help-26

[tawny axle]

sure?

[storm hearth]

if f(x) = ax+b
f-1 o f(x) = f-1 (ax+b)

Which is same as

x = f-1(ax+b)

Then just let ax+b = 13

No need to find f(x) or f-1(x) bro

f-1(13 is sufficient)

[tawny axle]

yeah

[storm hearth]

Try the method directly

[tawny axle]

Imma use the shortcut then

[storm hearth]

What is f-1 (3x+4) equal to

[tawny axle]

13?

[storm hearth]

Nope, it should be in terms of x

[tawny axle]

isnt f-1 o f(x) Ix

x

[storm hearth]

yes

I(x) is the identity (birim) function, I(x) = x

[tawny axle]

yeah

[storm hearth]

f(2x) = 3x+4 right

Retry

[tawny axle]

yes

2x = 3x + 4?

[storm hearth]

Put both sides of equation with composition to f-1 from left side

[storm hearth]

.

[tawny axle]

f-1(3x + 4) = 2x

[storm hearth]

Yes

To find f-1 (13)
Let 3x+4 = 13

[tawny axle]

yeah

x = 3

[storm hearth]

Then input the found x value to inverse of f

And obtain the value f-1 (13)

[tawny axle]

f-1(13) = 2x

6

thanks

f(x) = y is the same thing as f-1(y) = x right?

[storm hearth]

If function is bijective

[autumn geode]

Could you check my solution ?

[storm hearth]

f-1(y) can actually be a set if function is NOT bijective

Consider f(x) = x²

[autumn geode]

f(x)=3x+4/2
g(x) = 7 - 4x
fog = f(g(x))=
k = 3(7-4x)+4/2
2k = 3(7-4x)+4
2k = 21-12x + 4
2k = 25-12x
k = 25-12x/2
y = (37 - 3x) / 8
x = (37 - 3y) / 8
8x = 37 - 3y
3y = 37 - 8x
y = (37 - 8x) / 3
(fog^-1)^-1(x) = (37 - 8x) / 3

[storm hearth]

f-1 (-1) = {1, -1}

[tawny axle]

is it required to be onto function too?

[storm hearth]

f-1 (y) = { x | f(x)= y for some x in domain of f}

[autumn geode]

??

[storm hearth]

Consists of x points, described

[autumn geode]

isn't it f^-1^-1

[tawny axle]

isnt it 3

I did the long way

[storm hearth]

Bijective means injective and surjective

But f(x) = x² is not injective

[tawny axle]

oh there is no such a thing in Turkish

[storm hearth]

Because f(1) = 1 = f(-1)
But 1 is not -1

There is bro

Its bijektif

[tawny axle]

really?

[storm hearth]

Bijektif = birebir ve örten

[tawny axle]

They dont really use it in questions

[storm hearth]

Yeah i havent got to learn the term until university

[autumn geode]

why would you use that term in hs then

[storm hearth]

Nice to say 1 word only and skip 2

[tawny axle]

for real

[storm hearth]

The standart keeps rising each time

And i believe there are sources that use the term bijective

[tawny axle]

probably

[storm hearth]

Its not some advanced tech just 1 google search away

Im not asking you to learn a new tech that you wont use

Just teaching you a term that will make you think faster

[tawny axle]

It might be useful when Im in university

if I can get into one

[storm hearth]

Also isnt gof-1(13) = g(6) ?

[tawny axle]

why?

wait

isnt it equals to g(3/2)

[storm hearth]

Despite being a math major, im not good at mental math

So i wont waste time on verifying

You def found f-1 (13) wrong or i am tripping hard

[storm hearth]

You found it correctly here

[tawny axle]

I know it makes things harder but isnt f-1(x) (x-4)/3

if we put 13 in

I mean

its f-1(2x)

2x = 13

x= 13/2

[autumn geode]

I found it -67/3 somehow

you can try make sense out of whatever sorcery i made 😭

[tawny axle]

it should be 5/3

no

[storm hearth]

f(x) = 3x/2 + 4 so 2(y-4)/3 =x
and f-1 (x ) = 2(x-4)/3

[tawny axle]

(5/2)/3

[storm hearth]

This is why i told you to not bother finding the inverse of f

High chance to throw

I would suggest doing it easy way then hard way later

[tawny axle]

f-1(3x+4) = 2x

x= 3

so its 6

why was the first method giving me the wring answer

[storm hearth]

You found the inverse wrong

[storm hearth]

Here

[tawny axle]

I didnt convert it to x

[autumn geode]

me?

[tawny axle]

I kept it as 2x

[autumn geode]

i wanna know why i did it wrong

[storm hearth]

Also its getting pretty late, i would recommend not saying overnight for better memories + keeping fraction of what you learn

[tawny axle]

yeah I know

gonna sleep soon

[storm hearth]

Anyway i will dip now, gn

[tawny axle]

gn

thanks for the help

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[formal anvil]

i need help

[cedar wagon]

with what

[formal anvil]

why is 2 a special primenumber

[odd pagoda]

cause its 1+1

[desert atlas]

its the smallest

and the only even prime

[formal anvil]

is 51 a prime number

[desert atlas]

3x17

[formal anvil]

ohh thx

can you give me tipps for primenumbers and primefaktors

[odd pagoda]

what kind of tips do you want to hear

[formal anvil]

how can i now all prime numbers because i forgot all prime numbers from 1 to 100

[outer salmon]

There are infinitely many prime numbers, so you can’t „know“ them all

And there are no fast ways to check if a number is prime

[crude anchor]

You’re asking an age old question my friend

[outer salmon]

Except for small numbers

[odd pagoda]

you can just learn the first few

[crude anchor]

This question has stumped mathematicians since antiquity

[odd pagoda]

its like learning other things

[formal anvil]

thanks guys im not good at maths im at the 8 class

[crude anchor]

There are some tests you can learn for primality but they’re probably above the scope

[formal anvil]

yeah btw thanks for the help

[crude anchor]

If you want to learn more about primes you should look into number theory because that is one of the main questions of number theory

There is something called the miller-Rabin primality test but it will probably not make sense to you without a decent amount of background

So if you want to start somewhere in number theory I would recommend learning about modular arithmetic

You can probably find videos on YouTube or Kahn academy

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[formal anvil]

yess of course

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[radiant ledge]

hey guys
i understood everything except choosing the max part

[radiant ledge]

why do we do the max part?

cuz for epsilon-delta proofs not to infinity, we choose minimum

[mild hearth]

in this case, you need x > M_1 and x > M_2

if x > max{M_1, M_2}, then x > M_1 and x > M_2 as required

[radiant ledge]

is it because x > M
and if we set M1 = M, it could be possible that M1 is smaller than M2 so it wont satisfy both f(x) and g(x)

hence we choose the maximum of the two?

[mild hearth]

yes

if it was instead that you wanted x < M_1 and x < M_2, you would use min(M_1, M_2) to be sure

[radiant ledge]

hmmmm i see i see
cuz M is the threshold and we dont know if M1 or M2 is less
so we choose max

[radiant ledge]

thank you

i get it now

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[spare verge]

Help

[spare verge]

Is this correct? I actually used some unfamiliar method for proving this one

@golden blade

[pseudo sonnet]

still going at it?

[spare verge]

Another one

This one I did it a bit in a hurry since I gtg for my dinner

So l is a bit undefined there but just assumed it’s 0

I have never used this method though

[spare verge]

Knowledge is hard to come by

[golden blade]

This is why you will remain a homomorphism and never become an isomorphism

[spare verge]

What does it mean 😭😭😭

[golden blade]

Can you rate my proof

[spare verge]

What proof 😭😭

[golden blade]

Mine 😭

Be my hero too 😭

[spare verge]

yes be my hero please

And help me with that proof

[golden blade]

😭

[spare verge]

I still have to go to restaurant and my parents are almost gonna call me

And btw it’s my first topo proof though

[golden blade]

wow u are rich

[spare verge]

I mean it’s Christmas

[golden blade]

oh forgor

[spare verge]

That sounds very bad

I know my proof this time must contains a horrible deal of flaws

[golden blade]

Now I got hungry and it's almost 3am

[spare verge]

Go to McDonald’s

[golden blade]

I look at your proof

and

your proofs tend to get even more bizarre

[spare verge]

Like ridiculously wrong

😭😭😭

i will then use another method but I must improve my topology too

[golden blade]

nono

too complex for my brain

[spare verge]

But this is analysis

[golden blade]

I also dont what theorems you use

[golden blade]

so mega true

[spare verge]

For G n converges

I used monotone convergence theorem

Nested compact set property

And close set contains limit point

In essence heine borel theorem since A is bounded and closef its compact

Since dimension it’s still finite so bounded and closed is equivalent to that of a compactness

[golden blade]

so smart

oh now you also introduced big Omega

[spare verge]

Just for the sake of nice looking

[golden blade]

now you wan be real pro

[spare verge]

I am bad at math for real 😭😭

[golden blade]

Here is my attempt

Consider $f_n(x) = \arctan(nx)$ then it is $f'_n(x) > 0$ since $x \in (0, \infty)$ and $n \in \mathbb{N}.\$
Hence $f_n$ increases strictly.$\$
It is also that $\arctan(nx) < \frac{\pi}{2}$ for all $x \in \mathbb{R}$ which implies $f_n$ is bounded.$\$
By the monotony criterion $f_n$ converges.

[thorny flame]

𝔸dωn𝓲²s

[spare verge]

It’s about choose an N with special property right?

[golden blade]

i am not special

[spare verge]

And I also kinda want it to be phrased in topological language so that I can study both uniform convergence and topological at the same time

They are too daunting

[golden blade]

be more daunting

[spare verge]

For this N I ain’t so sure how to choose though which was why I used topological method as my initial attempt

[spare verge]

Then I will fail my Econ theories

[golden blade]

You just want to show it converges?

[spare verge]

Uniformly

[golden blade]

Not that it converges to pi/2

[golden blade]

oh i read to converge

ordinarily

[spare verge]

I was in a hurry so I think I forgot the word uniforms

[golden blade]

I think my proof is sufficient for pointwise convergence

[spare verge]

Yes indeed you took derivative

[golden blade]

and broke my neck doing that

(hehe i didnt really i just know arctan(nx) is strictly increasing)

[spare verge]

Is using topological language rigorous as well

[golden blade]

with the given conditions

[golden blade]

@pseudo sonnet

[pseudo sonnet]

sorry I saw analysis and ran away

[golden blade]

she says this is bad

[spare verge]

In essence we both use the monotone convergence theorem

[spare verge]

It’s a great one

[pseudo sonnet]

yo, I do not remember convergence theorems for sequences of functions

[golden blade]

hey it's still a sequence

[pseudo sonnet]

true but like

[golden blade]

Emmaaaa

I think I am having a breakthrough

[spare verge]

It’s about how to phrase it in topological language too so I can study more efficiently by studying scary uniform convergence and topology

[spare verge]

Yes? I am all ears

[golden blade]

You showed me nothing is impossible

[spare verge]

?😭😭

[golden blade]

🙏🏻 😭

[spare verge]

I am still have half a book of puzzles unsolved

Have

[golden blade]

idc

you even know quantifiers

[spare verge]

You learn that for cal c ap

[golden blade]

real

[spare verge]

Though basic ones

But you learn delta epsilon at high school

[golden blade]

so why are you so obsessed with uniform convergence

[golden blade]

genau

i wanna attend that high school

[spare verge]

No because it’s hard

Otherwise I will fail my coming real analysis 😭

[golden blade]

🙏🏻 😭 God bless kids in hs who dont know linear equations

[spare verge]

Come to USA

[golden blade]

cheeseburgers and fries

[spare verge]

I think I am like average for math at high though

[golden blade]

ohhhhhhhhhhhhhhhhhhhh

the modesty

crazy

[pseudo sonnet]

you're definitely stronger than most

[spare verge]

Really I was average like those taking AP cal C

[golden blade]

If I were to rate your math skills in terms of convergence, you would be uniform convergence 🙏🏻 😭

[spare verge]

That’s unrealistically high rating

I am not even studying math 😭

[golden blade]

[golden blade]

hahahahhahahahahaaha

What have you done today all day

[spare verge]

Like I just want to pass my RA final for next year

😭😭😭

[golden blade]

I understand 🚬

[spare verge]

I don’t even care about grade as long as I pass

[golden blade]

real

as long as you are stacking green later 🔥

[spare verge]

I am preparing it a year in advance

[golden blade]

wow pro

i prepare at most 1 month

[spare verge]

Just because I am bad at math

So I have to be the early bird

[golden blade]

The early bird catches the early worm

[spare verge]

Yes

[pseudo sonnet]

so you will obliterate it when the time comes

[golden blade]

You wannabe isomorphism

[spare verge]

In my dream yes

[pseudo sonnet]

no dreams, only reality

[spare verge]

But I am really just hoping for a pass

Stop my proofs aren’t even rigorous in term of formalism 😭😭

[golden blade]

You are more rigorous on yourself than on your proof so true

[spare verge]

I actually think MCT might not apply unless we assume knowledge of Lebesgue integration

In this case x is positive so fine but this part might not be covered by coursework

But rudins has Lebesgue integral and basic measure so I think fine

[golden blade]

Whats MCT

[spare verge]

Monotone convergence theorem

[golden blade]

how

[spare verge]

Because we have to define for some sequences to be integrable, this case is fine

[spare verge]

X is nonnegative and nice

The simplified version of MCT is like $\int \lim n = \lim \int n$

[thorny flame]

Emmaaaaa

[spare verge]

If it’s a sequence of function it will become something like

$\int\lim f_n d\mu=\lim\int f_n d\mu$

[thorny flame]

Emmaaaaa

[spare verge]

@golden blade like they are different right? I actually don’t know much

[golden blade]

me to

[spare verge]

I’ll just go to the restaurant, I am too dumb

[golden blade]

haha

[spare verge]

And btw just short one, how to rephrase the topo proof to make it actually rigorous I am sure it’s not rigorous

[topaz sinew]

@spare verge Has your question been resolved?

[topaz sinew]

@spare verge Has your question been resolved?

[spare verge]

Not until someone refines the topological language of mine

[golden blade]

⚔️

[spare verge]

Is it that I am thinking things or my help request actually gets less help😭😭😭

[sweet shard]

What even is your question now

[sweet shard]

This?

[spare verge]

Refine my proof I definitely know it’s wrong somewhere

[sweet shard]

That's some tiny handwriting

Uhhh learn latex

[spare verge]

I use latex though but handwriting faster

I want to use topological language so I can notbonly prepare uniform convergence but topology as well since it’s also daunting

[tawdry dagger]

try the ball pen

looks cleaner imo

[pseudo sonnet]

riemann bad eyesight

exposed

@spare verge you know, you can also ask your question in #real-complex-analysis

nobody there would mind

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[wraith zealot]

How to do question 17

[smoky sparrow]

Pythagoras in disguise

[wraith zealot]

but I haven’t learnt it

Is there other method?

Tangents to circles

[smoky sparrow]

You don't know what Pythagoras's theorem is?

[wraith zealot]

I know

[smoky sparrow]

Then you can use it

Diagram sucks but we are drawing a parallel line to AB through D

[wraith zealot]

OH

so the radius is 6

[smoky sparrow]

Correct!

[wraith zealot]

Thx

[smoky sparrow]

Np

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[smoky sparrow]

.close

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[olive charm]

a,b,c are positive real number, find the minimum value of P. Using derivative

Pls help me 😭

[long stirrup]

#help-32 is available

if you can't see it you need to unhide it in the settings idk

[olive charm]

ok thank you

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[cedar minnow]

If you have to find the center point of the quadric x² + y² + z² + 4xz - 6y + 8 = 0, how would you do that? I'm breaking my head over the mixed xz-term... ChatGPT told me you can group (x² + y² +4xz) together and rewrite it as (x + z)² - z². It sounded simple but I've been trying to understand how you can do that for the past half hour and it's driving me nuts. Any help would be appreciated!

[odd pagoda]

dont use gpt, its often wrong

(like here)

at least the idea is roughly correct

you'll want to complete the square

[deep forum]

(x^2+2xz+z^2)+(y^2-6y+9)=2xz+1
(x+z)^2 + (y-3)^2 = 2xz+1

is this approach wrong

[topaz sinew]

@cedar minnow Has your question been resolved?

[cedar minnow]

I thought about this but how do you reason the center point out of this? I get that you can determine that y = 3 is already valid, but from this how can you get that x = 0 and z = 0?

Btw the correct answer is (0,3,0)

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[tawdry dagger]

I have a problem that I got close to solving (I think) but I don't know how to finish it.
The problem is as follows:
There are 22 students that play 5 games of football. In each game, there are two teams of 11 students. If a team wins, every winning student gets 3 points, in case of a draw everybody gets 1 point. if youre on the loosing team you get 0 points.
Every student must be on the same team with all other students at least once.
Show that there are always five or more people with the same number of points.

My thoughts so far are that in case of a draw, everybody gets a point. This means that a draw "doesn't change anything". What I mean by that is if two people have the same number of points, they still will after a draw. This means that if two people have the same number of points, they must have won the same number of times.
Now there are five cases: no draw, 1 draw, 2 draws, ..., 5 draws.
In case 5, everybody has the same number of points.
Case 4: Only one game that is not a draw => 11 people win the same number of times
Case 3: In the first game (that's not a draw), 11 people have the same number of wins, in the "second" game there have to be 6 or more people in the same team that were in the same team previously at least once (can be shown with the pigeon hole theorem)
but then in case of one draw, two draws and no draw I don't know how to show that there are five people with the same number of wins.
Ive been told to draw a graph for each case but I dont know what that would look like

maybe you have an idea?

[topaz sinew]

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[tawdry dagger]

<@&286206848099549185>

[median latch]

how can i help you

[tawdry dagger]

do you have an idea on how to approach this

[median latch]

what is this?

[tawdry dagger]

.

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[cedar minnow]

If you have the quadric given by z = 2xy, how can you determine the axes of symmetry? I thought the following:

• F(x,y,z) = F(-x,-y,z) --> The z-axis is an axis of symmetry
• F(x,y,z) = F(-x,y,-z) --> The y-axis is also an axis of symmetry
• F(x,y,z) = F(x,-y,-z) --> The x-axis is also an axis of symmetry
  But apparently only the z-axis is an axis of symmetry. Why is this?

[topaz sinew]

@cedar minnow Has your question been resolved?

[topaz sinew]

@cedar minnow Has your question been resolved?

[topaz sinew]

@cedar minnow Has your question been resolved?

[mystic jay]

You don't have a function?

What is F(x,y,z)

[cedar minnow]

The quadric z = 2xy

By that I meant that you can substitute (x,y) for (-x,-y) without changing the function so that means that the z-axis is an axis of symmetry right? The same reasoning I applied to the rest

You can substitute (x,z) for (-x,-z) and the equation stays the same, so y-axis is also an axis of symmetry

etc

[thick lily]

I wouldn't describe any axis as an axis of symetry since I consider these to be arbitrary rotations around an axis. For instance z=x^2+y^2 has a z-axis of symmetry. z=2xy I would describe as having the planes y=x and y=-x as planes of symmetry.

but I suppose to give some benefit of the doubt, the z-axis is an axis of rotation by 180 degrees too so maybe I'm being too strict here

[cedar minnow]

Well the exercise I'm trying to make here goes as follows:
"Consider the quadric K: z = 2xy. Which coordinate axes are axes of symmetry? Which aren't? Rotate the axes so that this is the case for all 3 axes. (Matlab!). What does the equation look like after the coordinate transformation?" The answer to the first question is apparently the z-axis.
So the part I'm not sure about is how they determined that the z-axis is the axis of symmetry without plotting a graph and visually determining that it's the case.

[thick lily]

to answer that, the z-axis is all the points (0,0,z)

and these points are not moved by the transformation (x,y) |-> (-x,-y) which is quick to check

well that confirms it but how would we know from the transformation?

you have (x,y,z) |-> (-x,-y,z) so now you can set this equal to itself and solve x=-x, y=-y, z=z and this is what's forcing x=0, y=0

that make sense? I found the fixed points by setting the before and after equal to each other

[topaz sinew]

@cedar minnow Has your question been resolved?

[honest canyon]

-z = 2(-x)y
z = -2xy

the equation does not remain unchanged

y-axis isn’t an axis of symmetry

for z-axis:
z = 2(-x)(-y)
z = 2xy
equation is unchanged so z-axis is an axis of symmetry

likewise, for the x-axis:
-z = 2x(-y)
z = -2xy
x-axis isn’t an axis of symmetry

[topaz sinew]

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[neon iron]

how is the axiom of extensionality stated with FOL + ∈ but without =?

[neon iron]

without equality, it seems the axiom is itself the definition

[topaz sinew]

@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

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[velvet hatch]

Sorry if this isn't your typical help question but for the equation of a parabola in the form (y=ax²), why does the focus equal 1/(4a)? I can't find anything explaining this well online and it seems like the idea of substituting 1/4p (p being the focus) for (a) was just thrown at me with no explanation.

[quick dragon]

I think the best way to develop the intuition behind it is to reconstruct a parabola from the focus/directrix definition and see where the points lie basically

say we have a focus at (0, -1) and directrix y = 1

this is just going to end up being a vertical parabola

so (x,y) is an arbitrary point on that parabola

using the distance formula, the distance from (x,y) to the focus is $$\sqrt{(x-0)^2 + (y-(-1))^2}$$

[thorny flame]

vehnil

[quick dragon]

or $$\sqrt{(x^2 + (y+1)^2}$$

[thorny flame]

vehnil

[quick dragon]

and the distance from (x,y) to the directrix is $$\sqrt{(y-(1))^2}$$

[thorny flame]

vehnil

[quick dragon]

or $$\sqrt{(y-1)^2}$$

[wooden osprey]

if you think of the definition of a parabola in terms of focus and directrix, it makes sense. you know that the parabola is the locus of points equidistant from the focus and the directrix. let P be arbitrary point on the parabola, F = (0, f) be the focus and the line y = -f be the directrix. let D be the image of P on y.

it is clear that for P = (x, y), you have |PF| = |PD|, thus x^2 + (y - f)^2 = (y + f)^2. you shall see if you solve for y

[thorny flame]

vehnil

[velvet hatch]

Ok

[quick dragon]

or yeah more generally for any f, this is good

the last part of what I was saying was basically since you know these distances are defined to be equal in parabolas, you can set them equal to each other and when you solve it, you can flesh out the 1/4a relationship

[velvet hatch]

Ok that helps a lot, thanks

[quick dragon]

experimenting with different focus coordinates and distances between the focus and directrix can help flesh out that intuition if one example isn't enough

[velvet hatch]

Ok, I'll try a few different parabolas

[placid merlin]

heres a picture

hopefully that helps

[quick dragon]

but essentially what happens when you solve out the equidistant equation, y^2 cancels out so you can write y as a quadratic function of x

and the leading coefficient will always be 4( distance between directrix and focus / 2 )

[velvet hatch]

I guess I'll close this channel, I think I'll be able to understand it after looking at all of this and experimenting for a bit.

Thanks

.close

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[pure coral]

So what my way of solving was first choose 1 red and 1 blue from each bag (3c1x2c1)^ 4 and then multiplying it by 12c2 why am I overcounting? why does it matter from where the rest 2 are chosen

[pure coral]

one answer says I have to select balls in one go why is that

[ashen pecan]

My guess is that the way you're counting is over counting number of balls you pick from each box. It says each ball is distinct but not the boxes (semantics ik).

So let's say (2,2,3,3) says you pick 2 balls from box 1, 2 from box 2, 3 from box 3 and 3 from box 4. The way you count treats (2,2,3,3) different from (2,3,2,3) for example

Do you know what the answer is supposed to be

[coarse herald]

21816

this recently came in my test

[topaz sinew]

@pure coral Has your question been resolved?

[pure coral]

yeah

[coarse herald]

cases will be repeated here, if you label the balls Ar1, Ar2, Ar3... there is a possibility you first get Ar1, Ab1 then Ar2 there is also a possiblity you get Ar2, Ab1 then Ar1

in both the cases the same balls may be chosen

[pure coral]

like this?

[coarse herald]

Ar1 is a red ball from the first box

[pure coral]

.close

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[cosmic coral]

I need help fixing my proof since I don’t think it’s solid. I’m not really that good at proofwriting yet since I self-study, so please be patient with me.

[topaz sinew]

@cosmic coral Has your question been resolved?

[crimson nexus]

The issue with your proof of the direct implication (=>) is that you cannot take X to be the open neighborhood automatically, if f is continous then indeed you know that every point x has a neighborhood on which f is continuous however you have no control over this specific neighborhood so you cannot assume anything about it

[cosmic coral]

so it follows trivially?

[crimson nexus]

the direct implication is because the function is continuous each restriction is continuous

the converse seems right too

[cosmic coral]

I was less confident in the converse argument than the direct argument

[pseudo jetty]

this is not a correct statement

you have full control over the neighbourhood, since that is what you are required to prove exists

the existence is witnesses by X

[cosmic coral]

does that make my direct implication correct?

[pseudo jetty]

yes

[crimson nexus]

I don't think I get it, if you assume continuity and take x in X you know that there is Ux on which f is C0 but you don't have any control over Ux, you just know it exists

in the sense that you can't know if Ux = X

[pseudo jetty]

you've got it completely wrong

the assumption is that f is continuous

what you must PROVE is that there is a neighbourhood Ux around every point x where f is continuous

take Ux = X

[crimson nexus]

oh yeah right I don't know how I got this confused

so yes it really is immediate

[pseudo jetty]

the converse direction is the right idea but the notation is confused

[cosmic coral]

which

[pseudo jetty]

i can't tell what you've written here first of all

[crimson nexus]

it's the restriction of f to the union of those sets

but the Vx seems out of place right

[pseudo jetty]

oh lord okay that's a union symbol not a U

no the Vx is fine being there

[cosmic coral]

$f|{\bigcup{x\in f^{-1}(U)}V_x}$

[thorny flame]

sushi

[pseudo jetty]

but this is a big leap of faith

[cosmic coral]

why

[pseudo jetty]

do you have a theorem somewhere telling you that if f|Vx is continuous for all Vx, then f|their union is still continuous

[crimson nexus]

oh the Vx is also in the union

[crimson nexus]

I think that using the topological definition of continuity (open sets) is the right idea here

and you started in this direction

[cosmic coral]

I mean, their union is still an open neighborhood

[pseudo jetty]

yeah but you only know continuity on each one

i'm not saying it's wrong

it's just the statement itself requires a bit more thought than that

[cosmic coral]

but since the restriction of any open neighborhood is continuous and since their union is an open neighborhood, shouldn’t it be continuous?

[pseudo jetty]

no that's an incomplete justification

the restriction of a continuous function to any subspace is continuous

not just open ones

[cosmic coral]

but is the justification wrong?

[pseudo jetty]

it's incomplete

[cosmic coral]

what is the proof for this

[pseudo jetty]

it's by definition of the subspace topology

(f|S)^-1(V) = f^-1(V) intersect S

[cosmic coral]

oh I see

[pseudo jetty]

and the open sets of a subspace S are by definition opens intersect S

this is what you need to check essentially

that preimages of opens are open

[pseudo jetty]

which you haven't here

[cosmic coral]

yes

[pseudo jetty]

anyhow

you were looking to show that f^-1(U) is open

what you've got is quite fine

but now you need to say (f|Vx)^-1(U) is open in Vx

and since Vx is open in X, (f|Vx)^-1(U) also open in X

again, by the subspace topology

f^-1(U) = U_x (f|Vx)^-1(U), a union of opens in X

and so is also an open in X

essentially the same as what you've done, but slightly clearer

[cosmic coral]

alright

[pseudo jetty]

and dodging the need to prove that f restricted to the union is continuous

[crimson nexus]

here it works sushi because it's a union of open sets so it's open but it's a property of open sets, be careful with unions of closed sets for instance

infinite unions at least

[pseudo jetty]

exactly as deph says

[cosmic coral]

thank you

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[indigo willow]

this exercise is monstruous, please try to do it on paper if you're willing to help

[indigo willow]

i was thinking to basically say that there J=int(-a,a) of f(-x) dx and try to somehow add I+J so that i get something resembling the first equation

but to do that id need to write something like 1/alfa times the I integral, so that i got alfa f(x)

so it just doesnt work

if any of you got any ideas, im listening

[surreal mural]

what is $\int_{-a}^{a} f(-x) dx$, can you write it in terms of $I$?

[thorny flame]

rbit

[indigo willow]

no, i was thinking to call that function J

[surreal mural]

there's a simple way to get rid of the sign

[indigo willow]

how

[surreal mural]

u = -x

dx = -du

[indigo willow]

alright sure, but how would that help me

the problem of figuring out what I is, is that i gotta use that first equation

[surreal mural]

$= -\int_{a}^{-a} f(u) du$

[thorny flame]

rbit

[indigo willow]

ok and?

[surreal mural]

can you simplify?

[indigo willow]

simplify what?

the minus can change those a and -a

but again, i dont see how any of this helps

[surreal mural]

right, so just

$=\int_{-a}^{a} f(u) du$

[thorny flame]

rbit

[surreal mural]

but isn't this just I again?

[indigo willow]

exactly

oh wait

so you're saying that int fx = int f(-x)?

[surreal mural]

yes, but only on that interval

[indigo willow]

of course

so now what about alfa and beta?

[surreal mural]

now integrate $\alpha f(x) + \beta f(-x) = \gamma$ and solve for $I$

[thorny flame]

rbit

[indigo willow]

it was that easy?

thats insane

you are a wizard

i got another wierd exercise if you're up for it

[surreal mural]

yeah

[indigo willow]

sure 1 sec

2.1049

its not in english, but all you gotta know is that limit

[surreal mural]

ok let's see

my first thought would be to either let u = tx or do a step of integration by parts and see what happens

[indigo willow]

i was thinking to integrate by parts as well

but, i just didnt know what to do with int f'(x)sintx afterwards

[surreal mural]

oh you don't know if f can even be differentiated

so you need to integrate f instead

[indigo willow]

no no, it says in the problem it can

it said that the derivative is continuous

[surreal mural]

$\lim_{t \to \infty} \left(\frac{1}{t} \left[f(x) \sin(tx)\right]{a}^{b} - \frac{1}{t} \int{a}^{b} f'(x) \sin(tx) dx\right)$

[indigo willow]

yup

[surreal mural]

I thinkk this should be right

[indigo willow]

thats it

[thorny flame]

rbit

[surreal mural]

just intuitively, what should happen now when t goes to infinity?

[indigo willow]

i mean, the first thing from the parenthesis should be 0

right?

[surreal mural]

it should yeah

[indigo willow]

ok, so we can get rid of that

[surreal mural]

problem is the sin(tx) inside of both

but there is one property of sin that let's us justify it

[indigo willow]

well, the one outside the integral, we can deal with it easily

[indigo willow]

yea -1<=sintx<=1

and then devide by t

and the limits of the bounds are 0 therefore sin tx/t is 0

[surreal mural]

right

we can also see that $-\int_a^b f'(x)dx \leq \int_a^b f'(x) \sin(tx)dx \leq \int_a^b f'(x)dx$

[thorny flame]

rbit

[indigo willow]

holy moly

thats pretty cool

ok i see

im so impressed by the fact that this is lightwork for you

[surreal mural]

just seen all these tricks before

[indigo willow]

listen, i dont wanna push it too much, but i got a few more exercises that i dont understand

these are university entrance exam for computer science btw

and some of these are chinese for me

so thats why im like flabberghasted rn

[indigo willow]

so if you wanna help more id be so grateful

[surreal mural]

just make a new help channel at this point

I would

[indigo willow]

amazing, im making another help channel then

you better be there to help me

.close

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[indigo willow]

i need help with this exercise, i cant figure it out:

[smoky sparrow]

For 0 < x < 1, x^n tends to what as n goes to infinity?

[indigo willow]

idk, thats the exercise

i assume that it goes to infinity as well

[indigo willow]

so yea lets assume it goes to infinity

[surreal mural]

f is continuous and defined on a closed interval, what does that tell us?

[indigo willow]

ummm, are you talking about Rolles formula?

[surreal mural]

nono

[indigo willow]

it doesnt tell me much then

[surreal mural]

or maybe let me approach this from a different direction

$\int_0^1 x^n dx$

[thorny flame]

rbit

[surreal mural]

do you know what the limit of this could be?

[indigo willow]

0

[surreal mural]

right

so the only thing that gets in our way is the f(x)

[indigo willow]

yes

[surreal mural]

but can f(x) grow uncontrollably?

[indigo willow]

its constant right?

[surreal mural]

not per se

[crimson nexus]

Just to make sure because it can be done quickly if you know some measure theory, are you familiar with lebesgue stuff like dominated convergence ?

[smoky sparrow]

https://math.stackexchange.com/questions/3964939/pointwise-convergence-in-integral#3964943

But f(x) has to be bounded

That's the key idea

[indigo willow]

wow, im bombarded

let me check what yall are saying

[indigo willow]

keep talking rbit btw

[surreal mural]

but that's just skipping to the solution

[indigo willow]

yup, that s exactly the solution at the end of the book

[indigo willow]

yea, there must be a more methodical way to solve it

so teach me rbit

[surreal mural]

yeah as you already figured out, integral of x^n goes to 0

[indigo willow]

yessssss

hype

[surreal mural]

and the question is whether multiplying by the f(x) term really matters in the end

[indigo willow]

mhm

so i assume it doesnt

why doesnt it tho

[surreal mural]

it would matter if, say, f(x) = e^(1/x)

[indigo willow]

what the phaaaaa

[surreal mural]

this function will grow really large, outweighing anything that x^n can do

[indigo willow]

so the xth root of e

[surreal mural]

but remember, our function f must be defined on [0, 1]

[indigo willow]

mmmmmmmmmmmmm

interesting

ok

[surreal mural]

and e^(1/x) is not defined at x=0

[indigo willow]

ok but, i feel like we kinda pulled the xth root of e out of our asses

[surreal mural]

oh that was just an example

you can ignore it for the actual solution

[crimson nexus]

The idea is that if your function has too big of a growth it could overpower the x^n

that was such an example

[surreal mural]

just goes to show why it's important for our function to be defined on [0, 1]

[indigo willow]

okk

[surreal mural]

so maybe you happen to know of any theorems for continuous functions on closed intervals that could help

[crimson nexus]

my bad asking if you knew lebesgue dominated convergence theorem btw I didn't see your pre uni tag

[indigo willow]

yea i just thought you were making up words so its chill

[indigo willow]

no clue

[surreal mural]

ok it's that f(x) must have a minimum and a maximum on [0, 1]

[crimson nexus]

a function continuous on a compact is bounded and attains its bounds

[indigo willow]

oh wait

is it like Weierstrass's formula?

is that why you were saying its convergent

[surreal mural]

which formula

[indigo willow]

oki makes sense

[crimson nexus]

I think that you are referring to the bolzano weierstrass theorem right ?

It's used to prove this result

every bounded sequence has a convergent subsequence

[indigo willow]

yah

[crimson nexus]

in R

[crimson nexus]

yeah you're on the right track

well you don't have to prove the theorem again obviously

[indigo willow]

mhm

[crimson nexus]

just use the result

[indigo willow]

ok so i know that m<f(x)<M

right?

[surreal mural]

yeah

[indigo willow]

finallhy

[crimson nexus]

yep hence |f| <= max(m,M) = M'

[surreal mural]

but it shall be repeated it only works because f is on [0, 1], it wouldn't work if it was, say, (0, 1)

[crimson nexus]

continuity is also the keyword here

[indigo willow]

you wizards are waving your wands a lil too fast

[crimson nexus]

the reflex really is : continuous on a closed and bounded set (so here just [0;1]) => Function is bounded

[indigo willow]

so m<f(x) < M on any [a,b] interval, but [0,1] is important in our specific case

[crimson nexus]

[0;1] for the x^n

[surreal mural]

yeah if continuous

[indigo willow]

it couldnt be more continuous

so we are safe

ok, now what

[crimson nexus]

yep but really we only care about the continuity of f because that's the one we want to bound

[crimson nexus]

well you bound your integral above and below, and you use the squeeze theorem for instance

[indigo willow]

mmm

ok i get it

how old are you guys btw

[crimson nexus]

20

[indigo willow]

cool

im 19

[crimson nexus]

but you should thank rbit they did most of the explaining

have fun with math 🙂

[indigo willow]

i do thank him, hes helped me like an hour ago too

ight, cya

.close

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[indigo willow]

.close

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[vocal sorrel]

Is the series addition property valid if and only if both sequences converge?

[odd pagoda]

which property do you mean? that the sum of the two series converges to the sum of the limits of the two series?

it can also happen that both series diverge but the sum converges

see if you can come up with an example

[vocal sorrel]

Property number 2 for example

Is it only true if and only if both series converge?

[odd pagoda]

.

sum a_n, sum b_n both diverge but sum (a_n+b_n) converges

[vocal sorrel]

And how will we know, before using the identity?

What if one series diverges and the other converges

Can we also use it then?

Can we use it on all cases?

$\sum_{n=1}^{\infty}(a_{n} \pm b_{n})= \sum_{n=1}^{\infty}a_{n} \pm \sum_{n=1}^{\infty}b_{n}$

Why ś bruh

[odd pagoda]

no, if one converges and the other diverges then the sum also diverges

\pm for +-

it cannot happen that of the three series exactly two converge. either all of them do or at most one does

[vocal sorrel]

So does that mean we can use this identity every time?

$\sum_{n=1}^{\infty}(a_{n} \pm b_{n})= \sum_{n=1}^{\infty}a_{n} \pm \sum_{n=1}^{\infty}b_{n}$

[thorny flame]

xtra

[wispy pier]

What do you mean by "A series is convergent" ?

The sum of the series approaches a constant

So yes we can say, if two series are convergent, then their product, addition , subtraction will also be constant

Hence the new series will also be constant.

Is this understandable?

[vocal sorrel]

Yes

[topaz sinew]

@vocal sorrel Has your question been resolved?

[vocal sorrel]

My question remains tho...

[topaz sinew]

@vocal sorrel Has your question been resolved?

[topaz sinew]

@vocal sorrel Has your question been resolved?

[quick dragon]

No the statement doesn’t work in both directions

If series A and B are convergent then a series formed by their element wise sum, difference, product, or quotient (assuming the divisors aren’t 0) is also convergent

However that doesn’t mean the converse is true as well

Just because a series formed by the element wise sum, difference, product, or quotient is convergent does not mean both individual series are convergent

See if you can find a counter example to bolster the intuition for that

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[lunar nymph]

how can you rigorously prove that x^2is not a one to one function

[lunar nymph]

beside the horizontal line test

[odd pagoda]

1^2=(-1)^2

[hazy pumice]

you only need a single counterexample to rigorously disprovee something

[odd pagoda]

you just need to give a single example that shows that two inputs map to the same output

[lunar nymph]

thank you guys

close.

[odd pagoda]

.close

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[normal flint]

$$
\sum_{j=1}^{\infty}\left(1+\frac{1}{j}\right)^{j^2} \frac{1}{3^j}
$$

[thorny flame]

aName

[normal flint]

can anyone tell me whether this means?

[olive ridge]

wdym?

[normal flint]

my latex skills aren't that great but basically about the exponent of the first term: can it be simplified to j*j or 2j?

or rather which one of them is the right simplification in this context?

[pseudo sonnet]

the latter

,, (a^{m})^{n} = a^{mn}

[thorny flame]

OurFallenStars

[normal flint]

this is what it looks like here. So I can just put the extra parentheses there?

[earnest herald]

this is the actual question??

[normal flint]

I sent a screenshot of the problem here

[earnest herald]

ohhh okay

[normal flint]

I'm supposed to check for convergence

[earnest herald]

😭 how

[normal flint]

trying the quotient criterium right now

idk if it'll do anything

don't even know if I even understood the problem the right way because of the ambiguous parentheses

[chilly walrus]

recall that $\left(1+\frac{1}{j}\right)^{j}$ increases monotonically and is bounded by $e$

[thorny flame]

artemetra

[chilly walrus]

and $\left(1+\frac{1}{j}\right)^{j^{2}} = \left(\left(1+\frac{1}{j}\right)^{j}\right)^{j}$

[thorny flame]

artemetra

[chilly walrus]

can you see a sum comparison you can do from here? @normal flint

[normal flint]

@chilly walrus lim \left(\left(1+\frac{1}{j}\right)^{j}\right)^{j}$ = $e^j$ ?

[thorny flame]

aName
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[normal flint]

so then this would be wrong?

nvm this is the quotient

since it limit is 1/3 it should mean convergence

but e^j would not converge and the series would diverge even more

[chilly walrus]

sorta

[chilly walrus]

not quite

you need to do a comparison

[chilly walrus]

what i am trying to say by this is

$\left(1+\frac{1}{j}\right)^{j} < e$

[thorny flame]

artemetra

[chilly walrus]

and this holds for all j

[normal flint]

yes

[chilly walrus]

hence
$$\left(1+\frac{1}{j}\right)^{j^2} < e^j$$

[thorny flame]

artemetra

[chilly walrus]

so

[normal flint]

so the same goes for the series

[chilly walrus]

yep

[normal flint]

but wouldn't that then be a divergent majorant?

[chilly walrus]

nope and there's a very good reason why

[chilly walrus]

remember about the 1/3^j term

$\sum_{j=1}^{\infty}\left(1+\frac{1}{j}\right)^{j^2} \frac{1}{3^j} < \sum_{j=1}^{\infty} e^j \cdot \frac{1}{3^j}$

[thorny flame]

artemetra

[chilly walrus]

the series on the right ||conv||erges, because ||e ≈ 2.71828 < 3||

[normal flint]

sorry how do we know the right series converges?

ohhh

🤦

[high portal]

it is because (e/3) < 1

[normal flint]

yes

that makes a lot of sense

just saw that

so obviously that way of finding a convergent majorant is a lot more elegant... is what I did with the quotient criterium also okay?

[chilly walrus]

,w \sum_{j=1}^{\infty}\left(1+\frac{1}{j}\right)^{j^2} \frac{1}{3^j}

[thorny flame]

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Requested by arte_m_etra

[chilly walrus]

👁️

[normal flint]

🫠

[chilly walrus]

yeah idk what happened there

,w \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{n^2} \frac{1}{3^n}

[thorny flame]

Results provided by WolframAlpha

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

Requested by arte_m_etra

[chilly walrus]

tf?

lmao

it converges tho

dw

[chilly walrus]

this is correct

[normal flint]

how is this still saying it's diverging

that is so confusing

[chilly walrus]

correct and from here you need to take that term and take it to infinity

[normal flint]

yeah it makes a lot of sense the way you explained it

[chilly walrus]

and you'll still get a finite number which will mean the limit will converge

but in order to get the limit you'll still have to use the fact that the limit of (1+1/n)^n is e so ig you can't avoid that

[normal flint]

thank you so much!

[normal flint]

yeah I keep forgetting about recognizing series I already know 🙈

[chilly walrus]

ohh i get it, it accidentally interpreted the 1/3^n as being outside

,w \sum_{n=1}^{\infty}\left(\left(1+\frac{1}{n}\right)^{n^2} \frac{1}{3^n}\right)

[thorny flame]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by arte_m_etra

[chilly walrus]

yep

[normal flint]

🙌

cool

thank you!! this really helped a lot

[chilly walrus]

no worries :) if you are done, type ".close"

[normal flint]

.close

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[worn bison]

The drawing below represents a model of the stained glass windows in a church that need to be restored. A special glass paint will be used, which costs R$80.00 per can and each can covers approximately 100 cm2 of area. The dark area in the figure is the part of the stained glass window that needs to be painted with the special paint. Considering R = 40 cm and r = 10 cm respectively as the radii of the largest and smallest circles, and that the inscribed triangle is equilateral and the inscribed hexagon is regular, the approximate amount that will be spent on the restoration of each stained glass window is:
pi = 3
sqrt(3) = 1,7

i need help

[worn bison]

ok nvm i think i can just do it

.close

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[round pagoda]

can somebody check my answer:

[neon iron]

where did u get x from

remove it

it should just be 3u = 1

solve for u

[round pagoda]

u = 1/3?

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@round pagoda Has your question been resolved?

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[magic atlas]

Can someone explain to me what's wrong with this argument?

[magic atlas]

P is meant to represent a point on a Unit circle

and I tried to reason that the transformation from a rotation by pi/3 followed by reflection by y = x and then a roration by -pi/3 is equivalent to a reflection by y = x.

But this is wrong

They are not the same thing

Why is this wrong?

[torpid sparrow]

Last part looks wrong

You cant subtract -pi/3 after reflection

[magic atlas]

why?

[pseudo horizon]

you need to find the angle of P'' and subtract pi/3 from it

the angle isn't theta+pi/3

[magic atlas]

is the theta before reflection not the same as theta after reflection?

[pseudo horizon]

the variable theta is the same

but i'm talking about the angle OP'' makes with the positive x-axis (O is the origin)

moving counter-clockwise from the positive x-axis

[magic atlas]

Wait so what does theta represent after the reflection

[pseudo horizon]

the angle of P with the x-axis

the same as it did in the beginning

[magic atlas]

But i just cant change that angle anymore after I reflect it

?

[pseudo horizon]

the angle you want to change is the angle of P''

[magic atlas]

So the final answer would not be in terms of theta?

Or is there a way to relate it to theta

[sonic escarp]

if you wanna rotate by an angle y you need to express the point (here: P'') in a (cos x, sin x) type.

[pseudo horizon]

This should rotate (x,y)\
$(x \cos\theta - y\sin\theta , x\sin\theta + y\cos\theta )$

[thorny flame]

Axe

[magic atlas]

is what ThM said correct?

[pseudo horizon]

yeah

[magic atlas]

Wait but then, instead of switching cos and sin when reflecting by y = x. I let substitute new angle = pi/2 - theta

[sonic escarp]

exactly.

[magic atlas]

That should have the same affect as switching cos and sin but Im keeping it in form (cos, sin)

but if you do taht you get the same mistake as I did originally

[sonic escarp]

why?

[magic atlas]

We are talking about something like this right?

oops

I did it wrong

[torpid sparrow]

[magic atlas]

replace sin with cos

[torpid sparrow]

That is no replacing

[sonic escarp]

theta -> (rotation) theta+pi/3 -> (reflection) pi/2-(theta+pi/3) -> (reverse rotation) pi/2 -(theta+pi/3)-pi/3

[torpid sparrow]

It is switching

[pseudo horizon]

$P''=\left(\cos\left(\theta + \frac{\pi}{3} - \frac{\pi}{2}\right),\sin\left(\theta + \frac{\pi}{3} + \frac{\pi}{2}\right)\right)$

[torpid sparrow]

✅

[thorny flame]

Axe

[magic atlas]

I am confused, how have you gotten this?

[pseudo horizon]

using the formula:

$\sin(\theta)=\cos\left(\theta-\frac{\pi}{2}\right)$

[thorny flame]

Axe

[magic atlas]

oh

Yeah

My bad

sorry about that, wasnt thinking properly

[pseudo horizon]

no worries

[magic atlas]

Shouldn't your sin term be pi/2 - theta - pi/3 ?

[pseudo horizon]

it amounts to the same thing

[magic atlas]

Oh

Alright that makes a whole lot of sense

So you can only rotate the angle when you have cos in the x term

and sin in the y term

[pseudo horizon]

i don't know if that actually works

[magic atlas]

wait what

okay let me check some stuff then

and I will see if it works

[pseudo horizon]

i think it should work but i'm getting weird results, i must have made a mistake somewhere...

[magic atlas]

what do you get

[pseudo horizon]

it's rotating in the opposite direction

i think it doesn't work because the angles inside the cos() and sin() aren't the same

[pseudo horizon]

this should be the true way

[magic atlas]

hmm, alright

[sonic escarp]

whats about this?

[magic atlas]

So in order to rotate by an angle, it needs to be in (cos, sin) form and they need to have the same angle?

[magic atlas]

that's what we tried I believe

[sonic escarp]

is this a issue of believing?

[pseudo horizon]

that seems to work

[magic atlas]

oh, I thought that was what we were doing though

[pseudo horizon]

i used a different transformation to get (cos, sin) form

[sonic escarp]

a point on a unit circle is determined by its angle. then do the three operations (rotate, reflection, reverse rotate) and youre done.

[magic atlas]

But you get the same form at the reflection stage

[pseudo horizon]

no i got this

[sonic escarp]

you didnt to the last rotation.

[pseudo horizon]

yeah because the last rotation doesn't work on that one

[magic atlas]

so if it doesnt work on that then it doesnt work on this

Cause they are the same thing

[sonic escarp]

why not?

[pseudo horizon]

i don't know

[sonic escarp]

if you can rotate by adding +pi/3, you should be able to do the reverse rotation by adding -pi/3.

[pseudo horizon]

i tried it in desmos, it's not the same

https://www.desmos.com/calculator/hvg1uqtbcc

[magic atlas]

Also, if you do that method

It simplifies down to (sin, cos)

which is the reflecion by y=x

[sonic escarp]

why do you think it does not work?

or: what do you mean with "its nor the same"?

[pseudo horizon]

because the green point and the purple point aren't the same

the purple point uses your formula and the green point uses mine

huh?

at the reflection stage, i think they're the same point but the arguments aren't the same, so when you add something to the arguments it has a different effect

like sin(0)=sin(pi) but sin(0+k) doesn't equal sin(pi+k)

[sonic escarp]

lets write in degrees
starting with an angle 0 -> rotate by 60 -> angle 60, reflection by y = x -> angle 30, rotate by -60 -> angle -30. -> purple point

[pseudo horizon]

ok

[magic atlas]

I think the purple point is right.

[pseudo horizon]

it is

[sonic escarp]

you did:
starting with an angle 0 -> rotate by 60 -> angle 60, reflection by y = x -> angle 150 (which has the same cos), rotate by -60 -> angle 90 -> green point

[pseudo horizon]

not 150 but 330

for the cos

150 for the sin

i was only thinking to preserve the equality but your method of subtracting the angle from pi/2 actually preserves the angle

[sonic escarp]

no need for sin/cos beacause of this.