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03:34
Is there any errata page for Korner’s Fourier analysis book? Does anyone know about this? The problem is that: the theorem 3.1’ (ii) is clearly wrong (unless there is a typo) but I don’t see any flaw in the proof.
The theorem is wrong because there are counterexamples.
It may be irrelevant, but some authors assume that a Hilbert space is separable.
OH! Big thunder. Very close.
Here is a proof which is almost word to word taken from the book: math.unm.edu/~crisp/courses/wavelets/fall13/…
Yay for monsoon season!
we had a change of thunder, but nothign materialised
03:39
In the link, (1.2) is problematic but I’m not sure what’s wrong with the proof.
I drove through a lot of rain on Monday, but none fell in Holbrook. It started raining about five miles south of town (mostly just a drizzle, but it pounded down over 14 Mile Hill), and rained in Snowflake for much of the day as several storm cells blew through.
It rained most of the way home, until about 14 Mile Hill (which is, appropriately enough, about a mile less than 15 miles from Holbrook).
Okay... it's almost 9. I should go to bed. So much work to do tomorrow.
(1.2) follows from (1.1). Earlier, I somehow convinced myself that replacing <r gamma> in place of r gamma throughout will solve the problem. But it doesn’t!! Assuming that doesn’t prove (1.1) for the case when f is of the form e^{ist} with a non zero s.
04:13
koro if there are errata for the book they would be on the publisher's website or the author's home page (at least, he is still alive and used to be pretty good about updating things like that). a standard way of handling this would be to distill what might be a series of difficulties into a single, minimal question (minimal meaning, not requiring access to the book), focused on where you think an error might be, and with counterexamples included if they provide context for the difficulty.
i say this not because i have an opinion on whether there is an error in the book, but because in the course of that distillation you may be able to locate the error more precisely than now.
at least, i'm not going to wade into a proof that might have something wrong with it with no guidance other than "it looks wrong" or "there are counterexamples"
i don't see errata for the book on the author's webpage; he has material about a lot of stuff but not on that book, which he refers to the publisher for. he does have some notes on fourier analysis on that page that contain similar statements of weyl's theorem, which suggests either that there's less likely to be an error, or that if there's an error, at least it's being pasted consistently across multiple documents
@copper.hat Ok, that is a bit weird. Perhaps it's a bug, or a manual mistake by staff. I could understand losing points from years ago due to "User was removed", but not for "Voting corrected". And that last loss on Math.SE doesn't say what it's for, at all!
There's info about rep changes due to user removal on meta.stackexchange.com/q/126470/334566 & info on changes due to serial voting with no recent voting irregularities on meta.stackexchange.com/q/379345/334566 If you're concerned that it's a bug, you could ask on Meta SE, or contact staff directly: math.stackexchange.com/contact But make it clear you're reporting a potential bug.
@copper.hat this made me giggle
05:05
@PM2Ring Thanks! It is persists I will follow up.
05:25
@copper.hat No worries. Another option is to ask in the main Meta.SE chat, chat.meta.stackexchange.com/rooms/89/tavern-on-the-meta Some of the regulars there have a lot of knowledge on the fine details of the system, and several staff are frequent visitors of that room.
Much appreciated!
05:58
I'm not getting notification of being tagged in the chat unless I explicitly already have the chat open. Before I would be able to close the chat, come back and see where I was tagged. I only noticed I got tagged because I happened to be reading something on another StackExchange site.
06:22
I am studying deformation lemma, Palais Smale condition etc. in Calculus of Variations. I found that the deformation lemma is very similar to a theorem from morse theory. I watched this talk youtube.com/…. So I was thinking if anyone could give me references for this talk.
I know Milnor's book. I'm reading that book now. But it's very old so I was wondering if anyone knows any new books.
How do you generally read books written in a different language. I was reading a book and the author constantly refers to a french book.
do you mean math books specifically? a lot of folks are lucky enough that the math they want to read is written in languages that they already know for some other reason. for the rest of us it is a little harder but if you have some familiarity with the math it can be OK. it helps that math is not a kind of literature where the full spectrum of a language might be used.
i don't know that i would be able to learn a math subject from the very beginning in another language, but when i was in school, within the narrow confines of my subject area i was able to consult resources in in both french and german despite not really knowing either of those languages.
@leslietownes thanks. I hadn't seen Fourier analysis notes on his webpage earlier. I'll look for them now.
06:37
i don't know that they all include proofs (or include any proofs), but if the concern is only about the statements of results maybe they will help.
 
3 hours later…
09:16
I have this proposition in my lecture notes stating that if $c\geq 0$, then $\sup cA=c\sup A$. However, how should one interpret the formula if $c=0$ and $\sup A=\infty$?
@sunny assume $\sup A$ is finite or $c > 0$ for this one
also note that we can have $\sup A = -\infty$
yeah, my notes only assumed $A\subset\mathbb R$, which is somewhat weak
did they mention $c\geq 0$ or did they say "$c$ is positive"
if the latter than we can just blame it on them using different convention for "positive"
either way, I'd just replace $c\geq 0$ with $c > 0$ and end it at that
they state $c\geq 0$. It is these notes, chapter 2, proposition 2.23. However, they have not introduced the notition of $\sup A=\infty$ yet, that comes later, so probably $\sup A$ is finite here.
> Making a set smaller decreases its supremum and increases its infimum. In the
following inequalities, we allow the sup and inf to be extended real numbers.
No, they do consider $\sup A = \pm \infty$
09:29
yeah, I was wrong
I think it's just a typo because they consider $c < 0$ next
most likely, yeah
No wait they say it's obvious when $c = 0$ in the proof, so it's not a typo
they don't define what $0\cdot \infty$ is though
so it's still an error on their part
in fact, they mention it's ambiguous what it means
I would say so too, an error
10:09
Hi
Could you please help me with this question: math.stackexchange.com/questions/4754299/…
no folks around :-(
10:34
Statistics isn't my best field
11:07
statistics is COOL
12:06
I'm not denying that it can be
0
Q: Is there a modal modification of the law of excluded middle that may render constructive?

Dannyu NDosIntuitionistic logic rejects the law of excluded middle, and paraconsistent logic rejects the law of non-contradiction. I wondered whether the rejected laws can still be incorporated, if they're modified with modal aspects. I see an easy modification of the law of non-contradiction that is compat...

though I don't think I've seen a lot of cool stats
Constructivists help
12:27
@leslietownes resolved. f is periodic so f(2pi r gamma) = f(2pi {r gamma}); {,} is fractional part.
@PM2Ring SE introduced some new tools to detect serial voting and sock puppet voting. There are some old cases which are being dealt with. There is, for example, a user on DSP who lost over 30k XP because of this.
My guess is that some user who was engaged in inappropriate behaviour a while ago finally got caught, and their votes reversed.
30, 000 rep points?
Now ^that is worth a meta discussion
If I lost 30K Xp in a game I’d just delete it and destroy my computer
@XanderHenderson wow, and I was feeling bad about losing 50 points. ;-)
600 times more
12:40
@user726941 34770, if I recall correctly.
Yeah no I’d lose my mind over that
And this was on a site much smaller than Math SE, where the user with the most XP has around 80k.
No wonder stuff like the bit coin scam keeps getting bigger.
There's too many loop holes in the net.
Tell that to Sandra Bullock.
12:50
LoL true dat.
13:10
Am I missing something here? math.stackexchange.com/a/4754212/27978
@copper.hat it is well known the downvoter is always wrong unless they explain their downvote
Using the ring of real quaternions as a model, we define the quaternions over the integers $\text{mod}$ $p,$ $p$ an odd prime number, in exactly the same way; however, now considering all symbols of the form $a_0 + a_1i + a_2j + a_3k,$ where $a_0,a_1,a_2,a_3$ are integers mod $p.$
I have done a solution.
Here're the snaps of it.
@shintuku :-).
13:16
I'm not going to even try to follow that.
Write on only one side of the page. Or use a pencil.
latex or bust
@XanderHenderson I understand, but currently there's no more scratch papers left.
Is paper that expensive?
Scratch papers are!
13:18
@user726941 I just bought a box of paper from OfficeMax for $25.
Wow, that's great!
box has multible slabs of paper right?
5 reams. 2500 pages.
oh worth it
What the hell is that paper 🤣😂
13:20
Should I type the whole thing in latex? That'll be a tough and boring job tho but since I am the one seekin help so maybe it's worth takin the trouble...
Nothing is true until it has been TeX'd.
3
@XanderHenderson Ok, give me some time.
$neither is that statement$
I don't tex homework since this year. It saves a lot of time.
Wonder if I can ask ChatGPT to Tex it for me
13:31
it does support latex outputs
but it doesn't do good math
I started TeX'ing my homework as an undergrad. I don't think that not TeX'ing actually saves me all that much time, since I do my work on a backboard or whiteboard before copying it to something more permanent.
The time it takes to TeX is really no more than the time it takes to write it clearly by hand.
same all my draft stuff is on printer paper
I take notes on printer paper. I put an engineering pad behind it to keep my lines straight.
Or, at least, that is what I did when I was a student.
I don't have much call for note-taking anymore.
Did you ever take notes from a textbook?
@user726941 Yes.
You should always read with pen and paper.
3
13:38
Do calculations, draw sketches, and take notes.
@XanderHenderson as a moderator, what do you think about "include your results" for this type of posts? math.stackexchange.com/questions/4753940/…
@Jakobian As I've said about a million times in the past, I really don't think that "showing your work" or "including your results" does much by way of adding context. Accepting these kinds of things as context are a result of a compromise, which tolerates homework problems as long as they show some effort.
In that case, the question is pretty terse, and I don't think it's great, but I do think that an argument can be made for it providing sufficient context, without "showing work".
As I read it, there are several definitions embedded into the question, for example.
13:54
what if we divide mathSE into two. lower level vs higher level
lower level requires showing work and is for stuff up to mid undergrad
The major thing that is missing is motivation---why is this question interesting?
@shintuku It'll never fly.
You are not the first to suggest this.
re mathoverflow
All questions are held to the same standard.
yeah but mathoverflow is research level, there's stuff in between that just doesn't make sense subject to such stringent requirements
@XanderHenderson Or, perhaps, "where does this problem come from? why are you asking it?" Motivation.
13:58
and then there's the eternal contradiction between being a help website and being a database of math
Does being a mod here make you a mod on mathed.se?
Notes @XanderHenderson has no affiliation with mathoverflow :P
@shintuku that's true, we are at a weird intersection and everything is awkward
@user726941 No.
But Math Ed had an election last year, and I nominated myself in order to ensure that there were enough candidates to hold an election.
And then things went pear shaped, and I was appointed there.
I do have an MO account, but I don't use it all that often.
I work in a part of mathematics that doesn't actually get a lot of traffic on MO.
Fractals haven't been cool for 20 years.
Now its all category theory. :P
We need to work on changing that!
How about a fractal tag here?
I could ask a question about how to get high schoolers interested in it.
created 13 years ago
viewed 60 times
active 5 years, 3 months ago
5 years with no activity :(
14:21
I wish I was here much earlier
@user726941 There are several tags related to fractal geometry and analysis.
Hello
$$\frac{1}{Z(w)}\mathcal M \bigg [\sum_{r\in R}\Gamma(r,s)\bigg] (w) =\sum_{r \in R}\int_{\mathbb R^\times \cap ~(0,1)} |x|^r~f_s(x)~{dx\over |x|}=\zeta(1-w)$$
14:40
@冥王Hades When you decide to destroy your computer, I would be happy to adopt your misbehaving and traitorous computer and pay for the shipping cost to Canada. That way you are helping the environment too 😀.
I'm trying to iron out what this calculation means. On the LHS you have the Mellin transform of the Gamma factor (summed over a spectrum) premultiplied by a "Gamma factor $\frac{1}{Z(w)}$ (where $Z$ is quite literally made of Gamma functions)
and $f_s(x)$ is a handpicked Schwartz class
@XanderHenderson I have texed the whole thing.
@XanderHenderson That's a great idea.
@XanderHenderson if you have a moment please take a look at this question
0
Q: Obtaining the correct Gamma factor for this Schwartz function

geocalc33In this question I overlooked that the class $f_s(x)=e^{sT(x)}$ for $T(x)=\frac{1}{\log x},$ is Schwartz, on $x \in (0,1).$ Therefore we can obtain a functional equation, see this answer. However, I am having some difficulties obtaining the Gamma factor. I started by writing down the calculation:...

After posting "Can the Riemann Hypothesis be viewed geometrically?" I realized you can directly obtain the symmetric functional equation from any Schwartz
Using the ring of real quaternions as a model, we define the quaternions over the integers $\text{mod}$ $p,$ $p$ an odd prime number, in exactly the same way; however, now considering all symbols of the form $a_0 + a_1i + a_2j + a_3k,$ where $a_0,a_1,a_2,a_3$ are integers mod $p.$
I have solved this problem as follows:
Let $R$ be the set of all quaternions modulo $p$ where the quaternion modulo $p.$ The quaternion addition and multiplication are performed modulo $p.$

If $a,b\in R$ let $a=a_0+a_1i+a_2j+a_3k,$$b=b_0+b_1i+b_2j+b_3k,$ so $a+b=(a_0+b_0)\pmod p+(a_1+b_1)\pmod pi+(a_2+b_2)\pmod pj+(a_3+b_3)\pmod pk.$

If $a,b,c\in R$ let $a=a_0+a_1i+a_2j+a_3k,$$b=b_0+b_1i+b_2j+b_3k,$ and $c=c_0+c_1i+c_2j+c_3k,$ so $(a+b)+c=(a_0+b_0+c_0)\pmod p+(a_1+b_1+c_1)\pmod pi+(a_2+b_2+c_2)\pmod pj+(a_3+b_3+c_3)\pmod pk=a+(b+c).$
14:49
of course the so called "optimized" Gamma factor is when your Schwartz is Gaussian because it's self dual
Is this solution good to go?
and then it's funny because you take the Gamma factor sum over a certain spectrum and take the Mellin transform and you obtain $\zeta$ (this time it seems convergent only on a half plane).
@XanderHenderson Here it is. If you're interested, you may give your judgement and I'll be obliged.
Others may give their opinions as well. But I want no suggestions on any alternative approach.
I want a clear judgement about the validity of my solution.
For those who are wondering what this is all about: I posted a snap of my solution, but many suggested, I type out the whole thing Tex, and I did it.
Now, I anticipate some productive response. Thank you all!
@ThomasFinley too long didn't read
I actually anticipate a response 2
15:00
@shintuku Np
Also, one thing I missed is: The point of my post is to show that the only ideals of R are 0 and R.
15:12
This is for those folks who will write, "TL;DR". Just reposted my solution with snaps so that it doesn't look lengthy
I did my best.
And again: I wanted to prove the only idea of R is 0 and R itself.
@shintuku Ha ha, ig you'll be able to keep ur patience now, if ur interested...
@XanderHenderson they still have some physics cache if you look in the right places
Tho Hofstadter’s butterfly is pretty old now
(Also periodic billiards and the like)
@XanderHenderson Don't you mean, "Nothing is true until it has been $\TeX$'d"?
Trying to figure out how to formulate a proof for the following. Suppose I have a convex set S in R^2 with a line of symmetry L. Then the projection of S onto L is equivalent to the intersection of S with L
15:27
@robjohn brilliant
@robjohn $\TeX$ is wonderful. God bless Donald Knuth. I learned $\LaTeX$ for my CS papers. Never imagine it's still in use today.
Now clearly, the intersection is at least a subset of the projection. (The intersection projects onto itself.)
@Semiclassical Suppose that there is some point in the projection which is not in the intersection. The preimage of that point (with respect to the projection) will contain at least two points on opposite sides of the line of symmetry. The segment between these points will be perpendicular to the line of symmetry, but not contain the point on the line of symmetry. This contradicts convexity.
For the other direction, I want to say that it’s enough to observe that if $p\in S$ then its reflection $p’$ across $L$ is also in $S$ by assumption. Therefore convexity guarantees their midpoint is also in S, but this lies on L.
@Semiclassical Yeah, I just said that. :P
But the direct argument is nicer.
15:35
Hey, I was typing on mobile :P
Yeah, it’s just that I can’t quite convince myself it’s enough.
Hmm, maybe I do. The proof by contradiction makes more sense for some reason
Xander did you read it? Please don't mind, I was just curious. No issues, at all, if you aren't interested any longer.
what if xander did read it but has a suggestion on an alternative approach
I feel like my issue is that I’m forgetting whether my definition of “projection of a set onto a subspace” is equivalent to the usual one
8
Q: Can the Riemann hypothesis be viewed geometrically?

geocalc33In statistics, distributions in the exponential family are very natural to consider. One famous example is the Gaussian. It shows up in the heat equation, in probability theory, and in many other fields. While the Gaussian may be most important due to the central limit theorem, distributions in t...

No I am not claiming I've solved the RH - just exploring connections to Tao's de Bruijn constant example
I think it's similar to Tao's thing and there might be an analogous constant to de Bruijn
Mine is: a vector $v$ lies in the projection of S onto a subspace E when there exists $u\in E^\perp$ such that $p+u\in S$
15:51
@shintuku Huh?! But then he will oblige me, if he gives me a concise judgement about my approach.
He has priorities I think
He didn't respond to me either but I just am like the duck with water let it roll off me
@geocalc33 we're on the same boat ig :)
plus nobody wants to read about my crank Riemann hypothesis post
@geocalc33 a very possible reason...
@geocalc33 I don't know it, but I definitely heard a lot bout it. And I bet it's interesting! ;)
My reaction to RH Q&A is that I don’t have the expertise to distinguish valid math from nonsense, so I stay away from it
16:01
I probably need someone like Tao to answer but he's only on Mathoverflow. I'm cautious to migrate it over there
and on Mathoverflow it'll be answered in two lines like here read this 200 page reference
Can you have a look at this problem please: math.stackexchange.com/questions/4754299/…
Anybody with expertise in importance sampling monte carlo?
or Kcd might answer
16:26
@ThomasFinley yeah but what if your approach is not the one xander had in mind?
impasse
it is e.g. conceivable that your approach is less efficient and an experienced mathematician instantly recognizes the problem, and sees an alternative way of solving it without needing to work through a lesser method
Elitism based on efficiency of written solutions to exercises
or eboeowste, as an acronym
@shintuku Hmm...what about you? Do you think my procedure is a legit one?
Your proof of Riemann hypothesis? Yeah this line can be done so much more efficiently. Awful solution
i have an alternative method
16:47
@shintuku go on
I was planning to give a proof that $\eta_1$-sets contain a copy of $\mathbb{R}$, but the proof is too simple tbh
it helps to write things down though I guess
 
1 hour later…
18:17
@shintuku Approach for what? What am I supposed to know?
Suppose that we have groups $G_1 \le G_2$ and normal subgroups $N_1 \le G_1$ and $N_2 \le G_2$ such that $N_1 \le N_2$. Is it true that $|G_2 : G_1| = |N_2 : N_1|$?
I don't see why that would be true. Are there missing hypotheses?
Yeah, I don't see why either. I was hoping it was true. I don't think there are any missing hypotheses.
Then why do you ask the question?
What made you think that it could be true?
Just take $N_1=\{e\}$. Is it true?
18:24
Because I was hoping it was true. For the special case when we have a semidirect product decomposition of both $G_1$ and $G_2$ as $G_1 = N_1H$ and $G_2 = N_2 H$ for some subgroup $H \le G_1$, it is true. I was hoping it was true without the semidirect product decomposition (i.e., short exact sequences which don't necessarily split).
Oh, yeah...that's a good case to consider. D'oh!
18:43
$|G_2:G_1||G_1:N_1| = |G_2:N_2||N_2:N_1|$
the relation between $|G_2:G_1|$ and $|N_2:N_1|$ depends on the index of groups $N_1, N_2$ in $G_1, G_2$ respectively as well
@user193319 in your case we should have $|G_1:N_1| = |G_2:N_2|$
when those numbers are finite we'll obtain $|G_2:G_1| = |N_2:N_1|$
19:16
Welcome back to MSE @amwhy ! The ban was indeed really long and we really missed your presence here on MSE and meta a lot! We hope you are fully energetic now to make this place great again!!
M(MSE)GA= Make (MSE) great again
Today is truly a day to remember. We are finally going to be liberated!
No great again references, thank you. Irony drips.
19:49
how to find 3 linearly independent tangent vector fields on S^3?
I can find a non vanishing vector tangent vector field to S^{2n+1}.
I tried with the following: X1(a,b,c,d)= (c,0,-a,0), X2(a,b,c,d)=(-b,a,0,0), X3(a,b,c,d)= (-d,0,0,a)
But these are not linearly independent when a=0.
Use the fact that it is a Lie group.
is there any other way? As Lie groups haven't been introduced yet in the course, and I only know definition of Lie groups.
Yay! Only one of my classes got cut!
And it is the calculus class, which means that I am down to only one prep.
Though I would have rather kept the calc class. :/
cut in ...
I should have written a=x_1, b=x_2, c=y_1, d=y_2
it seems straightforward but I'm missing something: I just have to find 3 independent vectors in the tangent space of S^3 at a point p.
It is possible because the space is of dim 3.
but writing the basis elements explicitly seems ...
@newbie105 amWhy is finally coming back?
20:30
You don’t need much. You just need to think of $S^3$ as unit quaternions
The question is, @koro, why is your naive approach going to work only in dim 3.
@XanderHenderson I could never stand teaching two sections of the same course. The few times I substituted for someone who was ill it drove me nuts. I could never remember what I had said in which class.
@TedShifrin I've never had that problem, but you are not the only person who has had that complaint.
But I have a ton of committee work this semester, and am trying to get materials put together to teach this class sans book, so anything I can do to reduce my workload in other areas is good.
Hi guys.
hello Samir
Can anyone please have a look at this question math.stackexchange.com/questions/4754299/…
@Jakobian Can you have a quick look please and give me some hints?
@XanderHenderson I totally get this.
20:42
I think this is the same one from before? I don't think many people knowledgeable in statistics come around here... maybe copper or someone
For stat questions, you’re better off positing on the stat SE.
Anyone here with statistics background?
Our one stat-oriented person has not hung out here in years.
I mean technically your question probably boils down to some calculation
but I'm too lazy to try and have a go at it
I'd rather learn something useless
@Jakobian it is more intuition than calculation I believe
@TedShifrin bad luck of mine
20:48
in my statistics class it usually boiled down to few explicit calculations
you should make use of the fact that you know your variables are Bernoulli
of course, I won't help you with this
21:02
@TedShifrin Oh, it also turns out that one of my three precalc sections is asynchronous, which means that I only have to teach the material twice, then remember to post the videos to the third section.
@PM2Ring I don't really like the version by Tori Amos, don't get me wrong it's a piece of art, but it feels a bit too slow. The song by Patti Smith is I think, the same one I was thinking about
@PM2Ring yeah the guitar in this song is pretty good, and I definitely felt like the vocalist voice is a bit high
you can get used to it though
I like it
@Koro if R^n is a division algebra with two sided identity e, extending e to a basis v_2, ..., v_n, then x mapsto d/dt|_{t = 0} (x * (e + tv_j)) will be n-1 linearly independent vector fields on S^{n-1}.
where * is the multiplication that makes R^n a division algebra.
Ew... you're spilling algebra all over the place!
Needs more epsilon!
And it needs TeX
@PM2Ring not sure why they're singing about coffee, cats and people. Weird song
21:13
What I find funniest about that sketch is that, after seeing it, I can never unhear the cowbell in The Reaper. Never noticed before the SNL sketch.
@Jakobian This, I like. It isn't something I'd seek out, but I like it.
@Jakobian I feel a little more "Meh" about that one.
actually three songs are better so here
@XanderHenderson this one is more psychedelic, you might like this other piece by Coil:
@Jakobian Oh, very yes.
coil is more on the side of experimental music
@Jakobian That's not necessarily a bad thing, as long as it is not post-modernist atonal crap.
@Jakobian Hrm... so... I appreciate it (genuinely---there is some really interesting stuff happening in there---it has a very "found art" vibe in the way it uses samples, and it seems to find some musicality in some very strange and distorted sounds---and the sort of video-gamey melody early on, and the jazz trumpet later seem designed to evoke nostalgia)...
but it is not something that I want to sit down and listen too. I am glad to have taken the time to listen, but don't feel compelled to listen again.
21:52
I usually listen to this one if I'm in the mood for coil:
Going another direction:
I've posted way too many songs in the chat for now. Got to stop
@Jakobian I given to understand that one's musical tastes tend to stop evolving around age 30 or 35. I work hard to keep that from happening to me. So I generally appreciate links to music I've not heard before. Though, as you say, this might not be the right place to spam it. :D
@XanderHenderson I don't know how to describe it but I like it
those moments of breaks between notes create a sense of suspension
@Jakobian Most of what they do is interesting. In interviews, I've heard Katie Stelmanis (the woman you hear singing) talk about how (1) she doesn't really think that her voice is all that good in a traditional sense, and (2) the goal of their music is to create a sound or mood, and not to tell a story or any such nonsense. Hence the lyrics are mostly nonsense, which are meant to fill out a sonic palette, rather than convey any concrete idea.
Personally, I really like her voice, but I have a thing for rough and smokey altos, so that could just be a very personal bias.
22:02
I wonder what's the Kolmogorov complexity of the Universe.
@XanderHenderson Yes. I got stuck at folk songs.
22:15
> Example 10.12 The power series $$\sum_{n=0}^\infty (n!)x^n$$ has radius of convergence $$R=\lim_{n\to\infty}\frac{n!}{(n+1)!}=\lim_{n\to\infty}\frac{1}{(n+1)}=0,$$ so it converges only for $x=0$. If $x\neq0$, its terms grow larger once $n>1/|x|$ and $|(n!)x^n|\to\infty$ as $n\to\infty$.
I'd be very grateful if someone could explain why the "the terms grow larger once $n>1/|x|$"? Why is that?
if you let a_n = n! x^n, i think all they mean by that is that |a_n| > |a_{n-1}|, i.e., |a_n/a_{n-1}| is larger than 1, when n > 1/|x|. because |a_n/a_{n-1}| = n |x|.
and in general is it a good idea to replace what you are actually saying with statements about "the terms growing larger"? in my view, no it is not.
@sunny Because $n! > x^{-n}$ (approximately, when $x\approx 0$).
you should preface that with a medical warning: strobing edit effects.
Yeah, sorry---nothing is correct until TeX'd, because you see the errors when it's all typeset nicely. :D
which reminds me that i forgot to star that.
22:21
@sunny The basic idea is that once $n$ is larger than $1/|x|$, $n!$ has a bunch of factors which are (potentially MUCH) larger than $1/|x|$, while $1/|x|^n$ has only $n$ factors, all of which are exactly as big as $1/|x|$.
That's the intuition, anyway.
If you want to prove it, the ratio test should get the job done, I think...
and that's more of the substance of their second statement, that n! x^n goes to infinity. which is true but definitely does not follow from "the terms growing larger".
which is another reason not to replace what you are actually saying with prose approximations.
in conclusion, xander and i should have written Example 10.12.
thank you for your attention
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)! x^n}{n! x^{n+1}} = \frac{n+1}{x}. $$
This "clearly" diverges, since $x$ is a fixed constant, and $n+1 \to \infty$.
Slap some absolute values on there, if you want to be all rigorous and such.
thank you @leslietownes and @XanderHenderson :) I will have to process this
Or you can just prove Stirling, then use that to approximate $n!$, and write $n!x^n \approx \left( \frac{n}{\mathrm{e}x} \right)^n $, and again note that this grows without bound.
But that might be circular.
I don't remember how to prove Stirling right off the top of my head.
what does approx mean there? isn't there a bonus factor of sqrt(n) somewhere?
but now we're just freestyling.
22:26
@leslietownes Oh, right.
Yeah... I don't even remember Stirling, it seems. :D
$n!x^n \approx \sqrt{2\pi n} \left( \frac{n}{\mathrm{e}x} \right)$. I think.
Lemme Google...
I know there's pie in there somewhere...
n! > (n/3)^n is easy enough to prove with ratios and such without the full force of stirling. people always want to jump straight to stirling.
it's like l'hopital's rule, but for people who should know better.
@leslietownes Hey, my first instinct was the ratio test.
And I introduced Stirling as obvious overkill (and probably circular, anyway). :D
i'm trying to think of a more ornate approach than stirling (other than stirling with much more detailed asymptotics).
maybe there's some route by which GRH implies that n! x^n goes to infinity.
@leslietownes I love it when students try to prove that $\sin' = \cos$ by invoking L'Hospital.
@leslietownes If RH is true, [this proof] shows that $n! x^n \to \infty$. If RH is false, [this other proof] obtains the result.
@XanderHenderson close, $n!x^n\sim \sqrt{2\pi n} \left(\frac{nx}{e}\right)^ne^{-\frac{1}{12n}}$ iirc
22:39
@Jakobian I was giving only the asymptotic estimate. I believe that the extra factor you suggest gives an inequality which holds for all $n$.
That is, you can bound $n! x^n$ by $$\sqrt{2\pi n} \left(\frac{nx}{e}\right)^ne^{-\frac{1}{12n+1}} < n! x^n < \sqrt{2\pi n} \left(\frac{nx}{e}\right)^ne^{-\frac{1}{12n}} $$ for all $n \in \mathbb{N}$.
Which is even more swatting gnats with nukes than invoking Stirling in the first place.
The more salient error is that i put my $x$ in the denominator, because I was still thinking $1/x$. But no one noticed THAT.
@XanderHenderson let's add a few more terms just to be sure the fly is dead
It's dead, Jim.
23:39
@XanderHenderson Thanos: “I used the stones to destroy the stones”

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