help-36

[nocturne agate]

i initially had it set at 1 in 4000 which gave me a secret normal probaility of 6hrs

that was at a 0.025% chance

[slate narwhal]

the issue with splitting them up is that these times are asking for a secret of a given type

but your question, as i understand it, is just about getting a secret

do you care which type it is?

[nocturne agate]

yes

The type that can only be acquired during the special event

[slate narwhal]

because if you do, then what type are you looking for? that changes things a lot

so you want either a fire or candy type

[nocturne agate]

yes

[slate narwhal]

so now this is a trickier problem to solve

that means all of the normal period calculations don't really matter to us

(unless your game has a pity system - if it has, please reveal it now)

[nocturne agate]

none

[slate narwhal]

ok cool

[nocturne agate]

that i know of

[slate narwhal]

so our calcullations are going to be a little trickier

but we are sure that during special events, the probability for rarities do not change?

[nocturne agate]

Its only assumed. But based on my gameplay. it remains the same. It only becomes that much rarer to acquire the higher rarities because as you mentioned, the event only runs 15mins out of every 75. So the wait times are 5x longer just based off that?

[slate narwhal]

i don't think we can naively make that assumption. let me look this through.

because the spawn behaviour is deterministic during the event period, which will affect calculations

[nocturne agate]

so how would i turn the alternating spawns in the special event into math?

[slate narwhal]

i'm looking to see if 50/50 is ok. i did say it was earlier because i didn't know you were looking for types as well

[nocturne agate]

let me ask chat gpt lol

[slate narwhal]

how long have you waited for a fire secret?

(or event secret of either type, i'm consolidating them both since they are both event-exclusive)

[nocturne agate]

I have acquired only one after roughly 275 hours of gameplay. The game has a player time counter.

[slate narwhal]

ok so, i might be very mistaken

but from the looks of the problem, the average wait time between event-exclusive secrets (not just fire or candy, but both combined) looks to be about 122 hours

so i'd say given the number of uncertainties we're working with, your experience could very well be normal

remember that we don't actually know the secret drop rate, the actual rarity behaviour during a special event, and the effects of the timers you mentioned the game having to guarantee certain types or rarities

[nocturne agate]

Yes but those rarity timers only exist for the legendary, mythic and supreme rarities. Also i dont calculate them into any of the probabilites because they are guaranteed.

No timer exists for the secret type.

When it comes to the spin wheel it is guaranteed to change the pet in queue to the selected type but the type is still determined by the spinning of the wheel which is rng.

[slate narwhal]

does the wheel spin include event-exclusive types?

[nocturne agate]

yes

but you can only spin it once per hour

it even gives you the drop rates %

[slate narwhal]

so are we taking this into account? because i didn't

[nocturne agate]

Note: the secret listed there is for that specific pet. Not all secret pets.

[nocturne agate]

No i didnt either. I am only discussing it cuz it was brought up

[slate narwhal]

ok wait hold up. those rarity timers are for those rarities, but do those timers have an effect on events?

problem is, while those timers only guarantee that the next pet is one of that rarity, having a pet be guaranteed a different rarity can skew our chances away from secret pets

because every so often you're forcing a pet to be the rarity of the given timer

[nocturne agate]

Yes, if the timer counts down during the event. It will spit out a pet of that rarity. BUT only of a event type or normal type.

It will skew the numbers off a little bit just like the normal conditions but they are removed from the probability calculations because they are guaranteed

[slate narwhal]

they cannot be removed from calculations because they are not of the secret rarity

[nocturne agate]

i mean yes that makes sense

[slate narwhal]

if there was a secret timer then of course it can't be removed, but imagine that every so often you stop rolling the dice and immediately announce the result of the dice

that's exactly what these timers are doing

this will increase the average wait time by a little and i don't have the mental capacity to go through one more round of calculations with the timers, esp when i don't know how long the timers last

will the event-normal type rule still be adhered to if the next pet's rarity is guaranteed by the timer?

for instance, if the previous pet was normal, and a timer expires, say the supreme one, will the next pet be guaranteed a fire supreme?

[nocturne agate]

I said it earlier but only 5 pets are unaccounted for during the 1hour period. Around 25 seconds of unaccounted time. So overall i think the base probability chart would still be 99% accurate

[slate narwhal]

5 pets unaccounted for per hour over 122 hours or more is quite a lot

but like i said, the average wait time will increase probably by an hour or so, by a quick estimation

[nocturne agate]

but it would also screw up my probability if i factor them in because they are guaranteed? Thats not probability it is certainty.

[slate narwhal]

well again, yes, but you have to include them in regardless if you want accuracy

[nocturne agate]

like it would instantly double the supreme rate

[nocturne agate]

which rule?

[slate narwhal]

the event-normal alternating type rule

[slate narwhal]

it won't double the rate per se over a long period of time, but the rate will be slightly higher than at first apparent

[nocturne agate]

Yes it would be forced to be supreme because of the timer, and forced to be fire because of the alternating rule

[slate narwhal]

ok, fair

[slate narwhal]

same for legendary and mythic

[nocturne agate]

yes

[slate narwhal]

either way, if we do ignore all of these effects, i suppose 122 hours is the value i found for the average wait time

so if you want to factor in all of those other factors, i can only estimate for now

[nocturne agate]

Like for example there is currently an exploit with the spin wheel that uses the timer and pet in queue mentioned above. If you spin the timer at just the right time you can force the pet in limbo waiting to be dumped to be of the legendary, mythic or supreme variant because of the timers. and of the types on the wheel.

[slate narwhal]

that would complicate things even more, and i think i'll throw in the towel for this

sorry

[nocturne agate]

Yeah i know but the spin wheel is optional so i dont calculate it . The timers on the other hand are forced

[slate narwhal]

i think i'm still throwing in the towel for that

[nocturne agate]

So would you reckon i gather data including the timer pets?

[slate narwhal]

i suggest you gather data from multiple players, then keep a consolidated view of the overall probability

this helps you scale your data gathering faster

and gather data on every single forced event

timers, events, whatever else

[nocturne agate]

yeah no one is going to help me lol thats why i only did 2 hours

[slate narwhal]

maybe you can ask in a community server for the game

see if anyone's willing to help you out

or heck, maybe someone already figured out the exact drop rates

but yeah as a solo player, data collection is hard

unless you multiacc and somehow collect data off every account at once i suppose

[nocturne agate]

They banned me because i already proved the data for their other game wrong

they were really upset about that

[slate narwhal]

oh.

[nocturne agate]

I will atleast gather the rarity of pets independently on the event and see if they are conclusive with the probability during the normal condition.

[slate narwhal]

all the best

[nocturne agate]

I would have to do it for the same period as the normal conditions i gathered so 8 events worth.

how do i tell the bot we give up

[slate narwhal]

.close

[nocturne agate]

.close

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[loud moss]

hii kinda curious on why i cant rewrite (x^2 + y^2 -1)^3 = y by moving the cube power to the other side so that y becomes y^(1/3) and then solve for dy/dx. ik the typical and valid approach is to solve this via chain rule but js wondering since (x^2 + y^2 -1)^3 = y wouldnt x^2 + y^2 -1 = y^(1/3) be true as well?

[drowsy epoch]

why would they not be equivalent?

[loud moss]

uhhh not sure i differentiated x^2 + y^2 -1 = y^(1/3) and got a different answer

[drowsy epoch]

then you did something wrong (most likely)

[loud moss]

oh

omg lemme reattempt it 😭

[plucky rover]

Show us what you did

[loud moss]

it’s correct ig

when i apply chain rule the final answer looks different

so i thought it’s wrong

.close

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[vernal bloom]

Can someone help me out wiht this question

[vernal bloom]

I'm not sure where to begin for this problem

wait i got it

im sorry

u can stop typing

.close

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[ripe jewel]

does the book have an example of what they mean by "use differentials"

because im curious

[runic tulip]

i think they are using $\Delta$ and calling it differential

[soft zealot]

Wumpus Man

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[trail crest]

Is [ ] notation used for floor or just () ?

[final saddle]

@severe elbow Has your question been resolved?

[loud sundial]

Consider rationalising the denominator

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[cedar tendon]

Hi!! I'm here for my friend whos in a probabilities and statistics class and hes struggling with this question

[cedar tendon]

could someone explain all the steps

I am not in this class notably. I am a social work major lol.

[supple mantle]

Hmm

He can't access the server?

[cedar tendon]

i can add him!!

[supple mantle]

Please do :)

[cedar tendon]

doing so now 😉

why did it wink

[wary oak]

hHELLO

HELLO

HI

[cedar tendon]

THANK YOU FOR YOUR HELP GANG

THOS IS MY FRIEND

[wary oak]

i genuinely cannot wrap my head around how to do standard deviation for probability distribution

and i have to in order to do an exam thats due by 11:59 tonight

[proper dagger]

chiming in to say that the channel still belongs to andy, so it will close unless he also continues to talk here

[wary oak]

thank you i didnt know that one moment

[supple mantle]

Did you find the mean?

[cedar tendon]

I’ll let him make a new one :3 tysm

[supple mantle]

Ah alright

[cedar tendon]

he made a chat! #help-30

I’m gonna close this one, ty!!

.close

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[little cobalt]

How would I do this question? I'm in ap calc ab and we haven't learned L'hopitals yet.

[tiny gorge]

can you interpret this as a derivative?

[final saddle]

@little cobalt Has your question been resolved?

[tame gust]

The derivative idea is very nice

a different approach is noticing that your numerator is of the form $\sqrt[3]{a} - \sqrt[3]{b}$ which, when multiplied with $\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}$ yields $a-b$, which will take care of that pesky $x$ term in the denominator

[soft zealot]

Peter

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[rose agate]

Hi, is it possible to model the volume of ANY kind of shape? I know volumes of revolution but what about things that don't revolve? Like if I want to find the volume of something wonky like a plushie or a wine aerator, how can I do that?

[trim zodiac]

Use water and measure the displacement maybe

[worthy wren]

if you are able to describe the region of the 3d shape then yeah you can find the volume as a triple integral

[rose agate]

Neat

But I'd need the equation... right?

[worthy wren]

yes for the exact volume

[rose agate]

Okk

Is there some kind of regression calculation for that?

[worthy wren]

im not too sure but have you used solidworks before?

they directly calculate the volume for you once you build the object in their system

[final saddle]

@rose agate Has your question been resolved?

[deep condor]

i believe the most common is monte carlo integration, which never fails to make me laugh:

1. bound your shape in a cube of some kind
2. pick a random point inside the cube, and check if it is inside or outside the shape
3. continue doing this until you feel confident enough
4. find the percentage of points that are inside the shape, compared to the total (let's say x%)
5. the volume of the shape is then x% of the volume of the cube

[rose agate]

I haven't no

[rose agate]

How does that work?? I am so confused rn lmao

What if I had too little points in the cube

Wouldn't I need an infinite amount of points?

[deep condor]

the idea is: if 3/5 of the points are in the shape, then 3/5 of the shape is probably in the cube

[rose agate]

Ohhh

So it's likely

But not certain

Am I right?

[deep condor]

pretty much! from this graph, 10^4 points are needed to be within 0.01 of the true volume

[deep condor]

yeah, but you only need to go so far for, say, a metalworks project

[rose agate]

lmao 💀

I see

[deep condor]

this is why we try to integrate algebraically

normally, this means we break the shape up into easily integrable shapes, then add it all together

[rose agate]

I see

I think this is a bit out my league, so I'll try something else

Thank you!!

[deep condor]

oh i didnt realize this was for a project ._. most 3d modelling programs will have some kind of volume thing, like dj said (based on this algorithm prolly)

hope that helps ^-^

[worthy wren]

if this is for your IA, id recommend modelling a far simpler shape in that case, not a plushie

[rose agate]

lmao u were in IB too that's awesome

I am doing it for my iA yeah

I know, I was thinking if the math were less complex I could do that and it would be really cool

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@rose agate Has your question been resolved?

[final saddle]

@rose agate Has your question been resolved?

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@rose agate Has your question been resolved?

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[lost axle]

How to do such questions?

Ping when reply, ty

[odd seal]

Let the two numbers be a and b. If their HCF is 81, what does that tell you about the numbers?

[lost axle]

Both are divisible by 81

[odd seal]

yes, but mathematically, can you write that?

[lost axle]

Yea

[odd seal]

can you do it right here right now?

dont just yes/no

[lost axle]

Alr 1 min

[odd seal]

...

is that it?

[lost axle]

Idk what to do after it

That's why am here

[odd seal]

a is divisible by 81

so is b

[lost axle]

Is it really necessary to write all that if it's given in the questions already

[odd seal]

thats what I wanted you to write mathematically

[lost axle]

How do I write it mathematically

[odd seal]

no I dont want you to write the alphabets

[lost axle]

Idk how to use that maths bot if that's what you mean

[odd seal]

if some number n is divisible by say 10, then you can write n=10*m for some number m

thats what I want you to do here

[lost axle]

Ooooh

So the no. Will be 81a and 81b?

[odd seal]

you mean a = 81 * a ?

do you not see anything wrong with this?

[lost axle]

No let the no. Be 81a and 81b

[odd seal]

uhh, ok if you want it like that

so, can you use the other bit of information in the problem and use it with this?

[lost axle]

81a + 81b = 1215
81(a+b) = 1215

[odd seal]

yes

[lost axle]

A+b = 1215/81

[odd seal]

mhm, go on

[lost axle]

What will I do after this

[odd seal]

,calc 1215/81

[soft zealot]

Result:

15

[odd seal]

a+b = 15

[lost axle]

A+b = 15

Now?

There can be 6 or 7 different combinations I think

[odd seal]

now can you see what all natural numbers can a,b take?

[odd seal]

can you list them?

[lost axle]

(1,14) , (2,13) , (3,12) , (4,11) and so on

[odd seal]

are you sure all of these work?

[lost axle]

Yea

[odd seal]

ok, then lets try 3,12 shall we?

[lost axle]

Yess

[odd seal]

what would be the original numbers in that case?

[lost axle]

243 and....

[odd seal]

,calc 12*81

[soft zealot]

Result:

972

[lost axle]

Yes

[odd seal]

243 and 972 ok?

[lost axle]

Ye

[odd seal]

and whats their HCF

[lost axle]

81

[odd seal]

oh is it?

,calc 972/243

[soft zealot]

Result:

4

[odd seal]

seems to me its 243

[lost axle]

Oh...

So how do we know which combination works

[odd seal]

why do you think 3,12 did not work?

[lost axle]

OOOOOHHHH the numbers should be coprime

[odd seal]

yep

[lost axle]

How will I write it in exam can you guide me

Like once I get to a+b = 15

What after that

[odd seal]

you form all the possible pairs, and eliminate the ones that are not coprime

[lost axle]

No statement to be written?

[odd seal]

theres only (1,14) to (7,8)

[odd seal]

the statement necessary would need to be written earlier

[odd seal]

right here

[lost axle]

..... Where a and b are co prime?

[odd seal]

There you should write Let the numbers be 81a and 81b such that a and b are coprime natural numbers

[lost axle]

Alright thanks for the help 🙏

[odd seal]

in fact you can also add WLOG Let a<b

[lost axle]

What's that (haven't studied log yet)

[odd seal]

its an abbreviation

Without Loss Of Generality

[lost axle]

o

Alright

[odd seal]

coz it doesnt really matter if a is less than b or not, but specifying the condition lets you list the numbers properly

[odd seal]

so uhh, back to this

[lost axle]

So the no. We will get is 1,14, 2,13 4,11 7,8

[odd seal]

yes, those are all the possible pairs

[lost axle]

Alright, thanks alot

[odd seal]

np

!done

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

[lost axle]

.close

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[lost axle]

.reopen

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✅

[lost axle]

@odd seal will this question also require a similar approach?

Like let a = 12x, b = 12y

12(xy) = 1152

xy = 1152/12

,calc 1152/12

[soft zealot]

Result:

96

[lost axle]

xy = 96

Then I factorize 96

.close nvm did it

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[odd seal]

.reopen

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✅

[odd seal]

this is not how a product would be

12x * 12y = 1152 means 144xy = 1152

so xy would be 1152/144

,calc 1152/144

[soft zealot]

Result:

8

[odd seal]

and the LCM would be 12xy or 96

[lost axle]

Oh

[odd seal]

But you might wanna also know that HCF * LCM = product of the numbers = 1152, so you can directly do 1152/12 = 96 to get the answer

[lost axle]

Oh yea I forgot that

[odd seal]

.close

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[dry ledge]

help 😭

[odd seal]

start small, do you know triangle inequality?

[brittle shadow]

In any triangle, the sum of two sides is always greater than the third side.

[dry ledge]

yea

i do

[odd seal]

ok, so you got the sides that you need to compare

can you find suitable triangles?

[dry ledge]

is it

adc

[odd seal]

Yes, thats one way to do it

[dry ledge]

cause ac+dc>ad

or

abd

[odd seal]

yea, both are suitable triangles

its up to you to decide which one you wanna choose to solve the problem

[dry ledge]

okay the ab+bd>ad

[odd seal]

alr

[violet junco]

Take 2 triangles you'll get twice of everything then cancel 2 you'll get the needed I equality

[odd seal]

(tho keep the side names capitalized, like AB + BD > AD)

[dry ledge]

ooo okay my bad

[odd seal]

so, based on what the question wants you to solve, you have an extra BD and a missing side pair BC and CD

can you resolve that?

[dry ledge]

idk what to do w the Diagonals

construction?

[odd seal]

no need to construct anything

[dry ledge]

oh okay

[odd seal]

can you check the figure to see if these 3 segments have something special?

[dry ledge]

smth special like?

[odd seal]

special like, they also form a triangle BCD

they are not collinear or anything

[dry ledge]

yes

they do

[odd seal]

and does that help tell you something?

[dry ledge]

okay so

ohh

so

what triangle did we take

[odd seal]

BCD...

[dry ledge]

okay so

in triangleBCD

BD<BC+DC

[odd seal]

mhm

[odd seal]

and can you use that here?

[dry ledge]

Like add it here?

so it forms

AB+BD+BC+DC>AD+BD?

[odd seal]

you got BC + CD > BD
If you add AB to both sides, you can still write the inequality as,
AB + BC + CD > AB + BD

[dry ledge]

oh yesh

[odd seal]

Here you got both BD and BC+DC on the same side, but otherwise yea

[dry ledge]

yeah

[dry ledge]

OH

then

[odd seal]

so uhh, you got what now?

[dry ledge]

AB+BD>AD

it forms

AB + BC + CD > AB + BD>AD

[odd seal]

yea

and thats it, its solved now

[dry ledge]

yes thanksss

happy teachers day 😭

. close

wait what

.close

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[unreal anvil]

What have you tried?

[surreal bay]

the number could be square of prime number

cuz it have only 3 factors

[unreal anvil]

it is indeed a square of a prime number

[surreal bay]

yeahh

then 10001 pqrs * pqrs
73 * 137 = 10001

[unreal anvil]

but the hint will be:
How does one make "pqrs" into "pqrspqrs"?

[surreal bay]

but i am stuck at how can i find possible values now

[crude idol]

tbh listing could probably do

there isnt much prime numbers between 37 and 100

[unreal anvil]

so pqrspqrs will have prime fractorization: x² × 73 × 137

[surreal bay]

yeah frrr

[surreal bay]

only 14

but its wrong answer

[surreal bay]

yess

[crude idol]

thats my approximation tbh so ofc its inaccurate

[surreal bay]

its fine

[unreal anvil]

and buy Guy's suggestion, you can list out the possible factors

[crude idol]

is pqrs unique?

i mean like the 4 digits

[surreal bay]

yeah ig cuz it have only 3 prime number

[crude idol]

what?

i mean must p, q, r, s be unique

[surreal bay]

no its not

[crude idol]

listing make sense tbh so we could go with that, i cant figure a better way for this

[surreal bay]

i think i got it

[crude idol]

just do 10001*(the sum of every p^2)

[unreal anvil]

hint: list out all possible combinations of
x²(73)(137)

[surreal bay]

yeah this could be work

umm lets say
37^2 * 73 * 137 = (2+1)(1+1)(1+1) = 12

[crude idol]

i dont think factorizing 10001 could help

its probably better to leave it there since we dont utilize it anyway

[surreal bay]

and
lets 73^2 * 73 * 137 = 8

[unreal anvil]

note that there will be a special case when x=73

[surreal bay]

so all possible values should 20

thank you so much

[unreal anvil]

Cheers!

[crude idol]

im so sorry i just reread the problem

alright sorry for any damage i might have caused

[final saddle]

@surreal bay Has your question been resolved?

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[jagged flare]

i dont get the solution, like its relying on there already being elements in B, and whether to add it to B or not right?

but arent we starting with B not having any elements (empty set)?

tbh i feel like im misunderstanding the concept of the solution cause like almost every sentence doesent make sense to me 😭

[valid carbon]

whats MOSP

olympiad?

[jagged flare]

no idea either

lemme check

yeah mathematical olympiad summer program

[worldly mesa]

Its perfectly ok for B to be empty at the start. We only use it to count the number of elements of B which x divides

If B is empty then the count is 0

[jagged flare]

mmm

so since its the wrong parity, the largest x in X is in B?

[worldly mesa]

Yes they even wrote that in the solution

[jagged flare]

ohh wait ok nvm

whats the motivation to this?

[worldly mesa]

Are you asking why does this proof work?

[jagged flare]

im asking why would you think of the proof, or the idea rather

[worldly mesa]

The main observation is what they said about when introducing a new number to B, it doesn't change the parity for bigger numbers

Olymiad questions are built on observations like this. First you make the observations and then you glue them together for a final solutions.

[jagged flare]

hmmm ok i see

thank you very much!

.solved ❤️

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[stone wagon]

my guess is $|z| < 1$ but i am trying to justify that

[soft zealot]

artemetra

[stone wagon]

i have barely worked with complex series tho

by root test i get
$$
\lim_{ k \to \infty } \sqrt[k]{ \left\lvert \frac{2k+i}{k+2i} \right\rvert } \lvert{ z }\rvert = \lim_{ k \to \infty } \lvert{ z }\rvert
$$

[soft zealot]

artemetra

[stone wagon]

is this enough?

it's pretty obvious that it doesn't converge if |z| = 1

[spring haven]

Why is that obvious?

[stone wagon]

wait you are right

but hm

[stone wagon]

(2k+i)/(k+2i) converges in norm

[spring haven]

R = 1 is the radius of convergence, as you've found, but the root test fails if the limit turns out to be 1

[stone wagon]

indeed

which is why i am considering |z|=1 separately

[spring haven]

So we are certain the series converges if |z| < 1 and diverges if |z| > 1, but we don't know yet what happens when |z| = 1

[spring haven]

does it?

[stone wagon]

i think so?

doesn't it converge to 2

[spring haven]

so the terms of the series look like 2z^k for large k

[stone wagon]

yes

[stone wagon]

so |z| still can't be 1 right

[final saddle]

@stone wagon Has your question been resolved?

[stone wagon]

.close

[final saddle]

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[limber narwhal]

I was doing a problem involving a 50 ft long rope that weighs 0.6 lb/ft hanging off a 140 ft building and the work required to pull it and portions of it up. I managed to solve it but I am still very confused.

[limber narwhal]

the total work to lift the rope is 750 ft-lb

but

then it asks for the work requried to lift the top half of the rope and its a quarter of the work even though you are lifting half of the rope and wouldnt the rest of the rope weigh it down?

then it asks for the work required to lift the bottom half of the rope but the integral that is used is different

the function is 15 instead of 3/5x which is what was used to find the top half

why is this the case?

and when they add the two works together is comes out as 562.5 ft-lbs which is what im most confused about. Shouldnt the work required to lift both individual half of rope sum to the work required to lift the whole thing?

I am very confused with these problems

<@&286206848099549185>

[final saddle]

@limber narwhal Has your question been resolved?

[final saddle]

@limber narwhal Has your question been resolved?

[final saddle]

@limber narwhal Has your question been resolved?

[limber narwhal]

<@&286206848099549185>

[ivory island]

one second

whilst i read

[limber narwhal]

thank you

[ivory island]

lowkey this shit confusing me just a bit

for your last question

I can tell you that calculating just the work to lift the whole thing with one calculation

[limber narwhal]

ok

[ivory island]

isnt true

the more rope you lift

the less rope is hanging off

[limber narwhal]

yea

[ivory island]

meaning the less weight you pull

15 comes from .6x25

im still figuring out why though

[limber narwhal]

yea

i dont get why calculating the work to pull the entire rope comes out as 750 ft-lbs but when you calculate both halves and add them it is 562.5 ft-lbs

[ivory island]

so

think about it like this

lets say you have a rope hanging right (new scenario)

and it takes 3N to pull it up half way

just using newtons cause why not

bare with me

3N to pull it up half way

so you have the 2nd half still hanging

which means its only going to take 1.5N to pull that up half way

and .75N for the next half

thats whats happening

take 750/2

thats 375

divide that by 2 its 187.5

375+187.5 = 562.5

[limber narwhal]

ok

one sec

like this

[ivory island]

idk what that is

but basically

15 is the force needed to lift the rope up at the bottom of the first half

or start of the 2nd half

.6lb/ft x 25 ft = 15

the ft cancels and you are left with 15lb

[limber narwhal]

yea

[ivory island]

what are you confused on

[limber narwhal]

sorry i forgot to close

.close

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[oblique estuary]

This is meant to be basic review, but I have no clue what it is and have gotten like 10 questions in a row wrong

[oblique estuary]

I understand what like sin, cos, theta, and degrees are, I am just so confused by all this

[ivory island]

it should be -Asin(theta)

I believe

[oblique estuary]

that was right

but can you please explain why?

[ivory island]

took me some time

you can draw it as a square where vector A is the hypotenuse

[oblique estuary]

awesome

and is it just negative because the y is negative?

[ivory island]

yes

[oblique estuary]

alright Imma try the next problem, give me one sec

not sure why I got this wrong

lemme show my work, one sec

correct answer was Acos(theta) being negative

negative I get, it's negative x

but Idk why cos

[ivory island]

in the last problem

what you were solving for was the y value

do you know the unit circle?

[oblique estuary]

yes

[oblique estuary]

and in this one the x

[ivory island]

So you know that on the unit circle the y value is sine

and x is cos

its the same here

[oblique estuary]

wait one sec

because one time it wanted me to use sin for the x I think

which was got me all confused

[ivory island]

yeah you use sine for x depending on what angle is given

but in this case

the angle given is at the origin

just like the unit circle

so sine is y

and cos is x

[oblique estuary]

wait I figured out why

it's cuz of this thing

so like

one sec lemme cook

[ivory island]

that doesnt change anything

for the x value

[oblique estuary]

I cooked

right right

different problem

this diff one was for the y

I figured it out now, thank you

[ivory island]

just know that the -y changes nothing on the x problem

[oblique estuary]

right I know

[oblique estuary]

but this was for a y problem

[ivory island]

oh gotcha

you can just throw things onto the unit circle

for these problems

and its easy

[oblique estuary]

I am so annoyed one sec

I was on the last problem

got it wrong

and now I gotta do another 4

I thought since it's positive Y that it would be positive magnitude

and that it'd be sin because it's y

[ivory island]

think about it with sines actually function

its sine(theta) = opposite/hypotenuse

so look at theta

and whats opposite of it

[oblique estuary]

so it'd be cos

[ivory island]

yeah

[oblique estuary]

alright

[ivory island]

opposite in that case is x

for sine

on unit circle

it would be like 90-theta

and sine would work

[oblique estuary]

alr I got it all done

thanks for the help, that'll be all for today, physics done, time for english

then calc 1 HW tmmrw (kill me)

[ivory island]

I recommend you learn the unit circle really well

[oblique estuary]

I'm sure I'll be back then

[oblique estuary]

I've got it memorized

I just hadn't ever dealt with questions like these before

[ivory island]

If you need physics help at any point message me

i cant do english

though

[oblique estuary]

alr thank you

right on

it's just reading a book anyways

easy shit

[ivory island]

type shit

[oblique estuary]

.close

[final saddle]

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

@gentle zephyr Has your question been resolved?

[final saddle]

@gentle zephyr Has your question been resolved?

[dense badge]

the way i did it is to suppose || (x,y)-(-1,8) ||<=delta, this got me that |x+1|<=delta, |x-8|<= delta and some bound on |x| and |y|, then i showed |xy+8| < …insert magic… < delta times some constant using these previous ineq

[gentle zephyr]

dude

@dense badge @dense badge @dense badge

[dense badge]

if you want to play osu and chill tonight it’s also fine

[gentle zephyr]

no

[dense badge]

it’s late here

[gentle zephyr]

I wanted to say something too

absolute value is a specific case of the pythagorean distance

correct?

[dense badge]

if x=0 or y=0 in (x,y) you can compute it and yeah you will get |x| and |y|

[gentle zephyr]

you are not getting my point

||(x,y) - (-1,8)|| is always geq 0

and since delta is stricly greater than 0 we get
0 < ||(x,y) - (-1,8)|| < delta

any pointers ?

we get that f(x,y) = xy where lim xy = -8

from the precise definition of the limit

[dense badge]

my first message are the main pointers

[gentle zephyr]

why are you equating ||(x,y) - (-1,8)|| <= delta is a strict inequality

[dense badge]

oh that’s an error on my part but my proof remains valid

it’s <

[gentle zephyr]

🤔

[dense badge]

can you deduce from this |x+1| < delta and |y-8| < delta?

[gentle zephyr]

?

Which delta did you picked?

[dense badge]

min{1,epsilon/10}

[gentle zephyr]

how?

also dont spoil it

[dense badge]

suppose 0 < sqrt((x+1)^2+(y-8)^2) < delta

then i shower |x+1|<delta and |y-8|<delta

[gentle zephyr]

I dont think I follow? @dense badge

[dense badge]

the goal is to have some inequalities to play with and then try to bound |xy+8| by delta times some constant using those ineq

[dense badge]

those two were useful

the manipulation on |xy+8| were kinda tricky

[gentle zephyr]

I dont understand the triangle inequality in multivariable

[dense badge]

this is correct but the proof i got did not required it

like the ineq is true but might not be useful

[gentle zephyr]

how did you get to the point that you want to prove x - 1 delta y - 8 delta

@dense badge

are you using triangle inequality or not?

suppose 0 < sqrt((x+1)^2+(y-8)^2) < delta
then i shower |x+1|<delta and |y-8|<delta

for me that looks like triangle inequlity, no?

[dense badge]

yes but later, those two ineq are not from the triangle ineq, both can be deducedfrom sqrt((x+1)^2+(y-8)^2) alone

[gentle zephyr]

how

this bounding that you are doing

just I dont get it dude

[dense badge]

the end goal is to have delta as a function of epsilon here delta(eps) = min{1,eps/10}

to get that function, one strategy is to get |xy+8|<delta C where C is some constant because then you will be able to choose delta = eps/C to complete the proof

it’s thinking backward a bit

but it work

[gentle zephyr]

wait a second my dude

suppose 0 < sqrt((x+1)^2+(y-8)^2) < delta
then i shower |x+1|<delta and |y-8|<delta

how did we get here ?

why is delta in terms of epsilon?

[dense badge]

ok do you agree that |x+1| <= sqrt((x+1)^2) <= sqrt((x+1)^2+(y-8)^2)

the last ineq comes from the fact sqrt() is increasing and (y-8)^2 is always greater than 0

[dense badge]

ye

same thing for |x-8|<delta

[gentle zephyr]

oki

THAT IS CRAZY GOOD SHIT

[dense badge]

these ineq also make sense geometrically

if you think about your disk of radius delta centered at (-1,8)

[gentle zephyr]

ok ok

lets not get too ahead

anal is complicated it seems

[dense badge]

i would say it’s tricky math

there are a bunch of trickshots and techniques to bound stuff

absolutely terrifying exams where you need creativity

[gentle zephyr]

I need help with the magic part of |xy + 8|

[dense badge]

k the first trick is to add and subtract y as in |xy +y -y +8|

then you have to bound this to be able to use |x+1| < delta and |x-8| < delta

[gentle zephyr]

what do i do with 8 - y

[dense badge]

that’s -(y-8) and the absolute value can eat the -

[gentle zephyr]

dude im hardstuck lowkey

[dense badge]

triangle

[gentle zephyr]

yes but

[dense badge]

but what?

[gentle zephyr]

I am still hardstuck

oh, here is where absolute value kills negative

ohhh

I see it now

yes!

[dense badge]

unfortunately (|y|+1) is not constant 😭

[gentle zephyr]

we need to write |y| in terms of some constant

sum and subtract 8

and then bound by delta

haha! triangle inequality again

[dense badge]

i did smt different

but idk show and we’ll see

[dense badge]

yeah that works

i had this

but pretty much same result

[gentle zephyr]

yours is a much stronger statement

is crazy that we both conclude almost the same thing in diferent ways?

have you taken anal?

[dense badge]

yes of course

[gentle zephyr]

is it painful?

[dense badge]

i said it before, it’s this kind of tricky stuff

[gentle zephyr]

you say its tricky but you are very well versed in it

at least on basic calculus, u know?

[dense badge]

meh, at my uni we have like a sequence of 3 course on analysis then i took some extras because i kinda enjoy more this kind of pure math

like an intro to functional analysis and a intro to complex analysis, both went over my head but i mean it was cool while it lasted kek

[gentle zephyr]

nice dude

getting back on track, how to continue ?

[dense badge]

what do you get when you use |y| < delta + 8 in the sequence of inequality from before?

[dense badge]

that’s where the min(1, smt) came from

[gentle zephyr]

wdym?

basically we get that
|xy + 8| <= delta(delta + 9)

[dense badge]

we have to force an extra bound on delta to simplify this last ineq further

this can be done with a min when defining it

[gentle zephyr]

wdym?

[dense badge]

like wouldn’t it be nice if delta was ALWAYS less than 1

then we could say that delta(delta+9) <= 10 delta

[gentle zephyr]

|xy + 8| < 1 by assumption

or what?

why do you say delta always less than 1?

[dense badge]

because i can make this true when defining delta in term of epsilon

using delta(eps) = min(1,smt)

[gentle zephyr]

Suppose |xy + 8| < 1 , then we get that we can define a delta such that both conditions are covered

[dense badge]

that’s not the same as using this, btw you assume there that eps=1. Once you are convinced of my delta(eps)=min{1,eps/10} you can just plug eps in it.

[gentle zephyr]

how do I find that delta

[dense badge]

what i’m proposing is to defined delta so it’s always the case that delta < 1

so that delta(delta+9) <= 10 delta is always valid

[gentle zephyr]

no, I understand

I completely understand

we have
|xy + 8| <= delta(delta + 9) < epsilon
and we need to define a delta(delta + 9) < 1

well, we dont know that |xy + 8| < epsilon

but if we define delta correctly

we will get there

how do I find
delta(delta + 9) < epsilon?

delta = epsilon/(delta + 9)?

delta is positive so this never gets to division by 0

[dense badge]

delta can only be defined in terms of of epsilon here or else with what you have now it’s like recursive

in a way the trickery with the min can be avoided for this particular exercices because they only want you to get deltas for eps=1 and another one

[dense badge]

delta(delta+9) < 1 is a quadratic inequality isn’t it

[gentle zephyr]

whats your point

x(x+9) < 1

[dense badge]

you could just check when x(x+9) < 1

[gentle zephyr]

?

is that a typo?

i mean is tricky we dont need to use wolfram

cant we draw this inequality in the number line?

[dense badge]

yea

[gentle zephyr]

how?

I hate intervals

and inequality solving

[dense badge]

oh now that i think about it i would just throw in the quad formula get the roots and deduce the interval

[dense badge]

but yeah it’s kinda ass

the trick with the the min{} is cleaner in this regard

[gentle zephyr]

,w x(x+9) < 1

[soft zealot]

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[gentle zephyr]

lets go with the min{}

this inequality is ass

math is hard isnt it?

[dense badge]

k when you put 1 in the min when defining delta, you can use in your proof that delta <= 1 always

[dense badge]

welcome to the gulag

[gentle zephyr]

is not good enough because we need |xy + 9| <= delta(delta + 9) < 1

we are not allowed that delta = 1

[dense badge]

true but this is the constraint imposed by the min not the euclidean distance

[gentle zephyr]

if you let delta = min{1, smth} and smth > 1 then we get delta = smth
and since |xy + 9| <= smth^2 + 9smth < 1
we get an absurdity, no?

[dense badge]

maybe think of delta = min{1,eps/C} this way, i need delta to be less than 1 for this last inequality i.e. the delta(delta+9) <= 10 delta to hold AND i also need delta to be less than eps/C a certain constant to finish the eps-delta argument

[gentle zephyr]

why leq 10 delta

where did that came from dude?

[dense badge]

if delta <=1 the delta+9 <= 10

[gentle zephyr]

sure

whats your point?

if delta leq 1 then grass is greener?

[dense badge]

unfortunately no, but can pick delta=eps/10 to end the proof in that case

[dense badge]

go back to the main proof, this

[dense badge]

just the delta+9 you want to keep the other delta to transform it into a epsilon

[gentle zephyr]

dude what the fk is going on, I dont get it

[dense badge]

this is the strat i was yapping about at the start

[gentle zephyr]

delta = min{1, epsilon/10}

what the fuck is this sorcery

[dense badge]

it’s like 1:16 am here

so not the prime time for an eps-delta argument tbh

we can stop here, unfortunately the proof is like very close from complete

worst case you have the quadratic inequality from earlier

[gentle zephyr]

well I am trying but idk if |xy + 8| = delta(delta + 9) is possible

[gentle zephyr]

yeah I sort of understand but is tricky

[dense badge]

in general, i think a good way to do eps-delta proofs is to do them two part, 1 the scratch work where you get to the point of defining delta in terms of of epsilon and then a cleaner rewrite of the argument with the chosen delta(eps) to see it always work

[gentle zephyr]

I think we can leave it at here

[dense badge]

but you have most of it and imo you learned some tricks of the trade +smt -smt, the ineq with distance etc.

[gentle zephyr]

I just hope we dont getan exercise like this in my exam, I thought I liked hard problems in the exam

but anal is a beast on its own

also this is like 3rd week of uni and its eating me alive

killed my motivation that theres always someome who is stronger than u, in preuni it was different

[dense badge]

unsolicited advice here, imo comparing with others sometime makes you think you are better than you are or conversely that you are shittier than you are.

people have different backgrounds, tbh raw ability and exposure to this stuff

[gentle zephyr]

I appreciate the help, I will try to put all the pieces together tomorrow morning, is almost 3am here and I wasted almost completely my friday, I was supposed to be having fun in osu dude, not doing a epsi delta proof at midnight, you know what I mean, is not like I didnt had fun, but sometimes the creativity steps made it so tiring and we spent like at least 2 hours at this point in a single exercise dude, mostly because I didnt understood your hints, is crazy what undergraduate degree does to a human

[dense badge]

yeah mathcord is a rabbit hole on that kind of shit

i should have waited to see you tomorrow in the help channels on this one

anyways gn my dude

[gentle zephyr]

.close

[final saddle]

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[jagged flare]

$\lim_{x\to 0}\frac{e^{2x}-1}{x\sin(x)}$

[soft zealot]

ihave<skissue>

[jagged flare]

my intuition tells me this is undefined

i tried lhospital cause its 0/0 and i got

$\frac{2e^{2x}}{\sin(x)+x\cos(x)}$ which im not too sure how to find the limit to 0 either?

[soft zealot]

ihave<skissue>

[solid notch]

lim x-->0 (sinx/x) = 1

[solid notch]

use the fact on the question itself

[robust mulch]

you can lhop again

[solid notch]

wasnt that only for indeterminant form..?

[tired walrus]

what do the num and denom now approach

[jagged flare]

$\lim_{x\to 0}\frac{e^{2x}-1}{x\sin(x)}=\lim_{x\to 0}\frac{x(e^{2x}-1)}{x^2\sin(x)}=\lim_{x\to 0}\frac{e^{2x}-1}{x^2}=\lim_{x\to 0}\frac{e^{2x}}{x}$

[robust mulch]

ah shit yeah

[tired walrus]

[michael vsauce voice] or can you?

[robust mulch]

my bad

[jagged flare]

oh wait

[robust mulch]

dont listen to my bumbling ass

[soft zealot]

ihave<skissue>

[jagged flare]

oh and e^2x converges>0 while /x diverges, so its undefined?

[rustic wedge]

your doing limit!!! pog

[rustic wedge]

i think so

[solid notch]

yeah , you can also use expantion after the 3rd step , if you have covered it

[rustic wedge]

thats what i was about to type because thats what i did

[jagged flare]

taylor?

[rustic wedge]

yes

[solid notch]

snipe

[rustic wedge]

e^x = 1+x+x^2/2 + \dots but as x->0 its enough to consider only the first 3 terms

[solid notch]

yeah cause the denominator has x^2 , terms after x^2 in e^x , i.e. x^3 , x^4 , will just vanish

[jagged flare]

oh yeaa cause theres a x/x^2=1/x term

i seee

thank you!!

.solved

[final saddle]

Channel closed

Closed by @jagged flare

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

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[loud steeple]

pls help, I have no clue how to get the zeroes of this function

[edgy mauve]

Zeroes?

[loud steeple]

yeah, isnt that how I'd get the coordinates of the points that intersect the x-axis?

[edgy mauve]

Oh i got u

See on axis the value of y=0

Do we agree on that

[loud steeple]

yeah

[fiery bluff]

are you supposed to use a graphing tool for this question

[loud steeple]

I dont think so

not too sure tho

[fiery bluff]

well, you can always make your own graph by maybe plotting g at integers and sketching

[edgy mauve]

We can consider g(x)=y

[loud steeple]

that might take too much time, I get 1.5 minutes per question

[loud steeple]

yup

[edgy mauve]

So the whole expression=0

[loud steeple]

okay

[edgy mauve]

Can u factorize it?

[loud steeple]

I tried

didnt work

cant use grouping, nor can I take a common factor

[fiery bluff]

no linear factors over rationals

yeah evaluating small integers is quite helpful
g(-1)=3
g(0)=1
g(1)=-1
g(2)=-3
g(3)=19

i think basically you want to numerically get the roots

[edgy mauve]

Yea doesnt

[fiery bluff]

@loud steeple what does this series of calculations tell you where the zeroes are?

[loud steeple]

0 and somewhere between 2 and 3?

[fiery bluff]

whoops typo, i calculated g(0) wrongly

but yeah, the main idea is you can narrow down to find the roots

[loud steeple]

somewhere between 0 and 1, and 2 and 3

[fiery bluff]

it's multiple choice so that might be the intended method

alternatively, newton method out the roots, if you have access to a calculator

[loud steeple]

I do have access to a calculator

[fiery bluff]

then you can very well newton method out the roots

[loud steeple]

newton method?

[fiery bluff]

https://en.wikipedia.org/wiki/Newton's_method

not sure if you were taught this before

though i feel this might be one of the more difficult questions that you can afford to spend slightly more time on

at the cost of spending slightly less time on the easier questions

[loud steeple]

yeah it took too much time when I was doing it the first time I just had to guesstimate the answer and move on

[fiery bluff]

yeah so i would guess the two roots are near 0.5 and 2.5, and newton method it out, I would get...

,calc 0.5-(0.5^4-20.5^3-0.5^2+1)/(40.5^3-60.5^2-20.5)

[soft zealot]

Result:

0.78125

[fiery bluff]

,calc 0.78125-(0.78125^4-20.78125^3-0.78125^2+1)/(40.78125^3-60.78125^2-20.78125)

[soft zealot]

Result:

0.72352259659614

[fiery bluff]

,calc 0.72352259659614-(0.72352259659614^4-20.72352259659614^3-0.72352259659614^2+1)/(40.72352259659614^3-60.72352259659614^2-20.72352259659614)

[soft zealot]

Result:

0.72125934712853

[fiery bluff]

about 0.72 for one root, and you should be able to get the other root pretty fast

[loud steeple]

yeah that makes it easier

[fiery bluff]

with a calculator where you can iterate a calculation (there's certain scientific calculators that let you do that), it's much faster

[loud steeple]

yeah okay I got the answer as D, which definitely took more time than it should've but this new newton's method thing is pretty neat lmao

getting the first root definitely helped a ton

[fiery bluff]

when you are done with this channel you can .close

lemme check out the other root tho

,calc 2.5-(2.5^4-22.5^3-2.5^2+1)/(42.5^3-62.5^2-22.5)

[soft zealot]

Result:

2.371875

[fiery bluff]

,calc 2.371875-(2.371875^4-22.371875^3-2.371875^2+1)/(42.371875^3-62.371875^2-22.371875)

[soft zealot]

Result:

2.349262539011

[fiery bluff]

,calc 2.35-0.72

[soft zealot]

Result:

1.63

[fiery bluff]

yeah

[loud steeple]

thanks!

.close

[final saddle]

Channel closed

Closed by @loud steeple

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[woven ledge]

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[dry ledge]

2nd

ive stuck

[woven ledge]

write

[dry ledge]

well not midway exactly

[woven ledge]

*right

[dry ledge]

js at the beginning

[woven ledge]

well you can see sin and cos in the identity

so it would make sense to write tan theta as sin theta/cos theta right?

[dry ledge]

yes

idk what to do after that

[woven ledge]

well theres also p^2 and q^2 terms so better to multiply both sides with p/q

do you know about componendo and dividendo?

[dry ledge]

no-

[woven ledge]

hmm well

when we are given a/b = c/d'

this implies a+b/a-b = c+d/c-d

this is called componendo dividendo

can you see how we can use it here?

[dry ledge]

im not sure if im supposed to use that-

[woven ledge]

well lets first see what we have

sin theta / cos theta = p/q

[dry ledge]

yes

[woven ledge]

so p sin theta/ q cos theta = p^2/q^2

[craggy atlas]

or you can divide both sides by cos

[woven ledge]

right?

[dry ledge]

how huh

[woven ledge]

i multiplied both sides with p/q

[craggy atlas]

p tan theta - q / p tan theta + q when you divide both sides by cos

[dry ledge]

so p/q=p sine theta/ q sine theta?