help-43

[strange pendant]

{},{1},{1,2},...{1,2,3,4,5,6,7,8,9,10}

11 different equivalence classes

[{}] = {y ∈ P(A) | {} ~ y} ⊂ A

[{}] = {Y ⊂ A | {} ~ Y} ⊂ A

X ~ Y <=> |X| = |Y|, for X,Y ⊂ A

[{}] = {Y ⊂ A | {} ~ Y} ⊂ A

[{}] = {Y ⊂ A | |{}| = |Y|} ⊂ A

[{}] = {Y ⊂ A | 0 = |Y|} ⊂ A

[{}] = {Y ⊂ {1,2,...,10} | 0 = |Y|} ⊂ {1,2,...,10}

[{1}] = {Y ⊂ {1,2,...,10} : 1 = |Y| } ⊂ {1,2,...,10}

I get it I think

@random path

[{1,2}] = {Y ⊂ A : 2 = |Y| } ⊂ A

[random path]

hello

[strange pendant]

the problem with writing the set builder for the equivalence class

is that you need to know one representative

whatever I think i got it dude

[random path]

Alright

[random path]

Have you ever done modular arithmetic?

It's basically just equivalence classes on Z actually

And you use a representative from each class all the time

[strange pendant]

I haven't

but i think I will learn about it in this course

like diofantine equations

[random path]

Ooo

[strange pendant]

congruence systems

Chinese remainder

[strange pendant]

[random path]

Like the equivalence classes are based on the relation a ~ b iff n| b- a, where n is any integer greater than 1. Turns out that's an equivalence relation.

It's also the case that you can do arithmetic (addition and multiplication) with the equivalence classes and have it make sense regardless of whichever representatives you use

anyways imma try to sleep but idk how long that'll take. Just saying it in case I don't respond to any pings.

[compact pewter]

@strange pendant Has your question been resolved?

[strange pendant]

I'm so yird

tired

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[sick kiln]

need help dont know where to start

[twin anchor]

do you know what vieta’s formulas are

[sick kiln]

no

[twin anchor]

Using these, you can write 2 equations from the first quadratic and 2 for the second

[sick kiln]

ohhhhh

.close

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[tender pasture]

how did they do the green circled bit

[tender pasture]

how do i do 15/2/3 without calculator

[kind viper]

step 0 is to not write 15/2/3 bc nobody can tell wtf that means.

you should say 15/(2/3) here.

(15/2)/3 is not the same thing as 15/(2/3), yeah?

anyway, do you know how to divide fractions in general?

@tender pasture

[compact pewter]

@tender pasture Has your question been resolved?

[compact pewter]

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[winged lion]

. @eager pagoda

[quartz yoke]

uh

sorry did i take your channel?>

[winged lion]

nope

[compact pewter]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[quartz yoke]

ok

i thought of like

writing denom at

as

x(1+x^n) and then pfd

but i dont think it will end well

soo

idk

[keen granite]

pfd should work?

[\frac{1}{x(1+x^n)}=\frac{A}{x}+\frac{B(x)}{1+x^n}]

[boreal girder]

PajamaMamaLlama

[keen granite]

therefore (1=A(1+x^n)+B(x)x)

[boreal girder]

PajamaMamaLlama

[quartz yoke]

would it work

hmm

why not?

IDK

cuz u get

A=1

oh

wait

it works

[keen granite]

[quartz yoke]

okay let me write it

[keen granite]

[quartz yoke]

the x^n seemed scary

and i thought well maybe it doesnt works

by idea of pfd

B=0????

im not sure

Bx^2=0

B = 0 right?

[keen granite]

[kind crane]

,w partial fractions 1/(x * (1+x^5))

[boreal girder]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by rie.mann

[quartz yoke]

nah

crazy

[keen granite]

what lmao

[kind crane]

i don't think it's the right approach since x^5 + 1 has 5 roots

[keen granite]

oh I see the fraction bars

[kind crane]

and in general x^n+1 has n roots

[quartz yoke]

right

[keen granite]

,w simplify 1/x-x^(n-1)/(x^n+1)

[boreal girder]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by xxmrfancyu2xx

[quartz yoke]

i have an idea

[keen granite]

it works?

[quartz yoke]

multiply by conjugate?

uh idk

[keen granite]

just rearrange for B(x) with A=1

[quartz yoke]

how to do that

[kind crane]

i see. this isn't the traditional partial fraction decomposition though.

at least not the one listed here

[winged lion]

Probably you would end up having lots of constants being 0

[quartz yoke]

uh

should i look at like

indications

if there are

[keen granite]

.

[quartz yoke]

like this book has indications for some problems at the end

[quartz yoke]

idk how to do that

told you already sorry..

[keen granite]

oh mb didn't see it

we know A=1 so using that, solve for B(x) in terms of x and n

things should turn out nicely

[quartz yoke]

ok

[forest token]

take x^n+1 common

[quartz yoke]

x^n+Bx^2 = 0

Bx^2 = -x^n

B= -x^n-2

[keen granite]

where is x^2 coming form?

[forest token]

dont think pfd is required tho

[quartz yoke]

B(x)*X

[keen granite]

[quartz yoke]

therefore

[forest token]

||1/x(1+x^n) = x^-1/(1+x^n) = x^-n-1/((x^-n)+1) now take denominator = t||

[quartz yoke]

this is right?.?

ok

from the begginging

1=A+Ax^n +Bx^2

[kind crane]

start here again

[quartz yoke]

1=A+Ax^n +Bx^2

[kind crane]

no

[quartz yoke]

why

[kind crane]

you just copied it wrong

[forest token]

@keen granite

[kind crane]

where is the x^2 ?

[keen granite]

DMs

[quartz yoke]

is B(x)*x not Bx^2

[kind crane]

correct it's not

B(x) is function notation, like f(x). in this case, B(x) is a polynomial in x

[quartz yoke]

why

oh wait

really?

mb

i didnt know

[kind crane]

review https://en.khanacademy.org/math/cc-eighth-grade-math/cc-8th-linear-equations-functions/8th-functions-and-function-notation/v/understanding-function-notation-example-3

[quartz yoke]

ok then 1=A+Ax^n +B(x)*x
1=1+x^n+B(x)*x
x^n+B(x)*x=0
B(x)*x=-x^n
B(x) = -x^n-1?

[quartz yoke]

im fine thanks

[kind crane]

yes and it agrees with this here

[quartz yoke]

ok so

it turns on

that

ok i got that

from the picture

1/x +1/x(x^n+1)

first is ln

but second

pfd again??:(

[kind crane]

wut

[quartz yoke]

its okay

[kind crane]

you started with the bottom.

[quartz yoke]

i got it

i got it

thanksss

[kind crane]

the top is the result of all that algebra pajama did

[quartz yoke]

thank you all guys

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[pseudo wasp]

Help

[mild sky]

multiply in each fraction the denominator and numerator by the conjugate of the denominator

[pseudo wasp]

oK

.close

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[strange pendant]

the set made out of all the finite subsets of N is countably infinite?, what can i do about it? is the cardinality of this set 2^(Aleph null) or what?

[full moat]

No, its cardinality is ℵ_0

Else it would be uncountable

[strange pendant]

like usually the cardinality of a power set of a set S is 2^n where n is the cardinality of S

[full moat]

But what you're considering is not the full powerset

[ripe ether]

but power set includes infinite subsets, whereas here you are considering finite subsets

[full moat]

But actually a "small" section of it

Your set of finite sets is a countable union of countable sets(finite sets in this case)

[strange pendant]

how do you know the cardinality is aleph null

[full moat]

Countable union of countable sets is countable

(assuming ACC, but I doubt you'll be bothered by this)

[ripe ether]

🤔 why is it finite sets in this case? there are possibly infinite amount of finite set with fixed cardinality?

or did i just misinterpret

[full moat]

Each finite sequence is finite

[strange pendant]

yes infinitely many number of finite subsets of N

[ripe ether]

oh u r referring to the length of the sequence not the number of them

[full moat]

Yeah, the number of sequences is countably infinite

While each sequence is finite

But well, the important thing is

You can construct your wanted set(the set of finite subsets of N) as a countable union of countable sets

[strange pendant]

why do you mention sequences

[full moat]

Finite subsets of N "behave like sequences", but it's simply a habit, every time I've seen this exercise it mentions sequences, give it no mind

[strange pendant]

yes but as you said b its countably infinite, and you didn't explained why this subset of the powerset of N has cardinality of aleph null

[full moat]

Say you have a countable collection of sets A such that every set B∈A is countable

Then U(B∈A) B is countable

That's the theorem in question we are using

[strange pendant]

is this notation for index family?

[full moat]

Yes

[strange pendant]

which theorem

[full moat]

"A countable union of countable sets is countable"

[strange pendant]

how is the cardinality aleph null

[full moat]

Any countable set has aleph null cardinality

If you want to see it more "intuitively"

[strange pendant]

what, you serious?

[full moat]

n•ℵ_0=ℵ_0

[full moat]

Yeah

Well, countably infinite

Finite sets have finite cardinality of course

[upper bane]

if it can be mapped one-to-one to the naturals it's aleph 0 i think?

naturals*

oops

[full moat]

No, that's 2^ℵ_0

[full moat]

Yeah

[upper bane]

my bad

[full moat]

Well, essentially, ℵ_0 is the unique countable infinite cardinal

So given an infinite countable set A clearly |A|=ℵ_0

[strange pendant]

ok this explains a lot of my troubles i think, i appreciate it

any countably infinite set has cardinality aleph null

[ripe ether]

here is an example where I construct a bijection between Z and all finite sequence of naturals.

list out the primes: 2,3,5,....

for each number n, n is equal to, uniquely, as the product of primes raised to some power (possibly 0).

We can correspond n to the powers (over each prime listed above, which is a finite sequence of N). For example, since 4=2^2, then 4 maps to (2,0,0,0,...). SInce 6=2*3, we map 6 to (1,1,0,0,,,,) (its basically telling us what power we rise for each prime).

This correspondence is a bijection.

[full moat]

This is clever

Every time I had to prove this result, I simply constructed sets of finite sets that contained a specific natural, argued they were countable and proceeded to construct the set as a countable union of countable sets

[strange pendant]

this is pretty awesome dude... to say the very least

[ripe ether]

i 'm quite the opposite. I think of this as a proof that countable union of countable is countable.

[full moat]

Would such a construction be generalizable to any countable set, though?

Fairly sure if we went to weirder countable sets we'd need more machinery

[ripe ether]

you biject countable sets to N, N\times N, N^3,...

and since the union of that bijects with Z^+

[full moat]

Hm, so bijecting each set of the union with a power of N?

[lime juniper]

in general for all infinite sets S, S ~ SxS, and so on inductively

(this requires choice)

[ripe ether]

yeah, and you may probably need to use the Schröder–Bernstein theorem

[full moat]

I don't see the use of countable choice here, though

Which is odd, since the result is equivalent to it

[strange pendant]

how to prove the set made of all the finite subsets of N is countable?

[full moat]

This works

[full moat]

And this works

[ripe ether]

might be needed to show that there is a bijection between each countable set with N^k for some k ranging over all k in a formal way

[full moat]

The justifications are up to you though

[full moat]

Hm, well, yeah, I guess that ℵ_0^n=ℵ_0 requires choice

[ripe ether]

honestly idk. I just assume it

[azure vault]

If you know that a countable union of countable sets is countable, then just as a small hint, know that a finite subset of N (unless it is emptyset) has a maximum element

regroup the subsets of N into groups that share the same maximum element. How many groups? What's the cardinality of each group?

and done

[ripe ether]

i think that don't need choice but picking it over all n need choice.

[azure vault]

principle of induction anyone?

[full moat]

Actually, yeah, for finite n that does not need choice

It may be the assignment, yeah

Well, guess I'll just have to think about it

[strange pendant]

the cardinality of each group forms a one to one correspondance with the naturals

something like that?

[azure vault]

uhhhhhhhhhhhhhhhhhh

what

I just asked if it's countable, finite, uncountable

[strange pendant]

the subsets are finite, but they are infinitely many of them

there are*

[azure vault]

I don't get you

"the subsets are finite"

are you talking about the subsets of N themselves, the groups of subsets?

[lime juniper]

i dont think so becaue mathbb N is wordered

[strange pendant]

there are infinitely many finite subsets of N

[lime juniper]

so its inherently choicable

and N^n is wordered in the dictionary order

[strange pendant]

the groups of subsets with same cardinality are countable and infinite, no?

[azure vault]

so... this is not what I'm asking

[full moat]

Yeah, noticed afterwards

[lime juniper]

Renato I recommend reading the first chapter of Munkres's Topology

[azure vault]

so each group of subsets contains an uncountably infinite amount of subsets is what you're saying?

[lime juniper]

before he actually gets to topology

[azure vault]

you're wrong at that if that's what you meant

[full moat]

That's not where choice lies, they pointed out it's probably in the assignment of bijections, like, choosing the way to map each countable set to a power of N

[azure vault]

Take for example ${A\subseteq \bN, \max(A) = 4}$

[boreal girder]

Raphaelisius Maximus MMIII

[azure vault]

The "group" (set) of subsets of N with max = 4

[lime juniper]

i think you mean : or |

not ,

[azure vault]

actually it's an accepted one

[lime juniper]

also wtf you can just do \bN? Ive been writting \mathbb N for so long

[ripe ether]

same

i typed \rightarrow before knowing \to exists

[lime juniper]

$\bR \bC \bH$

[azure vault]

\bN is a shortcut for this bot, not on your overleaf (unless you define it)

[boreal girder]

gfauxpas
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lime juniper]

oh

[winged lion]

,, \N

[winged lion]

[strange pendant]

theres the group of finite subsets with cardinality 1, the group of finite subsets with cardinality 2, up to k, where k is the maximum cardinality a finite subset of N can have, this are countably infinite many groups of finite cardinality

[lime juniper]

k does not exist

well not as a number

so be careful how you're getting to k

[azure vault]

each group you define here actually have countably infinite cardinality

e.g the group of subsets with cardinality 1 has {1},{2},{3},...

[ripe ether]

so you are saying for each $i,j$ the collection of set with element bounded by j and cardinality bounded by i is finite?

[boreal girder]

qwertytrewq

[ripe ether]

there is no need for i in this case. all you need is j to say they are finite

but it works ig

[lime juniper]

actually even as a cardinal

it has a supremum but not a maximum

so your statement doesnt make sense sorry

[full moat]

A supremum in ON I assume you mean?

Because if we "limit ourselves" to N, then it has no supremum either

[lime juniper]

some set of cardinals large enough to cover the topics under discussion

[full moat]

Fair enough

[azure vault]

So just to say renato, there are many ways you can regroup the finite subsets of IN, I specifically proposed the "max" approach but you seem to either be doing "groups of subsets with same cardinality" or "groups of subsets with max, divided into subgroups with same cardinality".
The first approach is fine, but you have to show each group has countably infinite cardinality.
The second approach is okay-ish, even you didn't need that subdivision, but then there are "IN^2" groups instead of IN groups, which is fine since the former is also countable, but more convoluted.

[lime juniper]

what's IN?

oh, \bN

[ripe ether]

isn't the k they are talking about finite? and i thought finite ordinals can be intrepreted as natural numbers?

[full moat]

finite ordinals are natural numbers, the thing is that renato was talking about an upper bound for the cardinality of finite subsets of N

Which is outside of N

[azure vault]

that was my second interpretation of what that k meant, are they doing a subsubdivision for groups with max = k or something?

[ripe ether]

i thought renato is picking a finite k

ig that bit is unclear

[full moat]

"k is the maximum cardinality a finite subset of N can have"

[azure vault]

yes that bit made no sense

[azure vault]

@strange pendant did you mean that exactly? Or did you mean
"k is the maximum cardinality a finite subset of N with max = k can have"

[ripe ether]

ye i just striaght up assumed he meant: "for any finite k, we restrict to where~~"

[azure vault]

because you do understand set {1,....,k+1} has cardinality bigger than k right

and so there's no max cardinality for finite subsets

[strange pendant]

what's the difference? i think prim interpreted me correctly, sorry if it's wrong though

[azure vault]

If prim interpreted your sentence correctly

[strange pendant]

no?

[azure vault]

Then you believe there is some magic natural number k

such that EVERY finite subset of N

no matter which one

has cardinality smaller (or equal to) k?

Because

[lime juniper]

the famous biggest number

[azure vault]

As I literally just stated

{1,....,k+1} has cardinality k+1 bigger than k

no matter what your magic number k is

k+1 exists

[lime juniper]

the famous biggest number plus one

[ripe ether]

infinity +1

[strange pendant]

True, srry i guess i thought this was possible, because the subsets have finite cardinality

[full moat]

notice N is inductive, assume such k existed, k∈N, so k+1∈N, {1,...,k+1}⊆N has strictly larger cardinality than {1,...,k}

[strange pendant]

indeed, srry

[lime juniper]

don't feel bad for learning math btw

[azure vault]

Everyone is finite, but some are more finite than others

Animal farm noises

[full moat]

Oh, and in general, let (A,<) be a poset, if A has no maximal element or no minimal element, then A is infinite

[azure vault]

if you take a poset, then it's maximal and minimal elements, not maximum or minimum

[full moat]

Alright, I'll correct that

[strange pendant]

can you explain your idea? i misinterpreted it i think

[full moat]

I've seen the terminology used indistinctly though

[azure vault]

ok

Let's give some names to everything

I like to call $\operatorname{Fin}(\bN)$ the set of finite subsets of $\bN$

[boreal girder]

Raphaelisius Maximus MMIII

[strange pendant]

ok

[azure vault]

each one of those subsets

lest it be empty

each subset will have a maximum element

that maximum element, DEPENDS on the subset

(I'm clarifying this jic this was your source of confusion)

set {1,2,4} for example will have max = 4

but set {1,3,6,1999999375} will have max = 1999999375

the max depends on the subset, and can be as big as it wants, as long as it's finite

there is no "biggest" max

now that this is out of the way

Let's create our groups of subsets

$G_k = {A\in \operatorname{Fin}(\bN),, \max A = k}$

[boreal girder]

Raphaelisius Maximus MMIII

[azure vault]

and $G_0 = {\emptyset}$ because we need to have them somewhere

[boreal girder]

Raphaelisius Maximus MMIII

[azure vault]

G_... are the groups

[strange pendant]

yes

[azure vault]

There is G_0, and then there is G_k for any value of k natural integer

so how many groups are there?

[verbal rover]

so wait for exemple the cardinal of R is infinit right?

[full moat]

Yeah, but since it seems like this'll spark an entirely new convo

[azure vault]

a bit unrelated but R has infinite cardinality, specifically it's uncountable

[full moat]

You should probs go to #help-19 for example

[azure vault]

if you want to delve deeper, open your own channel

[quasi mountain]

I need help

With math

[azure vault]

!occupied

[compact pewter]

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).

[verbal rover]

ik i just saw the pinned messg

[full moat]

#help-37

[quasi mountain]

12 divided by 3

Please

I’m going to get kicked out of school

[azure vault]

12/3 = 4 days of mute, with ban possibly

[strange pendant]

for fuck sake

[azure vault]

good school

[quasi mountain]

Why

[azure vault]

alright dealt with

[azure vault]

sorry back to the subject

[verbal rover]

im confused about something we studied last week we have proven that there are series in Q wicch their limit goes to a real number so applying this we can have a bijective application going from Q to R so does that mean R and Q have the same cardinal?

[full moat]

Open your own help channel

I'll help you there if you want

[verbal rover]

alr mb

[strange pendant]

each group is a set made out of the maximum element present in each subset of finite cardinality of N, something like that?

[full moat]

Not really, it is a set of sets once again

[strange pendant]

i might have misunderstood this set builder

[full moat]

You're getting into a set all subsets of natural numbers that only contain elements lesser than a chosen k

[strange pendant]

ok

G3 = {{1}, {2}}

[azure vault]

G3 gathers all subsets of N with max = 3

The subsets you put in there don't

let's just start with G1

Find all the subsets of N

that have 1 as a maximum element

it means the number 1 has to be in there

and you can't have anything above 1

[strange pendant]

G1 = {{1}}

[azure vault]

yes

that's it

try G2

[strange pendant]

G2 = {{1,2}, {2}}

[azure vault]

yes, exactly

you can go for G3 if you want

otherwise if you noticed the pattern already it's good too

[strange pendant]

G3 = {{3}, {3,2,1}, {3,2}}

[azure vault]

almost

[strange pendant]

exactly 3 elements inside?

[azure vault]

you're missing a subset

[strange pendant]

3 1

G3 = {{3}, {3,2,1}, {3,2}, {3,1}}

dude you came up with this? is crazy

[azure vault]

I mean it's sometimes my approach to this problem

[strange pendant]

you have already tried to prove it?

[azure vault]

there are multiple proofs yeah

I have a latex I created on set theory w exercises

anywho

did you spot the pattern?

[strange pendant]

sort of

don't know how many elements Gn has

[azure vault]

well guess you have to try G4

but it's a bit tiring

let's try to force you to see the pattern

starting with G1, up to G3

I want you to copy paste the groups

but for each subset inside them

remove the max

for example, if G1 = {{1}}

I take the only subset in G1

I remove its max

{1} where I remove 1 becomes emptyset

[strange pendant]

yes

[azure vault]

G1 -> {emptyset}

[strange pendant]

G2 = {{1,2}, {2}}

= {{1}, {}}

[azure vault]

so G2 leads to (not =) {{1}, {}}

and G3?

you copied the wrong one

[strange pendant]

G3 = {{3}, {3,2,1}, {3,2}, {3,1}}

[strange pendant]

yes sry

G3 -> {{}, {2,1}, {2}, {1}}

[azure vault]

does that start looking like something

[strange pendant]

G3 leads to this plus {2,1} and {1,2} which is exactly G2

idk dude

[hallow fox]

when you do the ‘remove the max’ step, the groups are basically lining up with the collection of all finite subsets of the naturals. That whole collection is countably infinite, since you can match each finite set with a unique number (like by using binary strings). So it’s not uncountable like the full power set, just countable

[azure vault]

Don't you recognize G3 as the powerset of smthg...

[full moat]

This might help you a little

[hallow fox]

G3 itself is { {3}, {3,1}, {3,2}, {3,2,1} } = { {3} ∪ S : S ⊆ {1,2} }.
If you ‘remove the max’ from each set in G3, you get {}, {1}, {2}, {1,2} — that’s the powerset of {1,2}.
In general: removing the max from G_n gives the powerset of {1,…,n-1}.
So |G_n| = 2^(n-1). The union of all G_n (plus the empty set) lists all finite subsets of N, which is countable.

[azure vault]

way to jump the gun 😭

[full moat]

You're not supposed to give the guy the answer y'know

[strange pendant]

i didn't recognuzed the pattern ngl

[azure vault]

then rip

but if I told you that |G1| = 1

|G2| = 2

|G3| = 4

[strange pendant]

|Gn| = 2^(n-1) like the big pencil guy said 😭

|G0| = 1 aswell i think

[full moat]

We're not considering 0 natural here so

[strange pendant]

yes but G0 was not defined with the maximum element

G0 = {{}}

[strange pendant]

like from here

[full moat]

Oh, then yeah, you're right

[azure vault]

so you got spoiled but

you see that all the groups Gn are finite

and there's a countable (infinite) amount of them

[strange pendant]

May I ask why you say that all the groups Gn are finite? like {1,2,3,4} -> 1 < 2 < 3 < 4

[full moat]

Their cardinality is bounded above by 2^n-1(equal to be more specific , as was shown) for n the maximum chosen to define the group

2^n-1 for a finite n will always be finite, hence they'll bw finite

[strange pendant]

what is the point of doing this group association that are a set that consist of of sets with a maximum number n, for Gn to have finite cardinality that is?

[full moat]

Yeah, we want to construct our desired set from a countable collection of countable sets

In this case, a countable collection of finite sets, as you can see

[strange pendant]

the way the groups are constructed imply they are countable? because the cardinality is upperly bounded by a natural number?

[full moat]

Yeah

If you can bound the cardinality of a set upwards by a natural number the set is finite

[strange pendant]

then we take the union of this infinetely many countably finite groups and conclude the union is countably infinite?

sorry i might have confused things, is hard to keep track of everything

[strange pendant]

theres infinitely many possible groups

[full moat]

Yep, exactly

[full moat]

But the number of them can be put in bijection with N, map each group to the maximum element used to define it

So you have a countably infinite collection of finite sets

[strange pendant]

the union of this infinitely many groups is exactly equal to the

set which contains all the finite possible subsets of N

? i pressume then

[full moat]

Yeah

[strange pendant]

is my first time encountering this exercise on proving this set of finite subsets of N is countable

i didn't know it was famous or anything

[azure vault]

did you show whether P(N) was countable or not btw?

[compact pewter]

@strange pendant Has your question been resolved?

[full moat]

It's a common one, yeah

[strange pendant]

no, I haven't, but I presume some of those subsets are infinite,

actually this, showing that the finite subsets of N are countable was kind of related to an exercise on which i needed to count the number of distinct equivalence classes, (i think its equivalent of finding the cardinality of the quotient set, that is the set made of all the equivalence classes) for an equivalence relation on A, where A was the set of all the finite subsets of N, so because there were infinitely many equivalence classes I was trying to explain a friend that there was infinitely many equivalence classes for this equivalence relation on the set made out of all the finite subsets of N, but explícitly because the set made of finitely many subsets of N is infinite, so I was trying to explain a friend of mines this set this infinite but i think it turned out is actually countably infinite because you guys proved it forms a bijection with the natural numbers, explicity because the union of this groups is equal to the set of all finite subsets of N,

but coming back to your question, no i haven't showed P(N) is countable, but i guess that power set would contain not only infinitely many subsets but also, will contain subsets with infinitely many elements inside, which probably is enough to blow your head apart,

its my first introduction to proofs i just started university this week and I am already blown away by how difficult are things now, it will take time to get used to it most likely dude

my classmates, were duscussing that the set containing all finite subsets of N has cardinality 2^n where n is a magical biggest number, imagine bro

[strange pendant]

you also know it?

[full moat]

Yeah, though my solution was as I exposed above, simply considering finite sets with a specific element, much less refined

This one is much more elegant

[strange pendant]

@strange pendant

[full moat]

A lot of people struggle with the shift from computation-based exercises to proofs, yeah

[strange pendant]

i appreciate the help dude, sorry for the cringe I started university this week dude, I have too much stuff to learn

[full moat]

Nah, don't worry, nothing to be ashamed of in learning

I'm also quite the newbie btw

I did first year already but at a really bad uni, so I changed and I'm now practically going to repeat first year 😆

Though with some second year subjects mixed in

[azure vault]

P(N) is uncountable fyi

it's really fun to show

[strange pendant]

damn good luck with that dude, for me dude classes started for me this Monday and prof already left me shit tons of homework I have been doing them tho so Hopefully im not that behind the schedule, I wish you best of luck with your university dude

[full moat]

Thanks, and same thing man

And don't worry, if you're doing the exercises you're assigned, you'll probably do well

I'll be starting in some days still, though, my classes start the 8th of September

[strange pendant]

i appreciate the help really raph, will revisit this after I ask my prof how to do the exercise i mentioned, and ask her about the cardinality of this set because as you guys say, it has cardinality of aleph null, basically thank you for the creativity, of doing the help and reshaping the problem into another problem, like idk if i am explaining myself or whatever

[strange pendant]

you got this dude, try to pass all the classes you didn't this time you are more experienced

[strange pendant]

thanks, i hope so dude

we didn't covered continuum hypothesis i think so idk why the exercise uses a countably infinite set or whatever

[full moat]

The continuum hypothesis is much more advanced, doubt you'll cover that

[compact pewter]

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[strange pendant]

if you guys had a hard time explaining this to me you should try to explaining to my classmates i don't think it will be easy by any means

[full moat]

Well, its statement is not terribly complex, but proofs regarding it and the stuff that surrounds it definitely is

[compact pewter]

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[kind hollow]

So, I have been seeing this kind of posts for a good while now. I wonder if what they're saying is actually true or if they're making up stories that are so elaborate, no one thinks to fact check them lol

[upper bane]

https://arxiv.org/pdf/2410.09331v1

looks like this one is partly true

but only in context

[covert crater]

I'm going to guess it's more of pop science and heavily oversimplified

[kind hollow]

yeah thats what I think too

Even schroedinger's cat is just a saying. If what I heard is true, there never was a cat in the original experiment. It was just dumbed down for the oblivious humans like me to understand lol

I hate when they word this stuff like that

[covert crater]

tbh, your average layperson isn't going to want to read arxiv papers or any other advanced physics textbook that is maths heavy

[upper bane]

i mean if you go to wikipedia you would see the term hypothetical cat

[kind hollow]

then why should the matter be dumbed down? It loses its true significance imo

[kind hollow]

yeah but like 80% of the world thinks its a fact lmao

anyways sorry for ranting

[upper bane]

why do you think pop science and math pisses me off

but ig if pop science helps motivate the public to learn about the truth, sure

i mean, i was kinda motivated by pop science and math to begin with

was there once

[kind hollow]

yeah, I guess they could be learning worse stuff than that

[covert crater]

yeah I'm thinking it's more of providing motivation to do the actual stuff

like bill nye inspiring kids to do physics, biology, etc even if it's simplified for them

[kind hollow]

alright, question has been answered. It's just for plebs to think they know whats happening in the modern scientific front

[full moat]

Divulgative stuff is sometimes needed for two reasons:
1: to motivate the need for research on that discipline
2: to interest potential future students of the field

If I went up to someone and instead of the "some infinities are larger than other" phrase I started constructing the ordinals to them

I assure yoy they'd turn their brain off midway

[upper bane]

"you lost me at ordinal"

[covert crater]

Like for example, a lot of classical mechanics in high school have assumptions to make the calculations a lot simpler

[full moat]

So while I certainly find this stuff annoying at times, too, it is needed

[covert crater]

if you started assuming air resistance and other things, it gets complicated really quick

[full moat]

So real

[upper bane]

ig if it isn't outright false, it's fine

and as long as people realize there's more to it

[kind hollow]

I can conform to this perspective

[full moat]

That's more an approximation than anything else, though

[kind hollow]

.close

[covert crater]

yeah

[compact pewter]

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[full moat]

This particular phrase, for example, though

Every time someone writes it, a little portion of hell is let loose, the amount of misconceptions it produces is absurd

[compact pewter]

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[compact pewter]

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[old rivet]

\hr \ Let $f: \mathbb R^2 \to \mathbb R$ with $f(u(x), v(x)) = (\text{something})$. Why can we say [\frac{\pa f}{\pa x} = \frac{\pa f}{\pa u} \frac{\pa u}{\pa x} + \frac{\pa f}{\pa v} \frac{\pa v}{\pa x}?] It seems as if we would get the double on the right side (if we conceptually cancel)?

[winged lion]

What do you mean by double?

Also it helps to try small examples

[old rivet]

Derivatives in Leibniz notation usually behave like fractions, so if you think about cancelling du and dv, we get 2 * df/dx

(I know we can't actually cancel)

Is there some intuition to why the RHS matches the LHS and it's not 'LHS = 2 * LHS'?

[winged lion]

The cancellation doesn't make sense

,, \pdv{f}{x} = \pdv{f}{u}\f{du}{dx}+\pdv{f}{v}\f{dv}{dx}

[pearl burrow]

you don't really get to "cancel" with partial derivatives

[rigid perch]

you can always say that although you should be using regular derivatives for all derivatives wrt x

[old rivet]

Why do you make a difference between partial and d?

Can't we use them interchangeably here

(and should be using partial)

[winged lion]

because u and v are single variable functions

[old rivet]

ah

[old rivet]

Why does it not make sense?

[winged lion]

Although I realized, if you had something like f(u(x,y),v(x,y)) you could ask the question again

[rigid perch]

the whole "cancellation" thing with derivatives in leibniz notation works for ordinary derivatives but not for partial derivatives

[old rivet]

those are still 'infinitesimal changes in u' (intuitively) and we should be able to (intuitively) cancel

[old rivet]

oh

[stark marsh]

wait what happened here is some form of chain rule or what exactly ?

[old rivet]

Oh you're right I remember seeing something about having a triple derivative that feels like should cancel but gives -1

[old rivet]

Yes this is chain rule, I think you can rigorously justify it by writing out the total differential of f

Afterwards dividing by the differential of x

[winged lion]

Maybe it helps to think of it as this
[ \pdv{f(u(x),v(x))}{x} = \pdv{f(u,v)}{u}\f{du}{dx}+\pdv{f(u,v)}{v}\f{dv}{dx} ]
realizing they are not the \textsl{same} derivatives.

[old rivet]

Do you mean $\pa u$ versus $\dd u$ as the difference?

[boreal girder]

ILikeMathematics

[thorny kestrel]

example of derivatives not canceling as you'd expect they would:
suppose you have a relationship between 3 variables f(x, y, z) = 0) using implicit differentiation you can show that

[old rivet]

Right

[thorny kestrel]

yep

[winged lion]

Not only that but the right side is about different quantities

you are also holding one variable fix at one time, that should already tell you that they can't sum up to 2f_x

[rigid perch]

If you're familiar at all with matrix multiplication the familiar leibniz cancellation comes when multiplying the jacobians
[ \dv fx = \jdv{f}{u,v} \jdv{u,v}{x} = \pmat{\pdv fu & \pdv fv} \pmat{\dv ux \[1.5ex] \dv vx} = \pdv fu \dv ux + \pdv fv \dv ux ]

[boreal girder]

cloud

[old rivet]

Oh, that's the formal proof?

[rigid perch]

well you can frame all applications of the multivariable chain rule as multiplication of the relevant jacobians

and you do get the familiar cancellation property in the jacobian notation

[old rivet]

This is essentially this

[old rivet]

Ok I have a question

So we have d/dt f(x(t)) = f'(x(t)) * x'(t)

By the usual chain rule for total differentials

Now why is the jacobian of f'(x(t)) that matrix on the left

for x'(t) I agree with the vector on the right

[rigid perch]

the composition is hidden by the leibniz notation since it doesn't include where the derivative is evaluated

[old rivet]

Oh it's that

Thanks a lot!

.close

[compact pewter]

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[compact pewter]

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[paper marlin]

confused on what determines velocity and accelleration to be increasing/decreasing and positive/negative respectively

[paper marlin]

wait i think i got it, the velocity is directly correlated with the acceleration so if velocity is increasing then accelation is positive

.close

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[compact pewter]

Please don't occupy multiple help channels.

[hardy bramble]

my work (tentative)

[compact pewter]

@hardy bramble Has your question been resolved?

[hardy bramble]

<@&286206848099549185> anyone available

let me know

[compact pewter]

@hardy bramble Has your question been resolved?

[rough roost]

not checking ur differentiations this looks about right

i would justify some of your answers a bit more

but thats kind of a nit

u might want to justify that the higher derivs of a poly vanish

[compact pewter]

@hardy bramble Has your question been resolved?

[compact pewter]

@hardy bramble Has your question been resolved?

[compact pewter]

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[umbral sand]

can anyone teach me how to divide surd becuase i have a test in 3 days and that is like the only thing i dont understand

[covert crater]

do you mean this? $\sqrt{a} \div \sqrt{b} = \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

[boreal girder]

឵឵MxRgD

[umbral sand]

yea

[covert crater]

what part do you not understand?

[umbral sand]

most of it

[umbral sand]

or how to switch the surds and when you should do it

[covert crater]

you usually want to do it when you want to simplify the surd expression

[umbral sand]

okay becasue i had this math question that i was going to send to ask how to do it, but i left the page with the question on it at school

[covert crater]

sure, you can post the math question if you're stuck in it

[umbral sand]

i dont have it on me

[covert crater]

like here for example $\frac{\sqrt{27}}{\sqrt{3}} = \sqrt{\frac{27}{3}}$ and then you can see how you can simplify it

[boreal girder]

឵឵MxRgD

[covert crater]

oh

[umbral sand]

i left it in a folder at school

i thought i had brang it home

but i did'nt

[umbral sand]

could you get a question for me to do

[compact pewter]

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[summer crown]

I'm stuck at this question (I'm sorry for the rough translation)
In a space Oxyz, given A(1,0,0) ; B(0,1,0) and C(0,0,1). M is any point in that space. Calculate the minimum of P = MA^2 + 2.MB^2 - MC^2?

What I did is shown in the picture (I apologize for the poor quality)
Now how do I find the minimum of P? There are..3 variables.. I don't know how..

[random path]

is that 2*MB^2?

[summer crown]

yeah

[summer crown]

yep

[molten badger]

There's a trick to do these types of problem

and as soon as I saw the question I know where're you from

[summer crown]

how do i do it

q-q

[molten badger]

Choose a point I, then $P= (\overrightarrow{MI}+\overrightarrow{IA})^2 + 2(\overrightarrow{MI}+\overrightarrow{IB})^2 - (\overrightarrow{MI}+\overrightarrow{IC})^2$

[boreal girder]

Alexis_Fx

[molten badger]

expand this

[summer crown]

;o

so the P is minimum when MI is minimum?

cuz i think IA, IB, IC is uhm.. fixed

wait but what's the location of I?

[molten badger]

tbh yeah, that's what we want

[summer crown]

i mean uhm.. coordinates

[molten badger]

I would satisfy $\overrightarrow{IA}+2\overrightarrow{IB}-\overrightarrow{IC}=\vec{0}$

[boreal girder]

Alexis_Fx

[molten badger]

By choosing point I satisfy this we simplify it to the simplest form

[summer crown]

why is it have to be vector 0 tho-
and uhm how do you know the condition that I would satisfy (like uhhh.... how do you know I would satisfy IA + 2IB - IC but not anything else?) [If this is a dumb question to ya then im sorry, im just bad at math]

[molten badger]

can you like, expand this for me

expand everything in the parentheses

[summer crown]

uhm... we put I in the middle of MA? so it wil like MI + IA = MA

same goes with the rest

[molten badger]

no don't simplify it

[random path]

It can also be done purely algebraically. Just expand the original expression, simplify it, then complete the squares.

[summer crown]

q-q i dont know i only know that u put I in the middle of MA
idk the true purpose

[molten badger]

expand everything like you normally do, (a+b)^2=a^2 + b^2 + 2ab

[molten badger]

I don't want to torture myself lol but yeah

[summer crown]

i did, but the result is a func with 3 variable

and idk how to find the minimum of that func

[random path]

If you complete the squares, the minimum of the squares is 0

[summer crown]

oh-

but what about the -1?

[random path]

Well the leftover constant is your minimum :p

[summer crown]

wait what

[random path]

yup

[molten badger]

yeah he's right

[random path]

Heh I should've looked at all your work earlier XD

[molten badger]

tbh this way is just too long for me lol

but that works

[summer crown]

OOOOOOOH i understood

thank you guys

i rlly appreciate y'all efforts

:D

[random path]

np

[summer crown]

imma close ig

.close

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[rugged parrot]

I've got a function modelling a fountain canopy. I've determined the surface area (the first integral), volume (second integral) and width (third expression). My goal is to somehow find a way to minimise the width while keeping these integrals constant. Only issue is I'm not sure how to do that. Any ideas?

[rugged parrot]

heres a visual if it helps

[candid turret]

You're calculating the surface of a solid of rotation that has that parabolic (or close to it) profile

[rugged parrot]

yup

I need to keep the surface and volume of the solid of rotation constant whilst minimising the width

[candid turret]

What are you allowed to change

about the shape

[rugged parrot]

im pretty much allowed to change anything as long as the surface and volume stay constant

https://www.desmos.com/calculator/alb1w2ovdm

im only focused on that one stream that i have modelled, so the rest can be ignored

[candid turret]

i think that there's a crazy enough curve that will minimize the width but aren't you interested in parabolas

[rugged parrot]

not really, no. i just modelled it with a parabola right now because its what seemed easiest

[candid turret]

So the problem could be stated like this: you have a curve gamma1 whose solid of rotation (keeping the x axis as the rotation axis) has a fixed volume V and surface area S defined on a closed interval [a,b] in the xy plane. You call b-a = w the width. Does there exist another curve gamma2 with volume V and surface area S which minimizes the width? What is it?

[rugged parrot]

Yes the problem could be stated like that.

It's basically for a mathematical investigation im considering doing

[candid turret]

Btw why are you multiplying by s inside the surface area integral

[rugged parrot]

this is the formula for surface area right

i wasnt able to get ds on its own in desmos, so I just did it as sdx

which works because

[candid turret]

oh yup u right

[sly musk]

yes you are right

[rugged parrot]

any ideas on how to proceed with this?

[candid turret]

Firstly I'd try to think about how many solid of rotations that were made from different curve profiles have the same volume and surface area

Is it infinitely many? Is it none? Is it one?

If the problem has no restraints I think the answer is not easy to get

[rugged parrot]

what would make the answer easier to get?

[candid turret]

for example if we are looking only at a certain type of curves

[rugged parrot]

oh, thats an easy restraint. assume the family of curves as parabolic

[candid turret]

but idk maybe someone else will look into this and it ll be clearer i actually dont know if im right about what im saying

[rugged parrot]

thats what ive been doing and what I was planning to do

[candid turret]

yeah perhaps this

[rugged parrot]

im honestly just not sure what to even do next. ive got the volume, surface area and width. now how on earth do I minimise this width?

[candid turret]

assume you have y=ax^2+bx+c with width w and try to find a minimum for the width w

Calculate volume and surface area and set them equal to V and S

You'll have two equations with a,b,c,w, V,S. V and S are fixed, the rest are all variables you can change (in particular you are interested in finding for which a,b,c w is minimized)

[rugged parrot]

You'll have two equations with a,b,c,w, V,S
I don't get what you mean by this. What would those two equations be?

i can formulate equations for the volume and surface area - using integrals and a b and c. however im not sure where w would come into play

[candid turret]

It’s the bounds of the integral

From 0 to w

[rugged parrot]

OHHHH RIGHT

oh right. my current bound would also be equal to 0 to w, because its all just translations

good catch haha

so far these are the two equations I've formulated.

I've never really done optimisation with integrals before, this is quickly becoming over my skill level haha (im still in HS). What would I really have to do to proceed from here?

[candid turret]

Do the integrals

I see the problem is quickly getting complicated

I don’t know if this way will get you an answer

[rugged parrot]

i can plug the integrals into an integral calculator as long as i can support my working for later, so im not too worried. but the surface area integral doesnt seem solvable just of the top of my head

oh wat actually it might be with a u sub

right so this is the volume integral solved

and this is the surface area integral solved

[trail shale]

can someone plss help me with alternate angles idk how to get each side

[rugged parrot]

luckily seems to be fine for now. im still unsure on how to proceed tho. for the volume integral, it can just be written in terms of w since the lower bound of 0 cancels. is not the same for the lower integrla though

[rugged parrot]

#❓how-to-get-help

[compact pewter]

@rugged parrot Has your question been resolved?

[compact pewter]

@rugged parrot Has your question been resolved?

[compact pewter]

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[twin idol]

Did you try getting your hands dirty into the work?

[strange pendant]

no

im not sure on how to start

[twin idol]

Alright

First thing to do is to plug in z=-1 into all equations

[strange pendant]

why

[candid turret]

You’re taking a slice of the graph at z+1=0

[twin idol]

They ask you to draw the 2D curves for all the equations at z = -1, z = 0, z = 1 and x = 0

Those are level curves at those planes

Remember how you drew the planes with me?

They now want you to use planes to intersect them with those 3D shapes

[strange pendant]

so if i intersect the level curve with the first surface what do i get? a plane, no way should be a conic section

[twin idol]

It depends on the equations but yeah mostly it will be a conic section

[strange pendant]

something like a ellipse

[twin idol]

Yeah

[twin idol]

$x^2 + \frac {y^2}{4} = \frac 89$

[boreal girder]

VulcanOne

[twin idol]

After that what do you do to make it standard form?

[strange pendant]

idk

[compact pewter]

@strange pendant Has your question been resolved?

[twin idol]

Hmm...shouldn't you make the right side = 1?

[strange pendant]

wouldnt it be better we sketch the ellipsoid and the surface z = -1 separately to get an idea

[twin idol]

Sure if you find it easier that way

[strange pendant]

but we inserted z = -1 now is not equal 1

[twin idol]

Right side currently equals 8/9 right?

[strange pendant]

ye

[twin idol]

What do we do to make it = 1?

[strange pendant]

idk

[twin idol]

Don't you usually divide a number by itself to make it 1?

[strange pendant]

what's your point?

[twin idol]

Divide 8/9 by 8/9

Or multiply 8/9 by 9/8

Of course do that to the other side too

[twin idol]

Yep

Now can you identify the major and minor axes?

[strange pendant]

no

[twin idol]

Hmm

[twin idol]

$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \ \ \frac{9x^2}{8} + \frac{9y^2}{32} = 1$

[boreal girder]

VulcanOne

[strange pendant]

dude

[twin idol]

Wait based on what you turned the 8 into 2?

[strange pendant]

8/9 × 4

from basic arithmetic

oh no, i made a mistake

i am a monke

[twin idol]

Na you ain't

[strange pendant]

32/9

[twin idol]

Ye

[strange pendant]

xD

[twin idol]

What is "a" and what is "b"?

[strange pendant]

this will get nasty

[twin idol]

Don't write decimal values

[twin idol]

Alright

Now can you tell me which one of them is bigger, and thus which one represents the major axis?

[strange pendant]

,w sqrt(8)/3-sqrt(32)/3

[boreal girder]

Results provided by WolframAlpha

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

Requested by renatoxbz

[strange pendant]

b is the major axis

,w 2sqrt(10)/3

[boreal girder]

Results provided by WolframAlpha

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

Requested by renatoxbz

[strange pendant]

dude

[strange pendant]

i hate this ellipsoids

[compact pewter]

@strange pendant Has your question been resolved?

[potent berry]

@strange pendant

!status

[compact pewter]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

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[loud plume]

a word

[chilly basalt]

good

now post your question

[loud plume]

someone help me with pythagorean identity

[chilly basalt]

Fantastic 😄

[kind viper]

so do you know that identity?

[loud plume]

its like sin^2(theta) + cos^2(theta) = 1

[kind viper]

yes exactly.

[loud plume]

and you sub sin(theta) into the equation?

[kind viper]

yes, and find the value of cos(θ_1)

but before you take the square root you should stop and show us what you get

[loud plume]

does the quadrant matter tho?

[kind viper]

yes it does

which is why im asking you to stop before a specific step

[loud plume]

ok

one sec then

wait I gtg i'll be back later

[compact pewter]

@loud plume Has your question been resolved?

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[carmine garden]

Show if $P(A_i)=1$ for all $i≥1$ then $P\left ( \bigcap_{i=1}^{\infty} A_i\right)=1$

[boreal girder]

wai

[carmine garden]

Is this true

[eager thunder]

it is true

look at $P(\overline{A}_i)$

[carmine garden]

$P(A) =1 \iff S=A$

[boreal girder]

Arnavutköy

wai

[carmine garden]

where $S$ is the sample space

[boreal girder]

wai

[kind viper]

nope

just because P(A) = 1 does NOT mean A = S

[eager thunder]

counterexample:

[kind viper]

(and dually just because P(A) = 0 does NOT mean A = ∅)

[carmine garden]

that I know

[eager thunder]

wait never mind this is false

[kind viper]

anyway @eager thunder's idea is good: really do look at the complements.

[eager thunder]

counterexample: suppose A_i is true when $x\neq i$.

[boreal girder]

Arnavutköy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[kind viper]

what is $\overline{\bigcap_{i=1}^{\infty} A_i}$?

[boreal girder]

Ann

[carmine garden]

empty set

[eager thunder]

or like

as an expression in terms of individual not(A_i)s

(intutively: not everything being true is equivalent to at least one of them being false)

[agile ridge]

Remember $(A \cap B)^{\complement} = A^{\complement} \cup B^{\complement}$, wai

[kind viper]

it saddens me to say this, but [extremely loud incorrect buzzer]

[carmine garden]

$\bigcup_{i=1}^{\infty} A_i^C$

[boreal girder]

wai

[agile ridge]

[twin anchor]

you can rely fully on subadditivity on the complement sets here without worrying about what the sets actually look like

[carmine garden]

got ot

*it

tq

.close

[compact pewter]

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[compact pewter]

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[thorny kestrel]

,, \sum_{k=0}^n (-1)^{n-k} \binom{d+m}{k} \binom{n+m-k-1}{n-k} = \binom{d}{n}

[boreal girder]

bloubbloub

[thorny kestrel]

so uhhh

I've tried for small cases

it seems like it doesn't depend on m?

I'm looking for a way to prove this

maybe it has to do with an inclusion/exclusion principle?

[compact pewter]

@thorny kestrel Has your question been resolved?

[verbal rover]

induction?

[verbal rover]

did u try induction

[thorny kestrel]

no

over what even

[verbal rover]

n

d and m are constants

[hushed magnet]

induction over m could be interesting if you can show the sum for a specific value and then show that the sum doesnt change when you replace m by m+1

[thorny kestrel]

I was thinking about it actually

but I'd prefer a combinatorial proof over an induction one

[rotund sphinx]

Hint: Consider disjoint subsets $A$ and $B$ with $|A|=d$ and $|B|=m$. The right hand side is the number of $n$-element subsets of $A$. \ \ Now, consider using inclusion-exclusion to obtain the left hand side.

[boreal girder]

Civil Service Pigeon

[rotund sphinx]

Further hint:

,texsp ||Consider $\binom{n+m-k-1}{n-k}$ in the context of stars and bars||

[boreal girder]

Civil Service Pigeon

[rotund sphinx]

alright imma go poof but see what you can do with this

[thorny kestrel]

so uhhh I kinda went on wikipedia which told me that the stars and bars coefficients were the one of the generating function $\frac{1}{(1-x)^m}$

[boreal girder]

bloubbloub

[thorny kestrel]

so basically one can prove the formula by considering $(1-x)^{d+m} \times \frac{1}{(1-x)^m}$

actually no

actually yes

I'm still thinking about the combinatorial proof though

[compact pewter]

@thorny kestrel Has your question been resolved?

[compact pewter]

@thorny kestrel Has your question been resolved?

[compact pewter]

@thorny kestrel Has your question been resolved?

[compact pewter]

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[compact pewter]

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[molten badger]

Given x,y satisfying $5x^2+6xy+5y^2=16$ and a cubic function $y=f(x)$ with the graph as shown. Let M and m be the maximum and minimum values of $P=f\left(\frac{x^2+y^2-2}{x^2-y^2-2xy+4}\right)$, respectively. Calculate $M^2+m^2$.

[boreal girder]

Alexis_Fx

[molten badger]

the cubic function is $f(x)=x(x^2-3)$

[boreal girder]

Alexis_Fx

[compact pewter]

@molten badger Has your question been resolved?

[boreal nimbus]

i just use trig here

5x^2 + 6xy + 5y^2 = 4(x + y)^2 + (x - y)^2 = 16 then x + y = 2cost, x - y = 4sint etc

[compact pewter]

@molten badger Has your question been resolved?

[molten badger]

$x^2+y^2=\frac{(x+y)^2+(x-y)^2}{2}=\frac{(2\cos(t))^2+(4\sin(t))^2}{2}$

[boreal girder]

Alexis_Fx

[molten badger]

$x^2-y^2=(x-y)(x+y)=8\sin(t)\cos(t)$

[boreal girder]

Alexis_Fx

[molten badger]

Okay I get it

[molten badger]

I just don't know how can you come up with trig in the first place

[boreal nimbus]

5x^2 + 6xy + 5y^2 = C would be an ellipse rotated 45 deg so using x + y and x - y should unrotate it so u have smth like Au^2 + Bv^2 = C, then you know how to parameterize an ellipse, which makes extremizing f(x, y) a bit easier

[molten badger]

They haven't taught me that

[boreal nimbus]

oh maybe there is an easier motivation idk

ig from the form 5x^2 + 6xy + 5y^2 you can maybe guess it can factor as A(x + y)^2 + B(x - y)^2

[molten badger]

that seems like the case here

I guess I'm just bad

Thanks you

.close

[compact pewter]

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[compact pewter]

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[floral kelp]

guys i have a question for 1c