help-23

[lean geyser]

i dont understand man, could you please go through an example ?

[safe radish]

@lean geyser Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[safe radish]

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.

[white ocean]

im losing my mind yall

[white ocean]

i need to prove this by induction

but i always get stuck in the ^n mud somewhere

or even worse n!

[ornate skiff]

Show where you got so far?

[white ocean]

cant

[ornate skiff]

I suspect you're familiar with the steps... the supposition that P(n) is true, and what you have to prove?

[white ocean]

the notebook is in another place

i can do it again right now tho

[white ocean]

yeah

[ornate skiff]

P(k), albeit. I think that's how it's done at school

Then what will P(k+1) be?

[white ocean]

I have to pass through here

now i need to find something that is both less than n+1 ^n and greater than 2*2^n n!

but n^n doesnt work

or at least idk how to make it work

[ornate skiff]

are you sure in first line should be n! ?

[white ocean]

what do you mean

[ornate skiff]

I think you wrote there:

$P(k+1): (n+1)^{n+1} \geq 2^{n+1} \cdot n!$

[flat frigate]

Wild123

[buoyant dragon]

factor out n^n from the expression

[white ocean]

which one

[ornate skiff]

shouldn't have (n+1) factorial?

[white ocean]

the left side?

[buoyant dragon]

$(k+1)^k = k^k \left(1 + \frac{1}{k}\right)^k$

[flat frigate]

Shikhar

[white ocean]

yeah i did not know that equality existed

how do you get there

[buoyant dragon]

just factor out k^k from lhs expression

[ornate skiff]

ah, I see, you missed it on the 1st, but wrote it on the 2nd.

[ornate skiff]

k+1= k(1 + 1/k), and when you raise it all to power k...

you get that

[buoyant dragon]

$$(k+1)^k = \left( k \left(1 + \frac{1}{k}\right) \right)^k = k^k \left(1 + \frac{1}{k}\right)^k$$

[flat frigate]

Shikhar

[white ocean]

makes sense

my e senses are tingling

but i assume it doesnt matter here

since im not taking the limit as x → +∞

[buoyant dragon]

so if you aint taking limits.. you cant use this you mean?

[white ocean]

no i mean that this looks familiar

[buoyant dragon]

yea thats why we're using this

[ornate skiff]

if you can prove $(n+1)^n \geq 2 n^n$, seems solved

[flat frigate]

Wild123

[buoyant dragon]

cuz for all k>1.. this expression is always greater than 2

[ornate skiff]

I mean, a different method

Then you'd just apply the supposition p(n) (or p(k) )

[white ocean]

ok ok one step at a time, if i factor out n^n i get n^n*(n+1/n)^n >= 2*2^n*n!

now what can i say about this

[ornate skiff]

tbh, I realize we were both moving in same direction with the proof

you now have to prove that (n+1/n)^n >= 2

as n+1/n is clearly > 1, it increases as n goes up. It merely takes a check for the lowest n, to see it's true...

[white ocean]

i have to prove that it's greater than 2*2^n

what was the command for that again

my neck hurts

[ornate skiff]

you used something weaker than the hypothesis?

n^n was >= than 2^n n!, not just n!

$n^n \cdot (1+\frac{1}{n})^n \geq 2 \cdot 2^n \cdot n!$

[flat frigate]

Wild123

[ornate skiff]

the hypothesis, is, if I remember:

[white ocean]

ah i see

[ornate skiff]

$n^n \geq 2^n n!$

[flat frigate]

Wild123

[ornate skiff]

no?

[ornate skiff]

so, if you can prove the (1 + 1/n)^n is >= 2, it's done.

yuppie

[white ocean]

and....how do i do that?

do i have to use another proof by induction?

[ornate skiff]

I think it can work without, but it's fairly complicated, supposing you don't have a formal definition of 'increasing'. one option is to prove that the larger n is, the bigger the left side is (so, you only have to check for the smallest possible n)
Plan B is another induction

[white ocean]

well to prove that it's increasing is basically just induction again

i need to prove that f(n) < f(n+1) ∀n

[ornate skiff]

there's another option, where you consider it a chain, {a_n}, and do either:

$a_{n+1}-a_n \geq 0$ or $\frac{a_{n+1}}{a_n}\geq 1$

[flat frigate]

Wild123

[ornate skiff]

If you're familiar with such things

[white ocean]

is that not just this?

[ornate skiff]

sure, I didn't see it. mb

It's not quite induction though, as you don't have to suppose anything true, or all that

[white ocean]

sure

no idea what to do from here on though

[ornate skiff]

managed to prove it? probably works better with division

move right term with division to left

brb let me see if it works ok-ish

[white ocean]

i still cannot fathom how you're supposed to do this in about 10 minutes tops

i might be cooked

[plucky elk]

Homework rarely takes faster than exams

[ornate skiff]

true. same for difficulty, I'd say, in some highschools (might differ from country to country)

[white ocean]

tbh im not sure whether it's supposed to be a realistic example of what the exam looks like or just tedious homework

[ornate skiff]

probably a tough case of induction where it doesn't lead directly to result by just applying the hypothesis

I can only guess you might have 1 easier and 1 tougher induction on the test

Anyway, here the goal is to look at:

[white ocean]

biblically accurate disequation

[ornate skiff]

yea

work on that big fraction

hopefully it works to prove it's > 1

it will look pretty ugly if you open all the brackets and set a common denominator

got to $\frac{n^3+2n^2+2n}{n^3+n^2+n+1}$ ?

[flat frigate]

Wild123

[ornate skiff]

or something like that

[white ocean]

yeah

[ornate skiff]

we make it into a 1 and another fraction

[white ocean]

i only have (1+n)^2 though

in the denominator

[ornate skiff]

that is, we write the numerator as n^3+ n^2+n+1 + (something) which might be negative

[ornate skiff]

maybe I did a little oopsie. brb

[white ocean]

i am questioning the use of all this though

it's just tedious algebra

i know how that works

like the point of the exercises should be to get familiar with new concepts

[ornate skiff]

fairly certain it should be (n+1)(n^2+1)

which isn't equals to that

[white ocean]

i couldve just proved (n+1/n)n >= 2 by "trust me bro" and basically taken away the same thing from this exercise

[ornate skiff]

the first exercises are direct applications, the rest involve new approaches, ideas and such

[white ocean]

yeah but we've already done that by getting to that last step

it's proving that last step that only requires 3 hours of brute force algebra

[ornate skiff]

so, you know how to prove this?

Or... maybe there's something to gain from it 🙂

[white ocean]

you need to prove that f(n+1) > f(n) to say that it's increasing and to say that f(6) > 2

[ornate skiff]

I definitely think so

or ... induction again

[white ocean]

yeah

[ornate skiff]

to prove (n+1/n)^n >= 2

but you'll get to same thing, honestly

[white ocean]

but it would get me nowhere

yeah

[ornate skiff]

I still think this is the fraction

[white ocean]

i believe you lol

[ornate skiff]

$\frac{n((n+1)^2+1)}{(n+1)(n^2+1)}$

[flat frigate]

Wild123

[ornate skiff]

this is pre-expansion

Personally, I've preferred endless calculations over complex higher reasoning (albeit, not when it involved already having x idea / formula and applying it)

so, did you check your math? were you wrong? was I?

[ornate skiff]

top fraction would probably be (n+1)^2 + 1 / n+1, while bottom one will be n^2+1 /n, then we use the formula for dividing fractions...

[white ocean]

i dont think it's worth checking tbh

as i said there's not much to gain for me whether or not i get to the correct result

i'd rather revise taylor series or something

[ornate skiff]

that's probably tougher than this

Well, your choice either way

[ornate skiff]

this is basically 1 + (n^2+n-1)/(n^3+n^2+n+1), which can be said directly to be > 1, as 2nd fraction is > 0 (without further proving it, as n > 1, and clearly increasing)

[white ocean]

the point is that doing 2+2 a million times will not make me any better at doing 2+2 than i already am

[ornate skiff]

I guess. I don't know if you're in highschool, or maybe early undergrad studies, or self-taught

might be beneficial for highschool or so

[white ocean]

first year of comp sci

[ornate skiff]

Then I guess it depends on the teacher. Some do cut to the bone for highschool mistakes, even if the big reasoning is there

Anyway, think the exercise is clear somewhat?

[white ocean]

yeah i think i understand the gist of it

i'm decently familiar with induction by now

you generally want to get back to P(n) on either side of P(n+1) and go from there

I was able to prove this exercise by doing that

with the caveat that i didn't prove it for a=-1

but only a>-1

since at one point i divide by 1+a

[ornate skiff]

Is this a different exercise?

[white ocean]

oh wait yeah nevermind

sorry it's getting late lol

[ornate skiff]

npnp

if you need help, ask

[white ocean]

it was the previous one where you had to prove Bernoulli's inequality

[white ocean]

if you wanna try your hand at this go ahead, but you really don't have to i think i did it in a decent way

plus i asked on a whatsapp group and i got another satisfactory answer

[ornate skiff]

fairly tough

after further consideration

nvm. How so, how did you do it?

[ornate skiff]

After even more consideration, it's (once again) simpler than previous one we did

[white ocean]

sorry i was heating up some slop i missed this

[ornate skiff]

I got to a(1+a)^n >= a, which I used the hypothesis for again, and ended nicely

$a(1+a)^n \geq a$

[flat frigate]

Wild123

[white ocean]

how did you get there?

[ornate skiff]

if you want, you can share how you did it. I think this one can be done in about 4 lines

so, we need to prove

[white ocean]

it might be in the trash i gotta check

[ornate skiff]

$(1+a)^{n+1}\geq 1+ (n+1)a$, no?

[flat frigate]

Wild123

[white ocean]

yeah

[ornate skiff]

$(1+a)^{n}+ a(1+a)^n\geq 1+ an +a$ ?

[flat frigate]

Wild123

[ornate skiff]

took a (1+a) to the front in left side, separated it. opened the brackets on right side. is it ok?

then half of it is done, so all that's left is to prove

$a(1+a)^n\geq a$

[flat frigate]

Wild123

[white ocean]

i'm not sure about the left side

how did you get 1+a out like that?

[ornate skiff]

$(1+a)(1+a)^n= (1+a)^n + a(1+a)^n$ ?

[flat frigate]

Wild123

[white ocean]

i did (1+a)(1+a)^n

[ornate skiff]

so open first bracket...

[white ocean]

oh right

ok i see

[ornate skiff]

first term on left side and first 2 terms on right side are handled

by hypothesis

[ornate skiff]

so we're left with this. for which, we use it again

[white ocean]

(1+a)^n >= 1

[ornate skiff]

not sure we can solve that, but it's worth a try. In fact, it looks impossible...partly because, a could be negative

and switch the sign

it's a little easier than that, imo

a(1+a)^n> a(1+an)

using the hypothesis again

which is a+ a^2n...but a^2 and n are positive, so that is > a

$a(1+a)^n> a(1+an) =a+a^2n> a$, I mean

[flat frigate]

Wild123

[white ocean]

this stall is occupied lol

[ornate skiff]

fairly certain you should post in one of the open channels

[tawny cedar]

Oh sorry

[ornate skiff]

like 25, 28, 36, 39 or 44

[white ocean]

allg

[tawny cedar]

Sorry!

[ornate skiff]

npnp

[white ocean]

the thing is with a being potentially negative it might get pretty complicated

[ornate skiff]

what do you mean

[white ocean]

oh wait nvm

yeah that works

[ornate skiff]

not always, as a could be 0 doe

nvm, the sign was >=, not just >

$a(1+a)^n\geq a(1+an) =a+a^2n\geq a$

[flat frigate]

Wild123

[ornate skiff]

Otherwise it was wrong again

[ornate skiff]

had a different solution in mind?

[white ocean]

at this point i dont even know anymore lmao

im following 2 different lines of solving this at the same time

and trying to reconcile it with what i was doing

[ornate skiff]

we can try your method. show me what you got

[ornate skiff]

ikr. albeit, the calculations were the only part of uni I liked

[white ocean]

(1+a)(1+a)^n >= 1 + na + a ↔ (1+a)^n >= (1+na+a)/(1+a) ↔ (1+a)^n >= 1+na -(na^2)/(1+a)

and since 1+na-(na^2)/(1+a) is less than 1+na

[ornate skiff]

looks good

[white ocean]

cant find the final steps but you get the idea

[ornate skiff]

I think it works, for a=/= -1, like you said

[white ocean]

yeah

[ornate skiff]

The final step is the induction...

(1+a)^n >= 1+na which of course is >= 1+na - whatever

[white ocean]

no this was the inductive step already

oh ok i see what you mean

[ornate skiff]

sorry, yea, I mean using the hypothesis, thingy

well, a= -1 seems to be an interesting case now

did you solve that?

[white ocean]

i get 0^n

[ornate skiff]

it's all good

[white ocean]

im just gonna use proof by waving my hand

[ornate skiff]

lol. 0^n is 0, innit

[white ocean]

yeah

[ornate skiff]

either way, proof by "it's obvious" seems to work a lot

[white ocean]

subsitute -1 for a in the original disequation

[ornate skiff]

so you have $0 \geq 1 -n$, which means $n \geq 1$, which is true

[flat frigate]

Wild123

[white ocean]

oh yeah mb i removed the -n not sure why

i think it might be a sign that its time to call it

thanks for everything tho

[ornate skiff]

you're welcome. Good luck!

[white ocean]

.close

[safe radish]

Channel closed

Closed by @white ocean

Use .reopen if this was a mistake.

[safe radish]

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.

[thin lion]

Hello

[thin lion]

I was wonderign if my approach for these is correct

so for part i) I used the ratio test and found that it to be converging to 1/3

which is less than 1

so absolute convergence

[quasi bison]

are they making you do this with epsilon shit

please say they aren't

[thin lion]

no no

[quasi bison]

good

yes ok you're good on (i) then

[thin lion]

for the second one, Alternating series test tells me that its convergent, but taking the absolute value, for large n, that the series behaves like harmonic

so divergence

hence conditionally convergent?

but idk how marking is distributed here

.close

[safe radish]

Channel closed

Closed by @thin lion

Use .reopen if this was a mistake.

[safe radish]

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.

[broken hazel]

may I ask something about kripte's model in modal logic?

[languid ermine]

just ask your question! its faster for everyone :)

[mighty mango]

you may 🤩 and you also may ask your question right away next time

[mighty mango]

😡

[broken hazel]

okay

wait a minute. Im traslating it

[quasi bison]

!redir

[safe radish]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[mighty mango]

irrelevant.

[viral scroll]

sorry guys

[broken hazel]

Show if it is possible: a valid formula in a model which is not valid in a frame of the model

I do know the definition but I can not still find an answer to that

A formula α is valid in a model ⟨M, υ⟩ (⟨M, υ⟩ ⊨ α) if and only if α is true at all index of the model (for every index i ∈ M, ⟨M, υ⟩ ⊨ᵢ α).

A formula α is valid in a frame M (M ⊨ α) if and only if α is valid in all models based on the frame (for every valuation υ on M, ⟨M, υ⟩ ⊨ α).

[broken hazel]

I wasn't sure how was it realted with maths since modal logic its kind of philophical logic

[safe radish]

@broken hazel Has your question been resolved?

[gray mulch]

Are you struggling to find an example?

What are you stuck on?

[broken hazel]

Yep

[broken hazel]

Well, I cannot find a way to find it

I understand what is a valid formula in a frame and in a model

Idk how to answer your question

[gray mulch]

Imagine a frame with just one world $W = {w_1}$ and no relations

[flat frigate]

a handsome russian dude

[gray mulch]

Let $\alpha = p$ be a propositional atom

[flat frigate]

a handsome russian dude

[gray mulch]

Consider a specific valuation $v$ where $p$ is true at $w_1$. Since $p$ is true at the only index in the model $\alpha$ is valid in the model $\langle M, v \rangle$

[flat frigate]

a handsome russian dude

[gray mulch]

Now to be valid in the frame, $\alpha$ must be true for every possible valuation $v'$. If I choose a different valuation $v'$ where $p$ is false at $w_1$, then the formula $p$ is not valid in the frame

[flat frigate]

a handsome russian dude

[broken hazel]

Thank you so much

[safe radish]

@broken hazel Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[tender horizon]

help please

[tender horizon]

last problem in my homework but i don't know what to do

i was thinking maybe insert a random number for the radius of the circles and maybe i'll get somewhere but not really

[mossy lotus]

just let the radius be r

[dark elm]

The outer circle is twice the inside one

It won't matter for the question what the size is, as long as one is twice the other

[tender horizon]

ik

[dark elm]

The key part of this question is working out what proportion of the slice is blue

[tender horizon]

1/8

[dark elm]

That's of the circle

Of the x degree slice of the circle

[tender horizon]

the blue "adds" in 45 degrees but idk about the non-blue part

[dark elm]

Just because the blue area is 1/8 of the circle doesn't mean its 45 degrees

since the blue area isnt a slice

[tender horizon]

oh ok

i dont know how to continue

[dark elm]

Do you know scale factors/scaling

[manic radish]

do you know how to calculate the area of a slice?

[tender horizon]

yeah i know scale factors

[dark elm]

Since the slice of the big circle has 2x radius it's twice as big (in length)

So what happens to the area of something which is twice as big

[tender horizon]

its 4 times bigger

[manic radish]

no, it's 4x bigger in area

it's 2x bigger in perimeter

[dark elm]

So if the area of the slice of big circle (white+blue) is 4x the slice of small circle (white) what's the proportion of blue and white for the slice?

[tender horizon]

3/1

[dark elm]

So the blue area is 3/4 of the slice of the big circle

[tender horizon]

yes

[dark elm]

And the blue area is 1/8 of the whole circle so you can divide one by the other to work out what proportion of the circle the slice is, and then what x is

Doing the scaling and ratio methods may seem a bit weird when you could use formulas for areas using pi and stuff but doing enough of these problems this way gives you the ability to quickly solve these if you do enough of them and get the method

The reason for dividing is that 1/8 is 3/4 of what you want, which is (1/8)/(3/4)

[tender horizon]

so 60?

[manic radish]

yes

[tender horizon]

ok thx

.close

[safe radish]

Channel closed

Closed by @tender horizon

Use .reopen if this was a mistake.

[safe radish]

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.

[earnest mirage]

Hello, could someone check if this proof is ok please?

\begin{Definition}[Divisibility]
A nonzero integer $a$ is said to \emph{divide} an integer $b$, written $a \mid b$, if $b = ak$ for some integer $k$.
\end{Definition}

% --------------------------------------------------------------------------------

\begin{Theorem}
Assume $a$ is an integer and $p$ and $q$ are distinct primes. If $p \mid a$ and $q \mid a$, then $pq \mid a$.
\end{Theorem}

\begin{proof}
Let $a$ be an integer, and $p$ and $q$ are distinct primes.
By the definition of divisibility, $a = pk = ql$ for some integers $k$ and $l$.
Since $p$ and $q$ are distinct primes, then $l$ must have $p$ as a factor, so $l$ can be written as $l = pr$ for some integer $r$.
Then, $a = q(pr) = (pq)r$.
Therefore, by the definition of divisibility $pq \ \mid \ a$.
\end{proof}

[dark elm]

Theorem 2 it should say q | a right?

Also you wrote "Pove that" (See below)

[arctic raven]

"Prove that" still doesn't really belong in any statement that's called a theorem

[dark elm]

Language and grammar mistakes don't make it mathematically wrong so you could get away with it :P

[arctic raven]

I was going to recommend removing the spaces around the \mids so I'm glad you did that

[flat frigate]

Mor Bras

[earnest mirage]

Fixed

[earnest mirage]

Also fixed

[arctic raven]

the writing is fine for someone who has just started writing proofs

[earnest mirage]

<@&268886789983436800>

[dark elm]

<@&268886789983436800>

(Scam)

[earnest mirage]

Does the proof looks good?

[dark elm]

Don't see any issues with it

Maybe write p divides l

[earnest mirage]

So, "since p and q are distinct primes, then p must divide l"

[dark elm]

Actually i think im nitpicking over nothing

[arctic raven]

I mean I kinda agree, "p divides l" is more concise than "l must have p as a factor"

[dark elm]

I was thinking that since you didn't define what a factor is but did define "divides" you should say it that way

[earnest mirage]

I see

Thank you for your responses, I'd like one more to check this proof to close the thread

[dark elm]

You can ping helpers role

[earnest mirage]

<@&286206848099549185> Does this proof looks good?

[spice canyon]

looks good to me.

don't take my word though.

[dreamy urchin]

Logically, it is good.

[earnest mirage]

Thank you all!

.close

[safe radish]

Channel closed

Closed by @earnest mirage

Use .reopen if this was a mistake.

[safe radish]

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.

[worn lily]

(do you have a question to ask?)

[nimble vine]

wrong

[astral glacier]

<@&268886789983436800>

Just gonna close the channel cuz OP doesn't seem to be responding anyway

.close

[safe radish]

Channel closed

Closed by @astral glacier

Use .reopen if this was a mistake.

[hard crest]

left the server innit

[astral glacier]

Didn't check that lol

[junior smelt]

Seemed like that was all they had to post in the server when I checked their message history

[astral glacier]

Lmfao

[worn lily]

a valuable contribution

[safe radish]

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.

[thin bridge]

Were you provided notes in class?

Send a question you're struggling with atm

and we'll guide you through it,
and yes send the list of formula you have access to so we can see if it's sufficient for what you're doing

[kind seal]

Youre stuck on arithmetic progressions?

[robust palm]

Can anyone help me with probability 🥲

[safe radish]

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

[thin bridge]

ok, first page second section from the bottom
show a question you're stuck on and the issue you have with applying those formulae

this is gp not ap

do you know which equation from your formula sheet to apply?

yes

use that to get your first two equations

You're given info about
$$a_{\red {3}}$$
to apply that, simply substitute $n=3$ into that equation

[flat frigate]

ραμOmeganato5

[thin bridge]

and you're told the value of that is 1/8, so set that equal to 1/8

write a = sign followed by that value

to substitute a value
replace all of the same variable present in the equation (in this case n)
with the desired value (In this case 3)

e.g.
2x + 3 = x + y
to substitute x=5
You'd do
2 * 5 + 3 = 5 + y

All of the n

you're given $a_{\red{3}}$

[flat frigate]

ραμOmeganato5

[thin bridge]

and since the formula has
$$a_{\red{n}} = \dots$$
Using $n=3$ gives you information

[flat frigate]

ραμOmeganato5

[thin bridge]

No

one equation by itself doesn't give you much, from that you'll still have two unknowns

using info from n=3 gets one equation
and from n=5 gives another
and using these two equations, only then will you be able to start determining a_1 and q

Yes

did you make the substitutions I mentioned

if so, what are you're equations

[safe radish]

@mild ember Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[light tiger]

How do I solve b of question 10?

[last heath]

can you translate?

[light tiger]

There is given its translation

[last heath]

it’s written in india

[hard crest]

it's also written in english...

[light tiger]

Not it's not written in Hindi it's in Nepali

[hard crest]

anyway factor the denominators then find common denominator

[light tiger]

I tried but I couldn't go to the right direction

[hard crest]

what did you get?

[west oyster]

Hey Guys. I hope i don't crash this discussion! I need help and i need a person that is very good at math and at explaining. I am in 8th grade and i am bad at math. I am not the best at math but i wanna be cause i am interested in math. I wanted to ask if someone can give tipps how to get better at math?

[idle anvil]

do you know how you could factor x² -1 e.g. ?

[idle anvil]

go to discussion channel or so

[light tiger]

Yup it's very easy
It's( x+1)(x-1)

[idle anvil]

nice, this should help you with that problem

[last heath]

ah shit right

[light tiger]

There is no that type of form

[idle anvil]

you got p² - 1 e.g.

[light tiger]

Kindly check question b
a is very easy I have solved it

[idle anvil]

oh my bad

[hard crest]

oh i was also reading the wrong one

[idle anvil]

ok for b u also wanna simplify basically

[light tiger]

So check my rough

[light tiger]

It's not that much easy

[buoyant dragon]

multiply and divide first term by a, second by b and third by c.. it'll help

[idle anvil]

whats the command to rotate image again?

[buoyant dragon]

,rccw

[light tiger]

Btw I check my last part and that was not wrong mathematical
But that was not at the point for that question

[buoyant dragon]

@light tiger

[quasi bison]

I gane up

gane?

[buoyant dragon]

gave*

[light tiger]

Give up

[light tiger]

Kindly do this in mathematical 🥲

[quasi bison]

learn to write your v so that it doesn't look like n

[buoyant dragon]

$\frac{1}{a+1}=\frac{a}{a(a+1)}$

[flat frigate]

Shikhar

[buoyant dragon]

ah chill.. he's learning math, not english

[idle anvil]

tbf his writing is quite neat, owhat would Ann say about my Atrocity

[light tiger]

Yup even though I'm not English medium student.

[buoyant dragon]

still far better writing than me 💀 (english medium student)

[light tiger]

Oh sorry, I tried to write it fast

[light tiger]

Should I do it for a, b and c to the LFH respectly?

[buoyant dragon]

yes

[light tiger]

Ok lemme do

Hein after it what do I do?

[idle anvil]

see if u can substitute

[idle anvil]

,rccw

[idle anvil]

,rccw

[idle anvil]

,rccw

[idle anvil]

nice

If U see that whatever ur currently doing is way more complicated then it probably should be then its often helpful to take a step back adn see if theres a better approach

[nimble vine]

,rcw to rotate clockwise

[idle anvil]

ohh

thx

[safe radish]

@light tiger Has your question been resolved?

[light tiger]

No, it's already very simple

I was just on another direction of solution for this question 🥲

[safe radish]

Channel closed

Closed by @light tiger

Use .reopen if this was a mistake.

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.

[quasi bison]

. by the way there's also ,rcw for rotating clockwise

[earnest mirage]

Hello, could someone check if this proof looks good please?

\begin{Definition}[Divisibility]
A nonzero integer $a$ is said to \emph{divide} an integer $b$, written $a \mid b$, if $b = ak$ for some integer $k$.
\end{Definition}

\begin{Definition}[Congruency]
For integers $a$, $b$ and $n$, we say that $a$ \emph{is congruent to $b$ modulo $n$}, and we write $a \equiv b \pmod{n}$, if $n \mid (a - b)$.
\end{Definition}

% --------------------------------------------------------------------------------

\begin{Theorem}
Assume $a$, $b$ and $n$ are positive integers. If $a \equiv b \pmod{n}$, then $a^2 \equiv b^2 \pmod{n}$.
\end{Theorem}

\begin{proof}
Let $a$, $b$ and $n$ be positive integers and $a \equiv b \pmod{n}$.
Then, $a^2 \equiv ab \pmod{n}$ and $b^2 \equiv ab \pmod{n}$.
By definition of congruency, this means that $n \mid (a^2 - ab)$ and $n \mid (b^2 - ab)$.
So, by definition of divisibility, $a^2 - ab = nk$ and $b^2 - ab = nl$ for some integers $k$ and $l$.
Then, $ab = a^2 -nk = b^2 - nl$.
\begin{align*}
ab = a^2 -nk &= b^2 - nl \
a^2 - b^2 &= nk - nl \
a^2 - b^2 &= n(k - l) \
\end{align*}
Since $k$ and $l$ are integers, so is $k - l$.
Then, by definition of divisibility, we have that $n \mid (a^2 - b^2)$.
Therefore, by the definition of congruency, $a^2 \equiv b^2 \pmod{n}$.
\end{proof}

[glacial cairn]

It seems odd to me that you would use those definitions for that theorem instead of first proving that congruence modulo n is a equivalence relation, and so is transitive

But if that's how you're asked to prove it, then I suppose what you did is fine

In definition 2, it should be n | (a-b) instead of m | (a-b)

And the start of the proof should read "Let a, b and n be positive integers..."

[flat frigate]

Mor Bras

[earnest mirage]

Thanks for pointing that out

<@&286206848099549185> Hello, could someone else check if this proof looks good?

[marsh siren]

looks goodi think

[safe radish]

@earnest mirage Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[safe radish]

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.

[errant hearth]

i need help with discrete, set operations

[errant hearth]

specifically the generalized union thing

[delicate shore]

What do you need help with?

[errant hearth]

generalized union

i dont understand it tbh

[delicate shore]

What about it do you need help with?

Well do you understand what the union of two sets is

[errant hearth]

yup

aUb = a{both numbers of a and b}

[delicate shore]

Sort of

A ∪ B is the set of elements who belong to either A or B

[quasi bison]

(these elements do not have to be numbers)

[delicate shore]

Do you see a natural way to generalise this to 3 sets or more?

[errant hearth]

wdym

[delicate shore]

can you guess what A ∪ B ∪ C means for sets A, B, C?

[errant hearth]

means, all all elements of a mixed with b mixed with c

[delicate shore]

"mixed with" is not quite the accurate terminology but I suppose you have the idea

It is just the set of elements who belong to either A, B, or C

If I give you A = {1, 2}, B = {3, 4, 5} and C = {4, 8, 9} can you tell me what A ∪ B ∪ C is?

[errant hearth]

yes

{1,2,3,4,5,8,9}

[delicate shore]

Ok so you understand the union of 2 sets and the union of 3 sets, the "general union" is just extending this in the obvious way to unions of even more sets

Specifically, the union of a family of sets A_1, A_2, A_3, ... is just the set of elements who belong to at least one of these A_i's

Does this make sense?

[errant hearth]

no

lets say we have the sets A1 A2 A3 A4, B1 and we said A union

A union 5, will it take in the elements of A1, A2,A3,A4, and B1, or just the eleemts of the A sets

[delicate shore]

What is "A union"?

are you familiar with what an indexed family is?

[austere goblet]

you mean $\bigcup A_i$?

[flat frigate]

Hanako(x, y); ∂(fox)/∂x

[austere goblet]

or smth else?

[errant hearth]

like this

[delicate shore]

I think they are tripped up by the fact that we call the elements of the family A_1, A_2, A_3, etc instead of A, B, C

but it is just a notational convenience

[errant hearth]

but we only had 4 sets named A, and the fifth was named b1

[delicate shore]

Then what you wrote does not make sense because A_5 is not defined

[delicate shore]

Remember, this notation is just a shorthand for A_1 ∪ A_2 ∪ A_3 ∪ A_4 ∪ A_5

[errant hearth]

thats what am asking, will it only take the A sets

[austere goblet]

as written, your union includes A_1 to A_5, no mentions of B. but this union also makes no sense because A_5 is not defined, as plante mentioned.

[errant hearth]

even tho we have a b set

i think i got it

but

another thing is i see alot of times it says smth like this

what do we mean {n,n+1}

[austere goblet]

as written, each A_n has only two elements, n, and n+1

[errant hearth]

but then the first n would be stuck as 1

ih wait nvm

i got it

[austere goblet]

anything else then?

[errant hearth]

nope

ty'

[austere goblet]

!done, if so

[safe radish]

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

[austere goblet]

.close by OP agreement

[safe radish]

Channel closed

Closed by @austere goblet

Use .reopen if this was a mistake.

[safe radish]

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.

[winged flare]

huh

[winged flare]

not even sure if its in 1^infinity form

ok nvm googled it

.close

[safe radish]

Channel closed

Closed by @winged flare

Use .reopen if this was a mistake.

[peak estuary]

well the exponent goes to 0

[astral glacier]

But the thing being exponentiated goes to an indeterminate form no?

[winged flare]

yeah

actually .reopend

.reopen

[nimble vine]

-# lopital

[safe radish]

✅ Original question: #help-23 message

[winged flare]

for this

without cheating

[peak estuary]

what do you consider cheating

[winged flare]

8/2 + 8/3 is almost 6

[peak estuary]

isnt this like the arctan series or something?

[timber karma]

reminds me of arctan

[peak estuary]

I can never remember all those series

[winged flare]

ohg

thanks

.close

[safe radish]

Channel closed

Closed by @winged flare

Use .reopen if this was a mistake.

[winged flare]

just how many of these exist bruh

[astral glacier]

Too many

[safe radish]

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.

Channel closed

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

[quasi timber]

[safe radish]

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.

[errant hearth]

guys

[errant hearth]

i need help

oh

sorry revolvman

[marsh bloom]

its ok

[errant hearth]

wait here take it

[marsh bloom]

im just trying this out for the first time

[errant hearth]

!end

[marsh bloom]

ty

[errant hearth]

end!

[marsh bloom]

sorry i dont know how this works

[split kayak]

just ask the question

[marsh bloom]

Hi I have this logic (r→(s∨p))∧(r∧q)∧(p→¬q) and I am supposed to prove that it's true when using both the deduction and the reduction method for logic. I have made steps for both but im getting stuck on the later parts so I would like some help

[errant hearth]

.close

[safe radish]

Channel closed

Closed by @errant hearth

Use .reopen if this was a mistake.

[marsh bloom]

gimme a sec and i will show you my work

[errant hearth]

fuck did i break it

.reopen

[safe radish]

✅ Original question: #help-23 message

[astral glacier]

@marsh bloom claim another channel

.close

[marsh bloom]

ait

[safe radish]

Channel closed

Closed by @astral glacier

Use .reopen if this was a mistake.

[errant hearth]

sorry revolverman

okay so

[astral glacier]

No worries, we aren't running out

[errant hearth]

i need help with this

[lean otter]

Yall

[errant hearth]

1. Let A be the set of students who live within one mile of
   school and let B be the set of students whowalk to classes.
   Describe the students in each of these sets.
   a) A ∩ B b) A ∪ B
   c) A − B d) B − A

[astral glacier]

Oh you do need help?

.reopen

[lean otter]

Yall welp

[safe radish]

✅ Original question: #help-23 message

[astral glacier]

Claim another channel please

[errant hearth]

yes pls

[lean otter]

Idk how to

[errant hearth]

go to help 7

[astral glacier]

#help-44｜stanley-🌲-v2-dans

Just message there

[errant hearth]

and typep anything

[open wedge]

!occupied

[safe radish]

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

[errant hearth]

ok so

1. Let A be the set of students who live within one mile of
   school and let B be the set of students whowalk to classes.
   Describe the students in each of these sets.
   a) A ∩ B b) A ∪ B
   c) A − B d) B − A

[astral glacier]

It's pinned no need to repost

[open wedge]

aight

[errant hearth]

what does it mean describe the students in each of these sets

[open wedge]

basically it's just listing the conditions the set has

for example: in a)

[errant hearth]

like do they want me to type it in tahh

i forgot its name

[open wedge]

A is the set of students who live within 1 mile of school

and B is the set of students to walk to classes

[errant hearth]

in set builder notation ?

[open wedge]

then $A \cap B$ is the set of all students who live within 1 mile of school and walk to classes

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

im not sure about that

they told you to describe

[errant hearth]

yea but like what do they mean by describe

[open wedge]

probably say smth like $A \cap B$ are the students who ...

[flat frigate]

1 divided by 0 equals Infinity

[errant hearth]

ahaaaa

alr bet

ty

[open wedge]

!done

[safe radish]

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

[errant hearth]

.close

[safe radish]

Channel closed

Closed by @errant hearth

Use .reopen if this was a mistake.

[errant hearth]

!done

[safe radish]

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

[errant hearth]

.close

ok so back to the topic

.reopen

2. Suppose that A is the set of sophomores at your school
   and B is the set of students in discrete mathematics at
   your school. Express each of these sets in terms of A
   and B.

[safe radish]

✅ Original question: #help-23 message

[errant hearth]

what is the way to solve this

like i solved it but am not sure if my answers are right

[mighty mango]

you... shouldnt close the channel unless you are done

[open wedge]

new question

[mighty mango]

!clopen

fuck

!cloppen

fuck

[errant hearth]

nope same one, just want to be sure i solved it right

.reopen

.open

[open wedge]

he knows how to reopen dw

[mighty mango]

ok dont close the channel then reopen it

[errant hearth]

!.reopen

[mighty mango]

you opened it

[errant hearth]

sorry

[open wedge]

you already reopened your channel

[mighty mango]

sorry caps

[open wedge]

!occupied

[safe radish]

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

[open wedge]

YES

[lean otter]

Guys is there a shorter way

[open wedge]

!occupied

[safe radish]

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

[open wedge]

please don't interupt other channels

[errant hearth]

https://tenor.com/view/everyone-breathe-lets-all-breathe-nwabudike-bergstein-baron-vaughn-grace-and-frankie-gif-17555665

[open wedge]

chop chop or mods mods

[limpid lodge]

Meditate and let math go

[open wedge]

lmao

[errant hearth]

real, become shaolin monk and let go of all worldy rewards

[open wedge]

this is my favorite phrase rn

[limpid lodge]

Nah only math needs to be let go of

[errant hearth]

wish i can

anyways, whats the solving for this

2. Suppose that A is the set of sophomores at your school
   and B is the set of students in discrete mathematics at
   your school. Express each of these sets in terms of A
   and B.

[primal bone]

Express each of... what sets?

this question as phrased here reads the same as
"An apple costs $1 and an orange costs $1.20. Express the cost of each fruit."

[errant hearth]

wait sorry

[primal bone]

It's not really a maths question atm 😅

[errant hearth]

2. Suppose that A is the set of sophomores at your school
   and B is the set of students in discrete mathematics at
   your school. Express each of these sets in terms of A
   and B.
   a) the set of sophomores taking discrete mathematics in
   your school
   b) the set of sophomores at your school who are not taking
   discrete mathematics
   c) the set of students at your school who either are
   sophomores or are taking discrete mathematics
   d) the set of students at your school who either are not
   sophomores or are not taking discrete mathematics

[errant hearth]

discrete math

[primal bone]

(I mean at that stage it was just a comprehension question lol)

!status

[safe radish]

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

[errant hearth]

3

[primal bone]

And if it's 2-4,

!show

[safe radish]

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

[errant hearth]

i didi it on my note book tho

wait let me write them out

[errant hearth]

a)A⊂B

b)B-A

c)A∪B

[hard crest]

use of this gif may impede speed of help

[errant hearth]

sorry

d)B-A∪A-B

[safe radish]

@errant hearth Has your question been resolved?

[safe radish]

@errant hearth Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[safe radish]

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.

[craggy pasture]

im kinda confused whats going on here?

[craggy pasture]

how is derivative of g`(x^3) get u e^sqrt(x^3)?

what?

[plucky elk]

chain rule and FTC

[craggy pasture]

ok but why do we not end up with the same answer

[plucky elk]

did you plug in t

t is different from x

[craggy pasture]

huh?

the definite integral is not anywhere close

oh wiat so integral of derivative gives you back the antiderivative or the original func and derivative of the integral also gives you the original function evaluated at the boundary?

if you do the integral of the original definite integral is when you get the area?

[safe radish]

@craggy pasture Has your question been resolved?

[plucky elk]

yes definite integral gives you the signed area if the function is positive on the domain of integration then the signed area and area are the same

[craggy pasture]

ok 1 more question what does instantaneous rate of change exactly mean because that makes 0 sense

how can you have the rate of change at 1 instance you need 2 points

so there is no change at 1 point

like the definition here right this makes sense we have some tiny change - s(t)

divided by the tiny change width right

but as it approaches 0 we get the instantaneous rate of change but that makes no sense to me since if dt approaches 0 and is 0 there is no change so how can you have a instantaneous rate of change

[plucky elk]

the limit lets you define a new way to calculate slope at a point

read https://tutorial.math.lamar.edu/Classes/CalcI/Tangents_Rates.aspx

and
https://tutorial.math.lamar.edu/Classes/CalcI/TheLimit.aspx

[floral yew]

This is average rate of change in some interval

[craggy pasture]

yeah but we are just plugging in 0

as dt goes to 0 what we say

and get some value if continuous but that doesnt make sense to have some rate of change at 1 particular point

[floral yew]

you cant just plug in 0 it wont be defined you use limits for that

[safe radish]

@craggy pasture Has your question been resolved?

Channel closed

Closed by @craggy pasture

Use .reopen if this was a mistake.

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.

[copper shadow]

how can this be? (x^1/n)^ 1-n = x^(1-n)/n not x^1/n-1. Say we had x^1/3. its derivative is 1/3x^-2/3 which we get with the first one but with the second one we get 1/3 x^1/2

[tardy mango]

pretty sure it's supposed to be $x^{\frac{1}{n}-1}$

[flat frigate]

Civil Service Pigeon

[tardy mango]

aka $x^{(1/n)-1}$

[flat frigate]

Civil Service Pigeon

[tardy mango]

slightly bad typesetting (but also if there is more than one term in the denominator, there should be parentheses around the denominator)

[copper shadow]

oh I see. Thanks

[tardy mango]

[copper shadow]

.solved

[safe radish]

Channel closed

Closed by @copper shadow

Use .reopen if this was a mistake.

[safe radish]

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.

[eternal breach]

can someone explain to me what are surface integrals and what is the difference between them and regular contour integrals

[eternal breach]

I'm referring to this

[plucky elk]

https://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx

[eternal breach]

nooo look at the pic I sent

[dull radish]

what are regular contour integrals to you?

[eternal breach]

line integrals over a closed curve

[plucky elk]

read example 2

[dull radish]

okay, so the image you sent contains a surface integral over a closed surface and a volume integral over a "closed volume" (which is a bit hard to imagine)

[eternal breach]

putting the triple integral aside, how are you supposed to calculate a surface integral over a closed surface. I mean for contour integrals you basically calculate the line integral which is done by parametrizating the curve, but here how do you actually do that ? it's hard to picture

[eternal breach]

okayyy

[dull radish]

read what riemann sent to you

instead of parametrizing a curve, you parametrize a surface

[eternal breach]

ohhhhh I get it now

thanks y'all for your help !

.close

[safe radish]

Channel closed

Closed by @eternal breach

Use .reopen if this was a mistake.

[safe radish]

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.

[copper temple]

hi ,

[copper temple]

I didn't understand this proove

[buoyant dragon]

image sending?

[copper temple]

we have all x,y,x* ,y* in R

∆x=|x-x* | , ∆y=|y-y*|

But here he supposed x* ,y* in R * + witch is equivalent x* >0 ,y*>0

[buoyant dragon]

is it related to nnumerical approximations?

[copper temple]

The proove did'nt conviced me .

Yeah

∆(x+y)=∆x+∆y

[copper temple]

yes the proove did'nt conviced me

[obsidian oracle]

Honestly I don't see what the use for x* ,y* in R+* is other than the second part about relative error

[copper temple]

it's argument of (∆x=|x-x* | so ∆x>=|x-x* |) and (∆y=|y-y* | so ∆y>=|y-y* |)

[copper temple]

it's for absolute errors

[obsidian oracle]

are you sure? where did they use the hypothesis that x* and y* were positive?

[copper temple]

firstly they didn't supposed that x* , y *>0

then they supposed x* , y * >0

But's that's not a complet proove if they based on that because it's supposed to proove for x * ,y* belong to ]-infinity;+infinity[

[copper temple]

and then from that they deduced inequality

(x∗ +y∗) −(∆x+∆y) ≤ x+y ≤(x∗+y∗)+(∆x+∆y) here we have no problem

but when they did say that (∆x+∆y) =∆(x+y)

[obsidian oracle]

exactly. So I'm telling you don't need it

[copper temple]

here is the missing connection of argument .

[obsidian oracle]

once you get this

you exactly get that $|(x+y) - (x^+y^)|\leq \Delta x + \Delta y$

[flat frigate]

Rafilouyear2026

[obsidian oracle]

that's the definition of the absolute error of x+y

[copper temple]

Yes

[obsidian oracle]

$|(x+y) - (x^+y^)|\leq \Delta (x+ y)$

[flat frigate]

Rafilouyear2026

[obsidian oracle]

So Delta(x+y) and Delta(x)+Delta(y) are the same

[copper temple]

wait

∆(x+y)>=|(x+y)-(x* +y* )|=∆(x+y) because ∆(x+y)=∆(x+y) and (∆x+∆y) >=∆(x+y) so with antisymetrics of the relation

=

we get the equality ∆(x+y)=∆(x)+∆(y) .

[obsidian oracle]

Yeah I guess it works

[copper temple]

Thank's so much ,

I dont like the prooves that jump arguments .

Now it's completely certain .

No it's not certain there's a mistake :

[copper temple]

∆(x+y)>=|(x+y)-(x* +y* )|= ∆(x+y) and ∆(x+y)=<∆x+ ∆y

but it's missing :

∆(x+y)>=∆x+ ∆y in order to deduce by the asymetric of relation >= the equality : ∆(x+y)=∆x+ ∆y
so I'm not convinced

@obsidian oracle

(∆x=|x-x* | so ∆x>=|x-x* | so ∆x>=x-x* >=-∆x so ∆x+x* >=x>=-∆x+x* ) and (∆y=|y-y* | so ∆y>=|y-y* | ∆y>=y-y* >=-∆y so ∆y+y* >=y>= -∆y+ y * ) implies ∆y+y* > =y>= -∆y+y* and ∆x+x* >=x>=-∆x+x* it implies ∆y+y* + x * + ∆x>=x+y>=-∆x+x* -∆y+y* it implies

∆y+y* + x * + ∆x- x* -y* =∆x+∆y >= x+y-x* -y* >=-∆x+x* -∆y+y* -x * -y * =-∆x-∆y

it implies :

∆x+∆y >= x+y-(x* +y*)>= -(∆x+∆y)

[obsidian oracle]

I realized something, if your definition of absolute error is exactly ∆x=|x-x* | then you won't have equality

between ∆(x+y) and ∆x + ∆y

you can only have ∆(x+y) <= ∆x + ∆y

[copper temple]

Yes

[copper temple]

But a proove is maybe triangular inequality ?

[obsidian oracle]

yes triangular inequality is enough

$|x+y - (x^+y^)| = |(x-x^) + (y-y^)| \leq \Delta x + \Delta y$

[flat frigate]

Rafilouyear2026

[copper temple]

let's try a counter exemple :ah yes

[obsidian oracle]

x = x* + 0.5 and y = y* - 0.5

Absolute error of the sum is 0

sum of absolute errors is 1

because the errors counterbalanced

[copper temple]

Yes ,so they teach us something false

I will try to find a good textbook .

[obsidian oracle]

Yeah, the textbook you're looking at seems quite badly written too

latex-wise I mean

[copper temple]

I lost confidence of my textbook

[safe radish]

@copper temple Has your question been resolved?

Channel closed

Closed by @copper temple

Use .reopen if this was a mistake.

[safe radish]

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.

[sonic marten]

can someone help me find r im lost

[dusky gale]

Sure

Ok what do we know right now? What information is readily present?

[sonic marten]

like the right angles and the side lengths

idk

[dusky gale]

Ok yes that is correct, but is there anything else we can note? Particularly about r relating to the triangle?

[frigid spruce]

make use of the cartesian plane

[sonic marten]

like perpendicular bisectors ?

[sonic marten]

sorry i dont know what that is

[dusky gale]

Cartesian plane is the coordinate system

But first

Look at the circle. What do we know about it geometrically speaking that relates to the shape its inside of?

[sonic marten]

like 4 perpendicular bisectors

[dusky gale]

Hint: look from top to bottom

[sonic marten]

forms two squares?

[dusky gale]

Hint: try to derive the height of the left side of the shape

[sonic marten]

oh idk how to do that can you tell me

[dusky gale]

Nope. Just look at the shape and try to think about how you could measure its height, even if you dont know the absolute length

[sonic marten]

like use the radius somehow

[dusky gale]

Yes thats good

[sonic marten]

like 2r

[dusky gale]

Correct!

So now we know that the height of the shape is 2r

[sonic marten]

ya

[dusky gale]

Lets set up a mental image on the coordinate plane

Where would (0, 2r) be located if the bottom left of the shape was on (0,0)?

[sonic marten]

top corner

[dusky gale]

Correct

Now where would the top right corner be?

[sonic marten]

(3,2r)

[dusky gale]

Yup. Now the bottom right corner?

[sonic marten]

(7,0)

[dusky gale]

Awesome, now we have the shape on a coordinate plane

Now where would the center of the circle be?

[sonic marten]

r,r

[dusky gale]

Yeah thats right!

Now how can we use this mental image of the shape on the coordinate plane to find r?

[sonic marten]

how?

i was trying to do it with similar tryangles

i never learnt this coordinate stuff

[dusky gale]

So we have the top right corner, and the bottom right corner. That means we have a line.

What can we note about this line? Can you describe it?

[sonic marten]

slope is -r/2

[dusky gale]

How did you get that?

[sonic marten]

oh am i wrong sry

[dusky gale]

Are you?

Explain where you got that slope from

[sonic marten]

two points

y2-y1/x2-x1

[dusky gale]

Yeah thats right

[austere goblet]

to OP: it's best practice to also include how you get things unless they are rather blatantly obvious, or have been repeatedly stated

[sonic marten]

oh i missread you are right

[dusky gale]

In order to understand how youre thinking, I need to know what your process is.
I could just tell you if youre right or wrong, but it would be better if I let you figure it out for yourself.

[sonic marten]

true true thank you

[austere goblet]

most of the time the journey is more important than the destination

cuz sometimes you get the answer but used the wrong process. if the helper says you're right, you might think your process is right when it's not necessarily that

so yeah

[dusky gale]

Very true, thank you hanako.

Ok, now that we have the slope of that slanted line, what can we do with that?

[sonic marten]

i dont really know

[dusky gale]

Try to take a look at the center. Where is it located within the shape?

[safe radish]

@sonic marten Has your question been resolved?

Channel closed

Closed by @sonic marten

Use .reopen if this was a mistake.

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.

[cosmic sparrow]

hi, im making a game and i was wondering if someone could help me validate some functions propertys, the setup: We have an infinite chessboard-like grid of square cells (tiles). Some cells are blocked (walls), others are free (floor).

[cosmic sparrow]

the game is just the context, coders might understand it better, im just asking about the properties of the vector A to X+1 in this setup

[safe radish]

@cosmic sparrow Has your question been resolved?

[cosmic sparrow]

We have a floor made of square tiles like graph paper. Each tile is either blocked (a wall) or free (floor).

A player is not a point. The player is a 1×1 square (same size as one tile), always axis-aligned. The player’s position is described by the center of that square.

When the player moves from point A to point B, the center travels along the straight line segment from A to B. During that movement, the 1×1 square sweeps a “tube-shaped” region. A move is valid if, throughout the whole motion, the swept 1×1 square never overlaps any blocked tile.

Now suppose A* gives a path of tile-centers:
\ldots,\ X-1,\ X,\ X+1,\ \ldots
where each step between consecutive waypoints moves by at most 1 tile in row and/or column (8-neighbor steps).

Assume the straight move from A to X is valid (the tube from A to X touches only free tiles).

Also assume the last two A*(manager) steps point in the same 8-direction, meaning:
• the step from X−1 to X and
• the step from X to X+1
are the same direction (example: both are “down-right”).

Question:
Does it necessarily follow that the straight move from A to X+1 is also valid, except possibly for collisions occurring only at the new final tile near X+1?

thats question 1

question 2 Same grid of square tiles: each tile is either blocked or free.

The player is a 1×1 axis-aligned square. Position is the center of that square.

Moving from point A to point B means the center travels along the straight line segment A\to B. During the move, the 1×1 square sweeps a “tube” region. A move is valid if this swept region never overlaps any blocked tile.

Now suppose we already know that moving straight from A to B is valid (so all tiles the swept 1×1 square touches along A\to B are free).

Let C be a nearby target obtained by moving B by exactly one A* step in an 8-neighbor direction (up/down/left/right/diagonal by 1 tile).

Question:
Is it true that the straight move from A to C can only touch new tiles in a small “border strip” on the side corresponding to the direction B\to C?
Equivalently: do the tiles touched by A\to C differ from the tiles touched by A\to B only by tiles in a predictable one-tile-wide side band, rather than changing somewhere deep inside the already-cleared corridor?

sorry if some things are unclear! im not used to writing

[safe radish]

@cosmic sparrow Has your question been resolved?

[devout scroll]

No, imagine you have
A=(1,1)
X-1=(1,2)
X=(2,2)
X+1=(3,2)
And that the tile at (3,1) is a wall, but nothing else

[devout scroll]

Can you explain what exactly you mean by one tile wide band?

[peak estuary]

a picture could really help here

[wispy berry]

i am going to get banned

[safe radish]

@cosmic sparrow Has your question been resolved?

[cosmic sparrow]

il try to cook an image

something like that, theres a path from player to where you tapped by tile centers so each block is a tile, the player is the same as a cube

[cosmic sparrow]

like this is the player

if his not exactly in the center then his position can cross 4 blocks

[devout scroll]

So the width of the border strip is two tiles, right?

[cosmic sparrow]

yeah but im saying that it would always work except the last block(what you said) like if the block next to the x+1 block IS floor it will always allow it, or alternatively im saying that for the length of A to X we dont need to recheck that part for A to x+1 just the last one, so if the edge case of near the last block IS true, it holds true?

[cosmic sparrow]

2? a block is 1x1 and player is 1x1 the player can be anywhere so it can collide with up to 4 blocks, the walls stay gridlocked

from the example you gave it seemed you got it

lets say like if a to x spans x blocks qnd its true then for this length(excluding the neighbor of x+1) will x+1 also be true?

did i explain it well? im trying to

[devout scroll]

Hmm, I still think the answer is no

Consider

A=(0,0)
X-1 = (-1,10)
X=(0,10)
X+1=(1,10)
And there is a wall only at (1,5)

Then I think it doesn't work

[cosmic sparrow]

yes BUT you forgot the one assumption i had set, that the 8 way direction of X-1 to x is the same as x to x+1, here the direction is different

[devout scroll]

It's the same direction

[cosmic sparrow]

oh wait

yes your right.. but if we said the line is logical? like its the fastest path (it uses A star) so it wouldnt turn left just to turn right like that

[devout scroll]

It's still false,
(0,0)
(0,30)
(1,30)
(2,30)

Block at (2,25)

[cosmic sparrow]

yep but if 0,0 to 1,30 would have been true it wouldnt have gone from 0,0 to 0,30 to begin with it would have gone to 1,30 straight and the x there would have a different direction i think, because its like an 8 way direction in such a situation its first move if 0,0 to 1,30 before us even recieving it would have been B=1,1

im trying to abuse the properties of this function(fastest 8 way tile path)

im sorry im struggling to explain all the factors, is it somehow clearer? @devout scroll i appreciate the help

do you know a bit of code? if no i could generate a map and a point player and point goal, and you would choose from said generated path any x and x+1

im saying for ANY map with ANY goal and ANY player starting point for ANY 2 points chosen FROM the generated fastest 8 way path(that i can run on my end or you can draw) that are consecutive (x and x+1 ) AND have the same 8 way direction from x-1 to x==x to x+1 THEN aside from the last neighbor tile to x+1 the path A to X+1 is safe

[cosmic sparrow]

@devout scroll ?-?

[safe radish]

@cosmic sparrow Has your question been resolved?

[cosmic sparrow]

😭

[open wedge]

see path-finding algorithms

[cosmic sparrow]

what?

[open wedge]

But i don't really understand what OP's trying to ask here. Like i know the mechanics but just don't understand the question

[cosmic sparrow]

@open wedge here its still unclear?

im trying my best

[open wedge]

Hold uo

[open wedge]

What if your target destination is surrounded by 8 neighbouring walls

[cosmic sparrow]

it obviously wont execute, the pathfinding algorithm already works though thats not the issue

im just planning to smooth my entire path so it acts like float(shortest real path from shortest tile path) and i came up with 2 properties that if held true they make my algorithm like 3 times more efficient so im asking if their true

hello?

[cosmic sparrow]

World. Any grid map of blocked/free unit tiles.
Player. A 1×1 axis-aligned square, described by its center point; a straight move from P to Q is “safe” if the swept 1×1 square never overlaps a blocked tile.

Path. Let X_0, X_1, \dots, X_n be the optimal 8-way A* path from start to goal, with waypoints at tile centers and steps in the 8-neighborhood.

Pick an anchor. Let A be any waypoint X_i on that path (typically the last “committed/smoothed” waypoint).

Condition. Choose three consecutive waypoints X_{k-1}, X_k, X_{k+1} on the A* path such that the last two step directions are the same:
X_k - X_{k-1} = X_{k+1} - X_k.

Assumption. The straight-line move A \to X_k is safe.

Claim (what you want to test).
Then the straight-line move A \to X_{k+1} is also safe, except possibly for collisions occurring within the final “new” tile neighborhood near X_{k+1} (i.e., any newly introduced collision can only happen at the very end, not earlier along the segment).

is it understandable yet? anyone?

[sudden lotus]

weird flag tbh

dont know it ...