help-36

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[gleaming galleon]

Hey

[gleaming galleon]

<@&286206848099549185>

<@&286206848099549185>

Why is 5-4=3

<@&268886789983436800>

[steep hatch]

Wow they made this easy

[steep hatch]

"You have the rest of your life to be an asshole, why don't you take today off "

.close

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[fiery bluff]

welcome back, what have you figured so far

[warm python]

Typing that now

Let $a_n =x_{n+1}/x_n$. We then have $\forall \varepsilon>0 \exists N \in \N : \abs{a_n-L}<\varepsilon$. As $L>1$, we chose $\varepsilon= L-1$. So $-L+L+1< x_{n+1}/x_m< 2L+1$. From this we can conclude that $(x_n)$ is increasing.

[soft zealot]

wai

[warm python]

The issue here is it doesn't show the seqeunce is unbounded

[scarlet sequoia]

So the epsilon you chose is not a good enough one

[fiery bluff]

remember when Ann said "1 is scary"

it's for exactly this reason

[scarlet sequoia]

So now, either try again with a better epsilon

Or... use the fact that you just proved (x_n) is increasing

A certain theorem about ||monotonicity|| would be helpful...

[warm python]

hmm, for that I have to show (x_n)2L+1 is unbounded

[scarlet sequoia]

What type of proof could we do then?

[warm python]

contradiction

[tired walrus]

@warm python did you think about what we discussed before you went off to eat

[warm python]

I did ponder in the back of my head( As I ran into my friends)

[tired walrus]

ie what if your a_n (my y_n) was constant L

what type of sequence would that make x_n

it's sth that even jee students know one way or another

[warm python]

constant

[tired walrus]

nowhere near

after all, a sequence in which the ratio of adjacent terms is like... 2, for example

[warm python]

GP

[tired walrus]

that's not constant now is it

yes!!!

the idea in fact is that x_n "behaves like" a GP in the long run.

what this "behaves like" thing is for you to formalize,

but the idea will be to put a lower bound on x_n which itself is a growing GP.

[warm python]

I could also say x_{n+1}<2L(x_n)
but (x_n)>1, so 2L(x_n) is increasing an unbounded?

I suppose that has to be formalised and is the idea you were talking about

[tired walrus]

not quite

being bounded above by a seq diverging to +infty is no great feat

[warm python]

Let $(x_n)$ be convergent,. Then $(x_{n+1}) \to l; x_n \to l$., And then by the quotient rule $x_{n+1}/x_m \to 1$

[soft zealot]

wai

[warm python]

x_{n+1}/x_{n} \to l; l>1

this technically works I suppose?

though it doesn't give me any insights

[tired walrus]

yes and it also doesn't let you prove x_n -> +infty

and it also doesn't let you prove l ≠ 0

[desert mantle]

why should l>1?

[tired walrus]

you mean big L there

[warm python]

Oops

It's just non zero

And postive

[desert mantle]

maybe too big of a hint: ||suppose that L=2. then eventually x_(n+1)/x_n>1.5. what does that mean||

[desert mantle]

why

[warm python]

the seqeunce is of positive numbers and is increasing

[desert mantle]

(eventually increasing)

[warm python]

Well, it does doesn't it

The seqeunce is montonically increasing and doesn;t converge

Which means that it's NOT bound abobe

*above

[craggy island]

Wasn't the question asking you to prove that the sequence is not convergent?

By means of proving that it's unbounded above

[desert mantle]

the proof really manages to completely skip any possible insight and intution that you are supposed to gain from this exercise

[warm python]

Yea, I'm trying to formalise the GP argument now

[warm python]

Could I have some help with this

I do suppose I can ponder over this for a day

my RA exam is only next sat( so I have time)

[tired walrus]

pick your favorite value q strictly between 1 and L (for example (1+L)/2)

[warm python]

Should I do this

like let it simmer

[tired walrus]

show that x_n > c*q^n eventually

ponder this

[warm python]

okay, will do

tq

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[terse folio]

is there only 1 fourier series for this function?

[terse folio]

sorry i am new to fourier series

[tired walrus]

the fourier series for a function is unique if it exists yes

[terse folio]

the way this book does it is only consider f(x) from 0<x<pi but what about the pi<x<2pi part

[tired walrus]

they take the fundamental period as [-pi, pi]

[terse folio]

i see

still wouldn't we need to define the wave for (-pi,0)?

[pliant shore]

you don't need to, cause the individual cosine waves in the Fourier series are all even functions

hence the Fourier series will also be an even function

so it's enough that the original triangle wave is an even function, such that we only have to consider x > 0 to know everything about when x < 0

[tired walrus]

it's a reflection of what it is on (0,pi)

by evenness

[terse folio]

oooo yeah

lmao we meet again

[pliant shore]

lmao

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[tranquil lake]

I messed up with the top right corner (2x2), but not sure where

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[granite stump]

I completed every task except d. For a i got $9!$, for b i got $\frac{8!}{2!}$, for c i got $9!-\frac{8!}{2!}$. I have no clue what to do for d. I get the $7!$ for all the women and multiply that by 8 for the possible locations of the first man. What about the second man?

[soft zealot]

Gopher

[ebon agate]

In the situation of "atleast cases"

We usually take the opposite cases and subtract that from the total possible cases

[granite stump]

So i should find 9! subtracted by the cases where the men are adjacent and 1 apart?

[ebon agate]

Yes

You already found the adjacent case ig

Just need one apart case

[granite stump]

Yeah, thats the answer for c

[ebon agate]

b*

[granite stump]

Mb typo

Anyway for one apart, my brain just wants to go complete brute force but that wouldnt really be viable for exams

[ebon agate]

Yes

Try to see the (M1)(W)(M2) system as a whole unit

[granite stump]

Ooo

[ebon agate]

Find ways this can be changed internally

And then treat this unit as one person

And see how you can arrange this person with the other women

[granite stump]

So $7!\cdot2!$?

[soft zealot]

Gopher

[ebon agate]

Wait

[granite stump]

Okay okay wait i have an idea

[ebon agate]

Sure

[zenith cedar]

@granite stump quick question have you done stars and bars

[ebon agate]

How would stars and bars work here

[granite stump]

Since the woman in the middle can be any woman, then we can assume that the men are almost like a box surrounding a woman. Therefore we can keep the 7! for the women and multiply it by something else

[zenith cedar]

Uh we can keep 2 women in the middle then find the non integral solutions multiply by 7! 2!

[granite stump]

I have not

[ebon agate]

Ah ok

[zenith cedar]

Alright no problem it can make the question much easier though

[granite stump]

I am doing IBDP AA HL year 1 if thats anything thats useful

[ebon agate]

What's that 🥀

[granite stump]

AND THEN I MULTIPLY BY ALL THE POSSIPLE POSITIONS OF THE BOX AND THEN MULTIPLY BY 2 FOR THE TWO MEN BEING INTERCHANGABLE

[ebon agate]

Yeah effectively

[granite stump]

School system and course 🥀

[ebon agate]

Year 1

10+1?

Or 9th grade?

[granite stump]

So 7 possible locations of the box and then multiply by 2?

[granite stump]

10th grade but 11th grade in other schools

[ebon agate]

Yeah ig so

[ebon agate]

Fair

My way was

[granite stump]

Answer would be $7!\cdot7\cdot2$

[soft zealot]

Gopher

[ebon agate]

Choose a women between the two men

So we have 7 ways

Two men can interchange

So multiply by 2

Then we have 6 women and this unit we've created

Total ways to arrange this would be 7!

So yeah

[granite stump]

This is 70560

[ebon agate]

Same answer

[granite stump]

Ooo thats actually a lot simpler than my method

[ebon agate]

7!(72-14-1)

7!(57)

(5040)(57)

287280

[granite stump]

So the answer for d is 272160

[granite stump]

Uh oh diverging answers

[zenith cedar]

I am getting a different answer

[ebon agate]

What are you getting?

[granite stump]

Wait let me plug the whole equation into my calculator

[ebon agate]

Sure

[zenith cedar]

42*7!

211680

[ebon agate]

42?

[zenith cedar]

This

[ebon agate]

I'm getting 57

Oh wait

8!

Still 45

(5040)(42)

[granite stump]

Yeah i still got this when i did $9!-\frac{8!}{2!}-(7!\cdot7\cdot2)$

[soft zealot]

Gopher

[ebon agate]

Divided?

It should be multiplied right?

[granite stump]

OH WAIT YEAH

Sorry my brain was focused on choose rather than permutations

[ebon agate]

Fairsa

I'm also rusty rn 🥀

[ebon agate]

Ye ts is correct

[granite stump]

Yeah i got this

Hell yeah

Thank you guys so much

.close

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[spare scaffold]

How can I prove x^2<sinxtanx

[spare scaffold]

Derivative first and second fails??

[plucky rover]

On what domain are we working here

Cuz it definitely isn't true in general

[final saddle]

@spare scaffold Has your question been resolved?

[spare scaffold]

(-π/6,π/6)

@plucky rover

[bold zenith]

You could try finding lower bounds for sin and tan by making expansions up to x^3 term

You would need to show the higher terms don’t matter though

[final saddle]

@spare scaffold Has your question been resolved?

[tired walrus]

??

what.

uhhh do i even know you.

.............

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[tired walrus]

what was THAT all about.

[final saddle]

✅ Original question: #help-36 message

[tired walrus]

???????

you ping me, ask me to solve a problem i don't recall seeing before on the spot, and then cry "OH YOU CANT SOLVE IT" or some shit.

yeah idk <@&268886789983436800> deal w this guy please

[runic needle]

expansion of what

[tired walrus]

youre already getting into harassment territory tbh

[runic needle]

expand how

[valid carbon]

this is so funny

[runic needle]

taylor series?

of sinxtanx?

[tired walrus]

oh b& looks like

[valid carbon]

we all saw that

[opaque ember]

b&

[valid carbon]

yay

deserved

[plucky rover]

.close b&

[final saddle]

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[valid carbon]

why are we not just saying banned lol

[plucky rover]

Faster to type

[valid carbon]

eh

skissue

[plucky rover]

[tired walrus]

it's quite an old abbreviation

[valid carbon]

i seee

[latent dragon]

I love it when moderators delete the messages that break the rules and the people who talked about it look schizophrenic after.

[final saddle]

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[ocean sparrow]

Got a whopper of a problem boys. Been at this for 3 hours. Chegg, Symbolab, have not been able to get me to the finish line. The original problem is the integral of (5x^3+10x^2+17x-4)/(x^2+2x+2)^2

I have that broken down into 5 times the integral of x/(x^2+2x+2) added to the integral of (7x-4)/(x^2+2x+2)^2

The answer I have for the first part is 1/2ln|x^2+2x+2|-tan^-1(x+1)

the answer to the second part of the problem.. I'm having trouble with

[ocean sparrow]

if i take u=x^2+2x+2, du=2x+2dx, du/(2x+2)=dx, integral is (7x-4)/u^2 multiplied by 1/(2x+2) multiplied by du.. this is getting very messy and I'm not sure im working in the right direction

so my problem is setup is similar to the top equation.. but i'm really not quite sure.. how to make it to the bottom

this is my current answer after the work

original problem

forgot to carry a 5 in the beginning.. so should be 5/2.. other than that.. really not sure. I'm out of my depth

[final saddle]

@ocean sparrow Has your question been resolved?

[ocean sparrow]

no

[loud sundial]

ngl I'm too lazy to piece all of this together, but note that
$$\frac{7x-4}{(x^2+2x+2)^2}=\frac{7x+7}{(x^2+2x+2)^2}-\frac{11}{(x^2+2x+2)^2}$$
I assume you know how to integrate the first term. For the second term, note that
$$\int \frac{1}{(x^2+2x+2)^2} \dd{x}=\int \frac{1}{((x+1)^2+1)^2} \dd{x}$$
Setting $u=x+1$, we get this is the same as $\int \frac{1}{(u^2+1)^2} \dd{u}$. From here, you can either do a trig sub or integration by parts.

[soft zealot]

Civil Service Pigeon

[loud sundial]

[https://math.stackexchange.com/questions/170207/some-method-to-solve-int-frac1-left1x2-right2-dx-and-some-doubts?noredirect=1]

[final saddle]

@ocean sparrow Has your question been resolved?

[ocean sparrow]

woah.. yeah no i have the first time part of the equation handled.. this is definitely what I was missing. i burned through my submissions for the problem but I really wanna figure this one out. where did 7x+7 and -11 come from? i saw 11 as a fraction in some of the answers i got from calculators

[loud sundial]

$\dv{x}(x^2+2x+2)=2x+2=2(x+1)$, so write the numerator as $$\text{stuff where substitution is easy}+\text{the rest}$$

[soft zealot]

Civil Service Pigeon

[loud sundial]

"substitution being easy" meaning where u = x^2 + 2x + 2 makes it trivial

[ocean sparrow]

i believe i'm following. def felt a little underprepared for this. let me give u my notes 1s. so i'm not a complete lazy sob

[loud sundial]

you don't need to prove yourself to me lol

I am not your examiner

[ocean sparrow]

no i know. but u might spot errors. if it's not asking too much

i've been at this a long time and am def getting sloppy

[loud sundial]

I'm confused

are you trying to do integral of product = product of integrals

because that is not true

[ocean sparrow]

i am

[loud sundial]

that being $\int f(x) g(x) \dd{x} \neq \int f(x) \dd{x} \int g(x) \dd{x}$

[soft zealot]

Civil Service Pigeon

[loud sundial]

easy counterexample: $f(x)=g(x)=1$

[soft zealot]

Civil Service Pigeon

[ocean sparrow]

ah right. not product. not allowed at all

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[vernal bloom]

can someoen explaont o me

how the bounds for dy change

0 to x^3

i understand the 0 to 2

for dx

can someone illustrate 0 to x^3 how it became like that

[sick roost]

the boundary condition is given as $$0\le y\le 8\text{ and }\sqrt[3]{y}\le x\le 2$$

[soft zealot]

Cycadellic

[sick roost]

to change these bounds, we just need to chase the inequalities

so, how would you start?

[vernal bloom]

for the new bounds for x i set x = y^(1/3) and which led me to y = x^3 so therefore i now have (0 < x^3 < 8) then i cube root 1/3 to make the exponent go away leaving me with new bounds of 0 to 2 for x

idk how to do it for y tho

[sick roost]

this works, but i think its best to leave the <

but youll be doing the same thing either way

so, now you have bounds on x

[vernal bloom]

yes

[sick roost]

$$\sqrt[3]{y}\le x$$

[soft zealot]

Cycadellic

[sick roost]

and $0\le y$

[soft zealot]

Cycadellic

[sick roost]

how do i get the full y bound now?

[vernal bloom]

cube everything

y < x^3

[sick roost]

yup

and now youve converted the bounds

[vernal bloom]

can i ask where did this come from cause

[sick roost]

$$0\le y\le 8\implies 0\le y$$

[soft zealot]

Cycadellic

[vernal bloom]

$$\sqrt[3]{y}\le x \le 2 $$

[soft zealot]

Light

[sick roost]

it follows from here

[vernal bloom]

i see

idk thats hard to visualize fro me

but i understand

[sick roost]

thats why we stick with symbols

visualization is just so limited

[vernal bloom]

did you enjoy multivariable calc

when you took it

?

[sick roost]

oh yeah

[vernal bloom]

better than calc 2

?

,calc 2^4

[sick roost]

one of my favorites

[soft zealot]

Result:

16

[sick roost]

it generalizes beyond simply R and into R^3

[vernal bloom]

bro

[sick roost]

its still limited in many respects because general R^n dont behave so nicely

[vernal bloom]

youre a math major

fs

i dont know i guess its not too bad but its just that these assignments take all day

[sick roost]

for instance, you have to rotate about an entire plane in R^4, which breaks many surface elements

[vernal bloom]

ok in ur experience would u say

differential eq is easier

than multivar

wait between the two differential eq or linear algebra

[sick roost]

i had an easier time in calc than diff eq

[vernal bloom]

which one is easier

[sick roost]

imo
linear alg is the easiest
then vector calc
then diff eq is the hardest

diff eq is very rote until you learn laplace transforms

[vernal bloom]

ok well i guess im taking diff eq next semester

then linear alhebra

i know some people take them at the same time

but i hguess those people are just a lot more well versed

i cant do that

probably

[sick roost]

linear algebra is really fun, cause its your first real pure math course

lin al only helps in diff eq with cramers rule and partial fractions, its not super relevant otherwise

now, it shows up everywhere in calc 3

[vernal bloom]

icic

sounds like you dont struggle much 😭

[sick roost]

oh i certainly do

[vernal bloom]

but less than others

i wish i had a mind that could grasp things quickly

[sick roost]

i just learned this stuff years ago, and have to apply it regularly in the math i learn now

[vernal bloom]

youre still taking math

after years

of it

what are you doing

?

wait

that sounds mena

[sick roost]

self study mostly

[vernal bloom]

self study for what js wondeirng

[sick roost]

right now im learning type theory

modal logic etc

[vernal bloom]

all i know is like

abstract algebra is really hard or something but ill never have to take it

highest math i take is diff eq

and im done

[sick roost]

whaaat

you should!

algebra is so cool

[vernal bloom]

i dont wanna die

im sorry

[sick roost]

usually the intro courses only go up to posests and quotients, its not actually too bad

actually its cool cause number theory is everywhere

that aside, do you need help with anything else?

[vernal bloom]

mental thearpy with math?

[vital crag]

#study-discussion

[final saddle]

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[fathom sonnet]

im getting the wrong answer

[loud sundial]

discord.gg/physics

[fathom sonnet]

tyty

[sick roost]

i can help with this

if you just show me what youve done

[final saddle]

@fathom sonnet Has your question been resolved?

[fathom sonnet]

sorry im going to try this problem again if i dont get it ill be back

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[bitter crow]

.close

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[terse crypt]

uh huh

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[dawn plover]

Hey i just cant get this one

i just need help simplifying, all yall need to know is that b = 5a/13

[keen adder]

multiply by denominator/denominator

[dawn plover]

what is denominator

[keen adder]

mb I mean numerator

the top term

[dawn plover]

and how does that help me

i dont get it

[rocky talon]

to simplify the surds on the denom

difference of squares

[tired walrus]

denominator = bottom of a fraction

[rocky talon]

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

[tired walrus]

numerator = top of a fraction

[tired walrus]

use ^ for exponents. x^2 - y^2.

[dawn plover]

yeah no i know difference of squares

but im so lost on this one

literally tried resolving for over an hour

either im missing something or i havent learned a rule

[rocky talon]

try doing what Swindle told you

[dawn plover]

i dont really get it :/

[tired walrus]

$\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$

[soft zealot]

Ann

[tired walrus]

this is your original expression

[dawn plover]

yes

[plush portal]

hii

[tired walrus]

your idea will be to rationalize the denominator

by setting up a difference of squares in there

[dawn plover]

im a little slow on this one

[tired walrus]

hello and welcome to the server. this channel is currently occupied with someone else's question. please go to #discussion or #chill to chat.

[tired walrus]

was about to show you

$\frac{(\sqrt{a+b}+\sqrt{a-b})(\sqrt{a+b}+\sqrt{a-b})}{(\sqrt{a+b}-\sqrt{a-b})(\sqrt{a+b}+\sqrt{a-b})}$

[soft zealot]

Ann

[dawn plover]

how do you type it so fast

[tired walrus]

this time round it was some ctrl+c and ctrl+v

[dawn plover]

wait wait wait where did you get all this

[tired walrus]

i introduced a pair of equal factors to the top and bottom, as shown in red here: $$\frac{(\sqrt{a+b}+\sqrt{a-b}){\color{red}(\sqrt{a+b}+\sqrt{a-b})}}{(\sqrt{a+b}-\sqrt{a-b}){\color{red}(\sqrt{a+b}+\sqrt{a-b})}}$$

[soft zealot]

Ann

[dawn plover]

okay but why

[tired walrus]

you're about to see

[dawn plover]

i swear this is first time i see this

[tired walrus]

you've never simplified any fractions like $\frac{1}{\sqrt{2}+1}$ or sth?

[soft zealot]

Ann

[dawn plover]

with this i know there cant be root in bottom

oh

well that makes sense..

wait you would multiply by sqrt(2) + 1 right?

not just sqrt(2

[tired walrus]

wrong

[dawn plover]

how

[tired walrus]

or rather, misguided and over-strict

"there can't be roots in bottom" comes from a good place but treating it as a complete ban is just not good at all

[dawn plover]

well root of 4 can be of course

[tired walrus]

you'd multiply by (sqrt(2) minus 1).

[dawn plover]

but like root of 2

[dawn plover]

right

[tired walrus]

the idea with rationalization of square root shit is nearly always to set up a difference of squares

[dawn plover]

ok then no i havent ever had this kind of equation

[tired walrus]

$$\frac{(\sqrt{a+b}+\sqrt{a-b}){\color{red}(\sqrt{a+b}+\sqrt{a-b})}}{(\sqrt{a+b}-\sqrt{a-b}){\color{red}(\sqrt{a+b}+\sqrt{a-b})}}$$ simplifies as $$\frac{(\sqrt{a+b}+\sqrt{a-b})^2}{(\sqrt{a+b}-\sqrt{a-b})(\sqrt{a+b}+\sqrt{a-b})}$$ and then $$\frac{(\sqrt{a+b}+\sqrt{a-b})^2}{(\sqrt{a+b})^2 - (\sqrt{a-b})^2}$$

[soft zealot]

Ann

[dawn plover]

ok wait lemme write this down actually

[tired walrus]

and then $\frac{(\sqrt{a+b}+\sqrt{a-b})^2}{a+b-(a-b)}$ (we will get to the numerator but i am purposefully not touching it yet)

[soft zealot]

Ann

[dawn plover]

ok

[tired walrus]

these steps so far, do they make sense to you

[dawn plover]

yeah

[tired walrus]

ok

and do you see how to proceed?

[dawn plover]

i would apply sum square

or however you call it

[tired walrus]

square of a sum, i guess.

[dawn plover]

sure

[tired walrus]

but these identities kinda don't have established names in english

[dawn plover]

so do i not?

[tired walrus]

you do apply (x+y)^2 = x^2 + 2xy + y^2 yes

[dawn plover]

yea

2xy with these roots

do i add them in same root

or leave seperate

[tired walrus]

you don't "add" them

[dawn plover]

make them into one root

[tired walrus]

that term will be $\sqrt{a+b} \cdot \sqrt{a-b}$. you can combine them into one root as $\sqrt{a^2 - b^2}$. it's helpful but not \textit{absolutely} necessary.

[soft zealot]

Ann

[dawn plover]

bruh i forgot one part ok

[tired walrus]

(and the two just stays outside)

[dawn plover]

ok cool

now what

or do i make b as 5a/13 now

no i dont

[tired walrus]

can you show how much you've simplified the thing since my steps

[dawn plover]

i have

[tired walrus]

maybe send it on paper unless you yourself already know how to LaTeX

[dawn plover]

(a + sqrt[a^2 - b^2])/b

[tired walrus]

oh ok thats good

you can go a step further

break the fraction into $\frac{a}{b} + \frac{\sqrt{a^2-b^2}}{b}$

[soft zealot]

Ann

[tired walrus]

and in fact you can even simplify that fraction with the root as $\sqrt{\paren{\frac{a}{b}}^2 - 1}$

[soft zealot]

Ann

[tired walrus]

and the value of a/b is known

[dawn plover]

what'

[tired walrus]

$\frac{\sqrt{a^2-b^2}}{b} = \sqrt{\frac{a^2-b^2}{b^2}}$

[soft zealot]

Ann

[dawn plover]

oh

[tired walrus]

(a and b are both positive here so no worry about absolute values)

[dawn plover]

how does that help me

im getting lost again

[tired walrus]

(a^2-b^2)/b^2 simplifies as a^2/b^2 - 1

my idea is that the ratio of a and b is known, so if we could reduce the entire expression to be in terms of only that, then that'd help us a lot.

[dawn plover]

i dont really get how you simplify it

i probably shouldnt use it if i havent learned it

[tired walrus]

(x+y)/z = x/z + y/z

it is literally just this

like uhhhh

[dawn plover]

yeah i know this

[tired walrus]

ok actually here

lemme write it all out on paper in excruciating detail

[dawn plover]

it would be 1 over b^2 then

not just 1

or

nvm

i got it

b^2/b^2

[tired walrus]

would you still like a step by step breakdown of what i did

[dawn plover]

no no i got this part now

ok

um

how do i write it as 2 roots now

[tired walrus]

now your expression is just $\frac{a}{b} + \sqrt{\paren{\frac ab}^2 - 1}$

[soft zealot]

Ann

[tired walrus]

you should find a/b first (it will be a number)

[dawn plover]

yea

wait how

i cant

unless i get 5a/13

yeah i get 13/5

[tired walrus]

yes that

[dawn plover]

ok tysm

i solved it

that was hard

.close

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[jagged flare]

let $n$ be a positive integer such that $6\mid n^2-1$, prove that you can fit $\frac{n^2-1}{6}$ triangles with side lengths $3,4,5$ into an $n\times n$ square

[soft zealot]

ihave<skissue>

[jagged flare]

ok lowk i have no idea how to do this

the only sort of progress i got is that n is either 1 or 5 mod 6

@robust mulch

[tired walrus]

smallest value of n is 5

[jagged flare]

1?

[tired walrus]

then n^2 - 1 will be 0

try figuring out how you can make a 5×5 square out of an appropriate number of 3-4-5 triangles and a 1×1 tile

[jagged flare]

you can fit 0 3,4,5 triangles into a 1x1 tile

[tired walrus]

i mean sure you can

but im asking you to think abt the next value of n bc thats a lot less trivial than the 1×1 case

[jagged flare]

ok fair

[tired walrus]

basically instead of "fit (n^2 - 1)/6 3-4-5 triangles into an n × n square" think "assemble an n × n square out of that many 3-4-5 triangles plus a 1×1 tile"

[jagged flare]

cant this 1x1 tile be split into diffrent funky parts? it doesent say that the triangles must be on the grid

actually lemme make sure i translated this properly

[tired walrus]

eh i guess

for n=5 i can figure out a way without cutting up the 1×1

[jagged flare]

waittt

is the idea like making a sort of "windmill" for each possible shape

lemme try this

im struggling to find a solution for n=11

[severe verge]

Place 4 3x8 blocks like this

you should be left with a 5x5 square in the middle

[jagged flare]

oooo

ooh wait this beconmes induction then right

[tired walrus]

ish

[severe verge]

not sure

[jagged flare]

hear me out

oh wait not yet but almost

ok i think i patched it

ok im bullshitting

split the cases for n=1 mod 6 and n=5 mod 6, consider the n=1 mod 6. let n=1+6k, if k is even then we can construct a diagram something like this with 4x3j squares to create a new diagram for n+12 with center n-8, so like for n=25 we have 4x21 rectangles and theres the 17x17 center

i was thinking smth like this but theres a problem with even k

ah wait

for odd k, we can do 3x4j blocks with center n=1+6(k-1), so like for example n=19 we have 3x16 with center 13, and thats the previous case

ok actually let me tidy this up

[severe verge]

hmm you can do 4 8x9 blocks around to get 17

[jagged flare]

im p sure this is it?

and n=5 mod 6 should basically be the same method

and this cycles always makes smaller n so since n is a positive integer you will reach the smallest cases, which would be n=1 and n=5, both of which being possible

ok i think this works

thanks guys!

.solved

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[flint ibex]

i dont get this

[flint ibex]

like

if i do sinx=0

ull get three solutions and left with cosx^2=p^2

to be simple

ill jst draw that graph instead of square rooting

as idk if u get a +- if u square root the cos and its jst p

so if p is greather than 1 cos wont touch

so itll have 3 solutions

so its sufficient as im pretty sure n=3 can occour if p<-1 so yh?

and for II we can have p=0 so cosx^2=p^2 occours only twice

so then yh?

so B

but answer is C

[loud sundial]

"sufficent" means that if n=3, we can definitively say that p > 1

which given what you said here

means that n=3 is not sufficient to say p>1

hence I is false

[loud sundial]

II is saying that $n=7$ can only happen if $-1<p<1$, not that $-1<p<1$ $\textit{guarantees}$ $n=7$.

[soft zealot]

Civil Service Pigeon

[flint ibex]

ohhhhh thats right

yes

[flint ibex]

i dont get this though

i keep getting muddled with these neccessary, or sufficient questions

[loud sundial]

rip

honestly I do not have good advice other than "read carefully"

[flint ibex]

bro i dont get it though

its like the same thing

like n=3 is neccessary if p>1

is true right?

[loud sundial]

this doesn't make grammatical sense

are you trying to say that $n=3$ is necessary for $p>1$

[soft zealot]

Civil Service Pigeon

[loud sundial]

aka that $p>1$ can only happen if $n=3$

[soft zealot]

Civil Service Pigeon

[flint ibex]

yes

thats correct right?

[loud sundial]

yes

[flint ibex]

bro this is so like

im gna smash whoever made these questions bro

so like if A is sufficient for B

it means that if A occours B has ti happen?

[flint ibex]

well

technically cant p<-1

[loud sundial]

"A is sufficient for B": A implies B (aka if we have A, then we definitely must have B)

Ex. A quadrilateral being a square is sufficient to say that the quadrilateral is a rectangle
"A is necessary for B": A is required for B to happen, but we can't definitively say B will happen
A shape having four sides is necessary for it to be a square.

[loud sundial]

yes that's why I is wrong

[flint ibex]

so this is a memorisation thing right?

like u cant logically figure this out

[loud sundial]

if it's not intuitive then yeah

[flint ibex]

oh man

how did u naturally get that

like

[loud sundial]

uh

childhood trauma

tldr my mom would play word games with what I said when I was little even when I was distressed

[flint ibex]

😭

at least it allowed u

to be able to do things

others stuck on

[loud sundial]

[final saddle]

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

[flint ibex]

thanks alot

let me note down

ur explaination first

[loud sundial]

[flint ibex]

what

does definitively mean

[loud sundial]

guaranteed

[flint ibex]

i dont uderstand the google definition

[loud sundial]

if we have A, we are guaranteed to have B

or

if we have A, then B must happen

[flint ibex]

ok

so like

if we have a 3 sided shape its neccessary for it to be a triangle

[loud sundial]

correct

[flint ibex]

but if we have a 5 sided shage its sufficient for it to be a regular pentagon

as we can have irregular

its not neccessary

[loud sundial]

"A is sufficient for B": A implies B (aka if we have A, then we must have B as well)

does having five sides guarantee that the pentagon is regular

[flint ibex]

no

so its not sufficient

[loud sundial]

correct

[flint ibex]

im jst

so bad at this

its rlly annoying

man

jst gtta practice

[loud sundial]

@flint ibex anything else?

[flint ibex]

sorry

forgot tot close

.close

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[analog gazelle]

limit of (a^n-b^n)/(a^n+b^n)

[terse crypt]

As what approaches what

[analog gazelle]

infinite

n

a and b are both bigger than 0 btw

[terse crypt]

Given that ${a,b > 0}$, find
[ \lim_{n \to \infty} \frac{a^n - b^n}{a^n + b^n}]

[soft zealot]

k

[analog gazelle]

oh sorry didnt know you could do that

[terse crypt]

Hmm

Let me think abt it

[analog gazelle]

[ \lim_{n \to \infty} \frac{a^n - b^n}{a^n + b^n}]

huh

,help

[soft zealot]

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!

[analog gazelle]

,list

[soft zealot]

My commands!

Use ,ls to obtain a briefer listing, and use ,help <cmd>to view detailed help for a particular command, or ,help to view general help.

If you still have questions, talk to our friendly support team here.

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Page 1/4

[analog gazelle]

,autotex:

[soft zealot]

Unrecognised compilation level :.
See ,texconfig autotex_level for the valid options.

Oct 5, 2025, 4:44 PM

You have enabled personal automatic LaTeX compilation, with LaTeX recognition level WEAK.
Please be aware that automatic compilation may be restricted by guild settings.
See ,help autotex for more information about automatic compilation.

[analog gazelle]

okay

HOWWW

[warm shadow]

hm

when $a>b$, $\lim_{n \to \infty} \frac{b^n}{a^n} = 0$

[soft zealot]

Nab

[warm shadow]

does this help?

[severe verge]

divide both the numerator and denominator by b^n

[final saddle]

@analog gazelle Has your question been resolved?

[final saddle]

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[ivory island]

Can someone help me start this

[ivory island]

I've been just trying different things

but I can't figure it out

.solved

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[next maple]

So I'm basically supposed to

Solve considering all cases.

Why is my answer of

$-2 < x < 1$

Or

$x > 3$ incorrect?

I tried showing my work as well if you see it on the graph

This is

The answer at 5d

Why is it

$x < -2$ or $ 1 < x < 3$

[soft zealot]

Sep

Sep

[wraith glen]

yo

[next maple]

Wassup bro

[wraith glen]

where's the question at

original problem

[next maple]

5d bro

Should've posted it sooner sorry

Was trying to just write it but I guess image is easier

[wraith glen]

o

[next maple]

Yeah idk how to solve it

[wraith glen]

yeah ur graph and factorization is all good

what should you be looking at on your graph though

[next maple]

The issue is why did they answer it differnet though

Like I don't get it

[wraith glen]

yeah ik what u mean

i see your mistake

[next maple]

What did I do wrong

[wraith glen]

tell me this first

read the question out loud

like i'll read question 5a outloud

[next maple]

Wait bro

We doing 5d though

just so you know

[wraith glen]

"x squared minus 8x + 15 is greater than or equal to 0"
I'm thinking where is this true?

so read the question outloud for 5d now

maybe you will catch you mistake then

[next maple]

WAIT

It's less than zero

That's why it's the oppisite

I was doing the wrong thing

I was checking if it was above zero

not below

The question is

if x^3 -2x^2 -5x + 6 is less than zero

Not greater than

Dude thank you so much

You just helped me out so much

I can't believe I made such a stupid mistake as well

Bro just to let you know you're a great teacher, you thinking of being a teacher someday?

[wraith glen]

lmaoo naw bruh

[next maple]

lol

Bro trust me

[wraith glen]

i do want to tutor ppl though

[next maple]

You got it in you

Then maybe be a private tutor or some

or make some private school or something bro

[wraith glen]

im tryna master linear algebra so i can tutor at my univeristy

[next maple]

Ayyy

You in unI?

Good job man

[wraith glen]

yeah bro thanks

[next maple]

Of course, thank you

Now I can continue with my quesitons lol

[wraith glen]

i used to always slip up reading the inequality signs too

even at my level i still mess up

[next maple]

Guess it's common then

For many people

[wraith glen]

infact someone just now read it wrong and he was doing univeristy level math

[next maple]

Damn

[wraith glen]

he was in the help channel but we told him we read it wrong lol

[next maple]

Atleast I'm not the only one

lol

[wraith glen]

the more you do it, it will be easier

[next maple]

True that

Thanks once again bro

[wraith glen]

alright GL

[next maple]

Thanks, good luck on your stuff as well G

🙏

.close

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[last marlin]

i've read this a few times but i'm still confused, why can't a and b both be even?

i understand that if they are both even it would imply that a/b is not in lowest terms, but why does that matter?

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@last marlin Has your question been resolved?

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[tranquil pine]

can i get help with this one

[tranquil pine]

@sick roost yoo you free?

[sick roost]

no, im doing a simulation that doesnt want to work

due in 4 hours

gl tho

[tranquil pine]

sounds fun

good luck with that

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

<@&286206848099549185>

[livid solar]

what

[uncut aspen]

Maybe post in advanced math channels?

[livid solar]

what do u need help with

a-c?

@tranquil pine

[tranquil pine]

how do i draw the markov chain? the arrows are probabilities right?

im trying to build my understanding from this other example

the arrows here are probabilities ?

[livid solar]

yea so draw two nodes “A” and “Not A” with arrows labeled A->A 0.7, A->Not A 0.3, Not A->A 0.1, and Not A->Not A 0.9. the arrow labels are the transition probabilities

outgoing from each state sum to 1

[tranquil pine]

but those numbers are populations right

not probabilities

[livid solar]

no

[tranquil pine]

oh

[livid solar]

the arrow labels are transition probabilities

the populations go in the state vector

[tranquil pine]

howcome?

they resemble the population

like 70% OF students

thats an amount of people right

[livid solar]

ok

so

because the 70% on the arrow is conditional

it means 70% of the students who are in A now will be in A next time

whereas 70% of students after the first assignment is the current population mix

[tranquil pine]

wdym by conditional

[livid solar]

it means given the current state

0.7 means P(next=A | now=A) among students in A now, 70% stay A next time (not 70% of the whole class)

for example 1400 A, 600 notA -> next A = 0.7*1400 + 0.1*600 = 1040

[tranquil pine]

i just dont understand how thats a probability

thats a population number

like 70% of students will get this grade

but thats just a number of students

[livid solar]

70% is a per student chance

its not a class share

“for any student in A now, chance of A next time is 0.7”

[tranquil pine]

how

can you please explain more

i dont understand how thats 70%

chance

[livid solar]

hm

ok

[livid solar]

think of each student flipping a weighted coin based on their current group

if in A they have 0.7 to stay and 0.3 to move. if in notA they have 0.1 to move to A and 0.9 to stay

to get next counts split each group by those chances and add

next_A = 0.7*current_A + 0.1*current_notA and next_notA = 0.3*current_A + 0.9*current_notA

[tranquil pine]

its not clicking bro

[livid solar]

its a chance because the model defines a per student rule

given a student is in A now the next step is random with 70% chance to stay in A

those numbers are fixed transition chances of the process and only turn into counts after u apply them to how many students u have

[tranquil pine]

but why though

@livid solar i guess this might be a dumb question but

what does "x percent chance" for something to happen really mean

[livid solar]

Think of x percent chance as if u repeated the same situation many times under the same conditions the event would happen about x times out of 100

its a per case likelihood

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

so percentages are like :

out of a 100 , this thing happens a certain amount of time?

[livid solar]

yea

[tranquil pine]

so if for example

20% of trees get cut down = 20 out of 100 trees get cut down?

[livid solar]

yes

[tranquil pine]

but that also means 2 out of 10 trees get cut down too right?

why are they the same

[livid solar]

perrcent means “per 100” so 20% = 20/100 fractions that multiply or divide top and bottom by the same number are the same ratio, so 20/100 = 2/10 = 1/5 = 0.2

thats why 20 out of 100 and 2 out of 10 are the same share

[tranquil pine]

ohhh

yeah

thank you bro

[livid solar]

np

[final saddle]

@tranquil pine Has your question been resolved?

[livid solar]

wtf

[tranquil pine]

what happened

im just keeping the channel for consistancy incase something else pops up

i dont wanna scroll through a bunch of different channels

[livid solar]

oh

ok

[tranquil pine]

i need help again

how do i calculate the initial populations?

<@&286206848099549185>

[tranquil pine]

.

[final saddle]

@tranquil pine Has your question been resolved?

[gritty drift]

can you draw the state diagram?

[final saddle]

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[rain sentinel]

can someone tell me this topic and what the bit in red is

[rain sentinel]

is it a identity

[lone sky]

Heyyy

So

The topic here is about rewriting a wave function expressed as a sum of sine and cosine

@rain sentinel

(9 \cos(t) + 12 \sin(t))

[soft zealot]

YZYBlueBoy22

[odd seal]

yes, its sin(A+B), a pretty basic trig identity

[lone sky]

The red part shows the use of the angle addition formula for sine

[rain sentinel]

is there a video for this?

[lone sky]

I can help you out

[rain sentinel]

thanks

[lone sky]

@rain sentinel

[rain sentinel]

ill look that up

[lone sky]

You compare the coefficients of sint and cost from both sides

[scenic walrus]

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

[soft zealot]

Allen

[lone sky]

[
9 \cos(t) + 12 \sin(t) = R \sin t \cos \phi + R \cos t \sin \phi
]

[soft zealot]

YZYBlueBoy22

[lone sky]

Coefficient of (\sin t): (12 = R \cos \phi)

[soft zealot]

YZYBlueBoy22

[lone sky]

Coefficient of (\cos t): (9 = R \sin \phi)

[soft zealot]

YZYBlueBoy22

[rain sentinel]

.

so it would be this?

[lone sky]

You're close

But there is a small mistake in the second equation on the right-hand side.

[rain sentinel]

whcih one?

[lone sky]

[
R \sin(t + \phi) = R[\sin t \cos \phi + \cos t \sin \phi]
]

[soft zealot]

YZYBlueBoy22

[lone sky]

So your expression should be

[lone sky]

[
9 \cos t + 12 \sin t = R[\sin t \cos \phi + \cos t \sin \phi]
]

[soft zealot]

YZYBlueBoy22

[rain sentinel]

yeah

[lone sky]

Yes

Correct!

[final saddle]

@rain sentinel Has your question been resolved?

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[final saddle]

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[exotic rose]

Hello

[exotic rose]

Please help me understand the question

[bronze grove]

what do u don't understand

[exotic rose]

Firstly, what do surface equations look like?

[bronze grove]

well its a surface u get in x y z coordinates after mapping the values of x and y

if u want to visualise then you can use graphing site

[exotic rose]

Please help me understand...

The only surfaces that have x and y coordinates are x-y plane, and the planes parallel to it

The equation of xy plane is z=0. Other such planes' equations look like z=c.

But here, the equation is 2x+5y-3=0

How are the two ideas reconciled?

[bronze grove]

so u have the surface right?

you just need to integrate it

[pliant shore]

the only thing that matters is knowing what the path looks like: that's the standard equation of a circle

try subbing in $x = \sqrt2 \cos t - 1, y = \sqrt2 \sin t + 1$ for yourself

[soft zealot]

south

[pliant shore]

check that the equation of the circle is always satisfied for any t

[exotic rose]

These satisfy the circle's equation

It is true

When I integrate a surface over a linear path, what am I doing exactly?

[pliant shore]

you're finding the shaded area

for each point (x, y) on the path, there is a height S(x, y)

so you can represent the one-dimensional path on the x-axis using the variable t

and then the height of the function for some t goes on the y-axis

[exotic rose]

I am finding the area of a slice?

[pliant shore]

the area of the grey shaded region

do you see it?

[exotic rose]

It looks like a slice

A very thin vertical slice

[pliant shore]

yes, so that's the same as the blue shaded region

take a vertical slice, then translate it so that it goes onto the x-axis

since you're only moving all the pieces and not stretching them or compressing them

you do not change the area when you go from the blue to the grey areas

[exotic rose]

Are you sure? I feel like you are losing area in the curves

[pliant shore]

you take the projection of the curve onto the xz plane

[exotic rose]

A east to west curve would look like a zero area

[pliant shore]

indeed!