help-28

[quartz flare]

Like

$\int_{x_1}^{x_2}\frac{1}{\int_{3}^{5}1dx_3}dx_3 = \int_{x_1}^{x_2}\frac{1}{2}dx_3$

Oh lord the other one looks horrible

[solemn ermine]

yep! we may even leave it as (1/b-a) for an x that has the uniform distribution on [a, b]

[glossy valve]

Coolempire93

[quartz flare]

Ah now this is starting to sound like averages

[solemn ermine]

and if that's what happens when we're interested in just one random variable... what constant do we think might show up if we're considering 3 (independent, identically uniformly distributed) random variables?

.

[quartz flare]

Yeah I guessed something of the sort but I still can't explain it necessarily 🤔 when contextualized in that problem

It feels like I should then sum across all of the x1's and x2's in the interval

But I need to sum across the probability that each of

them lies in the interval

ah.

Trying to gain the intuition for it now

Okay

This makes sens

[quartz flare]

Looking at it in terms of average values the division is perfectly clear

Looking at it in terms of double counting I'm trying to get the logic now

[solemn ermine]

this fact may be helpful for intuition as well--we're scaling down the area we're integrating, because we'd like to make it so that the probability that a variable x with the uniform distribution on [a, b] is between a and b is 1

(if you didn't already come to this conclusion)

[quartz flare]

Yeah that's what I was thinking with the double counting argument

I think it's more succinctly treated with the bijection from before

[quartz flare]

In other words the integration variable undergoes a u-sub

[solemn ermine]

this is probably a more long-winded approach, this is true

[quartz flare]

$\int_{a}^{b}1dx$ becomes $\int_{a \over b-a}^{b \over b-a}1d(\frac{x}{b-a})$

And b/b-a and a/b-a are one apart

[glossy valve]

Coolempire93

[quartz flare]

That's kind of ridiculous 😂 but it makes sense

At the same time horrible

[solemn ermine]

you can definitely compact the long-winded approach into a pretty quick set of integrals if you feel comfortable with probability density functions--this is what jumps out to me as the "if you know about integration" argument

vvv one-liner integral approach

note that ||we multiply by 2 to remove the ordering of x1, x2||

[quartz flare]

Yeah I was going to bring up the idea of multiplication of the probabilities then I realized my integrals weren't multiplied 😂 and I didn't know how to take them apart

[solemn ermine]

we're allowed to multiply the probabilities, but only because x1, x2, and x3 are independent--"r.v.s are i.i.d." are some of my favorite words to hear in probability in stats

[quartz flare]

Right

Ah I can finally ask to learn what i.i.d means XD

[solemn ermine]

independent and identically distributed--basically, you have some number of random variables, and they both have the same probability distribution (identically distributed) and are unaffected by each other (independent)

[quartz flare]

But yeah one of my approaches was

multiply probability that x1 = x1 by prob that x2 = x2 by probability that x3 lies on x1 to x2

It looks like your integral does the inside out

Set x3 and find the probability that x1 is on the interval a to x3

And then find the probability that x2 is on the interval x3 to b

[solemn ermine]

(and then multiply by 2 to account for potential ordering)

[quartz flare]

And then nCr for because x1 and x2 are unordered

Yeah

That's a nice approach

I assume that came from the original question guy?

[solemn ermine]

yep! the question we're asking wants both cases when x1 < x3 < x2 and x2 < x3 < x1, not just one

[quartz flare]

Nice

Okay!

i feel like I learned a lot

Thank you!

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[zealous plank]

just need sm1 to help me check work and make sure i didnt get it wrong w/o using gpt

[onyx glen]

ok send your work then

[zealous plank]

yea 1s

Need to set p'(x) = 0 -> product rule to find p'(x)

how does one insert latex here

$\frac{d}{dx} 3x\sqrt{400-x}=3\sqrt{400-x} - \frac{3x}{2\sqrt{400-x}}$

[glossy valve]

Poyyo

[zealous plank]

set it equal to 0

$3\sqrt{400-x}= \frac{3x}{2\sqrt{400-x}}$

[glossy valve]

Poyyo

[zealous plank]

mult both sides by sqrt(400-x) to get 3(400-x) = 3x/2, div both sides by 3, 400-x = x/2

[bright bronze]

yea all good till here

[zealous plank]

400=3x/2, x =800/3

[bright bronze]

yep

[zealous plank]

anything wrong with it? option a?

[onyx glen]

,w d/dx (3x * sqrt(400-x))

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by ann_dec

[zealous plank]

im off by an order of magnitude which seems weird

[bright bronze]

you found x

[onyx glen]

x=800/3 is the correct stationary pt

[bright bronze]

what you need is P(x)

[zealous plank]

ohh

so find P(800/3) where p is the original func

[bright bronze]

yea

[zealous plank]

alr got it 1 sec

[bright bronze]

your question wants you to find the profit, not the amount sold

[zealous plank]

alr thanks

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[thick hedge]

Probaby a really silly question, but why does it say let $K_i$ be the elment of $RG$ that is the sum of the members of $\mathcal{K}_i$

[glossy valve]

wai

[thick hedge]

that is why is there only one such member

[fast peak]

why should there be multiple?

[thick hedge]

well, $\mathcal{K}_i$ can have multiple members

[glossy valve]

wai

[thick hedge]

say a,b. Then a,b, a+b all belong to RG

[fast peak]

and?

K_i is the sum of all elements of mathcal K_i

[thick hedge]

Okay, that's what troubled me.

Cool

Thanks

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[meager cave]

hi

[meager cave]

...

[gritty flax]

Ask your question and ppl comes to help

[meager cave]

k

[neon basin]

[meager cave]

my friend asked

[turbid badge]

😭 ⁉️

[neon basin]

I believe this is a family-friendly server.

[meager cave]

oh mb

[quartz flare]

<@&268886789983436800> I can't tell wbat to think

[meager cave]

lemme change the wording

[spiral vigil]

quick question why did you think this was a good idea

[meager cave]

its a reddit meme

[spiral vigil]

that does not answer my question and frankly only raises further questions

.close

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Channel closed

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Use .reopen if this was a mistake.

[meager cave]

.reopen

[full forum]

✅ Original question: #help-28 message

[meager cave]

...

[sacred yarrow]

Do you have any question?

[meager cave]

fuk

[sturdy valve]

What is bro sending

[neon basin]

Did you really give the image to ChatGPT

[meager cave]

yes

[sturdy valve]

Viscous force for a cylinder 💀

[neon basin]

Fascinating.

[sturdy valve]

If it was sphere u could use stokes law

For a cylinder I have no clue

[meager cave]

\textbf{Problem:}

Consider two long, vertical, concentric cylinders. The inner cylinder has radius $a$ and the outer cylinder has radius $b$. The space between the cylinders is completely filled with a viscous liquid of density $\rho$ and dynamic viscosity $\eta$.

Under steady conditions, the liquid forms a thin film around the inner cylinder and flows downward due to gravity. The downward velocity of the liquid at a radial distance $r$ from the center is given by

[
v(r) = \frac{\rho g b^2}{2\eta} \ln!\left(\frac{r}{a}\right)
- \frac{\rho g}{4\eta}\left(r^2 - a^2\right),
]

where $g$ is the acceleration due to gravity.

Answer the following:

\begin{enumerate}
\item Determine the viscous force per unit length exerted by the liquid on the inner cylinder.
\item Write down the integral expression required to calculate the volume flow rate of the liquid flowing downward along the inner cylinder. (You are not required to evaluate the integral.)
\end{enumerate}

[glossy valve]

andres

[meager cave]

better

[sturdy valve]

Dynamic viscosity 😭 what the hell

[meager cave]

\begin{tikzpicture}[scale=1.1]

% Outer cylinder
\draw[thick] (-2,-3) -- (-2,3);
\draw[thick] ( 2,-3) -- ( 2,3);
\draw[dashed] (-2,3) arc (180:360:2 and 0.5);
\draw (-2,-3) arc (180:360:2 and 0.5);

% Inner cylinder
\draw[thick] (-0.8,-3) -- (-0.8,3);
\draw[thick] ( 0.8,-3) -- ( 0.8,3);
\draw[dashed] (-0.8,3) arc (180:360:0.8 and 0.25);
\draw (-0.8,-3) arc (180:360:0.8 and 0.25);

% Centerline
\draw[dashed] (0,-3.2) -- (0,3.2);

% Labels
\draw[<->] (0,0) -- (0.8,0);
\node[right] at (0.8,0) {$a$};

\draw[<->] (0,-1.5) -- (2,-1.5);
\node[right] at (2,-1.5) {$b$};

\draw[<->] (0,1.2) -- (1.4,1.2);
\node[right] at (1.4,1.2) {$r$};

\node at (1.3,2.2) {Viscous liquid};
\node at (0,2.7) {Inner cylinder};

% Gravity arrow
\draw[->,thick] (3,2) -- (3,0.8);
\node[right] at (3,1.4) {$g$};

\end{tikzpicture}

[glossy valve]

andres

[sturdy valve]

This is a math server

But ok

[meager cave]

og diagram

now that's family friendly

[meager cave]

the phy server is dead

[meager cave]

pin this too

[meager cave]

ty

so

@silk sequoia

shit

<@&286206848099549185>

https://tenor.com/view/happy-halloween-gif-12849861831894812

y'all ghosting me

[neat sluice]

hmmm

physics

[sacred yarrow]

please be more patient as physics questions usually require more time to get a response

[meager cave]

physics has some math

[sacred yarrow]

It’s not a physic server after all

[meager cave]

ok

[meager cave]

[neat sluice]

i mean

this type of question requires littrelly propery physics

[neon basin]

!15mins

[full forum]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[neat sluice]

less of maths tbh

[meager cave]

its been 25 mins

[meager cave]

hmm..

[neon basin]

Read the first part of the sentence

"once"

[neat sluice]

?????????

wtf

was that

[meager cave]

there was hentai

[neat sluice]

yeah

isnt viscous force

applied dowward

[meager cave]

<@&268886789983436800> there was porn

[neat sluice]

the guy deleted it immediately

[meager cave]

yes

[neat sluice]

but how come its applying force on the

inner cylinder

[meager cave]

idk understand it was easier to visualize orginal

[meager cave]

coz it's moving

[neat sluice]

uh

[neon basin]

ironic

[meager cave]

i did once

[neat sluice]

sorry dude i am not gona be much of help maybe some other person will help u

or if u feel alot of time

ask in the physics server

[meager cave]

physics server is dead

[neon basin]

I don't know, seems more than once to me

¯_(ツ)_/¯

[neat sluice]

better post there too

[meager cave]

was wrong ping

helpful /= helpers

[meager cave]

k

[neon basin]

Ignorantia juris non excusat

[meager cave]

what?

[meager cave]

what law i ignored

[neon basin]

Not what it means

[neon basin]

Yes

[meager cave]

so

[willow sedge]

ong

[sacred yarrow]

So are you still having problem with a question?

[meager cave]

yes

[sacred yarrow]

This thread confuses me ngl

[onyx glen]

https://tenor.com/view/troy-community-room-fire-pizza-gif-5612111

[neon basin]

If only you'd seen the original question

[meager cave]

: heart

[sacred yarrow]

Let’s make it all over again. Just wait until someone responds to you.
Only after 15 minutes do you ping helpers.

[meager cave]

did chatgpt frame everything correctly?

[neon basin]

!status

[full forum]

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

[meager cave]

1

[delicate torrent]

mods have logs so don't ưory

[meager cave]

if this helps

so it's been 15 min

@helper

<@&286206848099549185>

[neat sluice]

o

i remember this formula

dv/dh

[meager cave]

hmm..

[neat sluice]

u can take x as hieght here

like as we go down ig

[meager cave]

😭

[meager cave]

k

[neat sluice]

lol

i thought there was no waiting before..

weird

[meager cave]

something like when you pull rod out gravity pulls liquid

so viscous force downward direction

[meager cave]

eh idk

[neat sluice]

first

diffrential the equation

[knotty grail]

why does gravity play a role here

[neat sluice]

wrt r

[meager cave]

wat

[knotty grail]

isnt the liquid layer quite light

[meager cave]

nah

[neat sluice]

witth respect to r

[meager cave]

it is

[neat sluice]

basically find dv / dx

[meager cave]

k

[neat sluice]

then hmm

[meager cave]

but r is radius of outer cylinder

[knotty grail]

b

[neat sluice]

no

its b

r is just general

[meager cave]

oh

[neat sluice]

radius

[meager cave]

mb

[neat sluice]

distance from centre

[meager cave]

ok

dv/dr

[neat sluice]

then replace r by a

to get dv/dr at r = a

which is at inner cylinder

[meager cave]

k

[knotty grail]

yeah what hericulum says

[meager cave]

lemme diffrentiate

[neat sluice]

then since force being applied maybe is by shear stress

u need to find shear stress as thats what the side force is on the wall

which is just dv/dr * n

by n i mean

eta

[meager cave]

hmm

[neat sluice]

η

coeff of shear stress

its asking

force per unit length

right

on the inner

[knotty grail]

eta is the coefficient of viscosity here...

[neat sluice]

sorry wrong term

i meant that only 🤣

you multiply it with the dynamic viscoscity or whatever

[meager cave]

[
\frac{dv}{dr}
= \frac{\rho g b^{2}}{2\eta r}

• \frac{\rho g (2r)}{4\eta}
  ]

[neat sluice]

idk i assume u did the diffrentiation correctly

[meager cave]

now

[neat sluice]

the shear force (viscous force) will be

dv/dr * eta

[meager cave]

k

eta cancels

[neat sluice]

eta will cancel out

[knotty grail]

definitely not the correct expression

[neat sluice]

:/

🤣 ig recheck

[knotty grail]

$\frac{d}{dr} ln(r/a) = \frac{1}{r}$

[knotty grail]

how did you even get 1/r^2

[neat sluice]

yea its wrong

[meager cave]

ah

right

[knotty grail]

🤨

[neat sluice]

you are also missing a

[glossy valve]

andres

[meager cave]

now

[neat sluice]

in the numerator

its still wrong

[meager cave]

wdym

[neat sluice]

when u diffrentiate

ln(r/a)

u get a/r

[knotty grail]

bingo

[knotty grail]

did you forget to chain rule maybe

[neat sluice]

bruh

oh shit

yeah

a cancels out

my bad lol

anyawys

so put a

in the equation

[meager cave]

a?

[neat sluice]

inner cylinder radius

[meager cave]

ok

now i multiply by eta

[neat sluice]

ye

[meager cave]

wait

[neat sluice]

u will get the equation of viscous force

[meager cave]

wat is A here?

[neat sluice]

Area

[meager cave]

oh

area of cylinder

i'm slow

[neat sluice]

we are first finding

the shear stress

[meager cave]

ok

[neat sluice]

stress has no area

in its expression

[meager cave]

aight

[neat sluice]

its asking hmm force per unit length right

so

[knotty grail]

area exposed to the liquid

[neat sluice]

once u get that

[knotty grail]

which translates to the surface area of the inner cylinder in this case

[meager cave]

wait so

[neat sluice]

multiply stress by circumfrence of inner cylinder

[meager cave]

whole surface area or

[knotty grail]

whole

[neat sluice]

F/L = stress * 2 * pi * a

[knotty grail]

of the inner cylinder

which is $2\pi a L$ here (L is the length of the cylinder)

[neat sluice]

i dont think

you need area

its asking force per length

[meager cave]

why

that is pressure

[neat sluice]

i mean

do you have

the answer key

[meager cave]

no

[neat sluice]

dang

[knotty grail]

pretty sure no

[meager cave]

[
\eta \left.\frac{dv}{dr}\right|_{r=a}
= \frac{\rho g b^{2}}{2 a}

• \frac{\rho g a}{2}
  ]

[glossy valve]

andres

[neat sluice]

yea

[meager cave]

so now

[knotty grail]

the entire expression becomes $F = \eta \cdot 2\pi a L \cdot (\frac{\rho g b^2}{2\eta a} - \frac{\rho g (2a)}{4\eta a})$

[knotty grail]

divide by L to find force per unit length

[neat sluice]

exactly

thats what i said

just

directly

[neat sluice]

^

[meager cave]

oh

ok

[neat sluice]

nice

[meager cave]

first part done

[neat sluice]

😢

[meager cave]

[
\eta A \frac{dv}{dr}
= \frac{\rho g a^{2}}{4}
\left(\frac{b^{2}}{a^{2}} - 1\right)
]

[glossy valve]

andres

[meager cave]

ikr

sorry to fuck your time

[neat sluice]

na dude

its a fun

[knotty grail]

poetic

[neat sluice]

i anyways not doing much xd

[meager cave]

:kekleo

[meager cave]

you are

[neat sluice]

2nd part is pretty simply ig

[meager cave]

ye

[neat sluice]

its just

integral (b to a) v.dA

it says u dont gotta evaluate

even better

[meager cave]

so A would be 2pir again?

[knotty grail]

to simplify even further, $\int_{r=a}^{r=b} v(r) 2\pi r dr$

[meager cave]

k

[neat sluice]

yea

[meager cave]

now i just substitute

[neat sluice]

2 * pi * r * dr

= dA

[meager cave]

tysm guys

you were a g

[neat sluice]

welc

[meager cave]

fuk them

[neat sluice]

lol

[knotty grail]

what crime did bro commit

[meager cave]

nothing

so @full forum we done

[knotty grail]

just .close

.close

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[meager cave]

.close

i can sleep now

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[green ingot]

can someone look this proof to me
i wrote it

[green ingot]

cyclic groups***

just a typo

[quartz flare]

The first paragraph looks good

Yep

Good

[zenith kernel]

Correct

[quartz flare]

I think it's proved a little more succinctly by taking the element first and looking at the subgroup it generates but that's a matter of semantics, what you have is good 👍

[green ingot]

i thought it is better to prove another statement which is "the only subgroups of a finite group with prime order are the trivial subgroup and itself" then to proceed the proof

maybe then i should keep them seperated?

[quartz flare]

You're good, what I meant is in mine I started with your end then proved that statement as a part of it

Like

Let G be a group s.t. |G| = p prime and consider g in G. <g> is a subgroup of G, so |<g>| | p by Lagrange's Theorem. Since p is prime, then, either |<g>| = 1, in which case g = e and <g> is the trivial subgroup, or |<g>| = p. But if |<g>| = p and <g> is a subgroup of G, then <g> = G and G is cyclic generated by g.

Fun way to say that for a group of prime order either an element is the identity element or it is a generator

[green ingot]

yeah exactly thats what the statement is saying actually

[quartz flare]

It's nice because then it's clear how they're all isomorphic to Zp and we could have proven it with a homomorphism instead

[green ingot]

yeah when i proved the general statement "every finite cyclic group is isomorphic to Z_n" i didnt consider that special case so i didnt notice that

true

[full forum]

@green ingot Has your question been resolved?

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[wind wedge]

I believe this is solved correctly. but how can I write this in a mathematical expression instead of just words? Part of that question also is why it is differentiable / non differntiable.

[wind wedge]

Especially the why is unclear to me

[quartz flare]

In terms of the why, are you familiar with the (limit) definition of the derivative?

[wind wedge]

barely. I believe this is the definition we're looking for?

[quartz flare]

Yes

[slate violet]

also, the slope of a straight line is constant

[quartz flare]

The meaning of that definition is that as we look at the change in the function between closer and closer points, we should find that the derivative gets closer and closer to a single number

The problem is with what south pointed out

[slate violet]

when they say 'limit', that assumes that the left and right limits exist and are equal to each other

[quartz flare]

Exactly, when we look at point x1, consider what the slope of the line on the right looks like compared to the slope of the line of the left

If they don't match (i.e. aren't equal) the derivative can't exist

[wind wedge]

so a derivative in a point only exists if the slope of the points to the left and right side are equal? I think I am not getting it correctly. Since from my understanding that'd mean we could only differentiate linear functions, which obviously is not the case.

[quartz flare]

Yes 👍 the first part is correct

[queen flame]

limit

[quartz flare]

The second part deals with this

It's not necessarily that they must be constant

But that on the left and right they come to the same value

In general we call this "smooth" function

https://tenor.com/view/secant-line-converging-to-tangent-line-animation-gif-18716967

Look at how as Q gets closer to P, the red line gets closer to the green line

Even though the function (in black) is curved

The 'slope at the point' (what we call the derivative) is reaching to a certain number

[wind wedge]

okay, I believe that made it clear for my part. Thank you so much!

.close

[full forum]

Channel closed

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Use .reopen if this was a mistake.

[quartz flare]

Great! No problem 🙂

[full forum]

Available help channel!

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[wintry smelt]

So I got this grade on my final mathematics exam.

The average is 72,42%
The median is 79,5%

But what are these lower and upper quadral?

[quartz flare]

So you know median is 50%

Quartile means 25%

Min = 0%
Lower quartile = lower 25%
Median = 50%
Upper quartile = 75%
Max = 100%

So 25% of people got scores between 10 (min) and 62.5

25% of people got scores between 62.5 (Q1) and 79.5 (Q2)

25% of people got scores between 79.5 (median) and 91.6 (Q3)

And 25% of people got scores between 91.6 and 100 (max), including you!

[wintry smelt]

LOL! so I got percentile of minimum 75?

[quartz flare]

Yep

And probably higher but we can't tell

[wintry smelt]

we can't calculate what centile i am?

[quartz flare]

Well based on the average actually yeah almost certainly higher

[quartz flare]

I don't think so without all of the data

[wintry smelt]

i mean its canvas so thats everything i get?

do you even have canvas

[quartz flare]

If you estimate its distribution you can guess a value but that's about it

I've seen it before and yeah if that's all you get then that would be all you could know

[wintry smelt]

So I can say that I got better grade than 3/4 (or possibly even more) of the class?

[quartz flare]

Yes 👍

[full forum]

@wintry smelt Has your question been resolved?

[wintry smelt]

@quartz flare also which is more accurate median or average

[quartz flare]

Depends on what you want to know

[wintry smelt]

like what grade getting below is considered as worse than others

[restive geyser]

so, by average do you mean "mean"

because the median is a type of average, technically speaking

[quartz flare]

Yeah the canvas thing says mean I think

[quartz flare]

Probably median? I meant to provide context but forgot to come back haha

[restive geyser]

oh shit it's a translation...

From Icelandic...?

[quartz flare]

Median gives a better understanding of how people did because it's resistant to the actual values of the scores

But mean will give you a better understanding of the 'expected grade', if you will (like "an average student in this class will have [mean] grade")

[restive geyser]

I don't want to sound rude by any means

[quartz flare]

If I'm wrong correct me please 👍

My statistics aren't super duper 😅

[restive geyser]

No this isn't directed at you

[quartz flare]

No clue what language that is

[restive geyser]

But I would feel that someone who can attain such a score here would be able to interpret this graph

[quartz flare]

Mm but what if it's a calculus class

[restive geyser]

This is precalc

[quartz flare]

It just says final exam (lokapróf)

Ah

[restive geyser]

It's a box-and-whisker plot, in English, typically

While they get re-mentioned in the UK during the A Levels (so around the same time as calculus), they appear earlier

[restive geyser]

But I mean - like, "top quarter of participants"

[quartz flare]

I forgot they made us learn all these names

Don't ask me what a stem and leaf is ❗

[restive geyser]

Well, I have to teach these concepts as well, so it's doubly critical for me to know what these are called

[quartz flare]

Oo a teacher

nice!

[restive geyser]

Well, a tutor, so slightly different

[quartz flare]

ahh

[restive geyser]

And by the nature of this server I've had to learn some in some other languages too

[quartz flare]

Thankful none of my tutees have come in for data representations

[quartz flare]

haha I've only been here 3 days and I feel that

I do kind of wish there was some support for other languages, even just a bit

[restive geyser]

There technically is; there's no regulation here that says help channels must be in English

[quartz flare]

One channel for it with specifically approved languages (since they're worried about the mods not being able to speak them) or such

[restive geyser]

Merely that it is advised to do so in English, just because it ups the chance that someone may answer

[quartz flare]

You're right

XD

I wish I could @ the mods to hear their opinions but I don't want to abuse it

[restive geyser]

I have considered making a request, but it is understandably still a moderation issue

Nonetheless:
@wintry smelt All that plot tells you is that you're in the upper quartile of students; unless you have over a hundred of them, it doesn't make sense to ask about percentiles, in any case

[quartz flare]

True

And even with over 100 percentiles can be deceiving 🥹

My 167 out of 170 points on the math portion of the GRE putting me in the 76th percentile goes to show that

[wintry smelt]

Ik what median and average mean

[wintry smelt]

Why do i need to be in a class of 100 students for percentile to work?

[wintry smelt]

Like being in 76th percentile means that 24% people did better than you

[quartz flare]

Not for it to work, but for it not to have gaps

If you have 20 people for example, the only possible percentile scores are 0, 5, 10, ..., 95

Second best person will only report as having done better than 90% of others

Which is true, but not necessarily telling the truth of the matter

[quartz flare]

Yes, which if I only told you the percentile score seems pretty mediocre

[restive geyser]

That is to say, it's functionally meaningless to work in percentiles with such a small number of people

[quartz flare]

But then when I tell you I only missed 3 points then it's like hm

[restive geyser]

To pull a Godot from Ace Attorney, it's like saying I've never lost a court trial as a prosecutor (while I have never been a prosecutor)

[quartz flare]

XD literally

Get that false implies anything

[restive geyser]

Veraciously speaking [i.e. insofar as how true this statement is], no problem; but functionally speaking... "what"

[wintry smelt]

You mean 95%?

[restive geyser]

Not really

[wintry smelt]

Anyways what percentile did I get?

[restive geyser]

If it's the 2nd best, there are 18 people who did worse

[wintry smelt]

75?

[restive geyser]

i.e. 18/20, i.e. 90 percent of people

[quartz flare]

Somewhere >= 75 and <= 99

[restive geyser]

Like mentioned earlier, there's no information to this extent as to what percentile you got, just what quartile you've got

[wintry smelt]

So I got into higher quartile

[restive geyser]

You're in the top quarter, yes

[wintry smelt]

Q1 - 10% - 62,5%
Q2 - 62,6% - 79,5%
Q3 - 79,6% - 91,6%
Q4 - 91,7% - 100%?

4th?

[quartz flare]

Yeah I think that's what I reported above

So yeah

[restive geyser]

yh

[quartz flare]

Which is what Waes said

[restive geyser]

(though, strictly, quartiles refer to the boundaries there)

[wintry smelt]

Why do you count 2nd quartile to the median?

[restive geyser]

(this is slightly wrong in phrasing, as a result)

[wintry smelt]

But not to the average

[restive geyser]

?? because the quartiles refer to the literal spread of the data points?

So, Q1 the lower quartile is the boundary between the bottom 25% and top 75%

Q2 the boundary between the lower 50% and upper 50%

Q3 similarly

Thus, Q2 is the median

Out of interest, how much statistics have you covered at school?

(because there are ways of attaining similar concepts with the mean, but it is rather high level, so I don't really want to encroach it if I can)

[quartz flare]

Yeah like the way I reported it originally ^

[wintry smelt]

I had statistics course actually in last year

[restive geyser]

But what was in that course?

(and what year-group was it in)

[quartz flare]

Maybe they did only the essentials of descriptive and skipped to inferential

I hope

🙂

[wintry smelt]

https://namskra.is/courses/538dd87f62933a13a10011de

[wintry smelt]

No.. Q1 - 0 - 25%
Q2 25%-50%
Q3 50-75%
Q4 75%-100%

[quartz flare]

0.0

[restive geyser]

Q1 is where the border between the first two quarters is

[quartz flare]

Maybe it's a language difference

[wintry smelt]

No

[quartz flare]

Then you should look it up

You're easily disproven 🧐

[restive geyser]

It isn't, the terms used for the quartiles are in the screenshot, used the same way I'm using them, i.e. as boundaries

[quartz flare]

There's a few things in the syllabus that suggest you should know this o.o and most all what I explained earlier

[quartz flare]

True I forgot

[restive geyser]

This course appears to be around the 16-year-old mark, so I'll go over it briefly

If you want to similarly compare the spread of data but using the mean, you need to calculate the standard deviation of the data

[wintry smelt]

Ik what that is

Lower quartile - 0%-50%
Upper quartile 50%-100%

[restive geyser]

...

you know the QUART- in "quartile" literally means "(one of) FOUR (parts)", right?

[quartz flare]

I think it's time to .close and go home 💀

[quartz flare]

Actually that's a great point

[quartz flare]

That was one of the first things I put forth

They must be baiting

[wintry smelt]

And why does second quartile end at average

[restive geyser]

So you really haven't been listening

The second quartile, Q2, is the boundary between the bottom 50% and the top 50%, i.e. precisely the median

It is NOT THE BOX of data between 25% and 50% of the data

Here's a very simple example of another box plot

Here, Q1 = 3.5, Q2 = 6.1 and Q3 = 9.6

[wintry smelt]

So Q2 ends at 6.1?

[restive geyser]

NO

Q2 IS THE VALUE 6.1

IT IS NOT THE BOX.

[quartz flare]

Should I call the mods before you have an aneurism

[restive geyser]

not just yet

[wintry smelt]

Here…

[restive geyser]

and even this example is telling the same thing as me

Q2 IS the number 36 here

[wintry smelt]

Q2 starts at end of Q1 and ends at median

[restive geyser]

No

Q1 IS the number 26 1/2

It's not a range of values up to that point

[wintry smelt]

Oh and if we divide the number set in 4 pieces then 2nd piece is at the median oh ok

[restive geyser]

Did you happen to take a finance course, by any chance?

Because you might be confusing Q1 - Q4 from that; in THAT context, they're QUARTERS

[wintry smelt]

And how do we know that lower quartile is 3.5?

[restive geyser]

because you're having to split the data between the value 3 and 4, to separate the bottom 25% and top 75%

So you take Q1 = (3+4)/2 = 3.5

[wintry smelt]

Huh

Why not 2+3

[restive geyser]

...

You see how there are 8 numbers in the data?

[wintry smelt]

Ya and?

[restive geyser]

How many of these would you need to make a quarter of the data, then?

[wintry smelt]

2…

[restive geyser]

Right

We take the lowest two data points (2 and 3) and put them in this quarter

The next point, 4, would be in the next quarter

The first quartile is explicitly, definitionally, the boundary between these two quarters

That means Q1 has to be between 3 (the largest value in the first quarter) and 4 (the smallest in the next quarter)

[shut barn]

this helped clarify a vocabulary question I've had for a long time, thanks math help stranger

[wintry smelt]

2,3 are in first quartile

[restive geyser]

Convention is, to let Q1 then be the mean of these two values, i.e. 3.5

[wintry smelt]

,w mean 2,3

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by fuzzials

[wintry smelt]

It’s 2,5

[restive geyser]

Are you reading what I am saying.

[restive geyser]

Is this not clear?

[wintry smelt]

Why

[restive geyser]

Because ^

[wintry smelt]

Oh that’s why Q2 is just the median

Oooh

.close

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Channel closed

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[cedar flint]

help please i dont know how to start other than to plug in random palindromes for b

[zenith kernel]

Find the minimal palindrome that >=2015. This you can’t avoid I guess

It should be in the form of 2xx2 right?

[cedar flint]

no? that wouldn't make sense

i think b is a 5 digit number

[zenith kernel]

But 2xx2 are all palindrome, 2112, 2222, 2332,…,2992

[cedar flint]

yes but a wouldn't be a palindrome

[zenith kernel]

Minimal one being 2112

[cedar flint]

the middle 2 digits of a 4-digit palindrome have to be the same

[zenith kernel]

English is a foreign language to me. Palindrome exact meaning please, a definition perhaps ?

[zenith kernel]

Yeah, 2xx2 satisfy this for any x

[brittle sun]

Nah you're good

[cedar flint]

but the middle digits of 2015 are 01 and if a were to a palindrome, b's ten digit would be 1 greater than its hundreds digit (considering b is in the form 2xx2)

[zenith kernel]

2112>2015

[brittle sun]

Oh his issue isn't with b, it's with a

[zenith kernel]

Oh, I meant b=2112

a being smaller is equivalent to b being smaller

[brittle sun]

We also need a to be a palindrome

[brittle sun]

So this doesn't work cuz a = 97

[cedar flint]

i think b has to be a 5 digit number

but i dont know what

[zenith kernel]

Jesus. I will never do math again without my energy drink.

Forgot a has to be that thing too

Really brutal force checking 2xyx2-2015= 7zz7?

Oh no, still 4 digits

2xx2-2015=7 x-2 7

x=your name

[cedar flint]

no offense but i think its best if we let a helper assist me on this problem

<@&286206848099549185>

[zenith kernel]

But it’s solved

(p,q,r,s) represents 1000p+100q+10r+s
(2,x,x,2)-(2,0,1,5)=(0,7,x-2,7) has a solution

Which is exactly your name

[delicate torrent]

hm

[zenith kernel]

(Second term of both sides need to be equal)

[cedar flint]

so youre saying a=7?

[zenith kernel]

I am saying b=2772

[zenith kernel]

Letters edited. Didn’t notice symbols duplicated

[delicate torrent]

hold up

assume that $b$ is a 4 digit, then $b = 1000b_1 + 100b_2 + 10b_2 + b_1 = 1001b_1 + 110b_2$

[glossy valve]

1 divided by 0 equals Infinity

[zenith kernel]

If you still don’t believe it:

,w calculate 2772-2015

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by cogwheelsofthemind

[delicate torrent]

🔥

[zenith kernel]

I summoned someone you might believe

A 200 million worth commercial tool

[drowsy finch]

so a is 757?

[zenith kernel]

Yeah

[cedar flint]

ok gracias

[zenith kernel]

(It’s just that a=2xx2-2015, last digit is 7, so a has to be of the form 7y7, actually y=x-2)

Np

[cedar flint]

.close

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Channel closed

Closed by @cedar flint

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[full forum]

Available help channel!

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[robust gorge]

could u guys check this one? i am not sure about the second option.

[robust gorge]

@floral scroll

[quartz flare]

Looks good to me

[robust gorge]

alright thanks cool like always

ill close the channel now

.close

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[full forum]

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[thin lark]

(μ u^2 + kux^2 - μxg) dx + μux du= 0

[thin lark]

is this supposed to have an integrating factor to become exact or some?

[drowsy finch]

no, it’s not exact and finding an integrating factor might be the way to go

[thin lark]

yea thats what im asking is there an integrating factor to make it exact

[drowsy finch]

we would need to find one potentially based on either x or u depending on the structure of the equation

[thin lark]

or both

[drowsy finch]

exactly

[thin lark]

yea whats the substitution though

[drowsy finch]

you can try substituting μ(x) or μ(u) and see if it simplifies the equation

[thin lark]

which μ(x)?

μ(x) = ?

[drowsy finch]

μ(x) = x^n

[thin lark]

with x

i get dM/du = 2μux + kx^3 , dN/dx = 2μux

[drowsy finch]

hmm

[thin lark]

also im not sure if its supposed to be ku^2 x^2 or kux^2

the problem says that its proportional to the square of the length and velocity

idk if this means square applies to velocity too

[drowsy finch]

if it's proportional to the square of both the term should be $ku^2x^2$

[glossy valve]

anflo

[thin lark]

when saying of the lentgh and velocity

does it mean the square of both

or square of length and proportional to the velocity

[drowsy finch]

yes

[thin lark]

ok

still its not looking great

wait ill send the whole problem incase i made a mistake

A homogeneous chain of great length, with linear mass density
𝜇
μ, is initially at rest on a horizontal table, coiled around a hole in the table.

At some moment, one end of the chain is released and begins to slide vertically downward through the hole.

The air resistance force is proportional to the square of the length
𝑥
x of the vertical part of the chain and its velocity

Find u(x)

ΣF = dp/dt <=> mg - ku^2 x^2 = dm/dt u + du/dt m <=> chain rule ... to the equation i sent

[drowsy finch]

Oh this is a variable mass problem

[thin lark]

yea

but u_a= 0

so dm/dt ua =0

but idk i cant really do anything with dp/dt cause there are 3 variables chain rule doesnt help so im exanding it

but nothing happens

this is supposed to be the answer

looks like its definetely came from exact equation but idk what a is supposed to be prob some big factor

[drowsy finch]

hmm

let me try solving

[thin lark]

alr ty

this is mechanics 1 class so it shouldnt be hard in theory

[drowsy finch]

okay i thinkk i got it?

don’t think of this as an exact DE / integrating factor problem that form is kinda misleading

everything depends on x not independently on t,u,x

[thin lark]

wdym

how did you solve it

[drowsy finch]

use [\frac{dp}{dt}=\frac{d(\mu x u)}{dt}]

[glossy valve]

anflo

[thin lark]

i did

this is d(mu)/dt = dm/dt u + du/dt m

[drowsy finch]

then apply chain rule

[thin lark]

i did

dx/dt = u

dm/dx = μ

thats how i got the original ODE

[drowsy finch]

use [\dot{x} = u, \qquad \dot{u} = u,\frac{du}{dx}]

[glossy valve]

anflo

[thin lark]

yea i did

thats how i ended up with only u,x

but i cant split variables

[drowsy finch]

so u get [\frac{d(\mu x u)}{dt}=\mu\left(xu,\frac{du}{dx} + u^2\right)]

[glossy valve]

anflo

[thin lark]

yes

this liturately goes to (μ u^2 + kux^2 - μxg) dx + μux du= 0

or (μ u^2 + ku^2 x^2 - μxg) dx + μux du= 0

actually

[drowsy finch]

yep, that DE won’t separate in u,x

set [y = \mu x u]

[glossy valve]

anflo

[thin lark]

ok

but i think this complicates more

wait i mreally close

integrating factor x^1/2 /u^2

you get μ x^3/2 g/u and 2/5 μ x^3/2 / u

so dM/du = 2g/5 dN/dx

can i get anywhere from here they are proportional

[drowsy finch]

hmm

what i did

[thin lark]

i feel like we have a mistake

because this ODE has like gaussian indefinite integral in answer

[drowsy finch]

$y = \mu x u$ $\frac{dy}{dx} = \mu \left( u + x \frac{du}{dx} \right)$

uh one sec

[glossy valve]

anflo

[drowsy finch]

and substitute this into the original equation

this is what i did it should simplify the terms

[thin lark]

how do i find dy/dx

[drowsy finch]

you differentiate?

[thin lark]

and then?

how do i sub dy/dx in

[drowsy finch]

ohh wait lemme show u

[thin lark]

its like du/dx = (dy/dx 1/μ - u) 1/x

[drowsy finch]

this leads to : $\mu x \frac{du}{dx} = \frac{dy}{dx} - \mu u$

[thin lark]

so i have to put this instead of du/dx

yea we have du/dx in it

[thin lark]

so i can change it like this

but cant see how this is not more difficult

[glossy valve]

anflo

[drowsy finch]

$\frac{y}{x} \frac{dy}{dx} = g \mu^2 x^2 - \frac{a}{\mu} y^2$

[glossy valve]

anflo

[thin lark]

yea ok

[drowsy finch]

you get this by substituting $u = \frac{y}{\mu x}$

[glossy valve]

anflo

[drowsy finch]

then you can just separate the variables and rewrite the denominator

[thin lark]

this the sol

but

we are supposed to get this

[drowsy finch]

oh

if we substitute y=μxu and proceed through the steps carefully the logarithmic term should naturally appear during the integration proces

[drowsy finch]

yeah i got this

[thin lark]

how

[drowsy finch]

i'll show u

[glossy valve]

Taebek

[thin lark]

no wait

[drowsy finch]

from this - $\frac{y , dy}{g \mu^2 x^2 - \frac{a}{\mu} y^2} = dx$

[glossy valve]

anflo

[thin lark]

$$
(\mu u^2 + k u^2 x^2 - \mu x g) , dx + \mu u x , du = 0
$$

[glossy valve]

Taebek

[thin lark]

we both have this right?

[drowsy finch]

yeah

[thin lark]

and the sub was what ?

y = μux?

[drowsy finch]

yea

is something wrong

[thin lark]

(y^2 - y)/μx^2 + ky^2 / μ^2 - μxg ) dx + y/μx dy = 0?

[drowsy finch]

you differentiate then substitute

[thin lark]

$$
\left( \frac{y^2 - y}{\mu x^2} + \frac{k y^2}{\mu^2} - \mu x g \right) dx + \frac{y}{\mu x} , dy = 0
$$

[glossy valve]

Taebek

[thin lark]

this ?

[drowsy finch]

.

[thin lark]

yea thats hwat i used

[thin lark]

and i ended up with this

[drowsy finch]

$u \frac{dy}{dx} - \mu u^2 = \mu x g - \mu u^2 - a x^2 u^2$

[glossy valve]

anflo

[thin lark]

whats this

dont you need to replace u

i replaced all u in terms of y and x

so we only have y and x

[drowsy finch]

okay wait i'll show u step by step

$\mu x \frac{du}{dx} = \frac{dy}{dx} - \mu u$ do you agree with this

[glossy valve]

anflo

[drowsy finch]

now multiply by u $\mu x u \frac{du}{dx} = u \frac{dy}{dx} - \mu u^2$

[glossy valve]

anflo

[thin lark]

y = μux => dy/dx = μ ( du/dx x + u)

<=> du/dx = ( 1/μ dy/dx - u) 1/x

<=> $$
\left( \frac{y^2 - y}{\mu x^2} + \frac{k y^2}{\mu^2} - \mu x g \right) dx + \frac{y}{\mu x} , dy = 0
$$

[glossy valve]

Taebek

[drowsy finch]

then u substitue to get $u \frac{dy}{dx} - \mu u^2 = \mu x g - \mu u^2 - a x^2 u^2$

[glossy valve]

anflo

[thin lark]

i also subbed u = y/μx

so i only have y,x left

[thin lark]

it should be the same as mine

i just went ahead and multiplied by dx

and subbed wherever there is u u=y/μx

[drowsy finch]

this will lead to complications

wait did you get this in the start $\mu x u \frac{du}{dx} = \mu x g - \mu u^2 - a x^2 u^2$

[glossy valve]

anflo

[thin lark]

yea

this is what i started from

and i just made sure i only have y and x left

[drowsy finch]

you substitute this into that

[thin lark]

yea i did bro

this is what i got after subbing du and u in terms of dy dx y x

[drowsy finch]

then u should get this wyf 😭

you there?

[drowsy finch]

this isn't wrong but when you substitute $u = \frac{y}{\mu x}$ into the term $\mu u x , du$ you have to apply the quotient rule to the differential

[glossy valve]

anflo

[drowsy finch]

it'll get pretty complicated

[thin lark]

yea sorry

i was checking if ther e is a way to solve

[thin lark]

yea i did

i solved for du/dx

[drowsy finch]

oh no worries i thought you were fed up of me 😭

[thin lark]

in terms of dy/dx

so i replaced what i got into du/dx

so the differential changes

and then u= y/μx everywhere

but the new ODE seems just as difficult

[drowsy finch]

the equation is linear in terms of $y^2$

[glossy valve]

anflo

[drowsy finch]

did your answer have an $e^{x^2}$ or an $\text{erf}(x)$ term in it?

[glossy valve]

anflo

[thin lark]

yea if we get there its fucked

we arent supposed to get indefinete gaussian integral in first semester

or is electrical engineering supposed to be like that

🥀

[drowsy finch]

if so then you're on the right track

[thin lark]

no i dont think we are supposed to get none elementary functions in first semester

its too much for it

the other problems were simpler

[drowsy finch]

your approach works but it's a bit more complicated

[drowsy finch]

this should be the easiest

[thin lark]

but im telling you

i think we have something wrong

it is supposed to be much easier

maybe there is an easier way or we forgetting some x etc that would make it exact

or we were supposed to not expand dp/dt

[drowsy finch]

its not that hard after that step

but if you so sure then hmm..

[dusky locust]

Is u dependent of x?

[drowsy finch]

yes

[dusky locust]

So this equation is something like
(μ u^2 + kux^2 - μxg) + μux du/dx= 0

Why cant we isolate u' and integrate?

ah

forget it xd

[drowsy finch]

😭

[dusky locust]

I mean, you can always say "By peano theorem there is a solution" and proceed by "If you want to find it use numerical analisys"

[drowsy finch]

hmm that's a way to look at it

i have to leave now but if you guys manage to get the solution please ping me!

[thin lark]

yea fuck that im moving on to work-energy

ive had enough of dynamics

[gilded mauve]

Yoo guys

[drowsy finch]

lmao but if u do find a solution pls tell me im curious to see how this turns out

[thin lark]

yea i asked in the uni group maybe someone from previous years knows

but im telling you its some minor mistake

this is the first semester

[drowsy finch]

this fried my brain so much

[thin lark]

bro wtf

[drowsy finch]

im thinking so too

[thin lark]

why is there grad div and curl in work in mechanics 1

[drowsy finch]

wildd

[thin lark]

this is first semester ffs

we have like calc 1

but its really calc 1-2

anyways yea ill dm you ig if i get a solution

can add me so i remember

[drowsy finch]

oh alright

thanks!

[thin lark]

wait uhm

you know how they changes dr to dS uτ?

oh its cause in the limit the curve and straight line collapse to the same line ?

[drowsy finch]

as the curve straightens the small change in position dr becomes the same as the straight line distance ds

[thin lark]

hey also

[drowsy finch]

😭

[thin lark]

why is Fx(x,yA,zA)

isnt Fx(x)

x direction F should only have x no?

[thin lark]

what

[drowsy finch]

$F_x(x, y_A, z_A)$ is the force along the $x$-axis with $y_A$ and $z_A$ held constant

[glossy valve]

anflo

[drowsy finch]

the same applies for $F_y$ and $F_z$ where the other coordinates are fixed.

[glossy valve]

anflo

[thin lark]

shouldnt it be Fx(x,0,0)

oh yea

the x axis can be moved

[drowsy finch]

yea

[thin lark]

alr thanks a lot ill lyk if i get anyone to solve it

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[dusty scroll]

guys i need help i dont understan whats going on

[tall onyx]

Copter just asked this question not long ago

What's your doubt

[dusty scroll]

Tbh, I don’t even understand the last sentence of the problem statement.

or.... I dont know how to prove that

[tall onyx]

Have you ever worked with sigma notation?

Uh, it's basically say here

You need to prove this, you can ignore what below

[grand hatch]

to prove it, use the fact that

1. for each i, xi + yi = n-1 -> justify
2. sum xi = sum yi = n choose 2 -> justify

[sonic stratus]

Lets start from the obvious question, do you understand what Xi and Yi represent to each player / ith player?

and knowing that, point 1) from what's should be a pretty obvious conclussion

[full forum]

@dusty scroll Has your question been resolved?

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[dusty scroll]

Thank you guys so much❤️ I can do it now

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[magic elbow]

i took the cases x >= 0, and x < 0 and formed simultaneous equations

[magic elbow]

$x^2 = px + q$

[glossy valve]

keqae

[magic elbow]

$-x^2 = px + q$

[glossy valve]

keqae

[magic elbow]

then, $0 = 2px + 2q$

[glossy valve]

keqae

[magic elbow]

$x=-\frac{q}{p}$

[glossy valve]

keqae