help-49

[midnight plank]

@near geyser Has your question been resolved?

[signal ibex]

there actually is a way to calculate that by hand but it is fairly advanced

[maiden bridge]

Do you even know how to divide on a paper?

[signal ibex]

[midnight plank]

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[twilit field]

Kind of unsure of how to do this

for context

so $L(\alpha, \theta)= \left( 1- \left( \frac{\alpha}{x} \right)^{\theta} \right)^n$

[grand pond]

wai

[twilit field]

and I need to maximise this wrt $\alpha, \theta$ right?

[grand pond]

wai

[twilit field]

I don't get the values of x given

Like do I have to maximise $\prod_{i=1}^{5} \left( 1- \left( \frac{\alpha}{x} \right)^{\theta} \right)^n$

[lyric charm]

\prod

[grand pond]

wai

[twilit field]

as in why they're given

.close

[midnight plank]

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[midnight plank]

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[nova lava]

Guys I need help with the following series :

[nova lava]

Limit as N approaches infnity Sum(F_n) where sum goes from n=1 to N

[lyric charm]

your series is $\sum_{n=1}^{\infty} F_n$ and you are 100% sure you wrote it correctly?

[grand pond]

Ann

[nova lava]

Where F_n is the nth fibonacci numbe

[nova lava]

yeah

[lyric charm]

not 1/F_n not anything else, just summing the fib numbers straight?

are you REALLY sure

[nova lava]

No no

sorry

the reciprocals

[scenic wyvern]

picture please.

[nova lava]

my bad

[nova lava]

The sum is correct , just the terms are reciprocal of the fibonacci numbers

This question came into my dream

Then tried to solve but couldnt, so asking for help here

[lyric charm]

$\sum_{n=1}^{\infty} \frac{1}{F_n}$

[grand pond]

Ann

[lyric charm]

do you want value or only convergence status

[nova lava]

Exact value

Btw which test would work if we wanna consider the convergence status?

[lyric charm]

ratio seems most straightforward

[nova lava]

This is the real challenge

[scenic wyvern]

no closed form

just a heads-up

[nova lava]

?

Where is this conclusion coming from?

[lyric charm]

yeah im wondering too

[scenic wyvern]

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

[lyric charm]

oh LMFAO

theres a wikipedia article about THIS EXACT series?

[scenic wyvern]

and an OEIS entry too

[nova lava]

Btw can you pls teach me to type in latex

[nova lava]

like you did here

?

[lyric charm]

from here

\frac{}{} makes fractions. first {} is the numerator, second is the denominator

\sum makes the big sigma symbol.

_ is for subscripts, ^ for superscripts, and by extension also gives you lower and upper boundaries on summations and integrals

if your sub/superscript is more than a single character long, you need to wrap it in braces

[scenic wyvern]

(also, recommending OP to visit #latex-help, esp. the pinned message there, for some LaTeX beginner guides and cheatsheets.)

[lyric charm]

e.g. $a_10$ vs. $a_{10}$

[grand pond]

Ann

[lyric charm]

also yes this

[nova lava]

Thank you so much for helping me

[scenic wyvern]

anything else? if not, you may close this channel, and see you around

[azure vapor]

thank you so much for thanking him.

[nova lava]

Thank you so much for Thanking me to Thank him

[azure vapor]

thanki you so much for thanking me when i thanked you for thanking him

[lyric charm]

... yall done?

[scenic wyvern]

okay folks, let's not

[slate ferry]

[azure vapor]

no

[nova lava]

Bro maintain the dignity of the group

[lyric charm]

btw neither hanako nor i are a "he".

[azure vapor]

ok

[nova lava]

You can yap in #chill and etc chats

[azure vapor]

it?

[lyric charm]

...

[slate ferry]

what

[lyric charm]

we have she/her role diamonds

[slate ferry]

do you know how to read

[nova lava]

You two are friends?

[lyric charm]

<@&268886789983436800> transphobia?

[azure vapor]

hello, omg see it is a @lyric charm !!

[slate ferry]

wow

[visual elk]

No way

That is rude as fuck

[lyric charm]

how is it that you see "we don't use he/him prns" and your default assumption is "it" of all things.

[azure vapor]

not rude

[scenic wyvern]

right folks, if there's no other question here, please just close this and move on, OP.

[azure vapor]

it is @lyric charm 's discord account

[patent snow]

That's rude, stupid and uneducated. Please be better next time?

[slate ferry]

-# hi bubblesssssss 👋

[azure vapor]

ok

[nova lava]

I also need help with the problem

[lyric charm]

progress so far?

[nova lava]

I started computing initial values

but struggled to find any pattern

[lyric charm]

what did you get for those values

maybe we can compare ours

[nova lava]

Then as fun I converted the inputs into Biryani numbers

[lyric charm]

biryani numbers?

[nova lava]

Then saw the the outputs are obtained by reversing the digits of the input

[visual elk]

Omg i love biriyani but what are biryani numbers

[nova lava]

Sorry my bad , Binary Numbers

[lyric charm]

wait did you autocorrect binary

ah

[lyric charm]

oh but then f(n)=n means n is a palindrome in binary doesn't it

[nova lava]

This called for Induction

I need help for the Induction

How to prove the conjectured hypothesis via Induction

[nova lava]

yes, you are right

[lyric charm]

think you want strong induction

[nova lava]

Then it boils down to a combinatorics problem

[lyric charm]

that, or prove that g(n) = binary reversal of n satisfies all the same properties that define f

[nova lava]

Can you elaborate on this?

[lyric charm]

define a function g by your conjectured rule

g(n) = number obtained by reading the binary digits of n in reverse order

[nova lava]

Oh got it, then we must prove that the function f and g are identical

Right?

[lyric charm]

show that it satisfies all the same properties that the question gives you about f

it'll follow that f=g

[nova lava]

Yeah

Thats a great approach

Thanks for your insight, I appreciate it

[nova lava]

btw are you into MO?

[lyric charm]

MO?

[nova lava]

Mathematics Olympiad

[lyric charm]

no

[nova lava]

You are a College Student?

[lyric charm]

wrong again

[midnight plank]

@nova lava Has your question been resolved?

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[inland patio]

Suppose A is an n x n Hermitian matrix of rank k < n. I know that it then has a k x k principal submatrix with nonzero determinant. I'm trying to figure out if it's possible to move this submatrix to the top left corner by consecutive row and column operations, and then also have the remaining entries in the matrix be 0 by consecutive row and column operations.

Thus I want A to be congruent (or equal?) to P^t UP where U has the principal submatrix in its top left corner and 0s everywhere else (and P is a product of elementary matrices).

[unkempt sluice]

Yes you can

[west iron]

P is a product of elementary matrices

Correct me if I'm wrong, but this is the same as just saying P is invertible, right?

[unkempt sluice]

Yes

[undone pagoda]

hi guys

im new to this server

[inland patio]

Ok. I understand the part of moving the principal submatrix to the upper left corner by consecutive row and column operations, but I fail to see how its possible to get the rest of the entries to be 0. I feel like there's a bigger underlying theorem/result that motivates what I want.

[lyric charm]

hello and welcome to the server. if you want to chat, go to #discussion or #chill. if you want help, then open your own help channel.

(see #❓how-to-get-help for instructions)

[undone pagoda]

can i like ask any level of math doubts ??

[inland patio]

yes

[undone pagoda]

alr

[west iron]

but not here

[undone pagoda]

ikik

[west iron]

This problem is beyond me

But I'll caution about applying any results that are actually about P*UP (conjugate transpose)

[unkempt sluice]

Applying permutation you have a matrix of the form (A,B;C,D) with A invertible, A can eliminate B,C and you will find the place where D was will be 0 by rank of the whole matrix being equal to the rank of A

[west iron]

Since evidently despite being a complex matrix were interested in P^tUP

[unkempt sluice]

U^h(A,B;C,D)U does make your matrix of that form where U=(I, -A^-1 B; 0, I)

So yeah that invertible matrix can be unitary (given that (A,B;C,D) is hermitian (A^h=A, C=B^h))

(A B; C,D)=Q^h (your original matrix) Q where Q is a permutation matrix, so it can be done by a unitary matrix, not necessarily a orthogonal real matrix

So not an orthogonal P, but a unitary P. Not P^t but P^h (conjugate transpose)

[inland patio]

Ok. Thank you. Since my matrix is Hermitian, we have B=C^h, right?

[unkempt sluice]

Yeah

[inland patio]

Are you sure that U=(I, -A^-1 B; 0, I)? I was looking at a decomposition involving the Schur complement, where we have the following formula:

Ah wait, I think it works.

[unkempt sluice]

Yeah. Works

D-CA^-1B =0 by rank on both sides being equal

to rank(A)

[inland patio]

yeah, I was thinking about why the Schur complement is 0

[inland patio]

so you mean D has the same rank as CA^{-1}B?

[unkempt sluice]

I didn’t mean that but it is obviously true

Since D=CA^-1B

What I meant was, rank(M, 0;0,N)=rank(M)+rank(N)

[inland patio]

And do we know that (A,0;0,D-CA^-1B) has the same rank as (A,B;C,D)?

[unkempt sluice]

rank(A)=rank(A, 0; 0, D-CA^-1B)=rank(A)+rank(D-CA^-1B)

Only zero matrix can have rank 0

[unkempt sluice]

Yes, multiplying by an invertible matrix doesn’t change rank

[inland patio]

Ok. 👍

.close

[midnight plank]

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[midnight plank]

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[buoyant linden]

kinda random question but is there a name for inequalities like nesbitt's inequality?

[buoyant linden]

like notorious/famous algebraic inequalities

i want to look into more of these inequality identities

[wary thorn]

Yes I think the name is Nesbitts inequality

[buoyant linden]

🥀

[late rover]

I mean

Any inequality have a name

[buoyant linden]

not sure wym by that

[late rover]

oh wait, you mean another name for nesbitt's inequality? or like a random famous inequality name

[buoyant linden]

yeah

[wary thorn]

Maybe you would say it’s a rational inequality

[buoyant linden]

2nd one

[late rover]

So, some famous one are AM-GM-HM, cauchy schwarz, titu's lemma

[buoyant linden]

yuh

those r like
lwk exactly what i've been looking into

[late rover]

cauchy schwarz has many variants

[buoyant linden]

https://en.wikipedia.org/wiki/List_of_inequalities
there's a wikipedia page but they're so egregiously out of the scope of what i'm learning that i don't think it's worth it to

like

go through it

[late rover]

Each has a name

[buoyant linden]

titu is derived with cauchy, right?

[late rover]

yeah

[buoyant linden]

oo

[unkempt sluice]

I suppose high school competition math has thousands of inequalities

[late rover]

It does

[buoyant linden]

yeah actually

i should look into IMO

[late rover]

I have learnt at least 20

[buoyant linden]

ur right tbh

they're pre challenging

[late rover]

I forgot all of their name btw lol

[buoyant linden]

alr cool
thanks guys

.close

[midnight plank]

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[midnight plank]

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[tardy bloom]

Integral 0 to infinity k e^(-a r⁴) 4 pi r² dr

[tardy bloom]

I don't need the exact value just need to get 'a' out of the integral

[lyric charm]

$\int_0^{+\infty} 4\pi kr^2 e^{-ar^4} \dd{r}$

[grand pond]

Ann

[lyric charm]

this?

wdym by "get a out of the integral" exactly...\

[tardy bloom]

Original question was
The number density of molecules of has depends on their distance r from the origin as n(r)= k e^-(a r⁴). Then the total number of molecules is proportional to
(a) k a^(-3/4)
(b) sqrt(k) a^(1/2)
(c) k a^(1/4)
(d) k a^(-3)

[tardy bloom]

I need to integral to be independent of 'a'

[lyric charm]

The number density of molecules of has depends on their distance ...

word salad

[dusty portal]

[lyric charm]

ok i think i see where you are going with this

you want to write the integral so that all a's appear only outside of it

u := ar^4 lmao

[dusty portal]

$ar^4=(\sqrt[4]{a}r)^4$, maybe let $u=\sqrt[4]{a}r$ then?

[grand pond]

;(

[tardy bloom]

Wait

[lyric charm]

eh

can do it either way

[tardy bloom]

I don't need to do all this

[lyric charm]

i was gonna do it in such a way that results in a Gamma integral

[tardy bloom]

Dimensionsal analysis works

[dusty portal]

wha

bro found a new method

[midnight plank]

@tardy bloom Has your question been resolved?

[midnight plank]

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[gaunt mauve]

Why the homomorphism function is like that

[gaunt mauve]

I mean i know homorphism is a tool to compare two groups structure

But how this definition make us deals with that (comparing two structure)

[lyric charm]

in words, the property given here means this:

if we run any two elements through the homomorphism, then operate on them in H,
the result will be the same as if we'd operated on them in G and then run the result through the homomorphism.

in other words a homomorphism is a function which allows this sort of "translation"

you may know an example of one of these by another name:

the exponential function x ↦ e^x is a homomorphism (an isomorphism, in fact) from (R, +) to (R^+, *)

[midnight plank]

@gaunt mauve Has your question been resolved?

[midnight plank]

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[midnight plank]

Please don't occupy multiple help channels.

[tawdry kraken]

<@&268886789983436800>

[midnight plank]

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[wooden badger]

Didn't u hv the newbie badge a day ago?

[tawdry kraken]

Yeah haha

😅

[wooden badger]

And you also hv the helpful role? :0

[tawdry kraken]

Yeah I've just been in the server a lot 😆

[wooden badger]

Wow

[west iron]

they're speedrunning owner%

WR pace run

[midnight plank]

Available help channel!

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[clear lion]

how can we study the monotony

[clear lion]

aka the derivatives

[midnight plank]

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

[runic hamlet]

well it surely doesnt hurt to compute the derivative

[midnight plank]

@clear lion Has your question been resolved?

[midnight plank]

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[urban ferry]

i got the y coordinate of P to be 5t

i dont know how to get the y coordinate of Q

[fallow scarab]

On Q, if x = t, what's y=?

[urban ferry]

How are you getting x = 5t?

[fallow scarab]

Sorry typo

[urban ferry]

how are you getting x = t

that isnt the x coordinate of q

[fallow scarab]

Yes it is

[urban ferry]

how do you know

[fallow scarab]

"parallel to y axis"

[urban ferry]

oh 😭 😭

they are on the same vertical line

ok ty

.close

[midnight plank]

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[maiden bridge]

@stiff bison I- didn’t do actually anything

[maiden bridge]

Bro just because I have a personality doesn’t mean I’m a robot.

Unlike other people here…

Yall expect people to be sad, depressed, calm and boring..

[fallow scarab]

What kind of math question is this

Take your drama to #discussion

[cedar mason]

💀 ?

[bold peak]

<@&268886789983436800> troll(?)

.close

[midnight plank]

Channel closed

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[maiden bridge]

AND AGAIN someone pinged mods for no reason on me.

NO IM NOT TROLLING.

[bold peak]

Well you're being an asshole

One way or another

[patent snow]

What's the matter? How about you both go on with your day/evening?

[cedar mason]

/night

[pearl hull]

You were either

1. Ragebaiting
2. Being a racist
   or 3. Being oblivious

[cedar mason]

go sleep xavier

[maiden bridge]

It’s Christmas and it’s a mess on this day.

[patent snow]

I suggest you direct any similar claims to ModMail where the team will investigate the situation
I wish y'all a Merry Xmas, and please move on now

[maiden bridge]

I needed help and they just closed my channel.. for no reason.

[fallow scarab]

.

[maiden bridge]

Fine.

[bold peak]

I don't see a question.

[maiden bridge]

#help-14 message

[midnight plank]

Available help channel!

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[twilit field]

Now this is equivalent to solving for $7x^2+5x-9=x$

[grand pond]

waimas

[twilit field]

ofcourse I could factorise as a product of two linear polynomials but that'd be pretty useless

But that's basically finding the solns

[solar hearth]

oh hi

[twilit field]

we won't require fixed point iteration anymore will we

[twilit field]

hello

[runic hamlet]

no

you are supposed to find the roots of g

[solar hearth]

brb

[twilit field]

mhm

right

ooh

[solar hearth]

you want to express g(x)=0 as x=f(x)

[twilit field]

so $7x^2+4x-9=-x$ for instance

[grand pond]

waimas

[runic hamlet]

no

[twilit field]

Like if I have $7x^2+5x-9=0$, I have $7x^2+4x-9=-x$ no?

[grand pond]

waimas

[runic hamlet]

sure but thats not what you want

[solar hearth]

i mean, sure, but does it converge?

[oak nymph]

A fixed point problem wants you to solve something of the form f(x)=x

[twilit field]

I don't follow

[oak nymph]

Why are you trying to solve something of the form f(x)=-x

[runic hamlet]

dont think about convergence for now

[twilit field]

yea, so here I have $9-4x-7x^2=x$

[grand pond]

waimas

[runic hamlet]

ok so there we go

thats one option

now find a different one

[twilit field]

alternatively $x= \frac{7x^2-9}{-5}$

[grand pond]

waimas

[twilit field]

cool

As for convergence you're talking about the derivative being less than 1 near the roots I suppose

[solar hearth]

sure but idt you can use this for fixed pt iteration

[oak nymph]

The question just asks us to reduce one problem to another.

[twilit field]

sure, but the problem asks us to reduce it to finding the fixed points of some other function

[oak nymph]

It doesn't say that the fixed points need to be found via some method.

[solar hearth]

"via fixed point iteration"?

[oak nymph]

Well, I completely forgot about that. In my mind the question started from the part about "in other words.." which just asks to reduce one problem to another.

[twilit field]

hmm

[midnight plank]

@twilit field Has your question been resolved?

Channel closed

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[twilit field]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[twilit field]

to verify that this works I have to show there exists an interval around the fixed point in whihc the derivative is less thn 1, right

.close

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[split ingot]

Am I supposed to convert the first matrix into the second one using row/column swap operations

[runic hamlet]

you could just draw the graphs

[split ingot]

proving that is a pain isnt it

i'd have to prove that the graphs are isomorphs

[oak nymph]

Sure, but it may give you reason to suspect non isomorphism, and it doesn't hurt to take a minute to draw it anyways.

[unkempt sluice]

Check whether two matrices are different up to row/column permutation

[split ingot]

can i use row and column operations on these

[unkempt sluice]

And pay attention to ranks

[split ingot]

i will only swap

[oak nymph]

Just swaps, yes.

[unkempt sluice]

Because rank(PAQ)=rank(A)

For permutation matrices P,Q

[split ingot]

alright i did it

took 6 swaps

[unkempt sluice]

Which one, b)?

[split ingot]

yea

1st is easy

two swaps

btw if i had adjacency matrices can i do the same

[unkempt sluice]

Good

Yeah you can

[split ingot]

but i have to preserve order

dont i have to follow every row swap with a corresponding colum swap

and vice versa

[unkempt sluice]

Yeah PAP^t

[split ingot]

wat is that 💔

[unkempt sluice]

rows and columns doing the same permutation for adjacent matrix case

[oak nymph]

PAQ is arbitrary column and row swaps, but PAP^t forced the row swaps to be followed by corresponding column swaps.

[split ingot]

how do i know when to stop

i mean i'd be doing swaps forever this way

if they are not isomorphs

[oak nymph]

That's why you draw pictures... You have argued non isomorphism before as well.

It will tell you what to do

[split ingot]

like the c part in this looks like they are non isomorphic

[bold peak]

You can also do some counting arguments I think

[unkempt sluice]

To save your time. b) check rank. c) check eigenvalues

[oak nymph]

I mean these methods are simply overkill here.

[bold peak]

Yes the first step should just be drawing the graphs

[oak nymph]

Sure, they are valid methods. I don't disagree

[unkempt sluice]

Oh, just want to save his time on b) , c), in case he tries dozens of permutation first

[split ingot]

man this is dumb

these graphs are squares or smth

herzog is right

i'll just draw them its easy then

[split ingot]

i will draw the graph

for c, first is a literal square and the second is a straight line

[unkempt sluice]

Sure. Whatever works

[split ingot]

yea ty

i will draw the graph if its a 5x5 or 4x4 matrix

thanks guys

.close

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[split ingot]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[split ingot]

alright isomorphism question

u2->u3 and v2->v3 are the only ones different here

how to write this in a formal way (for exams and stuff)

[unkempt sluice]

Observe what happens if you delete u4->u2

On both graphs

[sullen bolt]

?help

[unkempt sluice]

Said some dumb thing ignore this

[sullen bolt]

!help

[midnight plank]

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

[unkempt sluice]

Maybe try to show they might be isomorphic first?

I don’t know. Might make no sense. Because it’s like choosing two different vertexes of the same graph

[split ingot]

but the edges are different, can they till be isomorphic?

[shell granite]

can anyone help me at ( a )

[unkempt sluice]

Oh

[split ingot]

like all the other ones are same, its only the problem with a single edge

[unkempt sluice]

Both 2 circles

[split ingot]

@shell granite This channel is currently occupied by
another user, please take this to
another channel

so what about the question guys

[unkempt sluice]

Found the permutation

[split ingot]

what is it

[unkempt sluice]

PAP^t on the first matrix A where P=(0,1,0,0;0,0,0,1;1,0,0,0;0,0,1,0)

[split ingot]

are you talking abt the earlier question

[unkempt sluice]

I was talking about this

[split ingot]

so did you find an isomorphism

[unkempt sluice]

u 1,2,3,4 to v 2,4,1,3

Sorry should be the inverse

v_ 1,2,3,4 to u_ 2,4,1,3

[split ingot]

how did you do that

i drew thhe diiagrams of both and theyre identical

[unkempt sluice]

(Outer degree, inner degree)
Both graphs four vertexes
(2,1),(2,1),(1,2),(1,2)

So you only have four possible choices

By examining case by case exactly one worked

[split ingot]

wat abt dis one'

[unkempt sluice]

What do you mean

[split ingot]

nvm

so i have to consider both the in and out degrees for directed graphs right

[unkempt sluice]

Yeah

[split ingot]

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[split ingot]

for a) all ik is that n(n-1)/2 = 2m

2m is total sum of degrees

[lyric charm]

look at the degree of each individual vertex and not their sum

[split ingot]

its gonna be even

so n = 2m

[lyric charm]

in K_n, what's the degree of each vertex

it isn't n

[split ingot]

n-1

[lyric charm]

right

[split ingot]

so n-1 = 2m

so n is odd

[lyric charm]

indeed

[split ingot]

and K_n has a path

circuit mb

what if they asked me abt the path

exactly two vertices have odd degree

[lyric charm]

all vertices in K_n have the same degree

therefore if you are in a "2 vertices have odd degree" then those 2 must be the only ones in the graph to begin with

[split ingot]

and C_n has an euler circuit for all values of n right

C_n graphs start from n>=3

[lyric charm]

yes. it is its own euler circuit

[split ingot]

and W_n never have

circuits

alright

.close

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[lyric charm]

W_3 does

wait no

euler yeah never

[midnight plank]

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[split ingot]

isnt it exactly two in case of euler path?

[shell wigeon]

An Euler circuit is an Euler path

[midnight plank]

@split ingot Has your question been resolved?

[shell wigeon]

There's an Euler path if there are two or zero odd vertices (can't have an odd number of odd vertices)

In the zero case it's a circuit

[split ingot]

single vertice of odd degree is impossible?

alr yeah it is

handhsake lemma

[shell wigeon]

Yes

[split ingot]

so is just exactly two

ty

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[shell wigeon]

Exactly two for "path but not circuit"

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[split ingot]

in these problems, can i remove as many edges as i want (as long the graph stays connected of course)

[shell wigeon]

Pretty sure you can only remove a specific number of edges

All spanning trees have the same number of edges (one less than the number of vertices), so the difference is always equal

[split ingot]

well if i try to remove m-n+1 edges from a graph with m edges then i will ultimately end up with a disconnected graph

[shell wigeon]

No?

In (2), there are 5 vertices and 6 edges, 6-5+1 = 2, you can remove edges a-b and b-c for example

Still connected, and it indeed yields a spanning tree

[split ingot]

i think i meant more than m-n+1

that would give me a disconnected graph

m-n+1 is the number of edges i need to remove to convert it into a tree

[shell wigeon]

Well yeah, since that would be equivalent to removing an edge from a spanning tree

[split ingot]

yup

anyways thanks

i have the answer now

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[west sundial]

Hi guys, could someone help me as to how to label this diagram?

[west sundial]

I dont know where to resolve the forces

i think its something like this?

and then where do i add 4gcosalpha and 4gsinalpha

and mgcosalpha and mgsinalpha

[mystic condor]

You assume the system goes one way and put everything in terms of the accelerations

And because the string is taught they both have the same acceleration

So for one side you set the net force equal to tension minus gravity minus friction and the other side gravity minus tension minus friction

[atomic mist]

it would be better to resolve the tension into its sin and cos components rather than gravitational force in this case

[mystic condor]

I dont think so, since you need to resolve everything up and down the plane

[atomic mist]

i heavent done these questions in a while but cant you just resolve the tension, put the vertical component equal to the weight and find tension in terms of g and then as tension on both sides is equal you can just equate them?

[mystic condor]

Hmm yeah perhaps

[west sundial]

yea ok let me try that thank u

[mystic condor]

But you also need to deal with friction

[atomic mist]

isnt the surface smooth?

[west sundial]

but i need to resolve the weights into components

[mystic condor]

It mentions coefficient of friction at the bottom

[atomic mist]

oh nvm only one is smooth

[west sundial]

so mgcosalpha and mgsinalpha

and i dont resolve the tension right?

cuz its already along the plane

[mystic condor]

Yeah like @atomic mist said you can also resolve them vertically and use equalibrium but it works both ways

mgcosalpha for reaction force and sin for gravity along plane

[west sundial]

okk tysmm

so like for the yellow part

i resolved P particle

did i do it right

legt is 4gsinalpha

left

and down is 4gcosalpha

and then equate R = 4gcosalpha

[mystic condor]

In the yellow box the plane is the horizontal?

[west sundial]

Ohhh yess ur right wait

sorry like that?

[mystic condor]

Yeah

[west sundial]

yayy ok wait then where does my friction go

cuz ill do the same to Q wtva but frictionnn

this way?

cuz f = mew R

Then like it'll be 1/4 R

[mystic condor]

Yes

[west sundial]

Ur the goat

Tysmmmm so helpful!

[mystic condor]

Np

[west sundial]

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[lusty python]

🔥

what's your question?

!original

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[delicate meadow]

Let D be any point on BC. Draw DF and DE || AC and AB respectively. Extend EF to meet BC produced at G. Find any relation only between two or more segments of GC, which are GB, BD and DC.

That is, find a relation that consists of GB and BD, or BD and DC, or GB and DC or GB, BD and DC.

[tall lantern]

I guess you can use properties of transversals here?

and then find the length of one of the segments in terms of the other segment?

perhaps you could work with GB and BD since according to me, they seem the easiest

[delicate meadow]

Actually, only a relation between all GB, BD and DC is possible.

[tall lantern]

Are you trying to find a relationship between ALL of them or any two of them?

[delicate meadow]

I confirmed that it isn't possible to find one between only any two of them.

An equation which consists of only and all GB, BD and DC.

[late rover]

Just gonna ask lol

[tall lantern]

Ahh I see.

again, I thinnk using the properties of transversals would work

[delicate meadow]

Exactly what properties, because transversal simply refers to a line cutting a system of lines?

[late rover]

menelaus' theorem should do the trick, with similar triangles on top

[tall lantern]

Wow i've never seen that

[tall lantern]

I was just referring to the relations of the angles in a transversal

but I guess you could also try what @late rover is saying

[midnight plank]

@delicate meadow Has your question been resolved?

[lilac finch]

*cutting parallel lines

[delicate meadow]

Not necessarily.

[raw vector]

Seems like Menelaus and Thalès

[lilac finch]

ic but they are here

[delicate meadow]

Of course.

[maiden sedge]

i hv exam in january :(

[delicate meadow]

Ok.

!status

[midnight plank]

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

[delicate meadow]

Nvm. I got it.

.close

[midnight plank]

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[maiden sedge]

i am dumb

[lilac finch]

ok

[late rover]

well that's what I said

[raw vector]

Sorry i didn't read the whole chat, you get the credit

[late rover]

not that serious mate

but thanks

[raw vector]

[lilac finch]

https://tenor.com/view/dog-smile-gif-21913178

[soft stone]

out of curiosity did you find a cleaner approach in the end for that question

[raw vector]

Not yet, i took a break and did other problems. But i'm definitely not done with it yet

I will continue until i find a more elegant solution

[soft stone]

hmm it feels like there should be a more elementary route i’ll think about it as well

gl

[midnight plank]

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[zealous schooner]

For part (ii) two obvious candidates are [ \lambda\cdot z = \lambda z ] and [ \lambda\cdot z = \overline{\lambda}z ]

[grand pond]

kheer257

[zealous schooner]

Is it possible to find all such scalar multiplication maps?

[runic hamlet]

something something related to all automorphisms of C over Q maybe? which would be a lot

[delicate meadow]

?

[runic hamlet]

havent thought about such a question before

[zealous schooner]

How did you get this?

[runic hamlet]

if you set z=1 then z*1=phi(z) and from all the vector space axioms you should get some axioms for phi

that would be my first idea anyway

[zealous schooner]

is z the scalar here

[runic hamlet]

yes

whoops

[zealous schooner]

well then phi is additive and phi(1) = 1

so we just care about what happens at i

or some complex number with non-zero imaginary component

[runic hamlet]

what about multiplicative?

[zealous schooner]

it's not multiplcative

right?

[runic hamlet]

dunno

just throwing words at you

[zealous schooner]

we have z1 * (z2 * 1) = (z1 x z2) * 1

[runic hamlet]

it needs some sort of multiplicative condition

anyway gtg

gl

[midnight plank]

@zealous schooner Has your question been resolved?

[unkempt sluice]

I don’t think you can. Too huge, Hom_ring(C, Hom_Ab(C,C)). Probably why the exercise only asks you to show it has more than 1 element. You found two. In reality I think it’s infinitely many, and unaccountable

Other than the two you have found, the rest almost always require some set theoretical construction

[midnight plank]

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[split ingot]

[0] = {x | x belong to Q}
[1/2] = {x | x belong to Q} (same class as [0])
[sqrt 2] = {sqrt2 + k | k belong to Q}

is this right for 1?

[modest heart]

Yes

[split ingot]

btw here 1/2 is in the equivalence class of 0, then why do we have a separate class for 1/2?

is it the same class, but the representative has changed from 0 to 1/2?

[modest heart]

I think the question just wants you to realise [0]=[1/2]

[modest heart]

Indeed

[split ingot]

How would I state that in the solution?

[modest heart]

You don’t really have to

[split ingot]

[0] = [1/2] hence 1/2 has same class as0?

[modest heart]

What you’ve written here is basically a full solution

[split ingot]

also elements in [0] shoudlnt be in [sqrt 2]

but it doesnt look like that

[modest heart]

I mean look at the two sets you wrote down

Do they look disjoint?

[split ingot]

sqrt 2 is gonna be in [0]

[modest heart]

Why?

You said [0] is the set of rational numbers right

[split ingot]

oh alr

[modest heart]

Isn’t root(2) irrational

[split ingot]

so do these classes really make a partition

on R

how would P look like

[modest heart]

Well if you believe xRy is an equivalence relation then the answer is yes

Any equivalence relation defines a partition and vice versa

Regarding what it looks like, you don’t really get anything clean, each equivalence class is just a translated copy of Q

[split ingot]

all rational numbers have the class Q right

i mean the same class as [0]

and [1/2]

[modest heart]

All rational numbers are in the class [0]

[split ingot]

yeah

thats articulate

[modest heart]

But observe that say [sqrt(2)] is just the rational numbers translated root(2) units to the right

[split ingot]

yea

then where is the partition

[modest heart]

A result you can prove is that
[a]={k in Q: a+k}

[modest heart]

Idk how you want this answered

[split ingot]

is a rational or irrational

[modest heart]

a is an arbitrary real here

[split ingot]

ig it works for both

[split ingot]

got it

ty

what about the last question

are there gonna be infinite classes cuz there are infinite irrational numbers

[modest heart]

Hm idk what results around cardinality you know

Well to start, do you know what cardinality is?

Or is this just asking if the number is finite or infinite

[split ingot]

number of elements in a set

[modest heart]

Okay cool so you’re not doing anything like “it’s countable” or “it’s uncountable”

[split ingot]

well no ive not done countable or uncountable infinities

whats the answer to c then

[modest heart]

Okay that makes this easier

[split ingot]

ik aleph null is a thing but idk anythin abt it

[modest heart]

Anyway c you’re right in that it’s infinite, the way youd prove this is to provide an family of infinitely many disjoint equivalence classes

[split ingot]

cant i just say that there are infinite irrationals and each irrational has its own class

[modest heart]

Well it’s non obvious to me that these are disjoint

Like if I used [sqrt(2)] and [sqrt(2)+1] these are the same class right

[unkempt sluice]

Use part ii to obtain infinitely classes

[split ingot]

how

[split ingot]

yup

[unkempt sluice]

You did (2)?

[split ingot]

yeah

they are not in the same class

[unkempt sluice]

well, 2 and 3 are primes

[split ingot]

thats what i said

[modest heart]

Isn’t the result you need to do that non trivial (linear independence of square roots of square free integers over Q)

[unkempt sluice]

I don’t think so. sqrt(q)-sqrt(p) is rational then its square is rational

[modest heart]

Oh yeah you can do things pairwise

[split ingot]

how to prove infinite equivalence classes

[unkempt sluice]

So got it Sean?

[split ingot]

no 💀

[unkempt sluice]

There are infinitely many primes

[split ingot]

is being prime importnat

what about sqrt 8 or smth

that has its own class

[unkempt sluice]

No. Just make things simpler

Just like you proved (2) for 2,3 you can use the same proof for p,q

[split ingot]

i did not prove anything, i just stated that [sqrt 2] = {sqrt2 + k | k belong to Q} and [sqrt 3] = {sqrt3 + k | k belong to Q}

then i wrote that they have different classes so they dont belong to same class

thats it

[unkempt sluice]

You understand that x is rational then so is x^2 right?

[split ingot]

yea

[unkempt sluice]

square of a rational number is still ration
If sqrt(q)-sqrt(p) is rational, then its square is, impossible

(For prime p,q)

[split ingot]

so whats the method of proving infinite classes

[unkempt sluice]

[sqrt(p)] : p primes
These classes

[split ingot]

what about [pi]

or [e]

which class are they in

[unkempt sluice]

That’s a different question. So you understand (3) now? Then we move on

[split ingot]

well no but lets move on

tell me abt pi

[unkempt sluice]

Then you better finish (3) first

And π-e is rational or not is unknown to human at this stage

[split ingot]

btw does this (3) ask for the number of distinct classes?

[unkempt sluice]

Like double lance suspected, which I agree, (3) only wants you to answer whether there are finitely many classes or not

[unkempt sluice]

I am saying these are already infinitely many classes

[split ingot]

but are they talking about the distinct classes or just the classes

[unkempt sluice]

Using this

[split ingot]

distinct or not

[unkempt sluice]

Distinct

[split ingot]

if parent set has infinite elements then the equivalence relation on that set will have infinite classes?

right

[unkempt sluice]

No

[split ingot]

not the distinct classes

just the classes of every element

[unkempt sluice]

Trivial relation, any x,y x~y
Only one class

Back to (3)
{[sqrt(p)]: p prime} is a subset of R/~

Former infinite, thus latter infinite

[split ingot]

can you please clarify this confusion, if we are asked about the number of classes of mod 3 relation, would we say that there are infinite classes, or would we say that there are only three classes

[unkempt sluice]

Z/~ where x~y when x=y mod 3?

[split ingot]

the word 'distinct' is not mentioned in the question

[split ingot]

yea

[unkempt sluice]

Three

[split ingot]

but they did not mention the word 'distinct'

why are we stating only the distinct classes?

[unkempt sluice]

Definition of set

Elements of a set are distinct

Any set X and its equivalence relation ~, the quotient set X/~ is still a set

Whose elements are distinct [x] for x in X. We omit distinct because X/~ is a set. Definition of set already contain “distinct”

[split ingot]

so the x is just a class representative here right

[unkempt sluice]

Yeah

[split ingot]

the class is basically the same right

[unkempt sluice]

What do you mean

[split ingot]

in (3), what if i say that there are infinite irrational numbers, isnt that enough for proving that there are infinite equivalence classes?

[unkempt sluice]

That’s not useful, [sqrt(2)]= [100-sqrt(2)]

Two different irrationals

R-Q -> R/~, mapping x to [x] is not injective

{prime numbers}->R/~, mapping p to [sqrt(p)] is

Whatever works, any infinite subset of R/~

[split ingot]

but i can just say root 2 has diff class than root 3 and root 5 and root 7 and so on...
Thus there are infinite distinct classes, as this process can be continued for root n

[unkempt sluice]

That’s exactly what I was trying to tell you

Any two [sqrt(p)] and [sqrt(q)] are different

When p doesn’t equal q, for prime p,q

[split ingot]

cool

ty

got it

[unkempt sluice]

Np and

[split ingot]

so thats it

new question lmao

[unkempt sluice]

And we still don’t know whether [e]=[π]. So don’t give it too much hope that human can find out recently

[split ingot]

sure

[unkempt sluice]

What’s your result

[split ingot]

i have only proved reflexivity yet

[unkempt sluice]

Good

Other two will fail

So to save your time better start to find counterexamples

[split ingot]

(3,2) and (1,3)

not anti symmetric

[unkempt sluice]

Good

[split ingot]

(0,3) and (0,1) and (-1,2) for transitivity

isnt this gonna be transitive

[unkempt sluice]

That’s not a counterexample, but you don’t have to disprove transitivity anyway

[split ingot]

why?

[unkempt sluice]

You don’t have (5,5)R(1,3)

[split ingot]

i think it is transitive

[unkempt sluice]

However, (1,2)R(2,0), (2,0)R(0,1)

[split ingot]

what abt this

[unkempt sluice]

Good

[split ingot]

alright

how did you know it was not transitive

and not anti symmetric

[unkempt sluice]

I did them a little quicker

Just doesn’t look like they are true

[split ingot]

is it symmetric?

[unkempt sluice]

No

[split ingot]

a <= c or b<= d then c<=a or d<=b

how can we disprove this

[unkempt sluice]

(0,0)R(1,1)

By a counter example

[split ingot]

what if i do not realize that it is not true

and i end up wasting my time trying to prove symmericity

[unkempt sluice]

You will waste time

So I stand up to save them

[split ingot]

youre not gonna be there tomorrow in the exam hall 💔

btw this is not even reflexive right

[unkempt sluice]

Yeah

[split ingot]

alright

[unkempt sluice]

a word can’t come before its own

Dictionary in general doesn’t have exact duplicate terms for all words I guess, though no law prohibits it

[split ingot]

cool

thanks dude

[unkempt sluice]

Np good luck with exam

[split ingot]

.close

[midnight plank]

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[umbral tusk]

why is 2(ln2)^2 not 4(ln(2))^2?

[inner barn]

what do you mean

[slender walrus]

order of operations

[inner barn]

because it only squares what's in the bracket

[umbral tusk]

im a bit confused bc doesn't the power multiply the 2 infront and also the ln(2)

[slender walrus]

no

[zenith snow]

If it was ln(2²) then u could say 2ln2
But if it's (ln2)² then u cant move the exponent

[umbral tusk]

ohh

[slender walrus]

$ab^2 \redneq a^2b^2$

[grand pond]

ραμOmeganato5

[umbral tusk]

true

[zenith snow]

The exponent needs to be on the function on which the logarithm is being applied

Not on the whole logarithm

[umbral tusk]

ah okok so it actually becomes 2ln(2)^2

[slender walrus]

not recommended to write it like that

just leave it as 2(ln2)^2

[umbral tusk]

right okok

tysm!

.close

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[midnight plank]

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[ocean plaza]

In (c), how is omega defined on morphisms?

[lyric charm]

is A an order category here

can probably just do the inclusion map if you need one

[ocean plaza]

but as $\Omega$ is a presheaf, an arrow $a \to c$ in $A$ should be mapped to an arrow ${b : b \leq c} \to {b: b \leq a}$, and there are multiple ways of doing this mapping

[lyric charm]

do you mean \Omega

[grand pond]

Eduude

[lyric charm]

arrow a -> c means a ≤ c right

oh dear

[ocean plaza]

wdym?

[grand pond]

Eduude

Eduude

[midnight plank]

@ocean plaza Has your question been resolved?

[unkempt sluice]

b|-> b Λ a

Ω(c)->Ω(a)

[unkempt sluice]

And which book is it from?

[ocean plaza]

It's from sheaves in geometry and logic

[unkempt sluice]

Kashiwara?

[unkempt sluice]

Oh, thank you

[unkempt sluice]

And the answer to your question

[ocean plaza]

Thank you

[midnight plank]

@ocean plaza Has your question been resolved?

[midnight plank]

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[tawdry kraken]

Here for another proof evaluation
Excusing the abuse of notation with S, what do we think of this

\begin{proof}[\textbf{37.a)}]
  Let $x_1, x_2, \dots, x_n$ and $y_1, y_2, \dots, y_n$ be the nondecreasingly ordered sequences of positive real numbers as described, and let $S'$ be the sum of these in any ordering not matching $S = x_1y_1 + x_2y_2 + \cdots + x_ny_n$ (for example, we could have $S' = x_1y_2 + x_2y_1 + \cdots$).
  We seek to show that $S - S' \geq 0$ for all possible $S'$. 

  We can express the difference between $S$ and $S'$ as a permutation, keeping $x$ in order and permuting the locations of the $y_i$. 
  For example, transposing $y_2$ with $y_3$ corresponds to $S' = x_1y_1 + x_2y_3 + x_3y_2 + x_4y_4 + \cdots + x_ny_n$.
  Consider any transposition of $y_i$ and $y_j$ with $i < j$.
  Then $S - S' = x_i(y_i - y_j) + x_j(y_j - y_i) = (y_j - y_i)(x_j - x_i) \geq 0$ as $j > i$.

  We use induction to reduce every $S'$ to a transposition of $y_i,y_j$.
  Begin with $n = 1$. Then $S = x_1y_1$ is trivially maximal because no $S' \neq S$ exist.

  Now suppose $S$ is maximal when $n = k$ and let $n = k+1$.
  Since $S$ is maximal when $n = k$, we know that permuting the locations of any $y_i, y_j$ for $1 \leq i,j \leq k$ will result in $S' \leq S$.
  Therefore, we limit ourselves to transposing $y_{k+1}$ with a $y_i$.
  But we have shown that $S'$ under a transposition always has $S' \leq S$, so $S$ is still maximal in this case. 
  Finally, suppose we compose multiple transpositions involving $y_{k+1}$.
  This permutation is equivalent to taking some $S''$ with $n = k$ and transposing $y_{k+1}$ with a $y_i$, $1 \leq i \leq k$. 
  By the inductive hypothesis, $S'' + x_{k+1}y_{k+1} \leq S$, and since $i < k$, $S' \leq S'' + x_{k+1}y_{k+1}$.
  
  Therefore, we have that $S - S' \geq 0$ for all $S'$ and $n \in \NN$, and thus $S$ is the maximal.
\end{proof}

[grand pond]

Coolempire2026

[tawdry kraken]

Personally I don't like it, also induction hasn't even been introduced in the book but doing it without I couldn't prove that a product of transpositions had S' <= S

Originally I had planned to go transposition -> product of transpositions -> product of disjoint cycles = permutation

The fact that texit doesn't indent my paragraphs or space the lines more than half a pixel from each other doesn't do any favors for reading this either

[eternal pawn]

you could show that for any sum, you can always reduce its value by ordering 2 terms, thus proving that S is minimal

it's like "short circuiting" induction by considering a minimal element that satisfies some hypothesis, and looking whether the previous element satisfies the induction hypothesis

[tawdry kraken]

I think I've heard of this before

[eternal pawn]

it's very analogous and sometimes faster to pick a minimal element then find a contradiction or smth

[tawdry kraken]

Okay back let me read your feedback again

[tawdry kraken]

Oh yeah

Strong induction

[lusty python]

First time bro asks a question 🔥

[lusty python]

Im thinking of induction

[midnight plank]

@tawdry kraken Has your question been resolved?

[lavish venture]

rearrangement

[tawdry kraken]

Okay I was revising for review but I have to run so I'll leave now 😂

[tawdry kraken]

But your idea of strong induction basically yields this immediately I think

But yeah nex I'll come with polyomino tilings because I'm not sure how to start some of the proofs

Hopefully layla is on then

.close

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[maiden bridge]

You can’t do this right?

[mystic condor]

yeah thats impossible

[dawn dagger]

I cant

[void hollow]

from what i know yeah

[void hollow]

because |2x-5| is positive and -3 is negative

[deep vine]

You could make a new number though that satisfies |x| = -1

Lets call it σ

Then uh... you have an entire third axis

[void hollow]

I'm curious, as I don't know that much

still just a newbie at math

[mystic condor]

similar to how i^2=-1

[void hollow]

haven't learned about imaginary numbers yet, someday I will understand what that means u.u

[mystic condor]

ok cool then

[maiden bridge]

@mystic condor but is it impossible

[mystic condor]

yes its impossible

[maiden bridge]

But what number makes it to 3?

x=4

[mystic condor]

you can solve for that

[maiden bridge]

Ok ok

[void hollow]

ohhhhhhh wait I think I understood now what you both said gamer and oakley

[maiden bridge]

@mystic condor Are absolute lines considered as brackets?

[void hollow]

so if you had another dimension to it, you can satisfy it theorically, but not with just what's given right?

[maiden bridge]

You first calculate inside it and then change to positive number?

[mystic condor]

yes in order of operations

[void hollow]

if you add*

so if you add omega for example you could bypass the | | I guess? but wouldn't that change the equation? idk

[mystic condor]

yeah it just helps represent answers

instead of going "error"

[void hollow]

oh makes sense

[maiden bridge]

also x = 1

[umbral torrent]

for any real or complex number the modulus of the number is never negative

[umbral torrent]

small correction,

|2x - 5| >= 0

it can equal to 0 if x=2.5

[void hollow]

so basically here what's happening is that |2x-5| = -3 so |2-5| = -3 which means |-3| = -3 but |-3| is not equal to -3 because the | | makes the numbers inside always turn positive so it would be: 3 = -3 which is not right, because 3 is not the same (or equal) to* -3

[void hollow]

true, forgot about that, you are right!

[mystic condor]

pretty sure thats irrelevant for the point he was making

[maiden bridge]

Ikik

@mystic condor why do you multiply by sqrt3 - 1 not sqrt3+1 like it is?

[mystic condor]

to rationalise the denominator

you know difference of two squares?

$(x+y)(x-y)=x^2-y^2$

[grand pond]

ImOakley

[mystic condor]

when you have sqrt3 as x then when you multiply by the conjugate that sqrt3 will just be squared in the result

thus always making it rational

[maiden bridge]

Huh

Can you explain it more human language

[void hollow]

I don't know what this means, I tried to solve (x+y)(x-y) and I arrived at x^2 - yx + yx - y^2, ohh wait as I'm writing this I understand, -yx + yx annulates themselves so it's just x^2-y^2

[void hollow]

try to solve (x+y)(x-y) and you will arrive at that conclusion, in this specific case your (x+y)(x-y) is the (sqrt 3 +1) * (sqrt 3 - 1)

I'm learning new stuff here too, it's hard, but I think that's it?

[void hollow]

which solves into sqrt3^2 - 1^2

$(10sqrt3-10)/(3+1)$

[grand pond]

druiditbeau

[maiden bridge]

Bruh

[mystic condor]

@maiden bridge do you understand difference of squares

[maiden bridge]

Ya

[mystic condor]

So if the denominator is $\sqrt{a}+b$

[grand pond]

ImOakley

[maiden bridge]

But idk why that matches that

[mystic condor]

And you multiply it by $\sqrt{a}-b$

[grand pond]

ImOakley

[mystic condor]

Thats just difference of squares

So its $\sqrt{a}^2-b^2$

[grand pond]

ImOakley

[mystic condor]

Which is just $a^2-b^2$