help-4

[balmy monolith]

ok thank you

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[surreal sinew]

Hey everyone

[surreal sinew]

I need some help

There's this sequence:

$v_{n}=\frac{5n}{n+2}$

[rocky lotus]

stoicindiehacker369

[timber swan]

Ye

[surreal sinew]

I'm asked to find the minimum value of n for which v_{n} belongs to this neighbourhood:

[timber swan]

What needs to be done?

[pallid oasis]

Which neighbourhood?

[surreal sinew]

]5-0,1;5+0,1[

(Sorry if I type slow haha)

[pallid oasis]

You can write vn = 5 - 10/(n+2)
So -10/(n+2) needs to belong to the neighbourhood ]-0.1,0.1[. Is it easier now?

[stark wedge]

hoooold on.

that wording is suspicious.

i think there's a chance you're not translating it correctly.

[surreal sinew]

So basicaly, I have to solve this:

[stark wedge]

so i'm gonna ask for the original problem as a screenshot

just in case.

[surreal sinew]

Well, it's in portuguese 😅

[stark wedge]

send it anyway

i can figure it out

i speak french and have studied latin and if there's words i don't get i will ask you to translate

[stark wedge]

the notation's all gonna be the same-ish

okay right

[pallid oasis]

@river shale whats the problem?

[stark wedge]

so you don't want to find the lowest n for which |v_n - 5| < 0.1

you want to find the lowest n starting from which |v_n - 5| < 0.1

[river shale]

Nvm

[stark wedge]

bit of a distinction there.

cause like ignoring any properties of the sequence that we can prove

if we had something like... idk

[surreal sinew]

Isn't that the same

[pallid oasis]

Well, the sequence is monotone, converging to 5 but yeah we dont know this apriori

[stark wedge]

no let me explain

imagine we had something like
v_1 through v_9 are 0, then v_10 = 5.01, then v_11 = 6, v_12 = 7, etc. and it doesn't get back into the interval until v_30 and only then it stays there

[surreal sinew]

If n is the smallest p for which v{n} belongs to a neighbourhood, then, starting from there, all terms belong to that neighbourhood right

[stark wedge]

then the answer to "what's the lowest n such that v_n belongs to the interval" is 10
but "what's the lowest n starting from which v_n always belongs to the interval" is 30

[stark wedge]

nope!

you don't know this without calculations.

[surreal sinew]

Ok let me process what you wrote

[stark wedge]

i can try to make an illustration here

of the kind of situation im talking about

lemme go do that

also for the record, im not instructing you right now

im more explaining the reasoning and why this particular wording detail matters

[surreal sinew]

Ohhhhhh

Ok I understand now

But then, how can I find that value of n

[stark wedge]

ok im back here is my illustration

now

[stark wedge]

you do still have to look at this inequality

and it is actually very good that you already wrote it this way

because $\frac{5n}{n+2} - 5$ simplifies quite a lot

[rocky lotus]

Ann

[stark wedge]

it simplifies down to $-\frac{10}{n+2}$

[rocky lotus]

Ann

[stark wedge]

do you see how or do you want me to explain this

[surreal sinew]

So I just solve for:

[stark wedge]

technically correct but morally questionable

[surreal sinew]

Hm, how 😅

[stark wedge]

why not simplify |-10/(n+2)| to 10/(n+2)

then you get 10/(n+2) < 1/10

it's all the same inequality but you jump through 1 less hoop

[surreal sinew]

Oh yeah, I'd do that

I do that when the unkown side of the equation is negative

[stark wedge]

anyway from here it should be smooth sailing

since n is natural, n+2 is positive and so you can multiply by it safely

[surreal sinew]

Yeah

[stark wedge]

and in fact you solve the inequality outright, and it turns out that no weirdness will happen that i showed with my illustration

[surreal sinew]

Just one question:

Shouldn't it be:

[stark wedge]

,rccw

[stark wedge]

-5(n+2) ≠ -5n - 2

so no

[surreal sinew]

Oh

[stark wedge]

-5(n+2) = -5n + (-5)*2

[surreal sinew]

Damn sorry

🤦

Ok ok.

[surreal sinew]

When do I know if I am facing one of these cases↑?

Do I just have to try out values at random?

[stark wedge]

i do not know if there's an algebraic way to tell

like i think there is no workaround to just solving the inequality the honest way

[surreal sinew]

What do you mean

[stark wedge]

i mean that you can't know ahead of time whether you'll be looking at some weird case like that

maybe they won't ever ask you to do that explicitly... but you still need to be aware such sequences exist and there's nothing abnormal about them

[surreal sinew]

Well, if the sequence is defined by a polynomial, it would be one of those cases

[stark wedge]

ok let me be a bit more blunt

[surreal sinew]

Since a local minimum of the function might not be an absolute minimum, right

[stark wedge]

do not try to come up with heuristics for this

i can tell you it isn't worth it

you'll only confuse yourself

[surreal sinew]

Haha ok 😅

[stark wedge]

just do it the honest way and then look at the exact set of n's that you get and then answer the question(s) you're asked about it and be attentive

that's all i can say

[surreal sinew]

Ok

But of course, if it's a function that I know well, like a trig function or a polynomial with degree greater than 1, I know that it is one of the cases where it might be decreasing and belong to that neighbourhood and then increase

[surreal sinew]

Ok

Well, thank you so much!
I don't know why I hadn't simplified it like you did 😅.
Also, thank you for elucidating me on that specific case too

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[cobalt magnet]

Hi everyone! I had a question on probability and statistics, especially on the "Sigma-Rules" (i study at a german uni so i hope the term is known)

[cloud epoch]

Can you provide an example of thos "sigma rules"?

[fickle rose]

not familiar

[cobalt magnet]

it kind of looks like this

[fickle rose]

I know it by just the standard deviation??

what exactly is your question?

[cobalt magnet]

I will type it out in a minute i gotta translate

[cloud epoch]

oh standard deviation 🧠

[cloud epoch]

every language has different names tbf

[fickle rose]

yeah just wanted to make sure we were talking abt the same thing

[cobalt magnet]

We want to know if a random number generator generates the number 13 with the same probability as the other numbers from 1-20;
and to figure this out you do a "Hypothesis test" (literal translation of the german word)

In the picture X is the amount of times that 13 appears as a variable.
And after the first calculation you get 6,892 which is greater than 3, which is why you can use the laplace rules, and following that the "sigma rules"

My question here is where does that 2,580 come from??
Thank you!!

these are my teachers notes

If anything in here isnt clear i can translate or explain it further

<@&286206848099549185>

[cloud epoch]

What i Have seen before is that because you are playing with a finite sample size, the 2.58 comes from how sure you want to be of your result.
Basically, if you put 1 instead of 2.58 you're about 68% sure of your result, if you put 2, its about 95%, and with 3 about 99.7% or something, and so I guess your teacher had a specific target, maybe 99%, which is how he chose 2.58 accordingly

If I'm not mistaken of course, I'm not that good in probabilities/statistics..

comes from this basically

Hope that helps ! :)

[fickle rose]

Here we have (\alpha=1%). Meaning we need a critical z-value for (z^_{0.995}) for a two-tail test for this alpha level. And actually (z^{}_{0.995}\approx 2.58). So that's where it comes from

[rocky lotus]

PajamaMamaLlama

[cobalt magnet]

so what exactly does 2,58 mean then?

[cloud epoch]

Doesnt mean much itself

[cobalt magnet]

is it basically the middle between 2 and 3 here

wait no nvm

in this table have 2,583 written under 0,01

i am just kind of confused its hard to explain

[cobalt magnet]

okay rereading your answer here helped me

thank you

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[pure mirage]

How's the course called? We did those in the Leistungskurs in grade 12

[cobalt magnet]

Es ist Mathe für Informatik an der Uni und ich hatte keinen Leistungskurs Mathe

[pure mirage]

Ah, also kein Mathe BSc-Kurs

[cobalt magnet]

Nein gehört zum Informatik Bachelor aber bereits Mathe 2

Weißt du wo die genaue Zahl 2,508 herkommt?

habs eigentlich verstanden aber da ist halt diese exakte Zahl die ich mir nicht herleiten kann

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[pure mirage]

.Das wird numerisch hergeleitet, die Idee ist, dass P(mu - Konstante * sigma, mu + Konstante * sigma) immer das gleiche ergibt, wenn du Konstante variierst, unabhängig von mu and sigma

Also könntest du z.B. P(mu - 1 * sigma, mu + 1 * sigma) etwa 68.3% raus haben

[pure mirage]

Völlig unabhängig von mu und sigma

Das kannst du dann mit anderen Konstanten versuchen

1.5, 2, usw.

Wenn du z.B. 99% raus haben willst, versucht du dann halt so lange Konstanten, bis du das hast

Und scheinbar ist diese Konstante dann genau 2.508[usw]

[cobalt magnet]

also wurde in diesem fall "rückwärts" gerechnet weil ich bereits diese 99% hab?

[pure mirage]

Ja, du willst 99% und probierst so lange Konstanten bis P(mu - Konstante * sigma, mu + Konstante * sigma) genau 99% ergibt

[cobalt magnet]

okay das macht es endlich klar

Danke danke

[pure mirage]

Es ist egal was du für mu and sigma hast, kannst die Normalverteilung nehmen

[cobalt magnet]

ja verstehe

[pure mirage]

Es gibt dafür bei den meisten TR einen Knopf

Du kannst dir dieses 2.508 berechnen lassen mit einigen mehr Dezimalen

[cobalt magnet]

dürfen leider keinen tr nutzen in der Klausur

[pure mirage]

Naja aber wahrscheinlich habt ihr dann eine Tabelle

[cobalt magnet]

Ist bei dem Thema Stochastik sehr doof also bin gespannt was da kommt

[cobalt magnet]

ja das schon

[pure mirage]

Die haben das dann schon vorher mit dem TR ausgerechnet

[cobalt magnet]

ja genau

Vielen Dank nochmals

[pure mirage]

Kein Problem

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[midnight pier]

0 = 0 + 0 + 0 + ... = (1 - 1) + (1 - 1) + (1 - 1) + ... = 1 + (-1 +1) + (-1 + 1) + (-1 +1) + ... = 1 + 0 + 0 + 0 + ... = 1

where is the error?

[west cloud]

The problem is that you used to have as many -1s as 1s

back here (1 - 1) + (1 - 1) + (1 - 1) + ...

now, here, 1 + (-1 +1) + (-1 + 1) + (-1 +1) + ...

you have one more 1 than -1

[midnight pier]

but i have infinite number of -1 and 1

[fickle rose]

Grandi's Series does not converge

https://en.wikipedia.org/wiki/Grandi's_series

[zinc spear]

In general, you have to be careful when bracketing infinite sums

[midnight pier]

oh i see, so where is the error?

[fickle rose]

you assumed that 0+0+0+0+...=(1-1)+(1-1)+(1-1)+...

which is not true

in fact the series (1-1)+(1-1)+(1-1)+... does not converge to 0

but neither 1, because it doesn't lol

even though depending on bracketing it might

as pola_touche said

be careful when bracketing these infinite sums

[midnight pier]

yeah, i understand now

thank you all

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[midnight pier]

how can i show that given a triangle ABC with obtuse angle in A, the line CI, where I is the incenter of ABC, pass thru the center of circle ABI?

[west cloud]

!show

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

[west cloud]

@midnight pier First draw a diagram, and then post it here.

[midnight pier]

so far i can only show that JIE is isosceles

[west cloud]

I don't see vertices A, B, C here

[midnight pier]

its changed

to DEF

[west cloud]

alright, ic

give me a moment

[midnight pier]

i see a bunch of nice properties that i can use to prove it, like DH parallel to JE

[west cloud]

wait so <D is supposed to be obtuse? or do you mean <E?

[midnight pier]

E

[west cloud]

alr

ok so ABC is now what?

EDF? EFD?

[midnight pier]

EDF yeah

can be any

of these two

[west cloud]

Wait no

[west cloud]

it can't be

the order matters

I know that A is now E, but what is B? and what is C?

[midnight pier]

points

[west cloud]

right, but point A is now point E; what is point B now? and what is point C now? like, the new label?

[midnight pier]

can be any

D or F

in this case C is F

[west cloud]

ok, got it

ty

so you can draw the incircle

and the lines from the incenter to the tangency points on DEF

to prove that IF is the perpendicular bisector of JE

@midnight pier

[midnight pier]

of JE yeah

[west cloud]

wdym?

[midnight pier]

perpendicular bisector of JE

[west cloud]

ok, so try this

[west cloud]

sry typo

[west cloud]

and post it here

[midnight pier]

i dont think i need to draw the incircle

oh i think i proved

yeah

[west cloud]

great!

!done

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[midnight pier]

do you wanna see the solution?

[river shale]

Why not

[midnight pier]

angle JDI is equal to IDE, which is equal to IEJ and IJE because of cyclic quad, then the arc JI is the same length of arc IE, also, we have that the arc DI is the same length of HI because of the bissection of angle A, but since arc JI and IE are equal, we have arc HE same length as DJ, now angle FJE is equal to FEJ because they use arcs of same length, i.e., arc JE and DJ or HE, so JFE is isosceles, since FI is bissector we can show that FXJ is congruent to FXE, which implies FXJ = 90 degrees, so pass thru the center (also XJ = XE)

there's some useless segments in the diagram

[river shale]

They look kinda messy ngl

[midnight pier]

yeah i was trying to exploit some properties

[river shale]

Lemme check

[river shale]

Looks legit to me, how did you manage to come up with this in 3 minutes?

[midnight pier]

i was thinking about it 20 minutes earlier

[river shale]

Prodigy 🔥

[midnight pier]

ty!

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[rotund solstice]

hi guys, i need a little help, how i can find all possible combination for n where n 0<n<10.
so n cannot be - any number and not higher than 10 right?
so n is = {1,2,3,4,5,6,7,8,9}
so how do i find all possible combination for n ?
Ex n = 1 and so on till 9

[normal hollow]

2^n rule ,if i m not wrong

U do 1st element with the rest individually

As in
{(1,2),(1,3),(1,4)....

Till 9

Similarly foe each of 2,3,4,5,6,7,8,9

[rotund solstice]

bro nailed it

worked

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[slate basin]

if f(n) = f(n/2) and f(n) = f(n+5) everything is mapped to one thing except multiples of 5 that are mapped to the other thing meaning the range of this function f is only of length 2. Can u say this just by looking at it? And if yes then how

[slate basin]

Btw

[restive aurora]

Does n have to be an integer? If so I can probably explain why

[slate basin]

Yea more context was needed

Is is positive integers and the first condition is only valid for n = even and the second condition is only valid for n = odd

[restive aurora]

Alright, well can I ask where this question comes from? It kinda looks like a competition problem or a number theory problem

I wanna know what approach I should take

[slate basin]

It is a competition problem

Aptitude exam

[rough talon]

so f:Z+->Z+, f(n)=f(n/2) and f(n)=f(n+5), and your asking why there are only 2 possible ranges for f?

[slate basin]

Z+ yes

Wait so

[restive aurora]

Okay, do you know about modulo arithmetic?

[slate basin]

I am thinking it shud start from 1 right?

Positive integers

[rough talon]

probably just mod 5 the domain and manipulate it using the f(n)=f(n/2)

[restive aurora]

Sure, if you wanna show all numbers not divisibly by 5 are the same, 1 would be a natural place to start

[slate basin]

Yea so I did an example

And I got the result that everything mapping to one thing

And only multiples of 5 map to the other thing

But like without an example how can I say this

By looking at it

[restive aurora]

Well you can infer a few other rules

Like f(n) = f(n/2) -> f(n) = f(2n)

[slate basin]

Yup

I used this

[restive aurora]

Hmm... the fact that f(n) = f(n+5) only works for n odd is throwing me off

Are you sure that's right?

[slate basin]

Yea the example showed

The answer is 2 as well

[restive aurora]

Can you screenshot the original problem for me maybe?

[slate basin]

Uh it's on me book

I'm eating u will have to wait

[restive aurora]

Fair enough

[midnight pier]

!original

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

[restive aurora]

Well... lets start with the multiples of 5

[slate basin]

In a minute

[restive aurora]

Clearly we have f(5) = f(10)

[slate basin]

Yes

[restive aurora]

Do you know about induction in proofs?

I have a method with strong induction that should work nicely

[slate basin]

We also have f(20) from f(10) hence we also have f(15)

And so on

Maybe

[slate basin]

Nope

[restive aurora]

Well since we've got those first 4, lets pick an arbitrary m > 3 and assume we know that if n is less than m, then f(n5) = f(5)

So we're assuming we know that f((m-1)5) = f((m-2)5) = ... = f(5)

Now let's prove it for m5

No I actually mean less!

[slate basin]

Why did this file attach bruh

[restive aurora]

Lmao idk

[slate basin]

Oh ok

[restive aurora]

So there's two cases right, m is even and m is odd

[slate basin]

Yes

[restive aurora]

What do you think you can say if m is even?

[restive aurora]

Keeping this in mind

lmk if I'm not making any sense btw

[restive aurora]

Huh fair enough

Yeah that makes sense, I was just thinking about it from the wrong angle initially so the even/oddness threw me for a bit of a loop. I'm confident I know an answer now

[slate basin]

I am actually leaving for work

Are u gonna post the answer

[restive aurora]

I was planning on just guiding you to an answer, but if this is just a practice problem then sure

My general thinking was to show that past a certain point we could always get the n in f(n) to a smaller value

So if we already know that all multiples of 5 and all non-multiples of 5 less than n are the same, we can show that f(n) must be the same too (depending on whether or not it's divisible by 5)

[restive aurora]

What I was getting at here was if m is even, then f(5m) = f(5m/2)

[restive aurora]

And m/2 is clearly less than m, so this would apply

[slate basin]

No wait

[restive aurora]

Then similar for the odd case, with f(5m) = f(5m + 5) = f(5(m+1)) = f(5(m+1)/2), where (m+1)/2 < m for sufficiently large values of m

[slate basin]

Actually guide me to an answer I will close this rn and ping u later when I'm free

If that works

[restive aurora]

Eh, how long do you think you'll be? It's almost 9pm here

[slate basin]

I will open a thread u can reply whenever ur free

[restive aurora]

Aight sure

[slate basin]

Oki

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[sacred badger]

how would I go about solving this?

[safe fulcrum]

let's say I have a triangle with sides 3 and 5 instead

if you put the sides 3 and 5 on a straight line

what's the longest the 3rd side can be?

(yes you have a degenerate triangle, but you can bend the 3 or the 5 ever so slightly to make it a proper triangle)

[sacred badger]

anything under 8?

right?

[safe fulcrum]

yep! well it'd just be 8

so there's a technicality that the longest side can never be 8

but it can be 7.9
or 7.99
or 7.99999

we call 8 the supremum then

so yes you need the < sign, so that it becomes side < 8

[sacred badger]

so in the problem n would have to be 3x+5 or shorter

[safe fulcrum]

right!

and can you guess/see where x - 1 comes from?

[sacred badger]

subtracting x+3 from 2x+2

[safe fulcrum]

yeah

so the minimum is still going to be a straight line

but now the sides are pointing in the opposite direction

[sacred badger]

well wouldn't the answer be n<3x+5? Or is there a limit on how small n can be

nothing says n has to be the longest side

oh wait nvm

so it's x-1<n<3x+5

thanks!

.close

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[nimble cobalt]

the game pokemon tcg has mechanism where you opens packs and get random cards. some cards have a much higher chance than others. there are other games that probably have similar mechanism. what are some algorithms that could be used for a mechanism like this?

[nimble cobalt]

i know of random weighted sum. idk if that's the exact name, but it's where each item has a weight and you sum up the weight of the items and when you pick a random number between 0 and the sum and whatever item is that that random number is what you get

[sage glacier]

This seems more like a question for the programming server

[nimble cobalt]

there's some math behind things like so i thought i'd ask here

[sage glacier]

I used to play the tcg online and in person I think you'd need to present a more specific idea though, if you don't even have a specific problem in mind.

[nimble cobalt]

hmm yeah i guess i need to be more specific. i was wondering if there were just known algorithms for these gacha / roulette / rng mechanics

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@nimble cobalt Has your question been resolved?

[chrome beacon]

a lot of it is simulation

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[safe umbra]

I'm getting the answer of zero which is not in the options

[vale dock]

@safe umbra Has your question been resolved?

[gilded bloom]

@safe umbra

[gilded bloom]

this s the question?

[gilded bloom]

what exactly did you do after writing the expressing 1/(a+1)(b+1)...?

you took the lcm of the denominators?

[safe umbra]

I first multiplied the pairs (x-2)(x-3) + ....

Then added them

[gilded bloom]

yes

and you got 3x^2-8x+1

thats correct

after that

[safe umbra]

Then I used vietas equation for the roots

[gilded bloom]

yes

ok got it

you then got the values of sum of roots and product of roots

then solved the denominator

[safe umbra]

Yep

The ones encircled are the denominators

[gilded bloom]

oh

let me check your work

because you got the equation right

[safe umbra]

Ok thanks

[tidal terrace]

Like your work doesn’t look wrong
Question mistake ig

[safe umbra]

Could be i guess

[gilded bloom]

it is correct

your answer is correct

it is equal to 0

the options are wrong i think

@safe umbra

sorry i forgot to reply my bad lol

[jovial edge]

Indeed

[safe umbra]

Ok ok nice I guess the question is wrong then

[gilded bloom]

yes

not the question but the options provided are not correct

as u said

any other thing you need help ith

[safe umbra]

I hope the options are correct in the actual test lol, idk how I would solve this in 1min 30s

[gilded bloom]

its easy actually

if you can open those brackets quick enough

you can solve it in under 1.5 minutes

[safe umbra]

I guess I just need to solve more

[gilded bloom]

ok

let me know if you need help

feel free to dm or tag me

[safe umbra]

Sure thank you very much 😊

[tidal terrace]

This in 1:30 minutes damn

[gilded bloom]

ok you may close the channel now pls

[jovial edge]

ikr

[safe umbra]

.close

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[glass kelp]

wonderful

top one is in perfect square form

wait

lemme

[ I = \int_{-1}^1 \frac{\sqrt{x^4 + 2x^2 + 1}}{x^2 + 1}\cos(3x) \dd x]

[rocky lotus]

k

[glass kelp]

wonderful

now

[glass kelp]

.

hmm

ibp?

[lone furnace]

what did you just say

[glass kelp]

nah

doesnt seem to work

prolly feynamn tehcnique

lemme try

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[glass kelp]

wtf

i havent even finished

[lone furnace]

op trolled by asking kindergarden questions 😮

[glass kelp]

it is indeed not feynman

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[floral comet]

what does this mean

[strange thistle]

It means that the f'(X) only tells us about the local behaviour of y=f(X)

[safe fulcrum]

if you imagine the graph of the function f and then zooming in so that you can only see the neighbourhood of the turning point (and not anything else)

the derivative only tells you that the function is horizontal at the turning point

and that also implies that around the turning point, the slope of the function is close to 0

for a continuous function of course

but beyond the neighbourhood of the turning point, you can't know anything else about the function from f'(x) = 0

[floral comet]

wait wdym by “seeing the neighbourhood of the turning point”

[safe fulcrum]

https://www.desmos.com/calculator/lv0twgk9r8

if I have this graph for example

I can zoom in around the turning point to get something like this

[floral comet]

Ok so what about this

[safe fulcrum]

so just from this small portion of the graph

it's impossible to know if the function is lower anywhere else

if the function attains a lower y-value than this

[floral comet]

@safe fulcrum I don’t understand what it means by the local behaviour and global behavioir

[safe fulcrum]

global behaviour = the big picture, seeing the graph as a whole (the 1st image)
local behaviour = the smaller picture, focusing in on one particular area of the graph (the 2nd image)

[floral comet]

So what it’s saying is the first derivative is pretty much useless

In finding the extremes values

[safe fulcrum]

it's actually quite nuanced

[floral comet]

Also I’m confused why the first derivative doesn’t show the global behaviour

[safe fulcrum]

solving f'(x) = 0 tells you where the x-coordinates of the turning points are

but not the y-coordinates

say if I have x^2 + 5 and x^2 - 5

both of those have the exact same derivative (they differ by a constant)

so the derivative can't tell apart functions with different vertical shifts

can't tell apart min (y-value) = 5 from min = -5

[floral comet]

I don’t get this last sentence

[safe fulcrum]

the derivatives of x^2 + 5 and x^2 - 5 are the same

so yes, the x-coordinate of the turning points are the same (x = 0)

but the y-coordinates are very different: y = 5 and y = -5

so the derivative can't tell you what the y-coordinate of the turning point must be

[safe fulcrum]

.

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[lime bone]

Hello

[lime bone]

Excuse me

[west cloud]

Do you have a question? If so, post it here

[lime bone]

Okay

Two lines in a space that are neither intersecting with each other nor parallel

[noble anvil]

So skew lines

[lime bone]

All those way to find their distance is what I would like to learn

And compares each way about the efforts spent

[noble anvil]

There's a straightforward formula

End of 2nd page

[lime bone]

1: I can find the normal vector of one line, let’s name it vector A, then I find the vector, say vector B, that is parallel to the another line, which starts from a point of the line.

Vector B can be expressed as some variable and is actually moving point on the line

[lime bone]

The hell is that

No formulas

[noble anvil]

[lime bone]

No formula but simple understanding

Let’s go, drop your thoughts

.close

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[lime bone]

Im just wasting my time

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[crystal gulch]

2 lines in a graph? just use distance formula if that's the case (d= sqrt((x1-x2)^2 + (y1-y2)^2) to calculate the distance between point (x1;y1) and (x2;y2)

[noble anvil]

It's shortest distance

Not any distance

And it's in 3d not 2d

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@crystal gulch Has your question been resolved?

[crystal gulch]

hm okay i'm assuming they're talking about 2D cuz of this

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[fringe dune]

i need help for b)

i understand that if there is an x that is in the intersection that means it s divisible by all positive integers ( multiplying all integers with no end) so x must be infinite, but x is a positive integer ( it s and elemnt of An) so it must be finite. thus x cannot be in the interesection.

is this enough? i am unsure of the multiplying integers with no end part

[vale dock]

@fringe dune Has your question been resolved?

[pine prairie]

Ya I feel like that makes sense

Suppose a is in this intersection, then a is divisible by n for all n >=2, let n > a, a cannot be divisible by a number greater than itself, hence a ∉ A_{a+1} hence a contradiction

This would work too

[fringe dune]

this is so much clearer

ty

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[weary bloom]

Solve this AB+AC = A+BC , find smallest A+B+C

[wraith heart]

what are A, B, C

[weary bloom]

whole number

nothing else is given

[storm perch]

Oh, make it AB-A+AC-BC=0, and then factor

[wraith heart]

just plug small whole numbers

[storm perch]

Nevermind I'm tripping

[wraith heart]

(A,B,C) = (1,1,1) even works

[jovial edge]

Isn't a=b=c=0 smallest

[weary bloom]

Well how about you try smth like 0 and smth as it's written whole number

[marsh forge]

are they distinct, if not theres (0,0,0)

[wraith heart]

sometimes

[weary bloom]

Well the question is given this

[marsh forge]

sighh math just be normal for once

[wraith heart]

doesn't say distinct, so either 1,1,1 or 0,0,0

[weary bloom]

well mr , one is larger

how is one larger than 1

[wraith heart]

see

[weary bloom]

what

say smth more than see

@wraith heart

[wraith heart]

One of the numbers 1, 2, 3, ... (OEIS A000027), also called the counting numbers or natural numbers. 0 is sometimes included in the list of "whole" numbers (Bourbaki 1968, Halmos 1974), but there seems to be no general agreement.

[weary bloom]

Bruh ofc 0 is whole

[wraith heart]

alright go with that

but that's something particular to your teacher/class

[weary bloom]

Bruh...............

Solve the question

[wraith heart]

other people already gave you the answer

scroll

[weary bloom]

what...

show

[wraith heart]

.

[weary bloom]

Omg , not that , the answer

and the logical solution

damn.......

<@&286206848099549185>

damn do you all only deal with easy questions

[tidal swift]

i don't think there is any way to get a set of solutions where there is a largest whole number

[tidal swift]

i agree with this so far

wait

are you looking for the solution (2, 1, 0)?

actually this seems to work for any z >= 2 for solution set (z, 1, 0)

[next vortex]

it does

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@weary bloom Has your question been resolved?

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[swift flame]

i don’t understand this question… if anyone could help it’d be great :’)

[cloud epoch]

Which one, a, b, c?

all of the above?

[swift flame]

well yes

[cloud epoch]

ok

What is the product rule first of all?

do you know it?

[swift flame]

yea

[cloud epoch]

ok

so how would you use it on fgh?

[swift flame]

exactly how they show it in the statement

[cloud epoch]

basically you need to write fgh as a single product of two

[swift flame]

oh

of two

[cloud epoch]

so for example could you say like fgh = f x (gh) maybe?

[swift flame]

yea

[cloud epoch]

ok so try that on a piece of paper

and see if you get the answer

then you can move on

[swift flame]

…?

i’m confused

[cloud epoch]

Not exactly what you want to do

you want to do (fgh)'

[swift flame]

yea

[cloud epoch]

so what you need to do is do (f x (gh))' and consider gh as one thing

so you could name gh = l and do the product rule on f x l

do you catch my drift?

[swift flame]

is this what h mean?

u*

[cloud epoch]

yeah

the top right thing

[swift flame]

okok

[cloud epoch]

and now you can use product rule on gh

g x h

and do you get the result now?

[swift flame]

oohhh

yes

yes yea

[cloud epoch]

great!

[swift flame]

but can’t the product rule be applied directly like

(uvw)’ = u’vw + uv’w + uvw’

[cloud epoch]

thats a rule, but the product rule is just with two things

[swift flame]

oh ok..

okok cool

[cloud epoch]

like, you show this with the product rule of two things

just like we did here

[swift flame]

ahh ok if i have to show my work it’s thru that

okay🫡

[cloud epoch]

yup

ok so now b

try to do it, if you have difficulties come back here

[swift flame]

i don’t understand why there’s a f’(x) multiplied to 3[f(x)]^2 when it’s just the derivative of [f(x)]^3 when

and the question is asking for the derivative of [f(x)]^3

[cloud epoch]

write down the formula you found in b

with f = g = h

what do you get ?

[swift flame]

wait what

[cloud epoch]

In a i mean

lol

the formula you found in a

the derivative of a product of 3 things

apply to the case where f = g = h

[cloud epoch]

yep

so that's the same answer as they have

right?

[swift flame]

wait whar

oh yea

it is

[cloud epoch]

👍

[swift flame]

but how does the equation we found in a

= to [f(x)]^3

[cloud epoch]

the equation in a isnt = to [f(x)]^3

[swift flame]

oh

[cloud epoch]

the equation in a just allows you to write the first line you wrote here

[swift flame]

wait are we not proving that it’s equal to the equation in a

[cloud epoch]

its not what b asks

b asks to show that (f^3)' = 3f^2f'

which you did

[swift flame]

how is the equation in a (f^3)’ 😔

[cloud epoch]

its just that a tells you that (fgh)' = f'gh + fg'h + fgh'
So if you say f = g = h you get from a that:
(fff)' = f'ff + ff'f = fff' which is what you want

[swift flame]

OHHH

i get it now

[cloud epoch]

ok cool

[swift flame]

yes

[cloud epoch]

now c

[swift flame]

so that u don’t have to scroll

[cloud epoch]

yeah thank you :)

you can try to do it alone first

[swift flame]

the problem is i never know what they’re asking for 😭

whats part b

like the whole question b?

[cloud epoch]

what you showed in b

d/dx (f^3) = ...

[swift flame]

yes

[cloud epoch]

use that to differentiate e^3x

basically you need to find f = ? so that when you apply the formula you showed in b, it gives the derivative of e^3x

does that make it clearer?

[swift flame]

not rly i also think what i did is wrong…

wait no it’s e^3

[cloud epoch]

why is y' = 3e^x x 1

[cloud epoch]

yes

because d/dx (e^x) = e^x

[swift flame]

yesss

[cloud epoch]

ok so your answer is 3e^3 then

and you did use b right?

[swift flame]

wait why is it 3e^3

[swift flame]

uh…not rly

?

[cloud epoch]

why not?

[swift flame]

like this?

[swift flame]

is it not just e^3

[cloud epoch]

no not like this

i will write it doen for you so it is clearer

[swift flame]

okay😭😭

[cloud epoch]

Does that make it clearer for you?

We use the result shown in b applied to the function f(x) = e^x

and it gives us what we want

[vale dock]

@swift flame Has your question been resolved?

[swift flame]

sorry i didn’t get the notification for some reason 😔

[cloud epoch]

oh lol

[swift flame]

oh wait

e^x is f(x)

[cloud epoch]

yes

[swift flame]

not e^3

[cloud epoch]

yeah, that was your confusion, its because you want f^3 = e^3x

thats why you choose f = e^x

[swift flame]

how did yk that tho

[cloud epoch]

know what?

[swift flame]

😭😭 i never understand what the question is asking for

i thought it was just asking for the derivative or e^3x

[cloud epoch]

what you want is to apply the formula found in b to e^3x basically

[cloud epoch]

yes

that's the question

[swift flame]

but why isnt f(x)= e^3x

[cloud epoch]

and because thanks to question b you know how to find the derivative of a function cubed (f^3) and e^x is a well known function, you say f = e^x

[cloud epoch]

how do you differentiate then?

using b

[swift flame]

uhh

[cloud epoch]

this is the thought process wanted

Question b gives you the ability to differentiate f^3 if you know f and f', and since you want to differentiate e^3x = (e^x) ^3 and you know e^x's derivative, you say that f(x) = e^x

if that makes sense to you

[swift flame]

but if question c gives u y= e^3x does that not automatically make it the f(x) of the question..

[cloud epoch]

no because no one said y = f

[swift flame]

oh

[cloud epoch]

and btw, f is just a letter

i could very well name it john for all I care

[swift flame]

😭😭😭

[cloud epoch]

john(x) = e^3x is viable too yk, its just standard notations are easier for everyone to agree on and know what you're talking about

[swift flame]

okay i see

let me try to process this for a sec

[cloud epoch]

so in question c, you are the one choosing who f will b so that it matches the formula in b

question c could be formulated as:
What should f be so that we can use b's formula to differentiate e^3x

its just implicit

[swift flame]

oh okay

[swift flame]

maybe id understand the question if it was like this :’)

[cloud epoch]

that's something you'll get used to, understanding little implicit things like this

[swift flame]

um ok i was reviewing a and im a little confused abt that again

[cloud epoch]

what are you confused about?

[swift flame]

wait let me write it out one sec

[cloud epoch]

a is basically saying, we know how to differentiate a product of two things, now how do we use this knowledge to differentiate a product of 3 things

if that makes it any clearer to you

[swift flame]

yes but

i think i made an algebraic mistake or smthng

but i can’t get the answer anymore😭

[cloud epoch]

You write fgh as f x (gh)

[cloud epoch]

wow

i dont understand what you did lol

[swift flame]

😭ok wait

sorry

[cloud epoch]

you misapplied the product rule

[swift flame]

waittttt

oh

i see it now

[cloud epoch]

product rule: (fg)' = f'g + fg'
So (f x (gh))' = f' (gh) + f(gh)' and you wrote f' (gh)' instead

[swift flame]

oops

waitlet me try again

[cloud epoch]

alright

[swift flame]

ohh ok

i got it

[cloud epoch]

great :)

[swift flame]

do u have time for another calc
question :’)

[cloud epoch]

I want to go play a game rn lol, but im sure someone else will be able to help you

[swift flame]

ok tysm!

[cloud epoch]

you can start a new help channel

:)

[swift flame]

.close

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[midnight pier]

stuck on this, if we use "=" as an equivalence relation to prove that "~" is an equivalence relation, can we assume well-definedness from surjectivity or some other property? for example, if f is not well defined, f(1/2) might not equal f(2/4), making "~" not reflexive

[cloud epoch]

What do you mean f(1/2) might not equal f(2/4) ?

[midnight pier]

if f isn't well defined, we cant say ~ is reflexive

since ~ is defined by the image of f

[cloud epoch]

its a map of sets, so im not sure what you mean by well defined?

[midnight pier]

am i missing an axiom of set maps?

[cloud epoch]

Because every element of A has an image, and because f is surjective, every element of B is an image of an element of A by f

[midnight pier]

like are they well defined by definition

[midnight pier]

what about like f: Q->Z+ \ {1}

and f(a/b)=a+b

thats surjective

and f(1/2) != f(2/4)

since f(1/2) is 3 and f(2/4) is 6, yet 1/2=2/4

[cloud epoch]

yeah but then its not a map

because a same element has 2 distinct images

[midnight pier]

oh wait is map synonymous with function

like are all maps definitionally well defined

[cloud epoch]

map is basically a function

a map cant send one element to two different things

[midnight pier]

ok so i could have just taken that for granted

alright thats huge

idk why i thought i had to prove it

thanks

.close

[vale dock]

Channel closed

Closed by @alpine beacon

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[glad minnow]

is this good?

[quasi valve]

this doesn't look right

[glass kelp]

[ \sum_{i = a+1}^b i^2 = \sum_{i=0}^{b} i^2 - \sum_{i=0}^{a} i^2]

how did u

just apply the sum of square formula from i = a+1

seems sussy to me

[quasi valve]

should be $\sum_{i=a+1}^b i^2 = \sum_{i=0}^b i^2 - \sum_{i=0}^a i^2$

[rocky lotus]

Bungo

[glass kelp]

mb mb

[quasi valve]

second sum only goes up to a since you want to keep the a+1 term

[glass kelp]

why a not a+1?

[quasi valve]

you don't want to remove the a+1 term right?

[native plinth]

Mappings must be well-defined

[glad minnow]

i thought it is possible to remove 1 from the left and add it to the n terms on the right this way

[glass kelp]

oh shi u right

mb mb

[native plinth]

Oh, the question changed, don't mind me, sorry

[quasi valve]

nw

[rocky lotus]

k

[quasi valve]

ha np, it happens

answer's gonna be a mess haha

[glad minnow]

so this isn't valid?

[quasi valve]

no

[midnight pier]

The question before asked if certain functions were well defined it threw me off

[glad minnow]

got it, how do i use this?

[quasi valve]

you can't just substitute the formula for the whole sum from 0 to b, into the sum

[quasi valve]

you have formulas for both sums on the right hand side

[glad minnow]

this would give me an answer w both a and b terms?

[quasi valve]

yep

[glad minnow]

i see

[quasi valve]

answer's definitely gonna depend on both a and b

[glad minnow]

ill try it

[glad minnow]

what does the i = 1 look like?

[glad minnow]

what does the i = 1 version look like?

[glass kelp]

exactly like that

0 + anything = 0

[glad minnow]

is this for the right side?

[glass kelp]

ye

[glad minnow]

y doesnt it change?

[glass kelp]

.

[glad minnow]

oh is it bc 0 has no output?

or 0 gives back 0 for the 1st term

[glass kelp]

[ \sum_{i=0}^b i^2 = 0 + \sum_{i=1}^b i^2 = \sum_{i=1}^b i^2]

[rocky lotus]

k

[glass kelp]

the sames for a

[glad minnow]

got it, ty

[glad minnow]

this right?

[glass kelp]

why is it -6b + 6a

[glad minnow]

this is the original but its + not - made a mistake

[glass kelp]

so it should be

+6b - 6a, no?

[glad minnow]

yea

[glad minnow]

the qn also says a < b & a and b are +ve, what do i do with this?

[glass kelp]

its required so that the sum makes sense

ye i checked in wa

lookks good

[glad minnow]

thanks!

.close

[vale dock]

Channel closed

Closed by @glad minnow

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[inner mirage]

From this, I get the answer as 2

[inner mirage]

But upon looking at the question a little but more, I figured out that
$ln \alpha + ln \beta + ln \gamma = ln \alpha \beta \gamma = ln 45$

[rocky lotus]

Sarin

[glass kelp]

Can’t u do vieta

[autumn whale]

wow new profile picture k

[glass kelp]

Yup

From my trauma with chem

I hate chem

[autumn whale]

lol i havent been traumatized by it yet

im taking ap chem this year

[inner mirage]

Im curious to know how you get the other 2 roots. Can anyone explain that to me?

[wraith heart]

,tex .quadratic formula

[rocky lotus]

riemann

[inner mirage]

Yes I know that but here, t = e^x

So how do you get x

The other 2 roots are complex

[glass kelp]

There is no need for finding each root

[wraith heart]

x = log(t)

[inner mirage]

t is complex

[wraith heart]

log is the inverse of exponential function

[glass kelp]

The result from vieta suggests that the sum of the roots is -b/a

[desert dome]

im 14 and im trying to teach myself calculus its been a long road i finally made it to pre-calc but i'm so confused can i have some help

[wraith heart]

Oh yeah you're just looking for the sum anyway

[tidal swift]

!occupied

[vale dock]

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

[inner mirage]

Yes. I wanted to find out how to get x if x=log(t) and t is complex. But now I realize that it's probably a better idea to figure that out myself instead of wasting someone elses time.

While I'm here, Could I get some help on this question as well?

[glass kelp]

there’s no need to find each and every x

[inner mirage]

I was just curious

My working

[glass kelp]

Vieta again

[wraith heart]

Generally it's better to just open a new channel

[inner mirage]

Ok I will do that the next time

[inner mirage]

Im getting 2 solutions but its supposed to be just 1, 98

[glass kelp]

Inside bracket is Alpha + beta + beta + gamma

Beta + gamma is given

[inner mirage]

Oh

I am stupid

But wait

We still need to find lambda

[autumn whale]

that sgamma

wait no

nvm

[glass kelp]

Alpha * beta = 7
Alpha * gamma = 7+3lambda sqrt(3)

Use this to relate beta and gamma

And using that we can substitute it in

Alpha + beta = 4 lambda
Alpha + gamma = 3sqrt(2) + 2sqrt(3)

This whole mess is just vieta exercise

[inner mirage]

So divide the first 2 to get the relation btw gamma, beta and lambda, then subtract the other 2?

[wraith heart]

Just calculate and show

[vale dock]

@inner mirage Has your question been resolved?

[glass kelp]

fyi, $t_1 \cdot t_2 \cdot t_3 = 45$. So, $\ln(t_1t_2t_3) = \ln45 = \ln t_1 + \ln t_2 + \ln t_3$. Since $t = e^x$, $\ln t = x$. So, $x_1 + x_2 + x_3 = \ln 45$

[rocky lotus]

k

[glass kelp]

It’s not -b/a, it’s c/a mb

[inner mirage]

I'm trying k 's method of substitution, but I'm not really getting anywhere. What's wrong with the method I used here?

[glass kelp]

also [ \beta + \gamma = 3\sqrt{2}] with other cases too

[rocky lotus]

k

[glass kelp]

the calculation is so fucking tedious

i tried

this method isnt efficient

but gives u a soln at least

anyhow

[surreal drift]

There seems to be an incorrect simplification of those radicals, is the first thing I'm noticing

[inner mirage]

Which step?

[glass kelp]

(for mine)

[inner mirage]

Because I am getting the correct solution. But, according to the answer key, the answer is just 98, not 98 and 2

[inner mirage]

Did you get the correct solution?

Oh wait Im not supposed to ask for sols

mb

[toxic ridge]

what seems to be the problem currently

[inner mirage]

In my method, Im getting 2 answers

1 is correct

[toxic ridge]

the factorising one?

[inner mirage]

the other is not even listed a solution

[toxic ridge]

where you factorised the cubic into linear factors

[inner mirage]

This one

[inner mirage]

Nah that was a different question

[toxic ridge]

is this the current question

[toxic ridge]

this just looks like some vieta manipulation

[inner mirage]

Yes

[surreal drift]

Excuse the terrible handwriting, this is what I thought was the issue, I am only in high school, so I imagine this is way off

[glass kelp]

it is

we're wondering why lambda = -2 is invalid

in my working, i use 5 equations to remove lambda completely

which is very tedious

[inner mirage]

I mean if nobody has any ideas, I can close this channel and ask my maths teacher about it the next time I see him

Maybe the answer key is wrong

and 2 is valid

[glass kelp]

,w (sqrt(3) + 2sqrt(2))(3sqrt(2) + sqrt(3) - 2sqrt(2))

[rocky lotus]

Results provided by WolframAlpha

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

Requested by nuclearinteraction

[glass kelp]

,w (sqrt(3) - 2sqrt(2))(3sqrt(2) + sqrt(3) + 2sqrt(2))

[rocky lotus]

Results provided by WolframAlpha

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

Requested by nuclearinteraction

[glass kelp]

,calc 7 + 3(sqrt(2))sqrt(3)

[rocky lotus]

Result:

14.34846922835

[glass kelp]

,calc 7 - 3(sqrt(2))sqrt(3)

[rocky lotus]

Result:

-0.34846922834953

[glass kelp]

@inner mirage found it

[inner mirage]

?

[glass kelp]

if ${\lambda = \pm \sqrt{2}}$, ${\alpha\gamma = 7 \pm 3\sqrt{6}}$, right?

[inner mirage]

Wait but isnt lambda +- sqrt(2)

[rocky lotus]

k

[glass kelp]

ye mb

[inner mirage]

Yeah

[glass kelp]

and from our derivation

$\alpha\gamma = (\sqrt{3} \pm 2\sqrt{2})(3\sqrt{2} + \sqrt{3} \mp 2\sqrt{2})$, right?

[rocky lotus]

k

[inner mirage]

Yes

Oh I see

[glass kelp]

only lambda = +sqrt(2) agrees