it's a bit thundery and lightningy and rainy this evening.

I neglected in my math education to ever take an introductory course in graph theory, so now I skimmed through the beginning of a few books in an attempt to teach myself, and I have come to find that there are A LOT of variations in terminology and definitions among texts.

@leslietownes It is supposed to be that here soon.

I never took or taught graph theory. 🤷‍♂️

The most frustrating thing honestly is that there doesn't seem to be a middle ground between (1) a text that starts out informally and does not even assume familiarity with proofs and then develops the theory, and (2) a graduate text that assumes the reader has already been exposed to a survey in graph theory

That’s because lots of CS people take graph theory and are not mathematically sophisticated.

when i was an undergrad i think the only guaranteed chance of getting it was a 'discrete math' class that had absolutely no prerequisites. couldn't even assume linear algebra.

I don’t recall a graph theory course at Berkeley when I was there. It was perhaps in CS?

oh, they might have had their own. they didn't have a regular one in the math department. but, math 55 ("discrete math") usually did a tiny bit of it.

about as much of it as you can do without assuming students can read proofs.

55 was required for the CS major but not for the math major, at the time. i think they require it in the math major now.

Plenty of induction proofs, which CS people can handle.

math.berkeley.edu/courses/choosing/lowerdivcourses/math55 one week of it, in this outline.

that's.. not a lot.

Nope.

I'm in a linear algebra course, and was given a matrix A and vector b, and asked to determine whether "b is in the span of the rows of A". What is meant by this? I'm assuming they simply mean to "rotate" A (the first row being the first column, etc), however I'm unable to get ahold of the professor for clarification.

Found it: en.wikipedia.org/wiki/Row_and_column_spaces

yes, that's what they're asking. b is in the span of the rows of A if b is in the span of the columns of A^T. so you're checking if A^T x = b has a solution.

'transpose' might be the word that you find in the index of a textbook. it's kinda like rotating. kinda not.

i am always discreet about my mathematics.

no convex psqs today. the psq police seems to be in san francisco mode at present.

@copper.hat You’re rarely discreet.

Hello

Suppose we have a surjective map $\phi : X \rightarrow Y$. Let us say we have collection of $C^{\infty}$ maps on $X$ denoted as ${ p_{\alpha} : X \rightarrow (0,\infty) }$. We can use axiom of choice to define a collection of $C^{\infty}$ maps on Y right? That is a collection ${ p^{\prime}_{\alpha} : Y \rightarrow (0,\infty) }$.

@TedShifrin Just about my mathematics. I had to guess Wordle today.

anime: given your $p_{\alpha}$ and any function $g: Y \to X$, you can associate a family of maps $Y \to (0, \infty)$ given by $p_{\alpha} \circ g$. it might not be a family of smooth maps (maybe unless you use X and g to give Y a structure that makes it smooth, which might not be whatever structure it already has - if this is possible).

i'm not sure what surjectivity of phi has to do with it. if phi is surjective, you can use AC to produce $g: Y \to X$ with $\phi \circ g$ the identity on $X$, but there's no reason for such a $g$ to be smooth.

do you want something more than just a family of smooth maps on $Y$?

how should it relate to the maps on $X$?

@leslietownes I wrote it in the wrong direction

@leslietownes If I have a collection of $C^{\infty}$ families on $Y$ denoted as $p_{\alpha} : Y \rightarrow (0,\infty)$ then I want a collection of $C^{\infty}$ families on $X$.

surjection will allow that given any element $y \in Y$ we have at least one $x_y$ that maps to $y$ we can use axiom of choice to make a consistent choice of $x_y$ that maps to $y$.

then we can define $p_{\alpha}^{\prime} : X \rightarrow (0,\infty)$ I think this is also a family of smooth maps, do you agree?

@leslietownes do you agree?

once AC is involved i don't see how you could know that anything is smooth.

if $\phi: X \to Y$ is sitting around, you can certainly consider $p_{\alpha} \circ \phi$, and it will be a family of functions $X \to (0, \infty)$. it'll be smooth if $\phi$ is, and i don't think $\phi$ needs to be surjective in this case.

it seems like you don't just want some random family of smooth maps on $X$, but a family that somehow reflects or relates to some other piece of data in some way. i'm not sure i see what that is. this is not my area. at the moment i am just type-checking the input data and thinking about how you could combine it.

i have no idea what the goal is.

How do I break $\frac l{k+1}$ the right way that will serve the purpose of making RHS arbitrarily small?

here

:(

@leslietownes wait what?

1 sec reading

you keep asking if something is a smooth family of maps. suppose you had the smooth family of maps. what do you want it to do? how does it relate to phi? what is the application? that's what i meant by the goal.

@leslietownes 1 moment I will explain

I want to construct a smooth family on $Y$

given a smooth family of maps $X \rightarrow (0,\infty)$ I want to construct a smooth family of maps $Y \rightarrow (0,\infty)$

The reason I want this is because I want to construct some metric on Y

I am not really sure whether there are people around here who are interested in both philosophy and mathematics. Just in case somebody might want to have a look at some stuff posted in the Philosophy chatroom:

I've proved Collatz, would anyone like to check the proof?

There exists an elementary proof for it. I should put it on my blog, and then Vixra or something in order to timestamp it

If anyone wants to sponsor me as an accademic in order to get it published, we would split the prize $

They also can help me peer review and write the paper

It should be no longer than 20 pages

5 pages to derive the formula I use. I derived it using SymPy but of course we need to prove that it's true

Well I didn't prove the whole thing, but who knows how they're going to award the prize, it could be split

I proved that there are no non-trivial finite loops though

For the positive number case (the naturals) only.

So I don't study what causes the structure of the $-7, -5$ loop or the $-17, \dots$ loop

Wow, nice.

I want to ask a question regarding associativity of sets . it says (A n B) n C = A n (C n B) so my question is will (A n B) n C = (A n C) n B ? thank you

I mean when we interchange the last term with middle one , with Venn diagram I feel it is correct but just want to know a formal proof

You feel or you know?

@mohan10216 I know

Now I have a silly question that is

Don't Harmonic sequence violate monotone convergence theorem ?

or it can restated that "Is Harmonic Sequence Convergent but not Harmonic Series"

@RaMathuzen how so?

$H_n$ is not bounded from above so it doesn't converge to a finite constant.

Proof still holds, I wrote up a very rough draft and spotted some errors, fixed them. Everything still meshes!

Who'd have thought Collatz would have a relatively small proof after all the propositions proven about it so far. We will reference a few of them.

Well I want to note that I didn't prove boundedness. But maybe our technique can be extended. I proved no nontrivial loops for something like $N \geq 6$, but you only need it for $N \geq $ some large number (see Wiki article)

Where $N = $ length of an odd loop using accelerated Collatz function $f(x) = |3x + 1|_2(3x + 1)$ where $|

$|\cdot|_2$ is $2$-adic abs value i.e

Any accademics out there who can help us publish would get a share of the prize alotment

That would be: reviewing it themselves, helping to write, editing, trying to prove boundedness etc.

First thing I'll do is post to my blog for a timestamp and copyright

Guess what? I used SymPy for deriving part of it. lol

That part will have to be formally verified on paper but that only requires a plugging in of the loop solutions into a linear equation

It's probably my 20th approach over the past 3 weeks

I used to hate Collatz, then I heard about the prize money (recently although prize was opened last year). Someone should have told me about it earlier when I was in twin prime hell. :D

I did come up with a cool counting formula for twin primes, see my profile if interested

The proof is on MO, and got 1 upvote, then they banished me for another 6 months

For nothing!

I mean the proof of the twin prime counting formula in an interval, not Collatz conjecture, dare I say theorem

It's good that you made progress on this, but let's hold our horses. There might be more errors found, or just the things left to prove, as you said, may be difficult to prove.

Correct

I just want to move fast, in case someone tries to hack my internet connection and steal my ideas. I know they already spy on me

Blog => Vixra => NAU professors (Flagstaff) => paper copies to my town

Extraordinary results require extraordinary skepticism.

I will offer a prize splitting to anyone who gets it published with me

that's actually the hardest part sadly

@Jakobian Do you have a reference for the intermediate value theorem?

Wikipedia seems to be confusing for me :P

wikipedia I guess?

Of course they do, they're Jacobian

Why NAU?

Half the internet went missing or something, I noticed google sucks for the past year for searching relative to how it was before

No one there is particularly expert on the topic...

moving back to a town nearby there, I used to go to NAU

Oh? Which town?

A town dude, this is the internet!

You don't have to get offended---I list my location in my profile.

I ask, as it seems as though we might be neighbors.

And there aren't many towns near Flag.

Oh cool. It's Sedona, AZ

You live in AZ?

I'm in San Diego now

but moving soon

Ah... for the vortices, eh?

I am in Holbrook.

I love it there, and yes there is something in the EM fields there

@Prithubiswas This follows from any connected subset of $\mathbb{R}$ being an interval, and the fact that a continuous image of a connected set is connected

@Jakobian Well I haven't actually studied point set topology.

Xander, we are practically neighbors

lol

Indeed.

Okay, help us publish :D

No, thank you.

Not my field.

@PenAndPaperMathematics Have you verified the proof with someone?

hmm... well, I think proving it using topology is the easiest

is this generally true? if $P \in \mathbb{Z}[X]$ is monic, irreducible and all of its roots are roots of unity, then $P$ is cyclotomic?

@Prithubiswas I mean, things like connected sets etc. is very elementary topology.

@Jakobian I know it isn't too hard, but I always fear whether or not I am jumping into another rabbit hole.

@Prithubiswas only myself. Now twice. Working on thrice, but only as I draft out the paper. I will also demonstrate that my loop solution formulas indeed solve the loops for say $-7, -5$ and $-17, \dots$.

As far as the negative case goes, it seems like the positive case is easier, but maybe I'm wrong.

It's actually a proof by induction on $n$ the odd loop length.

Base case is $n = 2$ and that's provable by hand using this approach

And the only solution for $n =2$ is guess what, $(1,1)$!

the only trivial loop under the accelerated map's setting

I.e. there's no $4, 2, 1$ loop in the accelerated case, but proving the accelrated map case is sufficient. You always divided out all the $2$'s anyway, so why not rep them as powers of $2$.

I guess it also would count a 2-loop as a 4-loop $(x,y, x, y)$ but that's okay

I showed that no loops exist for $n \geq 2$ other than the $(1,1,1,\dots)$ i.e. odd loops

What's interesting is that I only work with one component of the loop, if one is integral then they all are, right?

Because the action of the Collatz map is integers to integers

It's an elementary application of Proof by induction once you set up the loop solution formula correctly using Linear Algebra. No body on the net seems to have done this direct approach, probably ignoring it because it's too obvious

Think, if you have a map $f(x)$ and it's the accelerated Collatz, how do I derive a linear system from that. That is an exercise to you.

You might come up with the same formulae I do, then we could say we each came up with it independently lol

Now we can say that the algorithm computing Collatz iterations is a Halting algorithm as long as it's bounded I think but might take more work after that but not too much of an argument. The map can't attain the same value again or otherwise it loops, but there are no loops. And since bounded there's only finitely many values to go to next

Thus we now have the study of algorithms that in fact do Halt, starting with maps of the form $f(x) = |ax+b|_2(ax + b)$ but each case $a, b$ may have loops or be unbounded. . There's a lot to discover

Replace $2$ by any prime

I don't know why they would multiply $|\cdot|_p x$ but it sure did come up with a long-standing problem

It's like the linear algebraic approach removes the recursivity inherent in the problem, and instead comes up with these huge solution formulas, of course they are symmetric up to a change in variable between all the components of a loop. We only need to show non-integrality for one component though. Then say "by symmetry, etc."

I'm wondering if you can apply this to the Twin Primes because that also has an algorithm that will "loop forever" if no more twins are found.

But the inherent formula is probably way different, so no

I have a set $$D = {\cos{\frac{2\pi}{n}},\sin{\frac{2\pi}{n}}| n\in\Bbb N}$$

can I call it open, closed or neither?

I think it will be neither just like Q is neither

I'm talking about R2 set

nvm, my question has a really easy answer

Consider $\mu , nu : \Bbb{P}^1 \to \Bbb{P}^2$ given by $\mu([x_0,x_1]) = [x_0^3,x_0 x_1^2, x_1^3]$ and $\nu ([x_0,x_1]) = [x_0^3,x_0x_1^2-x_0^3, x_1^3-x_0^2x_1]$, where $\Bbb{P}^n$ denotes the usual projective space over an algebraically closed field $K$.

Apparently, "the images of these two maps are both cubic hypersurfaces in $\Bbb{P}^2$, given by the equations $z_0z_2^2 = z_1^3$ and $z_0z_2^2 = z_1^3 + z_0z_1^2$, respectively; in Euclidean coordinates, they are just the cuspidal cubic curve $y^2 = x^3$ and the nodal curve $y^2 = x^3 + x^2$."

Exercise: Show that the images of $\mu$ and $\nu$ are in fact given by these cubic polynomials.

I'm having a hard time understanding what the exercise is asking me to show. E.g., the image of $\Bbb{P}^1$ under $\mu$ is literally by definition $$\mu (\Bbb{P}^1) = \{\mu([x_0,x_1]) : [x_0,x_1] \in \Bbb{P}^1 \} = \{[x_0^3,x_0 x_1^2, x_1^3] : [x_0,x_1] \in \Bbb{P}^1\}$$ So, it just consists of certain points in $\Bbb{P}^2$...What do these have to do with the above mentioned polynomials?

Perhaps the exercise is telling me to show that the images are related to the zero locus of those polynomials...or something like that; I'm currently reading through the textbook trying to make sense of the exercise...

@TedShifrin Hi! Do you have an idea about what the exercise is saying? I still can't quite make sense of it. I'm sure I'm just being a knucklehead!

@user193319 Yes that's what is expected as far as I can tell

@Astyx Just to be sure, so we want to show that $\mu (\Bbb{P}^1) = V(z_0z_2^2-z_1^3)$?

Yes

Okay, thanks! I'll give that a try.

@user193319 Yes, you want to show that both the nodal cubic curve and the cuspidal cubic curve are parametrized by $\Bbb P^1$ as given. What is interesting is that a smooth cubic curve cannot be so parametrized (i.e., the smooth cubic is not a rational curve). To prove this is nontrivial.

Hi, i'm struggling with $((1-3i)^8)^(1/4)$, using De Moivre and then 4th i can't even find the angle, squaring three times also gives something which doesn't help and you can't do $z^(8/2)=z^2$

the angle isn't "nice," but you can find it and give an answer in terms of it. although i do wonder if they meant 1 - sqrt(3) i.

$((1-3i)^{8})^{1/4}$ sorry for editing

wolfram says it's $2(4+3i)$

and i can't find a way to get there

$z=1-3i=\sqrt{10} e^{xi}$, where $\tan x=-3$.

i know how to do it using arc functions, but i don't think i'm allowed to use them

now try taking powers and find principal value.

@user379685 so what are you allowed to use?

De Moivre, euler formula, n-th root i guess

is this allowed: ((1-3i)^8)^(1/4)= rcos x+ri sin x?

yes

but the angle isn't solvable without inverse trig functions

using this approach

so you don't want to use inverse trig. functions? If yes, then why?

i want to obtain the anwser $2(4+3i)$

using the false relation $(z^{2})^{1/2}$ almost gives the correct result -8-6i but that doesn't get me anywhere

$(z^{2})^{1/2}=z$

reverse the roles of a and b. "between f(a) and f(b)" is symmetric so the result is the same.

@leslietownes But isn't [a,b] = {x : x ∈ ℝ ∧ a ≤ x ≤ b} ? Then How can we reverse roles?

if a > b, then let A = b and B = a and apply the result with capital letters. you learn that f takes on any given value between f(A) and f(B), i.e., any value between f(b) and f(a). the point is that "between" doesn't care which of f(b), f(a) is larger or if they are the same.

if you really wanna put this in symbols, put "between" in symbols. it's a symmetric relation. they're using "between" precisely because they don't want to get into the question of cases involving which of f(a), f(b) might be larger than the other one.

if f is a continuous function on a closed interval, then f takes on any value between the values of f at the endpoints of the interval.

this is the kind of detail that may only be bothering you because you are trying to work purely in symbolic formulas. suddenly choice of order matters, and what letters play which role matters. any closed interval can be written [a,b] with a <= b, so why would you even bother with a > b? it's certainly relevant to evaluating symbolic formulas, but doesn't affect the result.

i'll continue my crusade against logical symbols some other day :)

@leslietownes lol.

I think I have to go now, but thanks for sharing your thought on this =)

have a good one. i have many thoughts. more to come :D

@user379685 There is not a unique answer. Do not believe everything you see on WA.

But, yes, the fourth roots of a number are gotten by finding one of them and then multiplying by $1,i,-1,-i$. It's easy enough to find one. Just take $(1-3i)^2$.

then how do i go about finding $(1-3i)^2$

How does one show that $X^7+\theta\in F_2(\theta)[X]$ is irreducible, where $\theta$ is a root of $X^3+X+1\in F_2[X]$

using Kummer theory

@user379685 If you are asking that question, you are in the wrong course.

My idea is to show that F_2(\theta)(\sqrt[7]{-\theta})/ F_2(-\theta)$ is cyclic

but i'm not sure how to obtain a 7th root of unity in $F_2(\theta)$

Some mod needs to explain to this person that one doesn't delete as soon as one has been given the answer. @Xander

@TedShifrin Thanks for the heads up.

It had disappeared, obviously, so I could no longer comment to him.

i'm right here, ted. if you don't want to answer my questions about my proof of the parallel postulate from euclid's other axioms, you don't need to answer in the first place.

I think you should retract that statement.

i will never retract my proofs of the parallel postulate.

i'm like galileo, or the guy they made drink poison, if that wasn't galileo.

Defamation retraction.

haha, i like that.

that's better than witless for the prosecution.

You may borrow, but I expect 10 lesliecoin royalty each time.

Hi. From what follows that $\int_1^{\infty} e^{-(1+p)y} y^{q}dy$ is convergent iff $p>-1$?

Do you know how polynomials grow (going to infinity) compared to exponentials?

$\lim_{n\to\infty}\frac{n^b}{a^n} = 0$?

OK. So use that.

Can you please explain little bit more? I am not sure how to apply it here

Why don't you write it with the letter $y$ in there appropriately?

@unit1991 What about when $p = -1$?

Someone thinks they solved the Collatz conjecture on math.stackexchange. Amusing to say the least.

I haven't proven the Collatz conjecture

But if someone proved it I could maybe learn the proof and then I could say that I proved it

Can you have orthogonal complex planes?

They said that there are steps which they haven't gone through, so they haven't solved it

It's possible that they've solved a lot of it, but mayhaps the steps which still have to be done are potentially extremely hard

or, perhaps annoyingly, equivalent to the full conjecture

@geocalc33 Tooooo vague. Make your question precise.

Why is it that the mean things I say are the things that get starred on this chat.

What has this world come to

I was learning about symplectic reduction yesterday, and that was pretty cool and fun.

Yeah, it is a powerful technique. Not that I remember much. ... And it's only when you're mean that you have insight.

sick burn, ted

Well, you asked the question ...

Ted, does symplectic reduction appear when doing complex geometry like you were into? I could imagine it might have some use when dealing with Kahler manifolds, but I am not sure if Kahler quotients are something that people would stress over all that much.

No, not in my research, but in related things.

@anak the world has come to Canada declaring a state of emergency for at least the next 30 days...

Did they really?

yup

Oh this is the emergencies act thing?

Correct, mean while the UK has said the pandemic is now just an endemic

these aren't mathematical things, though @user4539917

Tell me of your mathematical intrigues

if the proof of the colatz conjecture is found in this chat, that would be intriguing :-)

it's not in this chat, so pick something else

🤔

@user4539917 what sort of mathematics do you enjoy? I vaguely remember the colour of your fancy profile picture, but I don't recognize the 7 digit number after "user".

If $L/\mathbb{Q}$ is a cyclic extension, then is $L(i)/\mathbb{Q}(i)$ also a cyclic extension?

Don't you have the same basis for the extension?

@anak almost mathematics

This name always makes me laugh

What name? What laugh?

Ted was named after one of the world famous chipmunks.

@TedShifrin can you elaborate?

Give a basis for $L/\Bbb Q$. Unless $i\in L$, won't it still be a basis for $L(i)/\Bbb Q(i)$?

I suppose this is sorta stooopid if $L=\Bbb Q(i)$.

Oh, I totally was not thinking of the Galois group.

that's what cyclic extension means, galois group is cyclic

So do the $\Bbb Q$-automorphisms of $L$ give $\Bbb Q(i)$-automorphisms of $L(i)$?

sounds like some sort of universal property... not sure

The only issue seems to be if $i$ is algebraically dependent on elements of $L$.

Take $L=\mathbb{Q}(\sqrt{2+\sqrt{2}})$ and $K=\mathbb{Q}$

Then there's no issue.

So in this case, $Gal(L/K)=Gal(L(i)/K(i))$..right?

But what if $L=\Bbb Q(i\sqrt3)$?

Yeah, works then too.

Sorry, I don't see how to establish an isomorphsim $Gal(L/K)\approx Gal(L(i)/K(i))$ for the example I gave

I haven't taught this stuff in about 8 years, so it's not on the top of my brain. The algebraists should have it immediately.

Because if you have a $\Bbb Q(i)$-automorphism, it must fix $i$. What I said is true: A basis for $L/K$ gives a basis for $L(i)/K(i)$.

Wait, that's just the universal property for simple field extensions, no?

I don't think about universal properties, sorry.

\o @copper.hat

If $\sigma:F\rightarrow \overline{F}$ is an isomorphism and $p\in F[x]$ monic irreducible polynomial with root $u$ and $p^{\sigma}$ (apply sigma to coefficients of $p$) has root v then there exists a unique isomorphism $\psi: F(u)\rightarrow \overline{F}(v)$ extending $\sigma$ @TedShifrin

and taking u to v

Oh, sure.

perhaps I should ask the following too.. is $\mathbb{Q}(\sqrt{2+\sqrt{2}},i)$ a simple field extension?

What does simple mean? Generated by a single element?

yes

Primitive element theorem

Just take the sum of the two elements, generically.

can be the single element be taken to be a root of unity?

Um, of course not.

not for every case, just in my example

I would bet plenty of money that the answer is of course not.

what's the intuition behind that?

I don't know too many roots of unity that have $\sqrt{2+\sqrt2}$ in them.

@mathsresearcher yeah, the 16th

Leaky should take over this entire conversation. Hands the microphone to Leaky.

nah i'm busy :)

Then you shouldn't have opened your big mouth!

but in short $\cos(\pi/8) = \frac{\sqrt{2+\sqrt2}}2$ and $\sin(\pi/8) = \frac{\sqrt{2-\sqrt2}}2$

Oh, yeah, I actually would have known that 8 years ago.

@TedShifrin u know, cunningham's law is irresistible

You always love correcting me, regardless :D

only because of the law :)

Nah, cuz it's me :)

you're wrong ;)

See?