help-0

[mossy locust]

(ln | ln (infinity) | ) - (ln|ln(2)|)

[buoyant saddle]

yep

which is

[mossy locust]

so then basically it doesnt matter anymore because the natural log of infinity is infinity and the natural log of that also happens to be infinity and it doesnt really matter what we decide to subtract from that at that point because hey its infinity it doesn't really care what we subtract from it as long as its not another infinity

[buoyant saddle]

yep

[mossy locust]

https://www.youtube.com/watch?v=uFlOCzv7PQY

that is also what the fine lady at Ms. Shaws Math Class said but my classmates seem to disagree with me

[buoyant saddle]

your classmates are fucking cornballs

it diverges sir

[mossy locust]

do you want to hear what this guy told me

right quick because he had ME fucked up

[buoyant saddle]

,w \sum_{n = 2}^{\infty} \frac{1}{n \ln n}

[ocean seal]

Results provided by WolframAlpha

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

Requested by kniefy

[mossy locust]

what is this

does it do math

[buoyant saddle]

it does

[mossy locust]

I've never been in this server before I was really twisted and had to come make sure that I wasn't crazy

[patent vale]

its crazy some of yall are just unaware of wolfram

[mossy locust]

but they were trying to tell me that you could say that it converged by comparison with

1/n^1

because 1/n * ln (n) is smaller than the function but hey guys guess what 1/n^1 doesn't converge so it doesn't really matter if its smaller thats true and great for it but it still doesn't make 1/n^1 converge

[buoyant saddle]

[mossy locust]

I appreciate the help there

It seemed pretty straightforward to me

[buoyant saddle]

you're welcome

it was

[mossy locust]

Little integral test

But now the answer key and my classmates are wrong

[buoyant saddle]

typical

[mossy locust]

I love high school

we almost out

[buoyant saddle]

nah lol

[mossy locust]

are you a genius how does this server work

cause chat gpt not very smart so I had to come over here

[buoyant saddle]

i just volunteer as a hobby when im bored

[mossy locust]

that's super cool

[buoyant saddle]

my brain isnt working rn

cant do my own math

[mossy locust]

valid

do other peoples math

what math are you in

like what is the highest course you've taken

[buoyant saddle]

real analysis

[mossy locust]

I don't want to take up too much of your time you've prolly got other people to help

real analysis

[buoyant saddle]

we somehow went from completeness axiom to convergence of fourier series and baire category theorem in the same semester

[mossy locust]

how many rings in the ladder would that be above a calc 2 class (not what I'm in) im in Calc BC which would be kind of part of Calc 2 but I'm not convinced there doing a good job at actually teaching much of it

[buoyant saddle]

i took calc bc last year

[mossy locust]

I can't wait to get here haha

(please make it stop)

[buoyant saddle]

so like if youre planning on doing a math major and go to a pretty good institution then you might do something similar

[mossy locust]

I'm planning to study mechanical engineering

so

[buoyant saddle]

ok then you dont have to cry

[mossy locust]

🎊

I don't

this is good news

[onyx swallow]

measure theory and metric spaces stuffed together in ur real analysis course?

[buoyant saddle]

you do not want to get here brother

[buoyant saddle]

yes dawg

[mossy locust]

I don't think that I do

so ME students don't have to take that stuff?

[buoyant saddle]

like tell me how we went from induction proofs on the first homework to fucking fourier analysis

😢

metric spaces wasnt too bad

[mossy locust]

I appreciate the help

I might be back this very evening

Idk why but I'm still not convinced its so annoying that the answer key is wrong

this is the second time

can we do a quick taylor polynomial

[buoyant saddle]

it diverges my friend

[onyx swallow]

measure theory tho so many bounds......

[mossy locust]

but like how did my teacher even think that it didn't

[buoyant saddle]

the actual measure theory course is in the fall

i cant imagine what that shits going to be

same professor too

[mossy locust]

alright so this taylor polynomial is going to be 3rd degree of

f(x) = ln (3-x)

about x=2

[onyx swallow]

gl soldier

[buoyant saddle]

i will need it

im going to guess hes going to want to cover big rudin

😭

[mossy locust]

f(x)=0 bye bye

f'(x) = -(3-x)^-1

f''(x) = -(3-x)^-2

f^(3) (x) = -2(3-x)^-3

[buoyant saddle]

i hate this book enough already

[onyx swallow]

[buoyant saddle]

first year course btw

take me back to sequences 🙏

[mossy locust]

then plug in at each of those for x=2 so

f'(2) = -1
f''(2) = -1
f^(3) = -2

so write the taylor polynomial

-(x-2) - 1/2 * (x-2)^2 - 2/3! * (x-2)^3

the answer key says that the sign of the 2nd degree is posotive

but I believe that the chain rule while taking the derivatives keeps it negative

[buoyant saddle]

why are we doing this?

integral test is perfectly fine here

[mossy locust]

nooo its a whole different question

[buoyant saddle]

whats the question

[mossy locust]

am I supposed to end the thing and start another sub thing

[mossy locust]

This

[buoyant saddle]

nah you're chilling

[buoyant saddle]

ok do you not know ln(1 - x)?

[mossy locust]

come again

[buoyant saddle]

common taylor series

[mossy locust]

yeah I hate to say that I don't

[buoyant saddle]

ln(3 - x) = ln(1 - (x - 2))

[mossy locust]

yeah so then just plug in

I know that I should be able to recognize them that will help for the test as well

[buoyant saddle]

the taylor series for ln(1 - x) is quite common

[buoyant saddle]

which ones were you introduced to in class

i mean

[mossy locust]

sinx cosx e^x

lnx I believe

[buoyant saddle]

tbf like ap calc only cares about sinx, cosx, e^x, and geometric

[mossy locust]

I'd have to break out the notes

yeah

I'm not sure if we hit lnx

but If I got the taylor series for ln(1-x) then how do you account for the 3

[buoyant saddle]

^

they said about x = 2 for a reason

the taylor series is -x^n/n

so just put x --> x - 2

-(x - 2)^n/n

[mossy locust]

therefore all negatives

the answer key says that the third term of the taylor series is

• (x-2)^2/2

but it is negative correct

[buoyant saddle]

if my memory is correct it should be all negative

lets derive it i guess

[mossy locust]

yeah I thought so because even if you do it by hand the chain rule keeps the derivatives negative

[buoyant saddle]

$\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^n$

[ocean seal]

knief

[buoyant saddle]

so this holds for |x| < 1

we can integrate both sides and this will indeed hold for |x| < 1

$-\ln(1 - x) = \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} \implies \ln(1 - x) = \sum_{n = 1}^{\infty} -\frac{x^n}{n}$

[ocean seal]

knief

[buoyant saddle]

then letting x go to x - 2 we get

$\ln(3 - x) = \ln(1 - (x - 2)) = \sum_{n = 1}^{\infty} - \frac{(x - 2)^n}{n}$

[ocean seal]

knief

[mossy locust]

always negative

[buoyant saddle]

yep

[mossy locust]

yeah this teacher has me twisted

but ay life rolls on

[buoyant saddle]

your teacher is fucking buns bro

hate to break it to you

[mossy locust]

seems as though I know semi whats going on

bro reads from the text book and thats about all

[buoyant saddle]

did you follow the derivation?

its literally a one liner

as an exercise, try to deduce the taylor series for ln(1 + x) given the taylor series for ln(1 - x)

[buoyant saddle]

which book?

[mossy locust]

but then I have these people that think they're geniuses who are using the comparison test in a way that doesn't even make any sense because they are trying to compare a diverging series to a smaller series and saying that it converges

[mossy locust]

one moment

[buoyant saddle]

they are not geniuses sir

almost no one is tbf

[mossy locust]

on baby

but I tried to tell them hey I think that doesn't make any fucking sense and they said no

so

[buoyant saddle]

everyone thinks they're of above average intelligence lmao

[mossy locust]

that's how I got here I thought there's gotta be someone out there better at math than me who can say hey no these guys are a bunch of morons

[buoyant saddle]

indeed they are

[mossy locust]

and I got a migraine right now but the mid term is tomorrow so

really no chance to wait now is there

anyways

[buoyant saddle]

midterm?

its may

[mossy locust]

I guess

final term

[buoyant saddle]

you have ap exams in a week

[mossy locust]

end of term

[buoyant saddle]

ugh

[mossy locust]

that one

[buoyant saddle]

i lowkey miss high school now bro 😭

ap week was peak

[onyx swallow]

ap aint rigorous tho

[buoyant saddle]

i know

thats the point

😭

[mossy locust]

we trying to avoid the rigor

[mossy locust]

okay this

wait no

[buoyant saddle]

calc bc is fun because you dont have to think

[mossy locust]

okay this

at x=0 right

[buoyant saddle]

yea

[mossy locust]

would you not just flip the signs

[buoyant saddle]

you would

[mossy locust]

ohh no cause the x is negative

[buoyant saddle]

ln(1 + x) = ln(1 - (-x))

take x to -x

[mossy locust]

wait a minute

[buoyant saddle]

you get an extra (-1)^n

which you can combine with the other - to get (-1)^{n + 1}

[mossy locust]

what do you mean by take x to -x

[buoyant saddle]

replace x in our known formula for ln(1 - x) with -x

[mossy locust]

yes

giving ln (1 - (-x))

[buoyant saddle]

but also in the series representation

[mossy locust]

but can you not just say then ln (1+x)

yeah I might be fried I didn't pay attention too well this last quarter here

is there a video you'd recomend for subbing in things to taylor series

[buoyant saddle]

uhhh

[mossy locust]

or just watch a video

[buoyant saddle]

im not sure, khan academy might have some

[mossy locust]

bet

[buoyant saddle]

bprp maybe?

let me find some

[mossy locust]

perfect

[buoyant saddle]

yea id just watch khan academy

i think its valuable to understand the nitty gritty of what these things are really telling you, like what really is a power series but in terms of computational problem you just need to do more problems to see the patterns

[mossy locust]

yeah I got you

I defintely haven't done the reps this quarter

[buoyant saddle]

exercises are the most important part of math

i laugh at people when they say they read math books then mention they dont do the exercises

like fym you read it

it isnt a novel

[mossy locust]

needed to hear this

it's obvious but especially senior year its easy to be lasy and not do them

[buoyant saddle]

ya know what they say

[mossy locust]

damn kids

[buoyant saddle]

what you read you forget, what you do you remember

[mossy locust]

I don't think my teacher did any of this

[buoyant saddle]

fair but like calc bc is the most important ap exam youll take

since youre going for mech e

you save yourself two semesters

hella money

[mossy locust]

we will see how I do

I'm definitely going to take studying seriously I've got about 2 weeks before the test

[buoyant saddle]

dialed

[mossy locust]

yessir I've got 6 more questions on these MCQs to get through here tonight I'm going to get back to cracking

I appreciate the help

[buoyant saddle]

no worries

[mossy locust]

I'll spin back if I get lost again or the answer key is wrong again

likely

how do I close the chat for rn

[buoyant saddle]

type .close

[mossy locust]

TYSM

.close

[lone heart]

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[alpine sable]

guys what the hell is this formula
i know of area of surface revolution but that contains 2pi
and this one doesn't
so that means we got height multiplied by length
thats area for sure
but not revolved i suppose?
why not just int a to b f(x) dx

[unique dune]

Lien integral??

[alpine sable]

ohhh wth

can u show like a proof that leads to that

or link to smth

i know of arc length

and area forced my revolving around the

arc of a curve

when i saw this problem my first thought was

its just area under the f(x)

but hell no

its not but should be solved with line integral

and im not very sure about that

why is that

waitttt if we don't revolve the arc but rather just multiply f(x) with ds

that would be area without revolution

is that formally said line integral?

[unique dune]

Line integral is the area of the surface formed by the line to the xy plane

In the simplest

[alpine sable]

so thats just

[alpine sable]

wait thats actually changing its path

omg

i was thinking it should be like going to be in same direction

so it seems like how line integral is different from area under the graph is that

it is over the plane

[lone heart]

@alpine sable Has your question been resolved?

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[midnight gale]

so i alr know that triangle klh in iscoscles

and kln is 45 degrees

but idk how to go one from there

[lone heart]

@midnight gale Has your question been resolved?

[midnight gale]

no one answered

bruh

<@&286206848099549185>

DJKSL;AFJLK;SAFDASFASF

hel p ehlp

[floral fiber]

yo

do u agree LKN = MNK?

[midnight gale]

yayy help

[midnight gale]

yeah

[floral fiber]

So i can draw like this?

[midnight gale]

yeah

[midnight gale]

yeah i drew like that too

[floral fiber]

If I put A, do you agree KH = AN?

can u find KH?

[midnight gale]

yeah

[midnight gale]

5

[floral fiber]

okay

With KH, can you find HN?

I highlighted for you to see the triangle in focus :0

[echo belfry]

Lh is root80

[midnight gale]

8

[midnight gale]

cool

[floral fiber]

yee

so from there, what trigo rule cna u use?

to find ur LKN

:0

[midnight gale]

im not sure

oh wait

lkn

[echo belfry]

It should be tan inverse root80/5

[midnight gale]

thats 45 degrees

[floral fiber]

HN = ?

[midnight gale]

8

[echo belfry]

8

[floral fiber]

by pyth, LH = sqrt(LN^2 - HN^2)

[echo belfry]

Lh is 5

[floral fiber]

So if u put in the numbers, u can find LH = 5

[midnight gale]

yeah lh is 5

i found that

[echo belfry]

So 45

[floral fiber]

from there if both lenghts are 5

[echo belfry]

Deg

[floral fiber]

the angle is 45

yee

:]

[midnight gale]

OHHHHHH

oops

.close

[lone heart]

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[junior lagoon]

why is this question wrong? the practice test explained it but it still doesn't make sense to me

[vale wigeon]

do you get why kj = -45

[junior lagoon]

yes

[lone heart]

@junior lagoon Has your question been resolved?

[junior lagoon]

.close

[lone heart]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[fervent jungle]

A hydraulic system has a mechanical advantage of 2.5 and is 90% efficient. Calculate the diameter of the input cylinder if the diameter of the output is 30mm

[fervent jungle]

can anyone do engineering

kinda confused on this

[lean tangle]

Efficiency formula:
n=MA/MA_ideal

So MA_ideal=n/MA

Then you need to calculate the area of the input piston.

Finally use this formula:
MA_ideal = Ao/Ai

[lean tangle]

You'll get Ao, which is Area of output piston.

And use that to find the diameter.

[fervent jungle]

aight

[lean tangle]

Yep.

[fervent jungle]

so i got 18mm

.close

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[lean tangle]

Correct.

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[late fjord]

ik im supposed to use completing the square but ym ans is kinda weird

[cobalt oriole]

what does the constant mean?

like

always real?

[late fjord]

what

[cobalt oriole]

oh i translated it

brb

[cobalt oriole]

do you have the answer?

is it uh

[late fjord]

.close

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[late fjord]

no but i got the ans alr, appreciate the effort

[lone heart]

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[devout helm]

Is this correct so far?

[vale wigeon]

nope

only the first step of factorizing the bottom is correct

it looks like there's a list of steps on the paper just above the example you're doing

are you following that or not?

@devout helm

[devout helm]

yeah but they only show 2 values in the numerator but this qn only has 1 numerator value

[vale wigeon]

ok, no, hold on.

can you type out what step 1 says to do

just type it out exactly as it is written on your paper

do nothing else except for what i ask

[devout helm]

step1. factor the denominator of the rational expression into linear factors

[vale wigeon]

ok

now answer me this:

have you already done step 1?

your answer must be "YES, I have done it." or "NO, I have not yet done it."

[devout helm]

yes

[vale wigeon]

ok

now type out what step 2 says to do

[devout helm]

step2. write one partial fraction for each of these factors, with unknown constant numerators e.g A and B

[vale wigeon]

ok

have you already done step 2?

[devout helm]

I tried but you said its incorrect

[vale wigeon]

what should the numerators be?

[devout helm]

the first is 1

[vale wigeon]

no

[vale wigeon]

read the step that you just typed out

it says something about what you should put in the numerators

[devout helm]

the first is 1(x+3)

[vale wigeon]

no

read. the step.

do you see where the word "numerators" appears in it

do you see it? yes or no.

[devout helm]

no

[vale wigeon]

right here.

do you see it now? yes or no.

[devout helm]

yes

[vale wigeon]

ok

tell me now:

the step tells you what the numerators should be.

what should they be?

not "what are they", not "what did you write on the paper".

[devout helm]

I dont know, I really dont know Im sorry

[vale wigeon]

dude

i am literally just asking you to read one sentence

i don't know if you're overthinking yourself into oblivion or having a massive blindspot or brain fart or what

but i am not asking you to do anything complicated at all

this is just a reading comprehension thing

step2. write one partial fraction for each of these factors, with unknown constant numerators e.g A and B

[devout helm]

oh its A and B

[vale wigeon]

third time's the charm.

ok, now write down the fractions as the step says.

[devout helm]

but there's only one constant here

[vale wigeon]

write one partial fraction with the letter A for its numerator and the other with the letter B for its numerator.

i think maybe you should come back to this problem when you are less brain-fogged because you are struggling with following simple instructions at the moment

[devout helm]

NO

[devout helm]

I did that

[vale wigeon]

...

ok yeah yknow what good luck i guess
ive had enough of this

[devout helm]

nooo come back, I need to learn this

[zinc bolt]

Nope

The reason it isn't

is because

[crisp turtle]

ok lets try again

[zinc bolt]

$\frac{1}{3} = \frac{1}{3} + \frac{0}{6}$ but this doesn't mean that $\frac{1}{3} = \frac{1}{18}$

[ocean seal]

@zinc bolt

[devout helm]

oh so how do I go about doing this

[tulip bone]

😭

[tulip bone]

watch a youtube video first

on how to do partial fractions

start with that

[zinc bolt]

You have the procedure just above the question

Start by doing that

[devout helm]

yeah but I dont seem to understand it, the first helper tried to explain it but I dont seem to get it

[tulip bone]

alrighty

well

[zinc bolt]

$\frac{1}{(x + 2)(x + 3)} = \frac{A}{x + 2} +\frac{B}{x + 3}, A, B \neq 0$

[tulip bone]

Wait wait

[zinc bolt]

This is step 1

[tulip bone]

I dont think he knows how to do

fraction addition

😭

[ocean seal]

@zinc bolt

[zinc bolt]

He should if he's doing factorization, right?

[devout helm]

ok I got this step

[tulip bone]

Why did he do this?

[zinc bolt]

Ok now from here try solving for $A$ and $B$

[ocean seal]

@zinc bolt

[devout helm]

split the constant

[zinc bolt]

Note that if $A = 0$ or $B = 0$ your solution will be invalid

[ocean seal]

@zinc bolt

[devout helm]

so I make it into one fraction?

[zinc bolt]

Sure?

Try doing it and I'll tell you if you're wrong

[tulip bone]

poor Ann

aint being paid enough for ts 🥀

😭

Just to put it out there btw $\frac{A}{B} + \frac{C}{D} = \frac{AD+BC}{BD}$

[devout helm]

so its Ax+3A+Bx+2B/(x+2)(x+3)?

[ocean seal]

Light

[zinc bolt]

Yep

[tulip bone]

doing so well so far good

[zinc bolt]

Put this equal to the other side

i.e.

$$\frac{Ax + Bx + 3A + 2B}{(x + 2)(x + 3)} = \frac{1}{(x + 2)(x + 3)}$$

[ocean seal]

@zinc bolt

[tulip bone]

yes

hiten what math are u doing btw

[devout helm]

so I cross multiply?

[zinc bolt]

yep

[zinc bolt]

?

[tulip bone]

what are you studying

[zinc bolt]

I am at presnet BTech student

[tulip bone]

cross multiply or just cancel out the denominators

[tulip bone]

why are you doing such complicated math then-

[zinc bolt]

?

I didn't do anything complicated here so far

[tulip bone]

$\frac{A}{B} = \frac{C}{B}$ is the same as $A = C$

[ocean seal]

Light

[tulip bone]

ik

but like

[zinc bolt]

if $A, B, C \neq 0$

[ocean seal]

@zinc bolt

[tulip bone]

i remember u asked a

combinatorics question

yesterday

[zinc bolt]

Oh yea

I am doing an optional course on Combinatorics

[tulip bone]

Did u get ur solution for that-

[zinc bolt]

Yes

[tulip bone]

Gotchu

That math seemed TOUGH

What did u do before btech to even understand that

😭

[zinc bolt]

Ok, if you wanna connect please dm me, let's not hijack this help channel for catch up chats

[zinc bolt]

I sent you a friend req, we can talk in dms as much as you want

@devout helm you there?

[devout helm]

yeah

[zinc bolt]

Did you try solving that equation?

How far did you get?

[tulip bone]

sure

[devout helm]

actually tho, am I allowed to eliminate the denominators

[zinc bolt]

You can state assuming $x \neq -3, x \neq -2$ and then proceed

[ocean seal]

@zinc bolt

[zinc bolt]

This assumption is fine because without this assumption the denominators are invalid

[devout helm]

what does this mean?

[zinc bolt]

It means that x's value can't be -3 or -2 because then the fracion would have a 0 denominator which is invalid

[tulip bone]

i don't think he needs to include all that yet

he's merely doing partial fractions

😭

but you are right either way

Basically that for ex, if you put in -2 and -3, you'll get a zero in the denominator since (-2+2) x (-3+3) = 0 and u can't have 0 in denminator

[devout helm]

but am I still allowed to do eliminate the denominators tho, so that I can get a clean polynomial

[tulip bone]

I genuinely don't remember doing partial fractions

[tulip bone]

that's what i did here

So yes, u are

[devout helm]

so Ax+3A+Bx+2B = 1?

[zinc bolt]

yes

Keep going

[devout helm]

Ok so I’m at this point now

[zinc bolt]

Have they not taught you how to find and use 0s of an equation?

[devout helm]

I guess not

[zinc bolt]

Hm...

[zinc bolt]

You remember this

[devout helm]

yeah

[zinc bolt]

Yea put $x = -3$ first and then once you have one variable for the other put $x = -2$

[ocean seal]

@zinc bolt

[devout helm]

but are you continuing from the step that I've done or

[zinc bolt]

Yes

[devout helm]

so whats the purpose od doing this then

[zinc bolt]

RN RN there is none

but which grade are you in

[devout helm]

polytechnic

[zinc bolt]

?

[devout helm]

its like a college

and I'm freshmen

[zinc bolt]

Oh ok

You'll require this for your calc-course

[devout helm]

calculus?

[zinc bolt]

Are you aware of what calculus is?

Yes

[devout helm]

yeah

[zinc bolt]

Yea this is something we need to do for integration a lot

[devout helm]

I'm taking engineering as my course next year and I definitely need calculus

[zinc bolt]

Yea

that's prolly why you're being taught this

GL for your future

[devout helm]

thanks bro

wbu?

[zinc bolt]

Let's do that in dms

!done

[lone heart]

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

[devout helm]

.close

[lone heart]

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[raw jetty]

the adjoining figure shows two intersecting chords in a circle, with B on minor arc AD.
Suppose that the radius of the circle is 5, that BC = 6, and that AD is bisected by BC. Suppose
further that AD is the only chord starting at A which is bisected by BC. It follows that the
sine of the minor arc AB is a rational number. If this fraction is expressed as a fraction m/n
in lowest terms, what is the product mn?

[raw jetty]

idfk status 1

[lone heart]

@raw jetty Has your question been resolved?

[crisp turtle]

fascinating question

i wont go to sleep tonight till i solve this

[verbal blaze]

me neither XD

[crisp turtle]

ok this diagram really sucks doesnt it

the radius of the circle is 5 but the chord BC is 6?

it should be more like 9

[verbal blaze]

missing a "(not drawn to scale)" XD

really, the system only has 2 degrees of freedom

here's a better construction
https://www.desmos.com/geometry/5yo6rdnada

[lean tangle]

I think I've seen this question before.

AIME question.

If not mistaken.

[empty mason]

Yep, it’s from the first year AIME existed (I think)

[lean tangle]

I see.

Was one of the question I learnt when I was preparing for national math contest.

Sorry not Olympic.

I'm not good enough for that.

[empty mason]

Oh ok. I happened to recognise it because I used to grind AIME problems. And the problem is the last problem of the first year AIME happened

[crisp turtle]

clearly im not smart enough to solve this

[lean tangle]

If I'm not mistaken, there was a lot of ways to solve this.

I think I recall one of them that uses coordinate geometry.

[crisp turtle]

my answer is

its one of 5, 10, 15, and 20

[lean tangle]

Nope.

[crisp turtle]

[lean tangle]

Hint.

[crisp turtle]

sobbing

[lean tangle]

It's larger than 100.

XD

Lemme have a look.

[empty mason]

huh it’s ||none||

[lean tangle]

XDXD

[crisp turtle]

lmao

[lean tangle]

[raw jetty]

oh ok found the question

[crisp turtle]

my method of solution was

[lean tangle]

Nice.

[empty mason]

Sorry, I just meant that it’s none of the options you gave

[raw jetty]

aw cmon trig really

[crisp turtle]

its the sine of an arc in a circle with radius 5
thus it must be one of 1/5 2/5 3/5 4/5

[lean tangle]

More than that.

[crisp turtle]

[empty mason]

Look at the bottom for some no trig solutions in aops

[lean tangle]

There are 8 solutions in total.

[raw jetty]

oh i thought i reached the bottom

[empty mason]

I can tell from the options you gave

[crisp turtle]

lmao

[lean tangle]

Go for solution number 7.

That one has no trig.

[empty mason]

There are actually more than 8. It’s just that 8 solutions happened to be listed out. But more or less most of them use trig

[lean tangle]

Yeah.

Personally I like the coordinate geometry one.

It inspired me to use coordinate geometry in the rest of my geometry questions.

[raw jetty]

whats a locus

[empty mason]

I think coord geo is good, but you shouldn’t rely too much on it in olympiads. You should stick the synthetic way always first

[empty mason]

A locus is a set of points that satisfy geometric condition/s.

[lean tangle]

Locus is the pathway of a moving point.

[lean tangle]

Lots of types can be extended from this.

[empty mason]

Btw the first solution is easiest to follow in my opinion. Just need to know your sine angle subtraction formula in hand and your hood

And btw I’m bad at trig

[lean tangle]

XD

[lean tangle]

Broski

+1

[empty mason]

I unfortunately cannot rely on trig in olympiads, especially memorising formulas. But I use it when times are needed

[lean tangle]

Yeah true.

I faced a trig question in my most recent competitions.

[raw jetty]

why is sin(AOB)=x/5

[lean tangle]

Which solution are you on?

1?

[raw jetty]

7

oh wait nvm

ok i see, thank you!

.solved

[lone heart]

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[latent needle]

Can somebody Substantiate why the two squares are squares (Pythagoras)

[edgy hemlock]

Does he mean the inside and outside sqaures?

[tulip bone]

the outer square was made from the inside square

[latent needle]

Yes

[lean tangle]

Yeah

[edgy hemlock]

Both have angles of 90°

[lean tangle]

Well I guess we need to prove it.

[lean tangle]

Alpha + Beta = 90°

[edgy hemlock]

Yeah

[lean tangle]

Since the inside square is equals to 90.

[edgy hemlock]

We can prove that through the • angle

It’s 90°

[lean tangle]

The exterior angle • is also 90°

So alpha+beta+90=180

Which leads to alpha+beta=90°

[edgy hemlock]

And because they’re alternate angles then they’re equal

[lean tangle]

I assumed that all grey triangles are equal.

[lean tangle]

Or congruent.

[latent needle]

Yes that’s true

[latent needle]

Yes

[lean tangle]

This side will be alpha.

[edgy hemlock]

And then you can do that to prove all the other angles of the inside square

[lean tangle]

Since alpha+beta is 90°

We know that it's a right angle

Do it on all sides.

You get 90° at all vertices.
Or corner.

And that's how we know that it's a square.

[latent needle]

Okay and how can I get the area of the outer square ? (2 different ways how to calculate)

[lean tangle]

4(1/2ab)=2ab

[edgy hemlock]

Actually the angles not the areas

Oh i thought the 2kgod was talking mb

[lean tangle]

Bruh

[lean tangle]

I was talking.

After the OP talked.

XD

Or a²+b²

[lean tangle]

a²+b² is the another way.

[latent needle]

Okay thank you guys !

[lean tangle]

Welcome.

[edgy hemlock]

No like i thought you went to finding the areas even tho he didn’t ask for them but he actually did💀

[latent needle]

How can I close this chat ?

[lean tangle]

XD

[lean tangle]

.close

[lean tangle]

Use this.

[edgy hemlock]

Bruh

[lean tangle]

XD

The second time.

XD

[edgy hemlock]

Doesn’t he need to sqrt that

[lean tangle]

No.

[edgy hemlock]

Yeah lmao

[lean tangle]

c²=area of large square.

[edgy hemlock]

Actually yeah

[lean tangle]

So why hassle to square root and square it again?

[edgy hemlock]

To find the length of the side to square it

[lean tangle]

I understand your idea.

@latent needle

Still here?

[edgy hemlock]

I understand yours as well

[latent needle]

Yes

[lean tangle]

Close it.

[edgy hemlock]

If you dont have more questions use .close

[lean tangle]

XD

[latent needle]

.close

[lone heart]

Channel closed

Closed by @latent needle

Use .reopen if this was a mistake.

[lean tangle]

.reopen

Bruh

I wanna reopen it.

.reopen

[latent needle]

Should I?

[edgy hemlock]

Send a msg in a few

It will reopen

[edgy hemlock]

No

[lean tangle]

Okay.

[vale wigeon]

no, you can only reopen your own channel.

[lean tangle]

Hmm...

So.

How do I send question?

[edgy hemlock]

Try another channel if you dont wanna wait

[vale wigeon]

take a channel that does not have someone else's name on it and send any message into it

[lean tangle]

Done.

[lone heart]

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[verbal timber]

but the answer is not a unit vector

[vale wigeon]

yeah it's BS

this is a perfectly serviceable proof that no such vector exists

precisely because the only vector you found that could possibly satisfy everything here actually violates the requirement of "unit vector"

[winter light]

But can't the vector be scaled and still have the same angle wrt those two vectors?

[vale wigeon]

no

the angles are wrong anyway because we assumed in the angle calculation that |r| = 1

when it isnt

[vale wigeon]

these four constraints:

• be in the XY plane
• be a unit vector
• make angle 45° with i+j
• make angles 60° with 3i-4j
  are impossible to satisfy all at once

[verbal timber]

the three conditions dont meet

[vale wigeon]

exactly.

[winter light]

Sorry, I meant in general
Here oc it doesn't work

[tight pier]

the scaling doesnt change the vectors angle to another

[winter light]

Yeah exactly

So why can't we scale the result so that it becomes a unit vector?

[verbal timber]

yeah,same concern

[stone coyote]

idts there exists a vector which make 45 with i+j and 60 with 3i-4j

like try to draw it out

45 with i+j means either x or y axis
and 3i-4j is making -37 with x

[winter light]

Yeah indeed

The problem is not being a unit vector, just those incompatible angles

[stone coyote]

yeah

[tight pier]

thats not the point of it

[verbal timber]

,w calculate angle bw (13/sqrt(170), 1/sqrt(170), 0) and (1,1,0)

[tight pier]

the vector being not a unit vector shows that it's not possible because it's a contradiction to the assumption we made

[verbal timber]

thanks everyone

.close

[lone heart]

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[jolly jasper]

Just a quick question. If A = {1,2,3} and R = {(1,1), (2,2), (3,3), (1,2)}, then is the relation R on A transitive?

[winter light]

What do you think?

[jolly jasper]

I don't think it is.

so... am i correct?

[tight pier]

Can you reason why tho

[jolly jasper]

there's no (2,3)?

[winter light]

I would have said it is

[jolly jasper]

i kind of understood the concept but when i put the question into chatgpt and gemini it says the relation is transitive.

[jolly jasper]

How?

[winter light]

Because for every element (1, 2 and 3) this condition is satisfied:

aRb and bRc => aRc

[jolly jasper]

but i dont see bRc

[winter light]

If there is not, the hypothesis is false. Therefore, this implication is automatically true

At least that's how implications in logic (truth tables) work

[winter light]

I mean, I'm not so sure but that could explain this

[jolly jasper]

I don't get it. if bRc then shouldn't there be (2,3)?

[winter light]

If it's not there you can't make any conclusion

If (2, 3) was there (but not (1, 3)) then it would not be transitive

[jolly jasper]

but we can prove using (1,2), no?

ohh wait. i get it now.

[winter light]

How can you write this if you don't have bRc? That's what I mean

[jolly jasper]

yeah there should be a (1,3) but then how is R transitive?

[winter light]

Again, I would like some confirmation by someone a little bit more fond of this

[tight pier]

ok i see my mistake

[jolly jasper]

???

[tight pier]

ok let me ask you this why did you say (2,3)

[jolly jasper]

bRC.

we have aRb so i thought we only needed bRc but then i realised that we would also need aRc.

[tight pier]

aRb and bRc are both required for you to look for transitivity, if bRc cannot be found then it's pointless to investigate aRc

for example for (1,2) you cannot find another pair like (2,3) therefore you stop there

for (2,2) it works and it implies (1,2)

[jolly jasper]

that's what im saying but chatgpt and gemini say that R is transitive so is the AI wrong?

[tight pier]

no what i am saying is just because there isnt (2,3) doesnt mean it's not transititve

[jolly jasper]

huh?

then is it transitive?

[tight pier]

yes

[jolly jasper]

how?

[tight pier]

Consider two cases first (a,a) where a in A

aRa and aRa implies itself trivially

the other case is with the (1,2)

[jolly jasper]

that's aRb, correct?

[tight pier]

1R2 and 2R2 implies 1R2

[tight pier]

yes for example

and the only match bRc is for example (2,2)

on that set

How about 1R1 and 1R2 ?

[jolly jasper]

quick question. how is (2,2) bRc?

aren't we supposed to take each element in set A as a, b and c?

[tight pier]

b=2 and c=2

a,b and c can be the same

[tight pier]

like here strictly speaking

i consider the case a=b=c

aRa and aRa => aRa is the same as aRb and bRc => aRc if a=b=c

[jolly jasper]

Wait so we don't assign each element in set A as a, b and c? like a=1, b=2 and c=3. so (1,1) is aRa, (1,2) is aRb and if we assume there's (1,3) then that would be aRc, no?

[tight pier]

nooo

[jolly jasper]

???

[tight pier]

a,b and c are not to be chosen constant

they are arbitrary

[jolly jasper]

I see.

Then how are we assigning those respective values?

[tight pier]

what you are doing is you are finding matching pairs

you can have (1,1) and (1,2) then a=b=1 and c=2
you can also have (3,3) and (3,3) then a=b=c=3 now

[jolly jasper]

Ohhhhh.... I see now.

[tight pier]

a,b and c are just place holders

[jolly jasper]

Wait. Let me try to find how this is transitive on my own if i still don't get it I'll ask for help. brb in 5 minutes.

TY.

Hello?

[median oar]

well you haven't said anything...

[jolly jasper]

is this correct? if i take (1,1) as aRb, (1,2) as bRc and (2,2) as aRc? so its transitive?

[tight pier]

well it's an example but not proof

[median oar]

transitivity is a for all statement

you need to show it's true for every aRb and bRc that it means aRc

not just 1 pair

[jolly jasper]

ohhh i see.

so (1,1), (1,1), (1,1) and (1,1), (1,2), (1,2) as aRb, bRc and aRc respectively?

[median oar]

there are some other ones

namely 2 more really trivial looking ones

looks like the (1,1), (1,1), (1,1) one

[jolly jasper]

(3.3) and (2,2)?

[median oar]

yeah

that's all the ones you can pick of the 4 that have the form aRb and bRc

oh there's also (1, 2) and (2, 2)

so 3 more you were missing

[jolly jasper]

so another quick question. so if R has (x,x) does that mean it's automatically transitive?

[median oar]

what does (x,x) mean

[jolly jasper]

any element.

i thought of putting "a" but decided not to.

like (3,3), (2,2) , (1,1) and so on.

Sorry my english is not so good :(

[median oar]

all of them

if all of them are in there it probably is a yes

[jolly jasper]

Okay. That will be all. TYSM.

.close

[lone heart]

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[faint spindle]

Hello, how do I factorise further?

[pale kestrel]

i would not keep that x(y+1) inside there

What is the overall goal?

[faint spindle]

To factorise it to 3 brackets

[pale kestrel]

if ur asked to factorise that entire expression, then u probably need to multiply it out first

[crisp turtle]

ok u need to resolve the - xy to do that

so u need to multiply it all out, expand it all out first

then factorize

[faint spindle]

Can I have step by step help perchance

[crisp turtle]

first step
multiply it all out

[faint spindle]

Like how?

I'm sorry I'm really stuck at it😭

[lone heart]

@faint spindle Has your question been resolved?

[faint spindle]

<@&286206848099549185>

[pale kestrel]

expand brackets

have you been shown how?

[slim kite]

Can u send a photo of the identities u were taught, maybe it would help in solving the problem

[pale kestrel]

(a+b)(c+d) can you expand this?

[faint spindle]

Perhaps

Nah nvm I think I found out

How do I close the question

[slim kite]

Type .close, but...

I'm curious

[faint spindle]

?

[slim kite]

Can u share the solution

[faint spindle]

. rotate

Oh wait

. close

.close

[lone heart]

Channel closed

Closed by @faint spindle

Use .reopen if this was a mistake.

[pale kestrel]

,rotate270

[slim kite]

Thanks

[lone heart]

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[heavy prairie]

Solve the equation:

[heavy prairie]

sending pic of what i tried

i thought i would have to divide into cases but apparently i can eliminate one of the two answers for cosx

but i dont understand why or how i can eliminate one

is it something about false roots?

i never understood false roots that well

[civic cedar]

try equation cosx with y in step 3 and then try solving. i think one of your roots is incorrect

[heavy prairie]

what y

[civic cedar]

any variable

the roots i got were 1/2, -2

[heavy prairie]

oh yeah whoops

[civic cedar]

and then it is easier to eliminate one of these roots

[heavy prairie]

my negative root should be -2 not -1/2

[civic cedar]

yup

so -2 is eliminated as cos x = -2 is not valid

[heavy prairie]

am i supposed to use that -1 <= cosx <= 1?

[civic cedar]

yes

[heavy prairie]

ok

i will try to continue from there

[civic cedar]

cool

[heavy prairie]

okay i still need help

i know that for cosx to equal 1/2, x would have to be +-pi/3 + 2pik, where k is an integer

do i just have to test what ends up equaling 0 in the original problem?

[pale kestrel]

goal is to write general form of solution

which is what ur doing

[heavy prairie]

but would that be true even for sin^2?

[pale kestrel]

you dont have to check if you got to your thing by a series of iff implications

using pythagoras is an iff

quadratic formula is also an iff

[heavy prairie]

im not sure what iff means

[pale kestrel]

sin^2 x - 3/2 cos x = 0
1 - cos^2 x - 3/2 cos x = 0

x is a solution for the 1st if and only if it is a solution for the 2nd

youve substituted in an identity that is true for all x

similarly, when you apply the quadratic formula to get cos x = +-1/2 (ill assume ur math is right), its the same story

[heavy prairie]

^ ah yeah it was wrong but the other person caught it

[pale kestrel]

If cos x = 1/2 or cos x = -1/2, then x solves your other equation too

[heavy prairie]

the negative root was cosx = -2

[pale kestrel]

ok, whatever it is but yeah

[heavy prairie]

i turned -8/4 into -2

but that eliminates that root

i guess i dont understand how one thing implies the next

the identity i used is called "trig-ettan" in my language

with sin^2 + cos^2 = 1

and i just rearranged it

to get a quadratic where i could solve for cosx

so i guess that far i follow

i didnt change anything

[pale kestrel]

yes theres no reason this process loses you or gets you extra solutions

[heavy prairie]

and it is still the same equation as the original

[pale kestrel]

you can draw this on desmos if youre not convinced

draw every new equation you get

[heavy prairie]

but then i lost a root because it isnt defined

[pale kestrel]

oh

[heavy prairie]

or is it called something else

than undefined

[pale kestrel]

cos x = 1/2 or cos x = -2

[heavy prairie]

sinx = 3 for example

what do i call that

[pale kestrel]

well yeah

no real solutions

[heavy prairie]

undefined or not possible?

oh ok

[pale kestrel]

latter

anyways

if uve done elementary logic at all

you would know that given x is real,
cos x = 1/2 or cos x = -2
if and only if
cos x = 1/2

iff is if and only if

the double implies sign

<=>

[heavy prairie]

ah ok i know about the equivelant sign but

i didnt know iff meant if and only if

my native language isnt english

[pale kestrel]

If your working goes from the start to the end using <=>, then you dont need to worry about losing solutions

or having extra

[heavy prairie]

i wrote an implication arrow

when going from

[pale kestrel]

yes thats not strong enough

[heavy prairie]

okay so i can put an equivelant arrow there

then i see that i wouldnt be losing any solutions or getting extra

[pale kestrel]

yes since its just quadratic formula

[heavy prairie]

ok thanks

both of you

.close

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[ebon anchor]

hello, I’m stuck on this problem.

I assumed that v1, v2, v3, p1, p2, p3, are all 0-forms, since they have no differential in them, but then I feel like every part of this question would have the answer of “0-form”, so I think I’m missing / misunderstanding something about p-forms, but I’m not sure what.

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[kind glade]

In general, one has better luck asking about things past computational courses in the advanced channels

#diff-geo-diff-top for instance

[rustic coral]

<@&268886789983436800> ^

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@ebon anchor Has your question been resolved?

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[coral rune]

How do i do this because I keep coming up with undefined as my answer

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Show your work, and if possible, explain where you are stuck.

@coral rune Has your question been resolved?

[coral rune]

just rewriting in a more readable way sorry gimmie a sec

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[violet jetty]

This is not more readable lmao

if I had to guess... you're getting undefined because you're trying to evaluate sin(x)/x at x=0, which is 0/0

[coral rune]

more readable than it was lmao 😭

[coral rune]

yea ik but i dont understand how they got an answer then

[violet jetty]

In that case you should just use the limit at that point no?

Because you're ultiamtely trying to evaluate the "height" of the function to get your average for Simpson's rule

since the true value does not exist at x=0 (undefined), you can still figure it out if the discontinuity is a removable one (ie. a hole)