help-4

[viscid hawk]

and then I don't, like today

(it is 5 am my time)

[proud leaf]

first derivative it then use y-y1 = m(x-x1)

[vapid sand]

I think the max I have ever really spent on a single thing of homework for a class is probably like 3 hours. Don’t get me wrong studying can go for much longer but I kind of go insane when doing it for too long

[viscid hawk]

question: how do I take the derivative when the Y isn't separated?

[vapid sand]

Do you know the calc 3 trick to doing this derivative fast

[viscid hawk]

Is it just... d/dx?

[vapid sand]

Okay you don’t

[proud leaf]

dy/dx?

[vapid sand]

Here I will teach it is very useful

Instead of doing it the regular way with dy/dx

You take the partial with respect to y and divide it by the partial with respect to x, and it gives you the derivative

[viscid hawk]

Yeah, I need meds to actually get basically... anything done, which is frustrating, but it keeps me on the task and not going insane

[vapid sand]

If you don’t know partials they are very easy to learn and then super useful for this

[proud leaf]

its literally make a derivative and simply make it = 0
then take a common factor and simplify it
then apply the (-1,1) to find m
and then use y- y1 = m ( x -x1)

[vapid sand]

Yes I know but the calc 3 trick is much faster

[viscid hawk]

also keeps me sleep deprived sadly because it lasts 14 (15 if you count the time to digest) hours and I need to eat a meal before I take the pill, which means I'll have to be awake for 2-3 hours before the thing even kicks in

[proud leaf]

like this

[proud leaf]

he still young for that twin

[vapid sand]

Bro I am four years younger than him

I am 15

[viscid hawk]

I'm not that young though

[proud leaf]

NO FING WAY

[vapid sand]

Yes

[proud leaf]

thats the slope btw

[viscid hawk]

I simply stopped doing math for like 4 years

Well, if you ignore chem, physics, biology

[proud leaf]

i do all of these

except biology

i hate it

[proud leaf]

i thought you were older

like 20 or sum

am 18 btw

[viscid hawk]

oh... so I was overthinking it.. alright

[viscid hawk]

my favorite lol

[vapid sand]

Also like I said this can be done super fast with the calc three method

Like in your head

[viscid hawk]

I love biology the most because it's just kind of... intuitive? both my dyscalculia and partial dyslexia can function in anatomy and biology classes

[proud leaf]

yup i know this

[viscid hawk]

or well, they don't hinder me nearly as much

[viscid hawk]

I can't do this, the brain fog can't do that lol

[viscid hawk]

but is it essentially the same thing? take dy/dx of both sides

[proud leaf]

brain rot

[viscid hawk]

I just can't do the whole in my head thing

I've had too many very simple errors from that

[proud leaf]

just use the accurate method for you

[viscid hawk]

Truthfully, I can't remember much of what I just consumed/heard/said, etc.

so, mental math isn't all that reliable for me

gotta write it down or I lose everything

[proud leaf]

so you wanna become a doctor or smth

[viscid hawk]

ehh not exactly sure, I love the medical field but I know I have my challenges

[proud leaf]

good luck then

am a software engineer major but i like maths and physics

[viscid hawk]

and it's not exactly safe to have a doctor who cannot read something new off of a paper and remember it without ample ability to rationalize what I read essentially

[proud leaf]

so i try to help you all

[viscid hawk]

hmm that's a risky major to take right now

unless you specialize in AI, perhaps

[proud leaf]

not really combine it with cybersecurity and you become valuable

[viscid hawk]

yeah, I was going to say that was your other option

[proud leaf]

i can literally get a salary of over 300k with these

[vapid sand]

Anyways I got to go to bed

[proud leaf]

ye twin gn

[vapid sand]

Gn

[viscid hawk]

I believe people who were generally majoring in that are funneling into one of the two jobs that most likely will not be endangered by AI

[viscid hawk]

good night

[proud leaf]

we cant get replaced with ai cause we make ai and someone gotta keep monitoring it

thats where software engineers shine

[viscid hawk]

I mean, if all else fails I go into real estate. However, it's unstable if the housing market crashes I'd be doomed.

[proud leaf]

am a real estate agent and customer service rep. ( i need money so i work and study)

[viscid hawk]

yes, you have to specialize in AI and AI utility or cybersecurity now. Those who make webpages, create new code for general management, databases, etc. would be out of a job

[proud leaf]

literally

[viscid hawk]

I know there's a lot more, however both my parents were coders and I am unbelievably awful at it, so I'm definitely not following in those footsteps now that it's even more unsteady

[proud leaf]

also i try to sell ai agents to local shops and companies and it really works
first payment and then getting monthly payment from monitoring them

its really simple

[viscid hawk]

HR is hell though, you run into the most incompetent people in high places.

[proud leaf]

try oil field engineer its really easy and profitable

[proud leaf]

i swear to god sometimes i wanna punch those jerks

[viscid hawk]

again, math isn't exactly my strong suit so I'll stay away from engineering lol

[proud leaf]

ah

not all engineering is math

[viscid hawk]

ehh, sort of, I can do basic construction math

[proud leaf]

there's bio engineer and chem engineer

enviroment eng

and etc

and they all pay well

[viscid hawk]

yeah, I was interested in that but getting a degree for environment is kind of scarce

not many places even teach it

with decent teachers and connections

[proud leaf]

kinda true but companies look for them alot tbh

[viscid hawk]

hmm never really thought of that esp with how my country is going in terms of pollution 💀

[proud leaf]

or study law if you love memorising

[viscid hawk]

law is fine, I just think I'd hate my job.

[proud leaf]

law is really beautiful

[proud leaf]

not if you work with the right people

i always wanted to become a lawyer but instead i chose engineering

[viscid hawk]

oh, huh.

Yeah, I guess my only significant interest has been in mostly biology.

[proud leaf]

then study it

[viscid hawk]

I already am lol

I need the calculus for a BA in bio

[proud leaf]

ah

[viscid hawk]

anyways, I have to go take care of my pet chickens and cook some breakfast, so thanks for the help :)

[proud leaf]

cook them

[bright wind]

Have the problem been resolved in this help channel?

[viscid hawk]

yes, idk how to make it unclaimed

[bright wind]

Type .close

[viscid hawk]

.close

[vale dock]

Channel closed

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[warm bloom]

wahts the calc three method

[vale dock]

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[next gull]

.close

hi

[honest stone]

huh?

do you have a question?

[next gull]

yes

one minute

[honest stone]

Just for future reference, can you please post the image/question as your first message so the bot doesn’t pin greetings or requests for help? This way, if the conversation continues for long, helpers don’t have to dig backwards to find the problem you are having trouble with.

[dreamy scroll]

you write this each time

[next gull]

how post the question? im new

[vale dock]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[keen tundra]

Picture p

[dreamy scroll]

you could type it or add a picture

[fleet grail]

uhh.... Upload a pdf, copy/paste.

[lyric sundial]

Oh lol my bad 🙈

[next gull]

im do the sistem of higher grade but I didn't understand

[honest stone]

no. I have it saved as a text replacement.

nohi
Creates that everytime

[dreamy scroll]

ic

where's the op btw

[honest stone]

no clue. Probably tryna figure out how to post an image

[keen tundra]

What is your question?

What do you need help with?

[next gull]

this

how do you do it

[umbral isle]

Can you find the solutions to the quadratic equation there

[next gull]

no if someone tell me how it did it

how its my question

[keen tundra]

Do you know any techniques to solve quadratic equations?

[next gull]

yes some tips

.close

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[autumn lantern]

[vale dock]

Available help channel!

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[spring lily]

I have this internal law on G = (-1; 1), x•y= (x+y)/(1+xy)

[spring lily]

i was asked to find 1/2 • 1/2 • …. • 1/2

i was not given the number of times i use the composition law

[stark wedge]

you were not even given it as a variable like n?

[spring lily]

if i was j was not given it

i assumed it’s just n

[stark wedge]

well that's the best you can do

[spring lily]

i did this

i’m unsure if it is correct tho

and if it is i’d like to know the proper syntax

to me it seemed that it approaches 1 as n approaches infinity

so it’s sort of like a limit

[dreamy scroll]

2nd terms wrong

[spring lily]

bro

wtf was i doing

[dreamy scroll]

the expression grows closer to (n-1)/n

[dreamy scroll]

happens to the best of us

[spring lily]

oh

okay but is there any way to find an actual formula

aside from whatever i did

like say i have n = 100

[dreamy scroll]

not rly I'm assuming they're just producing this function infinitly

so you can just find the convergence

[spring lily]

yea but the question was to find an answer for an n amount not just say it converges to 1

and i have no idea

for the actual formula

[dreamy scroll]

hmm

lemme think then

[spring lily]

i’m thinking it is something to do with powers and factorials

and the term below is just the same with +1

ig it does help that i know it converges to 1 but there is no other way of determining the formula other than just seeing it

i’ll try again at some point and correct the second term

thanks!

.close

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[opaque stratus]

any good resources of olympiad combinatorics for begginers

[vale dock]

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[dreamy scroll]

found it

[dreamy scroll]

shit

[dreamy scroll]

hey i found the patern to your question

took some time and inuition

.close

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[dreamy scroll]

mb

[dreamy scroll]

Xo here is 1/2

[vale dock]

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[tiny torrent]

if G is a group and H is a subgroup, are all the distinct cosets of H in G subgroups of G?

[tiny torrent]

or is H the only one we can confirm to be a subgroup

[quasi valve]

H is the only one that is a subgroup, the others are definitely not

because only H contains the identity, for example

[tiny torrent]

o ok

[quasi valve]

you can think of the other cosets as "shifted" or "translated" versions of H

for example, the x axis is a subgroup of the additive group R^2, and the cosets of the x axis are the horizontal lines

[tiny torrent]

mm I see that makes sense

thanks

.solved

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[tiny torrent]

ok im having trouble

[tiny torrent]

could someone give me a small tip

so the problem is "Prove that ifH has index 2 in G then H is normal in G"

so far

I have

[quasi valve]

do you know the characterization that H is normal if and only if aH = Ha for all a in G?

[tiny torrent]

its if aHa^-1 is a subset of H

for alll a in G

[quasi valve]

yes that's the definition

but the way i stated it will be helpful here

(and is equivalent)

[tiny torrent]

oh we learned what you stated as the definition and what

I said as the characterization\

I havent tried to use aH = Ha tho

[quasi valve]

ah i see

[tiny torrent]

let me give that a try

[odd jackal]

hint : eH is a coset of G, and eH is a normal coset ( show this)

[tiny torrent]

normal coset?

I havent seen that before

[quasi valve]

me either

[odd jackal]

H is a normal subgroup

my bad

[tiny torrent]

ok ill give it another go

[pallid oasis]

you can also get H as the kernel of a homomorphism from G to S_2

[odd jackal]

They're doing a book that hasn't introduced homomrphisms yet[ if the edition is the one I have]

[pallid oasis]

normal subgroups are modeled after kernels, and are exactly the kernels

[quasi valve]

yea and you need to do it this way if you want to prove the more general statement that if G:H equals the smallest prime dividing |G| then H is normal
but for G:H = 2, there's a simpler argument that works

[pallid oasis]

i mean sure, then look at both the left and right cosets of H

[vale dock]

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[sullen crater]

Can anyone explain me the last formula please

[honest stone]

Is this GPT.

[sullen crater]

Yeah

Okk

[honest stone]

Sorry, I just usually like to check when dealing with AI.

[sullen crater]

The 4th i know is something about binominals or something like that

[vale dock]

@sullen crater Has your question been resolved?

[distant galleon]

[https://brilliant.org/wiki/integer-equations-star-and-bars/]

[obtuse bobcat]

hi

[vale dock]

@sullen crater Has your question been resolved?

[sullen crater]

Hi

[obtuse bobcat]

my bad i didnt realize it was a help channel 😭

[sullen crater]

Ahh okk

No problem

[obtuse bobcat]

i could still try helping you depending what you need help with

[sullen crater]

its about the chart above

if you could explain me the formulas (4th, 5th and 6th) above

[obtuse bobcat]

what grade is it

[sullen crater]

12

[obtuse bobcat]

alright close enough

[silk hemlock]

Hi

[sullen crater]

hi

[silk hemlock]

Can I ask questions regarding alg word problems

[obtuse bobcat]

so first try finding out how many ways to place n amount of balls into k amount of urns

do you understand that part

also sorry if my teaching isnt very helpful (im in 10th grade but i still understand it)

[sullen crater]

okk thnx

[sullen crater]

kinda

[obtuse bobcat]

SSo basically n represents the balls, and k is representing the number of urns

is that more understandable

[slate herald]

#❓how-to-get-help

[sullen crater]

urns?

[obtuse bobcat]

yeah

[sullen crater]

I am not a english native speaker

[obtuse bobcat]

me neither

but

all you gotta know its that urns are being represented by k

is that better

[sullen crater]

Okk

yeah i get it ig

[obtuse bobcat]

ok

[obtuse bobcat]

ao now with the knowledge that n = the amount of balls, and k = the urns, try making sense of the first steps of the equation as its showb

shown

to find the possible ways to place n into k labeled urns

[sullen crater]

but i cant understand the last one

just not make sense to me yk

[obtuse bobcat]

which last one are you having trouble with

[sullen crater]

the last one in the chart

[sullen crater]

.

this

[obtuse bobcat]

combination with repetition?

[sullen crater]

yeah

the formula is very confusing

[obtuse bobcat]

all arrangements are just permutations of n+k-1 objects, of which n are stars and k-1 are bars

look for example

uh

suppose you have n = 12 balls and k = 5 urns

it would be (12 + 5 - 1)

5 - 1 on the bottom

its just like how its shown in the formulas

[vapid sand]

Partial derivative with respect to y divided by the partial derivative with respect to x, is equal to dy/dx

[warm bloom]

thats caluclus three? I thought thats calc 1

[vapid sand]

You dont learn partials until calc 3 bro

[warm bloom]

I dont even take calc like that

we just take calc 😭

[vapid sand]

Bro what do you mean

You always take Calc 1, 2 and then multi/3. Or BC then Multi/3. You don't really just take everything all together

[obtuse bobcat]

you dont take calc but you take calc? huh

[warm bloom]

no I live in egypt we dont get told which calc N we take

[obtuse bobcat]

ohhh

[vale dock]

@sullen crater Has your question been resolved?

[vale dock]

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[blazing ice]

Could anyone explain me this excercise?

[wicked zenith]

,rotate

[wicked zenith]

it wants you to find out how much larger the radius of the larger pool is than the smaller pool

1. use the given formula to estimate the area of the larger pool
2. use the given formula to estimate the area of the smaller pool
3. compare them

[blazing ice]

Thats it? Thankssss

[vale dock]

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[vale dock]

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[proper iron]

d/dx of x-3/x^2+1

[unborn sable]

if you just want the result you can use the bot in#bots

[proper iron]

is it good idea to move up the x^2+1 to numerator to do product rule or just use quotient rule

[unborn sable]

i wouldnt say either is necessarily better
but id probably just do the quotient rule

theyre doing the same thing really

[proper iron]

ok

ty

.close

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[hidden pendant]

any clue, i only know that the sum of all of the person on normal threatment is 40 ?

[hidden pendant]

<@&286206848099549185>

[untold tangle]

?

what

[hidden pendant]

help me solve the problem, please (?)

[untold tangle]

what type of math

[hidden pendant]

this one (?)

[copper stump]

you found a+b+c+d+e+f = 40, and you have maximum value of 3a+b+c+d+e+f as 81, so clearly, that tells you what the largest chosen value should be.

[hidden pendant]

damn, my approach is correct, gemini sucks, okay thanks

!close

.close

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[stuck socket]

Have a question

[stuck socket]

About optimization

When finding the dual of a primal, does the primal need to be in standard form?

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@stuck socket Has your question been resolved?

[vale dock]

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[orchid flower]

Can someone explain how did the left bottom just become the top right please?

[distant galleon]

https://www.wolframalpha.com/input?i2d=true&i=Divide[cos+A+cos+B+-sin+A+sin+B%2Csin+A+cos+B%2Bcos+A+sin+B]-Divide[sin+A+cos+B%2Bcos+A+sin+B%2Ccos+A+cos+B]\(44)+A%3DDivide[pi%2C3]\(44)+B%3DDivide[pi%2C4]

that's false

just ignore that top right line tbh

the rest makes sense if you omit that

[orchid flower]

Sorry ,i forgot to mention that it have 2 language , this is English only

[distant galleon]

I figured that out dw

[distant galleon]

I still stand by what I said here

[orchid flower]

Owhhhh

What shouldve happen is the second row of right side instead of the top one?

[orchid flower]

I see ,tysm♥️

.close

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[ruby pasture]

If I have S = Q Sigma Q^T and I am tasked with finding a similar matrix to S, what do I even do? I can come up with a Q and Q^T in my head but is there something I am actually supposed to do to find it

[ruby pasture]

If I have S = Q Sigma Q^T and I am tasked with finding a similar matrix to S, what do I even do? I can come up with a Q and Q^T in my head but is there something I am actually supposed to do to find it

[vale dock]

@ruby pasture Has your question been resolved?

[ruby pasture]

no

[amber blaze]

you should probably react to the bot. the bot can't detect messages telling it not to close.

[jade ivy]

If you have that then Sigma is already similar to S if Q is orthogonal

[ruby pasture]

i believe it has to be a non-diagonal matrix, but idk. I also have to do polar decomposition on it in a later problem so ideally something not insane

he hasn't really taught either topic so I am kinda SOL

[jade ivy]

I didn't say anything about the matrix being diagonal, in anycase if Q is not orthogonal you can use QR decomposition and absorb the Rs into Sigma

[ruby pasture]

Well wouldn't sigma be diagonal? I started with an S I had to create that couldn't be fully diagonal or the identity matrix. Then I basically had to break it down into S = Q Sigma Q^-1 - Now I am supposed to find a S' that is similar but idk where to even start. The powerpoint mentions something about a upper right triangle matrix and I think thats what we are supposed to be doing

[vale dock]

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[slate topaz]

why does the first one correct but the second not?

[steady charm]

how did you find side c?

[slate topaz]

57/12 should be c because that's the scale

but 12 isnt in range of cos theta

[steady charm]

wdym by "the scale"? c is the hypotenuse

[slate topaz]

because c*cosA = side length

[steady charm]

how do you know what cos(A) is?

[slate topaz]

tan

opposite/adjactent which equals to sin(A)/cos(A)

so cos(A) should be equal to 12, but its not in the range of the output of cos

[steady charm]

that's true, but it doesn't mean whatever is on the denominator is cos(A)

[slate topaz]

hm

ig i could just do pythagorean

lemme try that rq

[steady charm]

like 2/3 = 4/6 = 6/9 and so on

[steady charm]

that should work

[slate topaz]

😭

[steady charm]

5 and 12 aren't the side lengths

[slate topaz]

right

i must use b than

i have b and a

[steady charm]

for the pythagorean theorem you must use the side lengths you know, yes

[slate topaz]

so i should use those instead

https://tenor.com/view/packwatch-gif-22219180

this question is rip bozo ez

!close

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If you are done with this channel, please mark your problem as solved by typing .close

[jolly ledge]

it's .close

[slate topaz]

.close

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[elder fox]

"10-7+1" is the answer of "Between 7 and 10, how many integers are there?", but, why does it work?

[elder fox]

If you want some context im trying to learn math by the start, since im bad at it. What i truly want to learn is the logic behind it, the use of the (-), the use of the (+), why does it work, etc... also, excuse me if bad english, please correct me if you find any mistakes on grammar too, i'd be glad.

[modest kettle]

three units between 7 and 10

four demarcators

similarly n units need n+1 demarcators

[drifting hornet]

you can imagine 10 circles (or objects or whatever, I'll use O)
1 2 3 4 5 6 7 8 9 10
O O O O O O O O O O
Now if we subtract the first 7 of those circles away, we are left with

8 9 10
O O O

Now we just do +1 to add the 7th circle back

7 8 9 10
O O O O

We are left exactly with the numbers from 7 to 10

note that we could instead do 10 - 6 and get the same result. That could be written as 10 - (7 - 1)

common mistake would be to only do 10 - 7, this type of error is called "off-by-one error". To avoid it, it's good to try drawing the situation in some kind of diagram. If the numbers become too large (so that you cant draw it comfortably), you can draw a similar situation with smaller numbers to see how it works

[elder fox]

i know but, what i want to learn is the logic behind it, like, why does it work? how do i 'make up' an formula like that but more efficiently, without needing to decorate a lot of another formulas around there for diferent kinds of situation? sorry if i cant make myself very clear

like, if you got an question like that, instead of trying to remember that formula, how do you'd make up your own formula for it?

in the context of, instead of this question, being any other question

[drifting hornet]

experience should immidiately hint you to subtraction. After doing few problems like this one, you should start to get a "feel" for what subtraction really does. Then once you see a similar question, your brain should immidiately go "a-ha, imma use subtraction here"

in general, there is no easy way to solve any problem in math. Ideas often come from prior experience

[elder fox]

ooooh, thats exactly what i was trying to know. Does it work for harder questions? like those experienced ones?

[drifting hornet]

Yes, the difference is that you will often need to have more a-ha moments before u reach the solution

and sometimes the first a-ha wont be enough to solve it and may misdirect you to take a path which wont lead to the solution, in that case you'll have to give up on that path and try something else

experience is what makes you get better a-ha moments more quickly

[elder fox]

thats nice, how do i get more experience? doing more questions?

[drifting hornet]

exactly

doing more questions, learning more new topics

just like everywhere else, in math you get experience by doing math

[elder fox]

i understand, tysm, now i got it

[drifting hornet]

np

[elder fox]

.close

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[west shuttle]

Hello there

[west shuttle]

I struck myself in a dumb calculation

And I'm confused cause I can't understand it why

<@&286206848099549185>

[midnight iris]

can u show

[west shuttle]

Alr

[vale dock]

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.

[honest stone]

Please wait 15 minutes before pinging, and please send your question first next time.

[honest stone]

,rccw

[unborn sable]

cos, not sec

[west shuttle]

Oh, lemme re-do it then

Uhh...

Same thing happened

Lemme show

[unborn sable]

is this a right angle triangle?

[stark wedge]

9n triangle ABC

what's that nine doing there?

also yeah is angle B known to be right

[west shuttle]

Um im facing problems in the calculation, im confused in dividing the root to a number

[west shuttle]

Yeah

[west shuttle]

Nah it's 'I' pardon my bad handwriting

[stark wedge]

6/sqrt(3) is equal to 3.46 to 2dp by the way

,calc 6/sqrt(3)

[rocky lotus]

Result:

3.4641016151378

[stark wedge]

but you could also simplify it as 2*sqrt(3) if you wanted!

[west shuttle]

Yeah but how? Because im supposed to use my brain in the exam (no calculators allowed)

[unborn sable]

oh im a numpty

apologies for my incorrect guidance

[stark wedge]

wait what

[west shuttle]

It's alr

[stark wedge]

wait, you're expected to give approximations to 2dp WITHOUT A CALCULATOR?

[spark hamlet]

Nono I think he means rationalize the denominator

[stark wedge]

and they want specifically decimals?

[spark hamlet]

Wait what

They want you to approximate decimals(

?????

[west shuttle]

Im in 10th grade, basically we're allowed to use calculators in exams when we're in 11th grade

[west shuttle]

Yeah

THATS THE CONFUSING PART GUYS

[spark hamlet]

Wth bro

[robust wolf]

Indian standardized testing?

[spark hamlet]

Geometry test of utter doom and despair

[west shuttle]

Yeah Indian education system

Icse board

[robust wolf]

Oh cool

I did a year of CBSE in 7th grade

So glad Im not doing that shit anymore 😭

[stark wedge]

ok but wait hang on

[west shuttle]

You're grateful man😭

[stark wedge]

let me get this straight again

they expect: YES DECIMALS, NO CALC.

[west shuttle]

Yeah

Gives more marks

[stark wedge]

more like more marks to lose

[spark hamlet]

Do you get anything taken off for it?

[stark wedge]

cause that shit is absurd.

[spark hamlet]

Or is just random bonus points

[stark wedge]

i guess you'll have to learn how to take square roots on paper or something.

there is a way.

[west shuttle]

Damn

[stark wedge]

and for this you would do best to write $2 \sqrt{3}$ as $\sqrt{12}$ and take the root that way

[spark hamlet]

Bros js gonna pull out his compass and straight edge and get to work

[rocky lotus]

Ann

[spark hamlet]

What kind of absurd testing is this

[west shuttle]

Eh?

[spark hamlet]

It’s easier to just approximate root 12 then take root 3 and than multiply that abhorrent answer by 2

[west shuttle]

Answer is 3.46m

DARN IT IM SO CONFUSED IM JUS LEAVING THE SOLUTION TILL THE ROOT-DIVIDE PART

In exams

[spark hamlet]

Can you explain your process I just got here

[robust wolf]

Yeah Im in America and it's super easy 😭

[west shuttle]

Oh I assumed you would br from south India 😅

[stark wedge]

@honest stone why the sully react btw

[robust wolf]

I am

I moved to america

[west shuttle]

Ohh immigration

[robust wolf]

Yeah

[west shuttle]

That's cool, hope it's good there

[spark hamlet]

Yeah it’s not

[robust wolf]

The city is a 3rd world country

Yk how the absolute worst of Indian cities smell? Yeah same here

[spark hamlet]

At least the school system isn’t there are good parts

[west shuttle]

Education system is better there than here mine so indeed it would be

[spark hamlet]

My school js had somebody use chemical warfare on someone

[west shuttle]

Bruh I thought they maintained hygiene there

[stark wedge]

anyway uhhhh

[spark hamlet]

And they had to go to the er

[stark wedge]

!redir ?

[vale dock]

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

[stark wedge]

this has gone quite off topic

[robust wolf]

Ok

[west shuttle]

Oh alr

Less go to chill

[stark wedge]

like we seem to be done here regarding the question itself

!done

[vale dock]

If you are done with this channel, please mark your problem as solved by typing .close

[west shuttle]

I'll be back if I get doubts

[spark hamlet]

Alr sounds good

[honest stone]

cuz that’s horrible.

[vale dock]

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[polar surge]

hi

[polar surge]

im trying to calc a derivative and i am confused here

why does that thing in green go to the kronecker delta symbol

like im confused

[vale dock]

@polar surge Has your question been resolved?

[wraith heart]

the variables $\dot{q}^b$ and $\dot{q}^a$ are independent variables

[rocky lotus]

riemann

[wraith heart]

like in $f(x, y)$, $\frac{\pa x}{\pa y} = 0$ and $\frac{\pa y}{\pa x} = 0$

[rocky lotus]

riemann

[polar surge]

i see

yes

this is the definiton of kronceker delta

so how does this relate

[karmic forum]

,help

[rocky lotus]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[wraith heart]

plug in a=b and a not equal to b and see

[karmic forum]

do you know anythin about mine

[polar surge]

if a=b then it would just cancel?

ie its just 1

[wraith heart]

#❓how-to-get-help

$\frac{\pa x}{\pa x} = ?$

[rocky lotus]

riemann

[polar surge]

1

?

[polar surge]

1?

[wraith heart]

yes

[wraith heart]

repeat this for your independent variables

[polar surge]

it would be 0

since they dont depend on each other

i.e this as u said

[wraith heart]

great that should answer this question

[polar surge]

thx,. my final question on this would be, why does that green thing go to drj/dqa?

[wraith heart]

plug this in for delta^(ab)

[polar surge]

0 if a != b or 1 if a = b

<@&268886789983436800>

holy

that was bad

[polar surge]

@unborn sable i still dont get tho why the thing in my picture happens

[unborn sable]

there isnt really anything more to it

a=b is the only case where its nonzero

so the b became an a

[polar surge]

ohhhhh

ok lol

that makes sense

thanks buddy

and thanks @wraith heart

.close

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[vapid valley]

im unsure how to find and differentiate between a cusp and vertical tangent line within a function

i got the crit value but idk where to from there

[ivory valley]

Can you tell how f' behaves from both sides if you approach x=3

[vapid valley]

approaches infinity?

[ivory valley]

from the right side yeah

and from the left -infinity

so you could imagine that f having infinite slope at x=3 is akin to f having a vertical tangent line

and to tell if f has a cusp, you should see if f is continuous at x=3, we definitely know it isn't differentiable

[vapid valley]

so if the limits are opposed infinities would it be a cusp?

[ivory valley]

not always

for example 1/x²

we have no continuity in x=0

so that's why it plays a role

[vapid valley]

okay i think i understand, i have to go eat rn but im gonna practice abit with this and if i have any issues ill ask about it

thank you!

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[wraith heart]

<@&268886789983436800> spam

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[tiny torrent]

Want to make sure im reasoning about this right

[tiny torrent]

So if it didnt contain its supremum

then there would be infinitely many points in the set leading to the supremum

hmm

trying to see how that would cause a contradiction with compact

[frozen ledge]

what?

[tiny torrent]

oh is that wrong

😭

wait yeah I dont think that makes sense

[frozen ledge]

i mean yea

[restive aurora]

For a contradiction, you just need to show that if it doesn't contain the supremum, there there is some open cover that doesn't have a finite subcover

[frozen ledge]

[0, 1] has infinitely many points leading to 1

🤔

[restive aurora]

It's an easy mistake to make with this sort of question to think you need to show a contradiction for every open cover

But all you need is to show that there is at least one contradiction

[tiny torrent]

hmm I think im still having a hard time understanding what a cover is intuitively

its a set of intervals where A is the subset of their union

that makes sense but

[restive aurora]

Oh a cover is literally just a collection of sets that cover the thing

[tiny torrent]

and so a open cover is the same thing but the sets are open right

[restive aurora]

Like for the set {a,b,c}, a cover could be {a},{b},{c}, or {a,d},{b,c,e,f,g,h,i} or even just {a,b,c}

[tiny torrent]

can we say R is a cover for everything?

or does that not make sense

[restive aurora]

Yes!

Well everything inside R at least

But the fact that R is a cover doesn't add much here

[tiny torrent]

oh yeah ik I just want to make sure Im understanding cover in general

and so for finite cover

[restive aurora]

It means you've got a finite number of sets

Like {[0,1], [1,2], [2,3], ...} is an infinite cover for [0,infty)

(Notably one with no finite subcover)

Also worth noting is there can be overlap!

[tiny torrent]

ok I think im understanding it better now

[restive aurora]

{a,b},{b,c} is a valid cover for {a,b,c}

[tiny torrent]

ok makes sense let me take a look at this question again and see if I can see the contradiction

and this compact set can be any set?

like it doesnt need to be closed or open?

[restive aurora]

Well

I think right now you can just say it's any set that satisfies the finite subcover condition

[tiny torrent]

ok

[restive aurora]

But it does turn out any compact set is closed

(Though not every closed set is compact)

[tiny torrent]

I see ok thanks for the help ill come back after giving it another try

[restive aurora]

Good luck!

[tiny torrent]

hmm I think I have another confusion

a subcover is a subset of the cover right

so can the subcover be the cover?

[frozen ledge]

yea sure

[frozen ledge]

we generally say subcollection though

to be clear that it’s a set of sets

[tiny torrent]

hm

so this means every finite cover has a finite subcover?

So I probably dont want to use a finite cover for this problem

[frozen ledge]

well yea

[near ridge]

I think that's what it means to be a subcover. A subcover isn't just any subset of the cover, it's a subset that still covers the given set

[tiny torrent]

hm ok I see

[near ridge]

Otherwise, every open cover would have a finite subcover

[restive aurora]

Taking the cover {a,b},{b,c},{a,c} for {a,b,c}, we have that {a,b},{b,c}, {a,b},{a,c}, and {b,c},{a,c} are subcovers

[vale dock]

@tiny torrent Has your question been resolved?

[tiny torrent]

ok I think I have my solution

let me write it up rq and send it

[tiny torrent]

Does it need to be shown for the cases where it doesnt contain the infimum and the case it doenst contain both?

I think for it to be a full proof it would need to be shown, since we have "it doesnt contain its supremum or infinimum" and so we should show that each case leads to a contradiction

[frozen ledge]

you keep saying sup

instead of sup A

[restive aurora]

I think this is a solid proof that it must contain the supremum, so all you need to do is replicate it to show it contains the infinimum

[quasi valve]

for the infimum you can consider the set
-A := {-x : x in A}

(which allows you to leverage the result for the supremum instead of repeating more or less the same argument from scratch)

[tiny torrent]

ok thanks for the help ill fix it up

.solved

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[junior spoke]

how can you exactly infer that the tangent makes a 45 degree angle with the horizontal. I see 45, but it isnt obvious to me that this is the angle for the velocity tangent

[junior spoke]

I tried drawing a tangent to the curve

[steady charm]

given the red 90 degree angle (the tangent to a circle is perpendicular to the radius) you can find the blue angle, which is the angle between the tangent and horizontal

[junior spoke]

ohh ok, but im confused where is the hypotenuse

[steady charm]

the hypotenuse is just an imaginary horizontal line

you can draw it if you want

[hollow rune]

Tbh, there is nothing outright telling you that the road below and the upperside of the triangle made by the angle are parallel

But for the sake of ease, we just assume that is the case

[steady charm]

for physics problems horizontal lines can be taken to be actually horizontal and similarly for vertical

[hollow rune]

yes, ik

[junior spoke]

can I make that too, sorry its messy

[steady charm]

sure

[junior spoke]

ok thanks but i think i overcomplicated it, the way you did was simpler

thank you now i know you can do that

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[north scarab]

https://www.youtube.com/watch?v=75dMcyCUo2g&list=PLybg94GvOJ9GzztfmHCkH_5e6TLE9CfpK&index=3

I dont understand from 7:17

[stark wedge]

12/3 pi is the same as 4 pi, and that's twice all the way around, so we can subtract that from the angle to get 2/3 pi.

this line?

or is it this one:

When you have it, it becomes trivial to evaluate trig functions for any common angle.

@north scarab please pinpoint what exactly is said at the point where you want clarification.

[north scarab]

this

when he explains this

idk what he means

[stark wedge]

14pi/3 is bigger than 2pi and so doesn't appear directly on this unit circle. therefore we need to bring it back in range, somehow.

agree or disagree?

[north scarab]

how is 14pi/3 bigger than 2pi

i mean I can see it if i put it in my calc

but is it ok if i dont see it in first sight

[stark wedge]

14/3 is bigger than 2.

14/3 is bigger than 2.

[north scarab]

ok make sese

[north scarab]

agree

buut wdym bring it back in range?

back in the circle?

[stark wedge]

yes, back in the circle.

between 0 and 2pi so that we can locate it on there.

[north scarab]

but 14pi/3 doesnt fit between 0 and 2 pi

right?

[stark wedge]

yes that's what i said earlier

the angle isnt in range so we need to somehow bring it back in range

i would like you to not overthink this

the way we can bring this angle back in range is to notice that going all the way around (i.e. 1 full turn) takes us back where we started,
so we can freely add or subtract 2pi to the angle, as many times as we damn well please.

[north scarab]

ah, make sense

[stark wedge]

for 14pi/3 the magic number turns out to be subtracting 2pi twice to get it back

[north scarab]

iback to

back to 2 pi?

[stark wedge]

back between 0 and 2pi

[north scarab]

oh ok

yep I agree

alright thanks

.close

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[polar surge]

could someone explain to me plz how they go from that step to the next

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[true wedge]

.

This is kinda stupid but i need help

[stark wedge]

this equation looks broken

im going to try a wild guess here: did you translate it from Arabic, but then forgot to flip it to go from left to right?

[north walrus]

$1=\frac{3}{y}x\frac{1}{\sqrt3}$

[rocky lotus]

Allen

[north walrus]

?

[true wedge]

correct

mb

[lyric sundial]

!original

[vale dock]

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.

[north walrus]

so $\frac{1}{\sqrt{3}}\times\frac{-3}{y}=-1$?

[rocky lotus]

Allen

[true wedge]

yes sir

[true wedge]

^^yes

[north walrus]

you can multiply y on both sides

[stark wedge]

there are several things to do here

[north walrus]

and a negative

then rationalize/simplify?

[true wedge]

so Y equals

?

which is √3

ty

ty

solved

[spark sluice]

Put . before solved

[north walrus]

!done

[vale dock]

If you are done with this channel, please mark your problem as solved by typing .close

[north walrus]

use that

[tawny moon]

.solved by OP agreement

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[sterile mauve]

help

[north walrus]

$\lim_{x\to0}\left(\frac{\sin x}{x}\right)^\frac{1}{x^2}$

?

[rocky lotus]

Allen

[sterile mauve]

yes

[north walrus]

looks hard...

[sterile mauve]

ya and how would i know the graph without knowing the graph

[north walrus]

$=\lim_{x\to0}\frac{(\sin x)^\frac{1}{x^2}}{x^\frac{1}{x^2}}$

[rocky lotus]

Allen

[north walrus]

you just need to concentrate near x=0 actually

$\lim_{x\to0}\frac{1}{x^2}=\infty$

[rocky lotus]

Allen

[north walrus]

well thats annoying

good luck :)

im not good at limits

[sterile mauve]

you got any guess @north walrus

[north walrus]

it's not even an integer 😭

[merry crystal]

have u tried taylor expansion of sinx and applying ln to the expression

[sterile mauve]

i tried ln

[merry crystal]

try it with taylor series

[north walrus]

interesting

[sterile mauve]

i did

[merry crystal]

okay and what did u get

[sterile mauve]

lim0 1/x^2 . ln(1-x^2/3! ......)

idk latex

[north walrus]

my guess is sin(1 rad)

[sterile mauve]

its close

answer is e^(-6)

[merry crystal]

now try taking taylor series of ln(1+u)

[north walrus]

yeah

[merry crystal]

let u equal to the negative of allat aside from the 1

[sterile mauve]

uh

i cant

[sterile mauve]

how

[sterile mauve]

it's something very long

to write

[merry crystal]

just let u = (-x^2/3! +...)

also try adding at least one more term of the expansion for sinx so u can see the behaviour

[sterile mauve]

ya ya i did that

[merry crystal]

okay what did u get then

[sterile mauve]

u - u^2/2 ....
ln(1+u) expansion

i didnt get the rest

[merry crystal]

plug in u

[north walrus]

[\lim_{x\to0}\left(\frac{\sin x}{x}\right)^\frac{1}{x^2}]
[=e^{\lim_{x\to0}\ln\left(\left(\frac{\sin x}{x}\right)^\frac{1}{x^2}}\right)]
[=e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\frac{\sin x}{x}\right)}]

then idk

[sterile mauve]

plug in u where

[merry crystal]

into u-u^2/2

[sterile mauve]

how about rest of expansion

[rocky lotus]

Allen

\[\lim_{x\to0}\left(\frac{\sin x}{x}\right)^\frac{1}{x^2}\]
\[=e^{\lim_{x\to0}\ln\left(\left(\frac{\sin x}{x}\right)^\frac{1}{x^2}}\right)\]
\[=e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\frac{\sin x}{x}\right)}\]
```Compilation error:```! Extra }, or forgotten \right.
l.50 ...eft(\frac{\sin x}{x}\right)^\frac{1}{x^2}}
                                                  \right)\]
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.```

[merry crystal]

not needed but u can take as many terms as u want

the rest will go to 0

[north walrus]

idk good luck

bye

[sterile mauve]

lemme try

@merry crystal

second one

[merry crystal]

,rotate

[sterile mauve]

oh

oh

i got it now

[merry crystal]

yes

distribute the 1/x^2 term

[sterile mauve]

yea

-1/6

and ln

e^(-1/6)

.close

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[lofty crown]

if i prove that the dyadic rationals is dense in Q and i know Q is dense in R does that imply the dyadic rationals are dense in R

[stark wedge]

can we see your defn of denseness

[lofty crown]

We’re using Rudin so I believe it’s this one here

[hazy pivot]

Yes

Essentially if you show you can get arbitrarily close to any rational using dyadic rationals

And you can get arbitrarily close to any reals using rationals already

So you can get arbitrarily close to any real using dyadic rationals too

[lofty crown]

does it work with this definition

[hazy pivot]

Yup

In particular, being dense is a partial order

Although I imagine you'd have to prove that

Unless it's stated somewhere

[lofty crown]

i also guessed that being dense is probably transitive but i'm not sure how you would prove it

[vale dock]

@lofty crown Has your question been resolved?

[hazy pivot]

It's not too hard

Showing it for ℝ as the ambient set is trivial

You just have to use the definition twice

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[honest scarab]

hello guys i am seeking help understanding the boundaries of the Sequences

[upper ember]

Do you have a specific question about it?

[honest scarab]

yeah um i didnt understand that when we prove the boundary is less than < غ you can increase the value of the
Fraction

[upper ember]

This is not a specific question. It would be better if you show an example on which you don't understand (also, what is that symbol you just used?)

[rocky lotus]

Asteroid

[upper ember]

Oh, lmao

[honest scarab]

yes this thing (i am not english sorry)

thats on the solution paper

[upper ember]

Can you show the full question?

[honest scarab]

prove the boundry

[upper ember]

Can you share the original question please? Because ''prove the boundary'' doesn't really mean anything''

[honest scarab]

i translated it

i dont have the original question

[upper ember]

You can share the original question even if it is in a foreign language

[honest scarab]

1/3 is the boundary of this Sequence and i have to prove it

by doing |an-L|<ع

[frozen ledge]

you mean the limit

[frozen ledge]

it seems like you wrote out |a_n - 1/3| < epsilon

you can try simplifying whats inside the ||

but my preferred method would be two form a chain of inequalities

[frozen ledge]

but generally the method for these is just to get a common denominator

[honest scarab]

yeah limit is the word

[frozen ledge]

have you tried combining the fractions?

[honest scarab]

no i dont have a problem with these the problem is why can i increase the value of the fraction or the side against the epsilon and the answer is right

[frozen ledge]

can you provide an example of what you're referring to?

[frozen ledge]

,rotate

[rocky lotus]

Couldn't find an attached image in the last 10 messages.

Oct 18, 2025, 3:23 PM

[honest scarab]

here

[frozen ledge]

this?

ok yea so thats what i was saying with inequalities

we have

$\abs{\frac{n^2 - n + 2}{3n^2 + 2n + 4} - \frac{1}{3}} = \abs{\frac{-5n + 2}{9n^2 + 6n + 12}}$

[rocky lotus]

knief

[frozen ledge]

we want to make this < epsilon for large enough n

[honest scarab]

yes here is like the final steps when he finnished simplifying the left side he increased the value of the fraction by decreasing the bottom side of it

[frozen ledge]

right so do you first agree that if a < b and b < c then a < c?

[honest scarab]

ya

[frozen ledge]

now the fraction here is very complicated and finding an N so that for all n > N it is < epsilon would be quite difficult so we can instead bound it above by a simpler fraction that we can more easily find the value of N for

[frozen ledge]

then we apply this

[honest scarab]

ok so its not like one N that make it right

[frozen ledge]

yes, there is a smallest value of N but all we need is some N that works so taking any N larger than that smallest one will of course work

we don't really care what that smallest N is here, we just want to prove the limit is 1/3

[honest scarab]

okeyyy i get it now

thank you

[frozen ledge]

you're welcome

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[lethal lance]

How do any of these answers work? Doesn’t the modulus have to be positive? So this equation can never equal a negative number like -2?

[stark wedge]

who wrote that mark scheme

[lethal lance]

An AI

[stark wedge]

i mean like up to and including |x+7|=-10 it's correct but yeah the modulus does have to be positive or 0

[safe fulcrum]

DONT TRUST AI

[stark wedge]

oh that explains it

!noai

[vale dock]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[safe fulcrum]

also a really annoying thing about it is that AI always agrees with you when you correct it, and proceeds to then explain it wrong in a different way

[lethal lance]

How can modulus x+7=-10?

[stark wedge]

it can't.

[safe fulcrum]

the point is that |anything| = negative number is impossible

[lethal lance]

Oh I thought you said that part works

Okk that makes sense

Thankss

[stark wedge]

i said exactly when the AI slop stops making sense

[safe fulcrum]

.close

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[storm coyote]

How to solve, $\sum_{k=0}^{n} \frac{(n-k)(k+1)(\binom{n}{k})}{(k+2)}$

[rocky lotus]

Double_mytrouble

[storm coyote]

I used a longer (hella long) method and got the answer. (expanded it and then found each component by first integrating by multiplying x, and then diffentiating required number of times

but I need a shorter method

because that took 5.33 pages

Guys..... 🥲 anyone there?

[pearl nacelle]

I can try take a look

[storm coyote]

Thanks bro

[stark wedge]

i mean the obvious expedient seems to be like... considering something along the lines of $f(x) = \sum_{k=0}^n \frac{(n-k)(k+1)}{k+2} \binom{n}{k}x^k$ and then trying to do some calculus-y magic to it

[rocky lotus]

Ann

[stark wedge]

but then again that might've been exactly the thing that took you that many pages

[storm coyote]

yes!

I did exactly that

at last putting x=1

I mean can I change x^k+2/k+2 to a integration?

I tried that

then the summation becomes simplier without denominator and I integrate it later....

But idk if I can do that....

[stark wedge]

write x^2 f(x) and then take its derivative

$[x^2f(x)]' = \sum_{k=0}^n (n-k)(k+1)\binom{n}{k}x^{k+1}$

[rocky lotus]

Ann

[stark wedge]

uhhh

jeez i dont know how this will play out without writing it in full lol

[storm coyote]

no worries... you can send paper pic. till then I am trying it myself lol

I am also sending mine in paper if I get stuck 😭

[pearl nacelle]

Hmm well my initial thought was to make (k+1)=(k+2-1) then do 2 separate sums

One of the sums is just the sum over j of j*(n choose j) which is easy enough

[storm coyote]

Sorry to tag @stark wedge but is this same as what you intended?

[tawny moon]

,rccw

[keen tundra]

combinatorial proof?