help-36

[vague anchor]

Oh

Alright, talk to you later

<@&286206848099549185>

[vague anchor]

C'mon guys show some mercy on high school guy
Who has to do 18 more such questions

[candid pulsar]

why would abc > -1 need to be proved then

oh at least

okk

i think a proper fraction is a rational number between 0 and 1

[vague anchor]

Yes as Ann said

[candid pulsar]

ah

[vague anchor]

#unsolvedproblem
<@&286206848099549185>

[tiny harness]

..

[vague anchor]

<@&286206848099549185>

.close

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[short wraith]

Which one is greater 50^99 or 99!?

[final saddle]

Please don't occupy multiple help channels.

[short wraith]

Ahh

🙈

[untold wagon]

consder: $50 \prod_{j=1}^{49} \bigl(j(100-j)\bigr)$

[opal plinth]

Which is greater 50^2 or 49*51?

[soft zealot]

Rbit

[short wraith]

What you did?

[untold wagon]

pair factor 99! symmetrically

Clearly, you can observe: $$100j-j^2\le 100*50-50^2=50^2$$

[short wraith]

2500, 2499

[short wraith]

Please make it simple

[soft zealot]

Rbit

[untold wagon]

What do you not understand

[opal plinth]

Repeat for 50^2 and 48*52, and so on

[untold wagon]

Every pair j(100-j) is strictly less than 50^2

So 50^99>99!

[short wraith]

Could you guys please make it from basics

[short wraith]

2500, 2496

Do you multiply too?

[opal plinth]

?

[short wraith]

I meant 2496?

[opal plinth]

?

[short wraith]

I meant how do you multiply 48*52

Quickly

[opal plinth]

I don't

[short wraith]

Then?

[opal plinth]

You're only interested in which one is greater

50^99 = 50(50^2)(50^2)...
99! = 50(49*51)(48*52)...

[short wraith]

Yeah

[opal plinth]

... well, which is greater between 50^2 and (50-n)(50+n) ?

[untold wagon]

I.E: $$j(100 - j) = 100j - j^{2} = 50^{2} - (j - 50)^{2}$$, what follows from this

[soft zealot]

Rbit

[short wraith]

50^2 >(50-n)(50+n)

[opal plinth]

Right, so it should be clear which of these is greater

[short wraith]

Yeah

Thanks

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[tulip panther]

can someone walk me through this question?

[tired walrus]

well, you can rule out m=2 and m=4 immediately bc there aren't enough positive integers below these.

[tulip panther]

yep

[tired walrus]

i suppose that a more hands-on rephrase of this question is:

you have a set of five rods, each of which has length an integer number of centimeters. the largest of these is m centimeters long. but no matter which four rods you pick, you can't make a quadrilateral out of them that has nonzero area.

... what would even one such set of rods look like, i wonder?

i would imagine that chaining 4 of them together into a quad would mean they either don't join up or they JUST BARELY do, and when they do, there isn't any way you can flex the structure

and it has to be a single line if it even does join up

[tulip panther]

i see

[tired walrus]

oh yeah this will cook actually.

[tulip panther]

so one of the "rods" would have to have a length greater than the rest combined

or, equal to

as you said, that would form a single line in that case?

in this scenario, I can assume that m must be greater than 7, since if we take the second largest integer in S and call it x, the least x can be is 6 because it must be greater or equal to the sum of the smallest integers, 1, 2, and 3

am i correct to think this?

[tired walrus]

well let's take this a bit further

let's say our 4 smaller rods are 1, 2, 3 and 6

if you try to make a quad out of them, it just barely closes up with the 1, 2 and 3 adding up to the entire length of the 6

what could the fifth rod be to produce the same behavior

(or fail to close up at all)

[tulip panther]

the sum of all the previous terms, or larger?

[tired walrus]

yeah but we don't want "or larger" cause we need m to be as small as possible

also not all the prev terms

[tulip panther]

why not all the prev terms?

oh right

[tulip panther]

i realised, since only 4 elements are used

so therefore would it be the sum of all elements, not including the first element?

[tired walrus]

... sure

2+3+6

[tulip panther]

so the answer is d)

yeah it is

okay got it

thank you : )

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[radiant carbon]

i dont understand this

answer is B

[tired walrus]

what digits can you get from 8, 0 and 5 by removing up to three lights

well actually no

from just 8

cause it's only the first digit display that's faulty

[radiant carbon]

yeah uhh the thing is i don't really understand the premise of the question

so each player has a score of 5 digits?

[radiant carbon]

and what do you mean by removing up to three lights?

so would 8, blank, 0, blank, 5 count?

[tired walrus]

no

i mean literally look at the displays that they've printed out for you

[radiant carbon]

yeah

[tired walrus]

each of those white circles is a light that's on

[fathom lion]

i can ask something about set?

[tired walrus]

literally not even math just pattern recognition

[tired walrus]

!occupied

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[radiant carbon]

ah

[tired walrus]

start with the pattern of lights for digit 8

what digits can you turn it into by turning off up to 3 of those lights

[radiant carbon]

0, 2, 5, 6

[tired walrus]

ok

[radiant carbon]

9

[tired walrus]

and which of these could be the first digit of a darts score (which has to be between 002 and 501)?

(considering also that the last two digits are 05 and displayed normally)

[radiant carbon]

2

[tired walrus]

2 only?

[radiant carbon]

0

also

[tired walrus]

any others?

[radiant carbon]

505 > 501, 605 > 501, 901 > 501

so i don't think so

[tired walrus]

and you are correct

[radiant carbon]

okay i get it now thanks

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[snow bear]

Hello i need hekp with g

so far ive split ln2n into ln2+lnn and then i crossed ln2 out

and now i am stuck

[latent dragon]

Square roots grow faster than the natural log

Take that as a hint

[snow bear]

ok so 0

so i can submit that as the solution thats it?

[latent dragon]

Use l hopitals rule

Because you have ∞/∞

[snow bear]

thanks

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[crude robin]

Hi! I forgot how to do the top half
If someone could explain so I can continue the rest of the derivative myself

[sage osprey]

I would try to express the difference on the top half as one fraction

(Use a common denominator.)

[crude robin]

The common denominator is where I’m stuck

[sage osprey]

Multiply both halves of both fractions such that the denominators are equal

[crude robin]

If I multiply 5/z by h it would give me 5h/zh not z+h
I don’t understand how I would get to a common denominator by multiplying

[sage osprey]

I would multiply by the denominator of the other fraction

[crude robin]

Ok one sec

I’m assuming this is where u were trying to push me towards

[sage osprey]

Yep

Nice job

[crude robin]

Thank uuu

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[frank folio]

I dont understand the implication from the triangle inequality

[frank folio]

To start with $|s_n-s| = |s_n + (-s)| \le |s_n|+|s|$

[soft zealot]

BOSS

[frank folio]

How does this imply it is less than 1?

[vital crag]

what are you starting with

[vital crag]

great. that context was important

[frank folio]

so?

Is it because $|s_n| = |s_n-s+s| \le |s_n-s|+|s| < 1 + |s|$

[soft zealot]

BOSS

[vital crag]

yea that works

[frank folio]

tyty

idk why that was not written in the proof

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[jagged nacelle]

if the matrix is half empty does that mean it's all zeros or it can be any value

[vital crag]

just assume 0. the probably would be too hard otherwise

[royal gust]

These are 0s

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@jagged nacelle Has your question been resolved?

[jagged nacelle]

thank you

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[trail shale]

What did I do wrong (the drawings are the method I used)

[dawn merlin]

Is this correct?

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@trail shale Has your question been resolved?

[trail shale]

<@&286206848099549185> Hi if yall wouldn’t mind checking my question that would be great

[trail shale]

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[pliant shore]

yeah you need to apply integration by parts first

x^2 doesn't fit in with sec^2 or tan x (when you differentiate sec^2 x, you don't get anything with x^2)

[modest junco]

ok then

[pliant shore]

https://www.youtube.com/watch?v=2I-_SV8cwsw

[pliant shore]

given u = 2x^2 and dv = sec^2 x tan x
tell me what you get for du and v

[modest junco]

du = sec^2 x dx

[pliant shore]

okay you're doing the integral for dv I see

the problem is that you mixed your derivatives up

the derivative of tan x is sec^2 x

so that's not the right u-sub

[modest junco]

What I am doing is I am taking sec^2 x tan x integral as 1/2 tan^2x and then applyingi ntegration by parts

[pliant shore]

okay fair enough, that's correct

then yes, you need to apply by parts twice

[modest junco]

yeah thanks......I think I was making mistake in solving not logic 😅

[pliant shore]

so you should get $2x^2 \frac{\tan^2 x}{2} - \int \frac{\tan^2 x}{2} \cdot 4x \ dx$

[soft zealot]

south

[pliant shore]

which is why I recommended you this video

you still need to do the integration and so on

[modest junco]

yeah

[pliant shore]

but the table method helps you check sign mistakes

it's easier to get the correct signs, if that's your issue

[modest junco]

oh ok fine

[pliant shore]

alright no worries!

[modest junco]

Thanks!

[sonic crystal]

south with the clutch

[final saddle]

@modest junco Has your question been resolved?

[modest junco]

let me send you my solution

[steady wraith]

this isnt maths related but

why is this your username

bruh

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[twin pivot]

You draw a card from a deck of 40 cards, calculate the probability of obtaining a face card.

[twin pivot]

Its unclear

[craggy plank]

It is unclear

[twin pivot]

Hi @craggy plank

[craggy plank]

hello

Do you have the original question?

[twin pivot]

This

[scarlet sequoia]

screenshot, photo of textbook where you got it from, anything available?

[craggy plank]

Bruh

@scarlet sequoia since you’re here, I’m leaving

lol

[twin pivot]

I sent it wrong

oops

[twin pivot]

It's automatic translation

[craggy plank]

It’s alright, just send it anyway

[twin pivot]

@craggy plank

[plucky rover]

Send the original question, whatever language it's in

Someone here might be able to give a better translation

[twin pivot]

\documentclass[12pt]{article}
\usepackage{amsmath, amssymb}

\begin{document}

Let $\mathcal{D}$ be a finite deck of $N = 40$ distinct cards, each characterized by a face $f_i$ belonging to a finite set of values
[
\mathcal{F} = { f_1, f_2, \dots, f_{10} },
]
each repeated exactly 4 times in the deck.

Consider a random experiment $X$ consisting of drawing a card $C_1 \in \mathcal{D}$. Let $Y$ be the random event corresponding to drawing a card $C_2 \in \mathcal{D} \setminus { C_1 }$ such that
[
\text{face}(C_2) = \text{face}(C_1).
]

Determine the conditional probability $P(Y \mid X)$ that the second card drawn shares the same face as the first, formalizing the solution in terms of combinatorial arguments and/or hypergeometric distributions.

\end{document}

[soft zealot]

Richard Mullin
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tired walrus]

@twin pivot was original in English or no

[twin pivot]

No

[tired walrus]

what language then

[twin pivot]

Croatia

[tired walrus]

ok can you send the original Croatian version

right now i can say that the thing you call "face" is probably known in English as rank.

[twin pivot]

Izvlači se karta iz špila od 40 karata. Izračunaj vjerojatnost da će druga izvučena karta imati isti broj/lice kao prva.

[tired walrus]

ok, but our deck consists of 10 numbers rather than 13 as for a standard poker deck (but still each one is present 4 times)

is that correct

[twin pivot]

Poker ?

[tired walrus]

uh you know the most common type of playing cards

https://en.wikipedia.org/wiki/Standard_52-card_deck

[twin pivot]

I don't have good English

Lets me see

[tired walrus]

anyway

[twin pivot]

52 cards

How do I know the faces

[tired walrus]

whichever card you draw from your deck, 3 will remain that match its rank, out of 39 total.

[twin pivot]

But how do I know what cards the deck has

[tired walrus]

you don't actually need to know, do you?

you yourself said it's some ten ranks and each one is repeated 4 times.

[twin pivot]

4*10

40 times

-> 40%?

[tired walrus]

...

[twin pivot]

/52

0,77%

[true escarp]

Cards are drawn from a deck with 40 cards. Calculate the probability that the second card drawn will have the same number/face as the first

Translation

[tired walrus]

Špil se sastoji od 10 različitih brojeva i svaki se pojavljuje 4 puta (ukupno 40 karata). Jesam li u pravu?

[twin pivot]

?

Mmm

1/40

[tired walrus]

Želim znati jesam li te dobro razumila u vezi "špila od 40 karata".

nemoj me zasipati brojkama, molim!

[twin pivot]

40 card

[tired walrus]

fuck, i try to ask you things in croatian (with google translate and my own knowledge) and you still do not understand?

[twin pivot]

??

What are you asking me

Idk

[tired walrus]

Špil se sastoji od 10 različitih brojeva i svaki se pojavljuje 4 puta (ukupno 40 karata). Jesam li u pravu?

[frank ginkgo]

Hello, Richard

[tired walrus]

tell me what language you speak best. / reci mi koji jezik najbolje govoriš.

[frank ginkgo]

Hello, Ann

[twin pivot]

What does 4 times mean?

[tired walrus]

-> 🇭🇷 4 puta...

[frank ginkgo]

Can i take a look at the original problem?

[true escarp]

svaki broj pojavljuje se 4 puta

[twin pivot]

So 4/40

[tired walrus]

im trying to figure this out. the original is:

Izvlači se karta iz špila od 40 karata. Izračunaj vjerojatnost da će druga izvučena karta imati isti broj/lice kao prva.
which translates from Croatian as:
A card is drawn from a deck with 40 cards. Calculate the probability that the second card drawn will have the same number/face as the first

[twin pivot]

1/10

[frank ginkgo]

Ok, thanks.

[tired walrus]

but OP is unable to confirm any details at all

[frank ginkgo]

!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

[tired walrus]

sorry you're kind of getting in the way here

[tired walrus]

i try to ask you a clarifying question in english and you dont understand. i try to ask in croatian and you dont understand.

i ask you WHAT LANGUAGE TO SPEAK TO YOU and you dont answer

WTF?

[twin pivot]

Mmm

Croatia

[true escarp]

koj jezik hočes da pričaš

[twin pivot]

I told you before

[true escarp]

ali ti pričaš engleski

[tired walrus]

zašto me ne razumiješ kad te pitam na hrvatskom??

[twin pivot]

I understand

[frank ginkgo]

I'm writing a solution.

[tired walrus]

!nosols

[final saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[twin pivot]

I don't know these cards

Maybe that's the problem

[tired walrus]

i'm slightly frustrated that i never received any CLEAR answer of "yes you're right", "no you're wrong" or "i don't know" to my question

[twin pivot]

Ok

Sorry

[true escarp]

ajde da pričamo hrvatski, jel če to da bude lakše?

[frank ginkgo]

I'm walking them through the steps.

[twin pivot]

Yes

[frank ginkgo]

.

this is delightfully simple once you invoke a little Laplace and Kolmogorov

[twin pivot]

Ok

[tired walrus]

what the hell are you talking about

[twin pivot]

What Is laplacs?

[tired walrus]

<@&268886789983436800> this is definitely a troll.

[frank ginkgo]

Why?

Laplace’s classical principle (1812)

[tired walrus]

what business do these big names have in such a basic probability problem

[twin pivot]

@frank ginkgo can you help

[steep hatch]

Yeah I'm inclined to agree with Ann. Even if such a solution exists, its not useful to the helpee. Also providing full solutions is against server guidelines.

Please help in good faith, and strive to help the helpee learn by helping them

[twin pivot]

Meanwhile still 0 help

[true escarp]

Do you want help in English or Croatian

[twin pivot]

Try english

[steep hatch]

Did you answer the question Ann asked you?

[frank ginkgo]

just naming the principles

[steep hatch]

Cool, they're not helpful for something like this.

[true escarp]

If English, you need to be able to answer the questions clearly

[twin pivot]

Ok

[steep hatch]

We can't help you if you don't answer our questions

[twin pivot]

Ok

[true escarp]

Once again. The deck consists of 10 numbers which all appear 4 times, do you understand that

[frank ginkgo]

citing the principles helps contextualize why the formula works, not just what the answer is

[twin pivot]

Yes

[frank ginkgo]

it’s literally how probability is defined in the first place

[craggy plank]

It’s a pre-uni probability….

[steep hatch]

Look mate your options are either just accept that an elementary probability question doesn't need these principles and try to actually be helpful, or you can go brag about memorising 2 people's names in a different math server. Please don't reply to this message because I'd like this channel to get back on track

[true escarp]

Okay, assuming the cards are numbered 1 - 10. Whats the probability of drawing a card with the number 1

[twin pivot]

1/10

[frank ginkgo]

After the first card is seen, there are 39 equally likely possibilities for the second card. Yes ?

[true escarp]

Okay, what about two cards in a row with the number 1

[twin pivot]

0/10

0

[flint cipher]

Why

[twin pivot]

Because there can't be 2 ones in 10 cards

[true escarp]

There's 40 cards

There is 4 of each number

[flint cipher]

but then 3/40?

1 is done

r u removing after 1 came first time

[twin pivot]

1/40*1/40=1/1600

Wrong !

[flint cipher]

4/40 cuz thersthere's 4 of each

[true escarp]

these aren't independent of each other. You will have one less card to choose from. Also you start with four 1s

[flint cipher]

and total 40

[twin pivot]

So 1/10?

[flint cipher]

yea

[true escarp]

That's for drawing a single 1

What about two in a row

[flint cipher]

for atleast for 1

[twin pivot]

Thanks

Let's hope it's right

.close

[final saddle]

Channel closed

Closed by @twin pivot

Use .reopen if this was a mistake.

[true escarp]

what

we did not solve the question

@twin pivot

[gilded gulch]

well its not

[twin pivot]

What

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
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[twin pivot]

I handed in the answer

[twin pivot]

.

[gilded gulch]

...

[flint cipher]

waktwait

[latent dragon]

You opened another channel. Do you have a question?

[twin pivot]

No

[latent dragon]

Alright then. Use .close.

[twin pivot]

It's continuous

[latent dragon]

?

[bold turtle]

@latent dragon

[latent dragon]

My bad. I didn’t see that.

[twin pivot]

Lol

[flint cipher]

Did you want so its 4/40 then 3/39 or just like replace after you get

[twin pivot]

I have to draw 1 card

[flint cipher]

it depends first case 0.77

[frank ginkgo]

If an experiment has a finite sample space $\Omega$ with $(N)$ equally likely elementary outcomes, then for any event $E\subseteq\Omega$ with $|E|=m$,

[soft zealot]

Henry Whitmore

[frank ginkgo]

$$P(E)=\frac{m}{N}.$$

[flint cipher]

like just multiply 4/40 * 4/39

[twin pivot]

Maybe I can withdraw the answer

[soft zealot]

Henry Whitmore

[frank ginkgo]

Ok so far ?

[twin pivot]

Ok

[flint cipher]

fGeting twice if u reprepacereplace card iis444

[twin pivot]

Now I don't know if I can collect it later, let's hope we do well

[craggy plank]

okay fam, I’m pretty sure they have no idea what those symbols are. It’s not really helpful to bring those up. As lance said, OP is probably just solving an elementary probability problem.

[flint cipher]

1/10 x 1/10

is the answer if we replace

[frank ginkgo]

Ok, im leaving this.

[flint cipher]

0.77 or 0.01

[true escarp]

Neither of those are correct for the original question and we shouldn't just be giving answers

[twin pivot]

If I have to draw 1 time

[flint cipher]

can you repeat the qs

[twin pivot]

Then there is also the second draft?

[gilded gulch]

so first get the probability of drawing a card in 40 cards, then you remove that drawn card so we have 39 then find the probability that the drawn card is the same as the one drawn, and just divide the two probabilities

[twin pivot]

You draw a card from a deck of 40 cards, calculate the probability of obtaining a face card.

[flint cipher]

40 cards

?

[twin pivot]

So 1/40*1/39?

[flint cipher]

TNormal deck ks 5

normal deck is 52

12/52

for face card

[twin pivot]

Ok

How many faces are there?

So 12/52?

O.23

Right?

[gilded gulch]

is it 40 cards or 52 cards?

[twin pivot]

Normal deck Is 52

[gilded gulch]

yeah but what is stated in the question?

[twin pivot]

But they say 40 cards though

[gilded gulch]

this one right?

[flint cipher]

spanisspanish?

u mean spanish

my keyboard messes up cuz im on web

[twin pivot]

Idk

[flint cipher]

spanish cards?

[twin pivot]

It says onlyonly 40 cards

[flint cipher]

suppose it as Spanish then

[twin pivot]

Ok

[flint cipher]

but still dont you have any examples

for which might be the number kfof cards

for it

[twin pivot]

No

[gilded gulch]

so like there are total of 40 cards and 4 of which is the same face?

[twin pivot]

Yes

[gilded gulch]

so you were asked to find P(Y|X)?

[twin pivot]

The peobabiloty

[flint cipher]

getting twice or after 1 tryung

[twin pivot]

1

[flint cipher]

once u

eodo u remove it or add it

[twin pivot]

1 time

[flint cipher]

1/10

[gilded gulch]

yeah probability of getting the same face as the first card drawn

[twin pivot]

Ok

So 1/10?

[gilded gulch]

no

[twin pivot]

😭

[gilded gulch]

first find the sample space

so in the first draw there are 40 cards

and you draw one from that

[twin pivot]

1/40

[gilded gulch]

now you remove that card

how many cards are left

[twin pivot]

39

[gilded gulch]

okay so now what is the sample space?

[twin pivot]

1/39

[gilded gulch]

im asking the sample space

[twin pivot]

?

3/39?

[gilded gulch]

no

[flint cipher]

0/39

after removingremoving

[twin pivot]

Ok

[flint cipher]

if there is only 1 cafd

but u said 4

so

first

4/40

then 3/39

[twin pivot]

Yes

[flint cipher]

so 0.77

if u do

4/40

then agin 4/40

u multiple boboth

[twin pivot]

16/1600

[flint cipher]

giving 0.01

[twin pivot]

Nice

[flint cipher]

1/10 x 1/10

[twin pivot]

So this Is the answer?

[flint cipher]

kazkazuhakazuhakazkazuhakazuhakazkazuhakazkazuhakazuhakazkazuhakazuhakazkazuha

wht do u think

[twin pivot]

Yes

I'm going to send

[gilded gulch]

no

use the fundamental of counting principle

[twin pivot]

Wjy

Why

[flint cipher]

fovfavtotal

nownow jusjust do

new

[gilded gulch]

so there are 40 cards in the first one and 39 in the second one

[twin pivot]

Yes

[flint cipher]

yeayea

[gilded gulch]

so what is the total

[twin pivot]

79

[flint cipher]

thethen ktsit's 0.77

39

IsIs total

[gilded gulch]

you multiply it

[flint cipher]

3/39

[gilded gulch]

40x39

[twin pivot]

1560

[bold turtle]

Okay, @gilded gulch imma be real,. I have no idea what you're going on about

[flint cipher]

1/40 * 3/39

[twin pivot]

I know 3/1560

[gilded gulch]

now find the favorable outcomes

[twin pivot]

0,002

[gilded gulch]

so in the first one

you have 4

then you remove 1 in the next one so you have 3

multiplying it youll have 12

[twin pivot]

Yes

[flint cipher]

First draw 1: 4/40
Second draw 1 without replacing: 3/39
Multiply: 4/40 * 3/39 = 12/1560 ≈ 0.0077 (0.77%)

if not then

First draw 1: 4/40
Second draw 1 with replacement: 4/40
Multiply: 4/40 * 4/40 = 16/1600 = 0.01 (1%)

[gilded gulch]

just going from the basics

[twin pivot]

But I think I have to draw 1 time not 2

[gilded gulch]

now how many faces are there?

[twin pivot]

4?

[gilded gulch]

10

there are 10 different faces

[twin pivot]

Ok

[gilded gulch]

so you multipluy that 10 to 12

[flint cipher]

wanwant me to solvsolve for 10/40?

dont mind repetition

[gilded gulch]

and get the probability

[flint cipher]

mynmy keykeyboard

messed up

[twin pivot]

So what would the answer be

[gilded gulch]

.

[twin pivot]

10*12=120

[gilded gulch]

thats the favorable outcomes

[twin pivot]

120/1560

[gilded gulch]

okay so that is say P(Y) since we are finding P(Y|X)

[flint cipher]

nannah give ur qs

[gilded gulch]

and P(X) is true regardless

[flint cipher]

copy paste it

[twin pivot]

Ok

I sent

[gilded gulch]

so find the intersection of P(Y) and P(X)

[twin pivot]

Its wrong😭

[gilded gulch]

what answer did you send?

[twin pivot]

It was 12/40

<@&268886789983436800>

[tulip coyote]

Mind explaining the ping for me please?

[twin pivot]

Wrong answer

So trolling

[tulip coyote]

Well, in general, we don't give out answers here, but at least from what I've managed to see, it seems as if you haven't really been helping in any useful way @gilded gulch?

[gilded gulch]

im helping hes just not trying to understand it

[tulip coyote]

Seems that it hasn't been too clear though, as per here

I'm gonna have to ask you to leave OP be please

[gilded gulch]

thats not a clear basis but yeah okay

[flint cipher]

YohYou over complicated

just tell me fsvFav

fsvourable outcome

and total outcome

Deck has 40 cards, 4 of each number.

First card: anything.

Second card: 3 cards left with same number out of 39 remaining.

Probability = 3/39 = 1/13.

This then?

[final saddle]

@twin pivot Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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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.

[north aurora]

Hello, can someone help me?

[latent dragon]

Just for future reference, do you mind posting the image/question as your first message so the bot doesn’t pin greetings and requests for help? This way, if the conversation continues for long, helpers don’t have to dig.

[north aurora]

@final apex

[foggy token]

Hi

[final apex]

yo

[latent dragon]

please don’t ping individual helpers.

[final apex]

it's ok I asked him to

[latent dragon]

oh. Apoligies.

[final apex]

So, let's pick up from yesterday, did you had any idea ?

[north aurora]

It’s ok. Dw i should’ve specified

[north aurora]

I can’t really figure it out tbh

[final apex]

So the thing about your book telling you about right neighborhood, is just the same thing as "you don't need absolute value"

if i understand the sentence correctly

[north aurora]

So uh let me resend what the textbook said

[final apex]

sure

[north aurora]

Write and solve the inequality |f(x) - l| < epsilon, with epsilon > 0.

Observe that, for arbitrarily small values of x, the inequality is satisfied in a complete neighborhood of 0 and, in particular, in a right-hand neighborhood.

Well that makes sense

If that is satisfied in a complete neighborhood of 0

Then it’s satisfied in a right hand neighborhood asw

i mean its just logic

[final apex]

right

[north aurora]

But how am i supposed to do that

i need to isolate the x

Somehow

maybe i don’t need to solve it with delt

you know the method where you don’t have to use delta?

i can show you an example

lim x->x0 (f(x)) = l

[final apex]

Of course we don't really have to use delta, but I'm guessing that's kind of what you're training with no ?

What kind of class are you taking ?

[north aurora]

Can we do both? I’m struggling with both of em

[north aurora]

I’m in high school.

[final apex]

Oh okay

That's cool to do epsilon-delta in high-school 😎

I didn't

anyway, if you want for the epsilon-delta thingy we can break the problem down to two easy problems

[north aurora]

Lets do it without delta first

[final apex]

ok

[north aurora]

That’s not normal?

[final apex]

I think that depends on the country and the high school 😆 mine was not really high level anyway, and in france where I was it's kinda rare

[north aurora]

yes. I ‘m from italy, my teacher taught me that but its not on my textbook

So i think you’re not supposed to do it

But my teacher made me learn it anyways

[final apex]

I think that's very good training

[north aurora]

so

mhm

Ok, i’ll try

I wanted to use that thing u used yesterday

[final apex]

So, the usual method to find limit is very simple:

[north aurora]

On discord

U can write math terms on discord somehow

[final apex]

you have a function that's usually like the sum (or the product or whatevs) of functions whose limit you know

[final apex]

I can show you that too yeah

[north aurora]

Please, do.

[final apex]

so the idea is this ^ for example here we have

[north aurora]

I really just need epsilon

Ok

[final apex]

$f(x) = x + \sqrt{x}$

[soft zealot]

Twenty

[north aurora]

Yes

[final apex]

what's the limit of x when x goes to 0 ? (it's not a trap question)

[north aurora]

0

[final apex]

right.

now what's the limit of $\sqrt{x}$ when $x$ goes to $0$ ? (assume $x>0$)

[soft zealot]

Twenty

[north aurora]

0

[final apex]

correct

[north aurora]

Ok

[final apex]

Now add those up, what's the limit of $f(x)$ when $x$ goes to 0 ?

[north aurora]

0

[soft zealot]

Twenty

[final apex]

right

So, that's the "easy" method

[north aurora]

Oh

Well

That was easy

[final apex]

yeah, see ? 😁

[north aurora]

Ok

Let’s do the complicated method

[final apex]

but remark that we didn't actually prove that the limit of root of x is 0, we just "know" it, or feel like it's true

and to prove we need the epsilon-delta method

[north aurora]

Yes

That makes sense

Ok

[final apex]

So actually let's do that, we're going to prove that $\lim_{x \to 0^+} \sqrt x = 0$

[soft zealot]

Twenty

[north aurora]

Just sqrt x?

Without x+sqrtx?

[final apex]

We can do both, let me see

[north aurora]

$\lim_{\to 0^+} \sqrt x = 0$

[soft zealot]

Andrew

[north aurora]

Omg that’s so cool

$\lim_{\xto 0^+} \sqrt x = 0$

[final apex]

[soft zealot]

Andrew

$\lim_{\xto 0^+} \sqrt x = 0$
```Compilation error:```! Undefined control sequence.
l.49 $\lim_{\xto
                 0^+} \sqrt x = 0$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l```

[final apex]

you should have a space: x \to 0

[north aurora]

$\lim_{x \to 0^+} \sqrt x = 0$

[soft zealot]

Andrew

[north aurora]

Great!

Ok, let’s continue

[final apex]

Ok

Okay let's do it directly on f(x) I guess, the method is more or less the same

[north aurora]

Ok

[final apex]

So let's break it down

Let $\varepsilon > 0$ be any positive number. We fix it now and we don't change it from now

[soft zealot]

Twenty

[north aurora]

Alright

Is epsilon supposed to be very small

[final apex]

It can be super small yeah, but for all we know it could be anything

[north aurora]

Ok

I’ll let you talk

Sorry for interrupting

[final apex]

no no please interrupt 😄 if you have questions or remark it's ok to interrupt

[north aurora]

Okok

[final apex]

So the idea is that we kinda want to "solve" $x + \sqrt{x} < \varepsilon$

[soft zealot]

Twenty

[north aurora]

Yup

The absolute value goes away cuz both x and sqrt of x are positive (x is close to 0+)

[final apex]

Let's break it down this way: we're gonna solve $x < \varepsilon / 2$ on one hand and $\sqrt{x} < \varepsilon / 2$ on the other hand

[soft zealot]

Twenty

[final apex]

do you agree that, if we solve that then we won ?

[north aurora]

Why epsilon/2?

[final apex]

here's the trick

try to add up those two inequalities

[north aurora]

X + sqrtx = epsilon

[final apex]

ah wait

not =

[north aurora]

Oh right

<

[final apex]

what computation did you do ? (just checking your reasoning)

[north aurora]

Well
x < epsilon/2
sqrtx < epsilon/2
x+sqrtx < epilon/2 + epsilon/2
x+sqrtx < epsilon

[final apex]

perfect

[north aurora]

👍👍

[final apex]

so that's the trick, when you have sum of 52 terms, you just solve each term < epsilon/52

[north aurora]

Oh

I never realized that

52 is an example

Right?

[final apex]

yeah i just picked a random number haha

[north aurora]

Ok

[final apex]

So

Let's solve

$x < \varepsilon / 2$ on one hand and $\sqrt{x} < \varepsilon / 2$ on the other hand

[soft zealot]

Twenty

[final apex]

👀

[north aurora]

Ok

[final apex]

when I say solve, I mean find ONE suitable x that solves both equations

[north aurora]

Let me think

x < epsilon / 2
Sqrt x < epsilon / 2

X < epsilon^2 / 4

[final apex]

nice, I was expecting that epsilon^2 / 4

now if I ask you to choose only one x, but for which you know it satisfies both equations, what would you do ?

[north aurora]

Minimum of epsilon^2/4 and epsilon/2

[final apex]

[north aurora]

help im struggling to write that on the chat

[final apex]

You can type it like this

$\min(\varepsilon^2/4, \varepsilon/2)$

[soft zealot]

Twenty

[north aurora]

Yes

[final apex]

And boom

that's your delta

[north aurora]

yes

Ok

[final apex]

that's how you actually find delta in that kind of exercise

[north aurora]

So x < $\min(\varepsilon^2/4, \varepsilon/2)$

[soft zealot]

Andrew

[north aurora]

So thats delta

[final apex]

when you have a sum of terms you solve your equations for each term and you set delta to be the minimum

yep

[north aurora]

Ok

[final apex]

so we're basically done

because we found $\delta = \min(\varepsilon^2/4, \varepsilon/2)$ which satisfies: for all $x < \delta$, $f(x) < \varepsilon$

[soft zealot]

Twenty

[north aurora]

Ok

[final apex]

so now the interpretation of this is: we did this for any epsilon, without specifying anything

In particular this statement is true for epsilon as small as you want

[north aurora]

So both of those are verified

Then i sum em up

[final apex]

yep

[north aurora]

For this

After summing i end up with

[north aurora]

This

[final apex]

yep

[north aurora]

Ok

[north aurora]

Tbh i found the “easier way” to be harder bc i was trying to do it like this

Was trying to isolate x

[north aurora]

Wait is the method i used in this photo correct

[final apex]

Ah and in this picture you should actually look at $|x-1|$ and not just $x$

[soft zealot]

Twenty

[north aurora]

I was trying to get a neighborhood of 1

[final apex]

yep

[north aurora]

What do you mean?

[final apex]

I mean that your computation is correct but you need to see that

$-\varepsilon < x-1 < \varepsilon$

[soft zealot]

Twenty

[final apex]

(i just substracted 1 to your last line)

[north aurora]

Yes

What about it?

[final apex]

that's the same as

$|x-1| < \varepsilon$

[soft zealot]

Twenty

[final apex]

right ?

[north aurora]

Yes

[final apex]

so $\delta = \varepsilon$ does the job here

[soft zealot]

Twenty

[north aurora]

Oh

That would’ve been faster

[final apex]

I mean, you did all that you needed to do, you just needed that last conclusion ^^

[north aurora]

Alright!

[final apex]

to see that $\delta = \varepsilon$ is sufficient, you need that computation of yours

[soft zealot]

Twenty

[north aurora]

Ok

Alright that’s all. You’ve been very helpful, thank you so much!!!

[final apex]

happy to help

[north aurora]

Thanks for being patient as well

[final apex]

np 😌

you're a good student

[north aurora]

Ok how do i close this channel?

Thanks :)

[final apex]

you can type

.close

[final saddle]

Channel closed

Closed by @final apex

Use .reopen if this was a mistake.

[north aurora]

It says its still occupied

[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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[manic sun]

can someone explain how to solve these questions using the method shown

[manic sun]

i dont understand why you cant do
x - 29/1 - 29

what decides it has to be x - 1 / 29 - 1

[lucid nymph]

why are you given all that

[manic sun]

wdym?

[lucid nymph]

its just a horrible way to explain how to do it, theres no explaination

[manic sun]

yeah i know

[pastel iron]

you can but your expression would be equal to 2/(2+5)

you'll get the same answer

[manic sun]

wait so it doesnt matter which way it is?

the answer will always be the same?

i understand actually, thank you for your help

the direction you go in still gives the same answer, my teacher just explained it poorly

[pastel iron]

yes as long as you change the ratio it is equal to

[manic sun]

yes thank you so so so much

[pastel iron]

you should have the same answer

[manic sun]

.close

[final saddle]

Channel closed

Closed by @manic sun

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[frank folio]

anyone know what theorem is called and if i can find a video explaining it anywhere?

[frank folio]

The books proof is extremely difficult ot undersatand, and im having a hard time understanding (i)

is it just saying there are infinite points of the sequence close to $t$?

[soft zealot]

BOSS

[rocky tusk]

i don't know that this has a name

[frank folio]

hmm

Ok more important question, does this only apply to infinate sequences?

[rocky tusk]

yea it means that regardless of the distance you choose you can find infinitely many sequence members within that distance from t

[frank folio]

like for example if i take the sequence (1,2,3,4,5) then the subsequence (1,3,5) still converges to 5 but is very much finite

[rocky tusk]

contrast this with the definition of convergence which requires all sequence members to eventually be within that distance

this just says the sequence has to at least have some sequence members sticking around in the neighborhood

infinitely many, not necessarily all but finitely many though

[rocky tusk]

well this isn't really a "sequence" this is more just a finite list

[frank folio]

would it not be a finite sequence?

at least how my book defines it

[rocky tusk]

it makes no sense to talk about convergence for finite sequences

sequences are maps from N

[frank folio]

Ok yeah just making sure

Ok so for this condition (i)

supose i take $s_n = (-1)^n$

[soft zealot]

BOSS

[rocky tusk]

have you tried either direction?

[rocky tusk]

mhm

[frank folio]

Then we know there is a subsequence that converges to 1 and -1

[rocky tusk]

yes

[frank folio]

because there are infiniate points (in this case exactly on) those points

ok cool

[frank folio]

Yeah im trying them right now for (i)

[rocky tusk]

the for all epsilon is a bit awkwardly placed which is probably leading to confusion tbh

[frank folio]

for ii and iii I was curious

[rocky tusk]

i don't like the wording

[frank folio]

i guess its consistant with the definition for convergance and cauncy but its all just everywhere rn until i get more practice

[frank folio]

if any sequence is not bounded above, lim sup s_n is inf right

Like take $s_n = (-n)^n$

[soft zealot]

BOSS

[rocky tusk]

yes

[frank folio]

This goes to both inf and -inf (limit sup and limit inf )

so is ii and iii just saying we can take subsequneces of this one that go to either inf or -inf becasue its not bounded above or bellow

mind my spelling im on my phone lol

[rocky tusk]

yea it says that for an unbounded sequence you can string together an increasing sequence diverging to infinity

for bounded above

and a similar statement for bounded below

[frank folio]

ok but for any normal sequnce, if its not bounded above and its not monotonic, we cant say for certin that it goes to inf

unless lim sup = lim inf

[rocky tusk]

intuitively this should make sense, if we couldnt pick sequence members that get arbitrarily large than surely we must have some ceiling for the sequence

[rocky tusk]

well im not sure what you mean? if a sequence isnt bounded above and not monotonic there are many things that may be true

what goes to inf?

[rocky tusk]

if these are finite then the sequence converges to that value though

could also both be infinite

[frank folio]

the limit, im just stating that being not bounded from above, unless the seuqnce is monotonic, doesnt tell us much about the limit

or wondering

[rocky tusk]

i thought you said not bounded above

alright i have to go eat

[frank folio]

yes sorry, basicly if its not bounded above it doesnt nessisarly mean the sequence has a limit of infinity

[frank folio]

by inf i meant infinity (forgot there are two terms like that now)

.close

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[frank folio]

Hi, I dont understand how this would apply to a sequence such as $a_n = n-1$ where the infimum is defenately 0, but no sub sequence converges to 0

[soft zealot]

BOSS

[frank folio]

.close

[final saddle]

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[scarlet sequoia]

Not made of positive numbers

I assume that's why you closed the channel

[frank folio]

yes

I read it a bit more carefully

im guessing this sequence always has to be decresing monotonic if its aproaching 0

[scarlet sequoia]

Not at all

(2 + (-1)^n)/n is not monotonic

And still converges to 0

[frank folio]

given it is monotonic, it must be decreasing ****

[scarlet sequoia]

Well... yeah sure

[frank folio]

The subsequence is said to be monotonic in the theorem, i was just makign sure that its always decresing if a monotonic sequence of positive numbers is approaching 0

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[sand loom]

i need to prove that t is alway bigger or equals one

[scarlet sequoia]

More context?

[scarlet sequoia]

Because only given this, t = 0 works

Any t greater or equal to 0 works in fact

It's what's written

[sand loom]

yes but the question says prove it is in french hold on

i suppose i can solve each side ?

[scarlet sequoia]

What math module is this? Integrals?

[fickle bolt]

Study both sides
A ≤ B ≤ C
is the study of the signs of B-A and C-B

[scarlet sequoia]

We can also speak in french if its easier

[fickle bolt]

omelette du fromage

[scarlet sequoia]

Bruh

[sand loom]

actually i rather not english is better thank you tho

[sand loom]

is analysis from the french/moroccan/algerian program but they mean calculus i think

[fickle bolt]

The B-A case is a no brainer

Of course
(2+t)² = 4 + 2t + t² is bigger than 4+4t since t ≥ 0

[scarlet sequoia]

4+ 4t, not 4+t

[fickle bolt]

ye mb

still holds true

[scarlet sequoia]

And... not obvious anymore

[fickle bolt]

a simple 2nd degree study Sir

[scarlet sequoia]

The expansion you wrote makes it wrong

4+2t+t² is not bigger than 4+4t. However, 4+4t+t² is...

[fickle bolt]

aaaaah you got me so right

my time to sleep I may assume

[scarlet sequoia]

Anyways what I thought indeed is that they want you to prove this with integral comparison maybe? Since it's in calculus module

[sand loom]

would it make sense if we expanded everything and tried to get t alone in the center

hopefully the left side cancels?

[scarlet sequoia]

But to be honest the polynomial comparison approach is easier

[sand loom]

how more details please

[lime grove]

Since we know that the denominator has to be greater than 0, you can make two inequalities, then cross multiply through the inequality, then simplify

[scarlet sequoia]

Well multiplying by (1+t)² is enough

[lime grove]

for example, the left side inequality would be 4(1+t) <= (2+t)^2

(before simplifying)

[sand loom]

YOOO people sorry for wasting your time it looks like a beast but if you threat every inequality alone both sides simplify to sweet good old t squared

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[sand loom]

thank you

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[winged bobcat]

occupy

[worthy wren]

do you have a question

[winged bobcat]

yes

[signal vector]

You don't need to say occupy or anything, just send your question

[winged bobcat]

is there an intuitive way to think about the transpose map? i dont understand why we used the dual space at all for this

it seems so subtle

[worthy wren]

try playing around with the matrix representation of a linear map T:V->W and the transpose map defined by phi(T) where phi is a linear form in W

[signal vector]

It's a way of defining it in which the properties of the matrix transpose become apparent just by the definition of the transpose as a map between dual spaces

[winged bobcat]

I understand the matrix representation will correspond but I feel like I am not really understanding why it is defined this way

why the precomposition

[signal vector]

You want a map from dual space of W to dual space of V

If V,W are vector spaces over F, given a map T : V -> W, you want to get a map corresponding to this from W*-> V*. I.e if I give you something say phi in W*, you must give me something in V*, one natural way to do this is by taking phi o T which will give me a map from V to F which is something in V*
Hence the precomposition

[winged bobcat]

is there a geometric interpretation of this

i get the definition in the abstract sense

[signal vector]

There is a MSE thread containing more than any one person can explain which you may find helpful:
https://math.stackexchange.com/questions/37398/what-is-the-geometric-interpretation-of-the-transpose

[winged bobcat]

Thanks, I will look into that

I will keep this channel open for a little longer to see if someone else has a view

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@winged bobcat Has your question been resolved?

[winged bobcat]

I think I will close it then

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@torn crater Has your question been resolved?

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[twin pivot]

$\frac{sinx \cdot cosx}{cosnx}=six=6$?

[shell condor]

Whoops

[proper raptor]

stop trolling

[shell condor]

You could send an image

[latent dragon]

Have they trolled in the past? Im confused why you sent this. Jc

[soft zealot]

Richard Mullin

[desert mantle]

look at the post

its obviously troll

[twin pivot]

???

[proper raptor]

because I already knew what latex was going to make

[desert mantle]

you better get to your point fast

[shell condor]

Oh wow