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12:10 AM
Riemann Rock is a subgenre of indie rock that had a lot of underground popularity in the late 1990s
 
Ah yes, wheels
I don't think those are even in common algebraic structures at wiki or related
not here, nor on any of the sites wiki directs to on that page
 
"Unneeded" is an ugly word
 
but I did hear about them before. Just as a curiosity
and I do dabble a bit sometimes with more exotic structures as well as universal algebra
so I can see someone not hearing about it, definitely not common knowledge
 
12:33 AM
@EdwardEvans compsci is easy credits
And double the careerwise benefit
 
12:57 AM
george bergman has a good book on universal algebra. i recommend it if you haven't seen
 
1:12 AM
Is there an example that for integral domains R, S, for each prime ideal $I\subset R$, there are infinitely many prime ideals $J\subset S$ such that $I = R\cap J$??
 
I assume you want $R\subset S$?
consider $k\subset k[x]$
 
yes yes sorry
k is a field?
 
ye
 
Ah, k has only zero ideal so any prime ideal of k[x] satisfy that condition
zero ideal as prime
 
indeed
 
1:23 AM
Thanks!
 
np
 
@leslietownes I do know about it, but I read Sankappanavars book before I found it, and I wasn't that interested in reading it
but I heard Bergmans book is really good
 
1:39 AM
i used some ideas from it in my thesis. i was a teaching assistant for bergman a few times in the late 1990s. his students tended not to like him personally very much, but they loved his materials. the book is a good book.
 
Any ideas on a simple function $h$ such that $h(z+1) = z^{-1}h(z)$?
Kind of like $h(z) = e^{az}$ has $h(z+1) = e^a h(z)$
 
Sounds like $1/\Gamma$.
 
Does it? I figured out the previous question by the way, Ted. It was simple and I was overcomplicating it. All you needed to do was set $f(z) = e^{-az}g(z)$ and it works.
But now I am on a slight extension of this but I can't figure it out. :P
namely given a polynomial p(z), I want to find f such that $f(z+1) = p(z)f(z)$.
(entire f)
Constants are easy to do given what I wrote above.
 
Factor $p$ and then use a product of shifts of $\Gamma$?
 
1:57 AM
that's a clever idea
 
But I can't use the gamma function.
But I am guessing if it works for the gamma function then we can do it with my solution for f for my previous question.
In what way would I shift the Gamma function for it? I don't think shifting the argument works.
 
But why does one want this?
 
I wish I could tell you, Mike. I really do wish I could.
But exercises are exercises, and sometimes they just seem weird.
 
Fair enough, but not enough to hook me on the problem!
 
places Mike's favourite snack on top of the exercise.
By the way, what is everyone's favourite food? I need something new to cook.
 
2:13 AM
Roast duck :)
 
Have you always enjoyed roast duck, or is this a taste you acquired when you were older?
 
Always, and thai spicy duck curry, yum.
 
With a long family history of heart issues I think I'll pass on the duck most days.
 
Yeah, I have heart disease, so it's a rare treat. But, properly cooked, you cook off all the fat .
 
i like sashimi. not sure that it is something to cook. i am fantasizing about ordering it tomorrow
 
2:19 AM
Simple dish: poulet au diable.
You have to buy super fresh sushi-grade fish.
I just got some yellowtail tuna and cooked it rare.
 
that sounds delicious. i love yellowtail
 
Great fish in San Diego.
 
yes, san diego is great
bear flag seafood in newport beach is a good place. i don't know what it's like now. they made another one south of newport and another one in huntington. very high quality
i wish restaurants would come back. it's coming up on a year since i've been in an office
 
blanking on regular ole logic here @anakhro math.stackexchange.com/questions/4048001/…
 
I don't know anything about that question, sorry BigSocks. :(
 
2:33 AM
no stress- kind of a tricky one
 
Can anyone offer help on how to show that $a_n < 1$ where
$$a_n = \sum_{k=n+1}^{2n} \frac 1k.$$
 
It's easy. Do the most basic estimate.
 
2:47 AM
What is 'the most basic estimate'?
 
What is the largest term?
 
1/(n+1)
 
So what's the largest the sum could be?
 
1/(n+1) * (2n-n)?
 
Which is?
 
2:49 AM
@TedShifrin That works. You're right
@TedShifrin ...less than 1
 
Yippee.
 
Hello @Ted
 
For the same
$$a_n = \sum_{n+1}^{2n} \frac 1k$$
is it sufficient to say that $a_1=1/2$ and $a_n = a_1 + \text{positive terms}$ to prove that $a_n \geq \frac 12$?
 
3:12 AM
i agree that a_1 is 1/2. i don't see where you're finding a_n = a_1 + positive terms. i'd be tempted to estimate the thing with an integral. there must be something simpler
ok, a_n is a sum of n terms. the smallest is 1/2n. not a lot going on there
i'd still be tempted to use integrals
 
@leslietownes, well, the goal is to estimate so that we can use integrals to show that $a_n \to \ln(2)$ as $n \to \infty$.
So we can't really use integrals
 
@Jeff That's sufficient, yes.
Wait, hang on, I think I misread the thing.
 
@Fargle OK, waiting....
 
It's not true that $a_n = a_1 + \textrm{positive terms}$, but as Leslie noted, $a_n$ is a sum of $n$ things, each of which are at least as big as $1/2n$. Therefore their sum is at least as big as $\frac{n}{2n} = \frac{1}{2}$.
Like, $a_n$ is indeed bigger than $a_1$, but it doesn't break down into a sum of $a_1$ and other stuff in an obvious way. Leslie's proposed approximation is the right one here.
 
Oh wait, that's right. I was confusing $a_n$ with $\sum a_n$ (which is mentioned in the problem, but not relevant to this part).
So one thing I'm noticing is that $a_n = a_{n-1} - \frac 1n + \frac{1}{2n-1} + \frac{1}{2n} = a_{n-1} + \frac{1}{2n(2n-1)}$ (previous term, less the first fraction, plus the last two). So then $a_n is always increasing.
 
3:33 AM
Suppose $f(z) = ℘(z) − ℘(u)$, where $℘$ is the Weierstraß $℘$-function. Then $f$ has double zero at $z=u \bmod{L}$ if $2u \in L$, where $L$ is the lattice generated by $\left\{\lambda_1, \lambda_2\right\}$.
How do I construct an elliptic function with a double pole $z=0$ and simple zeros at $\frac{\lambda_2}{3}$ and $\frac{2\lambda_2}{3}?$
I think $f(z) = ℘(z) − ℘(\frac{\lambda_2}3)$ has a double pole at $z=0$ and a double zero at $z=\frac{\lambda_2}{3}$ and $g(z) = ℘(z) − ℘(\frac{2\lambda_2}3)$ has a double pole at $z=0$ and a double zero at $z=\frac{2\lambda_2}{3}$. Then maybe $\frac{f(z)g(z)}{(z-\lambda/2)(z-2\lambda/3)}$ is the required function?
 
@Jeff That would be sufficient for the reasoning as well.
 
@fargle Thought so. TY
 
In trying to remove the extra zeros I increased the poles I think.
I would appreciate if someone could give me a source that gives examples of how to construct elliptic functions from poles and zeros
 
3:54 AM
@TedShifrin: you're following me...
 
@robjohn Where are we going?
 
@TedShifrin I don't know. I floated into the room, and then you floated in after.
 
Ah ... Well, I don't expect to float anywhere.
 
Faculty and staff at UCLA have been green-lighted for vaccination.
one of my friends at the library got theirs today
 
Great!
 
4:00 AM
he had a bad reaction, but I bet he'll be better tomorrow
he had bad flu symptoms
 
Hi @TedShifrin
Hi @robjohn
 
hey there
 
my wife, also an educator, had some symptoms from the first round of the two-shot vaccine. mostly a very sore arm. apparently the second shot can be worse
 
Yes, second is typically worse. Sore arm is totally standard.
Hi Karim..
 
Any more interesting stories from the good ol' days at Cal between the three of you? Been fun reading the small anecdotes you guys have been discussing as I snoop around and eavesdrop...................
 
4:10 AM
@leslietownes My mom had the same reaction to the first shot, but not to the second.
 
Different Cals. Two Berkeley, two UCLA ....
 
@TedShifrin Really? That's opposite of my mom's experience.
 
What I said seems to be typical.
 
i've got one. grigory barenblatt, a student of kolmogorov and a gifted lecturer, has now passed on. he always reminded me of a james bond villain. one time i was in the elevator and he got on, and pointed at me, and said "your system is incomplete." i didn't know what he was talking about. i thought he was being very figurative. then my officemate told me that the top button on my shirt was undone.
 
I had chills, fever, headache after second.
 
4:11 AM
@TedShifrin I want to study something associated with smooth projective surface.
 
Lol
 
What is the best way to consider a family ? I didn't learn about this stuff yet. So far I have in mind hilbert schemes and spreading the variety
i.e varrying the coefficients and constructing $f : \mathcal{X} \rightarrow S$ where S is the original variety.
 
I don't know what you're talking about, karim. I've said that before. So you need other sources, not me.
Remember that in general deformations may be non-algebraic, even non-Kähler.
 
alright
 
did he pass on recently @leslietownes? or was that just stating where he is now?
 
4:15 AM
So just perturbing coefficients is not a good generic family.
 
right
yeah exactly this is something that I thought about. I want the best way to generate a family
 
he passed a few years ago. i just think of him whenever i fail to button my shirt all the way. someday i hope to tell somebody else that their system is incomplete
 
You can compute dimensions and see that.
 
right
 
This is 50+ years old stuff.
 
4:18 AM
I see
 
holy shit vaughan jones died. i somehow missed that. what an awful year
 
ok I have some idea
 
0
Q: Criterion of map $M_f\to Z$ to be continuous

love_sodamI was about to prove the following proposition A map $M_f\to Z$ is continuous if and only if the induced maps $X\times I\to Z$ and $Y\to Z$ are both continuous. Here, $M_f$ is a mapping cylinder of $f:X\to Y$ i.e. $M_f = X\times I\sqcup Y/f(x)\sim (x,0)$ for $x\in X$. I first let $p:X\to X\time...

Anyone can check if my proof is valid?
Bredon says 'the proof is an easy application of the definition of the quotient topology on a mapping cylinder' but my proof seems not very easy or trivial.
 
A while ago, @leslie.
 
4:33 AM
Is there any simper proof?
 
yeah i've been distracted by work stuff, it's like a whole year of news is still news to me. and none of it is good
 
5:25 AM
Better not to catch up!
 
my daughter has no idea how weird all of this has been. she puts on a mask and goes to day care and has never been in the same room as her grandparents
it truly is better not to catch up
 
 
1 hour later…
6:33 AM
if infinite monkeys are kept at a place with a typewriter
then atleast one of them will start by writing the proof of the riemann hypothesis
in fact infinitely many will start by writing the proof of RH
but not all
 
 
1 hour later…
7:37 AM
I am having trouble simplifying algebraic expressions in c++
my code is 150 lines long and it is still not completed :(
 
8:08 AM
Any hints guys!
 
 
1 hour later…
9:25 AM
Hello !! Which of the transformations rotation, translation, line reflection, glide reflection has no invariant points?
 
9:57 AM
Proving $M_f$ is T_2 when two given spaces are $T_2$ is harder than I first thought...
 
10:20 AM
@LeakyNun Hi I have a question about logic
 
just ask it
 
10:32 AM
can you give example of a theory
that is consistent , complete but not deductivly closed?
@LeakyNun I saw you on the logic forum haha that is why I'm asking you
am proving this statement
a theory is maximally consistent if and only if it is consistent , complete and deductively closed,
but in the reserve direction, there is no need for deductive closure
 
@JackOhara suppose T is a consistent complete theory. suppose T proves phi. By completeness, either phi in T or (not phi) in T. but if (not phi) in T then T isn't consistent. so phi must be in T. so T is deductively closed
 
11:05 AM
1
Q: Mapping cylinder is Hausdorff

love_sodam If $f:X\to Y$ is a map between Hausdorff spaces, then $M_f$ and $C_f$ are Hausdorff. Here, $M_f$ and $C_f$ are mapping cyliner and mapping cone i.e. $M_f = X\times I\sqcup Y/(x,0)\sim f(x)$ for $x\in X$ and $C_f = M_f/(X\times\{1\})$ I first tried to prove $M_f$ is $T_2$. Before, as $X$ and $Y$...

Anyone give me some hint of this problem?
 
11:28 AM
Guys . I am going to start making notes for physics and chemistry on IPAD using Apple pen. It saves a lot of time. But it will hurt eyes more. Do you think I should do it ?
 
@Sarabsrimt isn't there software that can take your handwritten notes into some better format?
 
@robjohn I didn’t get you exactly ?
@robjohn There is an app on iPad that is really good for handwriting notes
Also , I can arrange things better and write things that I may have left on the same using iPad
 
@Sarabsrimt yeah, it allows you to write your notes with the pen and then convert it to text so that it is easier to read
 
My only fear is that people say kids who wrote on notebooks are better in remembrance than kids on digital media.
Better in learning , thinking also
 
 
1 hour later…
12:37 PM
are we learning the same programm of math all over the world of the same years
 
No
People learn at different pace in different countries
 
Even in one country the curriculum can vary quite widely between institutions.
 
12:55 PM
Am I correct in saying this series converges only for real $s>0?$ $$ \rho(s)=\sum_{\rho\in\Bbb P} e^{-\rho^s} $$
 
man how should we know you're using two different rhos in that series and dont even specify what $\Bbb P$ is. Are those the primes?
 
I used one rho
 
yes, I mean semantically different rhos
 
$\Bbb P$ is the set of primes
 
is s real?
 
1:00 PM
$s \in \Bbb C$
If $\Im s\ne0$ then I don't think $\rho(s)$ converges
 
$s=a+ib$, $p^s = p^a e^{ib\log p}$, $||e^{-p^s}|| = e^{-p^a\cos({b\log p)}}$
I'm willing to bet that $\cos(b\log p)$ is dense in [-1,1] when $b\ne0$, which means the general term is not bounded
 
so when you want to see if you can extend a series such as $\rho(s)$ do you let $s$ be a complex number
and if $\rho(s)$ doesn't converge for $\Im s \ne 0$ then it can't be analytically extended?
$$P(s)=\sum_{p\in\Bbb P} p^{-s}$$
this function has a "natural boundary" at $\Re s=0$
and $P(s)$ converges for $s>1$
 
1:22 PM
Consider the Gaussian and Cauchy-Lorentzian distribution as defined in the screenshot:
They're defined in terms of the full width half maximum.
When only changing the full width half maximum, that is $\Gamma_{L,G}$, will these distributions still be normalized (so that the integral from negative infinity to positive infinity equals 1)?
 
@LeakyNun thanks but that is not what am trying to prove
in my prove no need for deductive closure is needed
i mean assuming complete and consistent we can get to max consistent , but not sure if there is something I missed or not
 
1:39 PM
Why do we call this Klein bottle even if it's not Klein bottle?
 
This is an immersion of the Klein bottle into R^3.
From this one can picture the construction (paste ends of a cylinder together in an orientation-reversing way).
Or, if you are willing to allow yourself color as an extra dimension, from this picture one immediately can see an embedded Klein bottle in R^4.
In addition usually Klein bottles are discussed for popular audiences. I am not certain you can convince a popular audience of an object which they cannot "see".
 
1:52 PM
Is the sum notation defined for $\sum_{k=1}^{-1}a$?
 
usually it's defined to be zero
 
okay. Thanks
 
I would say: $\sum_{k=a}^b f(k)$ is the sum "For all $a \leq k \leq b$, evaluate $f(k)$, then add up all the results."

When $a > b$ there is no such $k$. Therefore you are adding a sum over no numbers. Traditionally the "empty sum" is defined to be zero.
After all, if I have already added a bunch of numbers, and then I add no additional numbers, that's the same as adding 0.
 
@MikeMiller So say I have an isotopy MxI->N. Switch to the track map MxI->NxI, now this becomes an embdding, so, by keeping the first component fixed and varying the time t, we obtain a bunch of "flowlines" on the embedded copy of MxI in NxI. I wanted to extend these to a vector field on NxI, then flow along that and use that to construct a diffeotopy extending the isotopy, but I haven't managed to do so. Is this at least a sensible idea or am I completely off?
 
That's the right idea
Your field on F(M x I) subset N x I has a special property you need to demand on your extended vector field
But you've now managed to do the clever part
The rest is small technicalities
 
2:01 PM
hmm, I probably want to demand it is constant 1 in the TI factor of T(NxI)
that should give me control on the second component of the flow, so that I can forget about it again
 
You got it, the point of course being that $\Phi^t(m, s) = (\Phi^t(m), s+t)$ where your flow is already defined
 
@JackOhara what do you want to prove then
 
I answered a question in English Language & Usage and then I saw almost the same question in Mathematics. Am I allowed to answer the question in Mathematics in the same way as I answered the question in ELU?
 
2:34 PM
I don't see why not
 
@LeakyNun I think assuming the theory is consistent and complete is enough to show maximally consistent
sorry for late replies, somehow I don't get notifications when people text me
 
@JackOhara yeah deductively closed is redundant here. Suppose a theory is complete and consistent. Let $\phi$ be any sentence. Suppose $T \vdash \phi$. Since $T$ is complete, either $\phi \in T$ or $\neg \phi \in T$, so $T \vdash \phi$ or $T \vdash \neg \phi$. The second one is impossible by consistency so $\phi \in T$ must hold
 
Hey, I was just wondering. In Linear Algebra, why does one want minimum norm solution ?
 
for statistics estimators? for example
 
Yeah
Given a least squares regression problem
Why does one want the solution to the minimum norm solution?
 
2:47 PM
to have the best regression possible
 
Let us say $Ax = b$ , where $A$ is rank-deficient
But what if you minimize $||Ax-b||^2$ and you obtain an $x$ solution with a higher norm
Is it then a worse regression ?
Where there is an answer in the comments, but the answer says that there is a nice general explanation. So that got me wondering, does the minimum norm imply a better fit for a regression problem?
 
is there an s-dimensional ball in R^n
something that would be a natural set of Hausdorff dimension s
 
the unit ball from R^s?
 
3:03 PM
no
for any positive s<n
 
I dunno about "natural" and it certainly wouldn't deserve the name unit ball
But I agree it would be nice to see an explicit construction for all s
Especially if one can make it continuous in s
 
@user2103480 I see thanks ! I argued in a similar way
 
@Jakobian This would be a nice MSE question. Write C(R^n) for the closed subsets of R^n, considered as metric space under Hausdorff distance. (Hausdorff dimension is not continuous in this space but maybe there is a variant where it is.) Can one find a continuous map F: [0,n] -> C(R^n) so that F(s) has Hausdorff dimension s?
Another natural question: is there a good topology on [some kind of] subsets of R^n so that Hausdorff dimension is continuous? This is a soft question since someone dull might answer "the discrete topology" and the only thing this violates is that it's obviously not what I mean.
I guess smoothly varying submanifolds ought to define continuous variations in whatever this space is.
 
easiest thing coming to mind would probably be some Cantor-type constructions
 
3:20 PM
I'm not terribly familiar with Hausdorff dimension so I don't have a sense for how plausible this ought to be
Notice how easy it is to construct an example of dimension-dropping: take [0,1]^{n-1} x [0,t]. This is certainly continuous in the Hausdorff topology and in fact it's hard for me to imagine a non-stupid topology where it's not
But that seems more related to my lack of imagination than anything else
 
honestly all I was looking for is some natural example of set of dimension s, so that I can use it to show that the approximate Hausdorff measure doesn't satisfy the Caratheodory condition for Borel measures
the one which Hausdorff measure is limit of
 
Yeah but what does natural mean
I'm trying to encode that in terms of some continuous variation
 
just something I could work with
 
So more like "explicit" I guess
Anyway I was outlining a question above, not something I have any answers to :)
 
that was interesting though
 
3:27 PM
there are some topologies out there crying because they were just called stupid. i hope you're happy.
 
They probably needed a wake up call.
 
hausdorff and i share a birthday. with frege
 
I wonder who was the last person to not share a birthday with someone
 
3:42 PM
one time i was teaching probability and began with the birthday paradox. "for example, i'm born on november 8." by some strange coincidence, two people in the room were also born on november 8. i think this may have conveyed the wrong message re the birthday paradox. "in a group of 30 people, for any date you choose, there will be three people born on that date."
 
I remember one of my profs explaining the birthday paradox
 
@Astyx was "I don't see why not" to me?
 
Yes
 
Do you know where I can read a rule about that? I couldn't find any information about it.
 
You could have a look around math.stackexchange.com/help
As long as what you post is in good faith there's no issue AFAIK
If your posts are against the rules, someone will point it out and you'll know why
Don't stress about it
You could link to the original answer
 

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