help-27

[lament kraken]

Ok what else?

!xy

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[lament kraken]

If applicable

[barren quiver]

sorry it's kinda hard to take a picture and put it on discord

[lament kraken]

uh its ok

[misty crest]

maybe put roblox away

[lament kraken]

Just tell us the question

[barren quiver]

the question says find area of sector CAD which is an arc

[lament kraken]

mhm

[barren quiver]

and A is the center of the circle

[lament kraken]

Okay

[barren quiver]

12 is the radius

just found it

[lament kraken]

Okay

So do you know how to express the area of an arc as a ratio of the area of a circle

[barren quiver]

yes

[lament kraken]

Its 64/360 right

[barren quiver]

correct

[lament kraken]

Okay

Do you know how to find the area of a circle with the radius?

[barren quiver]

yes pie x radius squared

[lament kraken]

ye

[misty crest]

pie

[lament kraken]

Okay can you calculate that for me

(its π btw pronounced pi)

Note that once we have area of circle, we can multiply it by 64/360 to find the arc area

[barren quiver]

oh thankyou

the area is 452.4

and the sector is 77

thanks for the help

.close

[devout snow]

Channel closed

Closed by @barren quiver

Use .reopen if this was a mistake.

[lament kraken]

Np

[devout snow]

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[ancient pollen]

d

[coarse current]

Could someone please explain the continuity proof and how to construct similar proofs for other functions? This is on the exam

But I wasn't taught it

Oop someone beat me to it

[pseudo basin]

,rccw

[coarse current]

@ancient pollen what do u need help with

I'll try to help

[pseudo basin]

,rcw

[coarse current]

Is that tryna say the trapezoidal rule

[ancient pollen]

alr

[coarse current]

If so that's unfortunate, I hate that method

[ancient pollen]

yep

composite

[coarse current]

it's just boring lol

Haven't used it in yeaaaars

[ancient pollen]

ye 😔

but the question says “use composite trapezoidal rule”💔

[coarse current]

fair

[vague idol]

What help do U need with?

[ancient pollen]

i got 180

[coarse current]

For that stuff if you don't get it right, ask some AI like Gemini, they tend to be okay at this kinda stuff

Like checking it over and spotting errors

[vague idol]

!nogpt

[devout snow]

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

[vague idol]

or idl

[pseudo basin]

<@&268886789983436800> this user has been acting kind of spammy and disruptive in multiple channels

[coarse current]

I think GPT does suck tbh

I found AI to be okay in terms of like, Gemini

[vague idol]

Wth did I do

[coarse current]

For maths and coding

[vague idol]

I just said no gpt and I asked what help he need

or they

[coarse current]

Do you not agree AI like Gemini could be useful for numerical methods

[ancient pollen]

i used

[coarse current]

I don't trust it for getting concepts and etc

[ancient pollen]

pie radian

I think we use the actual value of pie

[pseudo basin]

pi, not pie

anyway it seems like you did the problem to completion

[narrow bloom]

seeing as they themselves have done AI-generated answers in these channels yeah

[pseudo basin]

was this an "i want my work checked" situation @ancient pollen

[narrow bloom]

this message contains this server's policy on AI, I don't think anyone denies that it can be an useful tool, now please stay on topic

[ancient pollen]

kind of

[ancient pollen]

alr 😭

pi

i used pi’s value and got it , i firstly used 180 cuz i though it was radian but nvm😭🙏

[pseudo basin]

i cannot see any mistakes except for a nitpick about your handwriting

pi is pi, and pi should not be replaced with 180 on a whim

[ancient pollen]

okay

[pseudo basin]

please do not call me "sir"!

thank you.

[ancient pollen]

mam😭🙏

.close

[devout snow]

Channel closed

Closed by @ancient pollen

Use .reopen if this was a mistake.

[ancient pollen]

.reopen

[devout snow]

✅

[ancient pollen]

Alr

f.
Composite trapezodial rule

[elfin pewter]

which one

[ancient pollen]

Progress:

[ancient pollen]

f.

It is writeen 1.0532 in the book😭🙏

idk if book is incorrect

[elfin pewter]

what is n there

steps?

[ancient pollen]

Yee

S

composite trapezodial

i am using pi radian

‘s value

[elfin pewter]

this is correct

[ancient pollen]

ahh mb😭

i forgot to multiply by “2”

AAAAH

thank you.

.solved

[devout snow]

Channel closed

Closed by @ancient pollen

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[devout snow]

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[ancient pecan]

Need help with a mathlimpics question i didnt manage to answer

[ancient pecan]

On the grid, the sum of the elements of each lane and column are the same, whats the value of A + B - C?

[pseudo basin]

$\bmqty{3 & a & 8 \ 7 & b & c \ d & e & 6}$

[woven radish]

Ann

[pseudo basin]

this is the grid, yes?

[ancient pecan]

yea

[pseudo basin]

how did a=5 happen

[ancient pecan]

3+5=8

[pseudo basin]

but that isn't what the problem says...

it says that the following sums:

• 3+a+8 (first row)
• 7+b+c (second row)
• d+e+6 (third row)
• 3+7+d (first column)
• a+b+e (second column)
• 8+c+6 (third column)
  are equal to each other

it does not say that the equality "first number + second number = third number" is true in any of these lines!

[ancient pecan]

OHHHH TBATS WHATS WRONG

ok so

6+c+8 has to be the same as 3+a+8 right

[pseudo basin]

yes

[ancient pecan]

its all a big trial and error imo

[pseudo basin]

i would not say so

[ancient pecan]

i got 3+3+8 and 8+1+6 which both come out to 15

now to see another lane and column to see what they make

[pseudo basin]

you can do some algebra here

[ancient pecan]

wait

no wiat i got that wrong

[pseudo basin]

and express all other letters in terms of a

[ancient pecan]

its 3+4+8 comes to 15

[pseudo basin]

you don't need trial and error here

[ancient pecan]

girl beside me got

3 4 8
7 7 1
5 4 6

[pseudo basin]

that's a possibility but it isn't the only one

[ancient pecan]

im a bit on the slower side of the human brain if you hadnt caught that by now

[pseudo basin]

i am not going to judge

but i will ask

what level of education are you on right now

middle school? high school? uni?

if you're in school you can tell me what grade you are in instead

[ancient pecan]

im gonna google what 1 year of 2 degree comes out to

cause im not americna

[pseudo basin]

really theres only one thing that's absolutely necessary to know

are you over or under 13 years old?

[ancient pecan]

freshman year is what comes out

over

[pseudo basin]

ok

so

are you familiar with algebra

like solving equations for x

that sort of thing

[ancient pecan]

i dont think its freshman year tho-

yea a tad

[pseudo basin]

if you are over 13 you should have already learned a thing or two about those, maybe in an earlier year or maybe this year idk

anyway

[ancient pecan]

16

[pseudo basin]

ok that's even better

you definitely need to be familiar with algebra by now

gonna bring the number grid back

$\bmqty{3 & a & 8 \ 7 & b & c \ d & e & 6}$

[woven radish]

Ann

[pseudo basin]

so

the sum of the first row is 3+a+8, or a+11.

the problem says that both other rows, and all columns, also give total a+11.

do you understand so far? yes or no.

(do not try to proceed on your own unless you are 110% confident that you are doing the right thing.)

[ancient pecan]

im gonna say, not the expressions

in english i mean

as in both other rows-

[pseudo basin]

the second row and the third row

you have been saying "lane"

[ancient pecan]

ohhh yea

[pseudo basin]

the right word in english for a horizontal line like this is "row"

[ancient pecan]

ive been saying like that csuse of the grid in plamts vs zomb8es

proceed

[pseudo basin]

so, do you understand what i said so far

[ancient pecan]

all of the columns and rows come out to a total being a+11?

[pseudo basin]

'default' is the wrong word

[ancient pecan]

but is it right?

[inland wadi]

.

idk

[pseudo basin]

not really, but if you replace it with a correct word like "sum" or "total" then it becomes right

all i did was repeat the statement of the problem

that all rows and all columns sum to the same total

and i just said that this total equals a+11

no rocket science here

but i want to make sure that i didn't lose you

@ancient pecan i will only continue when you say you're OK to continue

[ancient pecan]

ok

proceed

[pseudo basin]

ok. let's now look at the third column, because it has a letter that we care about (c) and two known numbers.

8+c+6 = a+11

in this equation, can you isolate c?

[ancient pecan]

c+14?

[pseudo basin]

not what i asked

[ancient pecan]

i generally dont know what people ask me about numbers in english

[pseudo basin]

the left side simplifies to c+14, yes.
but i want you to rewrite the whole equation so that it reads c = ...

hold on

what's your native language?

portuguese?

[ancient pecan]

por yea

[pseudo basin]

Resolva a equação c+14 = a+11 para c

[ancient pecan]

its gonna take a while

[pseudo basin]

it is one step

[ancient pecan]

grab some popcorn

[pseudo basin]

how long is "a while"

[ancient pecan]

im genuinely ashamed of how long this is taking me

im gonna go off on a limb and say

c = 11+ a -14

death to me😭

im cringing hard rn

maybe its killing my brain the fact im doing 2 assignments at the same time

[pseudo basin]

why are you doing 2 assignments at the same time?

[ancient pecan]

11 + 14 = a+c

[ancient pecan]

in school rn

[pseudo basin]

can this wait until you are back from school

[ancient pecan]

maybe

[pseudo basin]

on this server you're expected to devote your undivided attention when someone is helping you

ok then wait until school is over then come back here lmao

[ancient pecan]

okk

[pseudo basin]

i'm going to close this now.

when you get home from school again, take a new channel and post your problem there.

.close

[devout snow]

Channel closed

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[devout snow]

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[rough nova]

a,b,c are positive real numbers, prove:

[rough nova]

Or a³/b² + b³/c² + c³/a² ≥ a²/b + b²/c + c²/a when written out

This is probably some clever AM-GM

[solid osprey]

if you wanna get cheeky use muirhead

im p sure muirhead is just repeated am-gm lemme try to find it

actually nvm muirhead doesent work

[patent nexus]

did you try Cauchy–Bunyakovsky–Schwarz inequality?

[rough nova]

With which sets of numbers?

[patent nexus]

with ${\sqrt{\frac{a^3}{b^2}}, ...} and {\sqrt{a} , ...}$ for example

[solid osprey]

dont you need a+b+c for that

[woven radish]

ViNton

[patent nexus]

you can make another CBS

[patent nexus]

i mean after sets that i sent you can make another CBS to finish proof

[rough nova]

So i do this and hen i have to prove a+b+c ≤ a²/b+b²/c+c²/a?

[patent nexus]

correct

[rough nova]

Well a²/b + b ≥ 2a... So that leads me there

Thanks I guess, i found the problem in an introduction to Holder's inequality but i guess CBS is easier than that

Thank you @patent nexus

.close

[devout snow]

Channel closed

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[solid osprey]

i mean cauchy is holder just when p=q=1

[rough nova]

The introduction was mainly about p=q=3 so there's probably a very clever and neat solution

I'm satisfied with the CBS one though

[devout snow]

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[flint vine]

I NEED SOMEONE TO HELP ME BIG TIME

IN CALL

WITH INTEGRATING A CIRCLE PART

[brittle badge]

!da2a

[devout snow]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[flint vine]

higher outer circle equation:
(x-0.5)^2 + (y- sqrt(3)/2)^2 = 0.8

lower outer circle equation:
(x-1)^2 + y^2 = 0.8

integration limits:
upper circle (0 - 1.4)
lower circle (1.4 - 1.9)

[lavish nimbus]

So you're trying to make chatgpt logo with math functions?

[flint vine]

I made the logo but the integrating is killing me

[north roost]

polar coordinate?

[flint vine]

I just need help

[north roost]

?

[lavish nimbus]

You want to fill some part of the logo?

[flint vine]

yes

[flint vine]

how do I find them (out of my knowledge)

[lavish nimbus]

Then I guess you have to manualy take the area you want to color.

[north roost]

so u first convert the equations to polar coordinate

wait

is calculator allowed?

[flint vine]

yes

anything is allowed

just gotta show work

[north roost]

oh cool

[lavish nimbus]

What about taking the integral of the upper function minus the lower function (for area under x-axis, do the same)?

[north roost]

then find the equation in the form of y = something

save only the positive part

then integrate

[flint vine]

hmm

[north roost]

nvm

[flint vine]

would that find the area under it tho

[north roost]

i wont work

do polar coordinate

[flint vine]

and then how do I use them

[north roost]

so essentially u let x = costheta y = sintheta

[north roost]

let me try and graph that

gimme a sec

[flint vine]

can u call? if thats alr

[north roost]

sorry, no can do

[flint vine]

alright

[north roost]

so for the first circle

the bottom

equivalent to this in polar coordinate

[flint vine]

looks about right

how about the area tho

[north roost]

i simply integrate in the polar world

https://www.youtube.com/watch?v=GQ6cDvY8K9g

should look into this

[north roost]

not sure abt this, u should recheck with other sources

[flint vine]

alr thanks still

[north roost]

lemme share my desmos

https://www.desmos.com/calculator/9wbeccojpw

[devout snow]

@flint vine Has your question been resolved?

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[velvet jacinth]

I am trying to solve this integral using integration by parts and the answer I am arriving at is not matching the answer in my book

The book provided answer is $$(t^5 ln(t))/5 - (t^5)/25 + C$$

[woven radish]

columbian twizzler importer

[mellow willow]

I’m really confused because the amp is -5 but why are they adding and subtracting 5 for the max and min instead of -5

So now I am confused when I can use negative and when I don’t

[thick ledge]

You've also not multiplied by 4

[devout snow]

@velvet jacinth Has your question been resolved?

[velvet jacinth]

multiply t^4 by 4 ?

okay I see.

Thank you Rudy.

.close

[devout snow]

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[quick seal]

hello i know this is probably silly, but the video im watching the teacher said the yellow degree is 90 + teta/2
but doesnt that rule apply when it was bisector ? (i used translate im sorry)

[lavish nimbus]

Total degree of a triangle is 180. The equation of the big triangle is $$\theta + 2 \alpha+2 \beta = 180$$ for the smaller triangle, it's $$\phi + \alpha + \beta = 180$$ Substitube $\alpha+\beta$ from the first equation into the second, you'll get the result.

[woven radish]

Good

[quick seal]

woh i got it

im silly im sorry

thank you for helping!!

.close

[devout snow]

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[wooden citrus]

how would I simplify this?

[faint zinc]

You might want to rewrite the surds as fractional exponents first.

[wooden citrus]

not sure what surds means but you mean like x^1/2 and x^11/4?

[faint zinc]

$\sqrt{\cdot}$ is a surd

[woven radish]

OmnipotentEntity

[faint zinc]

and yes exactly.

[wooden citrus]

thanks for the help

.close

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[scenic shell]

Hi, I'm really confused on the b) part of this question? How do I find the standard deviation and interval estimate?

[hexed holly]

standard deviation is kind of tricky to calculate. do you have a graphing calculator?

[scenic shell]

Yes

[hexed holly]

a ti-nspire?

[scenic shell]

uhhhh

I think?

CS-121?

[hexed holly]

oh yeah

hm i have a cx 2

if you go into the calculator part and press the book icon, then press s and scroll you should find a stDev option somewhere

[scenic shell]

okay

[hexed holly]

oh wait im sorry ive totally misunderstood your question

[scenic shell]

oh its okay

[devout snow]

@scenic shell Has your question been resolved?

[devout snow]

@scenic shell Has your question been resolved?

[scenic shell]

.close

[devout snow]

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[tribal root]

Translation: Both squares are the same size. Show that the shadowed places in the squares has the same area

My work kinda messy

For the first square the area for the shadow part is

16r^2 - 4pi r^2

And the second square is

r^2 - pi r^2

[tawdry socket]

Here's how I'd start: $A_{1}=\left[Area\ of\ square\right]-\left[Area\ of\ the\ 4\ small\ circles\right]$

[frozen aurora]

de är inte samma r

[tribal root]

A shit sant

Hur fan löser man då

[frozen aurora]

det skulle vara bättre om du tog l som kvadratens sidlängd

[tribal root]

Okej

På andra circkeln är inte sidlängden 2r?

[frozen aurora]

eftersom l är samma för båda kvadraterna

[tribal root]

Förstår de har samma längd

[frozen aurora]

ja

eller

vänta

[frozen aurora]

tar du r som i den första kvadraten?

[tawdry socket]

$A_{1}=\left(x^{2}\right)-\left[4\left(\pi\left(\frac{r}{2}\right)^{2}\right)\right]$

$A_{2}=\left(x^{2}\right)-\left(\pi r^{2}\right)$

[woven radish]

hello

[tribal root]

Nej andra

[tribal root]

Why r/2

[frozen aurora]

okay maybe english would be better

[tawdry socket]

It's because each of the smaller circle's radius is half of the larger one in the 2nd square.

[frozen aurora]

yep precisely

[tribal root]

I get it

[tawdry socket]

Refer to the image you sent. You can visually check that.

[frozen aurora]

if you were to do this, you would get that
r_1 (first square) = l/4
and
r_2 (second square) = l/2
and you can plug those in directly

[tawdry socket]

My next step from there:
$A_{1}=\left(x^{2}\right)-\left[4\left(\pi\left(\frac{r^{2}}{4}\right)\right)\right]$

$A_{2}=\left(x^{2}\right)-\left(\pi r^{2}\right)$

[woven radish]

hello

[tawdry socket]

And then cancelling out the 4:

$A_{1}=\left(x^{2}\right)-\left(\pi r^{2}\right)$

$A_{2}=\left(x^{2}\right)-\left(\pi r^{2}\right)$

[tribal root]

I got there

X cancels

[woven radish]

hello

[tribal root]

And the pi r^2

[frozen aurora]

if the two expressions look the same then you are done

[tribal root]

Ahaaaaaa

Why didn’t it work to use r as the side length of the square

?

[tawdry socket]

The radius is not the diameter. The diameter would've been the side length of the square.

The radius is half of the square's side length.

[tribal root]

Isn’t the diameter 2r

[tawdry socket]

Yes.

[tribal root]

Is it possible?

In terms of 2r

[tawdry socket]

It is. Just remember to substitute "2r" into both expressions for the areas.

[tribal root]

Yeah I get that

Do you still have to do r/2

[tawdry socket]

Actually, the side length is "2r" for both squares

Sorry.

yes, it would work

[tribal root]

I still don’t quite understand the r/2 can you explain

[tawdry socket]

yes, you still have to use r/2 for each small circle's area.

Each of the small circle's radius is r/2.

"r" is the larger circle's radius

[tribal root]

Yeah the second circle/2

Oh yeah

Otherwise if you use r for both

It dosent work

Because they aren’t the same length

Ohhh alright thank you so much

[tawdry socket]

You're welcome.

[tribal root]

Tack så mycket @frozen aurora

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[sick bolt]

Did my teacher make a mistake? Our answers don't match, idk how she got that answer

[rough hollow]

\left( \frac{12.5 + 15 - 8}{5.25} \right) + \left( \frac{18}{3} + 2.4 \times \left(14.44\right) \right) - \frac{15.6}{3.9}

[potent tusk]

convert yds to feet bruh

units man units

[eager agate]

Or convert ft to yds.

[sick bolt]

Whar

[potent tusk]

im gonna be honest im not american so i dont know what the uhh conversion methods are

but you can probably google it

[tawdry socket]

Convert 200 yards to feet. 1 yard is 3 feet.

[rough hollow]

hi

[sick bolt]

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[rough hollow]

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[bleak tide]

Consider randomly selecting 4 cards from a standard 52-card deck. Let the random variable X denote the number of different card ranks in your hand. What is the expected value of X?

Is there a general setup for this? Say there were n possible ranks, m suits, and you were drawing k cards. Is there a good way to count the number of "different" objects selected in a sample, using some equivalence relation (in this case, suits)

[devout snow]

@bleak tide Has your question been resolved?

[faint hearth]

This problem can be solved just by knowing the definition of expectation:

Find the number of possible selections of 4 cards(52 choose 4).

How many of those selections have only one rank?(13)

How many have only 2 different ranks?(13*12*2).

How many selections have 3 different ranks?
(You find out, say there are x such)

how many have 4 different ranks?
(You find out, say there are y such)

Then the expected value of X will just be

[1*13+2*(13*12*2)+3*x+4*y]/(52 choose 4)

By definition of expectation

[bleak tide]

That's not what I was looking for but thank you anyways.

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[faint hearth]

Well what i am saying is that the general setup will also just be to go by the definition of expectation

The problem of counting different objects upto some equivalence can have varying levels of difficulty and there is no known "general method" to easily count them

I intended for my solution to be something from which you infer these facts

[faint hearth]

If you found my answer to not be satisfactory, you can reopen and wait for a more satisfactory answer

[bleak tide]

I realized I need to focus on finishing this assignment for now, I might try and figure out a generalized version later but I don't really have time now I think, and I don't want to clutter this up by leaving an open help thread

[devout snow]

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[upper forge]

I know how to do long division but I keep doing it wrong when its so many variables

[winter patrol]

do you mean terms? there's only one variable.
but that doesn't really affect the division algorithm. can you show your work?

[upper forge]

oh ya I meant terms

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[lilac vortex]

how do you do part d? trying to use the formula for the sum of a geometric sequence instead of just writing each term then adding it

[lilac vortex]

for part ii* i can do part i of course

[lilac moat]

add and subtract S_4 to both sides of part i

rather

if you take the sum on the left hand side of (i)

and add and subtract S4 from it

you're adding 0

[elfin pewter]

Brute force

[lilac moat]

but it allows you to use the geometric sum formula

[wind mason]

They said they wanted to use geometric series formula

[elfin pewter]

a

just plug in the values in the formula xd

a(1-r^n)/(1-r)

a is the first value

r is the ratio

[wind mason]

Adding on to what gfaupax said, in general $\sum_{k=0}^a=\sum_{k=0}^b-\sum_{k=a-1}^b$ (ignore the shit notation)

[elfin pewter]

in this case is -1/3

and n=4

[woven radish]

;(

[lilac vortex]

but does this not find the sum from the first term to n, whilst i want the sum from what... the 5th term to the 8th?

[elfin pewter]

first term(5th term) to next n terms

[lilac vortex]

but thats the formula for the sum of a geometric sequence... i dont understand why you say this wont get the sum from the first term

[wind mason]

You could try this out, as a more clear approach

[elfin pewter]

look at the screenshot I send

the answer is the same

try to prove the geometric formula, then you will understand

[lilac vortex]

i didnt know if youd have to try and find a where the sequence starts from n=1 (not 5 like the sigma notation). as in, youd have a * r^(n-1) so the first term would be a * r^(1-1) which is of course just equal to a. then youd do the s8-s5 using that initial a value. i kind of understand how this works now

[elfin pewter]

: D

does this solves your problem?

[lilac vortex]

it does. thanks mate

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[fast fable]

Find Area A2

My test is tomorrow I need help with this

[devout snow]

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[fast fable]

<@&286206848099549185>

[long trout]

Did you find the angles first?

[devout snow]

@fast fable Has your question been resolved?

[fast fable]

Yes, I was thinking about getting the area of the two circular sectors and rest them but I don’t know if that is the right way

[long trout]

It’s the right way. You draw a perpendicular from B onto AC, therefore creating two similar triangles (ones with same angles as original triangle). Then for one part of the sector it’s the area of the sector minus the area of the triangle and same with the other part of the sector. Then you add both together to find the area of A2

[long trout]

Does the problem specify that those curves are circular arcs? If so, then you can definitely do that way

[fast fable]

Thanks

A lot

[long trout]

No problem

[fast fable]

You just saved my exam

[long trout]

Happy to help :)))

[long trout]

You can type .close or press the tick here if you don’t have any more questions. Otherwise press the X if you have more questions to ask

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[brittle badge]

Find the maximum of $\frac{x!}{\left(x-k\right)!k!\left(x^{k}\right)}$ if k is a whole number and it ranges from 0 to x, where x is a (large) natural number.

[woven radish]

Philosopher King

[pseudo basin]

$\binom{x}{k} x^{-k}$?

[woven radish]

Ann

[pseudo basin]

call your expression $a_k$, work out the ratio $\frac{a_{k+1}}{a_k}$

[woven radish]

Ann

[brittle badge]

ok

,w substitute k = k + 1 in (x!)/[(x-k)!(k!)(x^k)]

[woven radish]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

May 22, 2025, 4:13 AM

[pseudo basin]

nope do it yourself on paper

it is not that difficult

[brittle badge]

(x!)/[(x-k-1)!(k+1)!(x^(k+1))] = a_(k+1)
a_k = (x!)/[(x-k)!(k!)(x^k)]
a_(k+1) / a_n = (x-k)/[(x)(k)]?

[pseudo basin]

almost

$\frac{a_{k+1}}{a_k}=\frac{x-k}{x(k+1)}$

[woven radish]

Ann

[brittle badge]

Yes.

[pseudo basin]

think about what you can say about this fraction

and what it has to do with maximizing a_k

[brittle badge]

(x-k)/[(k+1)x] = (1 - k/x)/[k+1]

This decreases clearly as k increases, since the numerator decreases and the denominator increases.

[pseudo basin]

why would i ask you to work out the ratio between two adjacent terms in your sequence?

can it be tied back to the increase/decrease behavior of the sequence a_k, and if so, how?

[brittle badge]

a_(k+1) = a_k * (1-k/x)/[k+1].
Let (1-k/x)/[k+1] = R_k.
R_0 = 1/1 = 1
So a_(1) = a_0
R_1 = (1-1/x)/(2) < 1
Similarly, R_2 = (1-2/x)/[3]
So a_3 < a_2 and so on.

a_0 = a_1 > a_2 > a_3 > a_4 > .....?

<@&286206848099549185>

[long trout]

Sorry to jump in here, but is it fine that we can prove that || C(x,k) ≤ x^k || by a combinatorial proof and then deduce when equality holds by providing examples?

[pseudo basin]

this is a lot of writing to say "The ratio is always <1 and so the sequence a_k (which is made of only positive terms) is decreasing."

[pseudo basin]

eh that works ig

@brittle badge anyway, now that you know a_k is decreasing: where will the biggest term of a decreasing sequence be found?

[brittle badge]

a_0 = a_1

= 1

Right?

[brittle badge]

idk

[long trout]

We can probably use a combinatorial proof here, but maybe first follow what Ann is doing which is more of the algebra proof

[long trout]

Yes that’s right

[brittle badge]

okay thankd

[long trout]

All good, do you want to hear the combinatorial proof as well?

[brittle badge]

yes

[pseudo basin]

yes

the biggest term of a decreasing sequence is found at its beginning

this, i would imagine, is not a difficult insight to make

[long trout]

Ok, so like Ann did we can first rewrite the expression as C(x,k) * x^-k. Now C(x,k) is the number of ways to select k distinct members out of a total of x members. x^k is the number of ways to select k members out of the total x members where you can pick the same member again (repetition is allowed). What does this tell us in terms of C(x,k) and x^k?

[devout snow]

@brittle badge Has your question been resolved?

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[opaque lodge]

I am watching Veritasium video and stuck at this part. Where is this formula or function came from?. I guess it is about axiom of choices and vitali set but i don't know exactly what it implies.

[pseudo basin]

the union of a countable number of disjoint copies of V contains [0,1] and so has measure at least 1, but is itself contained in [-1,2] and so has measure at most 3

that's what the formula is saying

[opaque lodge]

why most 3

[pseudo basin]

what's the length of [-1,2]

[opaque lodge]

so 3 mean 3 sets or what?

[pseudo basin]

no, 3 means 3

3 units of length, if you wish...

[opaque lodge]

=) ok

what is lambda(V)

and why we add it infinite

amount of times

[pseudo basin]

lambda() is Lebesgue measure

which is a generalization and formalization of the concept of length

the reason we add infinitely many copies of it is because we are trying to figure out what the measure of the Vitali set V should be

and we know something about the union of infinitely many shifted copies of V

iirc he specifies the axioms for a measure somewhere...

[opaque lodge]

ok thanks i think i need to learn about Lebesgue measure then.

[near jolt]

this is the measure (specifically lebesgue measure) of a set V, or, intuitively, length.

What is done here is they shifted this set V a lot of times: by how they defined V, these "shifts" are disjoint (meaning that , given two different shift, they do not have common elements).

What we want in a measure is for it to be "countably additive": that if we seperate a set S into different components (countably many), the "length" of S should be equal to sum of length of each component.

Another property we want is that if S has some length, and is a subset of T of some other length, then the length of S must be smaller than the length of T (so we don't want a set containing less elements to have more length).

Lastly, the length should be invariant over shifts. if you shift a set to the left/right, the length should be the same.

If you combine these three properties, The video showed that it is not possible for every set to have length. In particular, V cannot have a length: if it did have a length, then we can sum over all the shifts, it should have length between 1 and 3 (since it contains [0,1] but is contained in [-1,2]). But you can't have a number, added infinitely many times, to be between 1 and 3, contradiciton!

for "countably additive" I should emphasize that the components must be disjoint! otherwise you get weird stuff like length of [0,1] +length of [0,2] =length of [0,2] (1+2=2?absurd!)

[pseudo basin]

countable additivity (or at least finite additivity) is used all the time when you find areas of shapes in 2D, or volumes in 3D, by breaking up into pieces and adding up the area or volume respectively of each piece

[opaque lodge]

so it is related integral?

[near jolt]

there is a whole lot to talk about! given a way to measure things, you can define integral with respect to this measure.

[pseudo basin]

yeah it is but the idea of a measure itself is more basic in a way

[opaque lodge]

thanks a lot

[opaque lodge]

what do you mean when you write 1 year = 1 day = 1/365days?

1 year != 1 day?

what is 1/365 days? 1 day in a year?

[pseudo basin]

????

who said anything about years vs days

[opaque lodge]

uhm his bio

[near jolt]

oh that, it was me memeing

[pseudo basin]

[opaque lodge]

ok see you guys again.

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[devout snow]

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[wooden citrus]

in which cases would an answer be extraneous or no solution altogether

[polar chasm]

you shouldnt get anything extraneous here

[brittle badge]

2^2 = x^2 + 2x - 4

Exactly two values of x would make x^2 + 2x - 4, 2^2 = 4.

[wooden citrus]

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[sage tiger]

what's the math of gravity and why's the math of gravity

[sage tiger]

so like

u got a universe

empty

vector 0

nothing present

and u have an object A

in there

but A cannot realize itself without an object B, as to measure itself in terms, only can it do with something relative

so

A says

The standard measurement in this world is gonna be me

okay

and B arrives somehow

B says " I wanna measure myself"

A says " Oh, you're like 2 times bigger than me "

[drifting anchor]

It’s too ambiguous what you’re trying to say, something of a thought experiment?

[sage tiger]

and B says " oh so you're 0.5 B? "

[sage tiger]

yes but a meaningful thought experiment

to understand the universe and why gravity exists and how it is

[drifting anchor]

You cannot do that in terms of such thought experiments mate

[sage tiger]

and A says " nah we can't change everytime something bigger comes.
so let me be A as standard and you as 2 A"

[drifting anchor]

this is more philosophical than scientific

[sage tiger]

it's not a thought experiment

it's just asking how and why gravity

mainly maths

[sage tiger]

okay so now they have vectors and measurements and stuff and then time arrives after them realizing A turns around some in a motion

but

how does

gravity arrive

why is A attracted to B

what's the general formula to find it out?

then why's the general formula to find it out? like why's that, that the universe acts upon a certain situation where bigger mass attracts others

surely it cannot be just a sentence on a paper or a logic of "bigger means more powerful"
because universe doesn't have a mind to be biased

so

are we

all falling?

at the same time

in a direction

even so

how do bigger things attract each other here

so falling might not be true here

what if

the reason why bigger things attract more than small things and the reason things even attract each other

is because of something else?

something like a new constant of measuring like distance and time

we cannot know what that process is because we are limited to the matter around us

but we do know how that process results (i.e bigger things attract more than small things and all things attract each other)

maybe

just maybe

we can try to figure out that process using the constants we have over here

and not end up in a random theory but rigorous mathematical truth

hmm

how would we do it

let's bring some notations

To find process X-
X happens -(results in)- M{A|B}

[sage tiger]

process x, happens which results in matter's behavior of A- all things attract each other, B - bigger things attract more than smaller

how to find process X which makes this happen now

hmm

we need philosophy, maths and physics here

wish there was a chat which involved all 3

anyways

let me see what we have in our world rn

time, constant T

does time matter in gravity?

like does time affect the M{A|B}

hmmmm

yes but time is just a measurement not a process

earth might attract Mars in Z number of years

but Z cannot change without the process changing and Z doesn't matter here we can't change Z on its own without the process changing

because Z is a constant, every process has a certain fixed Z time resultance

okay

[sage tiger]

re arranging this

X -results- M{A|B~T}

the process happens in Time measurement T

okay

this constant wasn't really useful

now moving on

distance?

distance hmm

distance doesn't matter if u think about it

because

the farther u are, the smaller it is

and the nearer u are, the bigger it is

so distance doesn't change anything, and we find out that A and B are variables and not constant because of distance (I mean that they change even though their matter is same, so as B comes nearer A appears Bigger and so does B, vice versa for farther and smaller)

[sage tiger]

re arranging this

X -results- M{A|B~T (A>-<B ~ D-/ A<->B ~ D+)

[sage tiger]

so

Matters A and B attract each other during a resultance constant time T

and as distance reduces (D-) A and B become larger to themselves and each other If and only If

[rare kernel]

this is a math help channel dude

[sage tiger]

dude

I'm trying pure math

[rare kernel]

u are just yapping

[sage tiger]

no

I need help

with things like

giving me constants

etc..

while I figure out gravity

[rare kernel]

figure out gravity?

[sage tiger]

And as distance increases (D+) A and B become smaller to themselves on the perspective of each other

[sage tiger]

yea

[eager stone]

damn noble cause

[sage tiger]

like do u know why gravity's here?

I don't even think u know what happens if 3 bodies gravitate each other

[rare kernel]

do u know why anything is here

[sage tiger]

I'm tryna figure out that dude don't be a narcissist and stop something that's tryna reason things out

[rare kernel]

im just saying u have to accept some things and cant go around questioning everything

[sage tiger]

there's already been a three body problem which says even slightest of changes matter divergently

and double pendulum

example.

so

we need to understand

how gravity works and why is it here even with an unbiased universe

then we might be able to

make something work out and develop a more better way to

uh yk predict motion around and gravity

[sage tiger]

now let's not get diverted

so

we know where constant of time is and what it does here

we know what constant of distance does here

but

it all really is the same and follows the laws of;
all objects attract each other
Bigger objects attract more

[sage tiger]

let's try out on a new constant rather than distance or time

hmmmmmm

light? does light affect gravity

obviously not

but

not obviously, maybe?

because like

light does affect gravity

if u see something coming towards u

wavelengths of light are on u and that

but

hmmmmm

but like

it can emit light and stop, because of the distance we'd see it in past tense

okay

hmm

nah light is too insignificant here

although very much present

so

a new constant we need

time okay, distance okay

what about

speed

@eager stone does speed affect gravity

Hmm

wait we haven't thought of this

does the sun moving in a direction affect the pull of it on earth

dude can someone help me?

[eager stone]

bro 😭

https://youtu.be/XRr1kaXKBsU?si=Ew1b_lLfHVFC48-H

[sage tiger]

uh

what does he say

speed affects gravity?

dude

I don't want to be biased towards what he says

I want this to form of my own thesis of this

someone tell me

does speed affect gravity

🙏

O dear Lord provide me with a human to guide my noble causes
O generous behemoth grant me a user to answer my repenting question
and I shall be in your service for the rest of eternity

[soft sequoia]

bet

i got u

[sage tiger]

...

[soft sequoia]

what do u need @sage tiger

[sage tiger]

uh

does

gravity affect speed

no no I mean

does speed

affect gravity

[soft sequoia]

gravity is just 9.8 m/s

[sage tiger]

no like

[soft sequoia]

ultamitely gracity does affect the froce velocity

[soft sequoia]

ya

[sage tiger]

u moving does it affect your gravitational pull or something else pulling u

[soft sequoia]

ur weight times the gravtational force is equal to the force pulling you toward the earth

i meant to say mass

I don't think speed comes into the equation

[sage tiger]

but

why?

[soft sequoia]

that's how it works ig

[sage tiger]

the sun moving in a direction (typically around the milkyway)

can it not affect it's pull on us?

why not

[sage tiger]

do u know for sure

[soft sequoia]

not really

[soft sequoia]

read this

[sage tiger]

but

Acceleration

can come

without gravity as well

not of it only

what so then

[soft sequoia]

but ag is the symbol shown for accelaration of gravity

9.8 m/s

<@&286206848099549185> anyone wanna help this guy

[soft sequoia]

i agree with this kind of

[sage tiger]

Did we test if a pencil's pull on an eraser is the same as the same pencil's pull on a 1500km/h's eraser, vice versa

[soft sequoia]

is that a vector?
what direction is it going on

[sage tiger]

uh

north?

to the pole!

[soft sequoia]

ok no

away from the pencil?

or to the?

[sage tiger]

oh to the

away from the pencil

[soft sequoia]

then it will simply be the same

[sage tiger]

beritcally oppositley away

[soft sequoia]

then of course with that velocity it will go away

[sage tiger]

north of the pencil

[sage tiger]

yeah but

does the gravity of the pencil or eraser

differ

there

[soft sequoia]

no

there both being pulled to the centere of the earth

[sage tiger]

not center of the earth

[soft sequoia]

think about it

[sage tiger]

Forget earth forget university

[soft sequoia]

wtf do u mean

[sage tiger]

there's a pencil and an eraser in the universe alone

does the gravitational field of the pencil

which acts upon the eraser

equal to the gravitational field of rhe p ncil

which acts upon the eraser which moves?

[autumn girder]

$$F_G = G\frac{m_1\cdot m_2}{d^2}$$
$$g_{m1} = \frac{F_G}{m_2}$$
$$g_{m2} = \frac{F_G}{m_1}$$

[woven radish]

@autumn girder

[smoky gyro]

sir what is your question i dont think even you know what ur talking about

[sage tiger]

uh

[polar chasm]

the fields are not equal, the forces are equal if the masses are equal though

[soft sequoia]

ong

[sage tiger]

My question is

[polar chasm]

but the forces would be minimal

[soft sequoia]

.close lol

[sage tiger]

does the velocity of object B affect it's gravitational pull (considering both case scenario's mass is the same)

[polar chasm]

(equal in magnitude, not in direction ofc)

[soft sequoia]

sybau

[sage tiger]

uh

shit

I need to focus on better things

[sage tiger]

Can you

Explain me

These 3 things

welp

guess gravity is not my thing

how about something else that governs the universe

[smoky gyro]

sir no offense but if u dont know newtons law of universal gravitation i dont think you should be trying to reconstruct the idea of gravity from scratch without a physics or math background to support you

[sage tiger]

hey I'm a high school student (just started hs) stop calling me sir.
and what if newton didn't exist and I have to develop universal gravitation

[smoky gyro]

well newton did exist and you are not newton

[sage tiger]

well

does gravity have a "why" Yet?

[smoky gyro]

what ?

[sage tiger]

instead of just mass = pull

[smoky gyro]

why what

[sage tiger]

why is mass = pull

like why does

mass attract

each other

[smoky gyro]

because an object with more mass has a larger gravitational force sir

[sage tiger]

okay

so

why am I not

infinitely travelling through space

other atoms

attract me

[smoky gyro]

????????????

[sage tiger]

and if all my atoms

Are attracted by other atoms

then why am I not attracted towards those atoms

[hollow ice]

well, the force is distance dependent

[sage tiger]

and why don't I move towards it irl seemingly

[sage tiger]

I see

I see

well yes I know

but okay so that's a dumb question

[hollow ice]

closer objects, like earth, exert more force

[smoky gyro]

do u know how much an atom weighs

[sage tiger]

yea

it weighs a bit of the earth

[smoky gyro]

what

[sage tiger]

take earth as the measurement in terms of standard weight, no kgs no grams etc..

it weights about 0.0000 something etc.. Earths

[smoky gyro]

sir thats not how we calculate the weight of subatomic particles

we dont use earth as standard measurement

[sage tiger]

oh

what do you use the standard measurement as

okay

[smoky gyro]

kg

[sage tiger]

you use a standard measurement xyz

[smoky gyro]

or atomic mass unit

[sage tiger]

an atom weighs approximately xyz power n

[smoky gyro]

sir im going to stop u right there

[sage tiger]

y

[smoky gyro]

because i dont even know why youre trying to reinvent physics or change the SI units of mass

it makes no sense

[sage tiger]

im not

you are asking me

how much an atom weighs

[smoky gyro]

yes and i told u the si units

[sage tiger]

okay your si unit power n

is the weight of an atom

[smoky gyro]

what are you talking about "power n"

[hollow ice]

I think he means the orders of magnitude

[sage tiger]

supposedly, your si unit is x and it weighs 1 atom

[smoky gyro]

well then for kg its 10^-27

[sage tiger]

then the weight of an atom

[hollow ice]

An atom of hydrogen is 10^27 times lighter than a single Kg

approx

[sage tiger]

is xyz power n where insert your si unit which weighs 1 atom

so 1 power 1

[shrewd egret]

[smoky gyro]

sir

every element has differing atomic structure

[mystic scarab]

By the way @sage tiger

DO
NOT
WRITE
LIKE
THIS
!!

It's VERY DIFFICULT to read you

[dark dawn]

@sage tiger We don't understand the mechanism behind gravity, so there isn't really a proper answer to why it does what it does, but you can understand its behaviour through observation and depending on what big ideas you start with. The universal gravitational constant is really just a value we calibrate through observation to collapse the proportionality into an equality. Disagreements with observation probe new ideas which naturally led to treating gravity not like a force, but instead a curvature of spacetime :)) hope that clears up any initial confusion you might've had

[mystic scarab]

And also you'd better take some time to reformulate your thoughts in a concise manner, otherwise it's just a meaningless flow of consciousness
@sage tiger

[polar chasm]

this was a good read

[smoky gyro]

i think introducing the idea of spacetime curvature might be too complex for them

[dark dawn]

the maths yes but the idea no

[smoky gyro]

i hope they willl not be tempted to reinvent einsten's theory of general relativity

[shrewd egret]

https://tenor.com/view/woody-woody-toy-story-sora-kh3-kh-gif-27223516

[dark dawn]

[devout snow]

@sage tiger Has your question been resolved?

[shrewd egret]

vroski asking if gravity has been resolved 🥀

[sage tiger]

ah I see

[sage tiger]

hmm

yea

[sage tiger]

relativity

hmm

relativity

the idea that time is relative

and everything is relative

should have been invented

hundreds of years ago

in the 1800s

Alfred Adler

a psychologist

proposed a hypothesis

" All problems are Interspersonal relationship problems"

[smoky gyro]

sir u should stop while ur ahead 🙏

[shrewd egret]

I disagree. Keep cooking.

[sage tiger]

js wait 3 mins you'll make sense of what I say

[hollow ice]

relatives, gravity, surely there is some yo mama joke here

[sage tiger]

basically like, all problems are caused by humans and animals around us

but

u ask a question

" wtf how does me falling off a tree caused by some animal or human "

but

the thing is

[smoky gyro]

sir did an apple fall on ur head

[sage tiger]

you only have to do it once to

[sage tiger]

uh no

I don't eat apples

[sage tiger]

realize that it's pain

so hypothetically

in a world without any animals (large enough to see) and humans

[eager stone]

you climbed it in the first place

[sage tiger]

where earth and the universe is present

you'd eventually not do things which you know (have experienced) as painful

[smoky gyro]

u burnt the kitchen

stop cooking

[sage tiger]

so then, once u do enough things and only u exist

no one would give you pain other than your willingness

[shrewd egret]

Keep cooking. I repeat: Keep cooking.

[sage tiger]

you wouldn't survivalistically get much pain

[smoky gyro]

bro is never getting green role

[shrewd egret]

LMFAO