't think so

[open wedge]

for this one you consider 2 cases: one is if $\sin x = 0$, and the other is what they did

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

and for one other mistake

[hoary knot]

Oh wait I’m talking about the part about it

[open wedge]

try to use this condition

[hoary knot]

What’s wrong w it

Uh

[open wedge]

i took this one from the question

uhm here's the work

[hoary knot]

I don’t see the mistake🥲

[open wedge]

take a look when the student does $\cos x = \frac{3}{5}$ to $x = 53.1^\circ$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

adding in with this condition

[hoary knot]

Does it not go in the range?

[open wedge]

hm

close

but like

there's still missing something

[hoary knot]

Could there be other values

[open wedge]

when $-90 < x < 90$ then the angle has to be in the first or the 4th quadrant

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

yea

[hoary knot]

Is that the mistake

[open wedge]

ya

[hoary knot]

Ohhh thank you

.close

[safe radish]

Channel closed

Closed by @hoary knot

Use .reopen if this was a mistake.

[open wedge]

np

[upper forge]

today i got to know that cos53.1 is 3/5 not exact 53

[safe radish]

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[tranquil geyser]

Okay I realize i am really bad at this so it would be nice if someone can go through a few problems with me :)

[tranquil geyser]

Let $I$ be the set of irrational numbers, show $|\mathbb{R}| \leq |I|$

[flat frigate]

toast

[tranquil geyser]

so i would need to find an injection f: R to I

[upbeat swan]

Well first of all I is included in R

So now you just have to show that I is not less than R

Which can be done simply by proving that both I and R are uncountable

Or another way would be to find an injection as you proposed

[tranquil geyser]

how does both being uncountable mean they have the same cardinality?

[tranquil geyser]

ye and im stuck on this

[upbeat swan]

Yes you are right even if they are both uncountable they dont necessarily have to have the same cardinality

For example P(R) > R

So yes you would have to find an injection

[tranquil geyser]

ye

and what is the best approach for finding an injection?

<@&286206848099549185>

[safe radish]

@tranquil geyser Has your question been resolved?

[tranquil geyser]

nvm

solved it

.solved

[safe radish]

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[karmic plover]

!help

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[safe radish]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

@karmic plover Has your question been resolved?

[median sand]

Do you have a question?

Else you can close the channel.

[safe radish]

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[jade sundial]

help

[safe radish]

Please don't occupy multiple help channels.

[jade sundial]

ok

[outer pollen]

you already have #help-6!

[jade sundial]

i though it was closewd

.closed

[brave wolf]

you can use this one

[outer pollen]

you closed #help-14.

[brave wolf]

just ask your question already (this one isnt closed)

[austere goblet]

I will close #help-6 for you. stick to this channel

[jade sundial]

ok

[dusk halo]

@jade sundial , can you see all of the help channels, it feels like you cant see them so you don't know which ones are open

[austere goblet]

now please ask your question

[jade sundial]

what is Sine?

[azure harbor]

This

[jade sundial]

But why the sine,cosine and tangent have to stick with the angle?

[azure harbor]

Cuz its a function of the angle

[dusk halo]

They are based on the unit circle

[azure harbor]

Look at the image included

[azure harbor]

Well yes

[jade sundial]

It seems like unit circle is very important,it's like the key.?

[jade sundial]

what is it

[azure harbor]

The circle with radius 1, centered at the origin

[dusk halo]

So if you have a circle with radius 1 then pick a point on it, sin is the y coordinate of the point and cos is the x coordinate

[azure harbor]

Used to define sin,cos,tan and a whole lot of trig functions

[jade sundial]

Is there a formula for it?

[azure harbor]

... For what exactly?

[dusk halo]

x^2 + y^2 = 1 is the unit circle formula

[jade sundial]

So,actually,the sine is this formula.?

[azure harbor]

Nope

Its for the unit circle

[vague phoenix]

Comprehensively, it is the ratio between opposite side and hypotenuse.

[dusk halo]

I think we are overcomplicating things a bit, if you just need to find angles dont worry about the unit circle

[azure harbor]

I suggest reading this to understand what sine is
https://www.mathsisfun.com/algebra/sohcahtoa.html

[jade sundial]

So,what is sine without theta?

[azure harbor]

Its a function

sin(theta) is a function which takes the angle theta and outputs its sine

[upbeat swan]

again, lets not worry about the unit circle for now

[azure harbor]

I would also suggest reading about what a function is

[jade sundial]

So,what will sine do?what will this function do?is what does it calculate?I mean how how it calculates.

[upbeat swan]

it calculates opposite side / jupoyhenuse

in a triangle]

[azure harbor]

As i said read the article i sent

It includes what sine does

[jade sundial]

ok,i read frist

first

i finish

the

article

[upbeat swan]

[jade sundial]

In that example,I calculate the ratio.sin 30 degree equals to 0.5,which is Adjacent side divided by hypotenuse side,but and then I find out if I press the button sine on the calculator,and then I input the degree which is 30,thenand then I will get the ratio 1 over 2,right?but why I press sine 30,I will get this ratio?

[dusk halo]

1/2 = 0.5

its the same thing, but expressed differently

[jade sundial]

When I key in \sin 30^\circ on the calculator, it looks for the fixed ratio of the opposite side to the hypotenuse.

For any right-angled triangle with a 30^\circ angle, the ratio of its opposite side to its hypotenuse is always fixed at 1:2.

That’s why the calculator can directly show the ratio — the ratio never changes for the same angle.

is it correct

？

[upbeat swan]

yes

[jade sundial]

ohhhhhhhhhh,ok finally

.closed

how to closed a forum?

[upbeat swan]

this is a table if you want to memorize the values

type .solved

[jade sundial]

.solved

[safe radish]

Channel closed

Closed by @jade sundial

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[safe radish]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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• Be polite and have a nice day!

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

[tame atlas]

How can I solve this?

[stoic dune]

Take the left side, simplify

Take the right side, simplify

Show they are the same

Seeing as the right side has sin and cos, but the left side doesn't, you're going to need some trig identities

[safe radish]

@tame atlas Has your question been resolved?

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[sinful wyvern]

guys. is that variance/6 an application of Central Limit Theorem?

[sinful wyvern]

i'm so confused abt this CLT.

as long as number of sample is large, CLT applies. BUT if population has been distributed normally, CLT does not apply, regardless of the sample size?

is this correct

[stoic dune]

Yes. We need to know the variance of the sample mean, which the CLT gives us

[sinful wyvern]

but the sample is small, and it has been normally distributed, why do we still apply CLT?

[stoic dune]

CLT even applies if the population is distributed normally. The members can have any distribution they want as long as they're identical. That's what makes the CLT so powerful

[stoic dune]

However you're correct the sample is indeed small, and that can make the result unreliable. They're still using it anyways.

[sinful wyvern]

oh so even if the population is already normal, we still apply CLT?

why tho

isn't CLT like approximating a population distribution into normal?

why do we approximate a population that has been normally distributed into normal distribution?

[stoic dune]

The sample mean (that is, the average of the sample you've collected) is a random variable. It's random, because it depends on the sample you've collected.

Let me ask you this: what's the distribution of the sample mean?

Imo, that's a very non-obvious question

.
Like, you've got a sample of 6 pears and take their average weight. Do that over and over again with a different selection of 6 pears every time. Can you predict what that average is going to be, knowing what the distribution of the pears is? Can we give our average its own distribution?

Let me know if anything I'm saying is weird

[stone shale]

Help me

Help-10

[safe radish]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[median sand]

Just wait.

[sinful wyvern]

no we cant give their own distribution

but samples follow the distribution of the population?

[stoic dune]

Sure. But does the average of the 6?

That's what the CLT tells us. The sample mean is normally distributed. This is true independently of the actual population distribution.

Precisely the CLT tells us the average of the 6 pears has

• a normal distribution
• The mean is 45g
• The SD is √[52/6] g

[stoic dune]

So I agree with you it doesn't seem like we can. But we always can! CLT is insane imo

[sinful wyvern]

ohhh

uhhhhh

so CLT gives some form of confirmation...? 😭

gang i rly dont get this, it says one thing in the TB but it does the other thing in practice questions

what shld i follow for my exam

[stoic dune]

CLT gives us a distribution for the sample mean.

Make sure you know what the "sample mean" is, and why getting a distribution for it is a seemingly difficult task

[honest perch]

no, clt works for any distribution

as long as it has finite expectation and variance

[sinful wyvern]

wait why is it difficult?

i mean it seems understandable but i need to ensure i get the right understanding

[sinful wyvern]

ohh

[stoic dune]

How are you doing it? Genuinely, what's your strategy for getting the distribution of the sample mean, without CLT?

[sinful wyvern]

but why will it work for normal distribution? population that is already normally distributed do not need to be approximated to normal again, no?

[honest perch]

yeah, so then using clt is redundant. and your approximation is actually exact in this case

[sinful wyvern]

i mean it's usually given in the question 😭 im not sure what i would do if i were to take the sample myself

[sinful wyvern]

oh yes sometimes they would ask if CLT is necessary in the follow up question

guys do yall study this in university

[honest perch]

many people do

[sinful wyvern]

i think my understanding is not enough (in high school), i need this to be rereviewed later in uni

[honest perch]

yeah this is definitely beyond hs material so should be covered in uni if you take the right course

[sinful wyvern]

oh okay thats nice

[stoic dune]

Absolutely not. I expect any hser to understand this

[sinful wyvern]

oh wait that might not be nice

it says differently in the material im studying in hs rn

[honest perch]

i mean sure, but most people don't study this in hs

also like realistically, most ppl would struggle to understand this

[sinful wyvern]

guys how do yall eventually understand this topic

😭

[honest perch]

everything made more sense after i did measure theory

but that comes quite a bit later, and is pure mathy

[sinful wyvern]

wait so okay CLT can apply at any distribution, the problem is just whether or not it is necessary.

for population that has been normally distributed, CLT can still apply, it's just that it's not necessary.

now what if for small samples? can CLT still apply? bcs in my hs material, it says CLT only applies for large sample

[honest perch]

so clt gives u an approximation. it's not exact. do you know about limits?

it's just that the approximation is better the larger n is

[stoic dune]

What do you think the CLT is giving a distribution to?

[sinful wyvern]

what limits

[honest perch]

as n approaches infinity, you get the exact answer

[sinful wyvern]

yess

[honest perch]

ok it's something u might see in calculus

[sinful wyvern]

it's approximating to normal distribution

i saw the thing in my textbook where as n gets larger, the distribution becomes more normally distributed

like the normal curve becomes more normally distributed

[honest perch]

yes, this is what clt is saying

[honest perch]

not really

normal is normal

[sinful wyvern]

what about the already existing normal distribution? isn't it already in the perfect normal shape

[honest perch]

yes

[sinful wyvern]

approximation makes it even more perfect?

[stoic dune]

Like, no. Absolutely wrong. Not even close. Remove that.

The CLT gives the distribution for the sample mean.

[honest perch]

^ @sinful wyvern yes

[stoic dune]

I think the problem is that you've got a picture for CLT in your head, and rejecting other pictures.

[honest perch]

"more perfect"? no. normal is normal.

[stoic dune]

Take time to read what the "sample mean" is. I did detail it up here as well

[sinful wyvern]

how are we supposed to picture it

[stoic dune]

CLT gives a distribution for the sample mean

[sinful wyvern]

so approximating an already normal distribution makes the distribution more exact? (not perfect)

[honest perch]

idk what this means

ok suppose X_1, ..., X_n are iid rvs

[stoic dune]

I think you think that "CLT gives a distribution for the sample mean" is like saying "CLT approximates other distributions" and that's why you're ignoring me lol.

They're not saying the same thing.

[honest perch]

then Xbar = (X_1 + ... + X_n)/n is a rv

u wanna find the dist of Xbar

CLT gives u an approximate dist of Xbar

if each X_i is normal, then the dist that CLT gives is exactly the dist of Xbar

[sinful wyvern]

yes. what does random variable mean? my only understanding of 'random' is 'unbiased' but i think it's not that definition based on this context

[honest perch]

it's the thing before ~

like Y ~ N(0,1)

[stoic dune]

Oh, yeah we can talk about that. A random variable is a variable that can take random values.

For example "the result of a dice roll" is a random variable. Can take values from 1 to 6

[sinful wyvern]

ohhhhh thats literal like CLT rlly be the one giving the distribution?

[honest perch]

the sample mean has only 1 distribution

[stoic dune]

Random variables get distributions, which tell us specifically "how random it is".

Dice pick the values between 1-6 "uniformly". Dice give a uniform distribution.

[sinful wyvern]

so the Y is the random variable?

[honest perch]

yes

[sinful wyvern]

so it can only be normal if everything is normal? (i think i get this wrong)

[honest perch]

idk that's not too important

if i understand u correctly

[sinful wyvern]

ooooo yes

[sinful wyvern]

yes...

[honest perch]

ok look

just remember this

every random variable has exactly 1 distribution

the sample mean is a random variable

[sinful wyvern]

yess i agree

i get it i mean

[sinful wyvern]

yes

OH so is it CLT the one who gives or decide that 1 distribution

i think im getting this...

[honest perch]

yes

[sinful wyvern]

OOOOOO

so it's like

each thing inside Xbar might have different distribution

but since they are one random variable, they need to have only one distribution

[honest perch]

whatdatmean

[sinful wyvern]

CLT decides and gives that one distribution

[sinful wyvern]

i mean in samples there are alot of datas

[honest perch]

what is a "thing inside Xbar"

[sinful wyvern]

😭

the samples i guess...

[honest perch]

please use mathematical notation

[sinful wyvern]

i dont know any notation related to this CLT 😭

[honest perch]

what are the samples

[sinful wyvern]

uhhhh the proportion of population...

[honest perch]

https://tenor.com/view/stop-jk-simmons-fletcher-whiplash-cut-it-gif-18304600

[sinful wyvern]

representative of the population?

😭

[honest perch]

that's more my tempo

[whole hill]

Not my f'n tempo

[sinful wyvern]

what is that

i dont get whatever that is

[honest perch]

meme from a movie (whiplash)

[whole hill]

It's from the movie Whiplash

[sinful wyvern]

ok guys is it safe to assume that whenever i see the word sample, i js apply CLT

😭

[honest perch]

no

sample mean? possibly

[sinful wyvern]

yes sample mean

[honest perch]

you didn't say that

[sinful wyvern]

bcs CLT isnt restricted by sample size or population distribution right

[sinful wyvern]

yeah i didnt know there are other samples

[honest perch]

yes

sample size = n

[sinful wyvern]

yes

[honest perch]

each sample is a random variable

[sinful wyvern]

i genuinely thought it can only apply if sample is large and population distribution is not normal

the textbook makes it seem that way omdays

[sinful wyvern]

yes

[honest perch]

name them

[sinful wyvern]

name the sample...?

[honest perch]

actually idk what notation ur book is using

[honest perch]

yes

[sinful wyvern]

this thing

[sinful wyvern]

they have names?

what kind of names

[honest perch]

yeah whatever

ur book doesn't do that

gtg

[sinful wyvern]

anyways

[sinful wyvern]

thank you sooo much

[sinful wyvern]

heey thank you sooo muchh

thanks alot guys

my exam is tmrw

[honest perch]

welcome

good luck!

[sinful wyvern]

thankuuuuu !!!

.close

[safe radish]

Channel closed

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[tranquil geyser]

Prove that there is a proper class of ordinals $\alpha$ such that $\alpha = \aleph_\alpha = \beth_\alpha$. [Possible hint: Given any ordinal $\beta$, form a set ${a_n: n \in \omega}$ by $\alpha_0 = \beta$, $a_{n+1} = \beth_{\alpha_n}$ and look at $\alpha^*=\bigcup {\alpha_n : n\in \omega}$

[flat frigate]

toast

[tranquil geyser]

Hi, how do i approach this problem 😭

after defining these sets

[safe radish]

@tranquil geyser Has your question been resolved?

[tawdry plover]

show that its a fixed point of $\beth$

[flat frigate]

blanketism

[tranquil geyser]

what does fixed point mean?

For any ordinal $\beta$, we form sequences by induction., where $\alpha_0 = \beta$ and $\alpha_{n+1} = \beth_{\alpha_n}$. Now let $\alpha^* = \bigcup{\alpha_n : n \in \omega}$ \\
I first want to claim that $\alpha^* = \beth_{\alpha^}$. Observe by definition, since $\alpha^$ is an limit ordinal, $\beth_{\alpha^} = \bigcup {\beth_\gamma : \gamma < \alpha^}$. Pick any element $x \in \alpha^$. By definition, there exists, some $k < \omega$ such that $x \in a_k$. We also know that $\alpha_{k+1} < \beth_{a_k}$, so $a_k <\beth_{a_k}$ since our sequence is increasing. So $x \in \beth_{a_k}$ for some $k < \alpha^$, so $x \in \bigcup{\beth_\gamma : \gamma < \alpha^}$. Therefore, $\alpha^ \subseteq \beth_{\alpha^}$. Now, pick $x \in \beth_{\alpha^}$. Then there exists some $\gamma <\alpha^$ such that $x \in \beth_{\gamma}$. Observe that $\alpha^ = \bigcup {\alpha_n : n \in \omega}$, so there exists $k \in \omega$ such that $\gamma <\alpha_k$. Then by monotonicity, $\beth_\gamma < \beth_{\alpha_k} = \alpha_{k+1}$. So $x \in \alpha_{k+1}$, hence, $x \in \alpha^$. Therefore, $\alpha^8 = \beth_{\alpha^}$.

does this work?

[flat frigate]

toast

[tawdry plover]

that first sentence lol

[tranquil geyser]

oh

[tranquil geyser]

lmk if this is like really flawed

[tawdry plover]

sure

[tranquil geyser]

tysm

[tranquil geyser]

the aleph part is pretty simple

so i just care about showing alpha* = beth_alpha^*

[tawdry plover]

other than a’s becoming alphas it looks fine

[tranquil geyser]

kk

[tawdry plover]

a minor issue i have, shouldnt it be $\alpha_{k+1}=\beth_{\alpha_k}$?

[flat frigate]

blanketism

[tranquil geyser]

oh shit

yes

ur right

[tawdry plover]

thats it

[tranquil geyser]

kk

tysm

[tawdry plover]

np

[tranquil geyser]

.solved

[safe radish]

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[safe radish]

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[safe radish]

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

[jade shard]

1

[split kayak]

Well, to start with, you should be able to notice that there is a really small range of values where the equation even makes sense.

[open wedge]

there's an $\abs{x + \sqrt{1 - x^2}}$ so you should consider 2 cases to get rid of it

[flat frigate]

1 divided by 0 equals Infinity

[jade shard]

yes

[jade shard]

huh

[open wedge]

look at $\sqrt{1 - x^2}$

[flat frigate]

1 divided by 0 equals Infinity

[split kayak]

This part of the equation imposes a condition on x.

[jade shard]

yes okay

[open wedge]

but since we're working in reals

then $1 - x^2 \geq 0$

[flat frigate]

1 divided by 0 equals Infinity

[jade shard]

oooh

okok

[open wedge]

and solve for the condition

we get the small range that @split kayak said

[split kayak]

Theres also another condition we can impose btw

[open wedge]

ah ic

[jade shard]

oh ok yes

[split kayak]

,, \abs{...} = \sqrt2(2x^2-1)

[open wedge]

so this implies that $2x^2 - 1 \geq 0$

[split kayak]

Ignoring what is going on in the inside of the left side expression, we have this

[flat frigate]

1 divided by 0 equals Infinity

[jade shard]

OHHH

okay okay

[open wedge]

yea and solve for the condition for $x$

[flat frigate]

1 divided by 0 equals Infinity

[jade shard]

ohhhhhh

ohhh

icic

okay thanks

.close

[safe radish]

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[viscid marsh]

Translation: triangle ABC is drawn in a circle with the middle point M. The angle BCA = 30°, see figure.

Show that the circular arc (like the one between AB) is around 4.7% longer than the straight line AB

[viscid marsh]

How do i approach this

My first thought is to do this but idk where to go from here..

[glass carbon]

you must firstly find the measure of the central angle BMA (which is twice as large as the inscribed angle BCA subtended by the same arc)

[viscid marsh]

Can you explain to me why BMA is twice as largw as BCA sorry math isnt my strong suit and im a bit confused

[lavish hemlock]

There's a theorem

[stoic torrent]

right, well there is the circle property that angle subtended by an arc at the circumference is half the angle subtened by the same arc at the center

[viscid marsh]

WAIT RIGHT

I FORGOT ABT THIS

[stoic torrent]

okay so now try on your own
hint:

||hint: Do you see how you can find the arc length in terms of angle BMA and radius r of the circle?
After this you should also observe smth interesting about the triangle BMA from which you find side AB in terms of radius||

[viscid marsh]

Okay so i know that AB is the same length as the radius buti dont rlly get ehere to go from here?

Wait

[glass carbon]

You can divide both results to get rid of r (i.e. find the ratio)

What is your length of the arc AB?

(in terms of r)

[viscid marsh]

Wait is it like 60/360 * 2pi(r)

[glass carbon]

yes

so:

arc AB/AB = ...

[viscid marsh]

1.047

[glass carbon]

Convert it to percent

[viscid marsh]

4.7%

[glass carbon]

And you're done

[viscid marsh]

I have a question so is 60/360 bc of like the angle BMA

[glass carbon]

yes because BMA = 60 deg

[viscid marsh]

Ok yeah

I get it

Thank you!

I struggled w this a lot but now i understand! Thank you i apprecite it a bunch 😁😊🌷

.close

[safe radish]

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[safe radish]

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[barren fjord]

I think I’ve solved it correctly but I’m confused how to deal with the fact that it’s leaking out and is negative ?

[worldly lantern]

the rate of decrease in a quantity is always taken as a negative value

[barren fjord]

Does that mean dv/dt and dh/dt and dv/dh are negative ?

[worldly lantern]

yes

[barren fjord]

Oh so they just cancel out basically

[worldly lantern]

i cant necessarily comment on dv/dh

but anything with the word "decrease" next to it is negativw

[barren fjord]

Is that not decreasing

[worldly lantern]

does the question specify that volume wrt h is decreasing ?

[barren fjord]

If it’s leaking then the volume has to be decreasing tho

[safe radish]

@barren fjord Has your question been resolved?

[safe radish]

@barren fjord Has your question been resolved?

[safe radish]

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[safe radish]

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[hardy gulch]

Trying to approximate the direct integral of cos(x^0.5) from [0, 1] using the 4th Maclaurin polynomial for cos(x), then approximate the error.

[hardy gulch]

I'm being prompted to just use a calculator to find the actual value of the integral, which is fine, then state the error in the approximation.
I don't have a lot of examples, but I have misgivings about whether it would work to use the remainder theorem, since I've done step-by-step integration to to find the approximation.

This is (what I assume is) the closest similar example I can find from my text. When they mention the alternating series test at the end, what do they mean?

[safe radish]

@hardy gulch Has your question been resolved?

[tardy mango]

they're probably referring to the alternating series error bound

they're using the first four terms to approximate, so the error is bounded above by the magnitude of the fifth term

which is 1/216

[tardy mango]

Careful with your notation. (2n)! and (2n!) are not the same thing

the first is the factorial of 2n, the second is taking the factorial of n and multiplying by two (with a superfluous pair of parentheses)

yeah just do the bounds now

though for the error, I'd use the alternating series error bound as mentioned above

I suppose you could use the Lagrange error bound, but taking all of those derivatives would be very annoying

[hardy gulch]

Forgot to fix that, thanks for the reminder

[tardy mango]