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9:02 PM
the Fourier transform is applicable for many choices of $f(x)$ I think
 
that is a true statement
so infantile and random
 
what about a transform that is only well-defined for a few choices of f(x)
are they considered not as useful?
 
They are useful if they are useful
 
how do you determine what counts as 1 f(x)
like is $x$ and $2x$ different
 
That depends on the structure that you are interested in
 
9:07 PM
oh right
 
you can say that certain functions are equivalent in different senses. In measure theory, you often encounter the L^p spaces. These are spaces of functions f such that the integral of |f|^p over the measure space is finite
If we want this space to have a norm, we have to identify all functions f, f' such that the integral of |f-f'|^p vanishes
 
that is very interesting
 
this is equivalent to both functions being equal almost surely
 
equivalent with probability equal to 0?
 
equal, yes
not equivalent
And in topology, instead of considering all continuous maps between spaces, you can consider equivalence classes of functions (equivalence up to homotopy) to make life simpler
 
9:13 PM
oh. $X:=\{k/x : k \in \Bbb R \}.$ This is a space of functions, a "family". Is $g: k/x \to k$ a norm?
I have a feeling that families are not rich function spaces
in other words $X$ consists of 1/x,2/x,...
 
I thought k is a real parameter?
From its vector space structure, if k is real, this is just the real line
it's a 1-dimensional R-vector space
 
yes okay that makes sense, thanks for clarifying
 
so yes this is a vector space and no this is not a norm
but k/x -> |k| is
 
and what does this norm tell one
 
as much as the usual norm on R tells us
 
9:19 PM
it just induces the standard Euclidean metric on R right?
 
exactly
 
EM4
hello
 
EM4
how you doing
I need help with this cauchy integral
@TedShifrin , how have you been?
 
Hi, EM4. Just hear briefly — taking a break from cooking.
 
EM4
9:28 PM
can you look at this user answer on my post. I don't know why it is $i$ not $\frac{1}{2\pi}$
2
Q: Evaluating Contour Integral of square

EM4Evaluating $\frac{1}{2i\pi} \oint_C \frac{z^2}{z^2+4} dz$ where $C$ is the square with vertices at $\pm 2$ and $\pm 2 + 4i$ So, I rewrote the contour integral:$\frac{1}{2i\pi} \oint_C \frac{z^2}{(z+2i)(z-2i)} dz$ then used the Cauchy Integral Theorem: $\frac{1}{2i\pi} \oint_C \frac{z^2}{(z+2i)(z-...

 
Just finished my festivities. How is everyone?
 
Sad for Maradona
 
EM4
I am full and trying to find power to do some work.
and RIP to Maradona.
 
I'm good
 
Yeah, RIP Maradona
I'm feeling the effects of a food coma, myself. Everything is slowing down
 
9:40 PM
The solution is correct. I don't understand your issue. Look carefully at the Cauchy Integral Formula.
 
EM4
I know I did silly mistake, the pole is 2i and not -2i.

but do you multiply everything by $\frac{1}{2i\pi}$.
but should answer be 1/2pi?
 
About the fourier transform: there are two "most important" structural characterizations if I see right? One being the functor for locally compact abelian groups, i.e. pontryagin duality, and the other being the automorphism (linear homeomorphism?) on the schwartz space, and the induced automorphism on tempered distributions
 
@user2103480 Too fancy
It swaps product and convolution. Done
 
Yeah, but the schwartz space stuff also hinges on the relation to the differential operator
makes this more than a random cool property
 
I want to say that's the same thing actually
 
9:51 PM
I don't really understand the Runge approximation theorem
It's a complete mystery to me
 
@MikeMiller Is it?
Like, if you say you want an operator that does nice stuff to differentiation/convolutions, you get the fourier transform?
 
I'm saying that the convolution fact should imply the differentiation fact
Since F(x) = H you get F(xf) = H * F(f) which should be F(f)' by IBP I think
Something like this
@BalarkaSen I have been trying to prove that 1/sqrt(1-4x) = \sum C(2n,n) x^n combinatorially all day
 
why
 
Hobby
 
weirdo
 
9:57 PM
Probability is continuous combinatorics don't play this game
 
nah
explain to me why Runge's approximation theorem is really an $h$-principle in the complex analytic category
it says that if $K \subset \Bbb C$ is compact, $K^c$ connected, then $\mathcal{O}(\Bbb C) \to \mathcal{O}(K)$ has dense image
(much more, you get polynomial approximations, but yeah)
I have never understood this theorem. Total mystery
 
I don't know the theorem
 
@MikeMiller C(2n,n) is just the binomial coefficient?
 
Yes
 
@EM4 What everything? Write down what CIF says and look at your precise question. I already said the solution given is correct.
 
10:01 PM
It reduces to proving combinatorially that sum_{i+j=n} C(2i, i) C(2j, j) = 2^{2n}. Utter mystery
 
Pedro is a super clever combinatorial thinker.
 
You're supposed to start with a pair of subsets of the n-element set and produce a pair of splittings of 2i and 2j into complementary subsets
Not at all clear where this comes from
 
@MikeMiller so you also squared and multiplied by (1-4x)
 
No I squared and used the existing expansion of 1/(1-4x) to turn this into something that makes sense to prove combinatorially
 
ahh yeh
geometric series stuff
 
10:05 PM
about transforms. what does an integral transform tell you if every possible choice of f(x) just yields one function every time
 
@MikeMiller isn't this a convolution of you divide out the right way to get probability measures?
 
and B said this isn't probability
 
Could be a way to reduce this
 
I don't follow though
I have an approach now I think will work
 
@MikeMiller Count all self-avoiding walks of length $n$ in $\Bbb Z^2$ starting at origin
Open problem
 
10:17 PM
If you have independent discrete N-valued random variables X,Y, then P(X+Y=n) = sum over i,j with i+j=n of P(X = i)P(Y=j). One would have to normalize the sum to get a probability distribution, but that would involve taking sums of central binomial coefficients and I don't think that's a nice, often seen probability distribution
There's always some random walk interpretation for binomial coefficient problems haha
 
@BalarkaSen Shall we solve it?
 
Yeah lets do it
It'll probably get us the Fields medal
 
by PURE TOPOLOGY, I assume?
 
the corresponding problem for the hexagonal lattice is solved, by probability :P
 
@user2103480 Floer homology of random walks
 
10:21 PM
6
Q: Why does this weird iteration converge to the square root ??

mickLet $1 < x < 4$ , $a_1 = x$ and $b_1 = 0$. Now consider the (conditional) iterations if $a_n > b_n$ then $a_{n+1} = 4(a_n - b_n - 1)$ $b_{n+1} = 2(b_n + 2)$ else ( $a_n = b_n$ or $a_n < b_n$ ) $a_{n+1} = 4 a_n$ $b_{n+1} = 2 b_n$ Now consider $c_n = \frac{b_n}{2^n}$ Now define $f(x) = \lim_{n \to ...

any ideas ?
 
@MikeMiller It's always appropriate to throw some homology theory onto any problem
 
someone explain me Runge man
 
Isn’t Runge explained on wiki ?
 
10:39 PM
Mergelyans theorem is the ultimate theorem in complex analysis .

Agree ?
 
random wikipedia quote

Agree?
 
Why not post your question on main ?
 
my wish
 
So , do it ?
 
my wish is i won't post the question on main
 
10:43 PM
Why not ? Honor ?
 
whim
you have a problem with that ?
 
No. Just surprised
 
Why ?
 
posting on main is too tedious
 
I don’t see why not
 
10:47 PM
I don't see why
 
I guess you prefer too chat
 
I prefer to do whatever I want to
 
Lol
Not many people on chat I’m afraid
 
Why laugh ? Humor ?
 
Why live ? Life ?
 
10:52 PM
Well preferring to do what you want is like funny
 
Why ?
 
@Thorgott Why eat? Hungry?
 
You usually prefer what you want
 
So usual things are funny?
 
Nevermind
 
10:53 PM
Sun usually sets in the West. Is that funny?
Haha
lol lmao
 
lol
 
Well if you say you wish the sun would rise in the west they would be funny
 
Why hungry ? Lack of nutrition?
 
@Thorgott Why scared? Potter?
2
 
LMAO
 
10:55 PM
LOL
 
lol
 
Sun rises in east bro
 
I had to google that it's not the other way around in the souther hemisphere
 
lol
 
But that wouldn't make sense, something something equator
 
10:56 PM
algebra brain
 
no brain
 
flat brain lmao
flat brain theory
 
rotation: a geometric approach
 
f-brain theory
 
functors: a flat approach
 
10:59 PM
@user2103480 it only rises once at the poles
 
gold
 
that last one is what I think about when I see flat brain
 
my brain is euclidean
 
Flat earth brain haha
 
11:03 PM
oh wait it's actually semi-Riemannian. and when I think, I have to do Lorentz boosts
my neurons travel along hyperbola
 
How do flat earthers do spherical geometry? :)
 
I want my brain to be Banach
geometrically
the earth is flat Locallly
 
Not on a quantum scale
 
quantum mechanics is based on a flat background
okay I gotta get back to the math the magic
I have to go celebrate turkey day
stupid dumb idiot turkeys
 
@user2103480 it makes me sick
 
11:23 PM
better sick than brick
 
lol fair
 
O.o
 
i got quite into that game
 
ye was fun
 
11:25 PM
it was bad at first cuz I just couldn't lie properly but it went reasonably well towards the end lmao
 
It brings out the worst in you
 
totally
murder and sabotage and lie about it
 
The ratio of how long I procrastinate a calculation to the actual time it takes to do the calculation is way too large
 
@user2103480 I can't remember the last time I did a calculation
 
I'm glad that I finally need to do computations again
 
11:26 PM
i had to watch someone doing calculations with Haar measures today
 
It's possible I've been fed fake maths for the last 2 years given how seldom I check calculations
 
but thankfully there was no actual analysis involved
just pretend analysis
also idk if I said this earlier but literally nobody on the p-adic Hodge theory course knows what's going on
which makes me feel better about myself
 
lol
"You merely adopted p-adic Hodge theory"
 
the hard part is supposed to be doing stuff for p-adic Galois representations 2
but
even with $\ell$-adic representations nobody knows what's going on
aight I'ma go to bed, thanks for games
cya
 
gg cya
 
11:33 PM
bye bye
gg hf gl
 
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