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18:00
$g, g+g, g+g+g, \dots$ is the whole group then, right?
Let $M$ be an arbitrary real $2\times2$ matrix.

Show that if $x(t)$ $\land$ $y(t)$ $\in$ $C^1(\mathbb{R})$ and that if $a(t)$ $\land$ $b(t)$ $\in$ $C^0(\mathbb{R})$, then the following system of differential equations is linear:

$$\begin{bmatrix}
x\,'(t)\\
y\,'(t)\\
\end{bmatrix}=\mathbf{M}\begin{bmatrix}
x(t)\\
y(t)\\
\end{bmatrix}+\begin{bmatrix}
a(t)\\
b(t)\\
\end{bmatrix}$$

Hint: Show that the vector space transformation $f: C^1(\mathbb{R})\times C^1(\mathbb{R}) \rightarrow C^0(\mathbb{R})\times C^0(\mathbb{R})$ is given by $f(x) = x' - Ax$
@Astyx there is an operation that replaces the prefrontal cortex with MSE.
Haha :D
Hey @robjohn
@JessyCat Hey there. Did you overeat yesterday?
18:03
A little. But I went to work out twice this week. Today is a cleanse day (kind of). LOL
@Astyx so in $\mathbb{Z}/n\mathbb{Z}$, if $g$ is a generator, then $\mathbb{Z}/n\mathbb{Z} = \{g^n$ : $n \in \mathbb{Z}\}$
@JessyCat Yeah, I went on a hike earlier this week, and I will go on another today.
ok then
we know that $\phi(g^2) = \phi(g)^2$
and so on
Today, I'm trying to find a generalized Fourier series for a Sturm-Liouville problem I found the eigenfunctions for a few days ago.
Posted a question on MSE, but no bites yet.
@Alessandro Still studying for exams.
18:06
@Sophie if $P = \prod_{k=1}^n(X-\lambda_k)$, then the kth coefficient of $P$ is $$(-1)^{n-k}\sum_{I\subset [1,n]\\|I|=n-k}\prod_{i\in I}\lambda_i$$ (or something similar)
yes but I want that out expanded and it would take forever
@meow-mix that's right
ugh, can you give me a hint? i dont know how to prove that all the $\phi(g)^n$ generates $\mathbb{Z}/n\mathbb{Z}$
and $1+w+w^2+w^3+w^4=0$ so a lot of the terms are going to cancel out
oh wait; i realized
generators must map to generators
18:07
Good point
in order for it to be a homomorphism
Anybody on here know anything about Fourier Series or Sturm Liouville theory?
@Steve I'm not sure how to go with this sorry
the generator $g$ must map to another generator
@meow-mix Why? Why couldn't it map to 0?
18:09
@Balarka that's annoying, when are those exams? I have a few doubts about topological embeddings so I'll probably bug you when I put them together into a coherent (and hopefully interesting) question
if it mapped to $0$\
then $\phi(g^2)$ would map to $0$
@Alessandro Try me.
and thus
it would be the trivial homomorphism
not an automorphism
Alright, np. The problem is kind of useless IMO
Dunno why it was necessary
@Alessandro In about one and a half weeks. You should ask Mike.
18:12
I have to run away soon, I'll ask next time after pondering a bit longer
right, @Mike
Right
(even though I'm not Mike)
@Sophie If I were you I would try to find symmetries by plugin $x\omega$ instead of $x$ and so on
Not sure this is the way to go though
Is a real function that is continuous on $\Bbb Q$ necessarily continuous on $\Bbb R$ ?
I know the opposite is not true
$f(x)=0$ for all x except $f(\sqrt{2})=1$
Meh I'm an idiot
Thanks
I don't know anything about analysis so that might be wrong
18:19
Looks right to me
Oh yeah the question I wanted to ask was about a function that is continuous on $\Bbb Q$ but discontuous at every $x\in \Bbb R-\Bbb Q$
f(x) is 0 if x is rational and g(x) if it irrational where g is a discontinuous everywhere function
But then $f$ is not continuous on $\Bbb Q$
@Astyx How about this one then:
In a big auditorium, a massive metal ball is placed in the middle of the room hanging from a thin wool thread from the ceiling. The temperature in the surrounding room is assumed to be the same throughout the room and is a function $R(t)$. At the time $t=0$, the room temperature is $R(0)=R_0$.

At the time $t = 0$, the temperature of a metal ball is $M(0)=T_0$

The two differential equations are:

$1) d/dt(M(t))=-k(M(t)-R(t)), t\geq 0, M(0)=T_0$ (k is the heat capacity)
Unless I am mistaken
Solve $R$, then $M$ ?
@Astyx this is an attempt to use Lagrange's method on the quintic, just to see where it fails
18:36
@Astyx am i right in saying that if a generator maps to a generator, then its an automorphism?
Yes
@Sophie What do you mean ?
do you know Lagrange's method to solving the cubic and quartic?
oh, and the only generators of $\mathbb{Z}/n\mathbb{Z}$ are numbers relatively prime to $n$
Don't think I do
Not with this name
@meow Yup
ohhhh ok this makes sense now :)
even though i had to use generators
18:38
@Astyx you just take the roots, call them $x_1,x_2,x_3$, take the discrete fourier transform and then the answer drops out
@Meow now you should prove that a group is cyclic (with your definition) iff it is generated by a single element (which is the most common definition)
@Alessandro that should be easy lol
@Sophie Maybe then :)
Hi @semiclassical
How are you ?
I never said it will be hard ;)
18:49
eh, half-awake at the moment. post-lunch.
Whenever I browse through MSE questions, almost invariably I'll see exclamation points and get annoyed.
I love it when people give you answers that are really hints, and then don't bother answering follow-up questions when they write down something super confusing with absolutely zero explanation whatsoever!!
be patient, @Jessy
Not really the answerer's fault. You can't blame him
I know.
19:05
@Astyx so the free group $F_{\{a\}}$ consists of "reduced words"?
Yeah
ok, so a free group is a group isomorphic to $F_A$ for some $A$?
Yup
$F_{\{a\}} = \{a^k, k\in \Bbb Z\}$ if I'm not mistaken
It's basically $\Bbb Z$ if your alphabet only has one letter
that would be isomorphism, not equivalence, right?
Yup
19:08
ok
Wait, what is the difference between isomorphic and equivalent ?
now i have to prove that $F(A)$ is an initial object of $\mathcal{F}^A$
equivalent consists of the same objects
Is a constant function considered even or odd?
even
@JessyCat even
19:09
And odd if it's 0
Thanks @meow-mix
and @Astyx
isomorphic means there exists an isomorphism between them
@meow-mix I don't quite have the terminology. What is an object ?
A subgroup ?
sorry, i was thinking of this from a category theory perspective.
the elements of the groups are equivalent
I still don't get it
19:13
Two groups are isomorphic if there's a bijection between their elements that preserves the group law.
Yeah I know what isomorphic means
It's the other term I don't quite get
wait, hmm
i dont know how to explain it
@Semiclassical how would you explain equivalent groups?
Versus isomorphic?
I dunno. I'd need an example.
like, $\mathbb{Z} = \mathbb{Z}$
they're the exact same group
same elements, same operation
19:15
Eh, it's not a very interesting example.
Problem is, one usually needs to say "equivalent with respect to some condition"
anyways
the point i was trying to make is that
$F(\{a\}) \cong \mathbb{Z}$, but they are not "the same" because $F(\{a\})$ consists of words, and $\mathbb{Z}$ consists of integers
Is $\Bbb Z$ equivalent to $2\Bbb Z$ ?
If the only sense in which something can be equivalent is to itself, that's not a very interesting statement.
That's what I was thinking
I'm having a little bit of trouble dealing with basis transformation. Anyone know a good video or textbook that covers this well?
19:19
Have you looked through youtube ?
I don't know much more than that otherwise :/
Yeah, but it's a little more advanced than I'd like
I wonder how many mathematicians use SciRate. I forgot who linked it at academia, but it seems that at least in my area nobody I recognize seems to use it.
what is it?
a list of trending arxiv preprints?
@s.harp Something like that. You can apparently "scite" papers you like
and you can see who has scited a given paper
19:34
tbh I wouldnt use that
Yeah, I am also unsure if I see much merit to it.
19:47
Do you guys/gals sometimes get emails from Gabor Fekete?
Conjectured ?
Thanks
@s.harp No idea who that is (and no)
@TobiasKildetoft he sends mails telling people they are gonna go to jail because they do science and do not aknowledge his amazing contributions
@s.harp Sounds like fun
19:51
I started asking people at my institut and also the neighbouring ones, turns out everybody has been getting mails from him for years now
there even is a hastag on twitter: twitter.com/hashtag/gaborfekete
Bye
@s.harp I receive the occational crackpot mail (though surprisingly few, mostly I just get bogus conference and journal invites), but never from that guy
Psh
want the best crackpot article ever?
@SteamyRoot Yeah, stuff from that domain get linked on the main site occationally. It is flag-on-sight
(not sure if they have added that domain to their spam bots yet)
Indeed... I'm surprised his account hasn't been banned yet tbh :/
20:01
ohh, he makes new ones every time
That's the one I last saw him spam with
wait, gabor = thierno>
That reads like it was written by actual mathematicians to troll people
what do you mean>
my research area has to do with super-Klein, tangential, sub-linearly reversible isomorphisms
20:06
Yeah, okay, it's from a generator
Nice ^^
authors: superman
Definition 2.2. A naturally covariant factor equipped with an ultra-extrinsic domain Λ is tangential if u is right-canonical.
if you showed this to a physicist they'd believe it was a real paper
If that generator was somehow set to reuse terms and symbols more within the same section/environment, it would probably manage to fool more people
Do you think we could set up some system
where, if someone gets a mail from a predatory publisher
the generator automatically sends them a few hundred of those randomly generated papers
There was a "book" randomly generated by an algorithm into which a lot of mathematicals texts were fed
20:17
@s.harp yeah that's reasonable
Lemma 4.4.
Let $\Phi_{\gamma} = \tau$ be arbitrary. Let $S_R(\vec \zeta)\ne 1$ be arbitrary. Further, suppose we are given a measurable, co-unconditionally compact, contra-totally commutative plane $u$. Then there exists a hyper-essentially additive ring.
Proof.
This is left as an exercise to the reader.
@SteamyRoot this person is so cranky they could generate 60 watts of power
@s.harp this looks like the start of a joke "Boole, Hamilton and superman enter a bar..."
@Sophie you shouldn't overlook Z. MARUYAMA, he contributed quite heavily to the paper
In physics there is also a fun website snarxiv.org/vs-arxiv where it presents paper titles from the arxiv hep-theory section and randomly generated ones. You have to choose which one is real
6
a piece of math named after a japanese person means brace for impact because they are almost all 20th century mathematicians
@Sophie And so?
20:55
Admittedly, it usually does mean "I'm probably not going to understand what's going on here" to me.
Could one of you tell me what $\vec p \cdot \vec {grad}$ means ?
probably need \nabla in there i.e. $\vec{p}\cdot\vec{\nabla}$
If it's the other way around, being $\vec{\nabla} \cdot \vec{p}$, then it's the divergence
Yeah
20:57
I guess since the dot product is commutative it could mean the same thing
You do see that, though. It means you've got the differential operator $p_x\partial_x+p_y\partial_y+p_z\partial_z$
The context should make clear if it's an operator, then?
is using \lim_\limits{blah\to\infty} better than \lim_{blah\to\infty}? Someone suggested it in an answer. I'm not familiar with the suggested edits thing.
It's nicer because the blah \to \infty appears under the "lim"
But, at the same time, this may push the line below further down, which can look ugly.
I tend to just use \displaystyle for that
20:59
So if I understand correctly $(\vec p \cdot \vec {grad})\vec A = p_x\partial_x A_x \vec e_x+ p_y\partial_y A_y \vec e_y+ p_z\partial_z A_z \vec e_z$ ?
Then this ?
(edited)
It'd be $p_x\partial_x \vec{A}+p_y\partial_y \vec{A}+p_z\partial_z \vec{A}$.
Oh right
Thanks a lot
the free group of $A$ reminds me of the vector space whose basis is $A$ over $\mathbb{Z}$
21:03
Huh ?
for example, the vector space over $\mathbb{Z}$ (where vectors have components in $\mathbb{Z}$) seems to be isomorphic to the free group of a set with 2 elements
@meow-mix No, it is not
(and vector space over what field)
over $\mathbb{Z}$
That's a module to begin with, not a vector space
i said
21:05
Of dimension 2 ?
but, every element of the module can be represented as a combination of $\boldsymbol{e}_1$, $\boldsymbol{e}_1^{-1}$, $\boldsymbol{e}_2$, $\boldsymbol{e}_2^{-1}$, right?
@meow-mix What module are you talking about?
Yes but $ab\cdot a^{-1}b^{-1} \ne e$
hmm, what about the free abelian group of $\boldsymbol{e}_1, \boldsymbol{e}_2$
Can free groups be abelian ?
21:08
that is (basically by definition) isomorphic to the free $\mathbb{Z}$-module of rank $2$ (since $\mathbb{Z}$-modules are the same as abelian groups
Oh it's not the same thing
But right
@Astyx "free abelian" $\neq$ "free and abelian"
Yeah, great terminology :)
@Astyx it makes sense when considered from a categorical perspective
I'm gonna trust you on this one
21:10
@TobiasKildetoft sorry i didn't know the terminology
i knew it was a module over $\mathbb{Z}$
@Astyx one is a free object in the category of groups, the other is a free object in the category of abelian groups
Yeah I see why one would want to call them similar names
But I still find it confusing
but yes, i believe what you are stating is correct; the $\mathbb{Z}$-module of rank $2$ consists of elements of the form $(a,b)$ right?
well, vectors
if you will
definitely not vectors
they're not vectors, but if you think about the module together with the field of integers, they are the "vectors" of the vector space
21:15
Is it isomorphic to $\Bbb Z \times \Bbb Z$ ?
yes
@TobiasKildetoft so no, they're not vectors.
@meow-mix the reason I said not vectors is that this is not a vector space and the integers are not a field
hmm
right, they dont have multiplicative inverses
they do form a ring, correct?
How many types of free abelian groups are there ? Is it just $\Bbb Z ^n$ for all $n\in \Bbb N$ and $\Bbb Z ^{\Bbb N}$ ?
@Astyx afaik they're all isomorphic to $\mathbb{Z}^n$
21:23
Or $\Bbb Z^{\Bbb N}$ ?
@Astyx Depends on how you define it
$\mathbb{N}$ is a set
The set of integer sequences
oh, is that so?
Free abelian groups are a direct sum of some number (possibly infinite, but could be any cardinality) of copies of the integers
21:24
i've never seen that sequence
But I guess you could also take functions
Yeah right
@Astyx You need finite sequences to get something free
yes, given a set $A = \{1,\dots,n\}$, the free abelian group $F^{\mathsf{Ab}}_A \cong \mathbb{Z}^{\oplus n}$
Oh, so sequences that eventually stabilize ty 0
@Astyx Right
since you want the coproduct rather than the product of all those copies
21:27
And as a matter of fact the set of functions $f$ from any set $S$ to $\Bbb Z$ such that $S - f^{-1}(\{0\})$ is finite ?
@Astyx Yes, precisely
Nice, thanks !
@TobiasKildetoft im not understanding very clearly what it means for an object to be a "coproduct" or "product" of two objects in a category
i know the definition, just not what it really means
@meow-mix Yeah, the concept is probably better understood via knowing a bunch of examples
given two objects, does there exist only one unique object which is a product of them?
or is it possible to be more than one object; or none at all
21:35
@meow-mix the product is unique up to isomorphism
and when one includes the data of the various morphisms in the product, that isomorphism is itself in fact unique
this is because the product is initial in a suitable category
Back with a link to a follow-up question: math.stackexchange.com/questions/2030727/…
@TobiasKildetoft so there is no difference between, say, $G \times H$ and $G \oplus H$ besides that we use $\times$ to denote a product and $\oplus$ to denote a coproduct
in the category $\mathsf{Grp}$ that is
@meow-mix well, they are isomorphic (and in fact, we generally represent them in the same way).
But the latter is not the coproduct in that category
in the category $\mathsf{Ab}$?
the latter is not really a notation we use at all for nonabelian groups
right, there they are the same for any finite number of objects
the coproduct in the category of groups is the free product
21:39
ugh, im sorry, this is just not clicking
$\times$ is a product in $\mathsf{Ab}$, right?
yes
(and we may as well call it the product)
so then whats the point of using $\oplus$ instead of $\times$ in abelian groups?
i suck at math lol
@meow-mix because they become different when the number of objects becomes infinite
in what way?
take as an example the case from before where we have an infinite (say countable) number of copies of the integers
the product is the set of all sequences of integers. The coproduct is the free object which is the set of all finite sequences.
21:46
That was embarassing...I forgot to type in my Fourier coefficients :(
22:01
@TobiasKildetoft why would the coproduct be of finite sequences?
And...he only answers 1/3 of my question. If anyone wants to take a look at what I wrote, I have the link posted on here. Please do. Karma will reward you.
@meow-mix because that satisfies the definition of the coproduct
@TobiasKildetoft So why do we want to define the coproduct in such a way
@Krijn You mean the general definition?
@TobiasKildetoft Yeah
22:03
because it is a very natural type of thing to define.
(and fits with so many nice examples)
I'm reading up on Vakil's notes on Alg. Geom. and he's so informal about somethings, which I think is nice.
what is the product in $\mathsf{Ab}$?
@tobias
@meow-mix The product is just the cartesian product with pointwise addition
ok.
so $\mathbb{R}^{\oplus\infty}$ is the set of finite sequences of real numbers?
how is that an abelian group?
by adding things at the same position
22:16
ok
I am so stressed out.
@TobiasKildetoft oh, so $F^{\mathsf{Ab}}_A \cong \mathbb{Z}^{|A|}$ precisely when $A$ is finite?
@Tobias, do you do number theory as well? Couldn't find any real interest in that on your MSE-page
22:22
could anybody here answer a question regarding convergence of a Fourier series?
I.e. Algebraic number theory, class field theory,
@Krijn Not really, no
There is not much number theory going on in this chat, ever, I feel
Lots of topology
Probably thanks to Mike
Someone posted an answer but I don't understand it. They seem to think I should know complex variables, and have refused to elaborate on their answer in a way that would be meaningful to me.
If I could get a different answer, I'd probably upvote it and love you forever.
$x^3+x^2+x+1 $ is reducible in mod 2 because when x=1 there's 4 and 4 is 0 in Mod 2. I also was able to factor and found a root $x^2(x+1)+(x+1) = (x^2+1)(x+1)$
Irreducible means no roots
22:26
@Krijn Not much representation theory either
@TobiasKildetoft Ooooh I'd like to talk about representation theory
Not much PDEs either.
Of sl_3
@Krijn Sure
22:26
What is the average number of prime factors of a number between 1 and $n$ ?
Pdes -> torture
Let me first finish some work that I should have finished yesterdays
@Krijn I will probably have gone to bed then. It is getting late. But hit me up here any time you want to talk representations
hey I was wondering is the degree of $\mathbb{Q}((2)^{1/4}*i)$ over $\mathbb{Q}$ degree 4 ?
@TobiasKildetoft Will do
22:28
@TobiasKildetoft one more question
What time is it where you are ?
@Adeek Can you write $i$ as some combination of $2^{1/4}$?
@Astyx 23.30
Here also
@Krijn no
22:29
Weird time zone, here it's 23.28
so in my book they construct a category $\mathcal{F}^A$ whose objects are ordered pairs $(j, G_1)$, where $G_1$ is a group and $j$ is a mapping $A \to G_1$
It'ss only torture when you ask questions and people give you answers that you can't use because you don't understand them, and then refuse to answer follow-up questions that would make you understand them.
@Adeek Now the question is much better
@Krijn I made a mistake in my field
I said its torture because pdes are computation heavy
22:29
@Adeek You can try to find the minimal polynomial of that element, if you know the methods for that
so we have $x^4 - 2$ is irreducible over $\mathbb{Q}$ which satisfy that as roots.
I guess I don't need this intermediate field.
and whose morphisms are commutative diagrams like so:
@Adeek Yeah you're almost done now
I am trying to compute the galois group of the splitting field $\mathbb{Q}((2)^{1/4},i)$
Difference of squares? Doesn't that yield a square root of 2 which is irrational? Hence no roots and irreducible in Q?
22:31
where $\varphi$ is a group homomorphism
Let us denote that by $\mathbb{K}$
so, they define the free group $F(A)$ as an initial object of this category
What? No denote it by $G$ like any sane person :(
I have computed that $|Gal(\mathbb{K} / \mathbb{Q})| = 8$
Ahhh the field you mean, yeah okay
22:32
that means that the only homomorphism $\varphi$ satisfying $\varphi j_1(a) = j_2(a)$ is the homomorphism mapping every element of $G_1$ to $1_{G_2}$, correct?
@Adeek Do you think this group will be Abelian? If yes, try to prove that
there is 5 groups of order 8 which i have classified before.
@Krijn the morphism are we will have an element of order 4 and an element of order 2.
@meow-mix No, why?
$G_1$ is supposed to say $F(A)$
@Adeek So if it is Abelian, you would be done
22:36
I will check just a sec @Krijn
@TobiasKildetoft because if $(j, F(A))$ is an initial object of $\mathcal{F}^A$, then there exists only one morphism that commutes
wait, no
this is so confusing
hmm
if $(j, F(A))$ is an initial object, then there only exists one unique homomorphism in the commutative diagram, correct?
from $(j, F(A))$ to another object, say $(j_1, G)$
@tobias
@meow-mix Right, or in other words, once you have decided where the elements from $A$ (or rather, their images in $F(A)$) are sent, you have uniquely determined the morphism.
@Krijn so we have the following elements of the galois group
$\tau$ where how $\tau$ acts is $i \mapsto -i$ and fixes $(2)^{1/4}$and the other elements $\theta$ fixes i and sends $(2)^{1/4} \mapsto (2)^{1/4}i$ we can easily see that those two generate the whole group.
@meow-mix Anyway, I really need to go to bed now
as $|\tau *\theta| = 8$
22:45
Bye @Tobias
cya @TobiasKildetoft
Why do calculators and programming languages calculate chain-exponentiation operations from the end? For instance, 2 ^ 5 ^ 0.2 would equate to a long decimal instead of the expected 2. Where 2 ^ 5 = 32, 32 ^ 0.2 = 2...
I thought if operators have equal precedence operations are always calculated in a sequential fashion, so why is this?

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