help-17

[strong grove]

it ok :D

the black rectangle probably represents the 90 degree angle

[misty wave]

that would equal 14

[bronze osprey]

are you sure?

[misty wave]

omg bye

was it wrong

[bronze osprey]

this gives you 14 sqrt(2)

[misty wave]

how did we get the squared 2

[north apex]

sqrt(4^2 + 4^2). Then the answer of that * 7 / 2

[misty wave]

oh alright

got it

thank youu

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[hushed pewter]

!msgdel

[vocal sleet]

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[earnest marten]

Ehh okay

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[restive geode]

Just a quick question, the problem is whether this converges or diverges, and I thought it would converge because 1/n in infinity would be zero.
But it should be divergent according to the correction sheet and wolframalpha, the latter says it diverges to infinity which I dont get at all.

[tiny fable]

this problem is almost identical to whether the harmonic series converges

because it is the harmonic series minute the first 5 terms

harmonic series: 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 ...
this one: 1/6 + 1/7 + 1/8

so if the harmonic series diverges to infinity, this one must as wwell

[restive geode]

Yes I know but in the slides from my class there is an example with 1/n to infinity and there it does say it converges

so I dont know if I'm missunderstanding something or if either one is just wrong

[tiny fable]

well

uhh

they're both correct

if you look it says that

lim n -> infinity of 1/n = 0

does not imply that the sum from 1 to infinity of 1/n converges

actually it diverges

[restive geode]

so basically 1/n on it's own would be convergent but because it's the sum it diverges?

[tiny fable]

i don't understand what you mean

[warped thorn]

What about using some techniques like special test? It is clear to show it.

[tiny fable]

there's this elegent proof that it does diverge

harmonic series (H)
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9...

altered series (S)
1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16

now obviously, H > S, because all we did to make S is decrease some terms in H. But notice you can group terms in S as follows

1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16...
=
1 + 1/2 + 1/2 + 1/2...

this is the infinite series of 1/2, and trivially diverges

[restive geode]

I understand why according to this test it diverges but could you elaborate where the altered series S comes from?

[tiny fable]

uhh

it's the same as H but with all the terms lowered to the next power of 2

like 1/97 would become 1/128

[restive geode]

Yeah ok I understand, thank you for the help.

[tiny fable]

np!

[restive geode]

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[vast shale]

Let f be a real-valued function that is monotonically increasing on an interval (a, b). Let A = {x ∈ (a, b) : f is not continuous at x}. Prove that A is at most countable.

[vast shale]

I have already see that c(x) =< f(x) =< d(x) with c(x) is the sup f((a,x)) and d(x) is the inf f((x,b))

but idk how to prove that x in A, only if c(x) < d(x)

like idk how to write it out

my goal is to with x in A, I can pick out a g(x) in (c(x),d(x)) union Q
and then prove g is a injective function

and with g is an injective function hence A is countable

[vast shale]

but im stuck at this

[vast shale]

Can you show the question

[potent beacon]

It is here

[dim pumice]

couldn't we instead define c to be the left side limit at x, and d to be the right side limit at x? Then you just need to prove that these limits indeed exist, and it follows that if these limits disagree, it's discontinuous at x

[vast shale]

this is the qeustion

[dense eagle]

if we think about the image of f

it'll be like (f(a), f(b))

[potent beacon]

Im thinking basically to take "gaps" in discontinuities and use the fact that sum of them converge

[vast shale]

I dont get what u mean

[vast shale]

ah?

[dense eagle]

then like for each point of discontinuity, as u've shown i think, you have a jump discontinuity

so like there's some region around f(c) say like (f(c)-k, f(c)+k) that isn't in the image of f

so then you can map it to a rational point => countable

that's the rough sketch

[dim pumice]

$c(x) = \lim_{s \to x^{-}} f(s)$

[twin meteor]

rbit

[dense eagle]

in general, mapping to rationals is a good idea

[dim pumice]

it should come out to be the same as what you defined it as before, however the fact that if c(x) < d(x) it's discontinuous at x would follow directly

[vast shale]

you’re elaborating your own way right?

[dense eagle]

oh whoops nvm i thought u'd shown it had a jump discontinuity, mb

[dense eagle]

yeah ok ur method is fine, you just need to do this ^

[vast shale]

It has jump discontinuities

[vast shale]

I dont understand the "if these limit disagree it discontinious at x"

[potent beacon]

If right limit is different from left

The function is discontinous

At this point

[vast shale]

Since f is increasing the interval ix is disjoint for every interval of A

[dim pumice]

because if a limit exists at some point x, both the left and right side limits are equal to that

[vast shale]

Oh

Oh ok I understand

[vast shale]

LHL is equal to RHL

Right hand limit = Left hand limit

If they are not then they are discontinuities simple

Or discontinuous

but if c and di is limit how do I prove that c(x) =< f(x) =< d(x)

Easy

Just take out the limit of c on both sides and d on both sides

Listen f(x) is derivative so find the derivative of x

[dim pumice]

you will have to prove that $\lim_{s \to x^{-}} f(s) = \sup((a, x))$

[twin meteor]

rbit

[vast shale]

why is x^- on sup?

Im sorry, Im a bit confused atm

[vast shale]

You have to find left hand derivative

That's minus

[dim pumice]

what derivative?

[vast shale]

why is derivative here

[dim pumice]

no it's just a limit

[vast shale]

Ah nvm I see why it suppose to be x^- now

It is -x I guess

@dim pumice x^- elaborate it

[vast shale]

what would the purpose of proving that it discontinuous at x be?

cause my plan was initally trying to pick a g(x) and prove that it an injective function, I dont see the purpose of having discontinious at x

[dim pumice]

it's discontinuous at x iff c(x) < d(x), so the set A is equal to the set of all points where c(x) < d(x), and we can make this bijective to a subset of Q

[vast shale]

It is a monotonic function

Oh

Ohhh okay I understand it now

[dim pumice]

we could also try and go with your original c and d if you want

[vast shale]

It is injective subset

@dim pumice

[vast shale]

Im a bit confused here, since above I got c(x) =< f(x) =< d(x); what if c(x) = f(x) = d(x), that way it wouldnt be discontinious

[vast shale]

That is not possible because it has jump discontinuity

[vast shale]

can you explain that a bit for me, I dont get what u mean

[dim pumice]

I mean we can go with what you initially tried to go with, just trying to think of a plan to make it work

[vast shale]

yes please, Im really confused at this part rn

[dim pumice]

so let's say c(x) < d(x)

then pick some y such that sup f((a, x)) < y < inf f((x, b))

[vast shale]

You can see here that your function is monotonically increasing function so it has left hand limit less than right hand limit it has jump discontinuities

[vast shale]

You are saying it is equal which is not possible as it will make the function continuous

[vast shale]

yeah my bad, I forgot we had f is not continous at x since the beginning 😭

[vast shale]

You understood

hmm oka, I think I got the grasp of this now

thanks you two a lot

Welcome

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[uneven imp]

how do i access a statistic table about t student when it shows t(a/2), GL

[uneven imp]

i don't know what the gl means

[vocal sleet]

@uneven imp Has your question been resolved?

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[craggy sundial]

hello guys, with the graph sinx/x, why does y=1 when x=0?

[lunar totem]

It doesn't equal 1 it's undefined when x=0

[lunar totem]

Try to move the point until you get to 0

[craggy sundial]

sorry so let me rephrase

why does y->1 as x->0

[lunar totem]

Do you know l'hopital's rule?

[craggy sundial]

i do not

[dense cloak]

Because mathematical limit of the function sin(x)/x as x→0 which equals 1

[craggy sundial]

🤯

yeah why

[dense cloak]

Why? So you want me to proof it

In other words calling me a liar

[craggy sundial]

that's my question yes

[craggy sundial]

huh

[dense cloak]

Jk

[craggy sundial]

i've heard it has something to do with sin(x) tending to 0 faster than x does or the opposite way around

[dense cloak]

So

[dense cloak]

You don't know that

Rule

[lunar totem]

Me?

[dense cloak]

No

[dense cloak]

Mitesse I just used your comment

Was to lazy to write iz

It

[craggy sundial]

i do not

is it because at small values sinx=x

and therefore x/x = 1??

[vocal sleet]

@craggy sundial Has your question been resolved?

[warped thorn]

Yeah.

[vocal sleet]

@craggy sundial Has your question been resolved?

[craggy sundial]

new question

why does y->1 as x->inf??

[vocal sleet]

@craggy sundial Has your question been resolved?

[lilac pebble]

can we use a substitution?
u=1/x
the limit then becomes: lim(u->0) sin(u)/u

which is a classic limit, if i'm not mistaken

https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1

alternatively, write y as:
y=sin(1/x)/(1/x)
and use l'hopital

[lilac pebble]

maybe, but that isn't very rigorous

[lilac pebble]

l'hopital's rule uses derivatives. do you know derivatives?

[vocal sleet]

@craggy sundial Has your question been resolved?

[craggy sundial]

we call it differentiation

but yes

and i'm sure i know it just not by its name

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[wicked grotto]

My problem is how can i remove the () so the equation gets simpler, hope u get what i mean. pls help

[peak matrix]

i dont think it will get simpler if you remove it

[wicked grotto]

is there some german channels out here ? cause my task is in german

[peak matrix]

but if you want to do that anyway, you can split that fraction to -pir / 2 + 20/2 = -pir / 2 + 10

[peak matrix]

you can send it here

[wicked grotto]

in german tho ?

[peak matrix]

yeah, sure

[peak matrix]

there might be some german speaking helpers here

and if not, you could attempt to translate it

Umfang is area?

oh wait no

[wicked grotto]

in simple terms i gotta get the inner area from that thing and i got the "Umfang" thats the brown curved stick

brown curved stick is 20m

[peak matrix]

oh and you are maximizing that area

[wicked grotto]

ye i think so

[peak matrix]

then this formula seems kinda off

[wicked grotto]

ye maybe

[peak matrix]

so maybe lets start my getting the length of the brown line

can you express that in terms of r and h

[wicked grotto]

idk if its the right translation but the theme is
Extremal problem

2h * pir

[peak matrix]

yeah, you are probably supposed to make an expression for area and then maximize it using calculus

[peak matrix]

are you sure it's *?

[wicked grotto]

sorry

[peak matrix]

yep, so 2h + pir = 20

we will keep that in memory for now

and now try expressing the area

[wicked grotto]

1sec gotta screen it cause i alr got it

thats all that i had alr

[peak matrix]

are you sure it's * again?

[wicked grotto]

oh shit

its + again

[peak matrix]

indeed

[wicked grotto]

but how exaclty am i going on to get r ?

i couldnt figure it out

[peak matrix]

oh, sorry for late response

[wicked grotto]

no worries

[peak matrix]

so now you have area as $\frac{\pi r^{2}}{2}+2rh$

[twin meteor]

MæthIsAlwaysRight

[peak matrix]

it would be convenient to eliminate one variable

say h

[peak matrix]

which I see you already did heree

$\frac{\pi r^{2}}{2}+r\left(20-\pi r\right)$

so you should get something like this once you fix it

[twin meteor]

MæthIsAlwaysRight

[wicked grotto]

i needa sec to realize

[peak matrix]

we will have to use this

2h + pir = 20

so 2h = 20 - pir

[peak matrix]

and if you replace 2h with 20 - pir, you get this

= $\frac{\pi r^2}{2} + (20 - \pi r)r$

[twin meteor]

MæthIsAlwaysRight

[peak matrix]

is there something unclear?

[wicked grotto]

oh get it now

[peak matrix]

its essentially what you were doing before, just with + instead of *

[wicked grotto]

cause i divided h by 2 i got lil confused

[peak matrix]

ah i see

you could do this as well, but then you would multiply it by 2 again

so it would have no effect

[wicked grotto]

ye im kinda stupid

[peak matrix]

btw, were you taught derivatives or not yet?

[wicked grotto]

i needa google what that si

[peak matrix]

nah, you're just not used to this probably

[wicked grotto]

is

[peak matrix]

ok, so you werent

we can do it without them

[wicked grotto]

idk cause in german maybe its not called derivatives

[peak matrix]

it is the same in pretty much all languages afaik

or very similar

[wicked grotto]

ok

then i dont had it i think

[peak matrix]

it's that d/dx thing

or f'(x)

thats new to you, right?

[wicked grotto]

oh f'(x) i know

[peak matrix]

oh, really?

well, then we can use that

[wicked grotto]

its Ableitung in german

[peak matrix]

oh, interesting

[peak matrix]

okay, so now we have a "function" that gives us the area given just the radius

and we need to maximize it

so we can use take the Ableitung of that function and set it = 0

[wicked grotto]

but i never that to do that with such an complicated function

[peak matrix]

we can simplify it a bit

(20 - pir)r = 20r - pir^2

and you can also subtract the pir^2 from pir^2/2

[wicked grotto]

how u get the f'(x) or derivatives

[peak matrix]

I'd start by simplifying f(x) first

you can distribute the r into the parenthesis

[peak matrix]

.

and then you can also simplify pir^2 / 2 - pir^2

[wicked grotto]

where u got the pir^2 / 2 - pir^2 from. i mean pir^2 / 2 i know but the rest - pir^2 i dont get

idk if u understand what i mean

[peak matrix]

$f\left(x\right)=\frac{\pi r^{2}}{2}+\left(20-\pi r\right)r=\frac{\pi r^{2}}{2}+\left(20r-\pi r^{2}\right) \dots$

[twin meteor]

MæthIsAlwaysRight

[peak matrix]

Can you finish what follows?

there are still some steps left to be done

to fully simplify it

[peak matrix]

I got it from here btw

[wicked grotto]

i undestand but i cant finish the following part

[peak matrix]

$f\left(x\right)=\frac{\pi r^{2}}{2}+\left(20-\pi r\right)r=\frac{\pi r^{2}}{2}+\left(20r-\pi r^{2}\right)=\left(\frac{\pi r^{2}}{2}-\pi r^{2}\right)+20r$

[twin meteor]

MæthIsAlwaysRight

[peak matrix]

can you now simplify whats inside those parenthesis?

I just rearranged it in this step

[wicked grotto]

it looks easy but my brain is just hanging

[peak matrix]

what is half an apple - apple?

[wicked grotto]

-0,5 apple

[peak matrix]

yep

[wicked grotto]

i guess

easy

[peak matrix]

so just imagine pir^2 is apple

and it becomes apple / 2 - apple

[wicked grotto]

so in the parenthesis its -0,5 ?

[peak matrix]

-0.5 of what?

[wicked grotto]

of pir^2

[peak matrix]

correct

[wicked grotto]

-0,5pir^2

[peak matrix]

so -0.5pir^2

or -pir^2 / 2

$f\left(x\right)=\frac{\pi r^{2}}{2}+\left(20-\pi r\right)r=\frac{\pi r^{2}}{2}+\left(20r-\pi r^{2}\right)=\left(\frac{\pi r^{2}}{2}-\pi r^{2}\right)+20r=-\frac{\pi r^{2}}{2}+20r$

and that was the final step

[twin meteor]

MæthIsAlwaysRight

[peak matrix]

now it's simplified

you can take the derivative now (Ableitung)

and dont worry about the pi too much, its just a number like any other

[wicked grotto]

so the f'(x)
is -pir + 20

[peak matrix]

seems right

[wicked grotto]

if not ill go cry

[peak matrix]

what radius (r) will maximize the area (f(r)) then?

[wicked grotto]

20/pi

[peak matrix]

oh and one more detail

our function is actually in terms of r

so we should use f'(r)

i've done this mistake as well

[wicked grotto]

oh ye youre right

1 last question how old r u

[peak matrix]

17

[wicked grotto]

youre an genius ty for youre help. An 17 yo needs to help an 18 yo for homework. I appreciate youre help

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[peak matrix]

im not a genius lol

btw you still need to find the maximal area

[wicked grotto]

but genius enough for me

[peak matrix]

you only got the radius which maximizes that area now

[wicked grotto]

ye but that shouldnt be the problem

[peak matrix]

alright then

[wicked grotto]

.reopen

[vocal sleet]

✅

[wicked grotto]

@peak matrix i might need your help again

when i put the r = 20/pi into the equation -pi*(20/pi)+20 = 2h the left side is 0 so 0 = 2h but that shouldnt make any sense if im not stupid

[vocal sleet]

@wicked grotto Has your question been resolved?

[wicked grotto]

i might got the solution but that would mean that the teacher told us misinformation abt the task

.close

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[vital loom]

I forgot how to do this

I thought I just sub in 2 into the equaiton

and then sub that value into the xN+1 equation

[warped dragon]

That is the correct idea, yes

Can you find an expression for x_2?

[vital loom]

yeah its just 2p-6

but when i sub that into the equation

it wont give me what i need

yeah ive j made silly mistake

🤦‍♂️

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[dense jacinth]

how do you figure it out without knowing what the point from the beginning of the process was?

[dense jacinth]

like

without f(0)

how do u calculate it

like im pretending i dont have the point f(0(

[night zodiac]

like you don't have the point f(0)? you mean (0,4) in the table?

i don't understand what you mean

[dense jacinth]

is it possible to

know how tall the process was

without having the point f(0)

wai

[dense jacinth]

yeah 0,4

that point

[dense jacinth]

imagine it was like this

[night zodiac]

hm, no. you need that point.

[dense jacinth]

cant do that otherwise?

[night zodiac]

firstly, it's asking you to graph the function, so i don't understand what you're trying to do

[dense jacinth]

ywah my teacher is stupid

but

if the first question was

determine the logistic growth function : a/1+be^-kt

and i didnt have (0,4)

or the table

on*

could i do that?

or is that impossible @night zodiac

[night zodiac]

wait, this is an important question: does your teacher allow you to use a calculator to find the equation?

[dense jacinth]

yeah but not like an advanced one

[dense jacinth]

lets say if i have all the points on the table except for 0,4

then i can find the function? for the logistic growth?

[night zodiac]

it seems like it would be possible, then. you need to know what the midpoint, the growth rate, and the maximum value

[dense jacinth]

but if not then i cant?

[night zodiac]

graph the equation first, and you will find x0 and L.

yea, if you don't have the growth rate, then you can't

[dense jacinth]

yeah

but l is 15 right

or 14

[night zodiac]

you use a ti-84?

[dense jacinth]

no

some cassio

[night zodiac]

or does he want you to graph it by hand?

[dense jacinth]

ph

[night zodiac]

no, L is not 15.

[dense jacinth]

yeah i have the function to do that

[dense jacinth]

14?

[night zodiac]

okay, then you can choose to graph it by hand or on the calculator. Whichever one is easier for you.

What is L defined as in the photo?

[dense jacinth]

no its 20

[night zodiac]

Close

[dense jacinth]

19.9

[night zodiac]

yes.

[dense jacinth]

but

how can i be sure of that

cuz there is day 15

like x 15

not day 15

my bad

[night zodiac]

it's a logistic function, so there is a horizontal asymptote. If you look at the graph on your calculator, what do you notice is happening?

[dense jacinth]

wait im gonna go sleep im gonna paint it tomorrow

[dense jacinth]

but,

i would assume

its approaching the maximum value L?

[night zodiac]

wonderful

[dense jacinth]

also these functions dont have zeropoints right

[night zodiac]

no

[dense jacinth]

thanks alot

[vocal sleet]

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[worthy dew]

I was browsing old documents on my computer and I found a link to a set of corrupted images and some transcriptions. I forget what the document could be, however the transcriptions looked like set theory. Could anyone tell me what this means? (This is one of several transcriptions. Also if it helps, this specific text was marked under "Notation:".)

Ɐx,x ∈ Ωₙ₁ (k₁), Ǝy, y ∈ Ωₙ₂ (k₂), ∄x₂, x₂ ∈ Ωₙ₁ (k₁), //x₁,y//<//x,y//⇒
x(_M) (Ωₙ₁ (k₁),Ωₙ₂ (k₂))

//x,y//=ω+Δω

The // are replacements of || due to discord thinking I'm trying to do spoilers or smth
I'm fairly sure (_M) is supposed to denote subscript

Thank you

[vocal sleet]

@worthy dew Has your question been resolved?

[worthy dew]

<@&286206848099549185> ?

[vocal sleet]

@worthy dew Has your question been resolved?

[vocal sleet]

@worthy dew Has your question been resolved?

[worthy dew]

<@&286206848099549185> ???

[earnest nymph]

Yes

[worthy dew]

Is there any chance you could translate/solve this?

Or point me to somewhere where I could figure out what this notation is?

[clear shoal]

without context this is nonsense

[crystal lynx]

im like 99% sure this is corrupt gibberish

[worthy dew]

Really? After extensive reasearch I can read parts of it. Like: Ɐx (for all x),x ∈ Ωₙ₁ (k₁) (x is part of a set), Ǝy (there exists a y) y ∈ Ωₙ₂ (k₂) (y is part of another set). A lot of it seems to be comprehensible and I'm fairly sure that there is some sort of meaning behind it, though I really am not sure. If it really does not make sense, could you by any chance derive any sort of meaning from it or point me in the general direction of what it is talking about?

Context? There are some other transcriptions that I might be able to provide that might provide some sort of context, though most of them lable "Axioms:". I really am curious about what this could be so I am going to provide these tomorrow.

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[bitter pilot]

Hi, may somebody explain to me why
[ \mathbb{Z}2 \times \mathbb{R}{>0} \cong \mathbb{R}\setminus{0} ]
where $\mathbb{Z}2 \leftarrow +$ and $\mathbb{R}{>0} \leftarrow\cdot .$

[twin meteor]

𝔸dωn𝓲²s

[rough patrol]

oh um

do you know what needs to be true for an isomorphism?

an isomorphism $phi$ works when:

[twin meteor]

Arnavutköy

[rough patrol]

$\phi$ sorry

[twin meteor]

Arnavutköy

[rough patrol]

$\phi(1)=1$, $\phi(a)\phi(b)=\phi(ab)$

[twin meteor]

Arnavutköy

[rough patrol]

where $1$ represents the identity in the group

[twin meteor]

Arnavutköy

[rough patrol]

can you show these conditions hold

?

[bitter pilot]

give me a sec

[rough patrol]

additionally for an isomorphism you need to show injectivity and surjectivity

[bitter pilot]

So we defined that two groups $(G, \Theta_1)$ and $(H, \Theta_2)$ are isomorphic if there exists a bijective map $\Phi$ s.t.
[ \Phi(a\Theta_1b) = \Phi(a)\Theta_2\Phi(b) \text{ for all } a,b \in G]
then $\Phi$ is a group isomorphism.

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

I just realized Phi : G -> H

Ok Z_2 on + should be {0,1} right?

And R_>0 is just {x in R | x > 0}

[rough patrol]

yes

[bitter pilot]

I am wondering now for Z_2 x R_>0 what is the operation

[rough patrol]

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

read the definition

[bitter pilot]

oh right my bad

i funny enough i just wrote it down 5 ming ago

So it's *

So for two elements of G say a and b we get say a = (z1,r1) and b = (z2,r2)

If I apply the operation I get (z1z2, r1r2)

H was R \ {0} which is also multiplication I guess because 0 is not contained

[rough patrol]

yes

[bitter pilot]

Let $a,b \in G$ then $a = (z_1,r_1)$ and $b = (z_2,r_2)$ where $z_1,z_2 \in \mathbb{Z}2$ and $r_1,r_2 \in \mathbb{R}{>0}.\$
Let $\Phi : G \to H$ with $(z,r) \mapsto zr$.
[ \Phi(a\Theta_1b) = \Phi((z_1,r_1) \cdot (z_2,r_2)) = \Phi(z_1z_2,r_1r_2) = z_1z_2 \cdot r_1r_2 = \Phi(a) \Theta_2 \Phi(b) ]

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

Now I would need to show it's bijective

I am cooked

[hard atlas]

I would suggest to first biject Z_2 to {1,-1}

then stuff works out a bit more nicely

with the operations

[bitter pilot]

why {-1,1}?

I thought Z_2 = {0,1}

[hard atlas]

{1,-1} under multiplication

can you describe in words what your isomorphism is supposed to do

i.e. why these groups are isomorphic

in words

there is a very clear interpretation of this

[bitter pilot]

That was my initial question but uhm, they are supposed to have the same cardinality and maintain the algebraic structure but i am not really sure what i just said

[bitter pilot]

I am even more confused because I missed that lecture and a class mate wrote it like that like Z_2 is with +

[hard atlas]

Z_2 is usually with +

but working with {1,-1} under * is easier here imo

and thats isomorphic to Z_2

[bitter pilot]

so i can choose

sorry, it's just my first time doing proofs of this kind

ussually we just define stuff but never go that deeply

[hard atlas]

lets ignore the group stuff

on the left we essentially take half of the real axis, but twice

on the right we take the whole real axis except zero

how can you give a nice bijection between those

[rough patrol]

yeah what denascite is saying is good, try to think about this logically in terms of how any mapping would make sense. then worry about the exact details of the grouping

[bitter pilot]

ok give me sec to think and process

[bitter pilot]

So basically this with Z_2 = {-1,1} would form with R_>0 {y = 1 | y > 0} U {y = -1 | y > 0} visually

[hard atlas]

uhm

[bitter pilot]

like on the xy plane

[hard atlas]

lets call C_2 = {1,-1}

its not Z_2, just isomorphic

on the left we have C_2 x R_>0, that set has pairs of the form (1, x) or (-1,x) where x is a positive real number

yes?

[bitter pilot]

yes

[hard atlas]

on the right we just have real numbers x, both positive and negative

[bitter pilot]

yes

[hard atlas]

how can you pair those up

[bitter pilot]

I could multiply them

direct product?

[hard atlas]

the fact that you chose that word makes me a bit worried

what do you want to multiply

I said to forget that we have groups

[bitter pilot]

I could verknüpf them

?

[hard atlas]

ok dann wechseln wir zu deutsch

was willst du verknüpfen

[bitter pilot]

(1, x) mit (-1,x)

weil du hast gefragt wie man die paaren könnte oder so

[hard atlas]

wie kann man die linke seite mit der rechten seite paaren

links haben wir paare (+-1,x) mit positivem x

rechts haben wir reelle zahlen x ungleich 0

[bitter pilot]

achso meinst du

[hard atlas]

zu welcher zahl gehört das paar (r,x)

[bitter pilot]

ich weiß nicht, mein kopf sagt mir (r,x) ist doch ein punkt

[hard atlas]

ich will eine abbildung definieren von C_2 x R_>0 nach R^*

dafür muss ich jedem paar (r,x) eine reelle zahl zuordnen

wie könnte ich das tun

auf eine "sinnvolle" (also nicht "zufällige") Art und Weise

[bitter pilot]

skalar produkt?

ne

(r,x) -> rx?

[hard atlas]

ja

und fuck aber leider muss ich los

[bitter pilot]

ist gut

danke dir sehr

[hard atlas]

versuch zu zeigen dass diese abbildung ein isomorphismius ist

[bitter pilot]

wie oben?

[hard atlas]

ja

[vocal sleet]

@bitter pilot Has your question been resolved?

[hard atlas]

ok bin wieder da

wie läufts

@bitter pilot

[bitter pilot]

Ich schreibe gerade meinen beweis

[bitter pilot]

Show that $\mathbb{C}2 \times \mathbb{R}{>0} \cong \mathbb{R}\setminus{0}.\$
$\ \textbf{Proof.} \$
Define $G = \mathbb{C}2 \times \mathbb{R}{>0}$ where $\mathbb{C}_2 = {-1,1}$ and $H = \mathbb{R}\setminus{0}.\$
Let $\Phi : G \to H$ where $\forall c \in \mathbb{C}2, r \in \mathbb{R}{>0}$ s.t. $(c,r) \mapsto z \cdot r.\$
Then $\Phi$ is a homomorphism because for some $a,b \in G$ s.t. $ a = (c_1,r_1), : b = (c_2,r_2)$ where $c_1,c_2 \in \mathbb{C}2$ and $r_1, r_2 \in \mathbb{R}{>0}$ holds
[ \Phi(a\Theta_1b) = \Phi((c_1,r_1) \cdot (c_2,r_2)) = \Phi(c_1c_2,r_1r_2) = c_1c_2 \cdot r_1r_2 = c_1r_1 \cdot c_2r_2 = \Phi(a) \Theta_2 \Phi(b). ]

$\Phi$ is surjective.$\$
Let $y \in H.\$
Then $\forall y > 0: :\Phi(1, y) = 1 \cdot y = y \in \mathbb{R}{>0}.\$
For $\forall y < 0: : \Phi(-1, y) = -1 \cdot y =: \tilde{y} \in \mathbb{R}{>0}.\$

$\Phi$ is injective.$\$
Let $x_1,x_2 \in G$ s.t. $x_1 = (c_1,r_1) \text{ and } x_2 = (c_2,r_2) \text{ where } c_1,c_2 \in \mathbb{C}2$ and $r_1, r_2 \in \mathbb{R}{>0}.\$
Then [ \Phi(x_1) = \Phi(x_2) \Rightarrow \Phi(c_1,r_1) = \Phi(c_2,r_2) \Rightarrow c_1r_1 = c_2r_2 \Rightarrow \frac{c_1}{c_2}r_1 = r_2 ]
W.l.o.g. suppose $c_1 = 1$ and $c_2 = -1$ then $-r_1 = r_2$ is a contradiction since the left side is negative and the right side is positive.$\$.
This implies $c_1 = c_2$ for $c_1 = c_2 = 1$ or $c_1 = c_2 = -1$, $\$ which implies $r_1 = r_2$ and thus $x_1 = x_2.\$
Therefore $\Phi$ is bijective, and thus an isomorphism.

[hard atlas]

zuerst, tausche alle Z_2 gegen C_2 aus

Z_2 und C_2 sind nicht dieselben sachen

wenn y<0, dann ist Phi(-1,-y)=y, das meintest du

warum folgt aus der letzten gleichung dass z_1=z_2 ?

[bitter pilot]

wenn y < 0 z.b. y = -1 aber ich schreibe -y dann hätte ich 1

[hard atlas]

ja und so willst du es ja auch haben

in den paaren (r,z) darf z nur > 0

[bitter pilot]

aber kann ich nicht -1 wählen für z

-1 mal y = positiv da y < 0

[hard atlas]

du verwechselst wo zeug sein soll

das resultat von Phi darf positiv und negativ sein

[bitter pilot]

Wenn z_1 ungleich z_2 wäre also o.b.d.a. z_1 = 1 und z_2 = -1 dann folgt r_1 = -r_2 und dann wäre es nicht injektiv

[hard atlas]

Phi(....) soll gleich y sein, unabhängig davon ob y positiv oder negativ ist

[bitter pilot]

oh

[hard atlas]

ja gut aber du kannst ja nicht argumentieren "ja z_1=z_2 denn sonst wäre es nicht injektiv". du willst ja momentan zeigen dass es injektiv ist

[bitter pilot]

ja stimmt

ich brauch einen widerspruch

[hard atlas]

wenn du dir z_1/z_2 r_1=r_2 anguckst, dann ist die rechte seite positiv

also muss links auch positiv sein

[bitter pilot]

ja

widerspruch

weil -r_1 ist negativ

weil r_1 ist aus R_>0

[hard atlas]

gut man muss hier nicht mit nem widerspruch ran

da r_1 positiv muss dann auch z_1/z_2 positiv, also =1

oder man kann auch direkt von z1r1=z2r2 darauf schließen dass z1,z2 dasselbe VZ haben müssen

anyway, das ist nur ein bisschen nitpicking

[bitter pilot]

Ist das also ausreichend?

Show that $C_2 \times \mathbb{R}{>0} \cong \mathbb{R}\setminus{0}.\$
$\ \textbf{Proof.} \$
Define $G = C_2 \times \mathbb{R}{>0}$ where $C_2 = {-1,1}$ and $H = \mathbb{R}\setminus{0}.\$
Let $\Phi : G \to H$ where $\forall c \in C_2, r \in \mathbb{R}{>0}$ s.t. $(c,r) \mapsto z \cdot r.\$
Then $\Phi$ is a homomorphism because for some $a,b \in G$ s.t. $ a = (c_1,r_1), : b = (c_2,r_2)$ where $c_1,c_2 \in C_2$ and $r_1, r_2 \in \mathbb{R}{>0}$ holds
[ \Phi(a\Theta_1b) = \Phi((c_1,r_1) \cdot (c_2,r_2)) = \Phi(c_1c_2,r_1r_2) = c_1c_2 \cdot r_1r_2 = c_1r_1 \cdot c_2r_2 = \Phi(a) \Theta_2 \Phi(b). ]

$\Phi$ is surjective.$\$
Let $y \in H.\$
Then $\forall y > 0: :\Phi(1, y) = 1 \cdot y = y \in \mathbb{R}{>0}.\$
For $\forall y < 0: : \Phi(-1, -y) = -1 \cdot -y = y \in \mathbb{R}{>0}.\$

$\Phi$ is injective.$\$
Let $x_1,x_2 \in G$ s.t. $x_1 = (c_1,r_1) \text{ and } x_2 = (c_2,r_2) \text{ where } c_1,c_2 \in C_2$ and $r_1, r_2 \in \mathbb{R}_{>0}.\$
Then [ \Phi(x_1) = \Phi(x_2) \Rightarrow \Phi(c_1,r_1) = \Phi(c_2,r_2) \Rightarrow c_1r_1 = c_2r_2 \Rightarrow \frac{c_1}{c_2}r_1 = r_2 ]
W.l.o.g. suppose $c_1 = 1$ and $c_2 = -1$ then $-r_1 = r_2$ is a contradiction since the left side is negative and the right side is positive.$\$
This implies $c_1 = c_2$ for $c_1 = c_2 = 1$ or $c_1 = c_2 = -1$, $\$ which implies $r_1 = r_2$ and thus $x_1 = x_2.\$
Therefore $\Phi$ is bijective, and thus an isomorphism.

[hard atlas]

C_2, nicht \mathbb C_2

das verwirrt nur

[bitter pilot]

oh fuck

😂

[hard atlas]

(das C steht übrigens für cyclic, hab ich nicht komplett zufällig gewählt)

das mit y<0 musst du noch anpassen

[bitter pilot]

ups

[hard atlas]

es ist jetzt wirklich y, nicht \tilde y

das R_>0 muss weg

das willst du ja auch gar nicht

[bitter pilot]

wieso

[hard atlas]

y war in H beliebig gewählt

H ist R\{0}

nicht R_>0

[bitter pilot]

oh stimmt bin blöd

noch eine frage

[hard atlas]

happens

[bitter pilot]

wenn ich sage y < 0 wieso schreibe ich -y ist das nicht widersprüchlich

[hard atlas]

wieso

wenn y<0 dann ist ja -y>0 und darum ist (-1,-y) in G

und das brauchst du ja, ansonsten würde Phi(-1,-y) nicht definiert sein

[bitter pilot]

du hast so recht

ok danke

vielleicht verkacke ich Algebra klausur doch nicht

[hard atlas]

beachte, du bist noch nicht fertig

du musst noch zeigen Z_2 isomorph zu C_2

[bitter pilot]

[hard atlas]

und Z_2 x R_>0 isomorph zu C_2 x R_>0

(beide davon sind deutlich einfacher)

(und das erste davon ist damit das zweite einfacher ist)

[bitter pilot]

muss ich wieder so viel Aufwand betreiben

[hard atlas]

nein, das sind jeweils im grunde einzeiler

[bitter pilot]

Kann ich sagen Z_2 ist isomorph zu C_2 weil |Z_2| = |C_2| und beide sind zyklisch

[hard atlas]

den vorteil den du dir erkämpft hast ist dass du oben mit c1,c2 wirklich als VZ rechnen durftest anstatt immer bei 0 an positiv denken zu müssen und bei 1 an negativ

wenn ihr im kurs gezeigt habt dass es nur eine zyklische gruppe für jede ordnung gibt, ja

[bitter pilot]

Wir haben da einen Satz

Isomorphie zyklischer Gruppen mit Beweis im Skript

[hard atlas]

klingt gut

[bitter pilot]

Ist Z_2 X R_>0 isomorph C_2 X R_>0 das gleiche zu zeigen wie Z_2 ist isomorph zu C_2

[hard atlas]

im grunde halt ja

wie man erwarten würde, wenn G und H isomorph sind, dann ist GxK und HxK isomorph

versuch das zu zeigen

(die andere richtung gilt glaube ich nicht)

[bitter pilot]

einen moment

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

Sagt man richtig, Phi ist ein Isomorphismus für C_2 x R_>0 und R \ {0}

[hard atlas]

Phi ist ein Isomorphismus zwischen C_2 x R_>0 und R\{0}

[bitter pilot]

ahh ok

[hard atlas]

also den Beweis für die letzte Implikation würde ich schon gerne sehen

[bitter pilot]

ich auch aber leider muss ich gleich ein tutorium halten

[hard atlas]

das ist auch nochmal keine schlechte übung dafür wie man so isomorphismen kombiniert

und ich hoffe ihr habt gezeigt dass isomorph sein eine äquivalenzrelation ist

ansonsten musst du das auch noch machen

[bitter pilot]

ich kann alles gerade gebrauchen als gute vorbereitung weil wir machen halt nur sehr einfache sachen hab ich das gefühl

[bitter pilot]

ja

im skript

Ich denke ich lass das hier offen, nochmal vielen lieben Dank

[hard atlas]

gern

[bitter pilot]

Ich bin schon bisschen Stolz dass ich es iwie geschafft habe

so ein bweis ist nichtohne wenn man davor nie was gemach that

[hard atlas]

sich am anfang durch die ganzen definitionen und so zu kämpfen ist nicht einfach

wie bist du in nem algebra kurs gelandet ohne vorher beweise gehabt zu haben

[bitter pilot]

ist ein Wahlfach

war mir lieber als Finanzmathematik als Wahlfach

[hard atlas]

aber wieso kannst du sowas überhaupt wählen ohne vorher beweise gesehen zu haben

[bitter pilot]

Ich studiere angewandte Mathematik was das ganze erklärt denke ich

[hard atlas]

was hattest du denn bisher für kurse

ja ok

[bitter pilot]

Ich hatte viele Sachen schon aber auf einer oberflächlichen Ebene.
z.b. Diskrete Mathematik, Numerik, Differentialgeometrie, Software Engineering, Stochastik, Operations Research

aber gerade Differentialgeometrie oder diskrete Mathematik das kann man nicht vergleichen mit einer Universität im Vergleich zu einer FH

[hard atlas]

na dann, toi toi toi

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[jagged cargo]

i want to find a basis of U. to do that, my plan is to prove the list is LI, then argue that since the list length is equal to dimU, the list is a basis. but to show that, i need to find dimU, which is, uh, to find the basis. is there a way to find dimU without a basis?

[somber portal]

what is P4(F) here?

[jagged cargo]

set of all polynomials deg <= 4 with scalars either in R or C

[somber portal]

what you can do instead is prove that it's a generator system (idk if that's the term in english?) and then prove that they are linearly independant, thus making it a base and giving you the dimension of U

[jagged cargo]

can you elaborate on generator system?

[somber portal]

spanning set?

what i called generator system (again, not sure the english term) allows you to get any element of the vector space as a linear combination of its elements.
Additionally, if the vectors in the GS are linearly independant, you have a base

[hybrid flicker]

U is a hyperplane of P_4(F)

it's the kernel of a linear form

so dimension is dim(P_4(F)) - 1

[jagged cargo]

kernel?

[hybrid flicker]

you haven't seen linear maps?

[jagged cargo]

nope

[hybrid flicker]

yeah so there's no finding a basis using that

but I mean

you could argue the dimension is at most dim(P_4(F)) - 1

since U is not equal to P_4(F)

and then find an LI family with dim(P_4(F)) - 1 vectors

[jagged cargo]

ah, that could work

thanks

[hybrid flicker]

justify this though

by finding a polynomial that's not in U

[jagged cargo]

yeah that should be easy

1 should do the trick

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[jagged cargo]

.close

[vocal sleet]

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[hard atlas]

if you are willing to use the factor theorem you can show quite easily that the polynomials you wrote down span the space

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[novel hearth]

How can I factor out csc^2 x I keep getting a diffrent answer than the book

[novel hearth]

.close

[vocal sleet]

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[novel hearth]

.reopen

[vocal sleet]

✅

[novel hearth]

Nvm I have a other question sorry

cot^-1 (1) how do I type that into a scientific calculator?

I tried 1/tan^-1(1) but I got a diffrent answer than my book

[sly citrus]

i need help

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[violet fox]

hi

[violet fox]

here is my solution for x>0

how can I proof that f>0 for -1 <= x < 0

<@&286206848099549185>

[lilac pebble]

how did you get r(1+x)^(r-1)-r ?

[violet fox]

thats f'(x)

altough it should be c instead of x

where 0<c<x

[lilac pebble]

oh sorry

and yes it should be c

[vocal sleet]

@violet fox Has your question been resolved?

[lilac pebble]

how can MVT help you here?

it will only show existence of c, it won't tell you about the value of f at all points in the interval

[violet fox]

since x is greater than 0
r [ (1+c)^(r-1) - 1 ] is greater than 0 and r [ (1+c)^(r-1) - 1 ] is equal to f so f is also greater than zero

[lilac pebble]

i see

i think you forgot x in the denominator, but it still makes sense

maybe that x in the denominator will also help you with the case -1<=x<0

isn't it kicking the can a bit, to say r [ (1+c)^(r-1)-1 ] > 0 ?

[violet fox]

yes you are right

[violet fox]

it can solve the question

[violet fox]

what do you mean?

[violet fox]

we can say f(x) = rx[ (1+c)^(r-1) - 1 ]. the thing in brackets is negative and x is negative so f is positive

tysm

[lilac pebble]

you're welcome

[violet fox]

.close

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[ebon dune]

I already know the answer is 364, but is there a faster way

[wind geyser]

which way did you do it

[ebon dune]

I do it like this

[wind geyser]

uh

lol?

what did you do

[lilac pebble]

i think they counted the number of times each number 1 to 12 appears in the sum

[wind geyser]

ohhh

[ebon dune]

multiply the first half of the number by the other half, then multiply by 2

[lilac pebble]

um

[wind geyser]

uh

[lilac pebble]

ok maybe i was wrong

[ebon dune]

I use my logic🙏🏻🙏🏻🙏🏻🙏🏻

[wind geyser]

cute

[lilac pebble]

is that right though?

[wind geyser]

how i'd do it is quite standard

[lilac pebble]

oh you multiplied by 2 at the end

ok

it should be ok

[ebon dune]

yeaa

but it took quite a long time

and This can only be used if the number of numbers is even, what if it is odd

[lilac pebble]

the sum of triangle numbers is n(n+1)(n+2)/6

12*13*14/6 = 364

[ebon dune]

okay🙂‍↕️🙂‍↕️🙂‍↕️

thanks dude

.close

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[vocal sleet]

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[sly basin]

can someone check my work

[sly basin]

that was the question

[vocal sleet]

@sly basin Has your question been resolved?

[vast shale]

y>0 is given ?

[sly basin]

klind of

ye

because it would be undefined if it wasnt right

we would get imaginary

[vast shale]

$y=x$ and $y=-x$ both satisfy the equation $y^2 = x$

anyways

[vast shale]

you got it right

you could also do this by substitution

[sly basin]

substitution?

substitute what?

[vast shale]

nvm

e

[sly basin]

okok

can u also check this

[vast shale]

you asked this yesterday i think?

[sly basin]

i think so

but no1 was there

to help

[vast shale]

lol

[sly basin]

:((

unless u thinkin about sum1 else

i asked something similar

[vast shale]

i have no idea on how to do this

[sly basin]

okok

can u check this then

[vast shale]

i think its wrong

you obviously have to expand x(t)

its just 1500+ 96t

i suppose you get why that is x(t)

[sly basin]

but x(t) is always different depending on time

whereas v(t) we can someonewhat calculate it cuz we know volume and concentration

[vast shale]

this is correct

[vast shale]

even this depends on time

[sly basin]

do i have to solve for C for that one?

[vast shale]

did they give any condition ?

[sly basin]

ye

[vast shale]

what is it ?

[sly basin]

so i guess condition is anything after t=0

so i substituted t=0

[vast shale]

no no

[sly basin]

since i know thats initiatl

[vast shale]

fo this did they give any condition

[sly basin]

oh no

no other condition

[vast shale]

then you possibly cant solve for the constant term

[sly basin]

but i remember for other problems similar to this

dont we actually have to solve the differnetial equation

ohhh ur saying like

[vast shale]

you just did ?

[sly basin]

at y(0)

= 1

or sometihng

like that

right

[vast shale]

yeah

[sly basin]

ahh yee that make sense

then ye it impossible

idk how to solve this

[vast shale]

do you unbderstand why you have to expand x(t) though ?

[sly basin]

not really tbh

[vast shale]

could you tell me what x(t) is though

@sly basin

[sly basin]

well

i was thinking the formula is basically the concentration in times rate out.
This is the stuff we are given.
thats all rate in.
As for rate out.
we can figure out the rate out but we cant figure out concenntration out.
So to answer ur question. its basically concentration out = x(t) / v(t)
so x(t) would be total amount of salt divided by total volume?

obvsly its with respect 2 time

[vast shale]

yes

the total amount of salt

tell how much salt we had initially

oi @sly basin wake up

[sly basin]

so here it would be uh

its unknown

no?

wait no

3g

[vast shale]

its given 3g per litre ,it doesnt show the total amount salt

[sly basin]

ye

thats why i put it as x(t)

total salt with respect to time

[vast shale]

im asking the total amount of salt intially present in the container

since we had 500 litres of 3g/l the total amound of salt presset would be 500 *3

which is 1500

[sly basin]

Ohh

[vast shale]

get it ?

[sly basin]

yeee

Yeeee

[vast shale]

now

after time t

[sly basin]

because at time 0

[vast shale]

the amount added would be 8 * 12 g/min

do you undersyand why

[sly basin]

wait why

Ohh

yeee

Cin times rin

concentration in and rate in

right

i tried to use this guys method

https://www.youtube.com/watch?v=-Fd6Ates4Gs

[sly basin]

.close

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Channel closed

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[vast shale]

yeah

its the sum of the amout already present and the new amount ofsalt being added

[vocal sleet]

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[ember notch]

What's 4 times 4?

[bitter delta]

16

[copper mantle]

16

[astral pilot]

,calc 4*4

[twin meteor]

Result:

16

[astral pilot]

I can confirm that its 16

Also, if you're new to this server, don't post questions like these, they can be interpreted as trolling

[vocal pagoda]

the device they open discord on has calculator on it by default... it's hard to see it as more than trolling

[vocal sleet]

@ember notch Has your question been resolved?

[vocal sleet]

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[swift bane]

we're asked to find the radiuses (big circle radius = 6.25, small circle radius = 3), after that, we have to find the distance between their centers

but I don't know how to do that

how can we even find their centers in the first place?

[astral pilot]

The small circle is called the incircle

The big circle is called the circumcircle

The distance between circumcentre and incentre is $\sqrt{R^2 - 2Rr}$

[twin meteor]

Pro_Hecker

[swift bane]

I see

I thought that you'd have to find it manually 🤣

[astral pilot]

Do you know about this formula?

[swift bane]

nope

[astral pilot]

Then, you'll probably have to do it manually

Not a hard task since the triangle is isoceles

[swift bane]

are there scenarios where you have to do it manually?

or can you just use this formula every where

[astral pilot]

No this is a universal formula

[swift bane]

oh wow

[astral pilot]

Proven in all cases

[swift bane]

so works on every shape?

[astral pilot]

On triangles

[swift bane]

I see

alright, thank u so much 👍

.close

[vocal sleet]

Channel closed

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[astral pilot]

Hey, I mean you're not supposed to use that formula 😦

[swift bane]

🤣

why not

.reopen

[vocal sleet]

✅

[swift bane]

I actually don't know how to find this manually

and I'm pretty sure the drawing that I have made is inaccurate

the centers can not be the same

[astral pilot]

Nah good enough

[swift bane]

alright

[astral pilot]

Centres can be same in equilateral triangle

[swift bane]

how can you find the distance though?

if they're the same

[astral pilot]

Incentre is equidistant form all sides

Circumcentre is perpendicular bisector of all sides

You can use these facts

[swift bane]

so if I

drop a height that goes into the circumcenter

[astral pilot]

yeah

[swift bane]

it can be a bisector as well?

[astral pilot]

yeah

[swift bane]

equidistant?

[astral pilot]

equal distance

[swift bane]

I see

[astral pilot]

so, if draw a perpendicular form incentre to all sides perpendiculars will be equal

[swift bane]

where D is diameter

(of the incircle)

[astral pilot]

let's say I is the incentre

O is the circumcenter

Ok?

[swift bane]

uhhhhh

yeah I think i'm following

[swift bane]

d is the distance right

[astral pilot]

yeah

we need to find it

I think I saw a way

[swift bane]

is this even accurate?

the height can be a median as well

and you get that the diameter of the circumcircle is 8

via pythagoras

[astral pilot]

one quick question

Do you know the law of cosines?

[swift bane]

yea

[astral pilot]

then we can solve this

[swift bane]

we can use that there?

oh yeah

why not

[swift bane]

cosa = 6/10 right

[astral pilot]

yeah

[swift bane]

yea but

can't we find the diameter via pythagorean?

[astral pilot]

I don't see how we're going to do that,

[swift bane]

like the circumcircle's

,w a^2 + 36 = 100

[twin meteor]

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[astral pilot]

But you already know the radius don't you?

[swift bane]

oh yea

I found that via herons

ok my bad

🤣

I forgot

what are we trying to find via law of cosines?

[astral pilot]

We were going to derive a formula to find distance

between O and I

[astral pilot]

<BCI = C/2 (why?)

[swift bane]

is that the same triangle as the one I sent?

kind of interesting how u set the big circle

[astral pilot]

Yeah same one

[swift bane]

not sure

[astral pilot]

Incentre is intersection of angle bisectors

[swift bane]

so like

from point B and A