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8:03 PM
Need more answers here :)
1
Q: Waring’s analogue for fractions?

mickI was thinking about Waring’s problem and Also about fractions. So Naturally we can combine those. Is Every positive fraction the Sum of at most 4 positive fractions squared ? How about cubes ? Etc What is known about representations $$ \frac{a}{b} = \sum_{i=1}^k (\frac{x_i}{y_i})^n $$ ??

 
can somebody tell me why SL_n(R) is a manifold?
its a closed subset of M_n(R)
 
8:19 PM
Hi everyone :)
 
@JoeShmo Right. So why should it not be a manifold?
 
0
Q: A rational number is approximated to order 1 and to no higher order.

FaustI am working through my books notes trying to learn on my own and came across the following theorem where the proof is left ot the reader. Theorem: A rational number is approximated to order 1 and to no higher order. Definition: A number $\xi $ is aporximable by rationals to order n if there ex...

 
I do think that if the physics of our world were different, our math would be different.
 
anyone know anythign abotu rational numbers im missing something basic
 
@TobiasKildetoft well if it were open, it would inherit the same homeomorphism to R^(n^2) from M_n(R). Since it's closed, though, it doesn't. so what now?
 
8:23 PM
@JoeShmo Right, because it is of dimension one smaller
 
@Faust following my nose on that, take $\xi=1/2$
 
It is the preimage of a point under a very nice map
 
so if you wanted to prove it was approximable of order 2, you'd want solutions to $\left|\frac12-\frac p q\right|<K/q^2$.
 
@Faust, you've got a comment on ur question
 
right. SL_n(R) = det^-1({1}). Hence it's closed. What are you getting at?
 
8:26 PM
which rearranges to $K>q|q/2-p|$
 
@JoeShmo There is some sort of result saying that such preimages are manifolds
 
But then it seems like you could always just do $q=2p$ and have $K>0$ which has an infinite number of solutions...
Is there a requirement that $p,q$ be relatively prime?
 
hmm
 
@TobiasKildetoft the regular value theorem
 
doesn't look like it, though
But, see this:
 
8:29 PM
i realize my mistake in understanding the definition but still have no idea how to solve the question lol
 
@JoeShmo Something like that (I don't usually do manifolds and Lie groups, preferring the AG stuff)
 
@TobiasKildetoft AG? ew
actually, have you ever read any model theory in that context?
 
no, I don't go that deep
 
well, det(M) is a polynomial in the matrix elements
 
8:30 PM
This is my first time on chat. I don't really know if I have to wait for my turn to ask a question lol help me out here :3
 
I just do algebraic group stuff
 
so $SL_n(\mathbb{R})$ is some variety :)
 
@FᴀʀʜᴀɴAɴᴀᴍ just ask
If someone wants to help, they will
 
Did someone say model theory?
 
8:33 PM
me
im digging into the theory, and i just don't get the point sometimes. Typically in a lot of papers I read a lot of arguments are just rephrasals of existing arguments. So let's get bold -- what can I prove in model theory about, say, groups, that i can't prove with regular algebra?
 
@Semiclassical that's not a manifold
 
$ {\displaystyle {\hat {f}}(\xi )=\int _{-\infty }^{\infty }f(x)\ e^{-2\pi ix\xi }\,dx,} $
 
didn't say it was. just that it's a variety and therefore algebraic geometry
 
smh wait a sec gotta figure out how to get it to render properly
 
quick question
 
8:37 PM
algebraic topology > algebraic geometry
 
any1 has read the "proofs from the book"? book?
 
@JoeShmo you have to compute the derivative of the determinant function
 
i know
 
and then use the regular preimage theorem
then do that
 
rank(DdetA) = 1
hence SL_n(R) has codim 1
hence dim SL(R) = n^2 - 1
 
8:38 PM
$ z(t) = \int_{-\infty}^{\infty}R(\omega) e^{i\omega t} \mathrm{d}\omega $
 
what's the issue then
 
how do i know SL_n(R) is a manifold to begin with
but the regular value theorem gives me that
 
what?
it's a manifold because of the regular value theorem
 
thats what i just said
 
then you have no question?
 
8:40 PM
not anymore.
its so striking to me though. How much i like the topological definitions, and how much i dislike the analytic ones
for manifolds, say
 
if you dislike analysis, we can't be friends, sorry
 
i figured
can we bond over hating algebra?
i kinda like algebra..
 
I don't hate algebra
 
good. then we can be friends
 
no
 
There are some interesting results here. But I'm only beginning to learn model theory so I don't really know much about it
 
@AlessandroCodenotti oh. and here i was hoping to be a contrarian..
@0celo7 :(
 
@JoeShmo it's your fault
 
i was born that way
 
that's no excuse
 
8:46 PM
cmon. whats so hot about analysis
topology is illumniating
analysis is applied topology
 
I liked your username :) @orbit-stabilizer
 
@Leyla, me too!
 
it's just so aesthetic
 
you're a true artist
what can i say.. i find topology aesthetic
some algebra,
and some topics in analysis as well, to be fair. but on the whole i find the language and definitions to be clunky. something that i can say topologically in two words, will take many convoluted sentences in analysis.
 
these are incorrect opinions I'm afraid
4
this estimate is correct though, so hurray
(correct as of now, it might be wrong later)
 
8:58 PM
Ur clunky @JoeShmo
Don't insult my bootiful analysis
 
come on no topologists in the crowd?
 
@0celo7 have u got any updates on Chicago stuff
 
Let $X$ be a topological space, and let $Y$ be a subset of $X$. Let $A$ be a subset of $Y$. Is the topology of $A$ induced by $X$ the same as the topology of $A$ induced by $Y$? Let's call these toplogies $T_X$ and $T_Y$ respectively. I know that if $U\in T_X$, then we can write $U=V\cap A$, where $V$ is open in $X$. But then $U=(V\cap Y)\cap A$, so $U\in T_Y$. However, not sure how to go the other way around (if it is even true at all).
 
@EricSilva no, I need to email andre again for exact dates
but I don't want to bug him
 
I can send you an email with info abt securing living arrangements and whatnot, I've asked a few ppl but no one was sure about sublet needs yet
 
9:00 PM
yes please, thanks
 
Ok will do once I get to my comp
It's really a buyer's market so you probably won't have too much trouble if everything works out
 
ah, nice
I talked to a family friend in the area and it didn't seem as dire as I was fearing price-wise
 
Yeah it ain't that bad idt
Btw @0celo7 the mcf thing is starting soon, do you know anything that could be an interesting aspect to look at
Apparently I'm gonna have to present some kind of exposition of something for André
 
@EricSilva Is it immoral for me to suggest something relevant to my research that you can then teach me
 
Lol no
 
9:10 PM
the website is being a dick, one sec
This is a notoriously hard paper though
my professor and I were planning to take a while understanding this. The line "All the results of this paper also apply when K is a
compact region of positive mean curvature inside an arbitrary smooth riemannian
manifold. Only minor and straightforward modifications are necessary in the proofs"
is very mysterious because he uses translations in the proofs
 
sent you the email
 
I saw it, thanks
@EricSilva You should start with reading Mantegazza's notes.
 
yeah im reading those alongside ecker
 
9:44 PM
1
A: A rational number is approximated to order 1 and to no higher order.

Hurkyl$K(\xi)$ depends only on $\xi$ and $n$. Your error is that the choice you want to make for $K(\xi)$ depends on $p$ as well.

anyone see a less shitty way to do this?
 
Hi
 
Pretty boring :P @Faust
 
I have two determinants which are equivalent. I know they are equivalent because I have worked them out and the expression happens to be the same.
Here’s the picture of the two determinants (one above the other).
Is there some property of the determinants which makes me derive one determinant from the other (and vice-versa)?
Again, note, I know they are equivalent (i.e. same).
 
not really
just expand along the line of ones and show lhs = rhs sadly
 
@EricSilva those are the standard books
 
9:50 PM
I know how to expand determinants, as I have reiterated...
 
@nbro sure, column operations
 
@anon Can you be more explicitly? What are column operations?
 
i hate you
thats so painfully obvious
whacks head against wall
 
@nbro when taking the determinant of a matrix, you can add/subtract multiples of one column from another without changing the value of the determinant
 
take C_2 and C_3 subtract C_1 from both
then expand along the one 1 and the two zero bottom row...
its the same matrix in the 2x2
 
9:52 PM
is this too cheeky
 
@Faust Who are you talking with? Who do you hate ? :D
 
@Faust But why can you "make" a $3\times 3$ matrix a $2 \times 2$ as it pleases us, i.e. just for convenience. This doesn't seem like a proof.
 
anon for suhc a painfully simple solution
@nbro
take C_2 and C_3 subtract C_1 from both the means subtract column one from column two and column three then the bottom row is 1,0,0 so you know the determinate is 1 * the determinate of 2x2 matrix associated to it, but thats exactly the 2x2 matrix u wrote down...
the determinate of two identical 2x2 matrices must be the same...
you dont even need to calculate it
 
@0celo7 lol
 
><
 
10:03 PM
@0celo7 that is gold
 
prove that exist 2 irational numbers such that a^b is rational
does e^ln(x) counts?
which is x
 
take sqrt(2) ^ sqrt(2)
wait,
yup. or (sqrt(2)^sqrt(2))^sqrt(2) if that doesn't work
 
ye i know that sollution
but im asking if the others
are well
 
If $b$ is transcendental and $x$ is rational, then $\log_b(x)$ cannot be rational (else $b^{\log_bx}=x$ would imply $b$ was algebraic). So $b$ and $\log_b(x)$ work.
 
@0celo7 they're nice reading so far
 
10:09 PM
Ok, thanks
It would be nice to see a formal description of this invariance under column or row operations
Do you know of a good one?
 
"in other words the determinant is the unique function from n × n matrices to scalars that is n-linear alternating in the columns, and takes the value 1 for the identity matrix "
- Wikipedia entry for determinants
 
Yes, that's an explanation, but not a proof
I was looking for something like a proof
But, it's ok, I will have a lookt at it later
I am just in a hurry now :D
lol
 
@Daminark Oh my fucking lord, look at the new HowToBasic video
 
Face reveal?
 
you unironically watch that?
 
10:18 PM
@Daminark Yeah
It's great shit
 
@BalarkaSen hi
 
I'll do so tomorrow once I'm back in Chicago
 
Hey @Leaky
 
@BalarkaSen I still haven't gone through the proof that $H^\Delta \cong H$ lol
 
guys, how does an interval in $\mathbb C$ look like?
 
10:22 PM
Uh, what ordering?
 
lol
well ehm
the "usual" ordering? XD
I have no idea what that would be tho
 
@ShaVuklia there's no usual ordering
 
dayum
 
no linear ordering in $\Bbb C$ can be compatible with addition and multiplication
in particular, there is no consistent set of "positive numbers"
 
the usual one is of course the dictionary one
 
10:24 PM
hm I see
 
but $[a,b]$ is taken to mean $\{(1-t)a+tb \mid t \in [0,1]\}$
 
what is the usual topology on $\mathbb C$?
nvm
 
@ShaVuklia generated by open balls $B(x,r) := \{z \in \Bbb C ~:~ \|z-x\| < r\}$
 
oh yea, of course!
 
equivalently, $\Bbb R^2$ (product topology)
 
10:25 PM
omg, they're equivalent?
nais
 
what topology on R, leaky
 
@0celo7 the usual topology
 
the usual topology is generated by the usual topology
 
metric / order
 
very helpful
 
10:26 PM
hahahahha
hahahah omg, fantastic
 
Hullo
 
it's Astyxx
also, I'm off, thanks Leaky!
 
I'm that ugly ?
 
Apparently
 
Feels bad man
 
10:28 PM
Who likes integrals? Quick question about the notation
 
Engineers
 
@Astyx No, like, from a math perspective
 
i dont know i like them...
i onyl know two diffrent notations and there for two diffrent types of intergrals
 
Ask your question
 
is there really a buncha diffrent types of notation?
 
10:31 PM
question first, discussion later :P
 
maybe physicists use a W or something O.o
 
@Leaky!
one last Q
is there a usual order on $\mathbb R^2$?
 
@0celo7 could you please clarify what an incorrect opinion is?
 
no
 
@Faust Yes
 
10:34 PM
@Astyx is that directed at me?
 
maybe
 
lol wut:p
you have a pretty avatar
 
not sure what u mean by usual order there are a couple common ones?
 
Right, finished typing the Latex
 
I think there isn't
 
10:35 PM
For a function $f$ whose antiderivative is $F$, I always interpreted $\int{f(x) \mathrm{d}x}$ as canceling out the denominator in the antiderivative: $=\int{\frac{\mathrm{d}F}{\mathrm{d}x} \mathrm{d}x}=\int{\mathrm{d}F}$, and then summing up the differentials to get $F$. So, the integral sign is an operator that acts on differentials.
So, the question:
 
well what I mean is one that would be compatible with addition and multiplication?
 
Suppose we have a double integral $\iint{f(x) \;\mathrm{d}x \mathrm{d}y}$. Would it be correct to interpret this as: $\iint{\frac{\partial^2 f}{\partial x \partial y} \;\mathrm{d}x \mathrm{d}y}$, cancel out the denominator, and view it as summing:
$\iint{\partial^2 f}$?
 
@SirCumference thats really not how you should think about the integral cancels out the dx. but w.e on with your story
 
@ShaVuklia what is multiplication in $\Bbb R^2$?
 
it's component wise
 
10:37 PM
@Faust Ah come on, that makes intuitive sense. Summing up a differential to get the whole function
 
I don't think there is
You can try proving it :)
 
I mean I finished Calc a while ago, but only now am I considering the meaning behind these notations
 
@SirCumference but thats not what is actually going on and you shouldnt mix partials and whole derivatives there not the same thing
 
I'm actually now sure what this compatibility means, I'll have to think about it
anyhow, I have to go:( I'll come back to it tho!
 
@Faust I know I know, so what's a better intuitive way to view the role of the integral sign, and the differential, in doing an integral?
 
10:40 PM
Well here the intuition runs a bit counter to what's actually happening. Usually you interpret stuff as differential forms or just as measures
 
I mean, thinking of the derivative as a ratio of a differential change and the resulting difference makes it much clearer what's happening. Now I'm wondering the role of the differential in an integral
 
@SirCumference thats a hard to answer question its just notation for start here finish there in some sense it and the dx or w.e tells you which line your integrating along
 
@Faust That's a bit like saying "the Leibniz notation for the derivative is just notation" without appreciating why the ratio form is so cool/intuitive
Of course, the integral is an operator, but there surely is some logic behind its notation, right?
 
But those are two diffrent thing questions one is what the notation means and one is what does the notation says we are doing those are diffrent questions and the first isnt all that interesting. if your asking what an integral does then you can think of a integral geometrically as summing a bunch of lines from a point a to a point b along the line dx
 
@Faust Yes, I know. I'm wondering why there is a lone differential in the integral notation
 
10:45 PM
it just tells you which line your integrating along
 
Why is it written as a differential
 
I mean, if it's dx, we have x_i - x_i-1. If it's dF, then it's F(x_i) - F(x_i-1)
It's a difference in the sum
 
Gah, this freaking notation
There needs to be some logic in it
 
@SirCumference aren't you taking analysis
 
@0celo7 No, I'm taking abstract algebra. I'm thinking back on old calculus
Complex analysis is next semester
 
10:47 PM
in the end theres not a deep meaning behind the notation most of the time it comes down to the first person who wrote it down and that convention was adopted if they had a reason for choosing it you could ask them but most of them are dead...
 
What's wrong with the notation? It makes perfect sense
..
 
Some people are saying it's not meaningful, others are saying it makes perfect sense...
 
@SirCumference remeber integrals existed Before christ and derivates for basically as long they were not put together until newton which was 1500+ years later
 
There's no logical reason a multivariable function should have an upright d in its differentials, for iterated integration
Screw this
 
just go back to your physics
 
10:50 PM
so trying to force theres an logic behind the notation is well unlikely they were unrealted ideas for more than 1000 years
 
@0celo7 Math is supposed to be better than physics, i.e. more reasonable and rigorous
 
integration is incredibly rigorous
 
rigorous perhaps but if anything its much more abstract
 
It is rigorous. Doesn't mean the symbols we choose to represent the concepts can't be arbitrary.
 
Of course, but the notations are confusing and ruin any intuition, unless there's some logic
 
10:51 PM
how do they ruin intuition
 
There is!! What's not logical in integrating a function from R to R using standard integration notation!
 
its just something you have to get used to math notation is terrible
 
Obviously the derivative ratio notation makes complete sense, and gives an intuitive way to view differentials, the Jacobian matrix, etc.
But integration just seems to randomly have a differential involved
What other explanation is there than it canceling out the denominator and summing up the differential F?
 
Have you read chapter 6 in Rudin?
 
it's a small element of volume if you insist on being wrong
 
10:52 PM
You're not talking about singel variable functions?
 
physicists often modify math ntoation to try and make it make more sense but it usually just makes two sets of conventions and make everything worse.
 
inb4 differential forms
 
Perhaps the biggest pain is that people emphasize the distinction between total and partial derivatives, but are fine with including upright d's in iterated integrals
Like wtf
 
@SirCumference there is a rigourous approach that explains it inutivly but you need at least one real analysis course under your belt to understand it
 
differential forms are less fundamental than Lebesguen integration, so mentioning them is only something someone who just learned about them and wants to show off would do
 
10:53 PM
@0celo7 I don't know Lebesgue integration :/
That's analysis
 
i dont even really know what lebesgu integration is
other than you dont really need it
 
Actually, that's only taught in Honors Analysis here, not even regular Analysis
 
you can fix reimann integration w.o
 
Zee
@0celo7 lol at “differential forms are less fundamental “
 
@0celo7 plot twist: I haven't learnt differential forms
 
10:54 PM
I wasn't talking about you, Leaky
 
Iterated integrals make sense using the small d notation. You fix the other variables, so it's like you have a single variable function, hence the small d notation.
 
Sigh, so what exactly is wrong with the concept of canceling out the denominator? It makes intuitive sense
 
But I have seen people say "the dx is a volume form hehehe" and it's honestly retarded.
 
Zee
Differential forms are a research topic , Lebesgue integration is not
 
@0celo7 the dx is a volume form hehehe
 
10:55 PM
chromosome alert
 
What denominator are you cancelling? For dy/dx?
 
@SirCumference you can think about it that way if it helps you as long as you understand thats not whats happening...
 
@orbit-stabilizer I mentioned above the intuition I'd thought of for what's happening in integral notation
@Faust Why is that not what's happening
Forget it, everything in this notation is fucked.
I give up
 
beause you wrote something absurd like $ \int d\frac{F(x)}{dx} dx = \int dF(x) $ which is complete nonsense...
 
Zee
@SirCumference the dx in the Riemann integral is a signed measure
 
10:57 PM
by canceling the dx's
 
@Faust ??? I clearly stated the numerator had $dF$
 
so do i
 
You said $f(x)$
 
Oh I see
 
that what u want?
 
10:58 PM
@Faust $dF$, not $F$
Again, we're letting $F$ be the antiderivative of $f$
@Faust Why is $dF$ a function of $x$ in what you wrote
 

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