My idea is that if we take $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ to be given by $f(x) = (f_1(x),\ldots,f_m(x))$ and form the function $g(x) = (x_1,\ldots,x_{n-m},f(x))$ then, in order to use the inverse function theorem, we need to know that $ det g'(x) \neq 0$ for some $x$. But if we look at the $g'(x)$ the get a block matrix where the top left is the identity, the top right is the zero matrix, the bottom left is whatever, and the bottom left is that section I described above ( $D_jf_i$).

For $g'(x)$ to be nonsingular, we need just the top left and bottom right to be non-singular. Clearly the top left is since it's just the identity. We just need to show that the matrix formed by the partials $D_jf_i$ where $n-m+1 \leq j \leq n$ and $1 \leq i \leq m$ is nonsingular.

I think it's tantamount to showing that if we have a function $f: \mathbb{R}^m \rightarrow \mathbb{R}^m$ which is injective and $C^1$, then $f$ must have a point at which its Jacobian is nonzero.

Have you learned about measure zero in Spivak? Specifically, that the image of the set of critical points of a $C^1$ function must be measure zero?

No, not yet unfortunately. I think that's coming in the next chapter. This is chapter 2. No talk of measure at all yet.

Now that you say that though, yeah I that would be very useful hahaha.

In that case I could see it being constant rank theorem

Do you have that?

(If not, you could probably just prove it)

I actually saw some people mentioning that elsewhere. No he hasn't mentioned it actually.

I am not sure you can use the inverse function theorem to do it because I don't buy that injectivity implies there's some point where the derivative is non-singular

Or maybe that is true

[Random question]

Well, the problem was in two parts. It's in the Inverse Function Theorem section, and the first part asked to demonstrate the same for the case $n=2$ and $m=1$. The proof is basically the same idea, except that you don't really have to worry about determinants since our matrix of partials reduces to a single entry so that we get the annoying result that determinant is nonzero because the matrix is nonzero.

Yeah actually constant rank tells you it's true

What is an example of a named function that has jump discontinuities at points of the cantor set?

Well...

Yeah I'd go for just proving it

Thanks I'll see if I can find the statement of the theorem and take a stab at it that way.

Appreciate the help!

So let's say $U\subset \mathbb{R}^n$ is open and $f:U\to\mathbb{R}^m$ is a smooth map

($C^1$ probably suffices)

Such that $Df$ has constant rank in $U$

(Let's say the rank is $k$)

okay

i'm with you

Then for every $p\in U$, you can find some neighborhood $V$ of $p$ and $W$ of $f(p)$, and diffeomorphisms $\phi:V\to\mathbb{R}^n$ and $\psi:W\to\mathbb{R}^m$ such that the following holds

$\psi\circ f\circ \phi^{-1}(x_1,\ldots,x_n) = (x_1,\ldots,x_k,0,\ldots,0)$

(Where the zeroes fill up the $m-k$ remaining coordinates, though there may be none)

This can be proven using only the inverse function theorem and choosing your maps correctly

But do you see why this implies the answer?

@Secret Take the left derivative of the Cantor function, I suppose

Hmmm well first thing is that I'm not sure what a diffeomorhpism is. He hasn't mentioned it and I did a quick wiki for it and I see they're talking about manifolds, which he really doesn't get to until much later in the book. I'm sure the definition could be adapted to take the special case where we're just taking open subsets of $\mathbb{R}^n$.

Nevermind I was thinking of monotone functions

@Cryinshame diffeomorphisms are smooth functions with smooth inverses, not a priori a manifold thing

And smooth can be relaxed to $C^1$ if you like

Just found something in MSE, still trying to comprehend it:

I think diffeomorphisms need to be bijective also

I see I see. Give me a moment to think

@Leaky "smooth inverses"

alright

But yeah so I did look through my proof of constant rank and I don't think it requires more than $C^1$ so that's probably the way to go

@Daminark hmm... the link above me claims that it is true even for non-monotone functions...

It may be but I no longer wanted to assert that with confidence

The argument I use for monotone functions is very reliant on monotone-ness

but it's "jump discontinuity" instead of "discontinuity"

and for monotone functions, discontinuity is the same as jump discontinuity

@Daminark yes, mine too

Basically, every jump contains a rational number, and because it's monotone it has to be a different one for each

Countably many rational numbers qed

But this is sneakier

yes, same proof

Maybe there's a Baire Category argument for this?

Did they rule out the cantor set (or any uncountable sets with no intervals?)

If not, maybe it'd be something about playing that game on the domain

Like, I think can't have a sequence $x_n \to x$ such that $f$ has a jump discontinuity at each $x_n$

@Daminark hmm, I can believe this, and then?

ah, so it's isolated

right, then you can inject the points to the rationals

hmm, nice proof

I don't understand how it is shown to be isolated

oh he just asserted it and i just believed it lol

this is how math works isn't it

Yeah it's sketchy but I'm sorta inclined to say it's true

The heuristic being, after a jump, you can't exactly "plot the next point and then immediately jump again", it feels like there has to be a non-zero distance

hmm... I think I have to reread all of this later after my brain clears a bit, perhaps sketching a diagram. Everytime I saw more than 3 inequalities, my brain shuts down

Maybe someone can go through and try to prove that more carefully, right now I'm a bit tired and thus not in the ideal state of mind

@Dami

@Daminark Let me ask what is probably a trivial question: how do we know g will have a constant rank (or f if that is the proper substitution)?

@Daminark Take $f:[0,1]\to[0,1]$ with $f(x) = \left\lfloor 2^{-\log_2x} \right\rfloor$ and $f(0)=0$. It has a jump discontinuity at every $2^{-n}$. So much for that idea.

I'm talking about the g from the problem

RIP idea 2018-2018

My mistake I meant Df

@Cryin it won't be globally constant but it will be in some neighborhood of a point

@Daminark There's probably a simple explanation but I don't see it. Does this have to do with continuity of partials that rank doesn't change pathologically?

Okay so you can do this at a point where the rank is maximized

Let's say $p$ is a point where the rank of $g$ is maximized

Say that rank is $k$

The derivative of $g$ has an invertible $k\times k$ submatrix

By continuity and because the invertible matrices are open in the space of matrices, that $k\times k$ submatrix will remain invertible in some open neighborhood of $p$

So the rank doesn't decrease. But this rank is maximized so the rank doesn't increase. So it's just $k$ in that neighborhood

@Secret wrong trivial solution.

Hmmm i'm not sure I follow. I'll sit on it for a day or two and see if see if it'll hatch. These things come slowly to me.

@secret I meant an interval where C1 and C2 are zero. By producing linear combinations of x and x^2 with roots we can produce two roots and then on the interval between the roots set C1 and C2 as zero. That produces a nondifferentiable point at each root. Therefore it is not a solution that is minimally incorrect.

we can produce solutions with less nondifferentiable points

therefore other solutions fit the equation better

To be fair I'm also probably a bit of a shit explainer, but yeah see if that does it for you. Otherwise wait until you get Sard :P

assuming the imaginary equation has a y' in it anyways

@Daminark ?

@Typhon oh sorry I should've made it clearer, I was responding to CryinShame

ah

ok

Hey! Does anyone know this one: math.stackexchange.com/questions/2592692/…

@Programmerrrrr agh if only i remembered

Can you help me in to get to the right direction?

All the other questions in this sample assignment are related to linear system of ordinary differential equations

@Programmerrrrr what is your textbook

It is an online assignment and he gives his own notes.

I believe lagrange solved those sorts of equations so look for a section with his name in the title

therw is no class textbook

at all

you're joking, right?

nope. this guy gave his own pdf full of notes

ODE class

actually

these are recommended texts but he never really gave any readings or anything from there :/

I'll give them a look right now lol

mathworld.wolfram.com/LegendreDifferentialEquation.html

legendre

not lagrange

oops

wow this looks like some crazy shit. we never covered this in class :/

yeah that looks like it has to do with the first coefficient

1-x^2

rather this one is starting to make more sense: imgur.com/a/z0dNt

this is from wolfram alpha step-by-step solution thingy

ooh duh

im so stupid

the derivative of e^rx gives constants right?

so when you factor out e^rx you get constant

right

as in constant coefficients

yeah that makes more sense

what kind of functions have polynomial derivatives?

hmm

let me think

of higher or lesser order depending on the number of differentiations

hint: (x^2)' = 2x and (x^2)'' = 2

ohhhhh

ln?

wait

ammm i am just dumb i guess

fuck I am confused

nah

the solutions are most likely polynomials

I am confused about the question itself

oh yeah

my question?

ahh so that was a trick question? haha

no

i was serious

I am sort of lost in the conversation right now :/

the derivative of a polynomial is a polynomial

sorry I tend to make things more confusing that usual

right

so naturally you want each term to have a coefficient of x^2, right?

i thought you were asking some specific answer like ln or e or something that always gives a poly derivative

in the diff eq

yeah of course

so hence a polynomial is the solution

on that differentiated in those manners balances the terms

i.e. x ^ (5/14)

so basically this question was about simplifying the equation to that lambda equation

after that shit is pretty simple. hmm intersting

while you are here, can I clarify a few more things?

yeag

basicalky guess x^a

and solve for a

yeah. I wouldn't have thought of that had I seen this directly in the final haha

glad I am practicing

@Programmerrrrr solutions to diff eq are unique and complete so if you find a solution of that form by guessing you have all of them. Never be afraid to guess.

also never be afrais to algebraically manipulate the equation

such as by dividing by x^2

then it looks more like something that can be differentiated to balance out

dividing by x^2 is exactly what I thought but then that would have made the eq look even scarier

well think of it this way

wait how do you mean balanced out?

make each term a single term polynomial of the same order

so first derivative ought to be 1/x times somthing

second derivative ought to be 1/(x^2)

remember the whole partial fraction decomposition idea from calculus?

NOPE PDE: that's the horrible last topic of the course that went completly over my head during lectures

and the final is on sunday -.-

oh crap

PDE

we never covered that

ever

i took it a year ago

wow, so you learned that stuff yourself?

pde?

partial differential equation?

yeah

that was an ode

ordinary differential equation

ah wait, i thought you said partial differential

but you wrote partial fraction

lol of course I know partial fraction

that shit is easy

XD

> Many students multiply binomials using the FOIL method to avoid learning how the multiplication actually works.

@Programmerrrrr partial fraction basically relies upon a simple observance that coefficients of polynomials are unique. Therefore for the left hand side of that thing in your question to be 0, each term must be a polynomial of the same order (in our case -2) and they must have coefficients adding to 0.

so in your linked question i predict a solution of the form A ln(rx)

but i may be wrong

there are two solutions mind you and i dont recall the actual canonical solution. Im just guessing logically

Here is the next question + the answer I needed some clarity on: imgur.com/a/skASB

and to get back to your reply:

the same link has the final answer to the question I asked before

plus, it has the new question and it's step by step solution. THis solution is what I am confused about

What I understand so far:

actually to be more clear: what I don't understand is the second line. How did he get that x dot dot

wait

ln cannot be a solution

its not a polynomial

cause of y

so yeah guess x^a for those equations in general

yeah. I like the wolfram's approach and it makes full sense to me now too

i dont know if it is right or wrong

> A serious challenge for any author writing a math textbook is gathering many ways to politely say "I'm not actually going to prove this."

i dont know enough pde to make a decent conclusion

eliminating y makes sense

to turn it into an ode

i mean it seems logical but i cannot confirm it one way or another

@Programmerrrrr I imagine it means partial derivative or something

ive seen the symbol but i dont kbow ita meaning

that wasnt a notation when i had it

I am doubting there is pde involved here but ah well then I guess I will get to this question when I am at that topic

well id google what those dots mean

but i think this notation was for those linear system ODE questions where there were matricies involved

ive never seen then

ah it could be jacobian

math.ubc.ca/~njb/m256n4.pdf

you can see the two dots in these notes

Is empty set bounde or unbounded? this says it is bounded and this says that it is unbounded!

ok from that it definitely appears to be some kind of vector or matrix derivative

@Silent the first is about bounds in the real numbers. The second is about bounds in metric spaces.

the real numbers are built so that all finite sets have upper and lower bounds

metric spaces must not have that built into their construction

not a contradiction

Hey guys, real quick...what does "starring" a message in chat do?

I haven't starred any, but they still appear on the right of my screen.

just two different things to find bounds in

@CookieToast it marks it as something someone liked and displays it in a list for each chat room

other than that

nothing

its just to say you want everyone to see it

or you liked it

So its a global thing then? It's not just personally starred messages

no

@Typhon, but $\Bbb R$ is a metric space too, right? So, empty set is bounded, as well as unbounded in $\Bbb R$?

I'm glad you told me that before I started starring every single message I wanted to come back to later :P

Theres a lot of math in here I would love to revisit for future fun haha

you can unstar

@Silent i dont know what is and is not a metric space

hmm

i only know the axioms of the real numbers cause i just read them earlier today while checking out my intro to real analysis textbook for the coming semester

metric spaces are likely a different thing

@Silent in the second one the whole space is empty

@Silent well and just because null set is bounded in a metric space doesn't mean it is bounded in all metric spaces

not all fields have a lowest upper bound on finite subsets

Q doesnt

but R does

depending on context, it is sometimes by design/definition

so R may be a metric space

doesnt mean other spaces have null set bounded

@AlessandroCodenotti, Wow! Thank you so much!

@Programmerrrrr i just saw the other comment on your question

read it

it is a neat trick

and far easier

Usually the definition of a metric space requires a nonempty set, so you can just say that the empty set is always bounded

oh i'll check that. It's midnight here so Imma go to sleep. early morning class tmr but thanks a lot for the help dude!!

@Programmerrrrr you're welcome

also if it isnt due immediately

ask your professor in class

you cannot be the only one confused

I don't start class for another two weeks, so it always makes me feel slightly jealous when I hear other people mention class in the morning :)

@CookieToast hes still in the fall senester

about to take finals

Ooh, ouch

Jealousy $\to$ zero

@CookieToast indeed

pity $\to \infty$

Hi @Balarka

hey

hi

join our logic camp :P

Alessandro is in

@BalarkaSen

Not my cup of tea particularly. But thanks for offering

Balarka's a chemist, remember?

Why is this a meme now

C'mon dawgs

Transfinite metachemistry is a prominent area of research in logic, just fyi

@Balarka don't question what becomes a meme

@Alessandro kek

I want to read this paper on the chemistry of the propanoic acid.

Just as it's written "He who stares too long into the memes becomes the meme"

I'm confused, in the introduction written in Italian there's this sentence "This monograph is the extended version of a conference of mine held in Milan...", who's writing here?

a chemist who was attending the conference probably

:>

ヽ(´ー｀)┌

Another Italian! @Niki

tfw ignored

i have to study more organic chemistry...

not even ironically

fugg

Will you have an exam soon?

in about 3 months

There's no hurry then

you can do measure theory instead

Tru

But I love organic chemistry

I am an organic chemist, remember?

math is just a hobby

@Alessandro check Washington

hmm, is it even mathematically possible to violate the law of large numbers?

If $\sum_{k=0}^d a_kX^k$ is a polynomial with real coefficients and no real root, then the functional equation $\sum_{k=0}^d a_k f^{(k)}=0$ has no continuous solution. ($f^{(k)}$ denotes $f\circ f\ldots\circ f$)

someone knows how to prove this ?

@Secret the law of large numbers requires the existence of a first moment (integrability if you prefer)

Rudin in principles om mathematical analysis first proves $\lim _{n \to \infty} (s_n+t_n)=s+t$ and then states $\lim_{n \to \infty} (c+s_n)=c+s$ as another theorem! I can't understand why.

As a separate theorem or a consequence of the former?

see a and b

@AlessandroCodenotti

I guess that in the proof he just says that b is a special case of a then?

in b's proof he says that 'the proof of (b) is trivial.'

It is, after proving a

oh, ok :)

I don't know, I guess the author just wanted to highlight that in (a) a sequence can be constant

alright, thanks.

@Clarinetist hi

Hi @Gabriel

@Secret Yes, the Cauchy distribution has an infinite mean

@Clarinetist are you specialized in a subfield of statistics ?

@GabrielRomon Yes, I'm into stats, particularly machine learning, applied data analysis, and probability

@GabrielRomon Glad to see another ML/stats person here. Hard to come by in this chat room

Are you in the final year of your M.S ? If so, are you contemplating a phd ?

@GabrielRomon I'm actually doing a M.S. stats online, so it's not really your typical 2-year M.S. program. Actually, it's usually a 5 year program. I'm in year 2.5. Might get out in 4, not sure.

I don't have any plans on doing a Ph.D., but I might teach myself the material

ok !

Right now, I'm working full-time while doing the M.S.

which is why it's taking so long

ah I see

Can you tell me about the recruiting process ?

How about you?

@GabrielRomon For jobs?

I've seen much about interviews for coder and quants, but not much for data scientists/statisticians/machine learning practioners

yes, for jobs

@GabrielRomon Really, a lot of it is being lucky. The main thing I'd say is that it's really, really self-driven. Unless you're in a school where a lot of recruiting takes place, outside of that, your only option is really just applying to a company and accumulating more experience

I'm in year 2 out of 3 of my master's degree. I am applying to numerous programs in the US and UK to spend my final year there

@GabrielRomon I would say the most important thing is to make sure you have good programming chops

I use R every day

do interviewers test your math/quant skills ? or do they put an emphasis on coding ?

@GabrielRomon For most of the positions I've been in, it's been mostly coding, but they occasionally test statistical skills

Nothing too technical (yet)

Basically, can you write pseudocode and can you do an analysis for us, and explain some of the pitfalls of a method?

I actually have industry experience (I had a summer internship as a data scientist in a big firm) but the recruiting process wasn't the usual one (no technical interview)

@Clarinetist I see

@GabrielRomon Yeah, I'd say my best jobs have had a fairly technical interview

Are you a grad student?

Yes

What are you studying?

My major is statistics and my minor economics

Very nice

What classes have you taken so far in stats?

I've had courses in probability theory, convex optimization, a general stats course (it matches the content of Statistical inference by Casella and Berger), econometrics and stochastic processes

@GabrielRomon Very nice. I wish that I had a convex optimization course even available to me. I'll have to teach that to myself eventually

you can get some lecture notes from a course at CMU. IIRC they even videos of the lectures. The course is based on Boyd

@GabrielRomon Yeah, I have the Boyd text. Just haven't had time to read it

Right now, I'm prioritizing ML

Taking a course on it next semester

which areas of ML have you got into so far ?

I mean linear regressions, trees, ensemble methods, neural networks etc

stat.cmu.edu/~ryantibs/convexopt

@GabrielRomon Nothing really special. I have a surface-level understanding of cross-validation, $K$-means, Ridge and LASSO. I've done a little bit with decision trees but don't really know how they work. Neural networks I haven't even touched yet. I'm trying to learn a grad-level treatment of the material myself, and it's proving to be very difficult to get through the prereqs

and of course, I know linear regression and GLMs from stats

My work is letting me learn time series on the job

as well as Bayesian stats and causality

I definitely feel you !

Getting a rigorous understanding definitely takes much time

@GabrielRomon It's a lot of material to learn, but I think in the long-run, I'll be much happier learning this material rigorously, rather than, say, Ng's ML course -_-

yes, it makes a huge difference between those who blindly follow recipes and those who really understand what's behind a particular type of model

@user193319 why do you want to do a direct proof? A proof by contraposition is really quick and easy in this case

@MatheinBoulomenos Probably, but I'd still like to know where the proof goes wrong. I can't spot the error.

if $g$ is constant, then you can't conclude $|k_m|p=1$

But $|k_m|p=1$ follows from the initial assumption that $f=gh$; also $g$ isn't constant, but $\overline{g}$ is constant.

why do you know that $g$ is not constant? you're trying to proof that $g$ is a unit in $\Bbb Z[x]$, so $g$ better be constant

there are two cases: either $g$ is not constant, you showed that this leads to a contradiction. But the other case is that $g$ is constant

there is nothing wrong with the proof. You just haven't realized that you did a proof by contradiction

Ah! Okay. There's the problem. I didn't differentiate or recognize that there were two cases.

Thanks!

np

I am in the negative votes business with my questions. Now that I have received an answer, does it mean that my question will not be deleted by the community bot?

I placed an upvote, you should be safe