help-27

[sinful sand]

wai

[vernal monolith]

merhaba

[sinful sand]

nvm

ya turkish

?

anyways

[vernal monolith]

i am

[sinful sand]

ye i see it

but can we that 140

140+40=180

make that 140

[vernal monolith]

140 ı neden buldun

110 ve 40 derece var soruda

[sinful sand]

yardımcı paralel kenar hayali çizgi yani

[faint gorge]

X. E. on that

[vernal monolith]

paralelkenarı nereden oluşturdun? bu soru için paralelkenar oluşturmana gerek yok çünküü

not X.E. on this

[sinful sand]

k

[vernal monolith]

you just extend the parallel line

[sinful sand]

maybe

we could make a rule

roket yada kalemucu kuralı

[vernal monolith]

you mean use those that you typed?

[sinful sand]

ortaokulda bunu öğretiyorlar

cyrenux dm

[vernal monolith]

no, i would rather have it here

[sinful sand]

to talk turkish

actually

we cant talk turkish here

easily

anyways

lets contunine

look at it again

[vernal monolith]

you are allowed to talk foreign languages here as its allowed for educative purposes, just not in public channels like discussion

[sinful sand]

tamam

şimdi napıcaz

[vernal monolith]

soruyu çözmenin her yolunu mu arıyorsun?

yoksa ilk çözümünü elde ettin mi

[sinful sand]

Z kuralını

düşündümde

olmadı

3 açı beliriyor

M kural olabilir

diye düşünüyorum

110 ile 40 karşı çaprazı eşit olamazmı

[vernal monolith]

l paralel doğrusunu uzatıp açıları 110, 40 ve 30 olan bir üçgen elde edebilirsin

[sinful sand]

çizermisin

rica edersem

?

[vernal monolith]

yanımda çizecek bir eşya yok, ve serverin yardım anlayışına aykırı, helperların görevi helpee yi soru çözmek için guidelamak, soruyu onlar için çözmek değil

[sinful sand]

L tarafının üstünden mi altın dan mı çizeceğiz

[vernal monolith]

altına, aynı doğrultuda olacak

[sinful sand]

40 derecelik açının yanındanmı

@vernal monolith

cevap 110 mu

[vernal monolith]

110+40 derece

[sinful sand]

k ile l arasından

paralel çizsek

hm

neyse

.close

[devout snow]

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[devout snow]

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[simple grotto]

Okay, given a view point, a view plane, and an object in ND space defined by minimums and maximums extruded in all dimensions from minimum to maximum, projecting to the view plane as goes exactly through it to the view point from the shape all in Euclidean, Spherical, or Hyperboloid geometries, not mixed geometries, assuming we only see one like 3d, per such shape being projected, what is the maximum number of vertices, "planes", and maybe "shapes" needed to represent if I go by pixel in not Euclidean instead of triangles?

[simple grotto]

Should I just test it?

[faint gorge]

Yes X E.

[simple grotto]

Okay, anyone know enough to conjecture?

I am very inactive huh?

[devout snow]

@simple grotto Has your question been resolved?

[simple grotto]

This channel was hidden for a time until that bot. X E.

Also, if I only rotate in one not current 3d at a time then the max dimension of output is 4d?

[simple grotto]

I am mostly certain of this. X E.

[untold ravine]

can i ask what X E means

[simple grotto]

And it generalizes with more rotations. X E.

[simple grotto]

In bio no links. X E.

Just me being weird. X E.

[untold ravine]

cool

[simple grotto]

Also, cool, another Godot-er. X E.

I think this generalizes to if there is only travel in ND the results can only be of that ND regardless if there are more dimensions. X E.

So in short we can consider a 590d like 3d as long as our view is in that 3d. X E.

Smart 4d+, here I come. X E.

Or any arbitrary thing to some arbitrary number of dimensions. X E.

[simple grotto]

Is this too high level?

<@&286206848099549185> is this too high level, should I just test myself?

ND is any dimensional. X E.

Maybe not Josiah. X E.

Given a view point, a view plane, and an object in ND space defined by minimums and maximums extruded in all dimensions from minimum to maximum, projecting to the view plane as goes exactly through it to the view point from the shape all in Euclidean, Spherical, or Hyperboloid geometries. It is not mixed geometries. Assuming we only see one like 3d, per such shape being projected, what is the maximum number of vertices, "planes", and maybe "shapes" needed to represent? I may go by pixel in not Euclidean instead of triangles. X E.

[vernal monolith]

X.E. is a life style for us, its nowhere weird

[simple grotto]

I almost got a domain called moxe.plus. X E.

[simple grotto]

Should I just test it and stop bothering you?

[faint zinc]

@simple grotto if your plane is curved you can intersect a convex object in more than one "slice," so a projection may be disjoint

[simple grotto]

By the way, why is OmnipotentEntity not famous with their own 4d+, you know everything?

[faint zinc]

The audience for a 4d game is rather niche

I did publish a game with a few friends that did reasonably well, but it didn't have anything to do with higher dimensional geometry. I just like it

[simple grotto]

At least in spherical it works by the theory of rotations per amount and it being defined from one amount to another. X E.

Or I think it does. X E.

[faint zinc]

I'm not sure what you mean by this

[simple grotto]

Projection to any one point is a set of rotations in planes. If the object is defined as from a rotation to a rotation then this should be like the Euclidean case. X E.

[simple grotto]

Disjoint how, can I define the curve of slices?

[faint zinc]

Well. Ok I'm doing a little bit of sleight of hand here.

Because I specifically said curved plane and convex shape

If a shape is convex with respect to the metric then I don't think it's possible

But then again, there can be multiple paths in curved space that are equivalent, so convex might not be well defined in this case

Double check on that

[simple grotto]

There should be a way to mathematically define this even if it is like algebra. X E.

[simple grotto]

True but a straight line can only go two ways to any one point from one point. X E.

[faint zinc]

But that being acknowledged, what I mean by disjoint is that one object when sliced by a volume can create two disjoint areas where the slice occurs

This about it as sticking two fingers into water with the fingers separated, if we only consider the intersection of your fingers and the surface of the water, we have two disjoint regions

[simple grotto]

Seems like it should be possible to make a shell with outer parts as long as it is a solid n-cuboid. X E.

So with the disjoint parts, still all definable, max vertices, planes, and shapes needed?

[faint zinc]

For an arbitrary curved plane and n-cuboid? Probably no bound.

Or I guess curved intersecting n-volume

Note that intersection is different from projection

[simple grotto]

I guess we have to define that only a specific x and y on plane are viewable because otherwise we may have a circle. X E.

Like from min to max. X E.

If it were infinite or max sized it would have a maximum of the number of dimensions moving in dimensions. X E.

But yeah, if we have two possible opposite ways on a sphere then one object is in view twice. X E.

Possible to be disjoint that way but no other way I know of. X E.

[faint zinc]

With a hyperbolic surface you could intersect as many times as you cared to

I believe.

And if it's heterogenous then all bets are off

[simple grotto]

Nice to know. X E.

[faint zinc]

Like, how much have you played with hyperbolic spaces, actually?

[simple grotto]

None, just asked about them. X E.

[faint zinc]

It might help to get an idea how these spaces mess with ideas about straight lines, direction, etc. Try solving the sokoban puzzles here: https://sokyokuban.com/#0

[simple grotto]

Okay, tried that. X E.

As far as I can tell it is like Minecraft, infinite but you can move back with rotation. X E.

I still think that if defined correctly there is no disjoint. X E.

[faint zinc]

Essentially, the embedding of hyperbolic space into euclidean space is via a hyperbola. If you can imagine a hyperbolic 4d space as being a 4d hyperbola embedded in a 5d space, then you can treat your 3d volume as a plane that wraps as many times as you care around the hyperbola, perhaps self intersecting but not necessarily, and each time around intersecting the n-cuboid in question 1 or 2 times.

Each of these intersections, when "unwound" and projected into regular 3d euclidean space for rendering and later display would give one disjoint region

You can imagine this as roughly analogous to a line traveling along a hyperboloid in 3d space. And naturally in 4d you have two independent rotations axes, meaning that you can have two independent curvatures.

[simple grotto]

I thought hyperboloid meant a 2d hyperbola rotated around in 3d and then 4d is another rotation about a 4th axis in 5d. X E.

[faint zinc]

Are you aware of the idea of duoprisms?

[simple grotto]

Sort of. X E.

[faint zinc]

Essentially, you can construct a duoprism of two hyperbolas, that's 4d, and then rotate this to get to 5d.

[simple grotto]

Glad I am just doing spherical now. X E.

Oh, different. X E.

Thanks for talking. X E.

Well, other things to do now. X E.

Anything more before I close this?

I guess not. X E.

.close X E.

[devout snow]

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[devout snow]

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[marble hearth]

Where am I going wrong here?

[marble hearth]

There should be no solutions

[uncut crow]

one little algebra before we get into the logic

x - 24x = -23x

you have a 23x there though

fixing that, the bottom line should say x = 4

now what you have really shown is that for every x not equal to 4, x is not a solution

you can check that x = 4 does not make sense as a solution (because e.g. 1/(x-4) does not parse)

so yes there are no solutions

[marble hearth]

but aren't I doing $x-24x+96$?

[woven radish]

Vortac

[marble hearth]

when I move the 96 to the right hand side I get -92

[uncut crow]

and you should also have -23x on the left side

[marble hearth]

ohhh

ugh

long day

[uncut crow]

[marble hearth]

Why am I dividing 1 here?

[uncut crow]

well i’m just saying 4 is not a solution

[summer summit]

<@&268886789983436800>

[uncut crow]

1/(x-4) is an expression in the original equation

[marble hearth]

ohh

because (x-4) --> (4-4) = 0

[uncut crow]

sure

[marble hearth]

its not in the solution domain

I should've checked the domain first...

[uncut crow]

hmm

[marble hearth]

$\mathbb{R}-{4}$

[woven radish]

Vortac

[uncut crow]

that like… kinda misses the point

when you “do algebra” and find x = something, the something may or not may be a solution. what it does tell you is that everything that isn’t the ‘something’ isn’t a solution

and you can verify whether the something is a solution or not by seeing if it makes the equation true

or if you learn basic mathematical logic and understand implications and if and only if statements and what ‘algebra steps’ mean in terms of those, you can know without checking

[marble hearth]

thanks

.close

[devout snow]

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[devout snow]

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[glossy dew]

how does this go again

[glossy dew]

i tried writing the trajectory equation and solving for the coefficients

and just gave up after that

funnily ai also gave up after that exact step and pulled this

maybe even the derivation for what the ai said would help

[sand quarry]

Is this not correct

[glossy dew]

it is but it's so random

like google pulled it out of nowhere

i haven't seen anything close elsewhere

[violet wind]

surely i need more information

oh minimum speed

[sand quarry]

Ok wait conservation of eng

[violet wind]

well one guess is that you might want the object to peak at a height of 6m but i'm not 100% confident that's the optimal solution

We could parameterize something. My vote would be the horizontal velocity

oh i think i got it

[violet wind]

this should minimize both vertical and horizontal initial velocity components, thus total velocity as well

[devout snow]

@glossy dew Has your question been resolved?

[thick lotus]

I am sure I did this once

[shy osprey]

apply the projectile eqn twice

ull have two eqns

and your variables will br theta and inital speed u

[topaz axle]

why is that minimal horizontal

[violet wind]

because it maximizes the amount of time between the two points

[topaz axle]

i bet this has smaller horizontal

[shy osprey]

this doesnt seem right

the particle wont have this projectile motion if its gonna have minm speed

[topaz axle]

just talking about horizontal

[shy osprey]

Oh mb

[devout snow]

@glossy dew Has your question been resolved?

[violet wind]

hmm fair

ig that only applies to when they're on the same side of the vertex

if the pillars are really far apart (relative to their height) you probably want the peak to be really high for that reason then

Maybe parameterize w.r.t. the height instead

$v_y=\sqrt{\f h5}$

[woven radish]

Dreyuk

[violet wind]

$v_x = \f{(x_2-x_1)\sqrt5}{\sqrt{h-y_2}+\sqrt{h-y_1}}$

[woven radish]

Dreyuk

[violet wind]

Not pretty

[devout snow]

@glossy dew Has your question been resolved?

[thick lotus]

it might be inevitable that you have to bash the equation

[glossy dew]

doubtful man

[thick lotus]

u1 = ... >0
you would get a set of values for a

this shit is ugly

,w find minimum sqrt((10 * (2a) * (a + 4)) / (12 - 3a) * (1 + ((-3a^2 + 24a + 48) / (4*a * (a + 4)))^2))

[woven radish]

Results provided by WolframAlpha

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Requested by mis.ter.x

[thick lotus]

tha is so fucked up

[glossy dew]

seems to work 🤔

minima point is (1.25421, 11.83216)

almost 12

maybe theres a more effective way

[thick lotus]

maybe

[thick lotus]

ig you have to consider that peak height is 6m

[thick lotus]

there seems you can do something by converting the projectile equation into vertex form of a parabola

y = k + a(x-h)^2

$y = x \tan \theta - \frac{g}{2u^2 \cos^2 \theta} x^2$\

$a = \frac{g}{2u^2 \cos^2 \theta}$\

$y= x \tan \theta - ax^2$\

[thick lotus]

factor a out

[stoic cargo]

whose doing projectile motion?

[thick lotus]

$y = a (\frac{\tan \theta}{a} x -x^2)$

[thick lotus]

@glossy dew

[thick lotus]

hm..

ah^2 = k

[thick lotus]

looks like i messed up signs here

y = k - a(x-h)^2

need to express u^2 in terms of a and k

$$\tan^2 \theta = 4ak$$
$$\sec^2 \theta = 1 + \tan^2 \theta$$
$$u^2 = \frac{g}{2a \cos^2 \theta} = \frac{g( 1+ 4ak)}{2a}$$

[thick lotus]

now your vertex is at (h,k) , take the 3m wall as origin so your walls are at (0,3) and (4,6)

[thick lotus]

you sub y = 3 and 6 here and find the relation between a and h

from here you can try to bring the u^2 all in terms of one of the variables and this would become easier to minimize

[glossy dew]

ok

hmm

[violet wind]

Its not sadly

[devout snow]

@glossy dew Has your question been resolved?

[primal ferry]

I've managed to get down to a form in $u^2 = 80q + 90 + \frac{125}{16q}$ where

$q=\frac{g}{2u^2 \cos^2{\theta}}$

[woven radish]

sam beam

[primal ferry]

searching online tells me that this can be minimized by something called AM/GM which i am not familiar with

I usually avoid gpt but after having gpt apply AM/GM on 80q + 125/16q I got u_min = sqrt140 which is 2sqrt35 which by using the approximation gives 12

do you want me to share my work or would you rather i just outline the steps i took to get here

[glossy dew]

wait whaaat

[primal ferry]

would you rather i share my entire work or take you through my line of thought

[glossy dew]

maybe just the way to compress it to this form

it's easy after that

its hard to see

[primal ferry]

alright wait lemme take a picture i did it by hand

is the writing legible?

[glossy dew]

very

[primal ferry]

ignore the simply being cut that's usually how i write when im teaching online

i dont think ive made any logical slips here

[glossy dew]

uhhh that's a really weird simplification

wow

i don't think id have processed the equations that way

[primal ferry]

yeah it was definitely roundabout and i didnt do it alone my friendgroup is full of olympiad kids so i just took the question there and we banged our heads against it until i figured out you could use sec^2

[glossy dew]

hold on

cinema

[thick lotus]

🥀 true it comes out to be near 6.2

[primal ferry]

oh wait i dont even need AM/GM i can just differentiate and find the min point

[glossy dew]

ok so writing p in terms of q makes sense then

so i solve varying x and q

okkk thank you for your time

.close

[devout snow]

Channel closed

Closed by @glossy dew

Use .reopen if this was a mistake.

[tame sphinx]

$x-24x+96$?

$what$

[woven radish]

maso7

[thick lotus]

?

what are you doing

[tame sphinx]

$nothing$

[woven radish]

maso7

[thick lotus]

ok

do nothing in #latex-testing

or #bots

[devout snow]

Available help channel!

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[granite island]

can i pls have a hint for this problem

[granite island]

lowk dk wherer to start

[devout snow]

@granite island Has your question been resolved?

[granite island]

<@&286206848099549185>

[fair minnow]

what area of maths are you studying to come across this

there's a few different ways you can approach but it'd help to know the context of what you're learning

[granite island]

Math contest 😭

basically proof writing and induction but like yea besides that nothings

[arctic yarrow]

An induction approach seems promising

[fair minnow]

it's what i'd fiddle around with first but i'm not seeing an intuitive way to apply the inductive step

i think im veering into group theory trying to look at this ngl

[granite island]

oh hi again

[fair minnow]

which might be a bit beyond what it's asking(?)

[granite island]

whats group theory?

i haven't learned it before

[fair minnow]

going to try and solve it before i start rambling

give me a bit

[granite island]

ok

ill try it too ig

[arctic yarrow]

||Start by numbering the representatives of each country. You can build a string with the n+1 representatives of the first country and the n first representatives of the other countries that respect the condition.||

[granite island]

wait um doesnt each country only have n representitives not n+1?

[arctic yarrow]

yes this is an idea for induction step

[granite island]

yes but how do you get n+1 numbers if there are only n representitives

[arctic yarrow]

I assume the property for n and try to prov it for n+1

[granite island]

oh

i see mb sorry

ok so induction on n

that might work

ill think about it

hmm its a little hard to get from n to n+1 though

[arctic yarrow]

did you try this ?

[granite island]

like numbering every member?

[arctic yarrow]

yes and build a chain like i describe

[granite island]

can you pls reexmplain this part "and the n first representatives of the other countries that respect the condition. " again? i kinda dumb i don't understand

also how do u hide stuff like that like make the gray out

[arctic yarrow]

ok take your (n+1)^2 representatives that you put in a square. Each line is a country and each row is a number.

[granite island]

ok

like in a grid

[arctic yarrow]

Do you agree that if you remove the first line and the first row you get a square with side length n so we can use the induction hypothesis on it ?

[granite island]

yes

[arctic yarrow]

So you have to organize representatives of the first line and the first row so they satisfy the condition

[granite island]

but if we arranged them is a square how do we know who would be left of who

[arctic yarrow]

for the n by n square ?

[granite island]

yup

[arctic yarrow]

it doesnt matter who is where we just know by induction hypothesis that a configuration exists

you see ?

[granite island]

ok

yes

[arctic yarrow]

you sure ?

[granite island]

yes

[arctic yarrow]

ok so can you find a good configuration for representatives of first column and first row ?

[granite island]

get out scemer

[granite island]

i guess one country = one row and another country = another column

[arctic yarrow]

no i mean how can you put these people around a table such that you respect the conditions that are given ?

[granite island]

well isnt that true by induction step or the base case?

[arctic yarrow]

no what is true by induction step is that if you have n representatives from n different countries, you can put them around a table st they respect the condition.

Here you have n+1 representatives from one country and 1 from n other countries

so 2n+1 people that you need to organize so you would be able to use induction hypothesis on the others.

is that clear ?

[granite island]

so basically 2n+1 people need to be added to our square

[arctic yarrow]

yeah exactly but it is more convenient to build a chain with them separately first and then merge the 2 configurations

[granite island]

oh ok

[granite island]

ok so maybe we can chain these 2n+1 people together first and then add that chian onto the circle as a whole

[arctic yarrow]

yeah exactly

[granite island]

ok so the question is how to chain them

i mean prolly just alternate between people from the new country and people from the old country

[arctic yarrow]

try this

[granite island]

oh but that might not work at the ends

[arctic yarrow]

why ?

[granite island]

like every old country would have a new country representative to its right, so the person that is to the left of the person at the end of the line will match up with one of them

[arctic yarrow]

yes but is that a problem ?

one representative of a country can be to the right of a representative of the same country left ?

[granite island]

then the condition would be unsatisyed because then the person to the left of the person at the end would be the same person somewhere in the alternatinv sequence that has a person from the new country to its right

[granite island]

wouldn't the alternating sequence of n terms in sandwitched between the n+1 people from both countries so we would have to worry about both ends

like the ends where out chain meets the rest of the circle

[arctic yarrow]

Let’s not worry about the other chain for now

[granite island]

oh

but don't we have to add it on eventually

[arctic yarrow]

we will

[granite island]

but like in the sequence it works right

like just the chain before we add it on

[arctic yarrow]

If we can build a closed chain then we will be able to cut it where we want to merge

[granite island]

why is this?

also i may be dumb what what do u mean by cut it where we want to merge

like just pputting in the chain soemwhere

[arctic yarrow]

sorry if my English is not understandable haha

[granite island]

lol i think its just me being dumb

[arctic yarrow]

i mean if we have two closed chain then we chose where to cut them in order to merge them right ?

[granite island]

like we cut one of the chains and then put teh other one in it right?

[arctic yarrow]

yes but since both are closed we have to cut both of them

[granite island]

yea i guess

[arctic yarrow]

by closed i mean like a circle

[granite island]

so how should we cut them...

[arctic yarrow]

ok

so last step

In the final chain, we have to keep the property

Let me illustrate

Imagine this somewhere in the chain we built with A a representative of the n+1th country : ...ABAC...

We want to cut between A and B

But we have to keep the relation there is an A to the left of a B

[granite island]

ok so my plan is to pick a cutting point in the original chain. Then, we check what country is to the left of the cut line. then, we find that country in the second chain with 2n+1 countries. Then, we cut to the right of it (assume teh chain is already arranged in a line). Thus, we will not have an overlap, we can put those two chains seprately in the original chain

oh sorry mb

ill wait for u to finish typing first

[arctic yarrow]

So we will cut the induction chain somewhere there is a B to the right. so something like that ....XB....

then X can be anything since we destroyed a XB pair to create a new XB pair

[granite island]

wait so where do we cut

to the left of X?

[arctic yarrow]

between X and B

[granite island]

oh

ok

[arctic yarrow]

so in the end we lose AB pair and XB pair and we recreate AB pair and XB pair

so there is no change

so the property is conserved

[granite island]

ohh

my god

thats rly smart

holy orz

orz

I see now

omg thanks so much for the help

and thanks again

for the help

[arctic yarrow]

np thats a nice problem again

[granite island]

yup

one of those problms that have a slick ending

imm close it now, tanks for the help!

.close

[devout snow]

Channel closed

Closed by @granite island

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

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[proven anchor]

#help-23 message

[proven anchor]

@mystic scarab

z/z'=1 and z'/z=1

when z or z' are not equal to 0

[faint gorge]

me neither OP

can you post the original

[jagged harbor]

are you looking for another equivalent statement or trying to justify the book's statement?

[proven anchor]

(x^2-y^2)/(x^2+y^2)+2ixy/(x^2+y^2)

[jagged harbor]

which one?

[proven anchor]

this one

i am trying to make or connect this with argument of complex number

arg(z/z')

arg(z'/z)

[jagged harbor]

provided z (z') is nonzero, z/z' and z'/z respectively exist

[proven anchor]

if arg (z/z') is pi/4

which is one

[jagged harbor]

and you're simplifying [\frac z{z'}=\frac{x+iy}{x-iy}?]

[woven radish]

Flip

[proven anchor]

can I say numbers will be pure real

yeah

[jagged harbor]

this is not true

[proven anchor]

why not?

[jagged harbor]

3 is a real number

3' is just 3

[proven anchor]

3/3=1 no?

[jagged harbor]

3/3 is 1, and arg(1) is 0

or rather

[proven anchor]

ahha got it

[jagged harbor]

Arg(1) is 0 lol

[proven anchor]

pii

hmm

[jagged harbor]

the principle argument Arg(-) is 0 at 1; arg(-) is multivalued but returns Arg(-) + integer multiples of 2pi

[proven anchor]

x-iy/(x+iy)

if we look this arg

[jagged harbor]

you want to show that z (nonzero) has a zero imaginary component if and only if Arg(z/z') is 0?

is that what you're looking for?

or, do you just want to compute Arg(z/z') for arbitrary z?

[jagged harbor]

[\frac z{z'}=\frac{x+iy}{x-iy}=\frac{x^2-y^2}{x^2+y^2}-\frac{2xy}{x^2+y^2}i] has [{\rm Arg}\left(\frac z{z'}\right)=\arctan\left(-\frac{2xy}{x^2-y^2}\right)]

[woven radish]

Flip

[devout snow]

@proven anchor Has your question been resolved?

[proven anchor]

can it be 0?

[proven anchor]

when it will be purely real

[jagged harbor]

when will what be purely real? z?

[proven anchor]

z/z' is purely real

[jagged harbor]

z/z' is real if the imaginary component of z/z' is zero

@proven anchor

[jagged harbor]

@proven anchor so are you only trying to understand the original statement?

or, and reply only to the message that contains your motivation,

are you wanting to prove the same statement but by using an "arg"ument argument?

[devout snow]

@proven anchor Has your question been resolved?

[devout snow]

Channel closed

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[devout snow]

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[wooden trout]

can anyone send reosurces or teach me the techniques required to solve this question from the SMT Algebra problem 5 2023:

[devout snow]

@wooden trout Has your question been resolved?

[devout snow]

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[dapper fable]

Let $n\geq 2$ be a natural number. In a math class, there are $n$ students, each student has a card with the number $1$. The teacher has $n$ many cards, where the $i-$th card has a number $\frac{n+i}{i}$ for each $i = 1,...,n$. The teacher hands out all the cards; some students may get more than one card, or some students may not get any cards. Each student then takes the product of all their card numbers, and writes it on the board. Let $S$ be the arithmetic mean of the $n$ numbers written. Determine the arithmetic mean of $S$, considered over all possible ways the teacher can hand out cards.

[woven radish]

Copter

[dapper fable]

i dont know how to approach this at all😭

but apparently the answer is ||n+1||

[devout snow]

@dapper fable Has your question been resolved?

[cosmic pumice]

first, class average before any teacher card is handed out = 1

if we give $\dfrac{n+i}{i}$ to a uniform random student, this student has its number multiplied by $\dfrac{n+i}{i}$

[woven radish]

1048576Orz

[dapper fable]

yes

[cosmic pumice]

i.e. $\dfrac ni$ times more than the number he/she holds.

[woven radish]

1048576Orz

[cosmic pumice]

the difference from the expectation before and after is $\dfrac ni$ times the average of the number held by the people.

[woven radish]

1048576Orz

[cosmic pumice]

i.e. $E_{k}-E_{k-1} = \dfrac nk E_{k-1}$

[woven radish]

1048576Orz

[dapper fable]

and then we find En?

[cosmic pumice]

yes

as a product

this product is a easy application of telescope

[dapper fable]

im guessing its gonna be something like n^n/n! or smth

[cosmic pumice]

well, your inituition is right

$n+1$.

[woven radish]

1048576Orz

[dapper fable]

n+1 where?

[cosmic pumice]

notice, theres a $E_{k-1}$ in LHS so we must push it to RHS

damn sorry

[cosmic pumice]

this transition was wrong

$S_{k}-S_{k-1} = \dfrac nk E_{k-1}$

[woven radish]

1048576Orz

[cosmic pumice]

it should be this

[dapper fable]

whats S?

[cosmic pumice]

$S_k = n E_k$, the sum of the people's number

[woven radish]

1048576Orz

[cosmic pumice]

so it should be $E_k - E_{k-1} = \dfrac 1k E_{k-1}$

[woven radish]

1048576Orz

[cosmic pumice]

and here is the right formula where you gets $n+1$.

[woven radish]

1048576Orz

[cosmic pumice]

where do you stuck on

[dapper fable]

how would n+1 pop up again😭

[cosmic pumice]

did you understand this

[dapper fable]

telescope youd get En-E1 = something

[dapper fable]

yes

[cosmic pumice]

telescope for product

first push the E_k-1 to the RHS

$E_k = \dfrac {k+1}k E_{k-1}$

[woven radish]

1048576Orz

[dapper fable]

oh yeah mb

got it now

[cosmic pumice]

.close

[dapper fable]

wait actually

whys it 1/k

[cosmic pumice]

as $S{k}-S{k-1} = \dfrac nk E_{k-1}$

[woven radish]

1048576Orz

[cosmic pumice]

divide both side by $n$

[woven radish]

1048576Orz

[dapper fable]

oh oki

i c i c

thanks!

.close

[devout snow]

Channel closed

Closed by @dapper fable

Use .reopen if this was a mistake.

[wooden trout]

.reopen

[wooden trout]

.reopen

[dull parrot]

you need to make a new help channel

[devout snow]

Available help channel!

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[atomic idol]

How do I find the inverse of
f(x)= sqrt( (x2) / (1+2x) )

[atomic idol]

Attempt failed gloriously

[supple knot]

Did you learn quadratic formula

[atomic idol]

Not sure what that’s referring to

[supple knot]

,tex .quadratic formula

[woven radish]

pi_day

[atomic idol]

Nope never learned that

[supple knot]

[atomic idol]

I learned sth similar which is supposed to solve equations like this:

0= x^2 + p*x + q

[supple knot]

Okay yea that's the same with a=1

[atomic idol]

Kinda looks like:

x_1,2 = sqrt(p/2) +- sqrt(p/2-q)

[supple knot]

Yea move y^2 to the right side in your third to last line

[atomic idol]

Like this?

[supple knot]

Yea now you have this form

[atomic idol]

And how does that help with the inverse?

Wouldn’t it be useful for two specific x values

Wait nvm the parameter y just becomes my variable and im fine

I hope

Next question is, if my formula gives me two solutions how can that be?
This is injective and should only have one result

[devout snow]

@atomic idol Has your question been resolved?

[lunar harbor]

um

you're going to need to define it piecewise or smt

[atomic idol]

That looks concerning

Oh yeah mb the function is defined with the domain/codomain [0, infinity)

So it’s fine
Should still be injective

[lunar harbor]

Recall that the domain of the original function is the codomain of the inverse

that only leaves one of your cases as being possible

[atomic idol]

What case? Sorry I can’t follow

[lunar harbor]

Next question is, if my formula gives me two solutions how can that be?

[atomic idol]

Oh that’s what you mean by cases

Yes as long as there’s a positive and a negative result that’d work

But how do I know that

Ive been trying to prove it but it could technically be two positive results

In which case I’d have two solutions

[lunar harbor]

Recall that the domain of the original function is the codomain of the inverse
that only leaves one of your cases as being possible

If your domain is $[0,\infty)$, then your inverse should only spit out values in $[0,\infty)$.

[woven radish]

Civil Service Pigeon

[atomic idol]

Yes it should

But can I just use that to claim what I calculated is right

Or do I have to show that the result will always be one positive and one negative number

[lunar harbor]

?????????????

You calculated two possibilities for the inverse

My point is that you can eliminate one of those possibilities

thus leaving you with a unique inverse

[atomic idol]

How can i tell which one i need and which one to eliminate

[lunar harbor]

You have a function from $[0, \infty)$ to $[0, \infty)$

[woven radish]

Civil Service Pigeon

[lunar harbor]

So that means your inverse must also be a function from $[0, \infty)$ to $[0, \infty)$

[woven radish]

Civil Service Pigeon

[lunar harbor]

As I said earlier, one of the possibilities for your inverse spits out values that are not in $[0, \infty)$

[woven radish]

Civil Service Pigeon

[lunar harbor]

which violates the general rule of inverses

[atomic idol]

This is the part I didn’t get yet

I know the reasoning and why that’d be bad

But how do I know it’s violating the domain

[lunar harbor]

??????? Notice how I said "spits out" values, which refers to the values that the inverse should output (which are in the codomain).

You can't really do much eliminating off the domain since both of your possibilities are defined in $[0, \infty)$

[woven radish]

Civil Service Pigeon

[lunar harbor]

but you can eliminate based on the codomain, as I've been saying

[atomic idol]

Mb meant the codomain

[lunar harbor]

Mate

just pick a value in the domain like 1 and see what each possibility for the inverse spits out

Alternatively note that

,texsp ||$$\sqrt{y^4+y^2}>\sqrt{y^4}=y^2 \implies y^2-\sqrt{y^4+y^2}<y^2-y^2=0.$$||

[woven radish]

Civil Service Pigeon

[atomic idol]

That’s the prove I’ve been trying to get but I couldn’t think of it

[atomic idol]

I didn’t try that

Thats all

Ig I’ll just delete the one w the minus

[lunar harbor]

mhm

[atomic idol]

Ty ill close the channel now

.close

[devout snow]

Channel closed

Closed by @atomic idol

Use .reopen if this was a mistake.

[lunar harbor]

[devout snow]

Available help channel!

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[ocean haven]

claim

[ocean haven]

Tom plays a game, he is fighting the boss. Initially the boss has infinite hp, Tom has 100 hp. If Tom hits the boss once he gains 30 hp from the boss and if the boss hits him, he looses 10 hp. There is 50% chance of landing hit for both. If Tom's hp becomes 0 game over. Find probability that Tom looses

so 1) is 1/2^10

n ->0
1/2^(n/10)

@worthy tusk

[frozen solar]

idk how infinity works but can't the boss just tank inf hits from tom

[ocean haven]

idk

perhaps yes

[frozen solar]

so tom can never win...

[worthy tusk]

I don't understand that statement

Anyway

Do you see where this is going

[frozen solar]

think he meant n tends inf ig

but why it n/10

[ocean haven]

2nd N to 0

[worthy tusk]

Okay i get it

That's right

[ocean haven]

no

[gloomy aurora]

I mean since the boss has infinite hp, how the hell will it ever his/her hp ever become zero?

[ocean haven]

right

[worthy tusk]

The N->0 thing im still unsure

[frozen solar]

tom never win

[worthy tusk]

The dilemma is whether it can go on forever

[ocean haven]

the hp of tom N to 0

[worthy tusk]

Ok ok

Do you understand 3)?

[ocean haven]

yes

[worthy tusk]

So how do you deal with the following

[ocean haven]

idk

🥀

[worthy tusk]

So do you understand how to do 3) or are you struggling with 4)

[ocean haven]

yes

[worthy tusk]

My question makde no sense

So 3 or 4?

Basically lol

[ocean haven]

but there is an extreme case where the hp of tom always increase

[worthy tusk]

Yes

But the hp is capped at 100 isn't it?

Genuine question not a test

[ocean haven]

non fuck i forgot to mention

[worthy tusk]

Dw dw

If it is not capped it will in fact become easier

Ok

So now we assume it is not capped

And we also assume that the number of hp we wine or loose is the same for simplifications

We will also assume that the hp is not capped down at 0 so we can go negative

The idea is that if we ever reach negative we still loose

Okay

So do you have any idea how many situations of n turns will lead to hp reaching exactly 0?

With n being even and greater than 10

This is a very difficult question that we will cut into smaller question

[ocean haven]

even?

[worthy tusk]

Yeah

[ocean haven]

why

[worthy tusk]

Well

Can you reach exactly 0 in an odd number of turns (remember we count negative hp here)

[ocean haven]

we reach negative/ 0 on only even

?

[worthy tusk]

No

Okay let'ssimplify things

You can see toms hp as a paved road on which a man is walking

Ech turn he eithergoesforward orbackward

[worthy tusk]

<@&268886789983436800>

[ocean haven]

ok

[worthy tusk]

Imagine he isn't stopped at 0 on the right and could continue

[waxen carbon]

could i pls get help with this question

[worthy tusk]

He could reach 0 in 10 moves by going only right.
If he hesitated on one time and went back once before going on the right it would be 12

But like it couldn't be 11

Because for each step back you gotta go forwards once so it makes 2 more and stays even

[ocean haven]

oh

ok

[worthy tusk]

!help

[devout snow]

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

[worthy tusk]

So do you agree that to reach 0 in N moves, he needs to go left exactly 10 more times that he goes right?

[ocean haven]

yes

[worthy tusk]

Do you know how to translate that into binomial coefficient to describe how many different path there are to reach exactly 0 in 10 moves

How many possible path of lenght N that end on exactly 0 exists?

[ocean haven]

bruh

[worthy tusk]

<@&268886789983436800>

Annoying

Ty

[worthy tusk]

It's ok if you don't have the answer immediately

[ocean haven]

of length n meaning n steps or at nhp initally

[worthy tusk]

N step

The hp vision complicates things because it's 10 by 10, i'd rather go 1 by 1 first

[worthy tusk]

This is the important idea

[ocean haven]

so if man take A step towards increase no. of paths would be (2A+10)CA

few paths go below zero then come back

[worthy tusk]

Yeah

So let's see it like that: out of all the N steps the man takes, 10 of them have to be to the (N+10)/2 have to be to the left and (N-10)/2 have to be to the right (10 mor eto the left)

So out of N steps exactly (N+10)/2 have to be to the left, the number of possibilites for this scenario is exactly given by a certain binomial coefficient

[ocean haven]

idk

sorry

[worthy tusk]

Dw

Do you know that the number of ways to raise a combination of k fingers out of n fingers is $C^k_n$

[woven radish]

Almond

[worthy tusk]

Also written $\binom{n}{k}$

[woven radish]

Almond

[ocean haven]

yeah

[worthy tusk]

So here

You have to choose a combination of (N+10)/2 steps left

Out of N total steps

Similar situation right?

[ocean haven]

so $\binom{N}{\frac{N+10}2}$?

[worthy tusk]

Yes!

[woven radish]

ghost

[worthy tusk]

That's the number of ways there are to reach 0 in N steps

Exactly

Now

What is the probability of each of these combinations occuring

[ocean haven]

1/2^(N/10)

[worthy tusk]

Forget the /10

[ocean haven]

ok

[worthy tusk]

We'll go back to that later

You take N steps so yes

So what is the total probability of reach 0 in N steps

[ocean haven]

,, \frac 1{2^N} \binom{N}{(\frac{N+10}2)}

[woven radish]

ghost

[worthy tusk]

🔥

Yay we made big progress

[ocean haven]

tbh can you explain N+10/2 again

[worthy tusk]

Yeah

The idea was that to reach 0 you have to go exactly 10 times more steps left than you gosteps right

[ocean haven]

ok

[worthy tusk]

So if we solve the system right + left =N and left = right +10, we get exactly that

[ocean haven]

oh ok

[worthy tusk]

[ocean haven]

wait but does this get same

[worthy tusk]

Ngl idk

[ocean haven]

or it is the same just 2A + 10 = N

and A is steps taken to right

[worthy tusk]

Ahues it does work

Because it is the same to decide which steps you take left or which steps you take righr

It's complementary

[ocean haven]

[worthy tusk]

So im not sure how tomake a conclusion easily from there

[ocean haven]

welp

we arrived somewhere

to an upperbound where the paths may go below zero but end at 0

tysm almond

[worthy tusk]

I can tell you why i went from that

The idea is that on a 1 dimensional grid (jsut like here), the probability that you reach a given point (here 0) in an infinite time is 1

It also works in 2 d but not in 3d

[ocean haven]

i see

i search this it shows catalan numbers and dyck paths

does that apply ?

[worthy tusk]

Oh shit yeah but that was not what i was goung for

Ok ngl i feel like dyck paths are pretty hard

[ocean haven]

😭

[worthy tusk]

Doyou know asymptotic stuff withlog

Like log(1-x) when x goes to 0 is like like -x

[ocean haven]

lim x->0 log(1-x) ?

it is 0 ?

[worthy tusk]

Yeha but forget it

[ocean haven]

alr

[worthy tusk]

The idea is that the probability to reach 0 at some point should be $\sum_{N=1}^{\infty}\frac{1}{2^N}\binom{N}{(N+10)/2}$

[woven radish]

Almond

[worthy tusk]

No it's not true actually sorry

@ocean haven i think dyckpath might be better actually😭

[devout snow]

@ocean haven Has your question been resolved?

Channel closed

Closed by @ocean haven

Use .reopen if this was a mistake.

[ocean haven]

oh

.reopen

[devout snow]

✅ Original question: #help-27 message

[ocean haven]

no idea what shit is that

[worthy tusk]

well

dyck words are the number of ways to go up and down while staying above a certain treshold

so exactly what we want with hp

[worthy tusk]

this shit is tough sorry i don't have the energy, if you're still motivated tomorrow we we might see then😭

i hope you had least had fun learning some stuff here

but im sure the result still holds

[ocean haven]

i did

ill keep it open

[devout snow]

@ocean haven Has your question been resolved?

[devout snow]

@ocean haven Has your question been resolved?

[devout snow]

@ocean haven Has your question been resolved?

[shell hare]

It's going to br closed by timeout tho

[devout snow]

Channel closed

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[viral ledge]

What if there is a language which is a reglang A or reglang B depending on a condition which is uncomputable

[viral ledge]

What is that language classification

Is da bot ded

[jagged ledge]

you just gotta wait till someone can help

after 15 minutes you can tag helpers i believe

[viral ledge]

No my net was glitching

[viral ledge]

Yes

[jagged ledge]

can’t help but interested in the problem, whats it about

[viral ledge]

What r u confused about

[jagged ledge]

assume no knowlegde, what are languages

[viral ledge]

We can talk in dms if u want, don't want to clog the help channel

[jagged ledge]

ah alr, nevermind then

[viral ledge]

I mean I am happy to explain

Upto you

Reiterating the problem:
What if there is a language which is a reglang A or reglang B depending on a condition which is uncomputable
What is that language classification: decidable, recursively enumerable, uncomputable

Reglang = regular language

[jagged ledge]

are those the only classifications

[viral ledge]

If it can be classified more strongly then that'll be welcome

My intuition is that if that condition isn't dependent on the string being input

Then its truth value doesn't matter

The language is basically regular

And decidable

Whatever its truth value ne

Be

Irregardless of condition being uncomputable

I ... Think that has to be the case....

L

.close

[devout snow]

Channel closed

Closed by @viral ledge

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[vast quest]

I'm a bit confused on how this was factored out, how did they get x^(-2/3)(8x -1 )

[crystal dawn]

when you factor something out, you divide every term by the common factor

so, $x^{1/3} / x^{-2/3} = x^{(1/3) - (-2/3)} = x^{(1/3) + (2/3)} = x$

woops, poor bracketing, mb

[woven radish]

Hanako(x, y); ∂(fox)/∂x

[crystal dawn]

(as shown on the right)

and if you are asking instead how they thought to do so, the benefits of factoring here are twofold: besides factoring out a common factor so the main expression only has one x term, there's the added bonus of making that x term have a nice, integer power so you don't have to deal with any weird cube roots when you find x later.

[devout snow]

@vast quest Has your question been resolved?

[devout snow]

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[proven anchor]

|z1+z2|^2

[proven anchor]

|z1+z2||z1'+z2'|

how do i multiply now?

we have modulus sign between

[arctic yarrow]

$|z||z'| = |zz'|$

[woven radish]

Lin Xia

[proven anchor]

i see

so after multiplication i got |z1|^2+|z2|^2+z1z2'+z1'z2

[livid geyser]

is that all you need? Or is this for a larger problem?

[proven anchor]

last part should have a 2 or not?

[livid geyser]

yeah should have a 2 p sure

[proven anchor]

so my book is bad for sure?

damn

it is mentally torcher when they don't write properly

[livid geyser]

ikr

[proven anchor]

btw thanks

[supple knot]

You can also check this yourself by taking simple values for z

[proven anchor]

I will complaint it to principal today

[supple knot]

z1 = z2 = 1 gives (1+1)^2 = ...

[proven anchor]

that is amazing

thanks

.close

[devout snow]

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[devout snow]

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[proven anchor]

how can I prove it?

[proven anchor]

I checked this inequality with random complex numbers

[wicked turtle]

are you able to prove $|z_1| - |z_2| < |z_1 - z_2|$?

[woven radish]

Bungo

[proven anchor]

i can not..give me hints

and why did you remove equal sign?

[wicked turtle]

i did not

i removed the outer abs vals though

oh in the inequality

yea i meant <=

try writing $|z_1| = |z_1 - z_2 + z_2|$ and go from there

[woven radish]

Bungo

[proven anchor]

here

[proven anchor]

$|z_1| = |z_1 - z_2 + z_2| <= |z_1-z_2|+|z_2||$

[woven radish]

undercop

[proven anchor]

$|z_2| = |z_2 - z_1 + z_1| <= |z_2-z_1|+|z_1||$

[woven radish]

undercop

[proven anchor]

can I subtract?

[wicked turtle]

yep sure

[proven anchor]

$|z_1| -|z_2|<= |z_1|-|z_2|$

[woven radish]

undercop

[wicked turtle]

wait what

i mean that's true obviously

but not useful

[proven anchor]

yeah

did I make any mistake?

[wicked turtle]

$|z_1| \leq |z_1 - z_2| + |z_2|$, subtract $|z_2|$ from both sides

[woven radish]

Bungo

[proven anchor]

$|z_1|-|z_2| \leq |z_1 - z_2|$

[woven radish]

undercop

[proven anchor]

we need an extra modulus over left side?

[wicked turtle]

this is one of two pieces that you need

can you also show that $|z_2| - |z_1| \leq |z_1 - z_2|$

[woven radish]

Bungo

[wicked turtle]

if so then those combined imply what you want

[proven anchor]

i see

got it

thanks

.close

[devout snow]

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[proven anchor]

.reopen

[devout snow]

✅ Original question: #help-27 message

[proven anchor]

can I add one more thing with it?

$|z_2| - |z_1| \leq |z_1 - z_2|<= |z_1|+|z_2|$

[woven radish]

undercop

[wicked turtle]

sure, that's true

[proven anchor]

thanks

.close

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[devout snow]

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[kind terrace]

For question 11, one direction is extremely trivial, but I cant think of how to prove the converse
Question 12 is done
No idea how to start with 13 and 14.
Taken from Axler linear algebra 4th ed exercises for 3A

[devout snow]

@kind terrace Has your question been resolved?

[kind terrace]

Wow thanks i should quit mathematics and become a syncophant of misterbeess

[urban niche]

bro 😭😭

[kind terrace]

Yea whats up bro?

I saw a couple of people typing

Any leads?

[arctic yarrow]

for question 11, consider a basis (e_1,..,e_n) of V. Let S_ij be the endomorphism sending e_i on e_j and sending e_k on 0 for every k\neq i.

[kind terrace]

Im gonna have to go soon so ill close this

[arctic yarrow]

not sure if it is the quickest solution though

[kind terrace]

Idk what a endomorphism is

Tho so😅

[arctic yarrow]

an element of L(V)

a linear application from V to V in other words

[kind terrace]

And then whats rhe next step?

[arctic yarrow]

Test them on the identity ST = TS

And see what it gives

[kind terrace]

Isnt that the trivial direction

St=ts if T is a multiple of the identity

[arctic yarrow]

No we are doing the other

Suppose ST = TS

prove that $T = \lambda id_V$

[woven radish]

Lin Xia

[kind terrace]

Ok will do

Thanks

Ill just leave it to close on itself lest someone knows a thing or two about 13&14

[arctic yarrow]

For 13 you can consider a basis of U and extend it to a basis of V.
For 14 you can try to exhibit a free family from L(V,W) that is infinite.

[kind terrace]

For 13, I let u1..un be my basis of U and choose vi's such that u1..un,v1..vm is a basis of V. Then for any w belonging to V we can write w= Σaiui +Σbivi. Define then Tw=ΣaiT(ui) +ΣbiT(vi) and T(ui) = S(ui), T(vi)=0

like that?

What does a free family mean?

[arctic yarrow]

linearly independent

[kind terrace]

ok ok

[arctic yarrow]

Maybe the term free family isn't used in english

[kind terrace]

I don't understand

[arctic yarrow]

linearly independent you heard about that ?

[kind terrace]

what would a free family in L(V,W) look like though?

i have

[arctic yarrow]

Exactly like in an other vector space. For example $\phi$ and $\psi$ are linearly independent if there exists $a,b\in R$ st $a\phi + b\psi = 0_{L(V,W)}$ then $a = b = 0$.

[woven radish]

Lin Xia

[kind terrace]

maybe i need to know my theory better

this stuff seems like its coming out of someone's ass

[arctic yarrow]

lol yeah you should be familiar with this notion

[kind terrace]

I remember all the definitions and most of the associated theorems and their respective proofs so far

but then I can't seem to do any of the exercises

oh well not the hard ones anyways

what 'notions' are you referring to

[arctic yarrow]

free family of vector

you use that to define basis

[kind terrace]

oh i know what a linearly independent list is

[arctic yarrow]

and then dimension

[kind terrace]

thats when a linear combination is 0 iff all the coefficients are 0

[arctic yarrow]

oh okok

[kind terrace]

basis is a linearly independent list that spans

dimension is the length of the basis

and basis length is independent of basis

so the basis of V always has length V no matter how the basis is constructed