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1:33 AM
A line say $y=x$ partitions the plane $\Bbb{R}^2$ into open regions?
 
sure
 
 
2 hours later…
3:18 AM
Let $B\subset \mathbb R^n$ and $a\in \mathbb R^n$. If $B$ is open then $a+B$ open: This is because since addition continuous.
In above argument, we are saying that $+:(\{a\},B)\to a+B$ continuous, then how to conclude $a+B$ open?
 
@Silent $x \mapsto x - a$ is continuous
that's the preimage
a+B is the preimage of B under the map $x \mapsto x-a$
 
thank you very much!
 
 
1 hour later…
5:07 AM
$\vartheta \nu$
 
11 hours ago, by Alessandro Codenotti
@SirCumference It is consistent with $\mathsf{ZF}$ that $\Bbb R$ is a countable union of countable sets! It is also consistent with $\mathsf{ZF}$ that $\mathrm{cof}(\omega_1)=\omega$
and... you cannot construct a computable example of such $\Bbb{R}$
 
user131753
6:09 AM
in Homotopy Theory, 42 secs ago, by user 170039
In this SEP article on Category Theory, it is said that adjoint functors "can be thought of as being conceptual inverses".
 
user131753
in Homotopy Theory, 40 secs ago, by user 170039
While there already exists a question regarding the clarification of the meaning of "conceptual inverses" and an excellent answer to it, I am failing to understand what exactly is the concept with respect to which the adjoint functors can be considered as "conceptual invesers" of each other.
 
user131753
in Homotopy Theory, 40 secs ago, by user 170039
Can anyone clarify?
 
7:24 AM
Can we analyze the function $f(z) = \displaystyle \prod_{k=0}^\infty \left(1-z^{2^k}\right)$?
I know that the radius of convergence is $1$
where exactly does this power series converge?
It converges to 0 trivially whenever $z \in \Bbb C^\times$ is $2$-torsion, but I don't know about the other points
the 2-torsion points is dense in the unit circle btw
and if we do analytic extension, where do we get poles?
We also get interesting functional equations such as $f(-z) = f(z) \left( \dfrac{1+z}{1-z} \right)$ and $f(iz) = f(z) \left( \dfrac{(1-iz)(1+z^2)}{(1-z)(1-z^2)} \right)$
they might be helpful
I think $\omega$ is a pole, but I doubt whether it can be extended beyond the unit circle at all
if $\{z \mid \exists n, z^{2^n} = \omega\}$ is dense on the unit circle then I have a proof that it can't be extended beyond the unit circle
1 is definitely a limit point
this set is invariant under multiplication by the 2-torsion elements
therefore by translation every 2-torsion root of unity is a limit point
therefore the whole circle consists of limit points
therefore it can't be extended beyond the unit circle
 
 
4 hours later…
11:36 AM
Not exactly the same, but:
$$\prod _{n=1}^{\infty } \left(1-\frac{1}{c^n}\right)=\exp \left(\sum _{n=1}^{\infty } \frac{1}{n \left(1-c^n\right)}\right)$$
for $c>1$ or something. $c$ can be a complex number.
 
12:14 PM
So I wrote this way back in 2014. In the five years since, I have completely forgotten how I found this equation, or why it's true. So, um, if anyone could rediscover my proof, I'd be pretty grateful. — Akiva Weinberger 1 min ago
Lol, when I wrote that, I just kinda assumed the reader would know what the digamma function is
For integer arguments, it's $H_{n-1}-\gamma$, I think, where $H_n=1+\frac12+\dotsb+\frac1n$
I think it's the log-derivative of the Gamma function (so $\psi=(\ln\Gamma)'=\Gamma'/\Gamma$)
 
 
1 hour later…
1:17 PM
Do irreducible representations induce a transitive group action on the Grassmanians of the representation space?
For $Gr(1,V)$ it is clear, but I think it is false for higher grassmanians
 
1:35 PM
@s.harp the... trivial rep
 
is not irreducible
 
can you reduce it?
anyway how about the non-trivial irrep of $C_2$
 
2:03 PM
That representation is valued on a $2$-dim vectorspace, this means the only non-trivial grassmannian is $Gr_1$
ie the question is just the same as irreducibility
anyway, I noticed that $Gr_1$ is also does not necessarily have a transitive group action on it, so the question takes care of itslef
 
2:40 PM
@AkivaWeinberger As far as how one might prove your version, I agree with commentators that Gauss's digamma formula seems like the obvious approach
Main thing I'll add in that vein is that, in your formula, it isn't immediately obvious that $\psi(a/b)$ ought to be real-valued
But $\psi(x)$ is in fact real on the real axis. So I suspect that, if you take the real part of your formula and simplify a bit, you'll get the Gauss formula
 
@Semiclassical Also, $\sum\ln(1-\bar\rho)$ is $\ln2$, right? So I can probably simplify it by taking that out
 
dunno about that
but you can write that as
 
So, I'm trying to set up an exhaustive computer search for a very large space (currently, there are $(n-1)!(n-1)^{(n-2)^2}$ cases to check, but I'm trying to narrow it down further). Would it be better to set up something in MATLAB or mathematica or set up an executable via C or some other higher level language?
 
Oh no it's $\ln b$
 
$\rho = e^{2\pi i k/b}$ for some $k\in\{1,2\ldots,b-1\}$
 
2:49 PM
So we end up with $\displaystyle \sum_{\substack{\large\rho^b=1\\\large\rho\ne1}}\rho^a\ln(1-\bar\rho)-\ln b-\gamma$
I think
or $\displaystyle \sum_{\substack{\large\rho^b=1\\\large\rho\ne1}}\rho^{-a}\ln(1-\rho)-\ln b-\gamma$
 
$$\ln(1-\overline{\rho}) = \ln(1-e^{-2\pi i k/b})=\ln(e^{-\pi i k/b})+\ln(e^{\pi i k/b}-e^{-\pi i k/b})=-\pi i k/b+\ln(2i\sin(\pi k/b))$$
 
Oh right
 
note the presence of $\ln \sin(\frac{\pi k}{b})$ in there
 
2:51 PM
Gauss's formula has trig
so that's why you're doing that
 
yeah, and in particular it's got that $\ln \sin(\pi k/b)$ part
as for why the Gauss formula sums from $k=1$ to $k=\lfloor (b-1)/2\rfloor$
(in our notation)
 
Pair up the roots of unity by conjugate pairs
 
Right
 
what is the question talking about?
 
@Rithaniel more context?
 
2:54 PM
In the case of odd $b$, there's $(b-1)/2$ complex-conjugate pairs and $\rho=1$ (which you ignore)
 
I just in
 
In the case of even $b$, there's $(b-2)/2$ complex-conjugate pairs, along with $\rho=-1$ (which you use) and $\rho=1$ which you don't
 
@LarryEppes There's this thing called the digamma function, which generalizes the harmonic series function $H_n$ to noninteger values
$H_n$ is defined as $1+\frac12+\dotsb+\frac1n$
and $\psi(n)=H_{n-1}-\gamma$ for integer $n$
At one point I found a closed-form expression for $\psi(a/b)$ in terms of $a$ and $b$
 
ok, I've studied that before...
 
and I have since completely forgotten where that formula came from or why it's true
 
2:56 PM
@LeakyNun I'm searching the space of multiplication tables for magmas to find ones following a particular set of rules. I'm representing the table as an $n\times n$ array of arrays.
 
I'm quite sure $\psi$ isn't called digamma
 
Then you're quite wrong.
 
@Rithaniel what rules?
surely digamma is $\digamma$
 
In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function: ψ ( x ) = d d x ln ⁡ ( Γ ( x ) ) = Γ ′ ( x ...
 
that's why it's called di-gamma
 
2:57 PM
I dunno why people use $\psi$ for that
 
@LeakyNun I have no idea why it's not that
but it's written with a psi
But it's the log-derivative of Gamma
 
what heresy
 
(not to be confused with the constant gamma, which shows up here also)
 
I call upon the power of Zeus
 
(because $\Gamma'(0)=\gamma$ or something like that)
 
2:58 PM
"The digamma function is often denoted as $\psi_0(x)$, $\psi^{(0)}(x)$ or Ϝ (the uppercase form of the archaic Greek consonant digamma meaning double-gamma)."
from wikipedia
 
Well, there are two. First, for distinct elements $r,s$ we have that $ra=ar=s$ for all $a$ in the magma (easily controllable in the table, so I don't need to check for that). The other is $(ab)c=a(bc)\iff a=c$
 
see Zeus just fixed wiki
 
15
Q: Why is the Digamma function always denoted with the letter "psi"?

Elements in SpaceMy question is on the notation of the Digamma function. The Factorial function $n!$ (which is met in secondary school), is conceptually seminal to the Digamma function. The Factorial function is defined as: $$0!=1,\qquad n!=\frac{(n+1)!}{n+1}$$ This concept is extended with Gauss's Pi fu...

 
Correction: $\Gamma'(1)=-\gamma$
 
wow, I was also learned digamma is denoted by psi
yes
 
3:00 PM
Arright I need to take a shower
See ya later
 
ok,
 
the top-rated answer there is good
with the upshot being that the use of $\psi$ to denote the digamma function is actually older than the latter terminology
which is pretty funny
 
I guess $\digamma$ never caught on 'cause it's such an obscure letter
(and also is essentially an F)
Fun fact: the letter is also called wau
(Inb4 Vihart references)
 
At one point I was trying to figure out a way to get glagolitic script into latex so I could use it for variable names. (solely for my own entertainment)
 
i guess the 'digamma' terminology caught on because of it being related to the gamma function
but failed to displace the existing (and frankly more convenient) psi notation
 
3:05 PM
Γ = gamma. F = digamma. E = trigamma?
 
What about tetragamma, then? An underlined E?
 
I have a question about signed radon measure..
 
Pentagamma is double underlined E. Dodecagamma is is an E underlined by enough lines that they obscure following sentences.
 
What's the question?
 
3:10 PM
... my network is too slow... for a minute
 
Yeah I saw that image but what's unclear to you?
 
${\underline {\underline {\underline {\underline {\underline {\underline {\underline {\underline {\underline E}}}}}}}}}$
 
the 'delta' is dirac measure, i suppose.
how to get the final inequality
I think the inequality is reversed
 
Beautiful, Leaky.
 
is the measure non-negative?
 
3:15 PM
Are you sure $\mu$ is signed here?
 
no, its real
 
s n i p e d
it's wrong if you allow it to be signed
 
yes, I also think about that.
 
Then it's clear, by monotonicity the measure of a set $A$ is bigger than the sum of the measures of the $x_j$ contained in $A$
 
@Semiclassical do you like Lyapunov function?
you also think about that?
you mean you also think so?
 
3:18 PM
yes, its understandable if $mu$ is positive..
I'm reading "Evans - 1990 - Weak Convergence Methods for Nonlinear PDE.pdf"
in the theorem 1.4.2, It didn't say $mu$ is positive measure
 
I haven't ever had to deal with Lyapunov
which i'm quite okay with
never had a course in ODEs which was formal enough to address it
and in physics you'd only see it touched on in the context of chaotic dynamical systems
 
:c
 
In mathematics, the trigamma function, denoted ψ1(z), is the second of the polygamma functions, and is defined by ψ 1 ( z ) = d 2 d z 2 ln ⁡ Γ ( z ) {\displaystyle...
 
in which case "touched on" is about the most you'd do
 
"TRIgamma is the SECOND"
 
3:22 PM
I remember trigamma is the second derivate of logGamma function...
 
eh, it's understandable when you note that the gamma function isn't a polygamma function
not even in the sense of it being the zeroth polygamma
 
Still, "tri" and "second" are irksome.
 
@Semiclassical By the way
 
@Rithaniel yeah, it's not great
 
Say $\zeta_n$, $~n=1,2,\dots$, is chosen uniformly at random from $[0,2^{-n}]$
and they're all chosen independently of each other
What are the odds that $\sum_{n=1}^\infty\zeta_n<1/4$?
(The odds of it being less than $1/2$ is $1/2$ I think)
 
3:27 PM
Well, to make that less mysterious, let $\eta_n=\zeta_n 2^n$
Actually, to clarify
 
@AkivaWeinberger sure, it's either less than it or not
 
$\zeta_1$ is chosen uniformly at random from [0,1/2], $\zeta_2$ chosen uniformly from $[0,1/4]$, etc?
 
Ya
I will say that I don't know the answer, but I do know that the answer is known
 
Okay. So if $\eta_n=\zeta_n 2^n$ then each $\eta_n$ is drawn uniformly at random from $[0,1]$
 
3:29 PM
Mhm
 
and the question is the odds that $\sum_{n=1}^\infty \eta_n 2^{-n}<1/4$
 
More generally, we can ask for the odds that $\sum_{n=1}^\infty \eta_n x^n<1/4$
Which seems like a neat generalization if nothing else :p
 
so $P(X \le x) = \displaystyle \int_0^x P(X-\zeta_1 \le y) \ \mathrm dy = \displaystyle \int_0^x P(X \le 2y) \ \mathrm dy$?
 
@LeakyNun Looks right
And the reason it's $1/2$ for the odds that it's less than a half is that if we reflect it ($\eta_n\mapsto1-\eta_n$) the sum becomes 1 minus the original
 
3:31 PM
For instance, when $x=1$ we get $\sum_{n=1}^\infty \eta_n$ where each $\eta_n$ is chosen randomly from $[0,1]$
 
@AkivaWeinberger probably needs a few modification on the domains
 
In which case we're talking about a sum of iid random variables which seems handy
 
or else we would just get $f'(x) = 2f(x)$ or something like that
 
independent and identically distributed
 
3:32 PM
Ah
@LeakyNun $f'(x)=2f(2x)$
 
If I were to guess the best approach, though, I'd look at the characteristic function of $\sum_{n=1}^\infty \eta_n x^n$
 
oh
@AkivaWeinberger and that's the d.e. of your smooth-but-nowhere-analytic function?
 
by definition, the characteristic function of a random variable X is given by $E[e^{i t X}]$
 
This is the Fabius function, yeah
 
For simplicity tho I'm going to go back to just $\sum_{n=1}^\infty \eta_n 2^{-n}$
 
3:34 PM
All I know about it is the definition, that differential equation, and the fact that it's infinitely differentiable but nowhere analytic
 
So then $e^{i t X}=\exp(i t \sum_{n=1}^\infty \eta_n 2^{-n})=\prod_{n=1}^\infty \exp(i t \eta_n 2^{-n})$
And since the $\eta_n$ are all independent, the expected value is simply $E[e^{it X}]=\prod_{n=1}^\infty E[e^{i t \eta_n 2^{-n}}]$
I guess I'm commiting a sin here
I shouldn't be using $\eta_n$ for the value of the random variable and for the random variable itself
ignoring that, though: since those are uniform random variables, we have $$E[e^{i t\eta_n 2^{-n}}]=\int_0^1 e^{i t \eta 2^{-n}}\,d\eta = \frac{e^{i t 2^{-n}}-1}{i t 2^{-n}}$$
 
why don't we study diff eqs of the form $f'(x) = 2f(2x)$
 
Arright I need to go now
 
see you
 
3:40 PM
@LeakyNun Note that if you write $x=g(t)=2^t$ then $f(2x)=f(2g(t)) =f(2^{t+1})=f(g(t+1))$
So I suspect that differential equation is equivalent to a delay differential equation, and those are studied
 
ok nice
 
Continuing on with mine, I've got $$E[e^{i tX}]=\prod_{n=1}^\infty \frac{e^{i t 2^{-n}}-1}{i t 2^{-n}}$$
Which if nothing else is a pretty neat expression :P
moreover, though, one can then do inversion stuff to extract probabilities from that
 
@Semiclassical r.o.c.?
 
No clue.
 
:c
 
3:48 PM
can you take the natural log of that
 
you could, yeah. Not sure it helps tho
One thing you can note from that is that, as $t\to 0$, each factor converges to $1$
so you get $E[e^{0}]=E[1]=\prod_{n=1}^\infty 1 = 1$ as you should
nothing strange there, but a good check
you also can calculate expectation values of $X$ without too much hassle in this way
(and for that, taking the log is a very useful trick. Take the log of $E[e^{i t X}]$, differentiate, and evaluate at $t=0$)
from that i think you get $E[X]=\sum_{n=1}^\infty 2^{-n-1}=\frac12$
Of course, that's not too impressive. You can already get that from $E[X]=\sum_{n=1}^\infty E[\eta_n]2^{-n} =\sum_{n=1}^\infty 2^{-n-1}$
But a good check nonetheless
 
@TedShifrin Que signifie-t-il "soit" dans b) de Prop. 3?
 
Either or
as opposed to let ...
 
4:03 PM
je n'avais pas su cette signification de ce mot
 
je ne savais pas ... :)
 
yo @Eric
 
je ne sus pas
 
@Ted ive decided to commit to grad school finally
 
4:05 PM
@ÉricoMeloSilva congrats!
 
Mazltov!
 
@LeakyNun I should amend my "no clue" at least a bit. Note that $E[e^{i t X}]=\sum_{k=0}^\infty \frac{(i t)^n}{n!}E[X^n]$
 
that's hebrew?
@Semiclassical did you just take a fourier series or something
never mind
ok
 
Which means that the r.o.c. hinges upon the rate at which $E[X^n]$ grows
 
@TedShifrin let $\zeta_n \sim U(0,2^{-n})$. What's the probit. that $\sum_{n=1} \zeta_n < \frac14$?
 
4:08 PM
But the largest possible value for $X=\sum_{n=1}^\infty \eta_n 2^{-n}$ is $\sum_{n=1}^\infty 2^{-n}=1$
So $E[X^n]\leq 1$
Hence I think you'll have an infinite radius of convergence here?
 
agreed
 
@TedShifrin ended up choosing princeton for the confluence reasons we talked about
 
characteristic functions are typically pretty nice as far as analyticity goes
 
@Eric: I was pretty sure you would do that.
I'm sure you'll do great.
 
4:09 PM
yeah here’s to hoping! i’m definitely excited about the math
 
Hi, demonic Alessandro
 
Going back to my result from above for a bit
$E[e^{i t X}]=\prod_{n=1}^\infty \frac{e^{i t 2^{-n}}-1}{i t 2^{-n}}$
I suspect there's more I should be able to say about that
(Main virtue of this, btw, is that we've eliminated all the random variables in favor of a single complex $t$.)
 
nice job taking the Fourier transform anyhow
 
Yeah, it's a common trick
And that's indeed all it really is
it's a pity the numerator is $e^{i t 2^{-n}}-1=e^{i t 2^{-n-1}} ( e^{i t 2^{-n-1}}-e^{-i t 2^{-n-1}} )= 2i e^{i t 2^{-n-1}} \sin(t 2^{-n-1})$
 
that's a complicated expression
 
4:16 PM
nnnggg, where's my error
 
wut r u trying to do computing this expectation
 
lol where's the error
 
found it
 
top 10 mystetry
aha
double exponent
 
yeah
latex doesn't like double ^'s
anyways. that looks a lot like Vieta's infinite product
$\cos(\frac{\theta}{2}) \cos(\frac{\theta}{4}) \cos(\frac{\theta}{8})\cdots =\frac{\sin \theta}{\theta}$
But that involves cosines, not sines :(
 
4:37 PM
@akiva lots of values of the Fabius function here: arxiv.org/pdf/1609.07999.pdf
 
nice
 
4:59 PM
Here's a problem which seems pretty tough, inspired by my running some statistics on $X=\sum_{n=1}^\infty \eta_n 2^{-n}$
(well, strictly speaking on $X^{(m)}\equiv \sum_{n=1}^m \eta_n 2^{-n}$, since I can't have mathematica generate infinitely many uniform random samples)
What's the equivalent of the central limit theorem for $X?$
in other words, suppose I take $N$ samples $X_1,X_2,\cdots,X_N$ of $X$
 
what's wrong with the normal central limit theorem?
 
well, numerically, it doesn't seem to be holding
To be more precise:
If I generate X a whole bunch of times and compute the average value of that, I do get E[X]. So that much does agree with CLT
 
f'(x) = f(x-3), should this be very hard or very easy? aside from the trivial solution does it have any other?
 
But the CLT would moreover predict that $\sqrt{n}[(X_1+X_2+\cdots +X_n)/n-E[X]]$ should be normally distributed
 
@famesyasd delayed differential equation
 
5:05 PM
and I'm not at all sure about that
 
so very hard?
where can I read how to solve those
 
well I gave you a term
you can start by googling
 
I googled already and did not find much
nvm, what about y' = f(f(x)), does this get cover by any topic?
 
integration
 
so long as you don't mean y=f(x) as well anyways
@LeakyNun looks like my anxiety re: the central limit theorem wasn't necessary
i was just being silly
 
5:12 PM
oh lol
 
 
1 hour later…
6:13 PM
hi
@Semiclassical you know a bit of analysis right ?
 
6:32 PM
sin(x) is a solution for f'(x) = f(x-3pi/2). Big shock.
tldr'ed
 
7:17 PM
I'm having trouble proving that if a continuous function is constant over the rationals, then it is constant over the reals.
 
@Fuzzy what do you know about the rationals in relation to the reals?
 
Oh nevermind, I see it now
Without loss of generality we can suppose that our function is zero over rationals
Then if there exists $a$ such that $f(a) \neq 0$
We would be able to find a neighborhood of $f(a)$ where for all deltas we would be able to find a rational number that isn't within epsilon of $f(a)$
Since rationals are dense in the reals
Is there a simpler way to go about saying this?
@Daminark
 
7:33 PM
I mean that was pretty simple
You may prefer phrasing it as sequences?
One thing about continuous functions is that if $x_n\to x$, then $f(x_n) \to f(x)$, so you can say give me any real number $a$, choose a sequence of rational numbers $a_n$ converging to $a$, then $f(a) = \lim f(a_n) = \lim 0 = 0$
 
I'm afraid that would be much more complicated for me
Haha
I can't believe I forgot about this ...
 
I mean really that's pretty much the exact same thing as your proof
 
I blame linear algebra
 
Using the language of sequences vs delta epsilon but like, there's no difference in the content
Y tho? Linear algebra is great
 
Yes, so great it made me forget Real Analysis
 
7:37 PM
Ah, lmao, that's fair
 
Ransom question: Do you think one gets better at constructing mathematical arguments with excessive details and reasoning your way up from the assumptions in your problem, or is it accumulating a lot of "mini-propositions" in your head that let you operate at a "higher-level" ?
This occurs to me whenever I encounter the phrase "according to the non-trivial theorem..."
 
Some variant of the latter
I feel like as you get used to really basic stuff you begin to reference it second nature
You begin to say things like "By property X, we can do Y" when the fact that property X allows you to do Y is something that a few weeks ago would've been a non-trivial theorem to prove
But is now a fact that you can state very inexplicitly
 
I have a habit of obsessing over the basics for a long time, which holds me back from that... I'm afraid I might do some error and my memory is kind of weak
If can prove to myself everything I use then the probability of doing some faulty step becomes slim
But then I will need a long-long time to familiarize myself with everything I learn
 
So, if you actually are forgetful and liable to either not remember theorems you've already proven and reprove, or somehow think something's right when it isn't, then I guess it's something you'll need to contend with/work around. It's definitely gonna be an added difficulty
 
It's more of paranoia, fear that I might write something wrong on my exam paper.
Thanks for the talk !
 
7:52 PM
No problem, see you!
 
8:11 PM
Hey @Daminark, didn't recognise ya
 
hey @Daminark
@loch
hi
 
hi
 
I don't know if you remember me. I am Adeek.
we are both doing algebraic geometry :D
 
I think so - you did a masters (?) project on hodge theory or something related
 
yeah
 
8:15 PM
hows it going?
 
I am currently doing my PhD.
Last 5 month were shit though
I almost broke off with my wife because of math spending too much time in it
but now everything is back to normal :D so I am happy.
 
it is important to put time into non=mathy things too!
 
Hey igjo, Adeek, and loch!
 
Yeah
@loch math can get to the point of addiction
 
hi @Daminark
 
8:18 PM
How is everything going?
 
good @Daminark catching up to my classes
measure theory and mirror symmetry
I am enjoying mirror symmetry. I now know that I would like to put my effort into mirror symmetry, algebraic cycles, and Hodge theory for my next few years
 
Ah that sounds nice!
 
@Daminark it's going pretty well - other than things that im supposed to be reading, ive been trying to learn a little bit of probability theory for fun
 
Measure theory is soo *** technical
really
 
Lmao, yeah part of what I liked about my measure theory class was that we skipped certain proofs our professor felt weren't enlightening (e.g. Lebesgue measure is Radon)
 
8:22 PM
oh yeah that crap
@Daminark I am considering doing a youtube videos
on math in Arabic
 
Hmm, I could see a case for it, idk
 
why ?
 
Hey everyone!
 
There’s a very special kind of panic attack: “I think there’s a hole in the assumption I’ve been using in a paper that’s been in progress for months.”
 
8:42 PM
@Newbie How much algebraic topology would you say is used in mirror symmetry?
 
Mirror symmetry has so much from mathematics
 
I'm guessing you set up some cohomology theory and then make use of it and that's the extent of alg top one sees
 
@Newbie what kind of measure theory tho
 
yeah and covering spaces as well @Perturbative
up to Haar measure
 
Ahhh cool cool
 
8:57 PM
Okay
I think the problem I’m seeing, while it does affect the setup of my argument
It doesn’t actually change the actual conclusion
So...fingers crossed :|
 
9:29 PM
I'm back
I'm having trouble computing a polynomial approximation of degree 3 to $x^x$
 
1
Q: Is this is a well-known ring and a few other questions... (sum of ideal characteristic functions up to $\sqrt{n}$)

BananaCatsDefine $S = \{ s^2 : s \in \Bbb{Z}\}$ be the squares in $\Bbb{Z}$. Let $\chi_A(x) = $ the characteristic function of set $A \subset \Bbb{Z}$ which is another way of saying it's $0$ except when $x \in A$, then it's $1$. For ideals $(m), (n)$ in $\Bbb{Z}$ these functions have the basic property $...

Can anyone advise here?
It might be more complicated than that to prove ringness but the coincidences are too much for it not to be a ring... or pretty close
🍌😹
 
10:31 PM
I need to solve cos(z)=cosh(z) , z any complex number. I've reached to a system of equations { sinxsinhy=-sinysinhx , cosxcoshy=cosycoshx } , but I don't know how to proceed. Any hint ? Or any alternative solution ? . Thanks
 
10:48 PM
@JamesGroon try using the alternative definitions of cosh and cos
That is, like $\cosh(z) = \frac{1}{2}(e^z + e^{-z})$
This makes it easier.
 
11:06 PM
$\cosh z=\cos iz$, so you can rewrite the equation as $\cos z-\cos iz=0$. A certain identity then does the rest of the job.
 
11:34 PM
@anakhro I used that and thats how I got those equations
@Thorgott Im not sure which identity is on your mind.
 
one of the Sum-to-Product identities
 

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