1:25 AM
Have you ever come up with your own question, solved it, and then felt underwhelmed by the result?

Yes.
That's one of the many reasons I avoid coming up with my own questions; usually, I end up asking questions that throw me down a rabbit hole.

@user4539917 Thats the fun part
Of course not all you ideas/questions are gonna be good ideas/questions, but it is how you learn

Not in terms of time management.

That is a fair point

Rabbit holes can be very deep.
They usually test my confidence.
And, patience, of course.

1:39 AM

This is what research is like, if the question is substantial enough.

Sometimes the brown rabbit!

I just hope that the answer at least satisfies someone's curiosity even if it's.... a tad boring, or borderline trivial.

Borderline trivial may not be worth it.

1:41 AM
I'm a black cat hacker. :D
I mean categorically speaking

@TedShifrin It's too late I already did it
0

Suppose we have a sequence $a$ with a repeating pattern of length $k$. For simplicity, we can express it as: $$a=\{c_0,c_1,...,c_{k-1},c_0,c_1,...\}$$ Where $a_n$ is the $n$th term, starting from $n=0$, can a generating function be found for such a sequence?

I just wanted to know if we could guarantee that it existed!
And no one had written on it, so I figured why not

Someone voted to close.... that hurts :|

when one door closes another opens

LOL not

I figured it was egotistic of me to close it on my own answer anyways, but I think there's a timer before you're allowed to use your own.
@DLeftAdjointtoU Why? I wrote on something I hadn't seen anywhere else in case others had the same idea.

1:49 AM
I upvoted both, bro! What do you want from me? My soul too?
o__o

Yes, give your soul to math.

Anyone have a question to ask?

Want to study something?
What do you study these days?

2:06 AM
@DLeftAdjointtoU I recently study and think about how I see absolutely no way to foliate $(0,1)^3$ by once-punctured planes without singularities. But then I think how it's most likely possible because Inaba and Masuda from Japan proved something in 2021 for a much more weirder class of surfaces. So it's very likely there's a way. But I don't see it at all. And it's weird that I can't find any literature on this particular case so it sucks

@geocalc33 I know I could google this, but is it safe to assume foliation is... folding?

I might ask a question for resources

2:36 AM
@A016090 Not quite. It's more like layers. en.wikipedia.org/wiki/Foliation has some pretty diagrams. That article is probably a useful reference if you already understand foliation, but it's probably a bit confusing otherwise.
You don't need many triangles to make an ok-looking sphere. The three.js library has an animated demo here threejs.org/docs/#api/en/geometries/SphereGeometry with 32 subdivisions around the equator. That demo shows the raw triangles, but normally shading would be used to make it look curved, although you can still see artifacts in some situations, eg at intersections with other objects.
Of course, proper ray-tracing software can render curved surfaces more precisely. But meshes are very versatile, and GPUs can do the necessary coordinate transformations and shading very quickly for large numbers of triangles.
@A016090 Cute sequence. oeis.org/A016090

2:52 AM
I read Tasmania is the only political hold out down under.

Tasmania is a special place...

Its native devil thinks so to :-)

On the Australian mainland, we joke that Tasmanians are a bit weird, isolated, conservative, and inbred. But we still love 'em. ;)

3:08 AM
Is their accent any different?

Slightly, but it's hard for me to describe the differences. (I've lived most of my life in Sydney). And of course with modern media, accent differences are eroding.

3:20 AM
Speaking of accents, this just popped up in my notifications. Tatiana is from the French-speaking part of Switzerland, but she studied at the Sorbonne and now lives in New York. I love the clarity of her voice.

3:47 AM
Nice.
đź‘Ťđź‘Ť

6 hours later…
9:58 AM
peep

we could have gone 8 hours, but no....

Oops
I'll shut up next time

over 9,000 hours later...
we could have gone 9,000.999... hours

10:19 AM
but no...

I was about to ask a question on the main site but didn't because the question itself, I think, is easy but needs a lot of context...

@onepotatotwopotato If you can provide the context, and what you've tried, then it sounds like a good candidate for posting.

It's not something like exercise problems but some clarification of some statements. I'll just put everything for nobody

10:38 AM
This user is dominating academia.SE with over 6,000 answers at an average of 50
0

Dr Buffy is at 308,000 and counting!

11:08 AM
I decided not to post the question. I'll just skip that part. It seems it does not harm the understanding of the rest part of the book

what to do if my Fourier coefficient are becoming infinity
like I am finding Fourier series of x^3sinx and I got a_n = \frac{...}{(n-1)^3} + ....

11:25 AM
The group of bijection $G$ of the set $X=\{1, 2,3,..,n\}$ where $n\ge 3$ acts on $S$ via $i\to f(i)$ . Then find the number of orbits of the natural action of $G$ on $X×X×X$ defined by $fâ€˘(i, j, k) =(f(I), f(j), f(k))$

2 hours later…
12:58 PM
@PM2Ring Aww you caught me! It's my favorite sequence, I only discovered it after playing with members of its set that act like identities for unusual subsets of * mod n.
Just from trying to visualize a program that determines if a number can be an nth root based on k last digits.

1:25 PM
@A016090 I'm generally not very interested in "numerological" properties that depend on the number being represented in base 10, but that sequence is rather cute. :) And of course similar sequences can occur in other bases, so that property isn't just some peculiarity "caused" by the base 10 representation.
Similarly, we have cyclic numbers like 142857, from the decimal expansion of 1/7, but there are cyclic numbers from prime reciprocals in every base, and they're certainly worth investigating.

'foundation of hyperbolic manifold' << much more boring than I thought. It's really an encyclopedia
It's like dummit foote or Lang in algebra

I suppose binary is kind of privileged as a base. But it's rather painful looking for patterns in bit strings by eye. Human pattern recognition doesn't work well with long bit strings.
@A016090 A noble goal. It's annoying that it's not easier. Even detecting squares takes a bit of work. The usual technique is to find the root via Newton's method. I don't think there can be a faster way, in general.

I'm back

Some primality tests require you to check if the number is a square. But the square test is usually not the first step, other cheaper tests are done first. en.wikipedia.org/wiki/Baillie%E2%80%93PSW_primality_test

2:06 PM
prime numbers are the building blocks of the natural numbers
unintuitive that the building blocks should be somewhat random
I can't believe only 5 years ago I didn't know that a function might not have an inverse that is expressible in terms of standard functions

7

As Pete L. Clark mentions in his excellent answer the statement "the primes are considered to be the 'building blocks' of the integers" is a soundbite - a shorthand way to refer to a very important fact about the natural numbers and by extension, of the integers. The additive structure of the i...

"Finally, what makes the integers interesting is precisely the fact that we have both additive and multiplicative building blocks. Taking each separately on its own terms we have a completely transparent structure. But when we start mixing the two, we get tough questions alarmingly quickly"
love this..

2:23 PM
@geocalc33 I've never been totally comfortable with that statement. Sure, there isn't a simple polynomial to generate all primes, but that shouldn't be surprising. A polynomial that generates all primes somehow needs to incorporate all of the primes, and that seems to imply that transcendental (or at least irrational) numbers are required. So we need an infinite number of terms, and / or irrational coefficients. OTOH, the Sieve of Eratosthenes isn't exactly rocket science...
From en.wikipedia.org/wiki/Formula_for_primes#Mills'_formula W. H. Mills (1947), proved that there exists a real number A such that, if
$d_n=A^{3^n}$ then $\lfloor d_n\rfloor$ is prime for all positive integers $n$.

I'm trying to understand that "mixing the additive and multiplicative structures" part

2:39 PM
@geocalc33 A good starting point is modular arithmetic. For a given prime p, addition and multiplication mod p is well-behaved, and we can even define a multiplicative inverse for each nonzero element. But to understand arithmetic on Z we need to somehow combine all the primes, and that's messy.
If the modulus isn't prime, then we get divisors of zero, i.e., nonzero solution pairs to $xy\equiv0$, and they mess up the nice patterns.
Quick change of topic. Here's a cute pi thing from the HNQ. It's pretty easy, so please try to think about it before looking at the solution.
12

Take any scientific calculator and follow these steps: Ensure the calculator is in degrees mode. Enter some number of 5's (more than 4 is ideal). Take the reciprocal of step 2. Take the sine of step 3. You should find that the result is an approximation of $\pi$ (that becomes more accurate as m...

Oops. That got cut off. It should say "an approximation of $\pi$ (that becomes more accurate as more 5's are entered), multiplied by a negative exponent of 10."
Of course, the same trick works with tan.

3:30 PM
I'm studying regularity of linear second order hyperbolic pde $u_{tt}+Lu=f$.
I'm reading it from Evans. Since $f\in L^2$ so it's intuitive to use the pde in $\int_0^t\int_{\mathbb R^n}f^2 \,dx\,dt$ and check what regularity we may expect
Evans followed this approach in Elliptic and parabolic but didn't use this in hyperbolic. Why?

Consider the following exact sequence: $0\to \mathbb Z^2\to H_1(X,A)\to \mathbb Z^{k-1}\to 0$
How do I show that this is a split sequence?
If this splits, then I'll have $H_1(X,A)= \mathbb Z^{k+1}$
0

Consider the following exact sequence: $0\to \mathbb Z^2\to H_1(X,A)\to \mathbb Z^{k-1}\to 0$. How do I show that this splits? By definition of splitting, I should have $H_1(X,A)= \mathbb Z^2 \oplus \mathbb Z^{k-1}=\mathbb Z^{k+1}$. But I don't understand how to show that this is the case here. T...

3:48 PM
if the rightmost module is free then the given s.e.s splits iirc

what is free module ? :(
I'll check the definition.
I forgot this.

@Koro It is a module which is free.
Which is a bit redundant, since Abraham Lincoln emancipated the modules in 1863.

free of cost module

@Koro Ah, no! Free as in speech, not as in beer.
It is now considered unethical to buy or sell modules.

@PM2Ring Ohhh I haven't even thought of the other -aries!! I was just enamored with a friend's number theory question, I still haven't taken a course in that field, but it involved determining if a square root existed for an extraordinarily large number. When he showed me the idea of testing the terminal digits, it of course was a parlor trick, but it caught my imagination for if we could figure things out for n roots.

3:54 PM
Not for sale

It paralleled $\mathbb{Z}/n\mathbb{Z}$ of course when the number I was generating with was coprime to $10^n$, but I didn't expect to see cycles outside of that! When using things like <4> in (10) or <12> in (10^2) I saw cycles, but then for <4> in (10^3) it fails. In all cases they made cycles when the set that the cycle makes treats the nth term of A016090, where n is of 10^n as the identity.

4:28 PM
It's similar to the cyclic prime reciprocals I mentioned above. Eg $7$ doesn't divide $10^n$ for any $n$, but it does divide $10^6-1$. And we can show that for any coprime $a, b$ that $a|b^n-1$ for some $n<a$. That is, the representation of $1/a$ in base $b$ must repeat with a period $<a$.

5:24 PM
So, inversive transformations map lines and circles to lines and circles. Is there an equivalent type of transformation that maybe maps hyperbolas to hyperbolas? Or "hyperbolas and parabolas" to "hyperbolas and parabolas?"

5:44 PM
what is wedge sum of more than two spaces?
For two spaces, one common point is identified.
nvm

@Rithaniel I think you want to start by thinking about the projective plane.

6:02 PM
Well, when you invert the unit hyperbola with respect to the unit circle, you get a little figure-8. A parabola get mapped onto something like a cardioid. That kinda gives an intuition for what's happening at "north pole" point

Iâ€™m thinking real projective plane, not Riemann sphere. Nonsingular conics are all projectively equivalent. Lines are projective lines = circles.

6:17 PM
are there bigger | for determinants instead of matrix brackets
if so what is the latex command
trying to do determinants of matrices within a matrix

obliv: overleaf.com/learn/latex/Matrices provides a range of options in the amsmath environment. i think some of these are present in other environments too.

I found it, I think it's vmatrix
thank you

the manual addition of delimiters to a plain "matrix" environment is also always an option, but you have to follow the delimiters with indications of scale. which sucks. \left and \right attempt to scale automatically but sometimes give you stuff that's too big.

Not sure what $\begin{Vmatrix} a & b \\ c & d \end{Vmatrix}$ is supposed to be, the magnitude of the determinant?

no idea. my guess is someone just asked whoever was developing those environments to throw that one in there.
unless someone knows of some common use of that. it's pretty hard to reverse engineer mathematical content from options in a latex package.
needing to distinguish the determinant from the absolute value of the determinant is definitely something that comes up. i think it came up in here the other day. but, i don't know that "using unconventional notation that you will have to explain every time you use it" is a good way of resolving the ambiguity.

6:30 PM
I think some folks used that for the usual brackets.
I think I recall that in Shilov, perhaps.

yes, shilov does that. what a goof

0

Let $B$ be a standard Brownian motion. Define for all $n\geq 1$ the functions $B^{(n)}$ s.t. for all $t\in 2^{-n}\Bbb{N}_0$ $$B_t^{(n)}=B_t$$ and such that $B^{(n)}$ is linear on the intervals $\left[\frac{k}{2^n}, \frac{k+1}{2^n}\right]$ for all $k\geq 0$. I want to show that for a fixed $\epsi... can someone help me? math.stackexchange.com/a/967954/21813 <--- Does this answer make sense to you? I'm struggling to understand its 2 explanations. 7:39 PM The answer seems wrong to me. The book is garbage. Perhaps @leslie has a definitive legal ruling. same, looks like garbage The author doesnâ€™t think$\frac 0{0+1}=0$, apparently. Good grief. I have added a comment. 8:10 PM Hi! How can I write the floor and ceil symbol in a formula? Im tried it, but it dont seems to work? @PedroUrday \lfloor and \rfloor for$\lfloor x \rfloor$I write double dollar signs but it does format it \lceil and \rceil for$\lceil x \rceil$You may need to enable ChatJax: tinyurl.com/cfqcvpc @TedShifrin The question is nearly a decade old. I am guessing that everyone involved is long gone. Just now it work. It was a delay or a bug @Obliv Could be some kind of norm, e.g. mathworld.wolfram.com/FrobeniusNorm.html 8:19 PM @Xander Sure, but people like @ryang are apparently looking. 8:30 PM i am over a decade old and not long gone. 8:46 PM @PM2Ring You've given me quite a lot to play with next time I'm stuck in an empty tutoring room. Thank you! @leslietownes Munchkin would differ. she is almost half a decade old. think about that. Does a sequence exist that cannot be put to a recurrence relation? can you say more about 'a recurrence relation'? if you impose constraints on what you allow there, you may be likely to get constraints on the set of sequences. It's similar in definition to my first question on this exchange. Suppose any$n$th term of a sequence$a$can be expressed as a function of terms of sequence$a$whose indices are beneath$n$. 8:56 PM yeah, "a function of" is still too general. i had in mind things like en.wikipedia.org/wiki/… where the form of the function is not completely specified in that it has parameters that can very, but is specified to some extent. ie.)$a_n=f(a_{n-k})+g(a_{n-\ell})$(where you can, for example, see that solutions to these kinds of recurrences cannot grow 'too fast' with n) Right but those are purely linear, what about the spicier relations? well, say what 'spicier' means and who knows. eventually you get into logic territory, about what it means to define a function, or a sequence, or an infinite subset of N, in the first place. and hopefully you're stopping sooner than that. or you're not and there's some space age logic answer that is both super precise and un informative. The issue here is we can't focus on the relation before we have a sequence. Essentially, can you come up with a sequence$a$where there cannot exist a recurrence relation defined above some base case. @leslietownes Well since I said "spicier than linear" I suppose a safe definition of spicier would be nonlinear. 1 hour later… 10:13 PM 1 Numerical methods for approximating Pythagoras' constant$t =\sqrt 2$by fractions. (This is an idea from my mentor while he was barely$13$yo, as a response to a challenge). We all know Newton's method for finding$t$. It converges quadratically meaning like$o( C x^{2^n} )$where$C$is a cons... Made a huge edit. surprising patterns ! a016090: "any kind of function except a linear function" isn't the kind of narrowing down i had in mind (it still leaves, well, almost anything), but given any sequence a = (a_n), consider a new sequence F(a) defined by the sequence 0, a_1, 0, 0, a_2, 0, 0 0, a_3, 0, 0, 0, 0, a_4, ... and so on. note that if F(a) were to satisfy a recurrence with 'finite lookback' (i.e. there is some N and some function of N-tuples such that for all sufficiently large k, a_k is that function of the previous N a_i's) then the sequence a_n is eventually constant. because the function of (the sequence of N zeros) can only be one thing, and for all n sufficiently large, a_n appears in F(b) with at least N zeros immediately behind it. so if a_n is not eventually constant then the sequence F(a) as described would not satisfy any recursion of that particular form. Trying to figure out something which feels like it should be obvious. Suppose I have a complex Hermitian matrix P such that P and 1-P are both PSD. If there exists a vector v such that Pv=v, must P be the projector onto v? (More broadly Iâ€™m trying to find a simple premise to get$Pv=v \implies $projector) I guess another way to do this is to assume P is positive and has trace one, but the latter seems like a jump oh you mean you want P to be a projection onto a 1d subspace? both P and 1-P will both be PSD whenever P is orthogonal projection onto a subspace of any dimension. trace one certainly limits the rank. Yeah, why 1-dim image? Oh, trace 1 will do it. well, if you aren't assuming P^2 = P then a 2x2 with 1/2's down the diagonal would also fit the bill and not be a projection at all. ok, maybe iwth the Pv = vthing. hrm. anyway, you are going to want that trace condition. yeah, it's probably enough to assume the trace 1 condition and what you wrote above. 10:30 PM Yeah, one eigenvector with eigenvalue$1$and trace$1\$ simultaneously.
And PSD. All necessary.

Yeah. The tricky thing in my context is then figuring out how to motivate those assumptions
PSD is fine, and if I phrase trace as â€śsum of eigenvaluesâ€ť then thatâ€™s not horrid either
@leslietownes with Pv=v by itself the issue is ruling out stuff like P as the identity matrix
The trace condition rules that out ofc
Come to think of it, the identity matrix would seem to be a counter example to my P,1-P being PSD premise

just say if the trace is larger than 1, then there's a nuclear explosion. or alternatively, that the nuke doesn't go off. whichever outcome is the one that you want to rule out.

Lol
Here itâ€™d more or less be â€śif trace exceeds one, then you get negative probabilitiesâ€ť
Which is rather disastrous on its own terms

@Semiclassical You should be punished for even writing such a thing. NONSENSE!
It makes my head hurt. Probabilities are not negative. X(
(Which is your point, I think, but it still makes me mad.)

Re-asking: Did the driver get arrested multiple times?

10:42 PM
@TedShifrin Not once. Though one or two students made a really good stab at it. It was about what I expected.

@XanderHenderson yep. One of the exam problems in this area last semester amounted to: show that X is not a valid â€śprocessâ€ť, and the reason being that what it spat out wouldnâ€™t give valid probabilities

@XanderHenderson Rats.

@TedShifrin Yeah, though it is a very small class (only five students), so I'm not overly disappointed.

I wonder how it improves with the obvious diagram. Oh, really tiny.

(Formally it was showing that matrix transposition is positive but not 2-positive and therefore bad juju)

10:46 PM
positive maps that aren't completely positive are offensive to the universe.

Anywho... time to teach.

I donâ€™t understand this terminology.
Teach goodly!

11:00 PM
@leslietownes now this is what I'm talking about, thanks!