Mathematics - 2019-12-21 (page 1 of 3)

nbro

00:19

@Thorgott What does this have to do with pmf and pdf? You're talking about sampling.

Thorgott

I'm not talking about about sampling anywhere.

loch

He's giving you an example of a random variable which is neither discrete nor continuous

Thorgott

As far as pmf and pdf go, both examples have neither.

Semiclassical

Unless you act like a physicist and pretend that the Dirac delta distribution is an actual function

nbro

@Thorgott You're saying "first choose uniformly". Isn't this sampling?

Rithaniel

00:25

I would call it "sampling" if you were to do this several times, I think

But we're not technically performing this action even once

We're just describing the behavior of the action were we to actually take it

Thorgott

"Sampling" is something you do in statistics. I'm saying "choose", because I'm trying to describe the random variable as outcome of a two-stage random experiment, which I thought might be more intuitive to you. Formally, it would suffice for me to just tell you the cdf.

Semiclassical

It’s a random variable where the probability of being less than x is: 0 if $x\leq 0$, $x/6$ if $0<x<1/2$; $2/3$ if $x=2$; $2/3+(x-1)/3$ if $1<x<2$; 1 if $x\geq 2$

(Typing that out on mobile is a chore)

Anyways. The fact that the cdf is discontinuous at zero means it can’t have a pdf

Astyx

You haven't defined probability for 1/2<x<1

Semiclassical

And the fact that it’s not piecewise constant means it has no pmf either

Oops

I can’t seem to edit that on mobile, weird

Should have been $1<x<2$

Rithaniel

mobile is temperamental

Semiclassical

00:33

Also, should have been “discontinuous at $x=2$”

Rithaniel

you have to tap the message and then quickly tap where the "edit" button pops up at the top of the screen, when it appears for a tenth of a second (at least for me)

Astyx

You mean 0<x<1 ?

Semiclassical

Blaaaargh yes

Thorgott

Here's a picture

Semiclassical

back on laptop now, thank goodness

Thorgott

00:36

The jump discontinuity corresponds to a single point with positive probability ($1$ in this case)

That can never happen for a random variable with a pdf. (Why?)

Astyx

Also you wrote x=2 instead of x=1 I think

Semiclassical

boy

Astyx

Mobile is a pain

Semiclassical

well, tbf, i'm typo prone regardless

i can just cover it up much better on laptop :P

Ultradark

I have a hat

Astyx

00:40

Congrats

Semiclassical

neat

Rithaniel

Awesome hat

Thorgott

I should note that examples of random variables with distributions that are neither continuous nor discrete can, in essence, not get much more complicated than the above. More concretely, every probability distribution can be decomposed into a discrete and a continuous part.

Even better, it can be decomposed into a discrete part, an absolutely continuous part (meaning it has a pdf), and a singular continuous part (which is potentially ugly).

Semiclassical

The Cantor distribution being in the last category

(right?)

nbro

So, what the heck do people mean when they write p(x), p(X), p(X=x), P(x), P(X), P(X=x)? Why isn't P here a distribution?

Rithaniel

00:45

P(X) is traditionally the probability of the event X. It should be a number, if I'm not mistaken

Thorgott

Yup, the Cantor distribution is singular continuous

Well, what people mean when they write something is something those people should specify at some point

nbro

Well, they specify it as a distribution. In many many cases! Maybe I am just talking to you mathematicians, who are nitpicking

Semiclassical

there's a lot of cases where it's well-defined, yes

doesn't mean it's true for every case

and when it comes to proving things rigorously, you really do have to worry about that

Rithaniel

(Also, I'll go on a limb and say that pdf and pmf, if they exist, are just convenient functions that allow you to compute those numbers)

nbro

@Semiclassical If a r.v. is discrete, can I say p(X) is a distribution over X?

In other words

Thorgott

00:48

$\mathbb{P}(X=x)$ is the probability that a random variable $X$ takes the value $x$. That notation is universal.

nbro

If a random variable X is discrete or continuous, can I use the notation p(X) to denote the probability distribution over X?

Thorgott

Every other thing you listed above seems ambiguous

nbro

@Thorgott Ok, I know that, but I don't care about this. I am talking about distributions

Thorgott

What is a distribution "over" X?

Rithaniel

Yeah, probability theory has a lot of minutia to it

Thorgott

00:49

I don't understand what "over" is supposed to mean here

nbro

A distribution is usually represented as graph, where the integral of that graph is 1.

Semiclassical

yeah. it's not a useful probability in some cases, to be clear: the probability that a uniform (0,1) variable has X=1/2 is zero. not very interesting

nbro

X is the domain of that graph

Thorgott

Measure theory generally is a very technical field

nbro

I don't see the problem with that

Semiclassical

00:50

in the simple familiar cases, you can think of the distribution as a graph

Thorgott

No, you are talking about a pdf

Semiclassical

doesn't mean you can always do it

nbro

@Semiclassical Ok, in the continuous case, that's the case. But I know this. Just consider pmf

Semiclassical

(you can always think of the cdf as a graph, on the other hand.)

Thorgott

The "usual" way to represent a distribution would be the cdf

Cause every real-valued random variable has a cdf

(which can not be said about a pmf or pdf, as demonstrated by the earlier examples)

nbro

00:51

I know the difference between cdf, etc, etc

I don't give a damn about that

Rithaniel

Like, a distribution might be something like $N(\mu,\sigma^2)$, but, it's distribution is not it's pdf, and it's pdf is not the probability of a particular event.

nbro

Listen, consider a discrete random variable

that has an associated pmf

Thorgott

Ok

nbro

So, do not consider continuous r.v.s

Ok?

Thorgott

sure

nbro

00:52

Can we use the notation p(X) to denote the pmf, where X is the random variable? Answer: yes, we can!

Thorgott

Answer: no, we cannot

Semiclassical

typically I'd see that as $\{p_n\}$ where $\sum_n p_n=1$

nbro

And I don't see any problem with that

@Thorgott Why?

Semiclassical

though I guess mine has in mind a pmf with integer outcomes

nbro

p(X) would almost be interchangeable with what you call the probability measure

Thorgott

00:54

The notation $p(X)$ is used to mean a function $p$ to an argument $X$.

Semiclassical

wouldn't it be P(X=x) for that? It's a function of what point you choose

so $p(x)=P(X=x)$ would work

or $p_X(x)$

Thorgott

yes, that would be a valid way of defining the pmf

though the pmf in total is just "p" then, "p(x)" is a particular value it takes

nbro

P(X) simply means that the pmf is over some domain. P(X=x) is the probability (or mass, in this case) that X=x.

Semiclassical

@Thorgott Yeah

that makes it seem like P is somehow a function of X

Thorgott

the domain? but $X$ is random variable

nbro

00:55

@Semiclassical What are talking about?

Thorgott

a random variable is certainly not a set

Semiclassical

writing P(X)

nbro

I don't get it. You say you can write P(X=x), but not P(X)?

WTF

Rithaniel

like $P(X)$ might be "The probability that $X$ takes any given value", so maybe $P(X)=1$

nbro

Why can't P(X) mean also mean $P_X$?

Thorgott

00:57

why can't "apple" mean "banana"?

Semiclassical

because that's not how functions work?

nbro

@Thorgott Hahhhhh

Thorgott

no, I'm serious

that is exactly what you're asking

nbro

@Semiclassical How functions work?

Semiclassical

function notation, at any rate

nbro

00:59

Ok, so why is the notation P(x) ok, but not P(X)?

Thorgott

because $x$ (presumably) is a real number and $X$ is a random variable

nbro

And what the heck does P(x) mean?

Thorgott

very different objects

Semiclassical

probability that X=x.

nbro

So, P in P(x) is not a probability distribution?

Semiclassical

01:00

no.

nbro

What is it??????????

A fucking probability measure

Rithaniel

I wouldn't write p(x), personally. That's still a little ambiguous. I's say P(X=x)

nbro

So, what the heck is a probability distribution? Don't probability distributions integrate to 1???

Thorgott

No

pdfs integrate to 1

nbro

and pmfs????

Semiclassical

01:02

sum to 1

Rithaniel

pmfs sum to 1, but in measure theory you can equate the concepts of integration and summation

nbro

They should also integrate to 1, no???

Thorgott

however, cdfs dont integrate to 1

nbro

@Rithaniel OK, as if I didn't know this!!!!

Thorgott

their integrals always diverge

nbro

01:03

@Thorgott OK

So, what the heck is a probability distribution???

Thorgott

A measure on a measurable space such that the whole space has measure $1$.

nbro

Are you saying that a probability distribution is a probability measure???

Because I don't see any difference between your definition and what I recall to be a probability measure

Thorgott

Yes, these expressions are synonymous

nbro

Holy crap

So, a probability distribution has a measure of 1

But why doesn't it integrate to 1?

Semiclassical

if you're doing lower-level probability, you wouldn't use that definition of course

Thorgott

01:06

The probability distribution itself doesn't have a measure

nbro

Aren't these two expressions equivalent?

Thorgott

It assigns the measure $1$ to the whole underlying space

nbro

@Thorgott You just said that probability measure is a synonym for probability distribution and you said that a probability distribution is a measure

Thorgott

Yes

and I said "a measure [...] such that the whole space has measure $1$"

I did not say "the measure has measure $1$"

cause that makes no sense

nbro

So, the probability distribution IS a measure, but does NOT POSSESS a measure

What does it mean for a measure to have a measure of 1?

Thorgott

01:08

Yes

That doesn't mean anything; it doesn't make sense

Semiclassical

Mu (negative)

The Japanese and Korean term mu (Japanese: 無; Korean: 무) or Chinese wu (traditional Chinese: 無; simplified Chinese: 无), meaning "not have; without", is a key word in Buddhism, especially Zen traditions. == Etymology == Old Chinese *ma 無 is cognate with the Proto-Tibeto-Burman *ma "not". This reconstructed root is widely represented in Tibeto-Burman languages; for instance, ma means "not" in both Written Tibetan and Written Burmese. == Pronunciations == The Standard Chinese pronunciation of wú 無 "not; nothing" historically derives from (c. 7th century CE) Middle Chinese mju, (c. 3rd century CE)...

loch

lol i think the amount of time you spent engaged in this discussion couldve been spent on reading the intro to some rigorous probability theory and i think you'll end up being more clear on how things work

nbro

Ok, so what does it mean for "a measure such that the space to have measure 1"???

Thorgott

loch is absolutely right

nbro

I don't have time for that

I still don't get you definition

"A measure on a measurable space such that the whole space has measure 1."

Semiclassical

01:10

a measure is basically a function which assigns values to subsets in a consistent way. a probability measure is one which, among other things, assigns the value of 1 to the entire domain.

when you pick that as the subset to assign a value to

(don't ask me for details beyond that because I don't know them)

nbro

@Semiclassical Ok, this is consistent with my interpretation of probability distribution

I still don't get why I cannot say that a probability distribution integrates to 1

Semiclassical

because that's you assigning a value to the measure, not the measure assigning a value to a subset

Thorgott

What does it mean to integrate a distribution?

nbro

Isn't a probability distribution a function? If it's a function, shouldn't integration be defined for it?

Thorgott

Well, that's a difficult question. Disappointingly (for you), the answer lies in measure theory

Semiclassical

01:13

in the sense of measure theory, it's a function defined on subsets

not a function defined over the reals

nbro

@Semiclassical What do you mean? Do you mean that if I integrate something I am assigning a value? What's the problem with that?

Semiclassical

(the more technical phrase is that it's a function on some sigma-algebra over the domain. yay.)

Thorgott

You don't integrate measures, you integrate functions with respects to measures.

nbro

@Semiclassical So, it can be integrated??

Semiclassical

no. you integrate functions over the reals.

nbro

01:15

So, we cannot integrate a probability distribution

Ok!

This is a start to make things clearer

You can integrate the associated pdf or pmf, though

Thorgott

Yes, you can

nbro

Am I right?

Ok, so a probability distribution is a synonym for probability measure

Right?

Thorgott

(Well, you technically need to explain what it means to integrate a pmf, but this can be done with measure theory)

Yes

nbro

And a probability measure (or distribution) is a measure, of course, as the name suggests

Even though I am not sure what a measure is

Thorgott

Yes

nbro

01:17

But, in the case of a probability measure, it is required that "something" is 1

Thorgott

In terms of probability, the requirement is simply "the probability that anything happens at all is $1$"

nbro

A measure is a function over sets?

Thorgott

Yes, over a (specific type of) collection of sets

nbro

A probability measure (or distribution) is NOT a pdf or pmf

Thorgott

Yes

It is also not a cdf

However, the cdf, pdf or pmf (the latter only when they exist) each uniquely determine the distribution

nbro

01:22

A measure is not a function over reals. Right?

(Just to make sure)

Thorgott

The domain is not the reals, yes. The codomain is, however.

nbro

This is because a measure tells you the probability of a certain subset?

Semiclassical

a probability measure does, yes

nbro

Ok, I am writing down a sequence of statements to make some sense of all of this

Semiclassical

I have a book on measure theory and probability that I need to look at again over break

nbro

01:25

@Thorgott Why do you say "specific type of"?

Thorgott

Because you need to restrict yourself. Just trying to define the measure for the complete powerset is not going to work.

Semiclassical

the technical term is sigma-algebra:

Sigma-algebra

In mathematical analysis and in probability theory, a σ-algebra (also σ-field) on a set X is a collection Σ of subsets of X that includes X itself, is closed under complement, and is closed under countable unions. The definition implies that it also includes the empty subset and that it is closed under countable intersections. The pair (X, Σ) is called a measurable space or Borel space. A σ-algebra is a type of algebra of sets. An algebra of sets needs only to be closed under the union or intersection of finitely many subsets, which is a weaker condition.The main use of σ-algebras is in th...

Thorgott

The reasons for this are highly technical.

Semiclassical

you could probably make a decent enough argument for why those restrictions make sense for probability, but I have no idea what motivates them in measure theory

nbro

After having read the first sentence of that Wikipedia entry, it seems that a sigma-algebra is just the powerset (if I recall correctly what a powerset is).

Thorgott

01:28

There is no translation-invariant non-trivial measure on $\mathcal{P}(\mathbb{R})$ @Semiclassical

nbro

@Thorgott However, you say that you cannot define a measure for the powerset

Thorgott

Even worse, in higher dimensions, you run into the Banach-Tarski Paradox

The Wikipedia page for that has a paragraph on the measure-theoretic implications, I think

nbro

blissful ignorance!

Semiclassical

"If $X = \{a, b, c, d\}$, one possible $\sigma$-algebra on $X$ is $\Sigma = \{\emptyset, \{a, b\}, \{c, d\}, \{a, b, c, d\} \}$, where $\emptyset$ is the empty set."

from that wiki page

Thorgott

@nbro The powerset is a $\sigma$-algebra

Semiclassical

01:29

so that's a set of subsets of X which is a sigma-algebra but is not the power set

loch

Here's an example - throwing two coins
- Set (some people call sample space) $S = \{ HH, HT, TH, TT\}$
- A measure $\mu$ assigns a (non-negative) real number to every ("measurable" - in this case you can ignore this adjective) subset of your set. Morally, this corresponds to the 'probability' of that subset happening.

- For example (if the two coins are fair), $\mu(X) = 1$, $\mu(\{HH\}) = 1/4$, $\mu(\{HH, HT\}) = 1/2$ etc.

- A random variable is a (measurable) function $X: S \rightarrow \mathbb{R}$. An example of a random variable is "How many H's did you get". So $X(HH) = 2, X(HT) = 1, X…

Semiclassical

it's like rectangles and squares. not every rectangle is a square, but every square is a rectangle

sheesh. my brain today

Rithaniel

Or PIDs and UFDs. Every PID is a UFD, but not every UFD is a PID

Semiclassical

hmm, now I wonder

nbro

@Semiclassical So, a powerset is a sigma-algebra, but a sigma-algebra is not a powerset?

Thorgott

01:32

In probability theory, another motivation for $\sigma$-algebras is that they can be used to encode the amount of information one possesses

Semiclassical

the power set on X is a sigma-algebra of X, but there are other sigma-algebras on X which are not the power set

Thorgott

Also, there are, in general, a lot of measures defined on the entire powerset (discrete measures can always be extended to the entire powerset)

But on uncountable spaces, most prominently $\mathbb{R}$, this is usually not possible

Or rather, I'm not sure if it's impossible. The set-theoretic side of this is ridiculously complicated.

So let's just say you will not be able to construct them.

Semiclassical

there be dragons

nbro

Ok, I think I don't need all these details, but are you saying that you can define different measures for the same power-set?

Semiclassical

more like, you can define measures where their action on some part of the power set isn't defined

for...reasons?

i get the sense that the reason one picks these definitions is because they ensure you can't generate icky things

taking the example I quoted before, where $\Sigma = \{\emptyset, \{a, b\}, \{c, d\}, \{a, b, c, d\} \}$

nbro

01:37

@loch This seems valuable information, but, even though I was familiar with this sample space and the idea of a measure that assigns a probability to subsets of the sample space, I need a little more time to understand why a random variable is a measureable function and what is its relationship with the measure

Semiclassical

If I have a measure $\mu:\Sigma\to \mathbb{R}$, then it's meaningful to ask about $\mu(\{a,b\})$, but not about $\mu(\{a\})$ or $\mu(\{b\})$

that case is a bit artificial, though, for the reason Thorgott quoted before (discrete measures can always be extended to the entire powerset)

it's less artificial when my domain is uncountable, though?

nbro

@Semiclassical Because {a} is not part of the sample space?

Semiclassical

well, I wasn't insisting on this being a probability measure specifically

but if you specify that, then I think that's right?

hmm, no

the sigma-algebra would be the equivalent of the event space in probability

nbro

even space = sample space?

Semiclassical

no

if I roll a die, then the outcomes are getting 1 through 6

but there's a lot more possible events than that. one event would be getting an even roll

another would be to get 2,3,4

an event is a particular subset of outcomes

Thorgott

01:45

I finally managed to recover this statement about probability measures defined on all subsets of an uncountable set (with no single point having positive measure) from MO: "The existence of such a measure is equiconsistent to the existence of a measurable cardinal, one of the large cardinal notions, and if ZFC is consistent, cannot be proved in ZFC. (See the notion of real-valued measurable cardinal on the Wikipedia page.)"

Semiclassical

like I said, there be dragons

Thorgott

I don't know what that means, but it certainly means we shouldn't bother

loch

that's the technical part to do with things like $\sigma$-algebras. As an example of why the definitions there matter, consider the $\sigma$-algebra only consisting of the empty set and the entire space $S$.

i.e. your collection of subsets only has *two* subsets - $\emptyset$ and $S$. It turns out that the only measure you can define here has to be $\mu(\emptyset) = 0$, $\mu(S) = 1$.

now define $X$ as before. That's still a function $S\rightarrow \mathbb{R}$. However - it is no longer measurable, and you can kind of see why that's the case - We have $Pr(X=1) = \mu( \{HT, TH\} )$. But we n…

Rithaniel

Is there a term for a toplogy that is also a sigma algebra?

Thorgott

Discrete?

No wait

It isn't necessarily discrete

Rithaniel

01:49

Yeah, like the example on the wikipedia page

I thought it might be discrete, too

Thorgott

Well, a subset will either be clopen or neither closed nor open.

Not sure what to make of that.

Rithaniel

Well, actually, aren't all sigma algebras also toplogies?

Because the whole space and the empty set are included and you have countable unions and countable intersections (which includes finite intersections)

So a topology that is also a sigma algebra is just a sigma algebra

Thorgott

Topologies require arbitrary unions

Rithaniel

So more than countable, right

Thorgott

Yup

So, actually, I think a lot of $\sigma$-algebras fail this

Say, the Borel-algebra

Rithaniel

02:00

Well, perhaps we can say something about the cardinality of a sigma-topology, then

Well, maybe not, because arbitrary implies countable

Thorgott

@Rithaniel it seems this is what we're looking at: en.wikipedia.org/wiki/Alexandrov_topology

Rithaniel

(Terminology is one of the biggest hurdles in topology. A lot of stuff has been studied, but hardly any of it has an easy or predictable name)

Thorgott

I found it by googling "topology that is closed under complements"

Rithaniel

I was attempting "Sigma algebra with arbitrary unions"

Well, an Alexandrov metric space is discrete, I think. Check my logic, if you will:
every x in a metric space has arbitrarily small neighborhoods and every x in an Alexandrov space has a least neighborhood, and so this implies that every x in an Alexandrov metric space is a neighborhood of itself

Thorgott

02:20

Sounds right to me. Alternatively, consider $\{x\}=\cap_{n\in\mathbb{N}}B_{1/n}(x)$. So being closed under countable intersections actually suffices here.

Rithaniel

Ah yeah, fair enough

I think the least neighborhood characterization of an Alexandrov space is the most easily understood

Thorgott

It's a pretty interesting characterization

Semiclassical

02:37

here's a much duller question: How many sigma-algebras are possible for an n-element set?

found it: math.stackexchange.com/q/143796/137524

Shine On You Crazy Diamond

05:18

@Ultradark

I have to say guys I'm onna roll

0

Q: Equivalent statement to twin prime conjecture using elementary language.

Let $X_k^2 = \{ n \in \Bbb{N} : n^{2} - 1 = q_1\cdots q_k$ for some primes $q_i\}$. Supposing the twin prime conjecture false is equivalent to supposing that $X_2^2$ is finite. This is because $n^2 - 1 = (n-1)(n+1) = pq \iff n$ is the unique number in between $p,p+2 = q$. If $X_2^2$ is finite ...

Second day, another proof of Twin primes. Lol, I shit them out

Shootforthemoon

09:20

@Thorgott Ah, ok

However, if I'm not wrong, it shows how one cannot omit to check the assumptions before applying the theorem, cuz they may fail. Instead, I'm looking for an example of let's say a two variable function that can be rewritten as an explicit function of one variable but that does not satisfy the assumptions of the theorem

In fact, if the theorem only provides sufficient conditions, we should say that there exists some implicit function on which, however, the theorem may fail to be applied

Is this possible?

Adam

11:04

The twin prime conjecture question was deleted?

I don't know if trump being impeached is related to it I'm a believer in the idea that politics as we are informed is a puppet show performed according the wishes of a handful of billionaires, and people in public office actually hold little to no power over anything that happens

Thorgott

13:22

@Shootforthemoon I think you might want to take a look at chapter 5.4

Rithaniel

13:57

It's always difficult to search for things that don't fit a particular description

Like, right now, I want to know some ways to construct a ring out of two other rings without it just being the direct product of the two

Like, one I know is that if you have the integers and any other ring $R$ you can construct $\mathbb{Z}+R$ by letting $(n_1+r_1)+(n_2+r_2)=(n_1+n_2)+(r_1+r_2)$ and $(n_1+r_1)(n_2+r_2)=n_1n_2+(n_2r_1+n_1r_2+r_1r_2)$

But when I google "construct ring from two rings not direct product," I exclusively get descriptions of the direct product of two rings.

user131753

14:27

Let $M$ be a group. Let $N$ be a subgroup of $M$ such that if $a\in N$ then for all $g\in M$, whenever $g+a\in N$, $g\in N$. Does this type subgroup has a special name in literature?

Thorgott

That's true for every subgroup. If $a\in N$ and $g+a\in N$, then $g=(g+a)-a\in N$.

Ultradark

Is $\ln(x)\ln(y)=1$ a hyperbola?

I know that $xy=1$ is

Thorgott

Try graphing it

Ultradark

oh I think it's a hyperbola only in log-log space

$u=\ln(x)$ $v=\ln(y)$

user131753

15:08

@Thorgott Instead of $M$ being a group consider $M$ to be a monoid and $N$ to be its submonoid satisfying the property.

Thorgott

15:26

Same argument applies.

Adam

15:52

looking for a hint for proving:

$$\gcd \left( \left( n-2 \right) !-1,n \right) =
\cases{n&$n \in \mathbb{P}$\cr 1& $n \not\in \mathbb{P}$\cr}
$$

Shootforthemoon

16:08

@Thorgott Thank you!

The example is the inverse of $y=x^3$, right?

Because it fails in the requirement that $F_x$ should be different from zero at the origin

user131753

16:23

@Thorgott I don't understand. If $a\in N$ then $-a$ mayn't exist always.

Thorgott

@Shootforthemoon the inverse of $x^3$ is not differentiable at $0$

@user170039 sorry, I was thinking about modules

Shootforthemoon

@Thorgott mh, so even if the theorem cannot be applied, its consequence is not realized in the example

what about $(x-y)^2=0$ near (0,0) ?

as one user suggested

Thorgott

That should work

Shootforthemoon

=D

Ultradark

16:52

The map $(x,y)\mapsto(1/y,1/x)$ can be linearized by letting $u=\ln(x)$ and $v=\ln(y).$

Is it correct to conclude that $uv=1,$ in $\log-\log,$ space is a hyperbola, but in regular $x,y$ space it is not a hyperbola?

Shootforthemoon

@Thorgott i.sstatic.net/B4NZI.png would it be wrong to apply the formula for the derivative of the implicit function in this case, before verifying the assumptions? Cuz it would simplify. I know we would obtain a 0/0, which is undetermined, and maybe I'm silly making collapse centuries of maths, but in fact I see no reason why we couldn't apply such a procedure, especially if we think that derivatives are just limits for x,y approaching the point

We would also obtain - (-3/3), which is 1 and is true for both the cases y=f(x) or x=g(y)

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