could somebody who is good with ODEs look at my post: math.stackexchange.com/questions/3051716/…

thanks

there's dimension shifting for Ext as well?

I think the similar technique can be used

because like things have enough projectives

Ext of projective is zero

for $M$ let $P \to M \to 0$, then let $M^\ast := \ker (P \to M)$, so $0 \to M^\ast \to P \to M \to 0$, so $\operatorname{Ext}^i(A,M) \cong \operatorname{Ext}^{i+1}(A,M^\ast)$

oh wait no Ext is not commutative

ok let $0 \to M \to I$, and let $M^\ast := \operatorname{coker}(M \to I)$, so $0 \to M \to I \to M^\ast \to 0$

so $\operatorname{Ext}^i(A,M^\ast) \cong \operatorname{Ext}^{i+1}(A,M)$

not much chatting tonight

this happens often to me lol.

Well, it is the holidays and everyone is out partying.

What do you wanna chat about?

why are you not partying? :P

Well, I have nothing to celebrate.

:(

I would think the holiday parties are over

Well, one can have parties every day.

I have noticed that this chat room tends to focus on things like modern algebra rather then things like engineering math

Well, it depends on who is in this chat room.

@Bob: I don't think most people here are engineers.

And I think a lot are doing pure math.

@dair I agree with you

Analysis is a thing here but not much engineering for sure

Or I mean idk exactly what sophisticated engineering math looks like

Mean while, I have a pure math degree, but more programming experience than math lol.

Engineering math is mostly calculus.

But what most people mean when they say engineering math, like purely computational calc and ODEs

I can make webpages and do some undergrad math lol.

@Daminark Gotta whip out that matlab

I feel most folk just aren't that interested in it by itself, they want the theoretical content

sophisticated engineering math would be things like PDE

It is hard to compute integrals, but Chris's Sis does it really well.

I feel like Leaky Nun would be good at computational stuff, since some of the code golf stuff uses array based programming like Matlab but arguable much better.

I mean differential equations of any sort can go one way or the other

I consider differential equations to be practical math

Like, I haven't done much, ODEs came up for 2 weeks in undergrad analysis and 2-3 weeks in grad functional analysis

I remember there being a lot of issue with numerical stability...

But the stuff I learned feels like the kind of thing that would get me nowhere in solving an engineering problem

@user1357113 Hey! I see you changed your username, I changed mine too. Still waiting for your book. I hope you have a happy and successful new year! =)

did you learn the right stuff?

In undergrad we did proofs of existence/uniqueness and some stuff on Lyapunov functions which was okay but not quite my thing

Define "the right stuff"

"the right stuff" is the stuff you need to do your job

@Daminark Did you do any Euler Method? or Runge-Kutta method?

Like, I don't really care at all about the GRE type ODE stuff at all, to me that would've been a waste of time. Like oh here's how to solve ODEs of the form blah

For ODE just what I mentioned. Existence/uniqueness 3 or 4 different proofs

Gerald Teschl has a nice book: ODE and DS

You can get a preprint PDF on his web.

What does DS stand for?

discrete structures?

Dynamical systems, closely related to ODE

thanks

@Daminark I remember taking Linear Algebra/Diff Eqs class and the Diff Eqs portion was so tedious lmao.

I am signing off now. good night.

See you in your dreams @Bob

@JohnNash Wut...

Cya @Bob

@Dair My standard good night greeting.

Hmm... Maybe it's just me, seems like kind of a weird good night greeting.

Sorry, I am weird.

Could be just me though, I have terrible english especially for being a native speaker.

Which is why I said my standard good night greeting, not a standard greeting. =)

I feel like whenever I decide I'm going to work on math, my programming skills decrease lmao.

Programming is easy, right?

Yeah, except I kept getting stumped on like the easiest stuff today.

I was like: "I know how to do this... It's easy." 1 hr later "I've got nothing. Rip"

All you need is know the language, know what you wanna do, then just write it in that language. As easy as talking.

Yeah, like when if I want to program Game Of Life, I just write it like I would say it:

↑1 ⍵∨.∧3 4=+/,¯1 0 1∘.⊖¯1 0 1∘.⌽⊂⍵

No wonder my english is so terrible.

You play the clarinet?

I used to.

Are you looking at the Github page?

Yes. Now I remember talking to a clarinetist long ago, don't know if it is you.

I honestly don't remember.

Well, I can't play any instruments, though I sing.

Did you do your undergrad in Berkeley @Dair?

Yeah, it's been a while. I got wrecked by college and didn't really have time. The major issue is that I'm pretty competitive about it. If I were to pick it up again I would want to spend at least 4hrs a day on it. I mostly just mess around on the piano now.

@JohnNash Yeah.

@Dair Good for you. Are you gonna be a mathematician?

@JohnNash Atm, I'm applying for grad school in CS actually.

Good for you.

You applying for a grad school? Undergrad?

No, I am not applying for anything.

Math enthusiast? Current undergrad or similar?

I finished undergrad long time ago. Now I am just trying to get well from my mental illness and see what I can do in life if and when that happens.

Good luck! I believe in you!

Thanks, but I have lost faith in myself.

I chose this username to inspire me.

@Dair oh I trailed off earlier but yeah we did some Lyapunov stuff. It was kinda cool, like if you have some vector field and you can find a Lyapunov function then you can guarantee that an equilibrium point is stable. If it's a "strict" Lyapunov function you have asymptotic stability. Then there was some stuff about linearization

It was okay but not 100% exactly my thing. Maybe presented differently I'd prefer it though

But yeah stuff like oh you take this polynomial and you solve by doing this and pressing this button and blah blah, nope

@JohnNash Ah, well hopefully if John Nash can over come schizophrenia you can overcome what you're going through.

@Daminark For me math was pretty consistently interesting, but I had no idea what direction I wanted to take it for grad school. I did have a pretty good idea for CS rip. Forever a programmer.

Should have took math more seriously in high school lol.

@Dair Well you can work on the P versus NP.

Computability lies at the intersection of mathematics, computer science, and philosophy.

@JohnNash Haha, maybe some time in the future, I'm going to invest time in some less theoretical stuff atm. I'm also pretty interested in computer verification which is another intersection of math and CS.

Are you interested in artificial intelligence and machine learning?

@JohnNash Isn't everyone? :P

Not me. I know nothing about them. =)

Well, I need to do some work on some stuff, cya!

@Dair see you! And yeah I kinda wish I got into math earlier than I did. But I guess it's going okay now. As for future directions, I feel like I like algebra, number theory, and topology most at the moment? Though I still like some parts of analysis, esp functional, and harmonic seems cool

@Holo meme exists to spread the idea and multiply itself, jokes not necessary, as its I'm is to make people laugh

Can somebody tell me how that ^

can be broken down to this:

typo: I'm should be aim, stupid autocorrect

@Secret That is why I never use autocorrect.

@Startec partial fractions

$\frac 1{y(1-y)}=\frac {(1-y)+y}{y(1-y)}=\frac {1-y}{y(1-y)}+\frac {y}{y(1-y)}=\frac 1{y}+\frac 1{y-1}$ @Startec

That is wrong

$$\frac 1{y(1-y)}=\frac {(1-y)+y}{y(1-y)}=\frac {1-y}{y(1-y)}+\frac {y}{y(1-y)}=\frac 1{y}+\frac 1{1-y}=\frac 1y-\frac 1{y-1}$$

@Startec now its corrrect

@Secret I still don't get them :/

Well for mine: Yo mama so fat is a well known meme and joke, and my version of that is that she is so fat that she cannot be a set because a fat person is generally large in size

knowyourmeme.com/memes/your-mom-jokes

Knowyourmeme is a good database for learning about the origins and other details of internet memes

@secret I have a better punchline than you proper class one, "[...]that she proves the existence of large cardinals"

Well yo mama has simmered out but the occasional one still comes up.

Like a child who was adopted by gay parents was bragging about how he was immune to yo mama jokes.

"Yo mama so ugly you got two dads"

That was the last one I ever saw

@MohammadZuhairKhan How is it even work?

@Holo now that's superfat, though what kind of class theory is needed so that all large cardinals compatible with ZFC exists?

(I think that's all cardinals all the way up to but not including Kunen inconsistency, which is HUGE)

@Alessandro Heya

Gonna try and read something about amenable groups today

@Holo its basically stating that the mother was so ugly that the father went from straight to gay.

@Secret it is impossible to prove the consistency of even strong inaccessible cardinals in ZFC... So maybe no class theory will allow it

@MohammadZuhairKhan ohhh

Hi @BalarkaSen

I think I'll read about hyperbolic groups

Sounds like a plan

There's a type of fuel called hypergolic fuel

I wonder if it's just hypergole

is it possible to find that?

← 2 messages moved from CRUDE

Why is a matrix whose minimal polynomial has distinct roots diagonalizable?

i.e. if the minimal polynomial can be factored into distinct linear factors then the matrix is diagonalizable

Try to prove that means the eigenspaces are 1-dimensional

(Minimal polynomial, not characteristic polynomial)

@BalarkaSen I don't think that's true

A diagonal matrix whose diagonal entries are all nth roots of unity has a minimal polynomial that's a factor of $X^n-I$, but the eigenspaces can be whatever size you want by repeating diagonal elements

Ah, minimal polynomial. I misread, I am sorry

Oh this is the algebraic mulitplicity vs geometric multiplicity story that I don't understand well

I'll think about this maybe. Gotta run now

I looked at a proof and it's clever

Hi can somebody help me visualise the solution to the problem - "How many 1 x 1 square cells are cut by the diagonal of a rectangle of size a x b".

Consider the polynomials of the form $m_A(X)/(X-\lambda_i)=\prod_{n\ne i}(X-\lambda_n)$, the minimal polynomial with one of its factors missing

The minimal polynomial has distinct roots iff those polynomials are relatively prime to each other

so you can do Bézout

@BalarkaSen

@Apptica I imagine this depends on if $a$ and $b$ are coprime or not

If they have a common factor then the line will go through a corner of a grid square

Try drawing it with a 2x4 rectangle

I mean I would recommend drawing rectangles of different sizes anyway to see how this works

@AkivaWeinberger I tried and it is correct, i.e. the answer is a + b - gcd(a , b), but I was just wondering how to visualise it. I was heading it in this way - Consider the large a x b grid can be broken into some sub grids , all of which are of the same dimensions...

Wait hold on

@AkivaWeinberger just use JNF

@Apptica Right sorry you're gonna have gcd(a,b)^2 many subgrids

@AkivaWeinberger Can it be proved mathematically ?

alternatively the vector space is a module over C[minimal polynomial], and this ring factors to C x C x C x ... x C by basically CRT, and so the module factors similarly into k subspaces, each of which T acts by a scalar

@Apptica Well first prove the special case where gcd(a,b)=1

I think I would probably count the amount of times the line crosses a grid line

'cause each time it does so, it enters a new square

There are two types of grid lines, horizontal and vertical

It crosses a horizontal grid line a-1 times and a vertical one b-1 times (assuming this rectangle is a tall and b wide)

So it crosses a grid line (a-1)+(b-1) times

Add to that the square the diagonal starts in, and you get (a-1)+(b-1)+1=a+b-1

@LeakyNun I think that's probably equivalent to the Bézout thing in some sense

sure it is

but it offers you a higer level view

I see

I'm not all that familiar with modules but I think I get it

I never actually wrote the last line of the argument:

some permutation of my last message should yield a coherent sentence

@AkivaWeinberger So basically we can assume that for coprime a and b the diagonal will never pass through any corner point. Logically it seems to be correct (it is indeed), is there any mathematical argument that can contribute towards it?

@Apptica How many times will it hit a corner point?

(For arbitrary a and b)

Write $m_A/(X-\lambda_i)$ as $g_i(X)$. Notice that, for every $i$, $g_i(A)v$ is an eigenvector with eigenvalues $\lambda_i$.

Then by Bézout, $\sum f_ig_i=I$. Thus, $\sum f_i(A)g_i(A)v=v$, and we can write an arbitrary vector $v$ as a sum of eigenvectors, i.e. the eigenvectors span the space.

This implies that $A$ is diagonal, QED.

@LeakyNun The end of the proof ^

and i’ll emphasize again that bezout = CRT

Right

I don't think @Akiva is disputing that.. it ain't a competition lol

Cathode Ray Tube

/s

@ÍgjøgnumMeg He was just making sure we're on the same page

of the spectral sequence

Someday I'll learn what the hell those are

@AkivaWeinberger if the size of repeating grid is p x q then it will hit, a / p times = b / q times

If you can break it into a repeating grid, then it'll hit a corner. If it hits a corner, can you always break it into a repeating grid? @Apptica

@AkivaWeinberger No, not always

It's freezing, it's raining, it's even hailing a little, and I'm the idiot standing outside wearing sandals

and the bus won't be here for another eight minutes

@Apptica Remember that the line starts at a corner. Once it hits a corner, it'll look like the start again, just shifted a bit diagonally

@AkivaWeinberger Just remember to wear more clothes next time.

Hello @MatsGranvik still working on RH? =)

@Apptica Oh actually here's an idea

Say our thing is a tall and b wide again

The slope of the line is a/b

Suppose it hits a corner, p units vertically and q units horizontally from the corner of the rectangle

Then you can also measure the slope using that and it'll be p/q

Thus a/b = p/q

@JohnNash Today yes. But most of the time no. I am a dog walker now.

If a and b are coprime, then a/b is in reduced form

@MatsGranvik Interesting. I never had a dog.

It is a Jack Russell. Very friendly dog.

Yes like that, but more hairy.

I tried to make him look hairier but it just made him look like a cactus

:)

@AkivaWeinberger that slope thing is amazing :) thanks

Thoughts: Americans on busses in America don't yell "Driver" nearly as much as Israelis yell "Nahág"

insert "nag" pun here

Hmm.. I'm writing a scholarship application for a masters, do you think it's necessary to have a "reasonable" idea of what one would be interested in for PhD study afterwards? How would one define "reasonable"?

guys

I have a question

i'm trying to prove that a set C is open iff its complement is closed

What are your definitions of open and closed?

So, I can assume that a set C is open iff every point is an interior point so that that means that for any given point in C there exists a neighborhood completely contained in C and so that means that these are not limit points of the complement and that neighborhood is disjoint from any point in the complement

but i'm not sure how to proceed

@AlessandroCodenotti My definition of open is that every point is an interior point. Closed means that every limit point of C is a point in C.

Why exactly is relativistic gravity incompatible with quantum mechanics

@mathsresearcher A point $x$ is a limit point of the set $C$ iff every neighborhood of $x$ intersects $C$

(at a point other than $x$ itself, if $x\in C$)

If no point of $C$ is a limit point of the complement, then every limit point of the complement necessarily has to lie within the complement, no?

@Thorgott I agree

Aren't you already done then?

Every point in the universe is either an interior point of C or a limit point of the complement of C

Er actually that's not true because of the "other than $x$ itself" part of the definition of a limit point

For every point, either it has a neighborhood contained in C, or all of its neighborhoods intersect the complement of C

Thus, every point is either an interior point of C, or it's a limit point of the complement of C or contained in the complement of C or both

Suppose all of C's points are of the first type. Then all of its points are interior points of C, meaning C is open.

At the same time, this means that none of C's points are limit points of the complement of C (and clearly none of them are contained in the complement of C), and therefore all of C-complement's points are in C-complement, so C-complement is closed

I can say that: Assume C is open that means every point in C is an interior point so every point in C has a neighborhood completely contained in C. That means that it is not a limit point of the complement because it has a neighborhood

that doesnt contain any point in the complement of C. So if I consider x to be a limit point in the complement, then every neighborhood contains a point y in the complement such that x is not y. If I assume of the sake of contradiction that x is not in the complement then x has to be in C, and since C is open x has a neighborhood completely contained in C which is a contradiction, because I assumed that x is a limit point of the complement. Thus the complement is closed.

Am I right?

@AkivaWeinberger

@AkivaWeinberger but i'm also assuming that the complement has a limit point, why is that allowed?

You don't need to do it by contradiction, but that argument is correct.

@Secret

You can easily make that assumption as you just need to show the statement is true for all limit points

This doesn't require that one exists

E.g. finite sets are trivially closed

Thank you very much @AkivaWeinberger and @Thorgott

np

@Thorgott SO are you saying that a universal quantification of x doesn't necessarily imply the existence of x? so if I say for all x, P(x) is true, that doesn't mean that there exists an x?

I see, it makes sense.

@Thorgott how would I formalize it in terms of quantifierss?

@Thorgott could you please explain the logic of your statement?

I think it should be $\forall x(x\in (C^C)^\prime\Rightarrow x\in C^C)$. The idea is that the statement is a so-called "Vacuous Truth", that is, an assertion about all members of an empty set, which is true, because the empty set has no members.

What is the ' over the hypothesis?

Alternatively, if $C^C$ has no limit points, the assertion $x\in(C^C)^\prime$ is a falsum and the implication is true by definition.

$E^\prime$ denotes the set of limit points of $E$.

I see, thanks so much

@MatsGranvik ???

@JohnNash Hi Jasper! Nice to hear from you again, at least once in a while! I didn't forget you although I doubt you ever appreciated the calculations I'm interested in. Hope all is fine there and the new year will be great for you!

Hi @Mathei

Hi @Alessandro

my idea was nonsense btw: ultralimits are not shift-invariant

I know there should be a construction based on ultralimits

But everyone shows a Folner sequence and says "it can also be done directly with ultralimits" with no details

@LeakyNun Could you explain further, I don't currently see how a vector space over a field F could be a ring homomorphism, from one ring to another. Because I don't see how it makes sense to talk about the elements of the vector space, when it isn't even a set in that case?

@Perturbative if you have an $F$-vector space $V$ then you have a ring homomorphism $\varphi : F \to \operatorname{End}(V)$ defined by $\lambda \mapsto (v \mapsto \lambda v)$, while on the other hand if you start with such a homomorphism then you can give $V$ the structure of an $F$-vector space by defining $\lambda v := \varphi(\lambda) (v)$ (noting that $\varphi(\lambda) \in \operatorname{End}(V)$, confusing notation)

Oh, and in the "other hand" part, you're starting with an abelian group $V$

@ÍgjøgnumMeg Thanks for explaining that, I wasn't aware one could do that

the "equivalence" is in the fact that the existence of such a homomorphism gives you a vector space, and the existence of a vector space gives rise to such a homomorphism

I'll go over what you said in more detail once I'm off mobile. So then am I correct in saying that for a field $F$ an $F$-module $V$ is a vector space, but a vector space $W$ need not be a a module over a field?

a module over a field and a vector space over a field are exactly the same thing

Modules over rings generalise vector spaces

(note fields are rings)

so if $F$ is a field then an $F$-vector space and an $F$-module mean and are exactly the same thing, just different words

Okay I understand, but what if we take a field $k$ and consider a $k[x]$ module, which we'll call $V$. Then $V$ is a $k$-vector space but $k[x]$ is not a field. So $V$ in this case is not a module over a field, since $k[x]$ is at most a PID

If you have a ring homeomorphism $\varphi\colon A\to B$ and a $B$-module $M$ then it is also an $A$ module with $a\cdot m=\varphi(a)\cdot m$, so in particular a $k[x]$-module is also a $k$-module

Right, except @Alessandro means a ring homomorphism

lel

I've been doing too much topology :P

@Perturbative and, as you will note, a $k$-module is a $k$-vector space (because $k$ is a field!)

Ohh wow, so ring homomorphisms allow us to change the 'base rings' for modules?

That's pretty nifty

@ÍgjøgnumMeg I was about to type out what you just said, before @Alsssandro replied :p

Thanks @ÍgjøgnumMeg and @AlessandroCodenotti that cleared up quite a bit of confusion I was having!

you're not really changing the base ring, you just have an injection $k \hookrightarrow k[X]$ so you have a $k$-vector space sitting inside the $k[X]$-module. This is just the restriction of the $k[X]$ action to $k$.

I think

if that makes sense to anyone

It's the same thing as saying that a $\Bbb C$ vector space is also an $\Bbb R$ vector space

ye

@Perturbative also $k[X]$ is always a PID for $k$ a field (just spotted that you wrote "at most a PID" above)

@ÍgjøgnumMeg Oh when I said that $k[x]$ is at most a PID I meant that it would be a PID but not fully a field, sorry for the confusion

I see

Is there a way to measure how "close" an $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ map is to being affine?

i.e. its deviation, or distance to the "nearest" affine transformation?

Is there good reason to think that the zeros of the zeta are countable or uncountable?

err.. inclusion $\iota$ is continuous in the subspace topology because $\iota^{-1}(U)$ is just $U \cap \text{subspace}$... right?

yes

@Mason zeroes of a holomorphic function form a discrete set by the identity theorem

@MikeMiller. Ty!

I guess it remains why one expects infinitely many points at all, but I guess that if there were only finitely many zeroes the prime number theorem would be much stronger

www-users.math.umn.edu/~garrett/m/complex/notes_2014-15/…

Amoeba solves traveling salesmen problem interestingengineering.com/…