help-49

[floral apex]

yea

we can just use column ops on the right

idk why i didnt put it together

[twilit field]

It does?

[floral apex]

rank(A) = rank(A^T)

actually this is even better for us i think because it tells us what V is gonna be

[floral apex]

so we can get to full RREF

and we can swap around columns or rows

so we can for sure get to $\mqty( I & 0 \ 0 & 0 )$

[grand pond]

jan Niku

[floral apex]

I think this is enough

but feel free to tell me to fuck off

[twilit field]

I would never

I was in the toilet, hence I didn't respond

[floral apex]

ah

[twilit field]

sorry

[floral apex]

usually i only respond on the toilet

so cant relate

we can start with A

with some right multiplication

we can get to REF

then, some left multiplication

RREF

because A and B have the same rank, theyll have the same RREF

and, each individual elementary matrix is invertible

how do you feel about each of those pieces individually

[twilit field]

feels alright to me

[floral apex]

then good to connect?

[twilit field]

so eventually we can get it to $\begin{bmatrix} I&0\0&0 \end{bmatrix}$

[floral apex]

yup

plus some other junk

[grand pond]

Veni, vidi, perii

[floral apex]

idk what you wanna call it

lets say

[hard umbra]

you've sent examples from your textbook doing this in the past

[floral apex]

$A = E_R ^m \mqty(I & 0 \ 0 & 0) E_C ^n$

where R and C for row and column and 3 and 4 for square dimension

err

shit i should use m and n

[grand pond]

jan Niku

[twilit field]

That's using row reduction, we haven't done column reduction in class

[hard umbra]

no, you've definitely done this and last time you claimed the same thing

your textbook did row reduction on the transpose

which is just column reduction

[twilit field]

[floral apex]

its okay i believe you dont remember veni

[twilit field]

no like we haven't done in in our actual classes

there's a lot in our textbook we arn't covering

[floral apex]

IDK what you want us to say to that

[twilit field]

nothing

[floral apex]

ah thats easy

so what do you think of the steps from here

[twilit field]

first pre-multiply both sides by $(E^m_R)^{-1}$

[grand pond]

Veni, vidi, perii

[floral apex]

its not necessary

but sure

actually, i think this is gonna getcha

[twilit field]

and then post multiply by $(E^n_C)^{-1}$

[grand pond]

Veni, vidi, perii

[floral apex]

lets just say like

geez im running out of letters

I kinda wanna change the notation

lets just know n means columns

and m means rows

[twilit field]

I actually have a class in 5 minutes, shall I contniue this later. Also, I'll post in my thereads from now on

[floral apex]

$E^m_A \cdot A \cdot E^n_A = E^m _B \cdot B \cdot E^n _B$

[grand pond]

jan Niku

[floral apex]

we agreed both would have the same rref

thats all this says

moving forward from here is very easy

[twilit field]

Ah, cool

thanks

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[last slate]

Have I done correctly

The mark scheme says correct working

But doesn’t show any working

[fallow scarab]

Yup

[midnight plank]

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[trim pewter]

i am very confused

.close

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[twilit field]

So my idea was let U be a m by m square matrix,which is a product of the elementary matrices that gets us form R to A

[runic hamlet]

yes

[twilit field]

is that enough?

[runic hamlet]

yes

[twilit field]

Thanks

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[hidden sail]

can any one prove tan47/cot43=1

[midnight plank]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[novel lion]

can help tho

tan(90-x) = cotx

well, kind of already proved but yes

[midnight plank]

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[viral dagger]

let complex numbers $z_1, z_2, \dots, z_{2023}$ be the roots of $$P(x)=x^{2023}-x^{2022}+x^{2021}-\dots-x^2+x-1$$
find
$$(z_1^{4049}+z_1^{2024})+(z_2^{4049}+z_2^{2024})+\dots+ (z_{2023}^{4049}+z_{2023}^{2024})$$

[grand pond]

Skill_Issue

[viral dagger]

bruh who tf made this abomination 😭

[feral sedge]

oh this one is actually fine

[barren onyx]

Omg

[feral sedge]

looks very ugly at first though

[feral sedge]

i.e. notice what happens when you try to factor this

[viral dagger]

i tried to but i didnt get anything unfortunately

wait

[barren onyx]

what grade is this btw

[viral dagger]

no clue, olympiad

[feral sedge]

ye

[viral dagger]

wait which one do you want me to factor

[feral sedge]

oh the polynomial

[viral dagger]

$(x-1)(x^{2022}+x^{2020}+\dots+x^2+1)$

[grand pond]

Skill_Issue

[viral dagger]

this?

[feral sedge]

yes that would be right

ig this was noticeable at the very start but now to proceed it is clearer that || the thing on the right is a geometric series||

[viral dagger]

oh

unique question

[feral sedge]

it is unique yes i don’t think I’ve seen this setup appear specifically before

but the trick of rewriting polynomials using geometric series is a good tool to be familiar with

[viral dagger]

$$(x-1)(\frac{x^{2024}-1}{x^2-1})$$
$$\frac{x^{2024}-1}{x+1}$$

[grand pond]

Skill_Issue

[viral dagger]

soo x=/=-1, x^2024=1?

ig that clears the ^2024 part of what were looking for

how bout the 4049 part

thank god tbh i thought there will be some root of unity bs

wait i think i see the idea

say the $z_1^{4049}=(z_1^{2024})^2z_1=z_1$

is this corrwct

so it becomes
$$(z_1+z_2+\dots+z_{2023})+(2023)$$

and the first part you can get from vietas or smth

its 1 right?

its the negative of the coefficient of x^2022

is it 2024?

[grand pond]

Skill_Issue

Skill_Issue

[viral dagger]

whoops my dummy head used the wrong var :p

[feral sedge]

would say so

[viral dagger]

ty

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[viral dagger]

.reopen

[midnight plank]

✅

[viral dagger]

wait

am i stupid or is (x^2024-1)/(x+1) not equal to x^2023-x^2022+...

err ok stfu

.close im stupid asfk :p

[midnight plank]

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[feral sedge]

does this mean u see why it’s not

[viral dagger]

wdym

hold on

[feral sedge]

oh it was a typo

[viral dagger]

i meant to ask for (x^2024-1)/(x+1) cause using syntethic division i get something much diffrent, but i realized i tried to divide with x-1 instead

:p

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[red cave]

how do u do this

[hearty rune]

what are the domain and range of arccos(x)

[red cave]

domain is -1 <= x <= 1

range is 0<=y<=pi

[red cave]

what to i do next

[novel lion]

u got the domain

now for the range

range of arccos(x) is 0 <= y <= pi

so what will be the range of arccos(x)/2

[red cave]

um do you times the two sides by 2

0 <= y <= pi/2

[novel lion]

divide*

[red cave]

o

yea

[novel lion]

thats the range

[red cave]

how do i get the domain

[novel lion]

u already got the domain

[novel lion]

.

[red cave]

the answer says its -2 <= x <= 2

do i do this kind of method for arctan and arcsin as well

[novel lion]

oh yeah

my bad

i didnt notice its x/2

[red cave]

allg

[novel lion]

yeah so it'd be -2<=x<=2

since x/2

[red cave]

also i tried finding the range of arctan(x-6) + 3pi/2 but i got the wrong answer can u help me

[midnight plank]

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[primal tundra]

Hey

[midnight plank]

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[primal tundra]

Is 1 = 0.999999999999... ?

[feral sedge]

yes

[novel lion]

yes

[primal tundra]

But when u use lim it's not the same

[novel lion]

wdym

[primal tundra]

W?

It will gives u an epsilon when u use lim

Lim means that it's 1 but it's not

And here is correct too

[runic hamlet]

I think you misunderstood what lim means

[primal tundra]

Can u explain it for me pls

[runic hamlet]

what definition of lim have you seen

[primal tundra]

That there is a function near to our function that u never achieve it

Like lim of 0

[runic hamlet]

are you confusing it with an asymptote?

(thats also not the correct definition of asymptote tho)

[primal tundra]

So please explain it

[primal tundra]

For sure not

[runic hamlet]

while you not necessarily ever "reach" it, in the limit you are still actually equal to it

[primal tundra]

Unfortunately I can't explain lim in English I have to explain it in persian

[runic hamlet]

cant help with that

[primal tundra]

Exactly so lim of 0.999999... is one but there is an epsilon

[runic hamlet]

no

in the limit its equal

the ... itself stand for the limiting process

so we have the numbers 0.9, 0.99, 0.999, 0.9999 and so on

and you are right, they never equal 1

but we get closer and closer

and a limit is a way to make precise that "after infinitely many steps" we actually equal 1

[primal tundra]

I understand what u say but it's not okay that 1 is equal 0.999999...

[runic hamlet]

why not

[primal tundra]

Due to its never reach it( the definition of lim) but u can say that both r same

[runic hamlet]

so it is ok to say they are the same?

[primal tundra]

I say both are correct so we can say they are equal and they are not

But my idea can't be correct

[runic hamlet]

so we have two numbers which are equal and not at the same time?

[primal tundra]

Exactly what I mean

[astral garnet]

Schrodinger's number

[runic hamlet]

how could both of these be true at the same time

[primal tundra]

The problem is this

[primal tundra]

Or grassman number which said a×b=-b×a

[primal tundra]

Any idea?

[runic hamlet]

well they arent

the numbers are equal and thats it

[primal tundra]

No they aren't the same there is an epsilon between them

One another question is this interval correct? (0.99999...,1)

[runic hamlet]

its an emptyset

try telling me a number between 0.9999... and 1

[primal tundra]

I think it has two answers

[primal tundra]

We can't because it's an epsilon

[runic hamlet]

in the real numbers there are no such things

if you want to be in the hyperreals then go ahead

but you'll be lonely there

[primal tundra]

So maybe we have an imaginary number

[runic hamlet]

thats another completely different thing

[primal tundra]

As ur talking about hyperreals num we can believe that it would answer for imaginary mums too

[runic hamlet]

hyperreal numbers and imaginary numbers are very different things

hyperreal numbers are actually a way to have numbers "epsilon" like you want

[primal tundra]

But it can be it subset

[primal tundra]

Yeah when u have f(x)=f(ε+x)-f/ε

[primal tundra]

Is it correct?

[runic hamlet]

well ignoring the typo and that we are ignoring any semblance of technical accuracy, sure

before you try to work with hyperreal numbers you should first actually understand what real numbers are

[primal tundra]

Just one question r u a math teacher?

[runic hamlet]

no

[primal tundra]

Aha

U really helped me tnx man

[steep hinge]

however there are definitely faster methods to show that 1=0.999

[exotic pelican]

by definition yes, this dot dot dot notation, or the way I prefer to write it using a bar $1 = 0.\bar{9}$, is just some notation that refers to the limit itself, not the sequence terms that lead to this limit

[grand pond]

rbit

[primal tundra]

How?

It's the lim of x=>1

[exotic pelican]

the limit of the sum 0.9 + 0.09 + 0.009 + 0.0009 + ...

[primal tundra]

Okay

[exotic pelican]

to put it into more rigurous terms

[primal tundra]

Ah I got u

[exotic pelican]

if you have a sequence of digits $a_1, a_2, a_3, ...$ and some integer K, you can then define the notation $K.a_1 a_2 a_3 ...$ to refer the limit of the sum $K + \sum_{n=1}^{\infty} a_n \cdot 10^{-n}$ (which can be proven to always exist)

[grand pond]

rbit

[exotic pelican]

in our case, K=0 and a_n = 9 always

[exotic pelican]

exactly, plugging that into the definition leads to exactly that sum

[primal tundra]

Tnx man/madam

[exotic pelican]

so yeah in the end its just true by definition, you may argue whether this definition makes sense or not, but seeing as this gets used by almost all mathematicians that may answer it

[primal tundra]

R u an mathematician?

[exotic pelican]

anyone who does math

[primal tundra]

Aha in that case

[exotic pelican]

but ok, maybe I should have stated professional mathematicians

[primal tundra]

Mb

[exotic pelican]

theres a lot of unprofessional people that like to question it

[primal tundra]

Yeah

And ur not a teen I guess

[exotic pelican]

well more than that I means, they question it and then refuse to believe any evidence that is presented to them

[primal tundra]

Coz I'm teen but I enjoy studying math and helping others but other teen that I know hate math

[exotic pelican]

yeah, might have to do with the fact how math is being tought most of the time

[primal tundra]

Yep

[exotic pelican]

it can be much more than just boringly manipulate some numbers over and over again

[primal tundra]

We are at the same page

[midnight plank]

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[midnight plank]

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[stray totem]

hey does anyone know how to define $\exists!$ with only $\exists$ and $\forall$?

[grand pond]

wololo

[stray totem]

I don't really need any help I'm just really curious

[dreamy lichen]

Do you want a solution or just a hint?

[stray totem]

solution because I feel like I wouldn't even be able to find it myself

[dreamy lichen]

$\exists! x P(x) \iff \exists{x} \forall{y} (P(y) \iff y = x)$

[stray totem]

lmao

[dreamy lichen]

I hope it's correct

[stray totem]

how many edits was that

[dreamy lichen]

too many to count

It's not correct

[stray totem]

hm

[grand pond]

MæthIsAlwaysRight

[dreamy lichen]

now it should be correct

Now it guarantees that there exists x such that P(x) and that the x is unique

i need to get better at latexing

[stray totem]

oh

I kinda see how that works

thanks

[dreamy lichen]

btw, interesting fact is that you can even write \exists with \forall

[misty cairn]

I dont understand

what does p(y) <-> y=x imply?

[stray totem]

I think the for all mean that for every result you get with p means that the value used is x

so x is unique

at least that's how I understand it

[dreamy lichen]

Firstly, it gives us that if y = x, then P(y)

[dreamy lichen]

Or simply, P(x)

$\exists! x P(x) \iff \exists{x} (P(x) \land \forall{y} (P(y) \implies y = x))$

this would actually be equivalent definition

[grand pond]

MæthIsAlwaysRight

[dreamy lichen]

or even this

and the second part, that is P(y) => y = x tells us, that if P(y) then y must be x

meaning that there are no other y's, except for x such that P(y)

[dreamy lichen]

If you recall that
\forall is equivalent to \not \exists \not, then you can further rewrite this as:

$\exists! x P(x) \iff \exists{x} (P(x) \land \lnot \exists{y} (P(y) \land y \neq x))$

[grand pond]

MæthIsAlwaysRight

[dreamy lichen]

and now this literally translates to "There exists x such that P(x), and there isn't any other y such that P(y)"

[midnight plank]

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[last slate]

is $x^{\log_x m} = m$?

[grand pond]

anjali

[steep hinge]

yes

[last slate]

ok

[steep hinge]

may I close?

[last slate]

.close

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[high mist]

Helloo

[high mist]

I need some help solving this

I have a solution although...

Not quite sure if its correct

[fallen aurora]

erm it is not correct

[high mist]

Ooo ooo explain pleaseee

[high mist]

I see

I used the most bullshit solution ever

[fallen aurora]

it was cube root of y not x

[high mist]

Ahhh

Ohh I see

[novel lion]

even if that

how did y convert into x

[solar jetty]

Try rationalization
Times the denominator and numerator by √x³ + √y³

[high mist]

Brain damage math

[high mist]

Ahhhh

[midnight plank]

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[midnight plank]

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[slim eagle]

how to solve without using period properties

[slim eagle]

[.] represents floor function

[runic hamlet]

why would you

its basically just using the def of floor

[slim eagle]

just wondering

if it is possible

since it was given pretty early in the book so there might be i guess

i tried making series kinda thing but the last term was ugly

[tribal temple]

Why the hell is x both a limit variable and the variable for the integral

[quiet vale]

same question 💀

[midnight plank]

@slim eagle Has your question been resolved?

[steady oar]

bro how are you gonna integrate a function THIS discontinous😭😭

i mean i guess if the difference within bounds is less than or equal to 1 and doesnt cross an integer then you could just take the part you care about and analytically continue the rest to get a continious function for which the fundemental theoreom of calculus applies?

but if the difference of bounds is more than 1 then i dont think you can do a definite integral on it

so in this case x cant be greater than 1 or less than 0

(since if it is greater than 1 we will have an integration bound that goes through one of these jumps/asymptotes)

so in this case that means the functions domain is (i think) 0<= x <=1
(domain of the function that is the solution, not the integrand)

[feral sedge]

[steady oar]

so for all cases of x less than 1, the floor will just be zero, so itll just be an integral from zero to zero
but if x = 1 its gonna be the integral from 0 to 1 of 1 which is 1?

[midnight plank]

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[steady oar]

i feel like im doing something wrong

[steady oar]

real

[feral sedge]

(this is very much integrable)

[steady oar]

i feel like theres something wrong with the question

[feral sedge]

well aside from the poor choice of bounds it’s fine

[steady oar]

you can integrate a non continous function??

[feral sedge]

yes

[midnight plank]

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[novel lion]

bro?

[gritty hatch]

I am very confused how one would logically get to that conclusion.

[novel lion]

are you trolling or what

[gritty hatch]

I will do this for the rest of my life.

[novel lion]

we did this like 3 times already

[gritty hatch]

if it means I will understand.

[novel lion]

alright, what do you want to understand

[gritty hatch]

well its not as much that i dont understand

as much as it is

i dont get how someone would reasonably get to that conclusion

like

here ill format

[gritty hatch]

also astar i dont think we did this even once honestly

the phrasing in prior questions was so broken that i dont even think you mentioned the real problem tbh

or you did but phrased it in a way i wouldnt understand

[novel lion]

math

[slim steeple]

[gritty hatch]

1. Ok so this hurts my head the fact that they took 2 which was multiplication and put it inside of the 2nd parameter in order to create an equality where because they all multiply and are technically equal so they then undo this somehow by just adding and ignoring everything?```

so once they created the true statement they undid everything by just ghosting ^2 and adding 2?

[gritty hatch]

and remember the answer to erm

(x+1)^2

the entire answer was that there was no factoring there was no special operation that would solve (x+1) (x+1)

the only thing that mattered was that (x+1) is the sqrt of (x+1)^2

prior i also had not considered every logic state of the problem

i also did not take as much as a detail-oriented approach

so i mistook sqrt of each component as the same as the sqrt of the whole

leading to problems

[gritty hatch]

you there?

[polar mortar]

do you agree that 2y + y^2 = y(2+y)?

[slim steeple]

Buddy are you sure you are not trolling? Multiple people explained this to you multiple times

You keep repeating same and asking weird questions

[gritty hatch]

you seem not to understand thats how math is

[slim steeple]

You don't seem to understand that you don't want to understand

[gritty hatch]

its not exactly easy for me to just interpret something like that so give me a second

[gritty hatch]

believe that if you want shows how little you actually understand about students

[novel lion]

he's not making fun of you

you should practice factorization more

[gritty hatch]

i am

this is my practice

there is no other form of practice if you cant answer a single question

you cant practice solving it if you cant in the first place

you love to stray from the topic sigh

[slim steeple]

Take easy questions bro to practice

[gritty hatch]

i wish people were more efficient like me

and only focused on the solutions

[slim steeple]

If you aren't able to get this exact question

[gritty hatch]

i will ignore all responses from you guys and only take into account the problem from now on

[slim steeple]

[gritty hatch]

i am finished responding to distracting information that will not assist me.

[gritty hatch]

hm? What do you think that guy along with many others just presented?

what was i just doing

before you distracted me?

[slim steeple]

Like
x²+3x+2 = 0
Factorize this

[gritty hatch]

also one last thing before i answer this

i want to understand, likely even more than you ever have

all i do all day every day is focus on eliminating all thoughts within my mind in order to focus entirely on math

math is everything to me

[slim steeple]

[gritty hatch]

i spend every day focusing every single one of my thoughts trying to achieve 90% of the possible efficiency a human can output

[novel lion]

[gritty hatch]

you dont know anything about me or what i am doing

so stop commenting on it

[novel lion]

chill bro, its not that deep

[gritty hatch]

if you have an idea for an alternate route provide it and dont talk about me

[novel lion]

carry on with your question

[gritty hatch]

it is that deep

people get depressed when they dont care as much as i do because they live lives they hate

[slim steeple]

Buddy I care about people who are not able to solve things, that's why I am in this server 😅

[novel lion]

no way you arent trolling

[gritty hatch]

correct, now stop mentioning me or what im doing.

that will be all i will not respond to any other such topics

[novel lion]

alright alright

back to the question

so what is it that you want to understand

[gritty hatch]

I dont know maybe I mentioned that earlier somewhere.

making a new channel lets try this again

.close

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[midnight plank]

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[gloomy wigeon]

Hi

[novel lion]

hey

[smoky solar]

yo

[arctic lion]

a

[smoky solar]

question?

[gloomy wigeon]

How do I get a y=kx+b+c/x-x'

[last slate]

hi

[gloomy wigeon]

From

[delicate sage]

Hii

[delicate sage]

Earth

[midnight plank]

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

[gloomy wigeon]

y=(x^2-4x+3)/x+1

[gloomy wigeon]

The original is in Russian but I can rewrite it if u need

[delicate sage]

So long division polynomials

[delicate sage]

NP

[delicate sage]

Do polynomial division bro 🙂

[gloomy wigeon]

I need to check out what is this

[delicate sage]

| x
----------
(x+1) | x²-4x+3 |
-(x²+x)
-----------
-5x+3

Like this continue

Understood 🫠

?

[gloomy wigeon]

I guess I understood, will try to do this on paper rn

Will be back with a result in a sec

[delicate sage]

https://brilliant.org/wiki/polynomial-division/#:~:text=Division with polynomials (done with,a factor of the dividend).

[delicate sage]

There is also a faster way , look into it 🙂

[gloomy wigeon]

Like that?

Damn

I think I wrote the last one wrong, it should be 8/x+1 instead

[midnight plank]

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[quick pollen]

i've got the base sorted but i'm unsure if i carry out the induction as i would normally if the summation term was something like k^2? i'm mostly getting lost with the k2^k part 😭

[carmine sigil]

,rccw

[carmine sigil]

A minor point: we don't need the base case to be true in order to assume the inductive case (you just need it to be true in order for the inductive case to be useful).

[quick pollen]

oops, lol thanks for the correction

[carmine sigil]

So can you show your work on the inductive step?

[quick pollen]

well that's the problem is that i don't know where to start

like my main confusion is whether the 'k' in k2^2 is just disregarded (?; pretty sure i'm wrong 😭) and i would continue on with the induction or if there's a different procedure

[carmine sigil]

well, you assume it's true for n = k. And then you try to prove it for n = k+1. So any place where you have (k) you replace it with (k+1)

wait sorry

k is the summation variable.

well, you have it holds for n.

So you show that it also holds for n+1

which means every place where you used to have an n, you now have an n+1

\begin{align*}
\sum_{k=1}^{n} k 2^k &= (n - 1) 2^{n+1} +2 \
\sum_{k=1}^{n+1} k 2^k &= (n) 2^{n+2} +2 \
\end{align*}

[grand pond]

OmnipotentEntity

[carmine sigil]

The top line is what you already have (from the inductive assumption), the bottom line is what you need to show

[quick pollen]

okay so i sub n+1 wherever n is but then what do i do with the k? is it not important?

[carmine sigil]

the k is just the summation index, you don't need to touch it.

it's not not important

it's just unaffected by the substitution n -> n+1

[quick pollen]

ahh okie

so i just leave it and then continue with the induction and it's still proven?

[carmine sigil]

yup

[quick pollen]

omg i feel dumb 😭

but thank you so much i appreciate it

.close

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[twin kernel]

Did I do something wrong

[fallow scarab]

,w int log(x) / x^2 dx

[grand pond]

Results provided by WolframAlpha

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

Requested by rie.mann

[fallow scarab]

looks like just a sign error at the end

the second to last step is right

[twin kernel]

I don’t see where I did it wrong

[fallow scarab]

(-a - b) = - (a+b)

you did (-a - b)= a -b

[twin kernel]

Ok I forgot to add the negative sign from the front but how is the 1 positive

[fallow scarab]

because of this

[twin kernel]

👍

@fallow scarab for these problems can I use u sub

[fallow scarab]

no

[midnight plank]

@twin kernel Has your question been resolved?

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[next viper]

does this factorisation rule apply to function x^2 - 4

factorisation rule of polynomial?

[nova yoke]

sure

[next viper]

woah

[nova yoke]

can you see what x1 and x2 should be?

[next viper]

(x-2)(x+2) =

hmm

[nova yoke]

yep

[next viper]

wrong?

[nova yoke]

why wrong?

(x-2)(x+2) = x^2 - 4

[next viper]

because

[nova yoke]

is true

[next viper]

oh!

it cancel

yes

woah

insane

[nova yoke]

yep

[next viper]

[nova yoke]

it gets trickier if you have x^2 + 4 instead of x^2 - 4

[next viper]

😮

😮

[nova yoke]

because x^2 + 4 has no real roots

[next viper]

but

[nova yoke]

so it can't be factored as (x-x1)(x-x2) unless you allow x1 and x2 to be imaginary

[next viper]

the test maker will not be that much of as hole

?

so he give this nice example

[nova yoke]

beats me, i don't know your test maker haha

[next viper]

but then rule not aplpy?

i cant just

get x1 x2

and insert

[sudden yacht]

What do you mean with "test maker"?

[next viper]

the person who makes a test

in university

for mathematics

which i write in 2 days

if i pass im on top of the world

so currently i am working 24/7 9/11

around the clock

to maximise mental capacity and performance

[sudden yacht]

Still don't understand...

[next viper]

Exam

you dont use word testin US&A?

[sudden yacht]

Yeah (I'm Italian though)

[sudden yacht]

But I don't understand the context here

[next viper]

-b +- sqrt b^2 - 4ac / 2a

---> x1 = something , x2 = someting

[sudden yacht]

Yeah, and so?

[next viper]

these results used in factorising quadratic polynomials

a(x-x1)(x-x2)

[sudden yacht]

Yeah, I understand all of this

But I don't get how this has to do with the "test maker"

[midnight plank]

@next viper Has your question been resolved?

[midnight plank]

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[vestal sorrel]

I'm trying to think about this problem, and I'm stuck trying to figure out how to approach it. It feels familiar to Hensel's lemma as it's earlier discussed, but I'm not totally sure what to make of this problem

[vestal sorrel]

.close

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[cursive ginkgo]

show that the limit of the function in the origin doesn't exist.

[cursive ginkgo]

I don't know how to begin

[fallow scarab]

choose two paths that converge to different limits

[cursive ginkgo]

like this ?

path 1 and path 2

I get 0 for both I think

[fallow scarab]

no

a path is like y = 2x or y = -4x and you take the limit as x goes to zero

[cursive ginkgo]

like this

[jaunty canopy]

yeah that works

[cursive ginkgo]

since it's different , I can say that the limit does not exist

right

[jaunty canopy]

yeah

[cursive ginkgo]

thanks

you both

a lot

🫶

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[cedar urchin]

Im confused on how to put the positive part of the graph into notation

[cedar urchin]

I was thinking maybe (-inf, -1)U[-1,2]

[visual tiger]

-1 and 2 are included then?

Is it what you want?

[cedar urchin]

Im assuming so, because the question is asking to find what part of the graph is positive and put it into notation

[visual tiger]

But is the function positive at those values, -1 and 2 included?

[cedar urchin]

It would be neither because its at zero youre right

[visual tiger]

Yes

So how do you change what you wrote

To exclude those 0 values

-1 and 2

[visual tiger]

Like the same thing as this but without -1 and 2

[cedar urchin]

I could just not include them and do (-inf,-1) but then how would I mark the relative maximum on that side of the y axis

[midnight plank]

@cedar urchin Has your question been resolved?

[cedar urchin]

<@&286206848099549185>

.close

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[twilit field]

so U would be a 2 by 2 matrix here and V would be a 3 by 3 matrix ?

[twilit field]

so U for instance would be

[midnight plank]

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[twilit field]

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[leaden matrix]

A bit stuck, not sure how to integrate that RHS and not sure if what im even doing is correct

[midnight plank]

@leaden matrix Has your question been resolved?

[leaden matrix]

<@&286206848099549185>

[grand pond]

bacc

[dawn dagger]

Solve these instead

Then y(x) will be a piece wise solution and don't forget the IVP

[leaden matrix]

how do I make sure that the solution is continuous at x=4?

[dawn dagger]

You consider the LHS and RHS limit and equate them

You will need to find a constant C for your IVP and constant K for the continuity part

Both after integration

[leaden matrix]

as in equate the first equation and the second equation after writing it in the form y= ?

[dawn dagger]

Kind of

Plug in x = 4 basically and find K

[leaden matrix]

what would y be at x=4 ?

[dawn dagger]

idk

[leaden matrix]

can I plug something like 3.99999 into the first equation to find it?

or is that wrong

[grand pond]

bacc

[dawn dagger]

Thats why you consider limits

Yea one piece is undefined at x = 4 still you can see what would the limit be

and then determine a suiting constant K

[leaden matrix]

if I plug something like 3.999 into the first equation I get -9 but how would I find the limit from the right if I dont know what C is?

[dawn dagger]

since y(0) = -6 say constant C would be used for that in eq. 1 and constant K would be 2nd piece, so you would need to adjust the 2nd piece for continuity

[leaden matrix]

so would I be equating the 2 equations at x=4 ?

[dawn dagger]

basically

[leaden matrix]

thanks

.close

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[keen seal]

true right?

theyre both true right

[nova yoke]

yep

the second one says to show it though, it's not T/F

[remote tangle]

The first one wants justification too

[midnight plank]

@keen seal Has your question been resolved?

[keen seal]

oh okay ty

.close

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[left rover]

maybe a question that's a bit abstract, if I have X = X_1 + X_2 + X_3 + X_4 + X_5 (random variables, not necessarily i.i.d) and Y = X_1 + X_2, and I would like to add Gaussian noise(0, sigma) to X and Y (Y containing the sum of only a subset of X_i), would it make sense to just scale the respective noises?

That is, X = X_1 + X_2 + X_3 + X_4 + X_5 + Noise(0, sigma) while
Y = X_1 + X_2 + Noise(0, sigma * 2/5)

[midnight plank]

@left rover Has your question been resolved?

[midnight plank]

@left rover Has your question been resolved?

[uncut cloud]

Is there any more context to this?

(Note that, if this approach does make sense, you should scale the variance, not the standard deviation: sigma^2 and sigma^2 * 2/5)

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[spice scroll]

i dont understand what's a minor here

[spice scroll]

why minor Ai=1,2,3 , j=1,2,3 = 0?

[midnight plank]

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[onyx jolt]

how do i solve

[visual tiger]

suppose F is an antiderivative of f

what's the relation that links F and f?

[onyx jolt]

wdym by that

we differentiate F?

[visual tiger]

yes and what should you get if you differentiate F?

[onyx jolt]

xsin(x)

[visual tiger]

yes

the relation that links F and f is

F' = f

[onyx jolt]

but how do we find the constants

[visual tiger]

write this

and then you'll know

[onyx jolt]

alr

[midnight plank]

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[viral dagger]

find the largest natural number such that its divisible by 36, all digits are even, all digits are diffrent

[viral dagger]

is this 8640? thats the largest found

[visual tiger]

well it is a 5 digit number at most

[viral dagger]

yeah but i couldnt find one where the digits sum to 18

[visual tiger]

you also know the last digit can only be 0, 4 or 8

[visual tiger]

yep exactly

if you use all 5 digits, then the sum is 20

so can only be 4 digits

8642 is the largest 4 digit number you can make with "all digits even, all digits different"

since it's not divisible by 36, 8640 is the one

[viral dagger]

ty!

DAMN im fucked

my teacher game me an imo question

err

.close ty

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[viral dagger]

let S(n) be the sum of the digits of n, find all k that fulfills k+S(k)+S(S(k))=2025

[viral dagger]

2013 1995 1989
i got these from just guessing, are there any i missed?

[carmine void]

Did you check all n in 1959 to 2025?

[viral dagger]

not all, just ones that make sense

im pretty sure there isnt any less than 1989

for 199x 1995 was the only one, 201x it was 2013, i couldnt find one for 202x

[carmine void]

Ik 1959 was a pretty rough lower bound. 1985 is better.

[carmine void]

That's smart

[viral dagger]

do you think i missed any?

.close

[midnight plank]

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[carmine void]

.No

[next viper]

Is it generaly advised not to calculate the (x-1)^3 with the rule a^3 + 3a^2b+3a....

[next viper]

.close

[midnight plank]

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[next viper]

.reopen

[midnight plank]

✅

[next viper]

sorry i thought i understood it

but now i see like yo

i thought this cant be cancelled

cant comprehend this process

how cancel it just like that

is this all false?

[sudden yacht]

Cancel what? The (x-1)² you mean?

[next viper]

yesy

[sudden yacht]

That's completely legal, since it's common both in numerator and denominators

[next viper]

yes but im under arrest

this is how i did it

it not good?

starting from

so this was first derivative

and my solution is up

[sudden yacht]

I think you applied wrong the cube formula

[next viper]

😮

this 1 not good?

[sudden yacht]

(A - B)³ = A³ - 3A²B + 3AB² - B³

[sudden yacht]

That is correct, yes

But it seems you messed up some signs

[next viper]

hmmmm

but

generally

i should not have gone that way

calculating the cube right

cos could just have easily cancelled it

with the denominator?

[sudden yacht]

Exactly, with the quotient rule you should always try to see if there are possible cancellations, so that calculations become easier

But of course if you don't simplify you still get the correct answer, but in a "heavier form" with higher exponents

[next viper]

hmmmm

.close

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[hollow pier]

I am not convinced with any choice

[mystic patio]

i believe it's c

since the graph of the derivative is < 0 for x<0 and >0 for x>0 therefore the graph should have the trend

downward for x < 0 and upward for x > 0

[hollow pier]

the answer is indeed "c" but i dont get it , would u explain further

[mystic patio]

what do you know about a derivative?

[hollow pier]

don't worry about it

i see that the shown graph is 1/x
thus dy/dx = 1/x
then y = ln(x)

[mystic patio]

i'm not the best at explaining math things so just tell me if i confused u

first of all you can't decide f(x)=ln(x)

there are many other complicated functions that also have this derivative

now why is it c

a derivative shows the slope of the tangent of a function at x0

so if the derivative is positive then the original graph also have an upward trend

dy/dx

in a graph x also goes forward

so dx is always positive

now we have dy

[hollow pier]

YES , NOW THAT MAKES SENSE

thx man i got it

[mystic patio]

that's great

[hollow pier]

.close

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[midnight plank]

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[clear saffron]

!help

[midnight plank]

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

[clear saffron]

Why was this wrong?

[fallen aurora]

,w 4000*pi/6

[grand pond]

Results provided by WolframAlpha

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

Requested by seventysevensquare

[fallen aurora]

erm

[fallen aurora]

how did you got this result?

[clear saffron]

Is it not (4000tan(30))

[fallen aurora]

no

it is 4000 times the theta(in radians)

NOTE: Assuming that earth is spherical

[clear saffron]

Oh because we’re speaking in terms of (theta equals s/r)?

[fallen aurora]

yeah

[clear saffron]

Oh ok. I thought it was talking about sin cos and tan

I guess it never defined it as a right triangle

[fallen aurora]

distance between cities is generally surface distance

not the straight distant through the earth

(if my words are confusing then ignore it)

[clear saffron]

lol ok thanks

.close

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[stone sky]

let $x\pitchfork X$ be a relation that says if point $x$ lies on line $X$. prove that their exists a isomorphism between $S^3$ in which we define a point as a pair of antipodes, and a line as a circle through them, and define a point as a line, passing through 0, in $\mathbb{R}^3$ and a line as a plane in $\mathbb{R}^3$.

[grand pond]

bigpufik

[stone sky]

I mean clearly both spaces are proejctive, but I do not know what to do next

[midnight plank]

@stone sky Has your question been resolved?

[stone sky]

<@&286206848099549185>

[midnight plank]

@stone sky Has your question been resolved?

[midnight plank]

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[inland summit]

Help pls

<@&286206848099549185>

<@&286206848099549185>

[grave summit]

Oof this looks nasty

First instinct wld be to write it in summation notation and see if smth pops out

Doesn't seem like it 😭😭

[inland summit]

It didn't work bro

I tried

[grave summit]

Maybe there's some sort of combinatorial argument?

Spit balling rn

Wait wait wait

Maybe u can use pascals identity here

[inland summit]

I thought we can take like (1+x)^30 and (1-x) ^30

Write the binomial expansion

And multiply them

But i don't know what to do next

[grave summit]

A generating functions approach?

Intriguing

[inland summit]

It is an IIT JEE PYQ

[grave summit]

Yeesh

[inland summit]

Bro can you try

[grave summit]

I'm trying 😭😭

[inland summit]

And dm me if you get the soln

[grave summit]

🤭🤭

[inland summit]

I have to go sleep

[grave summit]

Aight gn gn

[inland summit]

Thank you

Good night/morning/afternoon/evening

[grave summit]

Evening for me

[inland summit]

.Close

.close

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[midnight plank]

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[lament fox]

How can i prove that for image taken of a sphere (from far away such that rays incdenct on camera sensor are essntially parallel) half of the sphere's surface area is visible?

[zinc spear]

Verbally, you can say that images are 2D pictures only, so whatever clicked in a camera will be a 2D plane with 2D shapes only, and a sphere looked from any angle is a circle in 2D space.
Experimentally, you can color half the surface of the circle and then.....
Idk if there are even more methods...

[latent elk]

Prove by projection?

Is this for calc 2?

[lament fox]

well sure but that doesn't prove that half the area is visible

if you are close to the sphere you will only see a portion of it

[lament fox]

how so?

[latent elk]

?

[lament fox]

not quite

[zinc spear]

What's calc 2...?

[latent elk]

Calculus 2?

[lament fox]

its for a math paper

[zinc spear]

Ok....

[latent elk]

In calc 2, u learn how to project a sphere

AKA describe by f(x,y,z)

onto a plane f(x,y)

[lament fox]

im cool with calc

[latent elk]

Then use that to prove this

[zinc spear]

Can't we use the fact that for finding a sphere's volume in 1 octant, we use integration in such a way forming a quarter of circle on the XY plane then?

[latent elk]

Since all point on the sphere (not inside) say with coord x1,y1,z1 when project onto the plane call xp,yp,zp always share the same xp,yp,zp with another point on the sphere with coord x2,y2,z2 except the ring

[lament fox]

yes...

but isn't that the case for any closed surface?

[latent elk]

Thus, half the sphere is visible

[latent elk]

Pretty sure for sphere only

[lament fox]

but for any closed surface a ray will intersect it at two points except for a curve where it is tangent

take a cone for example with its tips towards the camera

any ray will intersect two points except for the circle on the base

[latent elk]

Correct, but not "Half" of the surface area are visble

For the cone

[lament fox]

exactly

[latent elk]

The majority is visible then

[lament fox]

so thats why i dont think thats a valid proof

[latent elk]

The only hidden part is the circle

[latent elk]

Well

U then can prove the area?

[lament fox]

i guess i could that yeah

[latent elk]

U need 2 parts

[latent elk]

From the ring

U can prove the ring diameter = sphere diameter

Then thats pretty much it

[lament fox]

yeah that works i suppose

kinda drawn out 😭

but ig its fine

[latent elk]

To be more careful, prove that ring divide the shape into 2 equal parts

Thats easy

inversion, reflection, etc

Or, easier

Prove that the plane contain the ring, is the reflection plane

Thats way easier

[lament fox]

ahh true true

ok thanks

[latent elk]

QED

Done!

[lament fox]

.close

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[midnight plank]

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[narrow mural]

how

[frozen talon]

surely $\sum_{n=1}^{\infty}|b_n|^t$ converges for all $t\geq 2$

[grand pond]

Flip

[frozen talon]

if that's true, then maybe that helps?

[midnight plank]

@narrow mural Has your question been resolved?