help-28

[restive geyser]

Keine Problem, sind fast gleich da, glaub ich

[proven lynx]

so ein drecks thema um ehrlich zu sein

[onyx glen]

i see that my request not to call me "bro" was not acknowledged

[restive geyser]

Ist oft kaum gut erklärt

[safe nova]

GUYS don’t ignore this

[proven lynx]

HuhV

I didnt call you "bro"

[onyx glen]

right here, you did.

[proven lynx]

That was before you told me

Im sorry

[onyx glen]

[proven lynx]

I dont think its that big of a deal though

[restive geyser]

ANYWAYS

@onyx glen That WAS Desmos, right? that you used for the normaldist?

[gritty rose]

How about you just stop being an asshole

Not that hard

[safe nova]

We should respect how people want to be called

[proven lynx]

Alright my bad

I apologize

[onyx glen]

yes it was

[proven lynx]

I apologize for saying "bro" its just like a way of how i speak ill be careful about it in future

[restive geyser]

He apologised, drop it

[proven lynx]

Like i said, im sorry

[restive geyser]

Okay so I still get 8e-43

[proven lynx]

Hm yeah

[restive geyser]

OH

@onyx glen your bounds were off

[onyx glen]

the continuity correction barely makes a difference

[proven lynx]

Btw, when do i have to do -0,5 +0,5

is there a case where i do not have to do that correction

[restive geyser]

It's not the cont. correction

You're calculating from 700

[onyx glen]

oh 750 not 700

[restive geyser]

It's from 750

yh

[onyx glen]

lmfao

[restive geyser]

You're going from a binomial, with discrete data, to a normal approx., which defaults to a continuous variable as the assumed variable

This is when you need a +- correction

[proven lynx]

Oh

So if its not a bernoulli type thing

Then ik i dont need the correction

[restive geyser]

Sorta?

It depends on the data you're approximating

[proven lynx]

Oh

Oh yeha i remember

So like

Height for example wouldnt be needed

[restive geyser]

If you're measuring e.g. "the number of minutes" it takes for something to happen, that's a continuous variable

yh

[proven lynx]

A correction*

Damn okay

And when its not discrete i just do it the exact same way just without a correction?

[onyx glen]

#serious-discussion but yes i do.

do not call me "dude".

[restive geyser]

BRUH

[onyx glen]

but thank you ig.

[restive geyser]

but yh

[proven lynx]

Muss mir jetzt nur noch Hypothesentests und kombinatorik anschauen, denkst du es wird schwer?

[rare dock]

kombinatorik?

[proven lynx]

?

[onyx glen]

oh come the fuck on.

[proven lynx]

Like combinations

[onyx glen]

i told you not to call me "dude".

<@&268886789983436800> transphobia.

and/or deliberate boundary-stomping, idk.

[odd locust]

dont misgender people on purpose. take a day to reflect

[onyx glen]

psgwsp

THIRD TIME??

[restive geyser]

Schau mir mal was du mit Kombinatorik meinst?

[proven lynx]

alright drop it

[cunning sapphire]

"dude" again fr

[haughty ocean]

isnt dude gender neutral?

[cunning sapphire]

No, a dude is a man

[onyx glen]

you are a fucking rotbag

[safe nova]

If someone doesn’t want to be called a certain way, DON’T
(Just a reminder of some common sense)🙏🙏

[proven lynx]

Ummm

[onyx glen]

anyway

[restive geyser]

Depends on the speaker, but she did literally just stipulate not to call her that

[proven lynx]

Kinda messy

[onyx glen]

this is turning into a shitshow.

[haughty ocean]

Does not look like transphobia to me

[onyx glen]

!redir

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

[proven lynx]

can we calm down

[restive geyser]

@proven lynx

[proven lynx]

Ja zeige es dir gleich es eskaliert hier gerade 😭

[fading harbor]

i thought it was

[cunning sapphire]

Please remember this is a help channel, this isn't the place to discuss gender-neutrality of the word "dude". Please move on with actual maths or move elsewhere

[restive geyser]

Zeig's anyways

-# Chomsky's linguistics has entered the chat

[proven lynx]

genau sowas

Es gibt 6 verschiedene formeln wtf

Wie soll man sich sowas merken also ohne scheiß digga

Warum übertreiben die so

[rapid sable]

hi honey

[restive geyser]

These should really be explained one at a time tbh

[restive geyser]

!redir

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

[proven lynx]

Hmm yeah i guess ill watch a yt video on it

[restive geyser]

yh, best bet

[proven lynx]

Anyways ill look into hypothesis tests now thats also a fucked up topic

But tysm for the help

[restive geyser]

Wenn's dir nur Nilpferden-Testen fehlen, dann geht's okay, denke ich

Ich meine, that's the only one left to revise

Do still learn them lol

übrigens,

!done

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@proven lynx Has your question been resolved?

[steel solar]

wow this thread was a mouthful

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[spice grail]

im trying to solve this to get r

[onyx glen]

if the numbers were not scary would you know how to do it?

maybe you could even show the original formula that you had before you plugged in some measurement data into it

[spice grail]

the original formula is a statics formula

[onyx glen]

so far so good, you now have

(NUMBER) - r^4 = (NUMBER)

though some funny business is going on with bracketing

and your handwriting in general

[spice grail]

yh sorry about the brackets

i think somethings wrong here

[onyx glen]

why so

i mean you replaced pi/2 in the denom with 0.03 for no good reason ig

[full forum]

@spice grail Has your question been resolved?

[spice grail]

so it would be this?

[somber cloud]

yes

[spice grail]

this doesnt give an another though

[somber cloud]

another what

[spice grail]

i meant an answer

[somber cloud]

well the answer will be plus minus of this humongous value

[hollow aspen]

@spice grail never use radical symbol it's ugly

use ()^1/4 instead

[spice grail]

ok

thanks

[hollow aspen]

is that (0.03)^4 or 0.034

[somber cloud]

its the power

[spice grail]

(0.03)^4

its solved now

[hollow aspen]

always use brackets broski

[somber cloud]

r u sure its this big of a colculation

id be scarred by now

[hollow aspen]

man this looks ugly tbh

[somber cloud]

fr

[spice grail]

¬_¬

[hollow aspen]

this question is dumb quite frankly

they aren't testing anything

[somber cloud]

ye

[somber cloud]

fr

[hollow aspen]

can you show me your solution @spice grail

[somber cloud]

endless calc wont build aptitude

[spice grail]

its part of a statics problem

i got d = 0.0275

[somber cloud]

understandable

[spice grail]

thanks guys

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[lilac swallow]

topic: trig identities: the question asked to factor the expression and use fundamental identities to simplify

[lilac swallow]

im just confused as to why the answer key stopped at the answer i boxed

my instict would be to further simplify

[wintry cosmos]

what would you further do?

[wintry cosmos]

cot and cosec - are also considered to be standard functions persay

[lilac swallow]

i would put cot²x into its quotient identity and put csc²x into it's reciprocal identity and try to do something from there i think....

[lilac swallow]

what is standard function... 🤔 sorry...

[grizzled copper]

"standard" trigonometric functions really just demonstrate a ratio of sides for a given right angle triangle

cot and csc are good enough, because they are representative of a geometric ratio, which is why cot csc and sec are considered "standard functions"; a function like sinc(x), however, would not be

i think thats what he meant, and why its fine to leave in cot csc sec

[lilac swallow]

ohh okay thank you so much i get it now

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[glossy breach]

hello

[glossy breach]

can someone help me figure this out I'm having an existential crisis

[astral sinew]

!da2a

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Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

@glossy breach Has your question been resolved?

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[rich rose]

Im confused because I did 35 and got no where near the correct answer which was 1.125 rev/min, and I got 67.5. I have no idea where I went wrong, and I don’t even know how to tackle 37. 1st image is the problem, 2nd is my work, 3rd is my math notes

[rich rose]

Sorry first is upside down gimme a sec

[dire swallow]

Did you convert?

[rich rose]

Wdym?

[dire swallow]

Did you convert from per hour to per min?

[rich rose]

No idea let me look

[dire swallow]

Because it just so happens that your answer is 60 times of the actual answer

[rich rose]

So your saying I was one dig away 💔

[dire swallow]

One conversion away

The 67.5 was per hour

Per min, it would make it 67.5/60, which is 1.125

[rich rose]

Give me a sec

Im so upset but so happy

Can you help me with 37 please

[dire swallow]

Do you know the equation v=rw?

[rich rose]

Yea that’s what I learned yesterday

[dire swallow]

First, we need to convert everything into the same base units

So time needs to either be in min or s

[rich rose]

Minutes

[dire swallow]

Ok, so convert the speed into m/min

[rich rose]

Like this?

[dire swallow]

m refers to metres, not miles

[rich rose]

Oh

This good?

[dire swallow]

A minute is 60 seconds

So you have to divide by 60

[rich rose]

Wait that’s wrong

Is that right?

[dire swallow]

Yeah

[rich rose]

Ok I’ll solve the rest

Ok I did so wrong

The answer sheet says 6m

I got 0.003

Are you not supposed to divide 1.37 by 131pie

[dire swallow]

There's not supposed to be a pi

[rich rose]

What

[dire swallow]

This is revolution per min

[rich rose]

Im confused because my teacher said we are doing the work backwards from here, and this has pie

[dire swallow]

Ok I got it

So for each revolution, there is 2pi radians

[rich rose]

Yea

[dire swallow]

So for 131 revolutions, there are 262pi radians per minute

82.3 m per second means per second, its multiply by 60

[rich rose]

Ok I’m confused

Is their anay way you can draw what your saying

[dire swallow]

I can't really draw it

[rich rose]

Like do it on paper

Like what I’m doing

[dire swallow]

But think, if in 1 second, 82.3m is covered, then in 1 minute, you travel 60 times of that distance

Does it make sense?

[rich rose]

No but maybe I’ll understand it when I’m not tired

[dire swallow]

In 1 second, distance is 82.3
In 60 seconds, distance is 82.3×60

Right?

[rich rose]

Ohh yea

[dire swallow]

So you've got 82.3×60m/min as your speed v

And 262pi rad/min as your angular speed w

Then you apply v=rw

[rich rose]

Wait what

I thought it was linear speed

[dire swallow]

131 revolutions per minute refers to circular motion

In 131 revolutions, there's 262 pi radians

[rich rose]

Ok

[dire swallow]

So you've got your angular speed as 262 pi rad/min

[rich rose]

Ok

[dire swallow]

So you've got your linear speed v and your angular speed w

Now you just sub it in

[rich rose]

Ok I’m solving it now

Omg I love you

I solved it

I got the right answer

Thank you

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[spring pagoda]

How do I find the width of the rectangle to find the area

[dire swallow]

If the length is the diameter of the semicircle, the width is its radius

[shy bane]

do uk the diameter of the circle?

[shy bane]

u have diameter then how do u get the radius?

[spring pagoda]

diameter is 16

[shy bane]

then the radius is..?

[spring pagoda]

8

[shy bane]

so the width of rectangle is..?

[spring pagoda]

8

[shy bane]

so area of rectangle?

[spring pagoda]

the product of 16 times 8

[shy bane]

yes

now area of semicircle..?

[spring pagoda]

100.48

[shy bane]

ooh u have found it

so now area of shaded reagion?

[spring pagoda]

27.52

thank you

[shy bane]

np

[spring pagoda]

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[lean rock]

how to construct sorites argument for "is tall" ?

[wind quartz]

take any exemplary base step like a person,who is 7 ft. tall

do u think reducing his height by negligible amount will counter our point of him being tall

[lean rock]

yes, if done repeatedly, after some repetitions not tall

T: "is tall"
Tᵢ: T "is tall" at i seconds

T₀, T₀→T₁ ⊢ T₀, T₁→T₂ ⊢ T₂ … Tₖ₋₁, Tₖ₋₁→Tₖ ⊢ Tₖ

where k is some very large number

[wind quartz]

true, after repetitions we will reach a point here the example cannot be referred to as tall and hence we see vagueness in the word tall

[lean rock]

what if too tall, like for example we add height until someone is like reaches the next galaxy ? still tall?

[wind quartz]

yes, anything above our chosen point of 7ft will be considered as tall no matter the scale

[lean rock]

so not tall only works with negative direction?

[wind quartz]

we are just proving vagueness in the word tall, through induction we can say after like -0.001ft per repetition we get 4ft that should be tall by our intuition as the difference each step was negligible but the result isnt tall as 4ft is a major difference from our starting point of 7ft

[lean rock]

we are proving vagueness, nice

[wind quartz]

can be said to apply in negative only but if we also increase repeatedly it can be considered infinity. hence positive has its own problems

[lean rock]

vagueness is we cannot say what truth value is tall(7ft) ?

requires context?

Thank you for your help @wind quartz 🐬

[brittle steeple]

We can maybe say that that's true

But like

tall(5.9ft)?

tall(5.7ft)?

[lean rock]

there is a grey area

[brittle steeple]

Yeah that's what I imagine

[wind quartz]

we cant say a determined truth value, even tho we know the height. the predicate "tall" needs conyext to fix standard

so yes u are true here

[lean rock]

I mean tall(7ft) in general, requires context, like what is tall, relative to what?

[wind quartz]

ur welcome, i had fun learning with u

[brittle steeple]

ok fair enough

[lean rock]

yes, you change those values approach zero and then at what point is still tall?

T₁: If a sapling is 1 leaf tall, it is tiny → not tall 🌱

but then of two saplings we can ask which one is taller I guess

1. T₁: If a sapling is 1 leaf tall, it is tiny → not tall 🌱
2. T₂: If a tree with n leaves is tiny, then a tree with n+1 leaves is tiny 🌿
3. … Tₙ: Apply iteratively (n leaves) → still tiny / not tall
4. But at some N, the forest elder 🌳 is “tall”

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[astral yarrow]

im doing my homework and the relation R* was defined like so: $R = R\bigcap_{}^{}R^{-1}$
what does the symbol shown in the picture mean?

[glossy valve]

ori299

[fast peak]

that makes little sense

can you show the entire context?

[astral yarrow]

well its written in hebrew

il try to translate

for context: A is a non-empty set and R is a relation above A

we have the function pi: A -> A/R* (pi is just a symbol in this context)

and its defined like so: For every x E A pi(x) = [x]_R*

we are needed to prove that there exists a relation <symbol showen in the picture>

does that make any sense?

[fast peak]

is R supposed to be an equivalence relation or something?

otherwise I'm not quite sure what A/R^* is supposed to be

its certainly some kind of set of classes

so two elements are for example [x] and [y]

[astral yarrow]

R* is an equivalence relation

[fast peak]

and then you can define a relation [x]~[y] via xRy

maybe you are supposed to show that this works

and is well-defined

[astral yarrow]

im sorry i still dont get it

[fast peak]

the elements of A/R^* are classes [x]_R^*

between elements of a set we can define relations

we could define a relation [x]~[y] using the old relation R

namely [x]~[y] iff xRy in A

the problem with this is that if [x]=[z], then its not clear whether [x]~[y] if and only if [z]~[y]

aka, whether this definition depends on the representative

[astral yarrow]

i think i get it

[fast peak]

maybe you are supposed to show that

[astral yarrow]

regardless i missed some info

[fast peak]

but again, I dont know what you are supposed to do

this is just my best guess

[astral yarrow]

tysm

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[sinful charm]

This is not a specific q but can anyone help I'm very slow at trig even after practice if I can do a q that's fine but whenever I get stuck it can take me a lateral hour to solve

[sinful charm]

And it's just high school math can someone help

[neon silo]

Getting stuck on problems isnt that rare
guess you could try sending a problem you got stuck on so we can see where youre meeting difficulties

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@sinful charm Has your question been resolved?

[sinful charm]

I had to prove this

[hot herald]

did you copy the question correctly?

because that doesn't seem to be true

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[proven lynx]

Hello, im practicing hypothesis tests rn and im confused here

[proven lynx]

So h0 is 0,1?

or

This got me really confused i just finished normal distribution

<@&286206848099549185> sorry

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@proven lynx Has your question been resolved?

[proven lynx]

alright then nvm

i think i got it

i was confused cuz p1 isnt the complementary of p0

right

[ionic knoll]

tbh I'm not sure if this is a hypothesis test

Are you stuck on a or b?

[proven lynx]

a)

so p0 = 0.1 and p1= 0.4

[ionic knoll]

I haven't done a hypothesis test like this

I don't think it is?

[proven lynx]

yeah man wtf is wrong with my teacher he alwqys has to make some complex shit

[proven lynx]

it is tho

[ionic knoll]

We know everything about the random variables

[proven lynx]

yeah

[ionic knoll]

So there's nothing to hypothesise

[proven lynx]

i think it translated rly weird

yeah

100%

i just re read it

this is worded so weird in english

[glossy valve]

BBMaths

BBMaths

[proven lynx]

this is so weird

[ionic knoll]

Wait

I got the numbers wrong

[proven lynx]

so it says you take 5 screws from each box. If all but one of the screws are Alright, then its a first quality box or screw idk, else its 2nd quality. Now i have to calculate alpha error and beta error?

[ionic knoll]

Let's let:\

• $S_1$ be the number of good screws from the sample of the screws in the 1st quality boxes\
• $S_2$ be the number of good screws from the sample of the screws in the 2nd quality boxes\
  $$S_1\sim \text{Bin}(5.0,9)$$
  $$S_2\sim \text{Bin}(5.0,6)$$
  The probability you will make a misclassification of a 1st quality box is $\mathbb{P}(S_1<4)$\
  The probability you will make a misclassification of a 2nd quality box is $\mathbb{P}(S_2\ge4)$

[proven lynx]

wait why s1 <4

[glossy valve]

BBMaths

[ionic knoll]

Because if S_1 is < 4 then we guess it's 2nd quality but it was 1st quality, so we made a mistake

[proven lynx]

so if there are more than 4 good screws?

[ionic knoll]

4 or more good screws we guess it's 1st quality

So if S_2>=4 we make a mistake because we guess 2nd quality is 1st

[proven lynx]

and if more than 1 bad screw then its 2nd quality but

[ionic knoll]

Yeah

[proven lynx]

why dont we just say

actually

yeah

[ionic knoll]

So you will need to calculate binomial distribution

[proven lynx]

yeahh

yes so

For the misclassification of a 1st box

N is 5

p is 0.9 right

[ionic knoll]

Yeah

[proven lynx]

and k is uh?

[ionic knoll]

You can also do p=0.1 and work out P(B_1>=2) where B_1~Bin(5.0,1) is the number of bad screws

[proven lynx]

then its 1 - P(B_1>=2)?

[ionic knoll]

No

[proven lynx]

because atleast

[ionic knoll]

it's just P(B_1>=2)

Probability there are at least 2 bad screws

[proven lynx]

yeah so

thats the same as 1 - P(b1<=1)

right

[ionic knoll]

Yeah

[proven lynx]

so then that probabilty is the chance that we assume h1 is false even tho its correct?

[ionic knoll]

We want to find
$$\mathbb{P}(S_1<4)=\mathbb{P}(S_1\le3)=1-\mathbb{P}(S_1\ge4)=1-\mathbb{P}(S_1>3)$$$$...=\mathbb{P}(B_1\ge2)=\mathbb{P}(B_1>1)=1-\mathbb{P}(B_1<2)=1-\mathbb{P}(B_1\le1)$$
So there are multiple ways of doing this

[glossy valve]

BBMaths

[ionic knoll]

8 ways

[proven lynx]

wait why is P(s1<4) the same as P(s1<=3)

[ionic knoll]

Because s_1 is a whole number

$\mathbb{P}(S_1\in{0,1,2,3})$

[proven lynx]

omfg

sorry

my minds all over the place

[glossy valve]

BBMaths

[proven lynx]

so we assumed that the content of the box is 2nd quality even tho its 1st?

[ionic knoll]

Yeah

Try it and see where you get to

Just post a photo of the work

It's better to try things and check what you got

[proven lynx]

Will do

[full forum]

@proven lynx Has your question been resolved?

[proven lynx]

wtf did i do wrong 😭 @ionic knoll

,rotate

[proven lynx]

this doesnt even make sense

tf

my bad

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[proven lynx]

.close

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[plush ibex]

Have they done this wrong if so where

[plush ibex]

and also why don't i use the parallel component of gravity

[gritty rose]

Which part do you think is wrong

[plush ibex]

the bit where you solve for d

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@plush ibex Has your question been resolved?

[dark peak]

excuse mecan someone explain functions to me idk anything about it

[verbal crater]

#❓how-to-get-help

[gritty rose]

This step is correct

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[plush ibex]

isnt this wrong tho? or is there a reason they get rid of one of the g's

[plush ibex]

i understand this bit now tho

[sand escarp]

this is algebra right

[plush ibex]

mechanics

[wary condor]

they just added +gd to boith sides and took a common factor

[sand escarp]

yes, but the picture

I do think they're wrong tho

there's just a g missing

pretty funny mistake

[plush ibex]

ok thanks i was so confused

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[cinder thorn]

what's the radius of convergence (at x=0) of $\sum \frac{1}{\sin n \pi \sqrt{2}} x^n$

[glossy valve]

bloubbloub

[cinder thorn]

since the coefficients are bigger than 1 we have R <= 1

then I have $\left|\frac{1}{\sin( n \pi \sqrt{2})}\right| \leq \frac{1}{2 {\sqrt{2}n}}$ where ${x}$ is the fractional part

[glossy valve]

bloubbloub

[brittle steeple]

right, so you need to find a lower bound for {nsqrt(2)}

if {nsqrt(2)} were extremely small, that would mean nsqrt(2) was extremely close to some integer k

but squaring both sides, that would mean 2n^2 was extremely close to k

since both of those are integers, there's a limit to how close they can be (1 apart, since they can't be equal by irrationality of sqrt(2))

so you use that to find a lower bound for {nsqrt(2)}

[cinder thorn]

Yeah

The lower bound isn't quite clear

I think the argument should involve something like the "frequency of being close to 0"

[brittle steeple]

The test for radius of convergence just requires a lim sup, right

So you don't need to worry about frequency

[cinder thorn]

You can't have a uniform lower bound on |sin n pi sqrt2| though

[brittle steeple]

You can have one in terms of n

[cinder thorn]

Do you have any hint for me

[brittle steeple]

2n^2 ≥ k^2+1

k being floor(n√2)

[cinder thorn]

Lemme think

[cinder thorn]

Ok i got it thanks

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[rotund pebble]

how do we proceed

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[rich pier]

Can I get help please on this question

[rich pier]

I believe the answer is x has no solution and that x cannot equal 1

[rich pier]

But its saying its incorrect

[robust slate]

I think they want you to say that solving gives you $x=1$, but then later say that $x=1$ isn't possible (so put in $1$ for both blanks). \ \

[glossy valve]

Civil Service Pigeon

[robust slate]

Also, in the future, please make the question the first thing you send since that's what the bot pins.

[rich pier]

Got it

so I should say 1 and 1?

[robust slate]

Yeah that's what I said

[rich pier]

Thanks

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[robust slate]

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[dawn owl]

im having hard time how to find the answer. any help will be gr8

[dawn owl]

my answer was incorrect^

[stiff musk]

how did you arrive at that answer?

[dawn owl]

RREF to this matrix

any1 plz?

[full forum]

@dawn owl Has your question been resolved?

[queen gull]

you can find A^-1

and then compute B * A^-1

[coral ore]

is that the same doing T = W(output matrix) V^-1(input matrix)?

[dawn owl]

hey 🙂

ive finally solved it thx yall

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[queen gull]

it was just a calculation mistake no?

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[digital void]

,tex
Hello, i was wondering if the proof for the $s_n $ sequence being monotonous is good enough.
\
$a_n $ bounded sequence, and $s_n = \sup{ { a_k | k\geq n } } $
\
\
So, $s_{n+1} = \sup{ { a_k | k\geq n+1 } } \leq s_n $

[glossy valve]

fijokazż

[foggy vapor]

first of all, please don't line break with \\ in textmode. You can just line break normally as you'd do normally in a text passage

second of all, your intuition is correct but you should explain what that inequality holds

[digital void]

whats the normal way to line break?

[foggy vapor]

Hello, i was wondering if the proof for the $s_n$ sequence being monotonous is good enough.


$a_n$ bounded sequence, and $s_n = \sup{ \{ a_k | k\geq n \} }$

\medskip
So, $s_{n+1} = \sup{ \{ a_k | k\geq n+1 \} } \leq s_n$

[digital void]

could i explain that since we have fewer terms to choose from, then the sup is definitely gonna be at most equal to the previous one?

[digital void]

medskip?

[foggy vapor]

if a set $B$ is contained in a set $A$, then $\sup(B) \le \sup(A)$

[digital void]

very true

okay thats easy then

[digital void]

thank you Lex

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[foggy vapor]

.reopen

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✅ Original question: #help-28 message

[digital void]

yo

[brittle sun]

Non-consensual channel reopening is a big issue

[queen crater]

.close

[foggy vapor]

\e{tabular}{{|p{2.5cm}|p{2.5cm}|p{2.5cm}|p{2.5cm}|}
    \hline
    \textbf{smol} & \textbf{med} & \textbf{big} & \textbf{vs} \\
    \hline
    Text A \par
    \smallskip
    Text B 
    & 
    Text A \par
    \medskip
    Text B 
    & 
    Text A \par
    \bigskip
    Text B 
    & 
    Text A \par
    \vspace{2cm}
    Text B \\
    \hline
}

[foggy vapor]

you can use any of those

ok anyways

.close

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[brittle sun]

.reopen

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[frigid carbon]

occupy

[frigid carbon]

don't understand

@rare dock

she went to sleep

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@frigid carbon Has your question been resolved?

[frigid carbon]

<@&286206848099549185>

[full forum]

@frigid carbon Has your question been resolved?

[rare dock]

just woke up

i need to do something but i can explain in a bit (maybe like 30 mins if i don’t want to immediately fall back asleep

[frigid carbon]

good morning

[rare dock]

no heisenberg i don’t drink coffee

i am back from moving my car

[rare dock]

ok now about this i misread a < as a <= in your problem so i need to change some 10s to 9s

let me post a fixed version then we can talk about it

fix a positive integer $k$. by elementary generating function theory, the answer is the $x^{2(k-1)+10}$ coefficient in
$$F:=\prod_{i=1}^k\sum_{j=0}^{2(i-1)+9}x^j = \left(\frac{1}{1-x}\right)^k \times \prod_{i=1}^k (1-x^{2(i-1)+10}).$$
that is what we will find. let's put
$$G = \left(\frac{1}{1-x}\right)^k$$ and
$$H = \prod_{i=1}^k (1-x^{2(i-1)+10})$$
for reference purposes. lucky for us, the coefficients in $G$ are easy to find. the coefficient of $x^m$ in $G$ is the number of nonnegative integer solutions to $x_1 + \ldots + x_k = m$, which is ${k+m-1\choose k-1}$. just as a notational thing, we denote by $F[x^m]$ the coefficient of $x^m$ in $F$. now since $F=GH$ we have \begin{align*}F[x^{2(k-1)+10}] &= \sum_{m=0}^{2(k-1)+10}G[m]H[2(k-1)+10-m]\
&= \sum_{m=0}^{2(k-1)+10}{k+m-1\choose k-1}H[2(k-1)+10-m].\end{align*}
the problem is reduced to finding the coefficients in $H$. i'm not sure there is a nice way to find them, but we can just expand $H$ which is simple for a computer. also, since we are looking for $F[x^{2(k-1)+10}]$, we can work in $$\bZ[x]/(x^{2(k-1)+10}).$$ basically a 'polynomial system' where $x^{2(k-1)+11} = 0$, which is good because the larger terms don't affect the coefficient we care about

[frigid carbon]

Oh

[glossy valve]

slayla

[rare dock]

and the problem was:
fix a positive integer k. how many k-tuples (x_1, ..., x_k) of integers are there such that

1. 0 <= x_i < 2(i -1) + 10 for all i
2. sum over x_i = 2(k-1)+10

well that’s a rewording by me but hopefully what you meant

[fast peak]

https://math.stackexchange.com/questions/553960/extended-stars-and-bars-problem-where-the-upper-limit-of-the-variable-is-bounde see also this

[frigid carbon]

mm.. the condition changed a bit

[rare dock]

thx. i will check that out later

[rare dock]

.-. what changed?

[frigid carbon]

{x1...xk } denotes number of tails, bi = b(i-1) - xi + 2i -2, with b0=10 so b i> 0 for every i<k, ,for all xi belong to positive integers. bk=0

[rare dock]

hm ok that problem is kinda different

i can explain the generating function stuff if you wish but not sure how applicable what i wrote for your tuple problem will be

[frigid carbon]

ok, plz

[rare dock]

yay

ok let’s start with something simpler

i need to get an image from google for ts

https://thumbs.dreamstime.com/z/dice-roll-probability-table-to-calculate-probability-dices-vector-dice-roll-probability-table-to-calculate-probability-239664183.jpg

these are the sums of rolls when you roll 2 dice

if we wanted to know how many ways to get 4s or 8s or whatever we could count them from this table

but now i’m gonna explain another way to do this same thing

[rare dock]

imagine that we made a similar table but label the rows x^1 instead of dice showing 1, x^2 instead of dice showing 2, etc

well and the columns too

then entry (a,b) in the table is x^a * x^b rather than the die sums

since x^a * x^b = x^(a+b), do you see the table will look pretty much the same, just all the numbers will be exponents on x?

[frigid carbon]

yes

[rare dock]

ok great

so now let’s imagine the polynomial
(x + x^2 + x^3 + x^4 + x^5 + x^6) (x + x^2 + x^3 + x^4 + x^5 + x^6)

if you were to expand this with the distributive property, do you see that the results would be exactly like what is in the table?

we are gonna multiply each element in the first polynomial by each element in the second polynomial

[frigid carbon]

yes

[rare dock]

and all added together

,w expand (x + x^2 + x^3 + x^4 + x^5 + x^6)^2

[glossy valve]

Results provided by WolframAlpha

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

Requested by slayyla

[rare dock]

so we’d get an x^7 term 6 times

[rare dock]

agreeing with this

[frigid carbon]

right

[rare dock]

the coefficient of x^7 is the number of ways to roll two dice and get a 7

i used the same ideas to set up polynomials for your problem

the first value in the tuple needs to come from {0, 1, … 9}

so we have a polynomial x^0 + x^1 + … + x^9 for that

and so on

and the sum of the tuple needs to be 2(k-1) + 10, so we want the coefficient of x^(2(k-1) + 10)

[frigid carbon]

powers represent number of tails ?

[rare dock]

i am speaking only about this problem

[frigid carbon]

ok

but xi needs to be< x_i < 2(i -1) + 10 for all i

so should'nt we use from x^0 + ...x^9

for first tuple

[rare dock]

yes, fixed

[frigid carbon]

ohk

right

this makes sense

[rare dock]

already forgot about this. oops

i redid the computations with that fix last night. this is what the ‘golden ratio’ looked like

i am not 100% sure if it converges or not

[frigid carbon]

wow

hmm..

[rare dock]

or just grows extremely slowly

[frigid carbon]

i didn't expect golden ratio to come in

[rare dock]

well it’s not ‘the golden ratio’

just something analogous for ts problem

[frigid carbon]

y = mans wealth, these are all the paths the mans wealth goes 0 at end

but it goes negative at some points then back

[frigid carbon]

oh

ty ,
sorry to bother you it was amazing

[rare dock]

i am never bothered by anything generating function related

[frigid carbon]

-# it will take time to digest i'll go through again

[rare dock]

i am not sure how much larger i can compute. the numbers just get so enormous that their size slows down computation a lot even when it’s a reasonable number of operations performed

f(10000) took about 50 seconds

i wrote something for finding coefficients in H without expanding it (well that’s kinda what my algorithm does, but with dynamic programming specialized for this occasion rather than generic polynomial multiplication) which helped a lot

[fast peak]

wdym with "golden ratio" here?

[rare dock]

the value f(n+1)/f(n) converges to, if it even converges

i would like to know if it converges

[fast peak]

f(n) being the relevant coefficient?

[rare dock]

yes. the number of n-tuples such that yada da

[fast peak]

have you tried applying berlekamp massey on the first few 100 values or so to see whether they satisfy a small poly?

and then biggest root of that?

[rare dock]

no

[fast peak]

if it satisfies a small poly/aka small recursion then it should probably be doable to prove that by bruteforce symbolically

you might also be interested in the book A=B and its tools

which can also maybe prove a closed form

or at least find a recursion

which you could maybe prove

[full forum]

@frigid carbon Has your question been resolved?

[frigid carbon]

by inclusin exlusion

this should match

[fast peak]

dont ignore the bot

[frigid carbon]

ai

says does not converge

-# i don't trust

[fast peak]

what happened before that 27/4. interesting that it might be the correct value

[frigid carbon]

it was temp chat, let me try again

https://chatgpt.com/s/t_696509c59fa88191bef1d4e646e8e633

https://chatgpt.com/s/t_696509fbbea88191b8dc60f883e24155

[fast peak]

ok I havent looked at the problem properly but if the observation that most of the x_i arent actually upper bounded is true then yeah that seems reasonable

[frigid carbon]

Oh, yeah

[frigid carbon]

fix a positive integer k. how many k-tuples (x_1, ..., x_k) of integers are there such that

1. 0 <= x_i < 2(i -1) + 10 for all i
2. sum over x_i = 2(k-1)+10
   This is for generating function

[frigid carbon]

This is for original problem

[fast peak]

ok so the sum of the upper bounds is much much more than sum x_i

and the average x_i has basically value 2

ok so for asymptotics you really can forget the upper bounds

they really shouldnt make a big difference

[frigid carbon]

Hmm..

Ig i close

.ty

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[coarse pasture]

hello

[coarse pasture]

can some 1 help

[void magnet]

with what

[coarse pasture]

MATh

I got exam in 6 hour

[coarse pasture]

yo

help

[void magnet]

Yea

[coarse pasture]

Can we vc or smt

i actually need it so much

PLEASE

[void magnet]

So whats the probability that of event Y to be happening

[void magnet]

I dont spik inglis

[coarse pasture]

6/36

[coarse pasture]

6/36

[void magnet]

Yea 1/6

and whats the probability that white die rolls six after black has rolled 6

[coarse pasture]

wdym

[void magnet]

Lets say you rolles the black die

You got 6

[coarse pasture]

1/6

[void magnet]

yea

The probability of getting same number remains unchanged

[coarse pasture]

YOO

i kinda of get it

WOAHH

[void magnet]

:) yay

and now the second part

Whats the probability of event Z happening

[coarse pasture]

uhh 1/36

[void magnet]

Yea yea

[coarse pasture]

cuz only 6,6

[void magnet]

Yea only one case

And now lets say you rolled black die

You got 6

And now you rolled the white one

Whats the probably that the sum is 12

[coarse pasture]

1/6?

[void magnet]

yea! The probability of event Z is dependent on event X. Lets say you rolled black die and you didnt get 6 whats the probably of the sum being 12

[coarse pasture]

0/36?

[void magnet]

Yea 0

in the first part, even if X event hasnt occurred, the probability of Y would have remain unchanged

example- you rolled 5 on the black die, then the probable of event Y is still 1/6

[coarse pasture]

why tho

[void magnet]

its becuz the first throw has ntg to do with matching the number

the second die is the one who'll decide whether the number matches or not

[coarse pasture]

Wait then it is like this

wait

[void magnet]

huh

ok

[coarse pasture]

if x happen

and p of y would be 1/36

cuz only 6,6

right

[void magnet]

no

probability= favorable outcomes/total outcomes right?

[coarse pasture]

yes

[void magnet]

When the first die has rolled 6 it has reduced total outcomes

(6,1),(6,2),(6,3)... (6,6)

These are the only possible outcomes after event X occurs

[coarse pasture]

HUH??

[void magnet]

And the favourable one is (6,6)

[coarse pasture]

wait i am back to square one

[void magnet]

alr chill

[coarse pasture]

back to the start again pls

[void magnet]

Sure

When you roll the black die

lets say you get 4

What are the possible outcomes for second die after that

Is 6 or 36

[coarse pasture]

6?

[void magnet]

yes, u getting it?

[coarse pasture]

sure i guess?

[void magnet]

lemme give some other example to better understand

lets say you and ur frnd are competing on a test and you both decide to get the same score

ur frnd submits the test first and you get to know his result, lets say its 76/100

Now the pressure is on you to score 76

You are the one who'll decide whether you both will score the same score or not

[coarse pasture]

yes

[void magnet]

similarly when the first die rolled 6

the second die has the pressure to roll 6

the first die has done its job

that's why the probability is 1/6

no matter wht tge first die rolls

[coarse pasture]

ohh

p of x is 1/6

cuz

there are 6 outcomes and x is 6

so only 6 can happen 1 time

[void magnet]

Ye

[coarse pasture]

so x happen

so 1/6 happen

and p of y is

both number is same so

1,1 2,2 3,3

so 6/36

so 1 /6

[void magnet]

Ye

[coarse pasture]

and they are independent

[void magnet]

Ye

[coarse pasture]

because if x is no 6 then it doesnt matter to p of y

OHHH

[void magnet]

YEA! Two Independent event simply means that, no matter event one happens or not, the odds of the other remains unchanged

if black die rolls 1, odds of Y 1/6
rolls 2, still 1/6
rolls 3, still 1/6
...

[coarse pasture]

ehhh yeah

i think i get it

[void magnet]

Goodd

[coarse pasture]

can you give me similar like yk problem so i can solve and like you tell me if it is correct or not

[void magnet]

Questions on prob? Lemme think then

You have a deck of 52 cards

you pick 2 cards

X is the event of the first card that u selected is king

Y is the event that the second card that you selected is king

Are X and Y dependent or independent?

[coarse pasture]

dependent?

[void magnet]

Yea, but i want te reasoning behind it

[coarse pasture]

uhh because

because

there are 4 kings in a deck of card

[void magnet]

yea

[coarse pasture]

so p(x)=4/52

[void magnet]

Yupp

[coarse pasture]

and then after we do that

if we Do y

it would be 3/52

right?

[void magnet]

Slightly wrong

3/51

3/51 when event X occurs

[coarse pasture]

OHH RIGHT

[void magnet]

Yea :)

And whats the probability of Y if X doesn't occurs?

[coarse pasture]

so

4/52?

[void magnet]

4/51

We selected one card before

[coarse pasture]

heh?

OHH right

because Y is second

and first is already done

[void magnet]

Exactly

[coarse pasture]

so we remove 1 from total

[void magnet]

Yup

[coarse pasture]

YOU ARE A GENUIUS

thank you

[void magnet]

:) 🫂

[coarse pasture]

imma close this

so thank youuu

[void magnet]

Okk

[coarse pasture]

.close

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[wise herald]

Help me with this badass

[atomic hare]

the bottom looks like a square

[rare dock]

i love your banner

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[hushed barn]

Can someone help me understand the patterns that lie within the plot of the positive integer domain of of f(x) = (3^x-2^x)/(2^m-3^x) where m = floor(log(3^x)/log(2))+1? When I first look at the plot, I see some patterns but don’t understand why they occur. There tends to become a pattern emergent when you keep adding 12 to most values of x, and when you keep adding 53 to most values of x. Idk why that’s the case though

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@hushed barn Has your question been resolved?

[full adder]

It can be solved by partial fraction I think

X^6 and x^3 so it is similar to simplifying a second degree equation

I haven't tried but it looks like it

[dry arch]

you can find these lines when k*log_2(3) is close to an integer

[hushed barn]

Interesting. The orange graph is when you keep increasing x by 2, and the red graph is when you keep increasing x by 5 up to some point

I interpreted different lines though (ones with the x’s increasing by 12 and ones with the x’s increasing by 53)

[hushed barn]

I realised that if you changed the value of m for the orange graph, it categorises a set of lines that increase, and if you change the value of m for red graph, it categorised a set of lines that decrease

[hushed barn]

Could you describe what is happening here?

[dry arch]

so say x = 41 + 53k

[hushed barn]

yes

[dry arch]

then m = 65 + 84k for a while

so m = 65 + 84 * (x-41) / 53, you can substitute this value for m and get a curve

[hushed barn]

Yes, I got this by pattern checking but how would you derive this?

[hushed barn]

I understand this

[hushed barn]

And how did you get the constants 3/2 and 8/5 here?

[dry arch]

because

this will hold until the decimal part of (64.9834625296 + k * 84.0030125382) rolls over

[hushed barn]

Okay that makes sense

[dry arch]

it rolls over at k = 6

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[still current]

The population P(t) of a village is given by the rule

P(t) = 6530(3)^0.1t

where t is the number of years that have passed since 2011.

a) What was the population of this village in 2011?

[ripe hatch]

You just need to figure out the right value of t to plug in

If t is the number of years past 2011, and we want the year 2011, how many years passed

[still current]

hmm

wait

from 2011 to 2026

[ripe hatch]

where did 2026 come from

[still current]

nvm

sorry

[still current]

not sure

its like a number u put into t here

[ripe hatch]

Don’t overthink it

if x is a year, to find the population in year x we have to let t = x - 2011

that’s what the question is saying

it wants the population in year 2011

[still current]

ohh

but what do i put for x or t then

wait

i did a mistake while trasnlating

its like that

P(t) = 6530(3)^0.1t

[ripe hatch]

This is all the information you need

what would x equal to answer part a

[still current]

im not sure

wait

isnt it 6530

since its the initial value

[mental parrot]

Yes

Solve by P(0)

[still current]

that means t = 4519

[mental parrot]

Huh??

[still current]

6530 - 2011

[mental parrot]

No

6530 is people

[still current]

ag

ah

[mental parrot]

You’re substituting year from people

Which doesn’t make sense

[still current]

oh alr

then what do is it?

[mental parrot]

P(t)=6530 * 3^0.1t

Where t — is the number of years passed since 2011

[still current]

oh replace t by 0

[mental parrot]

Yes since the question asked in 2011

0 years passed in 2011 since 2011

Don’t overthink it

[still current]

oh alr

0,1(0)

gives 0

so its 6530(3) = 19590 people

ty

.close

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[brazen lark]

isnt 3⁰=1

[mental parrot]

Yes it is 💀

[silk galleon]

its just 6530

[still current]

oh alr

lol

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[gritty rose]

write 125 = 5^3 and use exponent rules

[undone radish]

express the roots with a 5 on the inside by splitting them

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@tulip kettle Has your question been resolved?

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[sturdy valve]

why cant i just do tan x = x
sin x = x so f(0) = 1

[grave elm]

Tanx - sinx is 0

You wouls be dividing by 0

The function is undefined at 0

[sturdy valve]

no but its cancelling out right

[split hatch]

sinx=x and tanx=x are only first order approximations

[grave elm]

(0 - 0) / (0 - 0) is simply undefined

[split hatch]

you can't just substitute them like that

[silk bridge]

it is of form 0/0

[grave elm]

Even if u took the limit, (x - x) / (x - x) would still bw undefined, so ur approximations arenf sufficient

But the question doesnt even ask about limit

[sturdy valve]

ohh

[sturdy valve]

?

[split hatch]

The question is a little ill-worded

[grave elm]

Oh gosh

"f is continuous at 0"

[split hatch]

the expression they gave is undefined at x=0 so as written, f cannot possibly be defined at 0

[silk bridge]

avg jee shit

[rigid crescent]

if it is continuous at 0, it is sufficient to find the limit to 0 for an answer

[sturdy valve]

ye

[grave elm]

Actually, there is a contradiction so the answer is simply 2pi

Or anything

[sturdy valve]

uhh its not

[whole lily]

Second order approximations do work

[grave elm]

Its just a wrong question.

[sturdy valve]

ok now that i got what i did was wrong can someone help solve it

[split hatch]

It probably meant to ask what value should you assign f(0) to remove the removable discontinuity at 0

[grave elm]

If we ignore that its wrong and instead try to find the limit, then try either 2nd order approximation or lhop twice

[sturdy valve]

what is 1st order and 2nd order approximation?

like expanding it to 2 terms?

[rigid crescent]

yes introduce a higher order derivativee

[sturdy valve]

bruh what

[grave elm]

Sinx is x - x^3 / 3!

[fast peak]

use more terms from the taylor series