help-39

[rough forge]

stop using text we know what u mean

[stoic imp]

maybe I just suck at tex

[rough forge]

you cant use \varphi inside text

just ditch the formalities and do the math

you forgot the sqrt

and the powers

[jolly parrot]

Renato

[stoic imp]

ponelee jajaja

what were the power of this shit?

@rough forge

[rough forge]

(3t²)²+(2t)² @stoic imp

[stoic imp]

poneleee jajajjajaja

,, \text{arclength of } \varphi(t) = \int_{-1}^1 \sqrt{(3t^2)^2 + (2t)^2} \ dt

[jolly parrot]

Renato

[stoic imp]

@rough forge like this

[rough forge]

yes

[stoic imp]

so we get that $\text{arclength of } \varphi(t) = \int_{-1}^1 \sqrt{9t^4 + 4t^2} \ dt

@rough forge @summer imp like this

[rough forge]

yes @stoic imp

[summer imp]

First term inside the root wasn't squared properly.

[stoic imp]

,, \text{arclength of } \varphi(t) = \int_{-1}^1 \sqrt{(3t^2)^2 + (2t)^2} \ dt,, \text{arclength of } \varphi(t) = \int_{-1}^1 \sqrt{9t^4 + 4t^2} \ dt

@summer imp @rough forge now its correct?

[rough forge]

i guess bro

[summer imp]

Well I can't see a thing but yes it should be t^4

[rough forge]

u been repeating it like 4 times now

[stoic imp]

edited dawg

[rough forge]

you forgot the t²

[jolly parrot]

Renato

[stoic imp]

now yes

[rough forge]

,, \text{arclength of } \varphi(t) =

you dont need that in the middle

[stoic imp]

@summer imp @rough forge now yes?

[rough forge]

@stoic imp

yes

[stoic imp]

care to elaborate on that

[summer imp]

On what? They just said that you don't need to rewrite the "arclength of phi(t)" part in your latex.

[stoic imp]

so what should i write then ?

@rough forge @summer imp

[summer imp]

Nothing?

[stoic imp]

what

[summer imp]

You just need to compute the integral

[stoic imp]

which integral

[rough forge]

How about you integrate it with the hint I provided instead of focusing on insignificant stuff

Is Dr. Strange erasing your memory every 10min or what's happening

[stoic imp]

what do you mean?

we had $\int_{-1}^1 \sqrt{9t^4 + 4t^2} \ dt$

@rough forge @summer imp

[summer imp]

You forgot t^2

And this is like the 5th time you rewrite this

You need to compute this integral, not rewrite it over and over

[jolly parrot]

Renato

[stoic imp]

@summer imp @rough forge its corrected now

[summer imp]

6th time

[stoic imp]

what do you recommend

[summer imp]

They gave you a hint, which was to factor t^2 out from inside the root.

[stoic imp]

sure

,, \int_{-1}^1 \sqrt{9t^4 + 4t^2} \ dt = \int_{-1}^1 \sqrt{t^2} \cdot \sqrt{9t^2 + 4} \ dt

[jolly parrot]

Renato

[stoic imp]

@summer imp @rough forge like this ?

[rough forge]

Ja

[stoic imp]

wunderbar

now what?

@summer imp @rough forge What should I do now?

[summer imp]

The integrand is even, so it's the same as $2\int_0^1 \sqrt{t^2} \cdot \sqrt{9t^2+4}\dd{t}$. Then what's $\sqrt{t^2}$?

[jolly parrot]

Azyrashacorki

[stoic imp]

this doesnt make any sense

care to elaborate?

[summer imp]

It's a standard fact that if $f$ is even, then $\int_{-a}^a f(t) \dd{t} = 2\int_0^a f(t)\dd{t}$. If you disagree, then instead recall that $\sqrt{t^2} = |t|$ and split your integral into two parts according to the piecewise definition of $|t|$.

[jolly parrot]

Azyrashacorki

[stoic imp]

@summer imp @rough forge wdym by even

[summer imp]

f(t) = f(-t).

[stoic imp]

how so ?

[summer imp]

The variable occurs everywhere as an even power.

You can check that f(t) = f(-t) directly.

[rough forge]

I feel like every new simplification just leads to more basic questions, maybe we could skip that?

I mean it's insane but still

[stoic imp]

im new to math, srry

[summer imp]

Well we'll need to deal with |t| one way or another.

[rough forge]

Indeed

[summer imp]

Let's do it like this instead. Do you agree, Renato, that the length of the curve x^2 = y^3 for x in [-1,1] is twice the length of the same curve in [0,1]?

[stoic imp]

where did u get the graph from

[rough forge]

a graphing tool

[summer imp]

You essentially have two choices :

• agree that you only need to compute the length between x=0 and x=1 and multiply that by 2 to get the length between x=-1 and x=1.
• unpack the definition of |t| in the interval [-1,1] in order to split your integral into two integrals.
  Which do you prefer?

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

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[rugged mulch]

"Mention the cases where (a+b)^2 = a^2+b^2" should i just brute force or like expand the (a+b)^2

[cinder flower]

context?

[static quartz]

is it a subquestion to a question or smth

[rugged mulch]

yes

[random ermine]

expand

[proper nova]

Expand the LHS

And cancel

[rugged mulch]

the thing in "" is the question

[proper nova]

Yeah we understand

[rugged mulch]

so like a^2+b^2= a^2+2ab+b^2

[random ermine]

safe to assume he's not talking about a field of characteristic 2

[rugged mulch]

wait hold up

[proper nova]

Yep

[rugged mulch]

i can cancel the a^2 and b^2

but then 2ab= 0

[proper nova]

Yep

[rugged mulch]

so does that mean a is 0 or b is 0?

[proper nova]

So?

[proper nova]

Correct

[rugged mulch]

okay thank u

[random ermine]

wait

you could've solved this without our help

just expand

if you don't know what to do, just try smth

[proper nova]

I mean we just said expand and he does all the rest of the work

[random ermine]

exactly

[proper nova]

So technically that's some help there

[rugged mulch]

eh

i wanted like some confirmation

[proper nova]

-# bruh

[rugged mulch]

but yeah that's all thank u

.close

[pearl pond]

Channel closed

Closed by @rugged mulch

Use .reopen if this was a mistake.

[static quartz]

howd u type like that yo

[proper nova]

-#

[static quartz]

-# hi

oh

and how to make it big

[cinder flower]

yea but like idk depending on the context “mention the cases where” could mean something other than “describe all real numbers where”

[proper nova]

Remove the -

[static quartz]

oh

ty

[rugged mulch]

yeah it actually said that i believe

[proper nova]

oh ty

[random ermine]

tru

[rugged mulch]

it's a question from my national olympiads that i didn't quite understand

[static quartz]

and howd u type like that

[random ermine]

like mention all fields where this is true for all a,b

[static quartz]

o

hi

[proper nova]

and howd u type like that

[pearl pond]

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[tired hemlock]

I need help with part d

[ivory basin]

You're gonna need to give us what you did for the rest, cuz otherwise I gotta solve it all myself lmao

[pearl pond]

@tired hemlock Has your question been resolved?

[pearl pond]

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• Be polite and have a nice day!

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[acoustic tangle]

Hi I’m really confused on a

[gritty notch]

I see

I would like to help

[acoustic tangle]

Yes please

I’m trying to resolve for I’m kinda confused

[gritty notch]

Yeah I see why you say your confused

Gimme sec

[acoustic tangle]

Thx

[gritty notch]

First, the forces on P

[acoustic tangle]

Yep

[gritty notch]

So P is sitting somewhere to the right of O, moving right. What forces are acting on it?

[acoustic tangle]

4v+8x to the left

[gritty notch]

The 8x force pulls it back toward O (so it’s in the negative direction)

[acoustic tangle]

So they r negative

Mhm

[gritty notch]

The 4v resistance opposes motion, so since it’s moving right, that’s also negative

Does that make sense so far?

[acoustic tangle]

Yes

[gritty notch]

So applying F = ma

[acoustic tangle]

Is the resultant force to the left ?

[gritty notch]

Yes! Great question.
Since P is to the right of O and moving to the right, both forces act to the left

[acoustic tangle]

I see

[gritty notch]

So the resultant force is indeed to the left

[acoustic tangle]

So I thought it would be f=4v+8x

Since they’re in the same direction

But it’s not

[gritty notch]

Same result! It doesn’t matter which direction you take as positive as long as you’re consistent throughout.

[acoustic tangle]

But then I get a-8v-16x=0

And not their result

[gritty notch]

You’re right, sorry! Let me redo this carefully.
Taking right as positive, the forces to the left are negative:

[warm tiger]

you have to yourself assume f = -(4v+8x), as they're applied in opposite direction of motion, i think

[acoustic tangle]

So the object is accelerating to the right?

[warm tiger]

no

[gritty notch]

That is their result! So your equation a - 8v - 16x = 0 was just a sign error somewhere. Where did you get the minus signs on the 8v and 16x?​​​​​​​​​​​​​​​​

[acoustic tangle]

I thought acceleration was in the same direction as the forces

[gritty notch]

Acceleration points in the same direction as the net force (not necessarily every force

[acoustic tangle]

But isn’t the net force towards the origin

[gritty notch]

Oh wait I made error agin I’m so sorry😭

[cinder flower]

[gritty notch]

Yes exactly! Acceleration IS in the same direction as the resultant force.
The resultant force is to the left, so acceleration is also to the left.
But x increases to the right, so if we take right as positive

is negative (it points left)
• Forces are negative (they point left)

[acoustic tangle]

So what’s the equation from that

[gritty notch]

From F = ma taking right as positive:

I’ll Send a pic

[acoustic tangle]

Thx

[gritty notch]

Then you mutilplay by 2 then rearranging that match the result

Does that help or?

[acoustic tangle]

Im still confused sorry

[gritty notch]

No that ok I’ll break it down

[acoustic tangle]

Do we just apply f=ma regardless of the way it’s accelerating

[gritty notch]

Yes exactly! That’s the beauty of F = ma.
You just write:

The signs on the forces tell you everything. Since both forces are negative (leftward), the acceleration d²x/dt² will naturally come out negative too, meaning it accelerates to the left.

[acoustic tangle]

Ohhhhh

So f is always positive

[gritty notch]

Yeah that ohhh told me everything 😭

! F takes the sign of whichever direction you define as positive.
In this case taking right as positive:
• 8x acts to the left → -8x
• 4v acts to the left → -4v
So F = -8x - 4v (negative because forces point left)
Then F = ma:

[acoustic tangle]

I see

[gritty notch]

The sign of F just depends on which direction the forces act relative to your chosen positive direction. Does that make sense?​​​​​​​​​​​​​​​​

Take all time you need!!

[acoustic tangle]

Hmm…

Let’s say towards the x axis was negative how would that work

[gritty notch]

So taking left as positive:
• 8x acts to the left → +8x
• 4v acts to the left → +4v
So F = +8x + 4v (positive because forces point left, which is our positive direction)

[acoustic tangle]

Ummm

But then u get ma-8x-4v=0

Which isn’t what we want

[gritty notch]

But now acceleration is also measured to the left, so a = -d²x/dt² (since x increases to the right, which is now negative)
So F = ma becomes:

[acoustic tangle]

Ohhh

[gritty notch]

Yeah sorry I’m slow typer

[acoustic tangle]

It’s ok

[gritty notch]

Here I’ll write the equation after multiplying the two

And rearranging

[cinder flower]

[acoustic tangle]

What?

[gritty notch]

Is everything ok🥹

Same equation either way! That’s why it doesn’t matter which direction you pick as positive, as long as you’re consistent

Is there anything else you wanna ask?

[cinder flower]

yes

[gritty notch]

Did she disappear or?

[cinder flower]

evidently

[gritty notch]

Well my work is done here?

I may need to get better on my breaking down.

[cinder flower]

it's ok you'll do great here

[gritty notch]

Tysm❤️ I already feel great here:)

[cinder flower]

yay

[gritty notch]

Well should close this channel or 😭

[cinder flower]

soon the bot will tag her and ask if she wants to close, and if she is still not back it will just close anyway

[gritty notch]

Oh! That cool this sever comes with there secrets

Well I will explore and answer some more questions

[cinder flower]

i could close it but i don't have any good reason to

[gritty notch]

True

BYE❤️

[cinder flower]

bye

[pearl pond]

@acoustic tangle Has your question been resolved?

[acoustic tangle]

Is it because

Acceleration itself has the direction

Not the force

Like in ma=-8x-4v a is negative

[gritty notch]

Yes exactly! That’s a really good insight.
• F = ma — you just plug in the forces with their signs
• The acceleration (d²x/dt²) naturally takes care of its own direction from the maths
So you don’t need to worry about the direction of acceleration beforehand. Just:
1. Pick a positive direction
2. Assign signs to your forces
3. Write F = ma

[gritty notch]

Yes exactly!

The maths handles the direction of acceleration automatically. You just worry about the forces 😊
Ready for part (b)?​​​​​​​​​​​​​​​​

[acoustic tangle]

I see

Oh I get the rest

Ty for ur help

[gritty notch]

Of course!

[gritty notch]

Would you like me too or no.

[acoustic tangle]

It’s ok I understand the rest

[cinder flower]

ves i love you but you shouldn't just post ai excerpts here 😭

[gritty notch]

Perfect!

[acoustic tangle]

I was thinking that omg😭😭

But I didn’t wanna be rude

Incase it wasn’t

[gritty notch]

No it Alr 😭

It late in the night

Like something

[acoustic tangle]

Yeah but I could just ask ChatGPT myself

In the future it’s better if you type yourself to explain it

[gritty notch]

I see thank you for the explanation. What I really do is since I want to be much more clearer my words are not that clear as if you know what I mean so I asked ai to expand it, so it makes more sense

[acoustic tangle]

I understand

[gritty notch]

Thank you❤️ but y’all are right first day on the job😭 so I’ll take that in mind

[cinder flower]

it's ok

[gritty notch]

Well is that it?

Very nice

[pastel umbra]

I will point out the reason why this server generally has a no-ai note in help channels is precisely because of situations like this. If you don't know how to explain something, you run the risk that the AI is bullshitting but you're not versed enough to recognise the BS

Now it can be the case that the AI's output is valid, but if someone didn't know better about the topic, noticing an AI's hallucinations is rather difficult

[gritty notch]

I see

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cinder flower]

just leave her be, you don't have to mansplain ai

[pastel umbra]

-# @cinder flower "why are you 'ing me I'm right"

Oh I'm sorry I'm being transparent on why the rules are here

[gritty notch]

Thanks waes

[pearl pond]

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• Type the command .close to free the channel when you're done.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gritty notch]

I’ll do better!

[smoky gull]

.close

[pearl pond]

Channel closed

Closed by @smoky gull

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[frozen lantern]

If $\lim_{n \to \infty} \Biggl( \frac1n \sum_{k=1}^n \ln\left(\frac{n^2+(k-1)^2}{n^2+k^2} \right) \Biggr)$ exists and is equal to $L$, the absolute value of $\floor{L}$ is:

I thought the numerator telescopes to ln(1/2)?
apparently, it doesn't
and i have to go via a riemann sum

can someone explain why it doesn't really telescope (other than 'smh smth infty weird')

[jolly parrot]

Περσυ

[frozen lantern]

ping me and @jovial spade

[frozen lantern]

indeed the actual answer is -1, and i absolutely fail to see why it'd be negative

[tulip ore]

divide the fraction by n^2/n^2, then split the log

[warm tiger]

I've seen the exact problem somewhere before 🤔

[frozen lantern]

eh sure yeah i can compute the riemann sum

im just wondering why its not fine to say that the numerator is just ln(1/2)

[tulip ore]

this is not a riemann sum, this is just a normal telescoping question

[frozen lantern]

oh the solution given here is via a reimann sum

[tulip ore]

its not a riemann sum because it telescoping doesnt exactly allow much room for riemann summing that isnt just "an integral subtracted by itself"

consider distributing the 1/n across both "riemann sums" then getting that, you have left - right

= 0

this limit -> 0

in case you didnt test it

[frozen lantern]

but why doesnt it telescope like-

the denom of term 1 cancels with the numerator of term 2?

am i cancelled out infinities?

[tulip ore]

what are you saying?

type it out

use latex

and remember, you can only cancel out fractions by multiplying/dividing both sides by the same number

[frozen lantern]

it goes like $\ln \left(\frac{n^2}{n^2+1^2} \frac{n^2+1^2}{n^2+2^2} \cdots \right)$ etc

[tulip ore]

yes that telescopes

telescopes to 0

[frozen lantern]

gah

[tulip ore]

try out the telescoping yourself to confirm

[frozen lantern]

what is it.

[tulip ore]

ideally by writing it as an actual sum instead of a product

your \left and \right need to be right up to the ()s

\left( and \right)

[frozen lantern]

ah

right

[tulip ore]

also, its \cdots

and also, its not n, its k

[frozen lantern]

i thought cdots was centre

[tulip ore]

no nvm thats stupid ignore that

[frozen lantern]

do i want centre?

[tulip ore]

well yea

youre multiplying

multiplying is with center dots

[frozen lantern]

i see, i thought it was also lower

[jolly parrot]

Περσυ

[tulip ore]

so you have n^2 / ............ / (n^2 + n^2)

[frozen lantern]

ok so why cant i just say ill be left with like

yeah what

[tulip ore]

this is ln(1/2)

[frozen lantern]

isnt that 1/2

[tulip ore]

then what?

youre forgetting somethign here

[frozen lantern]

then the denominator goes to infty, so i expect it to be just be 0

[tulip ore]

yea thats all

this -> 0

the partial sums are -ln(2)/n

or ln(1/2)/n

[frozen lantern]

the answer given here is 1…

with the limit being -1 apparently

[tulip ore]

go show the original problem and lets see if we can salvage it

[frozen lantern]

absolute value being 1

hold on

[tulip ore]

thats unfortunate

and the solution?

[frozen lantern]

its ugly so i retyped

hm

[tulip ore]

just screenshot it

[frozen lantern]

riemann sum

= < 0 so true

[tulip ore]

rancid

do not show these people 0.999999... = 1

[frozen lantern]

but why would it ever be negative

[tulip ore]

swap the limit and the floor symbol

only then would this be true

similar class of mistake in saying "0.9999 < 1, so 0.99999... < 1"

[frozen lantern]

i see

i still fail to see why it would be true there tho

uh

[tulip ore]

well we already simplified the partial sum as - ln(2)/n

[frozen lantern]

yes

[tulip ore]

now for really large n, this is a small negative number

now if you floor it, you get -1

so floor(1/n ∑) -> -1

[frozen lantern]

why are you called it a partial sum? im not sure i know this term well

[tulip ore]

a partial sum is a sum of only parts of the terms, not the whole thing

,,\sum_{k=1}^n

[jolly parrot]

mtt

[tulip ore]

this is opposed to the sum which would be a sum from k=1 to ∞

[frozen lantern]

okay i see

[tulip ore]

the nth partial sum ends in n (or has n terms)

[frozen lantern]

that i follow

i see

[tulip ore]

yea thats all

[frozen lantern]

and in our case it's 0 because the limit is taken first?

[tulip ore]

yes

[frozen lantern]

so this solution is just wrong?

[tulip ore]

yea

[frozen lantern]

i see, thanks mtt

[tulip ore]

I had you bring it up to see if we could salvage the solution they told you

np

[frozen lantern]

ill leave the channel open because my friend has questions

[tulip ore]

well technically, starseeker asking the question would be different

because now, you know the answer but starseeker does not

[frozen lantern]

eh she cant latex worth shit so i asked jointly

[tulip ore]

keeping a channel open on someone else's behalf does mean a whole lot of waiting

lets just see what happens

[frozen lantern]

alrighty
i think shes just busy being shy

ok there we go

[tulip ore]

consider looking through whats already been said, maybe that answers your question

if you do need clarification I can provide that

[jovial spade]

i dont get it why we cant just use telescoping series...

[tulip ore]

yea you can

the answer's just wrong

the series converges to 0

floor it, you still get 0

what they did instead was telescope it, see the terms are negative, then think "ah yes if you floor these terms you get -1"

in doing so they swap the limit and the floor operator, not part of the original question

the limit happens first, thats 0
floor(0) = 0

[jovial spade]

how is it becoming negative?

[tulip ore]

the series telescopes to -ln(2)/n

so youre just doing lim -ln(2)/n

1st term = -ln(2) is a number between -1 and 0
2nd term = -ln(2)/2 is a number between -1 and 0
3rd term = -ln(2)/3 is a number between -1 and 0
...

now theyre asking for floor(L) = floor(lim -ln(2)/n)

the correct way to do this is that L is 0, because thats what it telescopes to

then you do floor and you get 0 again

what they did instead was lim floor(-ln(2)/n)

they saw each term and thought "these are all negative numbers between -1 and 0"

so it gets floored to -1 instead

also going to bring up the nonsense involved in doing something like this

this is straight up wrong

you already brought n to infinity when you did the integral, you dont have an n afterwards

[frozen lantern]

okay as someone who's always a bit confused about riemann sums why's it nonsense?

i mean i certainly dont get it and i feel like it's janky but idk

[frozen lantern]

ah

[jovial spade]

how are they between -1 and 0

[frozen lantern]

so what would my bounds be?

is it even riemann integrable

[tulip ore]

0 and 1

probably not, I need to check

and I dont think they even riemann integrated it correctly

[frozen lantern]

smh i see

[tulip ore]

because of the circumstances, there are a large variety of possible riemann integrals

log(1 + x^2) is technically correct

[frozen lantern]

i expected better of jee advanced but omg it's still fucking jee i guess

[tulip ore]

but remember, this is just a left riemann - right riemann = 0

[tulip ore]

ln(2) is a number commonly memorized to be 0.693

so -ln(2) is -0.693, between -1 and 0

dividing by n (where n > 1) makes numbers closer to 0

so -ln(2)/2, -ln(2)/3, ... are all numbers between -1 and 0

all negative, and all > -1

[frozen lantern]

well they dont get a 0

because they secretly took a floor… somewhere in there… without showing it?

[tulip ore]

theyre not very secret about it

moreso they subtly made a mistake

[frozen lantern]

i dont see a floor symbol

[jovial spade]

okk yess

i forgot ln 2 value wow

[tulip ore]

there are many values to memorize

try 4 decimal places to be safe

but usually ln 2 = 0.69 is enough

also consider that ln 1 < ln 2 < ln e

so 0 < ln 2 < 1

[frozen lantern]

is it 0.6931? not sure

,calc log_e(2)

[tulip ore]

(ln is an increasing function, this is arguably more important)

yes

[frozen lantern]

oh ffs

[tulip ore]

0.6931

[jolly parrot]

The following error occured while calculating:
Error: Undefined function log_e

[frozen lantern]

alright texit

[jovial spade]

okk

[tulip ore]

√3/2 ≈ 0.8660
√3/3 ≈ 0.5774

[frozen lantern]

yeah okay, it's just that i dont even see a floor function anywhere there

[tulip ore]

other microcelebrity numbers

[tulip ore]

"greatest integer function" and "floor function" are the same thing

I prefer "floor" because its clearer

[frozen lantern]

yes i meant i dont see it in their integral

[tulip ore]

it was handwaved

its technically on the outside

[frozen lantern]

if they flipped the floor and limits it would be in the integral

[tulip ore]

no

[frozen lantern]

no?

[tulip ore]

remember, the 1/n and ∑ are part of the integral

[frozen lantern]

yes

[tulip ore]

but you could just place the floor on the outside of the 1/n and ∑

but inside the limit

[frozen lantern]

oh true

but the integral is still 0 then?

do we just conclude its not riemann integrable or smth

[tulip ore]

well now with that floor stuck inside,

you seem to be thinking its not riemann integrable for whatever reason

it riemann integrates to 0, as explained earlier

left - right = 0

you can split it to two riemann integrals, both of which are identical

f((k - 1)/n) is a left riemann sum

f(k/n) is a right riemann sum

[frozen lantern]

im a bit lost i guess

[tulip ore]

anyways with that floor stuck inside, you have to be a lot more careful about evaluating the integral

what you could consider is to first try it without the floor, see the behavior, then put the floor on

arguably you shouldnt do the integral at all if the floor's stuck inside the limit

but thats not what this question is asking

it put the floor on the outside

and so its an integral which is 0

floor(0) = 0

[frozen lantern]

like i understood you're saying $\lim(\floor{\text{the fucking sum}})$?

but if the integral remains unaffected and is 0, then dont we get 0 here too?

goddamnit

sfsdfsdfs

[tulip ore]

its almost impressive how much you both are getting lost over a single mistake that isnt even yours

[jolly parrot]

Περσυ

[tulip ore]

let me be clear on the summary

the Slop if you will

the Drama

first, the context that takes way too long to get through

[jovial spade]

ok...so we got ln(1/2) from telescoping series= (-ln 2)...so lim(-ln 2/n) n tends to infinity is between -1 and 0 so its floor is -1

[tulip ore]

thats not correct

the moment you do the limit, you just get 0

so when you do floor(L), thats floor of (the limit you got which is 0)

so you just get floor(0) = 0

if you want to floor each of the terms before doing the limit, thats different

thats limit(floor(sum))

since the sum is between -1 and 0,

floor() turns that to -1

then limit of -1 is -1

[frozen lantern]

im just lost on how they got whatever they did

when i asked earlier if flipping floor and limit wouldnt change the integral, you said it wouldnt

then its still 0 innit

[tulip ore]

oh I said that? thats a mistake on my end

you cautiously check if it would or it wouldnt

here though it just does

my mistake

[frozen lantern]

i might just be lost a bit but that's what i took from this

[tulip ore]

well who said it wouldnt change the integral

I just said thats where youd place the floor symbols

itd still change it to -1

thats the whole deal

[frozen lantern]

if its not in the integral-

[tulip ore]

I think I made a mistake somewhere around there

for some reason I was thinking the lim is not part of the integral

it is

[jovial spade]

how im confused

[tulip ore]

so inserting the floor essentially midway through the integral (which isnt really inside the integral) still breaks it as described

[frozen lantern]

yeah i understood the original question, the new thing im lost on which is unrelated to the original question is just what they're doing

but it might be futile to attempt to understand that anyway, it might just be slop

[tulip ore]

limits and floors are done in order, they didnt ask for floor of (-ln(2)/n), they asked for the floor of L
now L itself didnt have any floors, its just -ln(2)/n -> 0
so when they want you do find [L], or floor(L), thats just floor(0)
youre stuck to the order of operations here

they didnt ask for $\lim_{n\to\infty}\left[\frac1n\sum_{k=1}^n\ln(\cdots)\right]$

[jolly parrot]

mtt

[tulip ore]

they asked for $\left[\lim_{n\to\infty}\frac1n\sum_{k=1}^n\ln(\cdots)\right]$

[jolly parrot]

mtt

[tulip ore]

so [0] = 0 is what they shouldve gotten

but in their work, they get [-ln(2)/n] = -1 instead, which is incorrect

the answer they give doesnt match the question they asked, mostly because their answer is too handwavy to tell the difference

[tulip ore]

you need to loosen your standards on logic to see the mistake they made as a handwaving error

that =<0 is certainly something

[frozen lantern]

alrighty sure

[jovial spade]

wont the limit tend to a very small negative number? then its floor is 0?

[frozen lantern]

yeah it is, they have low latex budget

[tulip ore]

as a reminder, the limit being asked is labelled L

L is 0, yes?

[frozen lantern]

im getting them to hire me for latexing in the future

it's gonna be for the collective good of all humanity

[tulip ore]

they wanted [L], that would be [0]

limits are self-contained, like with most math operators

you do things step by step, inside first, outside last

remember, the floor was never part of the original limit

it was attached after the limit as [L]

but L itself, the limit, has no floors in it at all, its just a telescoping series, or a riemann sum minus itself

so L = 0 since the series -> 0

then you do [0] and you get 0

being able to move functions inside/outside limits only apply to continuous functions

floor is not continuous, so you cant do that here

[jovial spade]

yes im talking about the limit...wont it tend to a number just on left of 0?

[tulip ore]

remember, limits tell you what happen at the end

limits dont have the letter n when theyre done

the limit is not -ln(2)/n

the limit is just 0

so 0 is all you see when youre done with the limit

if for example you had 1 / L

that would not be 1 / 0^-

that would just be 1 / 0

because L is 0

then you get undefined

[tulip ore]

limits dont need to have anything to do with the terms that came before it

for example 0.99999... = 1

even though 0.9 < 1, 0.99 < 1, 0.999 < 1, etc.

similarly 3.14159265... = π
even though 3 is rational, 3.1 is rational, 3.14 is rational, ...

limits just need to be eventually close to the terms

[frozen lantern]

for exactly 500 milliseconds here i got confused too because 'it's 0 from the left!!'

this is like consuming math slop lmao

im getting dumber

/nsrs

[tulip ore]

youre making an implication there on starseeker's questions that I wouldnt prefer being said out loud

sometimes the problem is so poorly worded it really does just sabotage what we know to be true

thats what you mean, I believe, mb

[frozen lantern]

ah am i?

im just saying that i got confused the exact same way the solution writer presumably did

[tulip ore]

nvm the implication wasnt there

it does consider the question being asked as math slop

I wouldnt trust the approach this answer is taking, you need to consider it in its own separate compartment

as a reminder, the integrals are both identical, they both have bounds from 0 to 1

the bounds do not have n in them

[frozen lantern]

solution is math slop best i can tell, i dont understand why they have the bounds they do

but we've established that im thoroughly confused about the integral and ill pursue that when starseeker has caught up with us, i guess

i dont wanna have two people asking you different things so

[tulip ore]

the integral can be made in any number of reductive ways due to it ultimately being 0

this is a separate idea yes

[frozen lantern]

okay she dmed me and apparently the issue we have is the belief that limits are… not exact.

[tulip ore]

well alr thats separate

[frozen lantern]

indeed

[tulip ore]

this integral business is sort of fascinating

heres how you can get the integrals

[frozen lantern]

but yeah like that $\lim_{x \to \infty} \frac2x$ is 'some number that's almost 0 but not 0' or something like that

[jolly parrot]

Περσυ

[frozen lantern]

ill leave the rest up to her

[frozen lantern]

listening

[tulip ore]

now as a reminder $\lim_{n\to\infty}\frac1n\sum_{k=1}^nf\p{\frac kn}$ is a right riemann sum, it converges to $\int_0^1f(x)dx$

[jolly parrot]

mtt

[tulip ore]

$\lim_{n\to\infty}\frac1n\sum_{k=1}^nf\p{\frac{k-1}n}$ is a left riemann sum, it converges to $\int_0^1f(x)dx$

[jolly parrot]

mtt

[tulip ore]

left and right riemann sums both converge to the same integral

it shouldnt matter the specifics on the rectangles youre picking

[frozen lantern]

alrighty

[tulip ore]

now heres what we have:

,align
&\lim_{n\to\infty}\frac1n\sum_{k=1}^n\ln\frac{n^2+(k-1)^2}{n^2+k^2}
\&=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\ln\frac{1+(\frac{k-1}n)^2}{1+(\frac kn)^2}
\&=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\p{\ln\p{1+\p{\frac{k-1}n}^2}-\ln\p{1+\p{\frac kn}^2}}
\&=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\ln\p{1+\p{\frac{k-1}n}^2}-\lim_{n\to\infty}\frac1n\sum_{k=1}^n\ln\p{1+\p{\frac kn}^2}

[frozen lantern]

still following

oh hi yajant of all people

[jolly parrot]

mtt

[frozen lantern]

yeah and that's certainly 0

[tulip ore]

yep

this instead converges to:

,,\int_0^1\ln(1+x^2)dx-\int_0^1\ln(1+x^2)dx=0

[jolly parrot]

mtt

[tulip ore]

to be able to introduce a 1/n would be to ask: why isnt there an equivalent 1/n on the bottom bound of the second integral?

and also, the fact that theres no n in the thing that would make sense anyway

an integral uses n as a dummy variable then n is discarded when the limit is used

[frozen lantern]

yeah true what must have they done

is it gpt

[tulip ore]

chatgpt cant even make a mistake this bad

itd likely guess its a telescoping sum then... just do it

this problem is very standard

[frozen lantern]

yeah and ig it wouldnt write <=0

[tulip ore]

= <0

[frozen lantern]

oh worse

[tulip ore]

"= negative number"

[frozen lantern]

AH

i thought leq 0

lmao

[tulip ore]

thats <= 0 of course

not =< 0

and its in an equality chain anyway

[frozen lantern]

yeah i didnt realise they meant like that lmfao

[smoky gull]

-# mtt goated

[frozen lantern]

indeed

not mtt explaining a relatively silly mistake for like 20 minutes lmao

we love mtt

[tulip ore]

for around an hour it seems

[frozen lantern]

a fucking hour?

[tulip ore]

time flies fast

especially when you need to have everything repeated

[frozen lantern]

goddamn answer writer couldve just written the correct answer, i wouldnt have even checked the solution

ok well i dont know im content to close it @jovial spade stop being shy and ask if you have questions!

if not ill close it in a bit ig

or let it time out we seem to have plenty of available channels

[tulip ore]

will point out that, if the limit business is all you dont trust, thats technically a separate question

the question of what a limit exactly is

[jovial spade]

yess ig i ended up to that qsn

[tulip ore]

that deserves a separate channel

doesnt necessarily need to be me answering it though

alr that should be all then

[frozen lantern]

okay thanks mtt!

.close

[pearl pond]

Channel closed

Closed by @frozen lantern

Use .reopen if this was a mistake.

[frozen lantern]

@jovial spade just grab a new channel if you wanna pursue it ig

[pearl pond]

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[vernal hinge]

can some one help ques 6 i use root test , root of 2n +3, then i dont know what to do

[paper terrace]

Question 6 of what exercise?

[vernal hinge]

8.6

[rough forge]

I would use the ratio test personally

[vernal hinge]

i'll try

[rough forge]

\prpl I guess for [ lim_{n to oo} frac{sqrt[n]{x^{2n+3}}}{9 sqrt[2n]{2n+3}} ]
you can use that ${n}^{1/n} to 1$ as $n to oo$.

[vernal hinge]

what do u think

[rough forge]

this looks correct

[runic shoal]

sth looks wrong

yea your u_(n+1) is wrong

[rough forge]

x^(2n+5) and 3^(2n+2) so ending up with x²/9<1

somehow still got it right lol

[runic shoal]

2 wrongs make 1 right

chịu

[pearl pond]

@vernal hinge Has your question been resolved?

[vernal hinge]

sai à

check phát

đang định đi ngủ

có sai gì đâu

@runic shoal

.close

[pearl pond]

Channel closed

Closed by @vernal hinge

Use .reopen if this was a mistake.

[runic shoal]

u_(n+1) số mũ thg x là 2(n+1)+3 = 2n+5

[pearl pond]

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[flint basalt]

the opposite of squaring

[hoary relic]

The square root of x is the number that when squared, results in x

[flint basalt]

$x^2$, opposite is $\sqrt {x}$

[jolly parrot]

Krish

[hoary relic]

So sqrt(x)^2 = x

[pastel umbra]

well it's one of those things Minecraft hasn't implemented yet /s

[pastel umbra]

Don't use algebra yet, I believe that's already too much if someone's asking about square roots

[flint basalt]

the concept of "squaring" is multiplying a number by itself

[pastel umbra]

Firstly, do you understand what squaring a number means? @vast harness

[hoary relic]

https://www.mathsisfun.com/square-root.html how about you read this

[flint basalt]

two "squared" is 2 * 2, which equals 4

[hoary relic]

Hm? Whats wrong with it

[flint basalt]

not negative, but the opposite

if 2 squared is 4, then the square root of 4 is 2

[hoary relic]

If you refuse to read about it then how can you learn?

[robust kraken]

don’t waste ur time with this troll lmao

[pastel umbra]

bruh you're not gonna understand a short explanation if you're not gonna read that

[hoary relic]

Yea

[flint basalt]

yes

[modern zinc]

Hi guys

[hoary relic]

Yes

[modern zinc]

Um I have a question

[flint basalt]

!occupied

[pearl pond]

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

[hoary relic]

!occupied

[pearl pond]

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

[hoary relic]

Yes

[flint basalt]

yep

and we say "the square root of 25 is 5"

[modern zinc]

Ok

[hoary relic]

Stop spreading misinformation

[flint basalt]

highly recomment watching a video on this.

[pastel umbra]

okay, you tell us what doesn't make sense

[flint basalt]

okay, if you are going to be rude to helpers and keep swearing or mocking others then we won't help you

[pastel umbra]

It actually kinda is misinfo; it's against server rules

[modern zinc]

The diameter of a cone is 14cm and the vertical height is 25cm calculate the volume

[pastel umbra]

Again go to a separate help channel

[flint basalt]

open a new help channel please. this one is occupied #❓how-to-get-help

[modern zinc]

Umm ok

Ok

[pearl pond]

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

[hoary relic]

Erm please return to your question

[pastel umbra]

bruh you're questioning the concept of a square root let's focus on that first @vast harness

[hoary relic]

Please stop, otherwise we would have to ping mods

[robust kraken]

pls open ur eyes 😭😭

its a TROLL

[flint basalt]

no

[hoary relic]

Nope

[pastel umbra]

no

[flint basalt]

is 5 x 2 = 25?

[hoary relic]

Its x multiplied by itself, so x × x

[pastel umbra]

Are you trying to write $x^2$? because that's not what "x 2" means

[jolly parrot]

Waes (Wires)

[pastel umbra]

There is, it's ^2

5^2 = 25

Bruh it's just writing a fokken 2

[hoary relic]

Well firstly its shorter

[flint basalt]

what's the point of math?

[hoary relic]

Well its used to find the area of a square

[pastel umbra]

Do you know what a square, the shape, is?

[hoary relic]

... Its there for a reason.

[pastel umbra]

Okay, now, how do we calculate the area of a square? (given a side length x, say)

[hoary relic]

Okay please stop trolling, or else i have to ping mods

[pastel umbra]

Actually,

[flint basalt]

<@&268886789983436800> likely troll

[pastel umbra]

<@&268886789983436800>

[hoary relic]

<@&268886789983436800> troll

[robust kraken]

LMAO 3 pings

[pastel umbra]

Newsflash you're your own contradiction

[flint basalt]

being bad at math and refusing to learn are two different things

[robust kraken]

mods are gonna love this one

[pastel umbra]

Man thinks he's palpatine

[pastel umbra]

wait no that's Kenobi fok

Man thinks he's Kenobi

[slow oak]

.close

[pearl pond]

Channel closed

Closed by @slow oak

Use .reopen if this was a mistake.

[slow oak]

must've been the wind

[hoary relic]

Ye

[robust kraken]

alr it was NOT me

[zinc pewter]

Also bruh this is trolling it had already been pinged twice for about a minute 🌾

[pearl pond]

Available help channel!

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[desert cliff]

#help-7｜zen1thxyz message

[desert cliff]

Are you still available @wicked edge ?

[rough forge]

Can you post the problem

[fleet ridge]

I think there is a no ping rule. I forgot the shortcut for that

Oh yes

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[desert cliff]

https://cdn.discordapp.com/attachments/423244559682764800/1504485042385326242/754291311007236186.png?ex=6a072862&is=6a05d6e2&hm=c4f2934dbe5e0b70d5f5e6484b4f911d068ad688f126f5bb932db86cfb9b0667&

this is my problem

I did 2-sqrt(2)

That’s difference of squares identity

(a-b)(a+b)

[rough forge]

why did you do 2-sqrt(2)

[desert cliff]

Because the 2 doesn’t have an internal sqrt but the other 2 does

2*2=4

But sqrt(4) is 2

[rough forge]

what happens to the 2+sqrt(2)

[ prpl sqrt{2+sqrt{2}} times sqrt{2-sqrt{2}} = sqrt{big(2+sqrt{2}big) times big(2-sqrt{2}big)} ]

[desert cliff]

Ok I might be overthinking but I do understand it now

[desert cliff]

.close

[pearl pond]

Channel closed

Closed by @desert cliff

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[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[hoary relic]

[pearl pond]

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[valid nova]

Some bs that I tried

[dense jasper]

no way in hell that this question is well posed

take (a,b,c)=(3, 1/3, 1/2) and (1/2, 3, 1/3)

[jolly parrot]

Civil Service Pigeon

[dense jasper]

both triples satisfy both given equations

but 6a+5bc differs betweeen the two

[valid nova]

you are right

https://questions.examside.com/past-years/jee/question/pthe-sum-of-squares-of-all-the-real-solutions-of-the-equat-jee-main-mathematics-trigonometric-functions-and-equations-diwsxwjwh5b7tipg
question 2

the answer is another triplet

😭

a= 1/2

[valid nova]

wait

[valid nova]

these are 3 distinct numbers

gang

[dense jasper]

fine take (3, 1/3, 1/2) and (1/2, 3, 1/3)

the idea is the same

@valid nova

[valid nova]

hmm. you are right. my teach solved this quite sometime ago, i will ask him again

.close

[pearl pond]

Channel closed

Closed by @valid nova

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[pearl pond]

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[mystic geode]

a bit confused on b), do I take the sine fourier series or the even extension of the initial condition?

[hoary relic]

You could ask in #odes-and-pdes if no one responds here

[pearl pond]

@mystic geode Has your question been resolved?

[dense jasper]

You have
$$u(x,0)=\sum^{\infty}{n=1} B_n \sin (n\pi x),$$
so to find the $B_n$ that match $f(x)=x \sin (\pi x)$, you need to use the standard formula for the Fourier sine series:
$$B_n=2 \int^{1}{0} f(x) \sin (n\pi x) \dd{x}.$$

[jolly parrot]

Civil Service Pigeon

[mystic geode]

oh alr, just take the sine series

[dense jasper]

mhm

[mystic geode]

do you take the even extension to make a full fourier series from -1 to 1 ,
but here we have our fourier series, but just need to match the b_n terms

[dense jasper]

<@&268886789983436800>

[dense jasper]

the even extension creates a series with only cosine terms

to get only sine terms, the series implicitly depends on an odd extension of the initial condition

but here we have our fourier series, but just need to match the b_n terms
but yes to this

[mystic geode]

ye that makes sense

alr thanks Civil

.solved

[pearl pond]

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[dense jasper]

[pearl pond]

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[pearl pond]

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[swift spindle]

Can someone verify b) pls?

[viscid shale]

Well, we got something like this

Consider that this is all symbolic and not necesarilly close to what the trapezium really looks like

OY will be OC + CY

OC is by definition 4b

So we want to know what CY is

We know that CY = AC * 1/3

And we can compute AC

By concatenation, we have that OA + AC = OC
Which we can re arrange as OC - OA = AC

Which we we can rewrite as:
4b - a = AC

[viscid shale]

From here, we have that CY = (4/3)b - (1/3)a

[viscid shale]

So, if you go ahead, youll find that -(4/3)a is wrong

[pearl pond]

@swift spindle Has your question been resolved?

[pearl pond]

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[sharp galleon]

Yo guys

[dapper heart]

balriggt

Swiped my channel

[sharp galleon]

😂😂

Differentiation of y= smth

Is the gradient of tangent right

[dapper heart]

no idea

Good luck

[fleet ridge]

Yes, at that point

[pearl pond]

@sharp galleon Has your question been resolved?

[pearl pond]

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[odd kelp]

21st

[odd kelp]

21st question

@delicate temple @bronze glen

[hoary relic]

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[odd kelp]

they're my friends.....

i think i can ping my friends yea?

[hoary relic]

but why would you need to?

they might not even be able to do this

?

[odd kelp]

cuz i wanna ask help?

[odd kelp]

😭 ik them well dude

[hoary relic]

someone will come here eventually

[odd kelp]

sigh

[merry carbon]

.close

[pearl pond]

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[odd kelp]

21st question

[odd kelp]

moreover they dont even have the helper role...
so im good 😌

[hoary relic]

it seems like 15 minutes have passed since you opened the original question. you could maybe ping helpers now

[odd kelp]

nah....

the channel was closed by mods

[hoary relic]

idk

or you can wait another 15min and ping helpers

[odd kelp]

ill wait jst to be safe

[wicked edge]

Well one way to that is that notice AX,BY,CZ.

They are all perpendicular to OP.

What can you tell about their relationships?

[odd kelp]

I cant understand how they're perpendicular to OP

if O is the bisector of AB

then where should P lie?

[wicked edge]

They are given in the question.

[odd kelp]

yk I cant understand the diagram....

[odd kelp]

where should P lie?

[wicked edge]

On the line OP?

[odd kelp]

....

and where does OP lie?

like is it an extended line from A, B or C

[wicked edge]

Anywhere in the triangle ABC i suppose.

[odd kelp]

oh

[odd kelp]

inside the triangle

hmm

then?

[bronze glen]

yo mb for late reply

[odd kelp]

dw abt it

solve kr payega?

[bronze glen]

doing differentiation qs

didnt attempt

[odd kelp]

alrrr

[odd kelp]

<@&286206848099549185> 21st question

[wary shale]

oh

similiar triangle

[odd kelp]

can u help me out...

[delicate temple]

Hello

[delicate temple]

Yo ctpcm

I too have

[delicate temple]

Wsp bro I willtry

[odd kelp]

thnx

[wary shale]

im drawing stuff

[delicate temple]

Leme see

[wicked edge]

Well since AX and BY are parallel?

[odd kelp]

how...

[wicked edge]

Have you answered this?

Answer it then this question is solved.

[delicate temple]

Bro try giving Buddha a diagram

Like where is OP

[delicate temple]

I will try to help, I'm not sure to be able to do it tho

[odd kelp]

alr..

[wicked edge]

Do not mind about OP now.

Try to determine those lines relationship first.

That's not my question for you.

[wary shale]

oh shit

lol

[wicked edge]

Why did you answer?

Delete those comments.

[wary shale]

sorry

my bad

[wicked edge]

All good.

[wary shale]

well at least make a sketch bro

[wicked edge]

Like what I said, don't mind about it.

And if OP want, he can draw it.

[wary shale]

oh ye

[wicked edge]

We want to find the relationships between the lines first.

And sketching out here I think it is unnecessary.

[odd kelp]

hmm

[wary shale]

after drawing it is kinda easy to solve a and b

tbh the A wording is kinda stupid

[pearl pond]

@odd kelp Has your question been resolved?

[delicate temple]

@harsh adder

Do YK how to help here

Pls

Sry to disturb

Btw

[odd kelp]

tru

[wary shale]

alro

haven't done yet?

[harsh adder]

you will have 2 cases here

AB on the same side

and AB on opp sides

make sure you draw diagram correctly btw

[odd kelp]

yea im stuck ovr there

[harsh adder]

1s

ill show the diagram

back

mb i was a bit afk

[harsh adder]

basically its any line OP

and AX, BY, CZ are 3 perpendiculars to that line

note that O bisects AB

[pearl pond]

@odd kelp Has your question been resolved?

[harsh adder]

nope idk how to do this q

[wicked edge]

Still struggling?

Have you answered my question?

If not then keep struggling then.

[odd kelp]

Mb mb

[pearl pond]

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[wicked edge]

Good.

[pearl pond]

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[nimble kindle]

Question c pls, c1

[nimble kindle]

And c2

.close

[pearl pond]

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[pearl pond]

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[inland laurel]

how can I proceed further

[maiden cove]

do you still need help?

[inland laurel]

huh

?