Mathematics - 2021-01-23 (page 2 of 3)

kenkar

7:00 PM

am I wrong if I write $ 1\in \mathbb{N}^{*} $ ?

Sayan Chattopadhyay

It's pretty cool to earn Ted's approval

Ted Shifrin

@BigSocks: What happens if you try this with $3+4i$ instead?

What does $\Bbb N^*$ mean, @kenkar?

@Sayan: Well, Thor annoys me, though, by not responding to comments I make that he does not understand. So I have to repeat them and then deduce that he doesn't understand.

kenkar

$ \mathbb{N}^{*} = \bigcup_{i \in \mathbb{N}} \mathbb{N}^i$

Ted Shifrin

I'm not blaming him for not understanding ... just wishing he'd say so immediately.

By $\Bbb N^i$ you mean the $i$-fold cartesian product?

kenkar

Yes

Ted Shifrin

7:03 PM

OK, so isn't $1$ in $\Bbb N^1$?

Thorgott

I'll try to better myself

Ted Shifrin

Thank you, @Thor :)

kenkar

Yes it is, but .. huh

Ted Shifrin

I stand by my thin compliment (is that like a thin complement?).

Thorgott

I'll immediately be annoying again by pointing out that technically $1$ isn't in $\mathbb{N}^1$

but only a set theorist would care

Ted Shifrin

7:05 PM

BTW, kenkar, that notation is totally non-standard.

It's not a one-tuple?

kenkar

so $(1)$ would be correct ?

Ted Shifrin

Oh good grief.

Thorgott

Ok, I guess it depends on how you want to define tuples

Ted Shifrin

I'm not going to write $(a)$ whenever I have a real number $a$.

Thorgott

I take $\mathbb{N}^1$ to be the set of functions $1\rightarrow\mathbb{N}$

Ted Shifrin

7:06 PM

This is the sort of thing that makes me hate mathematics.

Thorgott

but as I said, it ultimately is hardly a matter of concern

also re:$\mathbb{Z}[i]$, there's a more general point to be made here about how these ideals work and what $2+i$ has to do with $5$, but I'll leave that for after Ted's $3+4i$ example

Ted Shifrin

I suspect your general point is related to my point?

kenkar

So, technically $\mathbb{N} \ne \mathbb{N}^1$ ?

idk how to latex not equal

Ted Shifrin

\ne

kenkar

thanks

Ted Shifrin

7:11 PM

This sort of subtlety just does not interest me. For any set $X$, of course $X$ is in bijection with $X^1$.

BigSocks

@TedShifrin you get $(3 + 4i) (3-4i) = 25$ so $25 \in (3 + 4i)$. Following Thorgott's proof, you do $f^{-1}$ which is the intersection $(3 + 4i) \cap \Bbb Z$ that should contain $(25)$. But in the last step $(25)$ is not a maximal ideal since $(25) \subset (5)$ so the rest of the argument can't follow

Ted Shifrin

But if you're being a set theorist about it, then ... shrug

@BigSocks, so can you conjecture a difference between $3+4i$ and $2+i$ as elements of $\Bbb Z[i]$?

Thorgott

@TedShifrin the category theorist now chimes in to defeat the set theorist by remarking that this bijection in fact gives a natural isomorphism

Ted Shifrin

ROFL

Nothing relating to intuition in this discussion, btw.

notes that BigSocks is still hanging menacingly close to San Diego

BigSocks

hahaha I'm just floating up here... thinking about the difference between these guys

anakhro

7:17 PM

ICK

Ted Shifrin

Question related to Thor's Riemannian question. If $\lim_{x\to 0^+} f(x)=\infty$ and $f$ is twice differentiable, must we have $\lim_{x\to 0^+} f''(x)/f(x)=\infty$?

anakhro

@TedShifrin Let $f\colon U\subseteq\mathbb C\to\mathbb R$ be a continuous function (U open). Is the "MVP" which is sufficient for being a harmonic function "For all $z_0\in U$ we have that for all $r>0$ with $B = B(z_0,r)\subseteq U$, $\frac{1}{2\pi}\int_0^{2\pi} f(z_0 + re^{i\theta}) = f(z_0)$", or does it suffice to only demand that there exists an r with $B\subseteq U$ with this property (for each $z_0$)?

Ted Shifrin

Oh, surely a single $r$ will never do it. But sufficiently small $r$ will do it.

BigSocks

I mean other than the obvious- one of their norms is a square in $\Bbb Z$ and the other is not... this doesn't speak much of a difference betwen them as elements of $\Bbb Z[i]$

Ted Shifrin

It suggests a difference, @BigSocks.

anakhro

7:22 PM

Okay, so it's small enough r. I see.

Thanks!

Ted Shifrin

@anakhro If you think intuitively, harmonicity's failing at one point would tell you by continuity that the Laplacian was positive, say, on a small neighborhood.

BigSocks

oh, also $3 + 4i = (2 + i)^2$ (missed that one for a bit)

but I don't think that's much of a conjecture. just kind of an observation

polite proofs

When doing limit proofs with two functions that we might need to relate in the proof, is it good practice to switch the variables that approach a value for two separate functions? For example, if we wish to prove that $\lim_{x \to p} f(x) = L_1$ and $\lim_{x \to k} g(x) = L_2$, should we instead prove that $\lim_{y \to k} g(y) = L_2$ as to not confuse ourselves?

anakhro

@TedShifrin That's true.

BigSocks

I guess $2 +i$ is closer to being "prime" in a way

Astyx

7:27 PM

in what way?

BigSocks

well at least it "complex conjugate" squares to a prime number in $\Bbb N$

more than you can say about $3 + 4i$ but I guess with the one observation $3 + 4i$ never had a chance of being "prime"

Astyx

What do you know about the modulus of complex numbers?

BigSocks

looked it up - that's the square root of the norm then huh

Astyx

Right, so what if that number is prime?

BigSocks

the "length" of the segment between $0$ and the number in the plane spanned by $\{ 1,i \}$

Astyx

7:31 PM

the square of that length

BigSocks

oh I must be looking at some nonstandard definition

ok square of that length is like the norm

Ted Shifrin

@polite: I suggested switching letters the other day because you were composing functions. If you're working with two unrelated functions at different points, who cares?

BigSocks

which I just remembered goes at the end of the field polynomial when you write that down

polite proofs

@TedShifrin Is that something you do?

Ted Shifrin

@BigSocks: Yes, @Astyx is asking you if you know about $|zw|$ for complex numbers $z,w$?

BigSocks

7:32 PM

so $5$ is the constant term for the field polynomial that has $2 +i$ as a root I guess

Ted Shifrin

@polite: Nah. I suggested it because you were doing the composition and you did, in fact, confuse things.

If we're doing two unrelated limits, why are you putting them together in the first place?

polite proofs

Okay, sorry

BigSocks

I mean, I guess yeah it's kind of like the norm, but I'm not 100% on a list of properties of that

Ted Shifrin

Really?

So the number theory norm is motivated by such things.

BigSocks

well I guess also 25 and 5 are the sum of 2 squares bc you end up with $1 + 4$ and $9 + 16$

idk if that helps, but you mentioned number theory and there are some numbers :)

Ted Shifrin

7:36 PM

OK, so this will ultimately be related to characterizing which primes are sums of two squares. But that's off on a tangent.

BigSocks

oh dang ok

Ted Shifrin

I'll give you a more direct hint: Is $3+4i$ divisible by $2+i$?

BigSocks

yeah $3 + 4i = (2 + i)^2$

I wrote it somewhere up there^

but I thought it was more of an observation... hmm I wonder how it is important

Ted Shifrin

It suggests that $2+i$ might be irreducible but $3+4i$ for sure is not.

Can you prove that $2+i$ must be irreducible?

Other than brute force.

polite proofs

Is there a way to see that $2 + i$ divides $3 + 4i$ without noticing the square?

Ted Shifrin

7:40 PM

Well, $|2+i|^2 = 5$ and $|3+4i|^2 = 25$ and $5|25$, so it suggests it's certainly possible. But, no, you have to check by dividing. The same numerical thing would have happened with $2-i$ and $2-i$ does not divide $3+4i$.

We had a discussion in here a few weeks ago about seeing the division algorithm in the Gaussian integers.

BigSocks

hmm I was thinking field polynomials maybe so you write $x^2 - Tr(a + bi)x + N(a + bi)$

For $2 +i$ you write $x^2 - 4x + 5$
For $3 + 4i$ you write $x^2 - 6x + 25$

Since the norm is the modulus and the trace is $2a$

but yeah idk, those don't factor so good

Ted Shifrin

I don't think that helps you relate two Gaussian integers.

BigSocks

nah :/ I guess I got stuck thinking about norms and stuff... let's see...

Thorgott

norms are a great approach

Balarka Sen

Norm Chomsky

BigSocks

7:44 PM

ok so you mentioned euclidean algorithm so maybe it's something like

$3 + 4i = a (2 + i) + b$ where $N(b) < N(2+i)<N(3 + 4i)$?

which is $5$

Thorgott

what are you trying to prove

BigSocks

idk, try to relate these 2 Gaussian integers in an interesting way

TedPuzzle

Thorgott

one is the square of the other, what more relation could there be

BigSocks

I guess for those two not a lot huh

Thorgott

irreducibility of $2+i$ is the more interesting Ted puzzle

BigSocks

7:46 PM

right yeah

Ted Shifrin

@BalarkaSen smack

BigSocks

If $(2 + i)$ is reducible then you can write it as a product of two other guys in $\Bbb Z[i]$

Thorgott

right

BigSocks

$(2 + i) = (a + bi) (c + di) = ac + adi + cbi - bd$

Ted Shifrin

weeps

BigSocks

7:50 PM

oh no, this product made Ted sad :?

Ted Shifrin

Ted did say without brute force.

Thorgott

Ted is a mathematical pacifist

BigSocks

people keep mentioning norms so I guess I'll take the norm of both sides for good measure

$5 = (a^2 -b^2)(c^2 - d^2)$ (pretty sure norms are multiplicative so you can do that without breaking the law)

Ted Shifrin

Generally opposed to violence (smacks don't count) ...

Why the subtraction, @BigSocks?

BigSocks

bc the lads had $i$'s

Ted Shifrin

7:52 PM

Hmm ...

BigSocks

but uh, since all of this is in $\Bbb Z$ now I guess one of those has gotta be $1$ and the other is $5$

Thorgott

that's not the norm, but go ahead and try to continue the argument. you'll note a + instead of a - there will be more desirable.

Ted Shifrin

<--- passes the baton to Thor.

Balarka Sen

As shown by Shifrin, brut al.

BigSocks

oh yeaaah, i just squared it

Balarka Sen

7:54 PM

Etale, et al.

BigSocks

should be a +, you're right

polite proofs

i.imgur.com/QMd7JvH.png How does the first line on page 145 follow?

BigSocks

So $5 = (a^2 + b^2)(c^2 + d^2)$

Thorgott

@polite sum of continuous functions is continuous

polite proofs

@Thorgott I meant inequalities, sorry.

Ted Shifrin

7:56 PM

Reread the hypotheses, polite.

Thorgott

re-read the first line of the proof

Ted Shifrin

That's always good advice.

BigSocks

And thinking about this for a while it works if $a = 1, b = 2, c = 1, d = 0$ or some permutation of these where the pairs match up

polite proofs

I know that $f(x_1) < f(x_2)$, but that shouldn't imply that $f(x_1) - k < 0$?

Ted Shifrin

Where did $k$ come from? That's part of the hypotheses.

BigSocks

7:58 PM

But I am guessing that none could be $0$ or something like that, and then that's the proof

schn

Anyone knows of a nice and calm proof of im($X^TX$)=im($X^T$)?

Ted Shifrin

Side question for @BigSocks: What are the units in $\Bbb Z[i]$?

I know lots of proofs, @schn. What constitutes "nice and calm"?

polite proofs

Never mind, got it

Ted Shifrin

Always reread ... sometimes numerous times, @polite.

polite proofs

But isn't this a fake proof in this case? $k$ does not have to be between the two values at a all. It could be way lower than either one, for instance. Or higher.

Ted Shifrin

8:00 PM

Try it out, @polite. What happens if so?

Thorgott

I tend to be precautious and reread already before my first read

BigSocks

well $1, -1$ are in there... I guess anything with norm $1$ so you toss in $i$ and $-i$

Ted Shifrin

Never hurts to get an early start, Thor.

schn

@TedShifrin Let’s say first course in linear algebra.

Ted Shifrin

Right, BigSocks.

BigSocks

8:01 PM

oh hold up, that means that one of the two things is a unit

so yeah that's why it's irreducible. because one of the factors will be a unit and the other will be I guess of the same norm as $(2+i)$

Ted Shifrin

Well, do you notice that $\text{im}(X^\top X)\subset \text{im}(X)$?

Hint @BigSocks (and then I really am done for a few hours): $5$ is prime.

user2103480

@Thorgott set theorist says this is just a class isomorphism

Ted Shifrin

You already noticed that.

polite proofs

Well, that means both inequalities are positive

If $k$ is smaller than both.

So it would be a fake proof.

Ted Shifrin

Right. So you can't get a sign change.

user2103480

8:02 PM

@AlessandroCodenotti you did a seminar on kunens inconsistency theorem right

Leaky Nun

is @TedShifrin solving 3 questions at the same time?

Ted Shifrin

Is it three? Yeah, I guess so.

Not solving. Asking questions.

Balarka Sen

I will ask a fourth if you want

polite proofs

But figure 3.8 illustrates my concern, how do we make it work?

BigSocks

@TedShifrin v impressive

Ted Shifrin

8:04 PM

I have to eat lunch and then interview a prospective MIT student in a few.

Alessandro Codenotti

@user2103480 I did

Leaky Nun

@BalarkaSen yes

schn

@TedShifrin Why is that? What does the linear transformation of $X^\top$ do that makes the image of $X^\top X$ a subset?

Ted Shifrin

@polite: You're not rereading hypotheses. One of the hypotheses is that $k$ is between the two values.

BigSocks

@TedShifrin wait this is for the interesting difference between $3 + 4i$ and $2 + i$?

Ted Shifrin

8:05 PM

It's true totally generally. $\text{im}(AB)\subset \text{im}(A)$ always.

So prove that.

Then I'm going to ask you about dimension.

(Well, I'm going to have to leave soon, so I'll give you the assignment, @schn.)

polite proofs

I don't like that word "between" there.

BigSocks

So I guess $2 + i$ is special in that it has prime norm, but $3 + 4i$ does not have prime norm

schn

@TedShifrin Do that :)

Ted Shifrin

Tough diddly whether you like it or not.

BigSocks

maybe it is closer to being prime because of this

Ted Shifrin

8:06 PM

The next step, @schn, is to look at nullity or dimension of kernel. I don't know what notations you're used to.

polite proofs

It can be interpreted as $f(x_1) < k < f(x_2),$ in which case it makes sense that $f(x_1) - k < 0$ and $f(x_2) - k > 0$

But if we don't interpret it as that, then I don't see how we can take care of the case which illustrates figure 3.8

Balarka Sen

@schn I hope you meant "I'll do that"

schn

@TedShifrin Kernel works.

@BalarkaSen Certainly.

Ted Shifrin

There are different ways to do it, but I suggest you notice that $\ker(AB)$ always contains $\ker(B)$. Then in your case I suggest you prove that $\ker(X^\top X)\subset \ker(X)$.

user2103480

@AlessandroCodenotti so what thor describes is basically an isomorphism of the universe

but the thing is that the elementhood relation is not the real elementhood

Alessandro Codenotti

8:08 PM

where?

Ted Shifrin

This shows that the kernels are actually equal, and that finishes the proof (with nullity-rank).

The hint for that last one is to take $x\in\ker(X^\top X)$ and consider $(X^\top X)x\cdot x$.

OK, I'm gone.

user2103480

@AlessandroCodenotti mapping x to the set of function from $\{ \emptyset \}$ to itself

Thanks.

Bye @Ted.

@Thorgott here

8:09 PM

cya @TedShifrin maytbe later tell me the cool thing!

user2103480

and for the actual elementhood this is not possible with the axiom of choice, right?

polite proofs

Well, could anyone else clarify?

Alessandro Codenotti

@user2103480 that's just a bijection, not an elementary embedding, right?

Leaky Nun

@BalarkaSen number theory is fun

Balarka Sen

seems believable

Thorgott

8:11 PM

set theory has been defeated

you guys should move on

user2103480

@AlessandroCodenotti it's clearly an isomorphism of structures that can interpret set theory, but no not an elementary embedding. That was basically my question, I'd think this is just because these are different sethood relations

@Thorgott to set theorists it's a subset of the universe that is basically an exact copy of the universe, so I think this is cooler than a natural iso :P

Thorgott

it's a subcategory naturally isomorphic to the ambient category

Alessandro Codenotti

I was thinking that sentences like "$\phi(y)=\exists x(x\in y\land x\text{ is not a function})$" are satisfied by many sets, but messed up by the map, but your point is also good

user2103480

well the sentence is true inside the structure :P

universe isomorphism

universe isomorphism

gtfo category theorist

I be taking ultrapowers of the whole universe while category theorist be drawing arrows between their arrows and feel like they reinvented the world

Leaky Nun

if L/K is normal and Ls <= L is the maximal separable subextension then is Ls/K normal?

BigSocks

8:20 PM

@user2103480 people.math.harvard.edu/~lurie/papers/Conceptual.pdf

uhhh

user2103480

also, category theorists feel accomplished when taking $V^1$ while set theorists take $V^{\Bbb B}$ for any boolean algebra $\Bbb B$

40

A: Why aren't functions used predominantly as a model for mathematics instead of set theory etc.?

Let me explain one sense in which using functions or sets provides exactly equivalent foundations of mathematics, in a way that is connected with some deep ideas in set theory. There is a translation back and forth between these foundational choices. For example, it is a standard exercise in set...

@BigSocks shush got luried again

why does this keep happening

thought after the HoTT course I'd stop getting luried

BigSocks

@user2103480 absolutely lost my sides

the ride doesn't end

Alessandro Codenotti

@BigSocks 147 pages? What is this, the notes Lurie scribbled on a napkin over lunch yesterday?

BigSocks

Such a great guy, trying to make it short sweet and to the point

user2103480

somebody make lurie interested in universal logic

@Thorgott shhh kid... want some logic?

reh.math.uni-duesseldorf.de/~arndt/research/…

Thorgott

8:31 PM

>a pushout of logics

user2103480

it's actually pretty smart from what I know

you can glue logics

so that you get from a common logic one that also has logical operators of both

and you can take "fibrings" to make them interact, so that out of first-order logic and propositional modal logic (say, S4 modal logic) you obtain first-order modal logic

Thorgott

modal logic went downhill after season 4 tbh

user2103480

yeah, all those new worlds

(in some forms, modal logic is basically "possible world logic")

Which is why they were/are of interest for theologists and philosphers

Balarka Sen

pushout of logics wtf

user2103480

if I had enough funds for myself I'd seriously just study universal logic

Balarka Sen

8:41 PM

this is the reason war on drugs cannot be stopped

user2103480

I can stop anytime

Balarka Sen

The dendroidal set of inferences In fΩ(L) of a Tarskian logic L is defined
to be the dendroidal set HomGenLog(Ω, U(L)). This defines a functor
In fΩ(L): LogT arsk → SetΩop
.

user2103480

and I think that's beautiful

schn

I have done ordinary least squares on the equation $y_i=a+bx_i+ce^{x_i}+\epsilon_i$, where $E(\epsilon_i)=0$. I’d like to know when the matrix $X^\top X$ is invertible. Is there a way to find that out by looking at $a+bx_i+ce^{x_i}$ and realizing it can have at most two roots, and then...?

Leaky Nun

a logic is just a topos

Balarka Sen

8:50 PM

topologic?

horrifying

we will end up completing the system

Thorgott

of german idealism?

Balarka Sen

yes

user2103480

@LeakyNun nah that's just higher order intuitionistic logic

@BalarkaSen well isnt it good that no one completed it

user2103480

9:24 PM

Ok so haven't started learning the cohomology chapter of topo II yet

anybody care to explain to me the general story? I saw that I have to learn about rings and ideals again

are ideals just the things you can quotient rings by while preserving the ring structure

Thorgott

Here's the algebra recap: Take the chain complex, dualize it, take homology of the dualized complex, stuff happens.

There's geometry beneath the algebra, but I'm not apt enough to explain it.

Astyx

@Balarka any hint on how to prove that identity for $\sum f(x,z)$ ?

Thorgott

@user2103480 yes

equivalently, they're precisely the kernels of ring homomorphisms out of the ring

Astyx

I feel like it's just combinatorics 101 but my brain sucks

Thorgott

and they're precisely the submodules of the ring when considered as module over itself

user2103480

9:35 PM

@Thorgott yeah that's sensible

is the kernel of a ring homomorphism an ideal

Thorgott

The two elementary reasons why cohomology is useful are 1. the cohomology groups naturally assemble into a cohomology ring and this ring structure carries more information than just the groups do, 2. Poincaré duality

user2103480

hm yeah those I know

okay it is

and so I need ideals to define cohomology rings or what

Thorgott

no

there isn't really any ring theory involved

polite proofs

Could someone review my proof please?

user2103480

okay

polite proofs

9:45 PM

Given a real-valued function $f$ which is continuous on $[0,1]$. Assume that $0 \le f(x) \le 1$ for each $x \in [0,1]$. Prove that there is at least one point $c \in [0,1]$ for which $f(c) = c$.

Define $g(x) = f(x) - x$. Then $g(0) = f(0)$ and $g(1) = f(1) - 1$. If $g(0) = 0,$ then $f(0) = 0$. If $g(1) = 0$, then $f(1) = 1$ and in either case this is satisfied. We can suppose then that $0 < c < 1$, and so $g(c) = f(c) - c$. Since $0 < f(c) < 1$, we have that $-c < g(c) < 1 - c$. It's clear that $-c$ and $1-c$ have opposiet signs, so by Bolzano's theorem, there exists a $k \in (c,1-c)$ such…

Astyx

What is it you call Bolzano's theorem?

polite proofs

Let $f$ be continuous at each point of a closed interval $[a,b]$ and assume that $f(a)$ and $f(b)$ have opposite signs. Then there is at least one $c$ in the open interval $(a,b)$ such that $f(c) = 0$.

Astyx

I don't know, you never define c in your proof

Thorgott

yeah, this is dubious

Astyx

weak IVT

user2103480

9:54 PM

ah, indeed it is

Thorgott

yeah, it's sometimes called Bolzano's theorem

not to be confused with the Bolzano-Weierstrass theorem

polite proofs

@Astyx I did define it, "suppose then that $0< c < 1$"

Thorgott

you're not applying Bolzano's theorem correctly

polite proofs

Why?

Thorgott

also $0<f(c)<1$ need not be a strict inequality on either side, though that's not a substantial issue

polite proofs

9:57 PM

Doesn't it?

We took care of the $c = 0$ and $c = 1$ cases

Thorgott

but $f(c)$ could equal $0$ or $1$ for some $c$ that isn't $0$ or $1$

@politeproofs You tell me why you think you're applying it correctly

Astyx

@politeproofs "We can suppose then that $0<c<1$" implies we already know what $c$ is, to my understanding

polite proofs

@Astyx We do, it's a real number between 0 and 1 :D

Astyx

"Let $0<c<1$"

Or even better "Let $c$ be a real number between 0 and 1"

polite proofs

Sure, but I don't think this is the main meat of the proof.

Astyx

10:00 PM

No, but the way you say it is very confusing

It seems like you're talking about the $c$ in the statement, and that you've deduced it's (strictly) between 0 and 1

Even though you still don't know it exists

polite proofs

I proved that it exists by taking care of the endpoint cases

And then the case that isn't the endpoint case

Astyx

No, you've only proved that either one of the endpoint works, either both of those are nonzero

polite proofs

Huh? It could be $0$

Thorgott

I've already said what the actual issue is

polite proofs

I want to give up on math.

user2103480

10:04 PM

mood

polite proofs

I have no idea why I am wrong in this case.

Astyx

I don't think you're wrong, I think you word it poorly

And wording is essential

polite proofs

Still don't understand.

Thorgott

again, specify how you are applying Bolzano's theorem

polite proofs

@Thorgott I don't know why you said that I am applying it incorrectly.

Which part was incorrect?

Astyx

10:09 PM

What is the $a$, what is the $b$, why are $f(a)$ and $f(b)$ of opposite sign?

Thorgott

If you think it is correct, you ought to be able to explain it to me.

polite proofs

I guess you are correct, it is unfinished

This kind of stuff is very demotivating

How do I finish it the proof?

Thorgott

Wanting to apply Bolzano's theorem is the right idea. Which function do you want to apply it to? What conditions do you need to apply it? How do you ensure these are satisfied?

polite proofs

10:24 PM

To $g(x) = f(x) - x$, and that $g$ is continuous on the interval $[-c,1-c]$ (true) and that $g$ has opposite signs at $g(-c)$ and $g(1-c)$

Thorgott

Is $c$ still supposed to be arbitrary? Surely not every $c$ will work.

polite proofs

No, $c \in [0,1]$

Thorgott

So it's arbitrary?

polite proofs

I don't know what you mean

Thorgott

I don't know what $c$ is and my point is that not every $c$ will work.

It's not hard to guess two points at which $g$ has opposite signs, for the record, think back to the start of your proof

polite proofs

10:41 PM

Oh.. $g(0)$ and $g(1)$ need to have opposite signs

Thorgott

yup!

polite proofs

But since $g(0) = f(0) - 0 = f(0), $ and $g(1) = f(1) - 1$, and $0 \le f(x) \le x$, $0 \le f(0) \le 1$

Ted Shifrin

@Thor: Did you and BigSocks get to the punchline, I assume?

Thorgott

well, of course they could also be $0$, but you already excluded those cases

polite proofs

Oh, but now what

They could potentially have different signs

Argh!

Thorgott

10:44 PM

why "potentially"? didn't we just say they always do (except in the 0 cases)?

@Ted we didn't, I think the conversation just kind of disintegrated

Astyx

Bigsocks went to bed I think

Ted Shifrin

Awfully early. BigSocks is somewhere in the western hemisphere :P

Anyhow, we were basically at the punchline with my interruptions.

Astyx

Ah, no, he just stopped talking about when you left

3 hours ago, by BigSocks

cya @TedShifrin maytbe later tell me the cool thing!

Ted Shifrin

Oh. I figured you guys would polish it off.

Astyx

I went to watch a movie with my family

Ted Shifrin

10:46 PM

A good movie?

Astyx

The best

Ted Shifrin

Zut alors. Lequel?

Astyx

A very silly french movie. La Cité de la peur

It's like a french monty python

Ted Shifrin

Ah, haven't heard of that. I'll look around.

Astyx

Don't expect anything too intellectual :p

Ted Shifrin

10:48 PM

Nor so violent as Godard?

Astyx

It's not really violent no, just very silly

Ted Shifrin

No luck, unless I go to France for Netflix.

I can get lots of stuff on DVD, but pas ça.

Astyx

Too bad

schn

@TedShifrin Sorry for the ping, but since we were talking about the proof of $\text{im}{(X^\top X)}=\text{im}{(X^\top)}$, do you know if this identity can be used to derive the normal equation in the method of least squares?

Ted Shifrin

Yes, it's relevant. You can actually see that in one of my YouTube lectures.

schn

10:53 PM

Do you have the exact link?

Ted Shifrin

Somewhere in this one. Somewhere around minute 22 is exactly the point you're asking.

schn

Will check it out. Thanks!

polite proofs

"You can see that in one of my YouTube lectures." Quite the flex

@Thorgott Yes, but potentially not. We need to prove that they do :(