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6:24 PM
@N.Maneesh constant
@AkivaWeinberger nice, i live here lol
 
akiva, I recall that you do some music. I am wondering whether you can help me to identify the genre or theme of this music:
2
Q: Trying to identify the genre (and if exists, source) of this music

Secret(Migrated from anime and manga upon advice) The following piece of music is extracted from some video on the internet (which the twitter link cannot be included here due to the content is rather violent): https://soundcloud.com/user-873168135/unknown-anime-crisis-type-music-extracted-from-anothe...

 
6:40 PM
Dunno if I can name a genre
 
do you know of any music that sound similar?
 
@Secret oh my god I have some good good information for you
@Secret have you heard of Frank Klepacki?
 
hmm?
 
Nothing comes to mind at the moment
 
nope
 
6:43 PM
@Secret Frank Klepacki wrote the music for the Command and Conquer series
if you look up like
Frank Klepacki Tiberian Sun there's a whole album of music that has a similar feel to the stuff you posted
 
@N.Maneesh one way to get a handle on this: there’s a biholomorphism from the upper half plane to the unit disk
 
@Semiclassical that's what led me to success yesterday :P
 
Since you’re interested in the right-hand side of the upper half plane, that biholomorphism gives you a map to half the unit disk
At which point you can ask the same question but with the domain being the half disk
 
@ÍgjøgnumMeg I should check that tomorrow (it's one hour and it is now 4:45 am) Will get back to you tomorrow
 
Cool :)
 
6:49 PM
@N.Maneesh $1/(x-c)$ where $c$ isn't in that region
You can do the same for any region whose closure isn't the plane
 
@topologicalmagician Take any element $g$ that isn't in $A_n$ and $g A_n$ is the other coset.
 
7:24 PM
can someone help me with solving $y-3x^2 y dx = ydy$ ? @Semiclassical maybe you ? :P
$y \ne 0$ ..
 
8:17 PM
@user123 are you missing a pair of parentheses?
i.e. did you mean $(y-3x^2y) \ \mathrm dx = y \ \mathrm dy$?
 
8:29 PM
If the last digit of this POSITIVE whole number becomes the first digit, the resulting number is exactly twice as large. My number is the smallest whole number with this property. What is my number?
example: 326578 -> 832657 which clearly isn't double of 326578
anyone wants to give this a try?
 
8:56 PM
A cheating example: if you double 052631578947368431, you get 105263157894736843
 
105263157894736842, 157894736842105263, 210526315789473684, 263157894736842105, 315789473684210526, 368421052631578947, 421052631578947368, 473684210526315789
what do they have in common, you ask?
they're 2/19 to 9/19
 
Ah dammit I just solved it
 
@LeakyNun copied of OEIS or something?
 
@Mathphile not really
just some semiautomatic calculations
 
@Mathphile If your number is abcd (doesn't have to be four digits, just using it for an example
 
8:59 PM
@Semiclassical you actually... don't
 
then let x=.abcdabcdabcd…
 
the former should end in 421
then the latter ends in 842
 
@LeakyNun well saw this on the internet with no solution
 
Since 2(abcd)=dabc by hypothesis,
We have 2x=.dabcdabcdabc…
and 20x=d.abcdabcdabcd…
 
Did I copy wrong?
 
9:00 PM
which you might notice is d+x !!
So 20x=d+x
and 19x=d, or x=d/19
If we compute 1/19, we see it has an 18-digit period
 
@AkivaWeinberger nice
 
Yeah, I did. Should be ...862
 
1/19= 0.*052631578947368421*
 
I guess we all love numbers
 
(the stuff between the asterisks repeats)
If you double 052631578947368421 you get 105263157894736842
so it does technically work
but we probably want something not starting in zero
so take d=2
 
9:03 PM
nice
 
2/19=0.*105263157894736842*
 
got it
 
Luckily, 105.... itself works
 
Double 105263157894736842 and you get 210526315789473684
 
yup
hmm
is there any prime number with this property?
 
9:04 PM
The smart way to get there is probably the fact that 2*10=1 mod 19
And 10^18=1 by Euler
So 10^17-2 is divisible by 19
 
Is the set of all function in $L^\infty(\Bbb{R})$ of bounded support closed in the norm $||\cdot||_\infty$?
 
hmm
 
You can also build 105263157894736842 one digit at a time, starting from the right, by writing …2 and then doubling the last digit you wrote and putting it next to it
and keeping track of carries
so …42
…842
 
okay so if the last digit is 1,2,3 the number will not be prime
 
…6842 (carry the one)
 
9:08 PM
@user193319 take $f_n(x) = \begin{cases} \exp(-x^2) & |x| \le n \\ 0 & |x| > n\end{cases}$
 
…36842 (carry the one)
…76842
etc etc
and eventually you get back to where you started
 
according to my calculations there cannot exist a prime with this property
 
Note that it you write the number twice
like 105263157894736842105263157894736842
then it also works
 
wait no a prime number might exist
@AkivaWeinberger that is a good observation
 
@Mathphile Fun fact: the first 6 multiples of 142857 are all cycled-around versions of each other
142857
285714
428571
571428
714285
 
9:14 PM
wow
 
857142
(Hint for why this is: compute 1/7)
(Note that 10/7=1+3/7)
Also: what's the seventh multiple of 142857?
7*142857
 
My post got DV'd up before I posted it
 
Downvoted up?
 
@LeakyNun Thanks!
 
This sounds like it might be an interesting topic in other bases.
 
9:18 PM
@AkivaWeinberger 999999 nice
@AkivaWeinberger do you think there is any prime number with this property?
 
@LeakyNun Wait, does $f_n$ have a uniform limit?
 
No idea
 
okay
maybe you can help me with another question
 
What motivates you @all?
 
2
Q: Can the product of $n$ positive integers, where $n \gt 5$ in A.P. be a palindrome?

MathphileReading the question can the product of four positive integers in A.P. be a square?, also made me question whether the product of $n$ positive integers, where $n \gt 5$ in arithmetic progression be a palindrome? Me and user Peter tried to find solutions for various $n$ in PARI/GP, and found that...

 
9:22 PM
@user193319 $\mathrm{exp}(-x^2)$
 
I want to upvote that post
 
@AlessandroCodenotti But isn't that a bounded function?
 
So it isn't a counterexample to my claim...
 
@user193319 but it ain't got no bounded support
 
9:23 PM
Oh, but it doesn't have bounded support...
Whoops, I forgot my own conjecture :P
 
do we fully understand the relationships that 5 has with other numbers?
 
That's a very nebulous question.
 
@user123 to second the point made: that equation only makes sense if there’s parentheses missing
 
5 is clearly the biggest number
 
@LeakyNun are you talking about my question?
 
9:27 PM
in base 5?
 
I don't know, Leaky, what about 5+5?
 
what about 5+5+5
 
@Mathphile no
 
(And now tetration blah blah blah)
 
9:29 PM
@Semiclassical see, you can't go higher than 5
tetration = 4
 
Hrm. You raise a point.
 
$\sum_{i=1}^5 \sum_{j=1}^5 \sum_{k=1}^5 5^{i^{j^k}}$
 
@Rithaniel ewww
 
So assuming we fully understand all the individual numbers independent of all the other numbers, it's the relationships between the collective group of (positive integers) that stump us?
 
You're right, I should have gone for 5 $\sum$s.
 
9:32 PM
I mean, this is like asking whether we can claim to know all properties of 2 if we don’t know the resolution of the twin primes conjecture
 
I mean I guess a number by itself doesn't even mean anything on its own
because there's nothing to compare it to
well maybe a number does mean something
like $5$ could represent that I have five things
but I feel like much more interesting stuff comes about when you have groups of mathematical objects and try to understand the relationships between them, not just the objects themselves
 
I've recently been interested in idempotent elements of rings.
 
@Rithaniel do you know that they factor through the nilradical?
 
For example, I know that if you have an invertible idempotent, it's the identity, and therefore every idempotent is either a zero divisor or a multiplicative identity.
factor through? I'm not sure what that terminology means.
 
What if the primes were completely periodic in a different setting
 
9:49 PM
You can easily define a setting which forces them to be periodic. For example, just define a base system where you add a digit whenever you hit a prime. Now every prime is of the form $10^n$
 
@Rithaniel it means, $e \mapsto \overline{e}$ is a bijection from the idempotents in $R$ to the idempotents in $R/\mathfrak N$ where $\mathfrak N$ is the nilradical of $R$
 
Please explain that more @Rithaniel
 
oh and the idempotents form a group with $e \star f := e+f-2ef$
 
I'm max{intrigued}
 
Leaky is over here blowing my mind
 
9:53 PM
leaky knows a lot of math
I probably know 1/10 out of the math that I'll learn in my lifetime
 
Also, I'll count from 1 to 15 in this "prime-th base:"
1, 10, 100, 101, 1000, 1001, 10000, 10001, 10010, 10100, 100000, 100001, 1000000, 1000001, 1000010
So, if I understand correctly, $R/\mathfrak{N}$ has the same number of idempotents as $R$?
 
@Rithaniel What I mean is that in some setting the primes are all right in a row, there are no numbers between them that create their disordered pattern wrt the other numbers
I think I can prove that that's impossible
 
I think you need to be more rigorous in what you mean by "setting."
 
yeah
I wonder if there's a smaller setting to embed the integers in, such that the primes have more of a pattern
Primes in N are interesting because of their algebraic properties, so it would seem the only useful transforms would have to preserve them mostly. One possibility is to embed them in a larger setting, like the Gaussian integers. Another idea is to change the metric, as we do with p-adic numbers
 
11:01 PM
0
Q: Showing the solution is unique or not for Poisson problem

johnny09Consider the following problem: $$\nabla^2u=u+\exp(xyz)$$ in the sphere $x^2+y^2+z^2<1$ (call this $V$) and $$\frac{\partial u}{\partial n}+u=5$$ on the surface of $S$ of the volume $V$. Question: How do we show the solution $u(x,y,z)$ is unique or not? My attempt: The above is in general a...

i'd appreciate any help/hints
 
11:13 PM
@TedShifrin are you familiar with the poisson problem or the laplace equation $\nabla^2u=0$?
 
11:48 PM
@johnny09 that’s not poisson’s problem. It’d Poisson if it were $\nabla^2 u=f$
Presumably your equation has a unique solution as well. But posing it as “Poisson’s equation” is more misleading than it is helpful
 
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