help-36

[onyx peak]

if u meant what u said

consider e.g. some oscillating bounded function defined on some open interval which oscillates faster and faster as u approach the endpoint (or oscillates faster and faster as u approach infinity)

[sick roost]

well, this counterexample isnt periodic tho

but yeah bounded + continuous does not necessarily imply UC

[eager shore]

Okay

Why is this true

Do I just say like

for any y, there exists an integer n such that y - np is in [0, p]

[sick roost]

atlases and whatnot

i think we just bash delta epsilon here tho

[eager shore]

atlases?

[sick roost]

oh wait you actually only connect at endpoints here, so not even that

lets focus on showing bounded on compact period interval first

[eager shore]

I've already showed that on a bounded domain, a uniformly continuous function is bounded in a previous exercise

[sick roost]

by what theorem tho

[eager shore]

No theorem, that was the exercise

[sick roost]

oh you cant do that

youll get circular

[eager shore]

Oh wait

I need to show that it is uniformly continuous

mb

Well uh

I think I can figure this out for myself

I'll try some stuff and show you my progress

[sick roost]

#❓how-to-get-help

[final saddle]

@eager shore Has your question been resolved?

[sick roost]

do the THIING house, dont forget to do the THING

[sick roost]

let R be a metric space is cursed

[eager shore]

It's how my prof abuses notation

[sick roost]

let 1=1

[eager shore]

How do I proceed then

I have my periodic function from [0, p] --> R

And I know its UC on [0, p] because of this lemma

So then if I have some point y in R \ [0, p], I can claim that there exists an integer n such that y - np is in [0, p]

Bing chilling?

[sick roost]

i think this works

now we need to show bounded

[eager shore]

Contradiction again?

[sick roost]

evt

[eager shore]

I don't know what EVT is

[sick roost]

extreme value theorem

[eager shore]

(canonically)

Do I have to prove another lemma now

[sick roost]

evt is completely standard and actually necessary for continuous integrals to be nontrivial

[eager shore]

integrals?

intervals?

[sick roost]

im just saying you can definitely invoke it

rudin does evt/mvt before UC

iirc

[eager shore]

i haven't been to class for 2 weeks so i wouldn't know

it doesn't seem to be in the lecture slides

[sick roost]

i think its necessary here

[eager shore]

i'll take your word

[sick roost]

continuous on a closed and bounded interval?

that just screams evt

[eager shore]

okay

EVT

I will look the statement up and see what can be done

[sick roost]

it just has a global min and max here

[eager shore]

I'll try formalizing this too

[sick roost]

actually really simple theorem

[eager shore]

that's guaranteed for us right

[sick roost]

yes

evt says continuous on compact set ==> global min and max on that set

[eager shore]

Hmm

okay

[sick roost]

ill give you the precise statement

EVT:

let $[a,b]\subseteq\mathbb{R}$ be closed and bounded, $f:\mathbb{R}\rightarrow\mathbb{R}$ continuous, then

$$\exists c,d\in[a,b]:\forall x\in[a,b]:f(c)\le f(x)\le f(d)$$

[soft zealot]

Cycadellic

[eager shore]

I'll just be slapping a |f(x)| =< M

¯_(ツ)_/¯

I think I'm gonna have to use like

[0, p + 1] for UC

or rather [0, p + p]

[sick roost]

nono wait

can there be a number below f(c)? what about above f(d)?

[sick roost]

for these f(c),f(d)

[eager shore]

uh

idts?

[sick roost]

so boundedness follows immediately from this

[eager shore]

just choose M = max{|f(c)|, |f(d)|}

[eager shore]

yeah

[sick roost]

were already guaranteed bounds by evt

[eager shore]

well yeah

now moving on to UC

[sick roost]

just prove they are actually bounds, and youre good

[eager shore]

life is too short, say trivial and move on

[sick roost]

true

but muh constructive formalism

so we have P the compact period interval, we know f is UC and bounded on P, and f is continuous and periodic in general

[eager shore]

I'm gonna have to dip for a bit

I'll come back in a few minutes

(10)

[sick roost]

well you can argue it in a few lines really simply tho
therefore f(c) is LB, f(D) is UB, (both by existential inst on EVT)
therefore f(P) is bounded (by bounded above and below)

[eager shore]

[sick roost]

it works

[eager shore]

Okay I'm back

Time to write stuff down

[sick roost]

from here, show UC + B

[eager shore]

@sick roost I'm kind of stuck

[sick roost]

The gist of the argument is ‘delta epsilon holds for the period, and periodic, so delta epsilon holds in general’
Ill be more specific in a sec

Fire alarms :Ppp

Worst part of living on campus honestly

[eager shore]

They shut the library here at 2 am

So I had to move out of there

Hence my delay

[sick roost]

Oh i see

[eager shore]

I think I got it

I can casework y'

And set it inside the epsilon

Let me try that

@sick roost

[sick roost]

works

[eager shore]

Gotta fix the grammar at the start

And also probably add that ep > 0 is arbitrary

[sick roost]

well

we need to let \epsilon>0 to prove the forall to begin with

[eager shore]

Yeah

But since I've declared delta explicitly

it's nice to redeclare epsilon with it

[sick roost]

just make sure you dont clash scopes

[eager shore]

I don't think I am

[sick roost]

you arent

[eager shore]

Right this took a while

[sick roost]

'redeclare' just made me raise an eyebrow

[eager shore]

But I think it's as hard as this sheet gets

These are the remaining problems

[sick roost]

lipschitz functions are fun

[eager shore]

Alright, I'm gonna get some sleep now, thank you for your dedication :)

.close

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[tranquil pine]

I need help solving for b here, the end goal is to derive an exponential equation that passes through both points listed [which are (-1,40) and (0,12)]

[tranquil pine]

this is for algebra 2 as well

I'm gonna cry

[elfin ingot]

Yo

[tranquil pine]

Yo

I'm like so stuck

[warm ether]

what is b^0

[elfin ingot]

Im sorry bro im just as new to this as you

[elfin ingot]

1

[warm ether]

there you go

[tranquil pine]

Anything to the powet of zero is 1

but what should I do with that

[warm ether]

now you can just solve for b

40=12/b

[tranquil pine]

How tho

[tranquil pine]

How did we get here

[warm ether]

b^0 is 1, that goes away

b^-1 is 1/b

thats all i did

[tranquil pine]

So I rewrite them as 1 and 1/b

[warm ether]

sure

[tranquil pine]

Fuck

I'm so dumb

I should have known that

I got it now

Wait

[tranquil pine]

Can we say a=12

So now I have both variables? And the equation that goes through the points is y=0.3*12^x

[warm ether]

it would seem so

you can check that if you need to

you seem to have switched a and b though

[tranquil pine]

Yep

It's 12*0.3^x

[warm ether]

aye

[tranquil pine]

Does this seem too hard for algebra 2

Or is this standard

[warm ether]

i have a different education system so i cant say, sorry

[tranquil pine]

Ahh

We use a new curriculum this year and it's so bad

🥀

.close

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[fresh axle]

Hey! I'm in high school and my math skills are really bad to the point where I don't even understand the basics from ninth grade. How can I improve?

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[rain sentinel]

hi so i dont need help answering the question i just have a question about it i.e. why do i only draw FBD of the ones stated

[loud sundial]

discord.gg/physics

[rain sentinel]

.close

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[plucky rover]

Fastest resolution in history

[runic tulip]

.reopen

[final saddle]

✅ Original question: #help-36 message

[plucky rover]

Oh nice reopen now links back to the original question

[cerulean coyote]

Hi

[runic tulip]

Is your question why we draw the fbd of C, D, E?

[gleaming anchor]

hi! if you are here to chat, please head to #discussion or #chill

[cerulean coyote]

Kk

Also, I’m a guy btw

Just sayin’

[plucky rover]

!redir regardless of your gender

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This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[craggy plank]

@cerulean coyote Can you please stop emoji spamming?

[runic tulip]

I guess I'll close

Op left

.close

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[plucky rover]

Tbf we should have tagged her

[runic tulip]

I reopened so yeah

Yeah

[gleaming anchor]

reopening doesn't tag the OP

[plucky rover]

@rain sentinel feel free to reopen this, we can still help you even through it's physics

[cerulean coyote]

Myb but these ppl are hilarious asf

[rain sentinel]

.reopen

[final saddle]

✅ Original question: #help-36 message

[rain sentinel]

.

[runic tulip]

Well my recommendation is to draw the fbd of all points

Then see which are useful to solve for the unknowns

[rain sentinel]

yes i did this

[runic tulip]

Send pic

[runic tulip]

Also share the question

You only started the answer

Shared*

[gleaming anchor]

was this not the question?

[runic tulip]

Try reading it

Let's say that the structure is in static equilibrium
Then we can see that pin A and B will experience some normal force due to the ground

There is no way of finding that hence we can't use those two pins to solve for the unknowns

That is why we use the fbd of C, D, E

@rain sentinel I hope this helps

[rain sentinel]

it does thanks

[runic tulip]

Nice

!done

[final saddle]

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[rain sentinel]

how do we know member BD is a zero force member

[runic tulip]

I don't know what "zero force member" means

[bronze creek]

make fbd at point B

vertical sum=0

[rain sentinel]

thanks

[formal tusk]

you use tan(theta)=opposite/adjacent

[rain sentinel]

.close

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[hearty swallow]

What is 1+1

[runic tulip]

<@&268886789983436800>

[steep hatch]

Please don't shit post in the help channels. Take it to #discussion or #chill

[formal tusk]

2

[steep hatch]

@hearty swallow

[rain sentinel]

,close

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[meager hamlet]

can someone explain how to aproach problems like these

[meager hamlet]

there's 12 problems here and im not sure how to aproach them

[shell condor]

Can you send images please instead of a file

[meager hamlet]

is a video fine?

[shell condor]

Yeah ig

[meager hamlet]

sry i just didnt feel like taking so much screenshots lol

anyways is there any strategies i should follow when solving these?

there's multiple questions types from what i see

[shell condor]

Alright so for continuity, x<3 and x>3 must be equal

Now g(x) looks like a graph of -x² moved up the y-axis by (approximating) 5.5 units

So what would be the equation?

You still here @meager hamlet?

[final saddle]

@meager hamlet Has your question been resolved?

[meager hamlet]

sorry i was eating

[meager hamlet]

-x^2+5.5?

[shell condor]

Yes

[shell condor]

No problem dude

Okay so g(x) is -x² + 5.5

What would g(x - 2) be?

[meager hamlet]

-x^2+3.5

4.5??

cuz subtract 2 from 5.5 and do lim x--> 1

[shell condor]

No no

for g(x-2) replace x with (x-2) in g(x)

So that'd be -(x-2)^2 + 5.5

Now expand it

[meager hamlet]

-x^2+4x+1.5

[shell condor]

Yup

Now that, is f(x) when x<3

For continuity, that and x>=3 must be equal

[meager hamlet]

-x^2+4x+1.5=x^2-k??

or do i wanna plug in 3

[shell condor]

Yes

Now you can plug 3

That's give you the value of k

[meager hamlet]

k=4.5?

[shell condor]

Yup

That's correct

Now B?

[meager hamlet]

wait, why is it ^2 over here

[shell condor]

Because originally, g(x) was -x^2 + 5.5

[meager hamlet]

oh nvm i see its x^2 and u put (x-2) for x

[shell condor]

Yes

[meager hamlet]

okay i see

ill need to step away for like an hour so ill come abck later to finish these

[shell condor]

Alright sure

Close this channel for now then, you can start a new one when you're here

[meager hamlet]

got it

.close

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[plucky pike]

can someone help understand why the 1/2pi pops out infront of the integral when representing X(omega) using parseval's theorem?

[proper raptor]

because $x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} , d\omega$ maybe ?

[soft zealot]

χ : s ↦ Tr(ρ(s))

[plucky pike]

im trying to find the power from 3 to 5 radians for , e^(-3t) for 0<t<infinity, and 0 for -inf < t< 0, and found that X(omega) = 1 / (3+ j omega)

[soft zealot]

χ : s ↦ Tr(ρ(s))

[plucky pike]

then how does it become a constant

infront of the integral

<@&286206848099549185>

[latent dragon]

!15min

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[plucky pike]

<@&286206848099549185>

[sick flame]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[ivory bear]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[sick flame]

lots of people with the 15m out here huh

[plucky pike]

its been 19

[ivory bear]

[static fractal]

I think it's just from the normalisation factor in the inverse fourier transform

[plucky pike]

F-1 { X(omega } ?

so what just substitude omega = 2pif

[static fractal]

I'm not super familiar with your notation, but basically if you expand |x(t)| to x(t)\overline{x(t)}, then substitute the IFT on the right

then you get a factor of 1/2pi there

and I'm not sure what the rightmost expression in the original picture is, sorry

[plucky pike]

its the representation of omega = 2pi f which is the convention, integrating with respect to frequency

[static fractal]

I think if you do that substitution though, it should remove the factor out the front, yes

just from chain ruling it

[plucky pike]

can you show me any working?

all good i solved it

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[radiant carbon]

idk what i did wrong

80g is the bigger force as the person is falling down

so i take away the other forces acting upwards (2Tsin60 + 20g) from the force acting down (80g)

but the solution does the opposite -> 20g + 2Tsin60 - 80g = 0, and idk why

[shell condor]

Shouldn't it be 80g - 2(20sqrt(3)*sin(60))

Because 80g is downwards

Oh wait you gotta prove that

[ivory bear]

and form your equations

[wise willow]

you inverted sin and cos no ?

[ivory bear]

oh yea mb mb

wait then

[wise willow]

You have 2T*sin(60) + 20g upwards

And 80g downards

[ivory bear]

here sir now try

i made an error on my end

[wise willow]

So you have 60g = 2Tsin(60)

T=60g/(2*sin(60))

$(80-20)g = 2*T*sin(60)$ => $T = \frac{60g}{2*sin(60)}$ => $T=20\sqrt{3}$

[soft zealot]

Semanteo

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[sonic crescent]

Hi, I am learning binomial distribution, and I was wondering how we should solve problems like P(X > 15) ?

[sonic crescent]

is the only way really to sum up P(X = 16) -> P(X = 20) ?

[vital crag]

yes

or use a calculator

[sonic crescent]

but that sounds so time consuming, what if you have to sum hundreds of values?

[vital crag]

lol

then you're doing something wrong

[sonic crescent]

Like take p = 0.1, N = 1000, X = number of successes

we want to find P(X >= 500), we have to sum P(X = 500) -> P(X = 1000) ?

[vital crag]

indeed

that's why you shouldn't make up your own problems and just do actual ones from your textbook or homework

[sonic crescent]

i understand i just cant believe it is really this inefficient

[vital crag]

yes that's why calculators exist

[sonic crescent]

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[robust horizon]

what exactly do they mean by this?

[robust horizon]

please ping if responding

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[robust horizon]

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[weak dune]

j

[stable ether]

can anyone give me tips, im learning algebra 2 but havent finished 1 at same time, any vids or resourceass

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[oak arrow]

I tried this out with a 2d case of C

[oak arrow]

if i understood correctly, essentially I did (dx(w^tCw) * (w^tw) - dx(w^tw) * (w^tCw), dy(w^tCw) * (w^tw) - dy(w^tw) * (w^tCw)), right?

but that doesn't exactly give an answer. idk im just really confused with this problem

i guess we could replace dx and dy with d1, d2, d3, ... , dN for all n variables

but still, im not sure if im on the right track

[heady otter]

$dx(w^tCw) * (w^tw) - dx(w^tw) * (w^tCw), dy(w^tCw) * (w^tw) - dy(w^tw) * (w^tCw)$

[soft zealot]

caspar

[blazing condor]

Are you you trying to compute the gradient of the quotient ? And setting it to 0

I think that’ll work

[floral nova]

I'm not really used to this notation, but the usual way to calculate this is to first define a function f_C(w) = w^{T}Cw and find its derivative, then use that to help you find the derivative of g(w) = 1/f_I(w) so that you can calculate the derivative using the product rule on f_C(w)g(w).

I don't see where your x's and y's are coming from unless you are just trying to assume the matrix is 2x2 despite the problem saying it is NxN

[oak arrow]

i was using a 2x2 as an example because i was just kind of lost. I was thinking it may help me generalize to nxn, but i don't think it did unfortunately

[blazing condor]

It’s not necessarily bad to start by evaluating a simpler case before abstracting the math

[oak arrow]

yes that's what im trying to do, im just having trouble because these are vectors rather than coefficients

[blazing condor]

OH I HAVE AN IDEA 💡

[oak arrow]

i also intially tried evaluating the gradient like this:

[floral nova]

If you follow what I wrote, you should notice that u with C as a parameter is just v where C = I.

[blazing condor]

Ok I’m to lazy to formalize this

But

1. notice g(w) is scale invariant for any nonzero scalar

2. conclude that because of that property it is sufficient to look for stationary points on the unit sphere = {w | w^Tw=1}

Which I think is just the same thing as finding the extrema of the numerator subject to the constant of w^T*w-1=0

Then this just becomes a Lagrangian multiplier problem .

[oak arrow]

like this?

[blazing condor]

This also makes sense and requires less observational tricks and hand wavy conclusions

[oak arrow]

i don't think my math skills are this high level 😭
I don't really understand 1. and i don't know lagrangian multiplier. I hardly even know lagrange error, if they're at all related lol

i found the gradient of that function to be this:

[floral nova]

I am saying that you have a function $f_C(w) = w^{T}Cw$ so that $d/dw f_C(w) = (C+C^T)w$. Similarly, observe that you can define a function $g_C(w) = \frac{1}{f_C(w)}$ and that by setting $C = I$, you get $\frac{1}{w^{T}w} = g_{I}(w)$. You can then get the derivative of $g$ in terms of $f$, then use the product rule.

[soft zealot]

JessicaK

[blazing condor]

How do you do the latex thing in this chat?

$f_C(w)$

[soft zealot]

Guido

[blazing condor]

Of we are cooking now

[oak arrow]

okay i get the first part (C + C^t)w. I don't get the part about defining the second function and setting C to the identity matrix. How do you know what C is? Can't C be anything?

i understand how if C is the identity matrix, then it just becomes 1/w^Tw, but i don't understand how you can make the assumption that c is the identity matrix

[blazing condor]

so $g(w) = \frac{w^{T} C w}{w^{T} w}, \qquad w \in \mathbb{R}^N, ; w \neq 0$ Let $N(w) = w^{T} C w, \qquad D(w) = w^{T} w$ therefore %\nabla N(w) = (C + C^{T}) w, \qquad \nabla D(w) = 2w%, using quotient rule $\nabla g(w) = \frac{D(w),\nabla N(w) - N(w),\nabla D(w)}{D(w)^2}$

[soft zealot]

Guido

[floral nova]

I don't really use this matrix algebra much, so maybe it's been a while, is it not the case that $w^T I w = w^Tw$? Maybe you need to more carefully define it in terms of a directional derivative or something to make it more precise.

[soft zealot]

JessicaK

[blazing condor]

$\nabla g(w) = \frac{D(w),\nabla N(w) - N(w),\nabla D(w)}{D(w)^2}.$

[soft zealot]

Guido

[blazing condor]

THERE WE GO

[floral nova]

The point I am trying to make is that you can differentiate the denominator in the same way as the numerator by treating C as a constant parameter

[oak arrow]

No I believe that w^tIw is indeed w^tw. I just don't understand how you can say that C is the identity matrix

[floral nova]

I'm not saying C is the identity matrix

I am defining a function with a parameter C

I then differentiate it to find the derivative of that. This itself gives me the derivatives of all expressions of the form w^TCw

in particular, it tells you what the derivative of w^Tw is by setting C = I

Sorry I need to go right now

[oak arrow]

this makes sense, but we are trying to find the stationary ponits. So it would be D(w) N(w) = N(w)D(w), which seems like a roadblock

[floral nova]

Imagine you have x^n and you differentiate

you have nx^(n-1)

and this lets you plug in n to get the derivative of x^2, or x^17

This is what I was doing with C.

Okay I need to leave for real now.

[oak arrow]

thank you for your help

[blazing condor]

no one are gradiants some are scalars

$D(w),\nabla N(w) - N(w),\nabla D(w) = 0.$

[soft zealot]

Guido

[blazing condor]

you cant cancel gradients like you do scalars

[oak arrow]

oh my bad, i think i misread it

[blazing condor]

you already computed all of those values in your note page

[blazing condor]

right here

[oak arrow]

So then the answer is abstract? It's just D(w)dwN(w) = N(w)dwD(w)

[blazing condor]

NO

you can do more

[oak arrow]

oh my bad

[blazing condor]

your good

[oak arrow]

ok im just going to write this out

[blazing condor]

the stationary condition after substitution I get : $(w^{T}w),(C + C^{T}),w ;=; 2,(w^{T} C w),w$

[soft zealot]

Guido

[oak arrow]

just continuing from the work i had earlier, i computed this with quotient rule:

that looks similar to what i have, but does the order matter since they are vectors?

or are both answers correct?

[blazing condor]

you have to do some symplification

but yes

start with that

set it equat to 0

[oak arrow]

okay so one thing i see here is that both sides can be multiplied by inverse w, right?

[blazing condor]

quantity (w^T*w) will never equal 0 since w = 0 so really just set the numerator equal to 0

then seperate at the subtraction sign

[oak arrow]

right, isn't that what i did up here?

[blazing condor]

yes

you did

[oak arrow]

oh okay good, just wanted to make sure.

[blazing condor]

you can divide by w^Tw on both sides since its not equal to 0

[oak arrow]

wait but i thought matrix division isn't clearly defined?

are you sure i can cancel those 2 vectors out like that?

[blazing condor]

so w^Tw is just a scalar actually.

its 1x1

[oak arrow]

big brain move

so now i just have this, right?

[blazing condor]

yep... now the right side simplifies further...

[oak arrow]

how can 2C simplify further? I was just thinking to multiply both sides by inverse (C + C^t)

[blazing condor]

ah i see i thought it was glitching and not showing your full work

matrix mult isnt associative

you cant cancel W^TCw /(W^Tw)

$(C + C^T),w = 2 \frac{w^T C w}{w^T w}, w$

[soft zealot]

Guido

[oak arrow]

ohhh i see. yes, i forgot how matrix mult isn't associative

[blazing condor]

using the Proffessor theorem Im reasonably confident this is right because the conclusion i got to this problem is narly

[oak arrow]

wait so that's the final answer? i don't need to multiply both sides by inverse (C + C^t)?

[blazing condor]

no look at what g(w) is defined as originally

then look at what i have writen down

we are heading towards an eigenvector relation

[oak arrow]

i never learned eigenvectors/eigenvalues 😭

[blazing condor]

JIMMINY CORNDOGS

[oak arrow]

so it doesn't need to be simplified any furhter?

[oak arrow]

we didn't have enough time in my previous lin alg class

very unfortunate

the prof omitted it from the final exam

[blazing condor]

what class is this for?

[oak arrow]

foundations of machine learning

[blazing condor]

Holy Guacafricken moly i assumed thatd just be some linear regression stuff

[oak arrow]

wait im sorry, im still confused on how this proves that it's the final answer

[blazing condor]

ok one sec

[oak arrow]

you're a funny dude

[blazing condor]

$\frac{(C+C^T)\omega}{2}=g(\omega)\omega$

[soft zealot]

Guido

[blazing condor]

by definition the left side is....

[oak arrow]

ohhhhhhhhhhhhhhhhhh

[blazing condor]

$C_{sym} \omega = g(\omega)\omega$

[soft zealot]

Guido

[oak arrow]

wait im sorry, i don't know what Csym means

[blazing condor]

all square matrixes can be seperated into a symmetrical matrix and a "skew matrix"

$C_{sym} = \frac{C+C^T}{2}$

[soft zealot]

Guido

[blazing condor]

a symmetric matrix is a matrix such that $a_{ij}=a_{ji}$

[soft zealot]

Guido

[oak arrow]

oh i see

[blazing condor]

anyways back to the whamy here

[oak arrow]

something like
1 2
2 1

[blazing condor]

yes exactly

For a symmetical C , must correspond exactly to the eigenvectors of C

if C is not symmetric, it corresponds to the eigenvectors of its symmetrical sub component

thats what the relationship i wrote out shows us

which is really cool and something i didnt know until this problem so thanks for teaching me something!

[oak arrow]

oh okay thank you

[oak arrow]

haha, don't be thanking me, i should be thanking you! Seriously, i really appreciate all your help

sorry for taking up so much of your time. i really appreciate it though!

[blazing condor]

time to go learn eigenvectors kiddo. they are super useful in a lot of things, defining steady states, finding perservitive mappings in transforms, System dynamics, solutions to ODEs etc. If your interested in graphics in computer sci its a tool you should put in your tool belt

I had fun dont worry about it

i bet 3blue1brown has a good video on it

[oak arrow]

yes, ill definitely have to check that out. Certainly it would be useful for AI concepts in compsci

thank you again!

[final saddle]

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[steel shore]

doing a math assignment, where im using calc to investigate volume of cake tins. the 2nd columm has the length 25 and the width 23, cutting away corners with length x gives me my intial volume formula i then find the derivitive to find the stationary points yk to find the maximum. BUT i also have a general formula shown above that i found by inserting in the variables into the quadratic equation (assuming r doesnt equal s). however when length 25 and width 23 is solved it doesnt have the the coefficient of 2 out the front of the radical. i dont know why? is my working out wrong or is it my general formula. (it worked for the other one)

[final saddle]

@steel shore Has your question been resolved?

[wary juniper]

Uh the 2 is put inside the square root to eliminate the denominator of 4

Both answers are the same

[steel shore]

denominator of 4?

isnt there only 24 or 6?

[wary juniper]

The fraction inside the square root

Would be 579/4 using the general formula

[steel shore]

ohh

i see

so could i pull that out and write is as 2sqrt (579/4)

because i need make it clear that the general form connects to an example

[wary juniper]

Yeah

[steel shore]

cheers

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[warm python]

{e}, Z_3,Z_5

[warm python]

n -generators is worrying me

my first thought was to use the euler phi function

and I suppose that I can say we chose the smallest k, such that phi(k)=n

but the range isn't N

[tired walrus]

does such k always exist

[warm python]

.

[tired walrus]

if not, when does it and when does it not

[warm python]

I have to prove that it doesn't, I suppose

[tired walrus]

this is now a number theory question

[warm python]

😔 , lemme try none the less

[tired walrus]

ie "what's the range of euler phi"

[warm python]

I suppose it's sufficient to show no cyclic group with 3 generators exists( for this problem)

[warm python]

Is that what I have to find here

or just show it isn't possible for all n

[tired walrus]

question wording makes it unclear whether a complete characterization is sought or just a single CE

[warm python]

I suppose a single counter example is fine.
Let there exist a group with 3 generators. Then there exists a number $n \in \N$ such that it has 3 numbers co-prime to it.
\
We then have $3=n \prod_{p \mid n} (1-\frac{1}{p})$

[tired walrus]

mm

[soft zealot]

wai

[tired walrus]

you have a few issues with wording and notation

well, one less now

it is probably best to simply say $\varphi(n)=3$ --- but i am also unsure whether this formula you wrote out will be convenient at all for the proof

[soft zealot]

Ann

[warm python]

I'm unsure if they even expect a proof out of us here tbh

[tired walrus]

here is a thing you ought to know: if p divides n then p-1 divides phi(n).

this has consequences for when phi(n) is odd.

[brittle atlas]

Answer if you need to view it:
||A cyclic group of order m has exactly Euler’s totient function(m)
To get a cyclic group with exactly n generators check whether n is a totient value
If n = 1 take Z1 or Z2
If n = 2 take Z3 Z4 or Z6
If n = 4 take Z5 Z8 Z10 or Z12
In general: if n + 1 is prime, Zn+1 works
But if n is some even number that is not totient (eg 14) then no cyclic group has exactly n generators||

[warm python]

so p|n \implies p-1|3 for us

and no such prime exists

[lofty sinew]

Does the group need to be of form Z_n?

[warm python]

Not necessarily

but my proof works for any cyclic group I believe

[lofty sinew]

Probably, I admittedly didn't read it too closely

[warm python]

like for examples I just chose the most obvious ones

[lofty sinew]

Hmm

Are you interested in another possible approach?

[warm python]

as long as it;s not too complicated

[lofty sinew]

The one I'm thinking of is pretty simple I think

Trying to decide how much I should outright say....

I guess... do you think for groups G and H, you can make a group out of GxH?

[warm python]

I haven't done group products :(

I mean I think I can learn it quickly

but I have an exam tomorrow so not now

[lofty sinew]

Fair enough

There's complex stuff like inner products and whatnot, but this is a very simple way of combining two groups

(g,h) * (g',h') = (g*g', h*h')

[warm python]

ah I've done this in my operator theory course

I guess its identical ( like a direct sum?)

[lofty sinew]

Yeah, it's essentially a direct sum

So if G has n generators and H has m, how many should GxH have?

[warm python]

nm

[shy arrow]

@hazy vine

[lofty sinew]

Sorry but I don't believe so

[warm python]

oops

right

[lofty sinew]

Like, if e is the identity of G and i the identity of H, then any (g,h) = (g,i)*(e,h)

[warm python]

yea, fair enough

[lofty sinew]

So if you can make g out of n things and h out of m things, you can make (g,h) out of ___ things

[warm python]

Is it fine if I don't do this now. I have my midsem tomorrow, and am just revising for now

[lofty sinew]

Fair enough

[warm python]

Like I'd honestly, and I don't mean this sarcatically love to do this( I want to do algebra), but, I have to do well in my exams too

,ti

[soft zealot]

The current time for math_rocks is 10:59 AM (IST) on Fri, 03/10/2025.

[warm python]

Thanks!

[lofty sinew]

yeye

[warm python]

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[split rivet]

Can I get help with rotation and translation and some others I forgot

[gleaming anchor]

you always can. please just send your question!

[split rivet]

I dont got math homework I just wanna know what to do with shapes

[gleaming anchor]

help channels work better if you have specific questions to use as examples to illustrate what you want to understand.

[split rivet]

Oh

Ok

I meant like with the shape gotta use a coordinate to transfer the shape to the other side and it could be flipped or the same

[gleaming anchor]

original questions work best, please.

[spark anchor]

we call it transformations

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[shy socket]

do i plug into the 1/f'[f^-1(x)] formula

[shy socket]

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[gaunt garden]

hi

[violet junco]

yes/

?

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[wary current]

If a function, f(x,y) = u(x,y) + iv(x,y) is analytic on open set G, does it mean u, v satisfy cauchy riemann equations on G

[violet junco]

!xy

[final saddle]

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.

[lone oxide]

So I've been working ona framework for 3 months. Its a physics engine primarily with suites of math, condmat, 🧲 s. Meta level red team orchestrstorn that maintains clay rigor. Anyway I solved a 100 year old problem tonight and want to share. This isn't typical ai garbage. I'm racing to beat the major companies to it.

Anyway, I'm finishing compiling a latex right now if anyone wants to see. Its going to naturenfirst unless someone can help me submit it somewhere more dedicated to mathematics. I built the engine entirely with a cellphone and the major llms by the way.

3D Ising Model Free Energy, 100 year old problem. Oct 3, 2025
https://drive.google.com/file/d/1_P-vMXbKuusG2DdVd6BDbAGMpq83wvag/view?usp=drivesdk

[violet junco]

BRO WTH

[wary current]

There isn't a question, this is just me trying to prove a part in the textbook

[violet junco]

oh

[wary current]

But in my proof, I realized I have to assume "if f is a function, f(x,y) = u(x,y) + iv(x,y) is analytic on open set G, does it mean u, v satisfy cauchy riemann equations on G
"

But I have no theorems in the textbook that actually says that...

[formal trail]

doesn't theorem 4 say that

[wary current]

wait a sec..

it does.

thank u

Just to make sure:
f(x,y) = u(x,y)+iv(x,y) differentiable at point z0 => Cauchy Riemann equations hold at all points at z0

However,
Cauchy Riemann equations hold at all points at z0 !=> f is analytic on z0
^ The converse is not true because u,v satisfying cauchy riemann doesn't guaranteed the first order partial derivatives are continuous

[formal trail]

yes, we need both continuity of partials and cauchy-riemann for differentiability

[wary current]

thank you for confirmation

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[manic leaf]

What is the best way of approaching this problem

[worldly mesa]

Just calculating the product

[manic leaf]

ah ok just directly

ill just try it out then

So wait, $P_{\sigma} = PQ_{\sigma}$

[soft zealot]

toast

[manic leaf]

Where $Q_\sigma$ is some permutation matrix?

[soft zealot]

toast

[manic leaf]

then you just expand out $P_\sigmaD_\sigmaP_\sigma^{-1}$

[soft zealot]

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

[manic leaf]

ok ye i see now

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[granite stump]

I got 9! For a, 8!/2! For b, 9!-(8!/2!) But i do not know how to do d

[granite stump]

I would really like a pointer rather than an answer

[wary juniper]

Lets say you have 7 women in a row
How many spaces are there where could you put a man in

[granite stump]

8 spaces for a man

But after the first man there sre uhh

It depends on where the first man is doesnt it

I guess i'll ask later

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[swift void]

I have one question, if i have two vectorspaces in $\mathbb{R}^4$
, $V$ and $W$ both with a set of rules like $y+z+u=0$ and the other with $x+y=0, z=2u$ if i want to find a base of $V\cap W$ do i substitute the rules?

[soft zealot]

jelly v20

[worldly mesa]

The intersection would be the subspace defined by these linear equations yes

[swift void]

Thank you

!help

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[swift void]

.close

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[stone wagon]

yo

careful with uppercase/lowercase

oh i read this as conditional probability lol

[worldly mesa]

P(Y=y) = P(Y^-1(y))

so yeah you are right

[stone wagon]

$P_X({x : h(x) = y})$

[soft zealot]

artemetra

[stone wagon]

i am at my job rn

procrastinating

bill gates? more like william doors

in any case what you wrote is correct

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[severe verge]

y = (x-1)(x-2)(x-3)...(x-2009). Calculate y'(1).

I have managed to get the correct answer by using the limit definition of the derivative. But this problem is in the "derivative of composite functions" part. So I wonder what's the book's intended approach here.

[scarlet sequoia]

y = (x-1) * [....]

And so I'm imagining it's refering to derivative of product of functions

y' = 1 * [....] + (x-1) * [....]'

and so y'(1) = ....

[severe verge]

it's not referring to the product. it's referring to f(g(x)) sort of stuff

[scarlet sequoia]

then... no it has not a lot to do with derivative of composition

[final saddle]

@severe verge Has your question been resolved?

[loud sundial]

anything is a composition if you try hard enough

maybe proving the generalised product rule by induction?

and that's their composition?

[severe verge]

might try later

[severe verge]

I did that and idk how that thing's got to do with the derivative of a composite function

it's just product rule repeated over and over again

after that I can just chuck in x = 1 and done

[loud sundial]

$g_{n+1}(x)=x \prod^{n}{i=1} f_i (x)$ and you want the derivative of $g{n+1}(f_{n+1}(x))$ idk

[soft zealot]

Civil Service Pigeon

[severe verge]

I'm starting to get a feeling that our uni is trolling us

[sick roost]

Youre going to want product rule here

[severe verge]

I'm just gonna close this I don't think there's any point in finding the intended solution for this.

[severe verge]

yes I know it works

[severe verge]

but this is the question I'm having

[sick roost]

I dont see how chain rule helps

Which is the composition rule of derivatives

[severe verge]

alr I'm gonna close

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[radiant carbon]

how is this possible without knowing the height of the lawn, the width of the middle veg section (we can imply it is 3m from looking at it, but is that really accurate?), width of the other 2 veg

also width of shrubs

i wanted to do (12 * 18) - area of all the box areas

but i can't work them out so it's cooked

[final tangle]

its implied that the plots are aligned

the text below also gives info not marked on the diagram

(paths are 1m wide exct for...)

[plucky rover]

Oh shit I didn't see the text

[final tangle]

(veg and shrub are 0.5m)

[radiant carbon]

how does this help me work out the height of the lawn for example

nvm

i see it

[final tangle]

you don't need the individual width,
if you can get the total, that'd be enough

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[silver quarry]

can someone explain why this justification is needed

nvm i misread .close

.close

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[static wind]

I am currently an adult student finishing my last semester of my bachelor's in data science. I have so far, gotten all A's and understand most of the statistical ideas and programming behind the concepts and modelling, but I am still rather shaky when I need to read a textbook and it is full of big formulas. Attached is just one example of the kinds of big formulas I am talking about, not the only kind of formula I am struggling with.

My issue is based on the fact that each individual piece of these sorts of big formulas, I understand. For example, I can follow along that these formulas are dealing with the population means, standard deviations, and giving outputs that are dependent on probabilities among classes. But the actual nuance of reading a formula like this, and perfectly comprehending what is happening without lots of struggle and confusion, I have not reached.

My question is: What approaches should I take to actually fully understand formulas like this when I come across them in a textbook? When I read aloud each component in english, I just get more confused.

[vital crag]

find other resources and do a lot of problems

this is better suited for #study-discussion

[static wind]

Gotcha, I can move the question there if that would be preferred

[latent dragon]

Dear riemann, why can’t I react to your messages? I can do it to everyone else…

[willow tiger]

i can

to both

[latent dragon]

My screen has a seizure and then removes it for some reason

only for riemann and no one else.

did he like block me or something

[willow tiger]

lel

[willow tiger]

!done

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[static wind]

.close

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[sage phoenix]

dude this is brutal

[sage phoenix]

do you basically just make 1 of this

[craggy plank]

presumably you're looking for the solution of x, right?

[sage phoenix]

x^2 + x − 2 > 0

[sage phoenix]

yeah solutions and interval

[vital crag]

set it equal then factor or use quadratic formula

[faint locust]

You can multiply through

[sage phoenix]

x1 = −1, x2 = 2 ⇒ L = (−1, 2)

[sage phoenix]

these

[lucid nymph]

its absolute value so you need to be careful

[vital crag]

oh yes you're right

[sage phoenix]

oh nvm

i just looked up the solutions its like crazy

[willow tiger]

Das ist nicht sehr schwer

[tired walrus]

why are they re-using x_1 and x_2

would they not want to name the sols of the second case x_3 and x_4

or is that not bürokratisch

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@sage phoenix Has your question been resolved?

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[loud sundial]

,w 3z+1=(8-2z)/i

[soft zealot]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[loud sundial]

,w -3b=8-2a, 3a+1=-2b

[soft zealot]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[loud sundial]

there's an error?

[round oar]

i think so

[loud sundial]

why

both a=1.5b+4 and 4.5b+12+1=-2b are satisfied by (a,b)=(1,-2)

[round oar]

the problem is in the last row

thanks anyways

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[mental gazelle]

how do i go about solving this problem,
at the very least I know that x < 9 but thats all i got

[vital crag]

x can be 0

[mental gazelle]

what about division by 0 error?

[vital crag]

what's the denominator when x=0?

[mental gazelle]

oh I mean I know x can be 0, but i mean the end result cant be 0 which is where i got x less than 9 from

[vital crag]

yes 0 < 9

x = -9 is less than 9

but is x=-9 in the domain?

[mental gazelle]

no

wait

nevermind thought i had something

[lapis sky]

ok so the denominator cant be 0 right

[mental gazelle]

right

[lapis sky]

ok so how do you write it, formally?

[mental gazelle]

formally? i wouldn't know

[lapis sky]

like x^2 - 81 is different from 0

thats what I meant

right?

[mental gazelle]

right

[lapis sky]

so that equals that x^2 is different from 81

right?

[mental gazelle]

right

[lapis sky]

so what are the only real numbers that make x^2 = 81 ?

[mental gazelle]

9,-9

[lapis sky]

exactly

so you cant have those two numbers on your domain

right?

[mental gazelle]

yep

[lapis sky]

so that leaves all real numnbers except -9 and 9

so you'll have the intervals )-inf, -9( U )-9,9( U )9, +inf(

and these are all open sets because -9 and 9 are not in the domain

does it make sense to you?

[mental gazelle]

i guess i dont get what the middle set is there for, and what is getting the numbers inbtwn -9 and 9

[lapis sky]

because the numbers between -9 and 9 are in. the domain

only -9 and 9 aren´t

[mental gazelle]

the U symbol is union right? so from my view it looks like:
(-inf, -9) including (-9,9) including (9,inf) but obviously -9,9 are excluded, and the way i have it written excludes the numbers between -9, and 9 because they arent included by anything above

[lapis sky]

the U is union yes

(a,b) this is an open set. It´s the same as ]a,b[

so ]-9,9[= (-9,9), which are all the numbers from -9 to 9. It does not include -9 or 9

[mental gazelle]

ok

i think i get it thanks

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[upper nimbus]

Given $P(S)=40%$ $P(S^C)=60%$ $P(P|S)=80%$ $P(P|S^C)=30%$

I am asked to find $P(S^C|P)$ and $P(P^C)$

I have 36% and 50% respectivley, is this right?

[whole halo]

correct latex uses \% instead of %

[soft zealot]

Player_X_YT

[upper nimbus]

ty

[whole halo]

% by itself is a "comment" operator that treats the rest of the latex as a "note from the developer" which is why most of it disappears

for example if you had something like
\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6 % Basil problem solved by euler

,,\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6 % Basil problem solved by euler

[soft zealot]

mtt

[whole halo]

yes, its correct
P(P & S) = P(P|S) P(S) = 0.8 * 0.4 = 0.32
P(P & !S) = P(P|!S) P(!S) = 0.3 * 0.6 = 0.18
P(P) = P(P & S) + P(P & !S) = 0.32 + 0.18 = 0.5
P(!P) = 0.5
P(P) = 1 - P(!P) = 0.5
P(!S | P) = P(!S & P) / P(P) = 0.18 / 0.5 = 0.36

[upper nimbus]

cool, just wanted to double check bc i couldn't find a good calculator online

thanks

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[warm python]

$0≤ \frac{n^2}{n!}≤ \frac{ n(n+1)}{n!}$

[soft zealot]

wai

[warm python]

Does thiswork for squeeze theorm

[true heath]

im a bot from the future, friend me

[pliant shore]

indeed, nice idea

I think you need a bit more work to turn it into the product of fractions *whose limits to infinity are all less than 1 however

[pliant shore]

oh wait, you don't need to, cause $\frac{n + 1}{n - 1} \to_{n \to \infty} = 1$ already

[soft zealot]

south

[pliant shore]

and then the next fractions are $\frac{1}{n - 2}, \frac{1}{n - 3} \cdots$

[soft zealot]

south

[warm python]

yea,pretty much

[warm python]

or alternatively 1/(n-3) ->0 and the rest is bounded

[pliant shore]

yeah basically the first whatever terms are bounded

so you don't have an indeterminate infinity * 0 form

[warm python]

let a_n= 2^/n!

I can show |a_{n+1}/a_n| <1

so a_n->0?

[worldly mesa]

Yes that works

I think

[warm python]

cool, lemme just make sure there's nothng wacky when I do that

2/n(n+1) is what you get

which does indeed go to 0

[worldly mesa]

Without the n

But still goes to 0

[warm python]

hmm?

yea

This looks kinda fun

eh, I suppose it's direct enough

My bad :(

[worldly mesa]

Very direct

[warm python]

yea, just realised 😭

I'm just bad at math

$|x_n-x|<x/2$ for some K by defn

[soft zealot]

wai

[warm python]

so x/2<x_n<3x/2

[worldly mesa]

Similarly you can prove x_n eventually will be in any open interval containing x

[warm python]

yea

I suppose this is direct enoughtoo

a^{n+1}/(a^n+b^n}->0

hmm, I'm trying to find an all encompassing soln

[worldly mesa]

Ignore

[warm python]

okay, so my idea works right

I think it converges to b

[worldly mesa]

Yeah

[warm python]

cool :D

Can I post a few more ( fairly easy questions ) for verification here

is there any quick way to do this

I could rationalise it and do stuff

but

[worldly mesa]

Am gm

[warm python]

idts, -n≤0, no

[worldly mesa]

??

[warm python]

AM-GM requires all term to be +ve

[worldly mesa]

Am gm for the sqrt

[warm python]

ah

right

got it

tq

[worldly mesa]

This gives a lower bound

Maybe completing the square

(n+a)(n+b) is a quadratic

something like = (n+(a+b)/2)^2 + ...

[warm python]

got it

tq

[worldly mesa]

What I did is basically proving AM Gm

Because I realised we are in a case very close to equality of AM GM

[warm python]

I think I can do this later?

I'll thin about it and see if there are any more powerful results that can be used

Tq!

I'm just a bit down now , sorry :(

.close

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[warm python]

I'm just wodnering How i'd go about this

[tired walrus]

let $y_n = \frac{x_{n+1}}{x_n}$ for convenience

[soft zealot]

Ann

[warm python]

I was thinking of chosing eps carefully

[tired walrus]

you're told that $y_n \to L$ and that $L > 1$, and yes indeed a good choice of $\ep$ exists.

[soft zealot]

Ann

[warm python]

`okay, I can work from there I think

[tired walrus]

think about what lim y_n = L says about the behavior of x_n

[warm python]

like I had the same idea but wasn't sure if it would work

[tired walrus]

would you like a hint

[warm python]

in 5 minutes?

Lemme write something rq

$1<y_n<eps+L$ eventutally then we have increasing seqeunce,with no upper bound?

[soft zealot]

wai

[tired walrus]

this is like vaguely in the right direction but 1 is scary

[warm python]

1<y_n?

[tired walrus]

i cant rly say more without spoiling

[warm python]

you said 1 is scary, what do you mean by that

[tired walrus]

i cant say more without spoiling

except if you want me to give that hint i offered

[warm python]

as L>1 , 1<y_n eventually

[warm python]

sure

[tired walrus]

what if $\frac{x_{n+1}}{x_n}$ were \textbf{exactly} $L$ for all $n$? what could you then say about $(x_n)$?

[soft zealot]

Ann

[warm python]

x_{n+1}=Lx_n

so if x_n were to converge

1=L

[tired walrus]

[medium-volume incorrect buzzer]

forget about x_n converging or not

[warm python]

is this worth thinking over, if so I;ll eat and come back to this

[tired walrus]

yes it is

[warm python]

okay

will do that then

Tq!

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[formal trail]

<@&268886789983436800>

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