help-19

[frozen musk]

-(cscx)^2 or (-cscx)^2

nvm

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[keen pike]

herrooo im looking for how to find the volume of a meniscus (the curve in the upper surface of a liquid close to the surface of the container or another object, produced by surface tension(wikipedia)) and preferrably a method for how to derive a general equation for any graduated cylinder radius

[odd edge]

@keen pike Has your question been resolved?

[lyric marlin]

Can I ask why? (I don't know of such a formula and deriving one seems to me as a non applied mathematician like a pretty complicated physics problem)

[keen pike]

mathematical exploration as a part of school

[runic cypress]

The markings on the graduated cylinder account for the meniscus so a priori it seems easiest to me to take that measurement, find out the true volume of the graduated cylinder up to the top of the meniscus, and take the difference

"A general equation for any graduated cylinder radius" is kinda nonsense though, can't help there

[keen pike]

i want to mathematically determine the volume of the meniscus

[runic cypress]

Do you have a reason to believe that there's an easy way to do this?

[keen pike]

not really no itll probably be stupid difficult but i dont even know where to start

ive gotten the advice that the horizontal cross section is kinda like a subtended arc (if thats the name i forgot)

and then i can take the rotational integral about the y axis

and subtract from cylinder height

but i wanna accurately graph the meniscus itself

[runic cypress]

Is this actually true?

Why should it be a circle?

[keen pike]

no it isnt

no no no

like wait lemme send an image

Under the line

That segment

[runic cypress]

Yeah, it certainly won't always be like this

[keen pike]

Exaclty that’s why I came here

[runic cypress]

I mean this isn't really a math question

[keen pike]

idk what to say to that brodie

[runic cypress]

I am unconvinced there's some magical formula which parametrizes the surface of a meniscus or something, but if you find out a way to actually describe it then helping you find the volume is something we can do

[keen pike]

yeah but there has to be some sort of magic formula bruh thats what im banking on

just some sort of way to graph the meniscus

[runic cypress]

Why do you think so?

[keen pike]

i know that for example chains under slight tension can be graphed by hyperbolic cosine function

like hanging chains p much no tension

[runic cypress]

Yes, some things have nice models

Not everything does

[keen pike]

that is true

[runic cypress]

If physics stack exchange is to be believed, the situation is not so great https://physics.stackexchange.com/a/156632

[keen pike]

looks pretty grim

ngl just opened the link nd it looks like theyre incorporating density n stuff

if you have the density n stuff predetermined by lets say using jsut water can you make it easier

[runic cypress]

Not really no

The density is just a constant factor

[keen pike]

yes exactly

so if the density is a constant

cant u eliminate multiple like differential equations that come up regarding density

[runic cypress]

It's already a constant, so having it be specifically water's doesn't really do anything

[keen pike]

o yeah thats right sorry i got my variables mixed up

in the article they r finding like the forces exerted on the liquid and other things but not directly the shape of the meniscus

[runic cypress]

They're finding a diffeq for the height at radius r

[keen pike]

only at the start because then it goes fully into the forces exerted on the liquid etc

or nah i was lying my bad

deepest gratitude to you mr hornet

very informative stack exchange articles out there

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[crystal trail]

Hello. Is this correct? (I marked the steps I followed)

[indigo dirge]

hi! what's the original question?

[crystal trail]

A number that when divided by 7 has a remainder of 2 and when divided by 5, a remainder of 4.

[indigo dirge]

also, for step 2 and 3, k might not be the same for 7 and 5

so i think it should be x=7a+2 and x=5b+4 (or whatever variables you would like)

they could be equal

but not necessarily

btw, if k is the same for both, it would be uniquely determined by setting the two equations equal to each other, therefore uniquely determining x

[sturdy cape]

tf? Clear working-out in the MathCord server!?

Unthinkable

Given the context, I presume you might have run across the Chinese Remainder Theorem?
https://www.math.cmu.edu/~mradclif/teaching/127S19/Notes/ChineseRemainderTheorem.pdf

[Context - first Google search result]

Wait, that said, you don't have an undergrad role

Well, with the mods being as small and few as they are, try listing out the sequences 7n+2 and 5n+4

If there's a number in common, it'll be within the first 5 numbers of the first sequence (and the first 7 of the second sequence)

(and then adding/subtracting multiples of 35 will give you more solutions)

If there isn't, then this is unsolvable

[crystal trail]

Ooh, thanks, that's helped

[crystal trail]

Thanks, I'm just starting to study number theory and I found a video on modular arithmetic where that exercise appeared.

[crystal trail]

I will investigate more about this.

thanks to both

Bye

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[kind solstice]

I’m practicing index laws rn since they’re a weakpoint for me and I’ve lost a load of marks on em, I’m doing this question rn and Ik I probably have to factorise it I just don’t really know how to factor the y^-5

[kind solstice]

My guess was to just rewrite it as 1/y^5 and then just factorise that but Idk how that’d work either

[vivid wyvern]

Do u k what happens to the powers of same variable when they r multiplied

[kind solstice]

You mean smth like

[vivid wyvern]

Yep

The powers sum up basically

[kind solstice]

Is this any close to what I have to do?

[vivid wyvern]

So here u got x and y variables and they r in multiplied form ,so u could add their powers

[kind solstice]

Wait fuck I forgot the ^7 on the first y

[vivid wyvern]

U dont need to factor out

[kind solstice]

I can just do 5x • 3x and y • y^-5 and mush it together?

[vivid wyvern]

Yes

[vivid wyvern]

It's in multiplied form after all

What u were doing would hv been effective in case of addition or subtraction i supposed

[kind solstice]

alr yeah that makes sense

Okay thank you very much I can sort it out now

[vivid wyvern]

Great

[kind solstice]

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[lean ivy]

The 2nd question.
Here,A={1,2,3,4,5,6,7}
How can it be true for all values of x?

[orchid torrent]

$8-x \leq 7 \implies -x \leq -1 \implies x \geq 1$, which is satisfied by all $x \in A$

[clever fjord]

Civil Service Pigeon

[orchid torrent]

Also, in the future, please refer to the question by the marking the paper (so iii here) instead

B/c it took a minute to realise that you were counting the cut off but at the top as a question

[lean ivy]

Oh,my bad

I misread it as 2

[lean ivy]

But when I take x=7,the condition does not get satisfied

[orchid torrent]

What

8-7=1

1 is less than 7

[lean ivy]

Wait......

😭

Damm

I think I may have dyslexia

[orchid torrent]

rip

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[lean ivy]

I am sorry

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[civic jetty]

I need help with part b of this question, thanks

[wooden python]

have you done a?

[glacial hearth]

have you learned like the general form of the answers to a trigonometric equation

ultimately doing part A then part B would be easier but in case you're supposed to be using general form

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@civic jetty Has your question been resolved?

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[deep mason]

if we have a density matrix $\rho$, we can consider its projection on a subspace using a projection operator $P$. the projection of $\rho$ is given by $P \rho P$.

we can also consider the partial trace of $\rho$ to obtain the reduced density matrix for a subsystem, $\rho_A = \Tr_\text{(not A)} \rho$.

is there any simplification that can be made to $\Tr_\text{(not A)} (P \rho P)$?

[clever fjord]

back to ginger

[odd edge]

@deep mason Has your question been resolved?

[faint knot]

shouldnt there be something in the superscript for one of the Ps?

[deep mason]

my prof wrote it with a ^-1 but that doesn't seem like it's right

projection matrices aren't invertible

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[deep mason]

oh shit

.reopen

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[deep mason]

if it was a ^\dag, i think P^\dag = P

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@deep mason Has your question been resolved?

[faint knot]

you have to reply ❌ to keep the channel open, otherwise the bot will close the channel due to you not watching it closely enough

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[frozen barn]

Translation is basically telling you to find the m
I literally don't know wtf should i do with ts
Like is it even possible🥀

[frozen barn]

,help

[clever fjord]

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!

[frozen barn]

,rccw

[frozen barn]

Help pls

[tacit haven]

you can probably take a guess that 5/3 = 20/m or something

[frozen barn]

Take a guess🥀
This ain't art
Prove why the small trapezoid is similar to the medium trapezoid
If you can't then the question is prolly impossible from lack of info

[tacit haven]

im just showing you that the sol is probably not impossible

[snow breach]

qouldnt m be 20

or am i slow

dont the 2 arrow things on the lines mean they r the same length

[frozen barn]

Yeah it prolly there's literally no way (from my perspective) to prove that smol trap is similar to med trap

[tacit haven]

parallel

[snow breach]

yeah

[frozen barn]

Parallel bruh

[snow breach]

oh

[frozen barn]

Except if u can prove that the trapezoid is a square or the 🔶 or rectangle

[snow breach]

ohhhhh

mb all

[frozen barn]

Yeah pluh i don't think this question is possible🥀

[tacit haven]

you can

[frozen barn]

How

[tacit haven]

you can find this side in terms of variables right

you should probably start by naming some points

[frozen barn]

Ok hold on

,rccw

[frozen barn]

Yo ass is not slick bruh undelete that one message💔

[tacit haven]

we know EG/EB = 3/5, can you find the ratio for DH/DC?

[frozen barn]

No

[tacit haven]

sob

[snow breach]

😭

[tacit haven]

it would help to draw EH and GC probably

[frozen barn]

,rccw

[snow breach]

what grade is this

@frozen barn

[frozen barn]

9th

[tacit haven]

now show EGH similar to GBC

[tacit haven]

crazy

[snow breach]

what grade level of maths should this be

copter

[tacit haven]

7-8th

i dunno i did this way back

[frozen barn]

💔

[tacit haven]

anyways can you show this?

[frozen barn]

I can prove angle EGH and GBC are equal but idk bout the rest

[tacit haven]

<GCB = <CGH = <GHE implies similarity

[frozen barn]

For CGH=GHE, GC must be parallel to EH
That can happen when EG and HC is the same (i used basic logic to figure ts out)
Or am i slow

[tacit haven]

true

[frozen barn]

Yeah but how is EG=HC🥀 there is literally nowhere in the question that implied this

[tacit haven]

so from this EGH is similar to GBC and you can just solve for m

[tacit haven]

triangles EGH, GHC share a side GH

[frozen barn]

Bruh how is eg=hc

[tacit haven]

leads to something called congruence

[frozen barn]

Yeah but they need to be similar first
And the CORRESPONDING sides needs to be equal

[tacit haven]

yeah

[frozen barn]

How EGH and GHC similar and how does that make EH parallel to GC

ok nvm about the how does eh parallel to gc part

I get it

But how is egh and ghc similar

[tacit haven]

<CGH = <HED = < GHE (because of our original pair of parallel lines) directly imply EH // GC

oops

[frozen barn]

Just say the question is impossible bruh the mofo writing this book (sukino) made at least 40 unsolvable questions in this book💔

[tacit haven]

its not

[frozen barn]

Ok

[tacit haven]

plus youre doing extra harder things instead of solving this question

we only need EGH and GBC similar to find m

[frozen barn]

Ik but there is literally no way we can make these 2 triangles similar we've gone through this

[tacit haven]

???

[tacit haven]

i literally just told-

[frozen barn]

Bruh am i slow or sum💔

[frozen barn]

FOR THAT TO HAPPEN EH AND GC MUST BE PARALLEL WHICH MEANS EG AND HC MUST BE THE SAME LENGTH

[tacit haven]

combined with <EGH = <GBC that you found

[frozen barn]

Cro🥀

[tacit haven]

what makes you think they cant?

[frozen barn]

they maybe are but we need to prove it to use it

[tacit haven]

also what i said doesnt use those two facts you listed at all

so we dont need to prove that ,theyre just implications from this

[frozen barn]

Sorry i don't understand

Can you explain in a more simpler way?

[tacit haven]

uh one second

oh wait

yh mb bro

ts horribly constucted i think

we have <BGC = < GCH

[frozen barn]

Dw bro i can read it

[tacit haven]

ignore what i wrote in red i dont think its right

[frozen barn]

k

[tacit haven]

what book is this even from bro

[frozen barn]

Yeah bruh i literally think we're just wasting time together here
You keep saying that CGH=GHE even though for that to happen GC MUST BE PARALLEL WITH EH, also forgot about eg=hc so gc parallel to hc since that's completely wrong, and bruh i hope you realize the only reason gcb=cgh is cuz gh and bc are parallel💔
Imma just close it here

[frozen barn]

.close

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[tacit haven]

yea vro

there arent official solutions at the back or something??

or online

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[frozen barn]

Nah the no solution in the back and no solutions online
There is a qr code to the answers page but it's literally broken sob
Like even my strongest technique "the way of staring" didn't work😭

[frozen barn]

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[runic cypress]

You assumed $x\in A\setminus B$?

[clever fjord]

Buzzing Hornet

[umbral jolt]

Im crying

Your prof was very nice

[runic cypress]

Let's not mock helpees

[umbral jolt]

Sorry

I'll focus

@tacit jacinth so the thing is

why did you assume?

and once you've defined x is an element of $A\setminus B$

[clever fjord]

Licht

[runic cypress]

Anyway Licht is right in that this is sort of a mess, it doesn't really follow any of the proof structures you've presumably been taught

It might be a good idea to think very carefully about the overall structure of your argument

[umbral jolt]

you cant furthermore tamper the definition of x, other than directly deriving it from the way you've defined x

Its like saying you have two dads

sorry💀

[runic cypress]

Unless, I'm missing something, we shouldn't even be starting with this

[umbral jolt]

Yeah

[runic cypress]

In what way?

[umbral jolt]

no no dont misunderstand me

if you define x twice

thats what my analogy referred to

[runic cypress]

And why's that like having two dads?

[umbral jolt]

biological of course

[umbral jolt]

because biologically you can have only one

[runic cypress]

Ok I will hash this out with you later

This is off-topic

[umbral jolt]

yeah not now

@tacit jacinth you can open up

I'll try my hardest

[runic cypress]

Sorry about that! Anyways, at this stage you should be thinking explicitly about questions like "Am I using proof by contradiction" or "Am I using proof by contrapositive"

[umbral jolt]

no wait, either way he made one assumption which he didnt even use

[runic cypress]

My point is that he's starting in the wrong place

Thinking about these kinds of things will at least avoid starting one's proof with "let x be in A setminus B"

[umbral jolt]

youre doing it like assume all girls are not blonde. Now we asume the sky is brown. But the sky is blue. hence all girls are blonde

[runic cypress]

Yes this looks right

Good job

[umbral jolt]

it still isnt

[runic cypress]

Hmm? what's wrong with this proof?

[umbral jolt]

no

ots ok

[runic cypress]

Then wtf does "it still isn't" mean lol

Anyways yeah I think you've got it, if you don't have further questions you can close the channel

It's a little messily written but I don't see any mathematical errors

Minor actual quibble is that "x in A and x not in B" is redundant, you only need one, and in fact in part two you only say one

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[late sinew]

@silent pumice I don't accept dms, but you can ask your question here- this channel is yours now, ping me when your question is resolved.

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@late sinew Has your question been resolved?

[copper quarry]

this channel is now mine

[late sinew]

Lel imma close it seems like they're afk

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[copper quarry]

.reopen

[late sinew]

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[mystic saffron]

Can I ask how does d/dx(ln y) = 1/y * dy/dx? I dont get how dy/dx spawns out of no where

[glass vault]

Its in terms of x

So really lny is ln(f(x))

So following the chain rule you have 1/y multiplied by the derivative of y with respect to x

[mystic saffron]

I understand this but dont understand why is it dy/dx

[amber veldt]

(f(g(x))'=f'(g(x))g'(x)
D_x[fog]=D_g[f]D_x[g]
dy/dx=dy/dg dg/dx

[mystic saffron]

I understand the last line but not the first 2 libes

[glass vault]

You take the derivative of the outer function times the derivative of thd inner function

[celest osprey]

The first two lines is chain rule

[glass vault]

And the derivative of y is dy/dx

[amber veldt]

same thing in 3 different notations

[mystic saffron]

Oh u see

I*

[amber veldt]

fog is the composition of f with g

try using 3 notations with a simple example

like

$$e^{\sin x}$$

[clever fjord]

gfauxpas

[amber veldt]

let y=sin x

so e^sin x = e^y

what would the chain rule say?

[mystic saffron]

Wait but differentiation haven't started

But what happends next

[amber veldt]

i mean to differentiate e^y wrt x

ill.do the second notation

D_x[e^y]=D_y[e^y] D_x[y]

your turn, do the first notation

[mystic saffron]

Give me abit im trying to process

[amber veldt]

np

[mystic saffron]

Im new to this

HAHAHHA

Sorry my head kind of hurts :,)

[mystic saffron]

Where did dx spawner from tho

Spawn*

[amber veldt]

on the rhs or lhs?

what is the derivative wrt x of e^sin x? forbet the notation for a moment ,whats the answer?

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[mystic saffron]

.reopen

[odd edge]

✅

[mystic saffron]

Both

But rest has one dy

Which somewhat violates my understanding on the equal sign

[mystic saffron]

Derivative of that hmmm, tbh I haven't learnt that

[amber veldt]

its not dy

its Dy

[mystic saffron]

Oh

[mystic saffron]

But its still something that spawns out of nowhere

[amber veldt]

its part of Dx

[mystic saffron]

Oh

[amber veldt]

like in

[mystic saffron]

Wait I'm abit confused

[amber veldt]

dy/dx=dy/dg dg/dx

[mystic saffron]

Because tbh I haven't learn anything about logorithmic calculus

I just searched it up today

[amber veldt]

nothing to do with logs

[mystic saffron]

I see

[mystic saffron]

Oh yes

[amber veldt]

ahyway you need to learn how to differentiate

d/dx e^sin x

i have to go now though

[mystic saffron]

Sorry but is it possible if you do it step by step so I can see hehe...

[mystic saffron]

Oh I will try

[amber veldt]

soz

[mystic saffron]

Wait is it just sin x

Or idk

[amber veldt]

let y=sin x

then e^sin x = e^y

and

[mystic saffron]

So e^sinx

[amber veldt]

d/dx[e^y]=d/dy[e^y]d/dx[y]

[mystic saffron]

e^x is always e^x right?

Or I forgot

[mystic saffron]

Ohhhh

Damn I didnt know that

[amber veldt]

(f(g(x))'=f'(g(x))g'(x)
D_x[fog]=D_g[f]D_x[g]
dy/dx=dy/dg dg/dx

anyway gl

[mystic saffron]

Okiee thanks

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

e^sin x
d/dx (e^sin x)
= -cos x e^(sin x)?

But i dont know how does this relate to the chain rule, is this even correct? TAT

[plush spear]

i think you differentiated sinx incorrectly

also the chain rule states that (fg)'=(f'g)(g')
if we let f(x)=e^x
we get that e^(g(x))=e^(g(x))g'(x)

so the e^g(x) rule is just a special case of the chain rule (where f(x)=e^x)

[mystic saffron]

I see

[mystic saffron]

But but what is it that i did incorrectly

[plush spear]

what's the derivative of sinx

[mystic saffron]

-cosx

[plush spear]

not quite

but close

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

Oh damn

I thought it was always flipped signs

Im self studying, preparing for my first semester

So kinda tusty

rusty

HAHAHA

But thanksss

[plush spear]

gl
and rmb to close the channel if u don't have other qs

[mystic saffron]

Okiee

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[fallow idol]

$x^4-8x^2+1=0$

[clever fjord]

Tan

[fallow idol]

let x^2=y

y2-8y+1=0

using the quadratic formula

$y=4\pm sqrt15$

[proven harbor]

$\sqrt$

[clever fjord]

Tan

[lean yew]

backslash

[indigo dirge]

use $\sqrt{}$

[clever fjord]

Frances

[lean yew]

HAHAHAHAH

[indigo dirge]

uhm

[fallow idol]

so x^2=√4+√15

[oak crown]

That looks right so far

Wut

[fallow idol]

i dont get it

[indigo dirge]

whered the root 4 come from

[lean yew]

huh

[oak crown]

$y=4\pm\sqrt{15}$, so $x^2=4\pm\sqrt{15}$ is what you mean?

[clever fjord]

;(

[fallow idol]

$\sqrt{4\pm\sqrt{15}}$

[clever fjord]

Tan

[fallow idol]

yes

[oak crown]

Oh, ok

[fallow idol]

so x would be this right

[oak crown]

[fallow idol]

but thats kida weird

so idk if im right

[oak crown]

Well, with another plus minus, yes

You can simplify it; I don’t think it is too hard (don’t trust me, I haven’t tried it yet)

[fallow idol]

in my textbook

it says the correct answer is

[proven harbor]

you can simplify this

[fallow idol]

+- 0.456

+-2.81

[wooden python]

so it just gives decimal answers

[oak crown]

skul

[wooden python]

well then you dont even bother simplifying

just toss that shit into your calculator or w/e

[lean yew]

calc bashing time

[oak crown]

Just put it into a calculator

[lean yew]

or WA

[fallow idol]

i did

[proven harbor]

8 + 2root(15) = (root(5) + root(3) )^2

[fallow idol]

and i got the same answer

but if im not allowed a calc..

then how do i do it

[proven harbor]

.

[indigo dirge]

well i dont think they would expect you to calculate that so

just give it in the exact form

[fallow idol]

,w solve \sqrt{4\pm\sqrt{15}}

[clever fjord]

Results provided by WolframAlpha

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

Requested by tan211111

[fallow idol]

im pretty sure they do

cuz in my textbook

[indigo dirge]

uhm

[fallow idol]

its factorized

can you teach me

Frances

or Ann

[indigo dirge]

i mean it is possible to calculate square roots to a certain accuracy

[proven harbor]

^

[indigo dirge]

but its quite tedious

[fallow idol]

should i just write down this

[indigo dirge]

do they provide solutions in your text book or

just the answers

[fallow idol]

just the answers

[indigo dirge]

hmm

on exams, i think they would either allow calcs and ask you to round or just say state exact answer

[fallow idol]

we have 2 papers

one with calc

another without calc

[oak crown]

It isn’t too hard, like I said

[indigo dirge]

just in case, lets say you want to calc sqrt 15

[oak crown]

Just let $\sqrt{a}\pm\sqrt{b}=\sqrt{4\pm\sqrt{15}}$ and solve for $a$ and $b$

[fallow idol]

is it like $4+\sqrt{15}\\sqrt{16*15}$

[clever fjord]

Tan

[fallow idol]

ive heard this somewhere

[clever fjord]

;(

[oak crown]

Not quite sure what you did there

[fallow idol]

multiply by 4

outside and inside the root

but somehow there is another 4 in the root

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[frosty coral]

How do i find the vertical asymptote for this equation

[frosty coral]

A video told me this but I wasn't sure

[wooden python]

do you know where the VAs of just tan(x) are

[frosty coral]

uhh

[wooden python]

and/or do you know what the graph of y=tan(x) looks like

[frosty coral]

its what makes tan 0

right?

oh

I know that the graph is like a bunch of lines going up

repeating

[indigo dirge]

what is the definition of tan(x)?

[frosty coral]

y/x?

[indigo dirge]

in terms of other trig functions

[wooden python]

@indigo dirge can you take this over i gtg

[indigo dirge]

yep sure

[frosty coral]

What is the definition?

[indigo dirge]

do you know tangent in terms of sine and cosine

[frosty coral]

I don't think I do

[indigo dirge]

hmm

have you heard of SOH CAH TOA?

[frosty coral]

yes

[indigo dirge]

ok, so if sine is $\frac{O}{H}$, cosine is $\frac{A}{H}$, and tangent is $\frac{O}{A}$, what do you know about tangent

[clever fjord]

Frances

[indigo dirge]

how would you manipulate the first two to get the third

[frosty coral]

you put the sine on the top and cosine on the bottom?

[indigo dirge]

mhm

[frosty coral]

I dont understand what you mean by manipulate

[indigo dirge]

can you express tan in terms of sin and cos given this? ^^

[frosty coral]

no

sorry I'm a little lost

[indigo dirge]

issok

how bout this

what happens if you take sin/cos?

[frosty coral]

that makes tan?

[indigo dirge]

express it in terms of O, A, H

[frosty coral]

that makes O/A

[indigo dirge]

mhm!

[frosty coral]

oh is it because H is canceled out?

[indigo dirge]

which is the def of tangent

alright

now when do vertical asymptotes occur?

[frosty coral]

I think it has something to do with 0?

when the denominator is 0?

[indigo dirge]

yes!

when the function is undefined, right?

[frosty coral]

yes

[indigo dirge]

so if the denom is cos(x), when is it 0?

[frosty coral]

90 and 270

wait is that

[indigo dirge]

in general

[glass vault]

If you think about it when tan is sinx/cosx, there will be an asymptote when the denominator is 0 since its impossible to divide by 0

[indigo dirge]

bc cos x has infinitely many roots (aka when it is 0)

[frosty coral]

2pi + kpi something like that?

that might be for something else

im not sure

[indigo dirge]

close! i think thats for sin

but cosine is just a shifted sin

[frosty coral]

im not sure

how to find the numbers that makes it 0

[indigo dirge]

hmm

think of the unit circle

[frosty coral]

oh wait the degrees

are 0

and 180

i think

in the unit circle

[indigo dirge]

you had it here ^^

[frosty coral]

oh

[indigo dirge]

just generalize it

[frosty coral]

I don't really know how to generalize it

[indigo dirge]

ok uh

can you give me another value when cos x is 0?

[frosty coral]

uhh

how would I get another value

[indigo dirge]

hint: it doesnt need to be in between 0 to 360

[frosty coral]

so like 270+180?

wait

no

[indigo dirge]

yes!

[frosty coral]

oh

so

would that have to do with Z

because its like

all integers

[indigo dirge]

alr take a look at this

can you generalize the roots now?

hint: all roots are the same amount apart

[frosty coral]

pi/2 +k pi?

[indigo dirge]

yep!

and therefore ||these are the vertical asymptotes for tan(x)||

[frosty coral]

ah

and my question doesn't have a b value

so is the answer just pi/2+kpi?

[indigo dirge]

yes

[frosty coral]

I remember asking about the question but for sec

and someone said you have to add k E Z

[indigo dirge]

ah well

[frosty coral]

nevermind I dont think he said you have to

he just added it

[strange aspen]

this to restrict k to the integers bcz if k is something like 7.3 it would mess up the solution set

[frosty coral]

so it is only for certain questions?

[strange aspen]

If the the interval is given no need to find a k value
If not then restricting k to the integers is the correct way to do it
Because as I stated it’ll mess up the solution set

[strange aspen]

Like here if k is in R rather than Z then k can take a value like 917.528 and ofc it’s not in here

[frosty coral]

so for my question would it be neccessary to add the restriction?

[strange aspen]

Yes because given there is no interval
k should be in Z

[frosty coral]

im sorry but what is an interval

[indigo dirge]

its just a range of numbers

[frosty coral]

oh

alr I think I understand

thank you

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[tiny pendant]

why

[tiny pendant]

why do we only need to test the first floor(sqrt(p)) positive integers

(question-prove that 37 is prime)

[quasi sparrow]

a * b = n implies a or b is less than or equal to sqrt(n)

[tiny pendant]

yes

why does that matter

[quasi sparrow]

that answers your question?

[tiny pendant]

i still dont see it

[quasi sparrow]

what is it?

[shell haven]

For all the pairs of factors of n,a and b, such that a*b=n, either a or b will be less than or equal to sqrt(n)

So like if we can't find any factors under sqrt(n) we know there aren't any pair of factors so the number is prime

[tiny pendant]

oh so if both of those factors will always be less than root p floor the max possible value will be root p floor

[trail horizon]

if both a and b are $>\sqrt{n}$ then their product will also be greater than n

[clever fjord]

DaveyLovesSocks

[tiny pendant]

am i right?

[shell haven]

I mean a factor can be more than root p floor but then the other one will be less than root p floor

[tiny pendant]

but how

its not possible its given

[shell haven]

For all the pairs of factors at least one of the factors in the pair has to be less than floor(root p)

Like 10=5*2

floor(root 10)=3

5 is greater than 3

But 2 isn't

[tiny pendant]

so the max possible value will always be root p floor?

[shell haven]

I mean in a way

[odd edge]

@tiny pendant Has your question been resolved?

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[soft terrace]

I'm trying to understand my textbook, and some confirmation would be much appreciated. Suppose I have the differrence equation
$$g(k) = p g(k+1) + q g(k-1) + 1$$
where $p \ne q = 1 - p$. We notice the homogeneous case can be refactored to
$$\Delta g(k) = (q/p) \Delta g(k-1)$$
which has the general solution $\Delta g(k) = \alpha' + \beta' (q/p)^k$. We then notice the telescoping sum
$$g(k) = g(0) + \sum_{i=0}^{k-1} \Delta g(k) \equiv \gamma + \alpha k + \beta (q/p)^k$$
If we then try a particular linear solution, we see $k / (q-p)$ is a solution.
\par
My question: Can this particular solution \textbf{replace} the linear term of the general solution, or can it just be absorbed?
$$g(k) = \gamma + \frac{k}{q-p} + \beta (q/p)^k \qq{OR} g(k) = \gamma + \left( \frac{1}{q-p} + \alpha \right) k + \beta (q/p)^k$$

[clever fjord]

Shuba

[soft terrace]

Never mind, I'm being silly.

Late night maths again

😮‍💨

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[pastel orbit]

soft question

[pastel orbit]

we say that a map F: X \to Y is proper if for every compact K in Y, its preimage F^-1(K) is compact in X

I'm curious as to why we call it that adjective

what is so "proper" about these maps, which aren't even necessarily continuous?

[odd edge]

@pastel orbit Has your question been resolved?

[gritty spire]

why do we call an ideal ideal if theres nothing ideal about it

[amber veldt]

park in a driveway drive in a parkway

[gritty spire]

anyways this property seems to stand out.
A topological space is compact if and only if the map from that space to a single point is proper.

[pastel orbit]

I'm not exactly sure why this gives it its name though

[gritty spire]

thats like a very cool property

[pastel orbit]

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[edgy holly]

What variables to replace here? a=b, or a=b+c does not work

[quasi sparrow]

what is the factor property

[edgy holly]

factor property is like if by replacing any variable a with any other variable b., the determinant becomes 0. then (a-b) is a factor of determinant

kind of self explanatory lol

[quasi sparrow]

[edgy holly]

<@&286206848099549185>

Help

[golden sonnet]

better go for row column transformations, like try r1->r1-r2, etc

try to just make as many zeroes for easier expansion, i'd say

[edgy holly]

we are asked to use factor rule....

[golden sonnet]

hmm

mb then

[edgy holly]

I just tried a=-b, I think it's working

[umbral jolt]

the degree checks out, its gonna have 3 factors at max.

[golden sonnet]

but it is quite absurd that they wanted us to solve it by such an unknown method

[umbral jolt]

try a=-b

[edgy holly]

yes

[golden sonnet]

quite weird

[umbral jolt]

and the symmetry is obv

so b+c and c+a are factors asw

write it as k(a+b)(b+c)(c+a)

now put a=b=c=1 in both sides, find k

should be 4

[umbral jolt]

its not, infact it works better than row and column operation sometimes

[edgy holly]

a=-b is not working......

[umbral jolt]

it should

lemme try

[golden sonnet]

i never learnt it tbh. it always seemed as high staking and absurd, so i relied on row column transforms, and your point is valid too

[edgy holly]

oh yeah it worked!!

Thanks a lot

[umbral jolt]

nps

determinants are annoying

and I dont like dealing with their evaluation

[edgy holly]

same, no need to learn it. I just discovered it last year by accident while doing the triangle collinearity type shi and this year it's apparently a part of our syllabus

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[leaden karma]

#1398993464338808872 message

[leaden karma]

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[indigo dirge]

how would you solve this using pigeonhole principle? i literally don't know where to start

[indigo dirge]

(or any method really, we were just learning about pigeonhole)

[ivory grove]

You can first look at any two points of these 5 points. These two points define at least one great circle, so choose one of them

and then pigeonhole the remaining points

[indigo dirge]

ohh ty marianne!!

alr i think i got it!

close

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[ivory grove]

[sturdy cape]

[I think they uniquely define a great circle, except if they're polar opposites]

[but the rest of the argument still applies]

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[cobalt owl]

hi

I got 2x-1 for [g o f]

not 2x + 1

im a lil confused

g(x) = 4x -2 + 4 / 2

4x - 2 / 2

2x - 1

[manic sleet]

-2+4 is -2?

[cobalt owl]

oh shoot

bruh

ty

[indigo dirge]

Do you need your answer checked?

Or do you need help with solving it?

[cobalt owl]

No i just figured it out

i have a question abt this one tho

would you say this is one to one

the only thing im confused abt is the origin

i cant tell if it passes the horizontal line test

[odd edge]

@cobalt owl Has your question been resolved?

[indigo dirge]

Hmm

This looks like an odd-degree function, no?

Of course, if you don’t have the exact function, you don’t know for sure if it’s one- to-one

[manic sleet]

though it appears to pass through (0,0) and (1,1), most logical guess is f(x)=x^3

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[candid sparrow]

Hi I need help with drawing the reciprocal of these graphs I know its 1/f(x) but as I drew them they started looking wonky

[candid sparrow]

for reference these are the graphs I got after

[odd edge]

@candid sparrow Has your question been resolved?

[odd edge]

@candid sparrow Has your question been resolved?

[candid sparrow]

[mental lotus]

The others looks alright, but I would like to understand your view on why you drew this one the way you did

[odd edge]

@candid sparrow Has your question been resolved?

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[dim igloo]

finding the s.a

i got 1392.7 when the answer is 1192.7

i dont know where i went wrong?

[hidden token]

The SA of the cuboid is not what you think it is

[dim igloo]

what do umean?

[hidden token]

I mean that while its TSA is what you have written, not all of it has to be included when calculating the SA of the entire shape

[dim igloo]

tsa means total surfacea rea?

[hidden token]

Yes

[dim igloo]

also, i got the answer, but i dont understand why

from my equation in the pic

i minus'd (20x10)

in the equation im sure i didnt double up on that already

[hidden token]

Look at the shape carefully, there is one side of the cuboid that shouldn't be included

[dim igloo]

oh the one touching the cylinder

but like when calculating the s.a for the half cylinder

there wasnt any working for the rectangle of the half cylinder

[hidden token]

Why do you need to include that rectangle?

[dim igloo]

oh wait im so dumb

[hidden token]

Its inside the shape so it shouldn't be counted as surface area

[dim igloo]

ur right

its internall

damn i understand

thanks bro

i been thinking of these problems in parts lol

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[shrewd trellis]

what did I do wrong

[narrow crypt]

hi hell

[shrewd trellis]

helloo

[narrow crypt]

,w 100 - 80x + 12x^2 = 0

[clever fjord]

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Requested by nuclearinteraction

[shrewd trellis]

i got a correct

[narrow crypt]

yup

[shrewd trellis]

not sure about b because my x is -5/3 which is

below 0

[narrow crypt]

it seems u've solved it incorrectly

[shrewd trellis]

i used the

complete the square method

did i do a mistake there?

[narrow crypt]

why is 20/3x suddenly positive

[shrewd trellis]

oh damn

[narrow crypt]

,w (x - 10/3)^2 = 25/9

[clever fjord]

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Requested by nuclearinteraction

[narrow crypt]

shouldve kept the negatiev sign

[shrewd trellis]

yep

is that the only mistake?

[narrow crypt]

,calc -80 + 24(5/3)

[clever fjord]

Result:

-40

[narrow crypt]

u need to use the original eq to find hte volume

other than that and the current one, nah

[shrewd trellis]

did i not use the orignal equation

its the same thing but

expanded

[odd edge]

@shrewd trellis Has your question been resolved?

[odd edge]

@shrewd trellis Has your question been resolved?

[hidden token]

From what I can see, you found the second derivative, plugged in your answer of 5/3 into that and then substituted that result into the first derivative, incorrectly

Which is not what you're meant to do

[odd edge]

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• Show your work, and if possible, explain where you are stuck.
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[ashen hatch]

Given AB=8 CD=5 BP=6 and angle APD being 60°, find the square of radius of the circle

[ashen hatch]

I forgot the chord theorems or whatevs its called

[tacit haven]

power of points?

[velvet gull]

PB * PA = PC * PD

[ashen hatch]

if I construct line AD, would it form a right angle triangle?

[tepid jolt]

Hello everyone! i am new here

[ashen hatch]

#discussion

[tacit haven]

not unless you prove it must be, no

why do you think it would be?

[ashen hatch]

looks close enough

[tepid jolt]

guys of what class this question is?

[ashen hatch]

idk man im just doing this for fun

[tepid jolt]

ok.. i love maths a lot

good to find people like me

[tacit haven]

yeah you cant just do that lol

[ashen hatch]

yeah no i proved a few circle chord theorems for fun but i ain't doing this

[polar whale]

can anyone help me with this

[ashen hatch]

make your own channel

💀

[lean yew]

!occupied

[odd edge]

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).

[lean yew]

please use an open channel

[odd edge]

@ashen hatch Has your question been resolved?

[odd edge]

@ashen hatch Has your question been resolved?

[glossy basin]

4²+a²=2.5²+b²

10²+a²=(x+2.5)²+b²

you can get x from here then use it to get r

@ashen hatch

this looks correct to me, tell me if you see anything wrong with this

[ashen hatch]

.close

[odd edge]

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[odd edge]

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[mystic saffron]

yo

[mystic saffron]

the answer is 2 sqrt of x lnx - 4 sqrt of x

what did i do wrong

[nimble blaze]

what are you doing in those last three lines

[mystic saffron]

rearranging

oh shit

[vernal yacht]

Long time no see

[mystic saffron]

i forgot about the right side

[mystic saffron]

😂😂

damn been a while

this is for my uni entrance exam

[nimble blaze]

oh, that third last line was two separate steps

[mystic saffron]

wait let me try and fix it

[nimble blaze]

you multiplied by half instead of dividing by half

[mystic saffron]

yup

[nimble blaze]

also just simplify each term individually

instead of factoring

[mystic saffron]

yea i havent solved math for a while and forgot 1/0.5 is just 2

made it way easier

thanks

how do i close

.close

[odd edge]

Channel closed

Closed by @sly lily

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[odd edge]

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• Show your work, and if possible, explain where you are stuck.
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[sage helm]

While reading a book in Linear Algebra, I noticed this passage:

[sage helm]

A dot product between a vector and a vector space.

Questions:

1. Is that common notation, or is that particular to this author.
2. I assume that means the following:

[low locust]

uncommon notation

it means what you assume

[sage helm]

Have you personally seen that notation before, or is it new to you, as well?

[weary pelican]

there's a lot of similar instances

for example, $a + B = {a+b\mid b\in B}$

[clever fjord]

Raphaelisius Maximus MMIII

[weary pelican]

so it's natural to assume $\vec w \cdot S = {\vec w \cdot\vec v\mid \vec v \in S }$

[clever fjord]

Raphaelisius Maximus MMIII

[sage helm]

Well, that latter I do understand. But, the result of that is a set, not a number.

[weary pelican]

then "0" would actually stand for {0}

it's a common abuse of notation

[sage helm]

The strictly proper would be this this, then?

[weary pelican]

that's correct

[sage helm]

cool, thanks.

.close

[odd edge]

Channel closed

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[odd edge]

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[tame echo]

i cant draw tgis

[tame echo]

may u b my mth techer

[stark heart]

Have you tried solving for the roots?

[tame echo]

roots were 1 and 0

[stark heart]

They are not

[tame echo]

1 nd -0

[stark heart]

You got the y-intersect correctly by substituting x=0 to the formula

Now substitute y = 0 to get x-intersects, i.e. the roots of the expression

[stark heart]

What do you even try to get these answers?

[tame echo]

i sued quadratic equation

i used calcualtor

i thnik clacluter uses quadrtic equation

calculaor said x1 and x2, 1 -0 respectively

[stark heart]

How did you use the calculator
Because it's wrong
You know that you need to rearrange the formula to
-x^2 + 4x -5 to get the correct answer?

[tame echo]

i did tha

??

confused

i am geting

[stark heart]

The solutions are
2 - i and 2 + i

[tame echo]

2-i

what does that even eman

[stark heart]

Which means they are complex solutions i.e they don't touch the x-axis

[tame echo]

oh

[honest turtle]

You're supposed to use the derivate.

[tame echo]

i shud first use discrminant

[tame echo]

huh

[stark heart]

I don't think so
Using the derivative won't help that much in sketching the original function

[honest turtle]

At least you get the max point, I guess that helps.

[tame echo]

it does aha

[honest turtle]

You're just sketching the function, other than that doesn't matter.

[tame echo]

did u see mty sketch'

its so difficult

scheme said shape down coz -1 and use y intecept

[honest turtle]

Sketching function is to be honest substituting values, plotting points and then connects them.

[tame echo]

turning point, x intercept, y intercept

[honest turtle]

Make a table for, say -2 -1 0 1 2, then you pretty much good to go (or find the critical informations, intercept and max/min point).

[stark heart]

The final 2 thing you can do are
1- get the turning point of the parabola and sketch it
Because the co-effecient of x^2 is negative then it would be the max point
2- substitute different values of x and sketch them on the graph

[tame echo]

ok

see

its not workign

[tame echo]

..

[tame echo]

.

[honest turtle]

Where's your table?

[odd edge]

@tame echo Has your question been resolved?

[tame echo]

i thought..what about using turning point, y intercept and x intercepts if they exist, you mean they will automatically be there?

when i use table

[tame echo]

<@&286206848099549185>

[odd edge]

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[gray hearth]

hello :3
i am trying to find a 3d point between two other 3d points, and the rotation to align an object along the line the two points make
i'd perfer a set of euler angles, but a straight answer for quaternions would work too.

[odd edge]

@gray hearth Has your question been resolved?

[ripe lynx]

Line from x to y will have direction of y-x, as a vector
Just normalize it and apply formulas to calculate pitch/roll/yaw, they should be easy to find online

[gray hearth]

that doesnt really explain much and everything im finding online isnt in the right format for what i need,

[ripe lynx]

what format do you need though

[soft terrace]

the vector direction x-y is the same, regardless of your axes. the pitch, yaw, and roll depend entirely on how you set up the axis.

are you trying to find the midpoint between two other points in a 3d space?

[gray hearth]

XAngle, YAngle, ZAngle, euler angles

or a quaternion

[soft terrace]

again, it depends on what axis you're using

[gray hearth]

ok let me rephrase my question

[soft terrace]

Moreover, you can't totally define a rotation by one vector, you need the forward and upward vectors.

[gray hearth]

i want the x, y, and z rotation that points from the origin to another point

[soft terrace]

there's an infinite number of these vectors - consider a plane pointing in that direction whose roll is undefined.
you need an "up" vector as well.

[gray hearth]

"up" is my Y Axis

[soft terrace]

Let $v = p_1 - p_2$ be your forward vector, and $y$ be the y-axis (your upward vector).
\begin{enumerate}
\item Normalise the forward vector
$$z = \frac{v}{|v|}$$
\item Compute the right vector
$$x = \frac{y \times z}{|y \times z|}$$
\item Compute the corrected upward vector
$$y = z \times x$$
\end{enumerate}
Now $(x,y,z)$ is your rotation matrix. Notice this only works if your forward vector isn't the y axis (i.e. the player isn't looking directly upwards)

[gray hearth]

so then how do those translate into rotations?

[clever fjord]

Shuba

[gray hearth]

do those numbers represent values between 0-360?
like am i getting degs? rads?

something else?

[soft terrace]

(x,y,z) is the rotation matrix. Take any point p, and (x,y,z) p is your point rotated in the direction of the forward vector v.

[gray hearth]

ok
and |v| means normalizing not absolute value?

[soft terrace]

$$|v| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$