Is there any name for this curve? I searched but I can't find.

something between epicycloid and hypocycloid

The main site is currently in read mode. Is anyone else facing the same problem?

meta.stackexchange.com/questions/370331/…

it is for planned maintenance. no announcement on the stackstatus twitter for some reason

@hyper-neutrino: thanks for the link. It appears the announcement was given on a short notice.

yes, it was quite short, and it was only featured like, an hour ago

@ParamanandSingh yes, I'm facing the same issue.

it was indeed quite short, but it seems like for the first few minutes the chat link led me to area 51

area 51 stack exchange, not the actual place

Let $X$ be a metric space. Does connectedness of $Y\subset X$ depend upon the space, it is embedded in?

@Koro The topology is induced from $X$, so yes.

I tried to parametrize the above picture chat.stackexchange.com/transcript/message/59254376#59254376 , I let the radius of big circle $R$ and small circle $r$, black point $P$, center of the small ball $C$, the origin $O$ and the angle that $C$ rotates from the $x$-axis $\theta$. I get $\vec{OP}= \vec{OC}+\vec{CP}$. I think $\vec{OC} = (R\cos\theta,R\sin\theta)$ and $\vec{CP} = (r\cos(\theta+\alpha),r\sin(\theta+\alpha))$ where $\alpha$ is an angle that $P$ rotates minus $\theta$.

Now I tried to remove $\alpha$ and represent the curve only on $\theta$. To do that, I need some relation between $\theta$ and $\alpha$ but I can't see.

You haven’t remotely specified how the little circle moves/turns. It is no longer rolling without slipping along the large circle. Very ill-defined.

if all we have is the graph to go from, looks like it might rotate three times when you go around theta once. so maybe try alpha(theta) = 3 theta + [some fixed phase]?

random guesswork really

Boo @leslie

I think that's nice deduction. I just draw the graph taking $\alpha = 3\theta$ and it shows the right picture

wolframalpha.com/input/…*cos%284*t+%2B+pi%29%2Csin%28t%29%2B%281%2F3%29*sin%284*t%2Bpi%29%29

ok, that somehow breaks its URL detector

yeah chat syntax is... questionable at best

tinyurl.com/yuv87x35

you'd probably be better off just titling the link using [](), or ^ is probably a better idea :p

i'm feeding "parametric plot (cos(t) + (1/3)*cos(4*t + pi),sin(t)+(1/3)*sin(4*t+pi))" into wolfram alpha, if you fiddle with the parameters you could tune it more finely

Anyhow, totally ill-posed question, @love_sodam.

although i agree with ted, i think the problem becomes somewhat posed if one assumes the goal to be only finding a parametric curve that has a similar-looking graph.

but we all know what happens when i assume.

Hypo- and epicycloid are defined by rolling without slipping and respective radii.

i didn't think this had anything to do with that. earlier someone was asking about epicycles, which i think are mere sums of circles of different radii.

How can I describe the above curve then?

or components of such sums. center of the smaller circle just travels the path of the bigger one.

love, what do you want to describe it for?

No clue. You have to specify the linear and rotational motion of the little circle.

i hope we aren't landing a spacecraft with this.

How about just saying that every time the center of small circle rotates $2\pi/3$, $P$ rotates $2\pi$ with constant velocity

what criteria are you using to assess the adequacy of that option over others?

it's not clear to me whether that is an intended definition of the curve (with only a parametrization to be found in terms of geometric inputs), or a candidate for a curve that is supposed to model the picture in some way

Professor Ted, my confusion arose from answer to this question: math.stackexchange.com/questions/4263028/…

If such continuous $f$ exists then range f=$\mathbb Q$, and $Q$ is connected in $\mathbb Q$? so no contradiction?

I think if connectedness of Y depends on space, it is embedded in then the answer makes sense only if Q is regarded as a subset in $\mathbb R$?

It’s totally disconnected. Where do you get connected?

The topology is the induced or subspace topology, not the indiscrete topology.

Definition of connected set that I am using: A set $Y\subset X$ is said to be connected in X if Y is not a union of two non empty separated subsets (A and B) of X,( i.e. $A$ and $B$ are such that $A\cap cl (B)=\emptyset$ and $cl(A)\cap B=\emptyset$). With this definition, Let X=Q with the usual metric $d(x,y)=|x-y|$. Now if $Y=Q$ suppose that Y is connected in X then there exist A and B (subsets of X) such that $Y=A\cup B=Q$ and $A\cap cl (B)=\emptyset,$ and $cl(A) \cap B=\emptyset$

So if Q is not connected in Q then how can I get a contradiction here?

professor Ted, sadly I don't know about induced/indiscrete topology yet :(

Do I need to know those for knowing that Q is connected in Q or not?

By the symbol $cl (T)$, I mean union of U with set of all limit points of U.

I made a mistake after "Now". I mean that: suppose on the contrary that $Y$ is connected in $X$. It follows that Y can't be a union of two separated sets.

The answer here (math.stackexchange.com/questions/4263028/…) is deleted now.

You can take $A = (-\infty,\sqrt 2)$ and $B=(\sqrt 2,\infty)$. Ergo, disconnected.

@TedShifrin I taught discrete math for years; indiscrete math sends shivers down my spine...

yes, it works for any irrational actually. Thanks a lot professor Ted. :)

quick question....what would be a standard basis for $\mathbb{C}^2$?

dc3rd: {(1,0), (0,1)}

nooooo

?

I think that's basis of C

for $\mathbb{C} ^ 2$?

not $\mathbb{C}$.

that's the one people have in mind when they say 'the standard basis of F^n' with F a field and n a positive integer

{((1,0),(0,0))... etc. dc3rd

i'm assuming you're regarding C^2 as a complex vector space here

if you're regarding it as a real vector space, a common basis would be (1.0), (i,0), (0,1), (0,i) but i don't know that you'd use the term 'standard' basis in that context

Basis of C^2 ={((1,0), (0,0)), ((0,1), (0,0)), ((0,0),(0,1)),((0,0),(1,0))}

assuming over field R

right?

So I had intially thought of using $e_{1}, \cdots e_{4} $

ok, they're looking at C as a complex vector space there

one hint that this is the case is that g takes values in F and by definition g is complex valued

C^2 is a two-dimensional vector space over C

actually where you talked about it in terms of a real vector space I was thinking about it like that, but then I couldn't see how that could be a basis....

and over R, 4 dimensional right?

ah I see what you're saying leslie. because in this case my "sclalar" is just complex values

valued

(1,0) and (i,0) are not linearly independent elements of C^2 (regarded as a complex vector space) because (i,0) = i(1,0) is a scalar multiple of (1,0)

yes, exactly

you mean over R, 4 dimensional. is my understanding correct, Leslie?

oh.....I just glanced and my eyes fell in the trap of thinking basis vector 2 and 4 were the same.

i'm not talking about C^2 as a real vector space right now, but its dimension is 4 if you regard it that way

:)

thanks a lot Leslie :)

at the risk of belaboring the point, (1,0) and (0,1) are C-linearly independent elements of C^2 and they span C^2 because (z,w) = z(1,0) + w(0,1) for all z, w in C

weird that they even specify C with (b); you could pose the same question for F^2 and g: F^2 to F given by the same formula

so in looking at $\mathbb{C}^2$ as a complex vector space I'm .................you just read my mind....I was thinking about how the "i" showed up

no need to get vulgar in here

had to change that "e" to a "w" in "how"

the question above it is $\mathbb{R}^{3}$ so in me trying to "outsmart the book" I had wondered what would happen in the complex case and in turn over thought it

it certainly doesn't help that complex numbers are themselves often written as ordered pairs

yeah...that's where I initially was taking things to try to figure out the basis.

but it turns out one needs to be very specific in what one means.....shocking...(sarcasm)

so something like ((a,b),(c,d)) = (a,b)(1,0) + (c,d) (0,1) makes perfect sense when interpreted properly but looks insane because (a,b) on the RHS is an element of R^2 aka C while the (1,0) next to it is an element of C^2

i should write a book that blends the worst notations together so that every equation is an invitation to insanity

make sure to include $i,j,k$ for cross products

i think i'll also use j for the imaginary unit like some engineers do

maybe italic j

hi grandad

hope the patient is doing well.

cast might come off next week. she's now got a bad cough, waiting on some results to see what that might be. still has enough energy to antagonize the cat.

poor thing :-(

i read this question (math.stackexchange.com/questions/4111639/…) but is it actually true what OP writes? $g(B) = \Delta$? Since actually shouldn't $g(B)$ be the set ${(x, x) | x \in X, x = f(x)}$?

I have a short question that's probable not suitable for SO. Suppose a fair coin is tossed until the 2nd head appears. I'm asked to find the probability that the experiment concludes with at least 5 tosses. My approach is to first find the probability that the experiment concludes with fewer than 5 tosses, call it "p". My wanted probability is then given by 1 - p.

I know that the probability of getting 2 heads in n tosses is given by (n choose 2) * 0.5^n, so I thought to simply add ((2 choose 2) * 0.5^2) + ((3 choose 2) * 0.5^3) + ((4 choose 2) * 0.5^4) to get the probability of 2 heads in fewer than 5 tosses. This gives 1, however, which clearly isn't correct.

there are repitition here.

like according to the formula you are using, for example, in 2 heads in 4 tosses case you are also taking head head tail tail case which will come under 2 head in 2 toss case already .

.

im sorry i am bad at explaining not sure if you can understand it

@robjohn haha :')

nyes

@BannedUser yno

what's the fastest way to show $(1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1) \ge n^n$? (for $n >2 \in \mathbb{N}$)

clearly all terms taken individually are equal or greater than $n$, and I've proved it by showing the middle term in the even and uneven case is greater than $n$, but is there a faster way?

$(n-k)(k+1)=n+(n-k-1)k\ge n$ for $n\gt2$ and $0\le k\le n-1$?

oh that second equivalence makes it obvious

thanks a lot!

did you have a way to read it off $(n-k)(k+1)$ or did you expand it into $n + (n-k-1)k$ before being able to see it?

@shintuku there are n terms in the product, so a good guess is proving that each term is $\ge n$

yep that's what robjohn's answer does

write down the general term: $(n-k+1)k$ and give it a try :p

thanks for the tip!

If we need an explanation or elaboration on a specific aspect for an answer on existing problem on math.stackexchange, should we comment in that answer itself eventhough answer is pretty old? what should we do?

@shintuku if you want to prove that $(n-k+1)k \ge n$, you can simply prove that $(n-k+1)k - n \ge 0$; here, I used $1 \le k \le n$. it goes in the same way rob did

@shintuku it is a parabola, and concave and equal to $n$ at the endpoints, so greater than $n$ in between. The expansion was just to make it clearer.

@LucasHenrique did you mean $(n-k-1)k$?

woah that is very insightful, thanks

People, I think I solved my combinatorics problem

@user69608 try commenting there, first. If the user is not active, there might be nothing you can do other than asking for opinions here.

@robjohn alright,thanks.

Give them a day or two to respond. Not everyone visits here every day.

yeah,okay.

@LucasHenrique Oh, your $k$ goes from $1$ to $n$, where mine goes from $0$ to $n-1$. Sorry.

is there any relationship between compactness and finite-measure in general measure spaces?

or at least Borel measure spaces

the statement of the Egorov theorem makes it seem that "compactness $\approx$ finite measure" in Borel measure spaces

probably not since compactness is not close at all to Heine-Borel in infinite-dimensional Hilbert spaces

yeah i don't know of one. note that the uniform convergence you get out of egorov's theorem is on a set that might be look really bad and not have any interesting topological properties. i think it's just a measurable set, at least by the usual proofs.

topological groups have invariant measures (haar measures) that are unique up to scaling. these measures are finite measures iff the underlying space is compact

cool!

is bounded curvature useful?

hi, could someone help me out with this? Say $X$ is a compact riemann surface with universal cover $\mathbb{D}$, (and suppose the genus of $X$ is $\geq 2$), I want to prove that the deck-transformation group $G$ of $\mathbb{D}$ has the following property: For any $d \in \mathbb{D}$, given $\epsilon > 0$ is chosen small enough, there is some positive integer $m > 0$ such that for all $x \in D$, $|G(x) \cap B_{\epsilon}(d)| \leq m$

I know that in this situation , the deck-transformation group $G$ (equipped with the induced topology from the space of continuous maps $\mathbb{D} \rightarrow \mathbb{D}$ with the compact-open topology) is discrete

is taht enough to show this?

i've tried with no success

Your genus assumption is redundant, @porridge. That follows from the fact that the universal cover is $\Bbb D$.

thats true

Yes, this should just follow from discreteness and a local compactness argument. I don't think you need to bother putting a hyperbolic metric on everywhere, although you could do that.

alright, ill try to fiddle with that more then, i was worried it might require the hyperbolic metric

It can't hurt to use it :P

You can always ask @Balarka if you see him :)

true, im sure there is a slick argument involving it, its just im a little brain fried at the moment and the hyperbolic geometry related stuff I learned about a month ago has disappeared from my brain :(

Hi @Ted long time no see

Hi @Lukas! Glad to know you're still around. I am leaving in a few minutes, though.

I've changed my undergrad thesis topic from basically all of Langlands for $\mathrm{GL}_2(\Bbb Q)$ to some basic stuff about group schemes

much more manageable

though I do want to point out that I found an error (and know how to fix it) in SGA

it's just a small thing

hm. we can try to reverse engineer the intended distribution (which really ought to be a part of the problem statement) from what the answers are, if you have them. do you know what the 'right' answer to (a) is?

so yeah, that does appear to be the poisson distribution. "lambda" is 7.5

oh i see.

maybe you're inputting that to the calculator without enough parentheses?

ok

I'll test this

(7.5^2)/(e^(7.5) * 2) is indeed 0.01555...

I see

I used my calculator in wrong way

sorry for my fault

hahahahaha

and e^(-7.5) + 7.5 e^(-7.5) + (7.5^2*e^(-7.5))/2 is .02025something

well you had the right idea, which is the important thing

if it hadn't turned out to be a poisson distribution i had no "plan b"

understood

ty

they should have made that problem more historically accurate and had it be about people being kicked by horses an average of 7.5 times per month

Don't forget those horses are from the Prussian brigade...

a police horse in new orleans bit my hand once.

thankfully i have never been kicked by a horse, mule, etc.

were you feeding it?

what is meant by a "faithfully flat cover of $\operatorname{Spec}(A)$"?

where $A$ is a commutative ring

It is a cover of $\operatorname{Spec}(A)$ which has accepted Jesus as its personal lord and saviour, I think.

:P

does there exist a real analytic algebraic variety in $(0,1)^3$ which is also a real analytic manifold (by deleting 2 singular points) - whose underlying topological structure is equivalent to the twice pointed sphere - and whose geodesics from (0,1,1) to (1,0,0) correspond to algebraic subvarities (forming the surface)?

it's a flat cover of spec(A) that dated mick jagger in the 1960s

@P-addict My recollection is that "faithful" means "injective" (in the sense of functors).

But I don't remember what any of the rest of those adjectives are.

@XanderHenderson lmao

@XanderHenderson yeah i think it's related to this - i know "faithfully flat modules" are a thing but i'm not sure what it means when we start talking about spectrums

@P-addict it just means that we have a scheme $X$ together with a faithfully flat morphism (of schemes) $X \to \mathrm{Spec}(A)$

Yeah, I don't really remember what the spectrum of a ring is. For me, the spectrum of an operator is (essentially) the set of eigenvalues and "generalized" eigenvalues.

@P-addict I can explain the connection. Do you know what a scheme is?

@XanderHenderson the spectrum of a commutative ring is the set of prime ideals which comes equipped with a topology (and a structure sheaf)

uhhh something about taking the zariski topology and turning it into a locally ringed space? i can't promise i understand the structure sheaf completely (i think restrictions on open sets excluding the roots of a polynomial $f$ have something to do with adjoining $f^{-1}$ to the global sections)... i'm not very good with schemes

i'm also not very good with category theory @_@

@LukasHeger Yeah, but where is the epsilon? If we aren't saying something about $\varepsilon > 0$, then it just isn't math! :P

@XanderHenderson actually, there are "infinitesimal arguments" in algebraic geometry which work in $k[x]/(x^2):=k[\epsilon]$or more generally just nilpotent elements

@LukasHeger NOOOOOOOOO!

@dc3rd No. I was not bothering the animal at all.

beware false epsilons

actually spectrums of rings and operators might be related...? mathoverflow.net/questions/70522/…

idk how

I sat in on a class in algebraic geometry years ago. It was not really my cup of tea. I have come to appreciate the abstraction of a sheaf in recent years (it provides a really good language for talking about meromorphic functions), but I wouldn't claim any kind of knowledge or expertise. I'll stick to analysis, thank you very much. :P

@P-addict probably through Banach algebras

there's the Gelfand spectrum

@P-addict an important example of schemes are affine schemes which are just given by $\mathrm{Spec}(A)$ for some ring $A$. If you have a morphism of affine scheme $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$, then this is basically just a ring homomorphism $A \to B$ and the morphism is faithfully flat if $B$, considered as an $R$-module is faithfully flat

so in the affine case, "faithfully flat covers" are just something you already know if you know faithfully flat modules

yes, there's lots of realizing operator algebras as rings of functions on abstract stuff. these are true epsilons.

$\epsilon$ is the length of time between when the light turns green in Greece and the driver behind you honks.

where a ring homomorphism $A \to B$ is flat if $B$ is a flat $A$-module, as you might have guessed

now that we have the notion of a flat morphism, we can just define a faithfully flat morphism as a morphism which is both flat and surjective (on the underlying spaces)

@copper.hat They wait that long?!

it's a nice exercise in commutative algebra to show that this agrees with the characterization for affine schemes I gave above (iirc, it's one that is given in Atiyah-Macdonald iirc)

(you don't need any algebraic geometry for the exercise, too. It's just saying that if $B$ is an $A$-algebra, then $B$ is faithfully flat iff it is flat and every prime ideal of $A$ is the contraction of a prime ideal of $B$)

@P-addict are you following?

typically it is safer to honk before the light turns green, that way you can be assured you honk at least once

Start honking as soon as the light in the opposing direction turns yellow.

here's a nice exercise in special relativity: how fast do you have to drive towards a red light so that it appears green due to the relativistic Doppler effect

@LukasHeger Yes!

I love that question!

Although I fear that the answer is "Fast enough to get a speeding ticket, whether or not the cop buys your 'But the light was green!' excuse."

I also recall a QM exercise about arguing with the police that due to Heisenberg, they surely can't know exactly where I was driving how fast

i didn't pull over because i can't see blinking infrared lights

@LukasHeger sort of… give me a sec to try and digest it (also I’m at school rn and going to lunch, I’ll grab some paper soon)

@XanderHenderson :-).

@LukasHeger Ostensibly, the only physics I know is QM. But, somehow, I feel like my knowledge of unbounded operators on Hilbert spaces is not very practical.

@XanderHenderson some physicist told me that I know QM because I know some rep theory

I'm not convinced, though

Heh.

but it's neat how he explained that figuring out elementary particles in a QFT boils down to figuring out irreducible representations because just like every particle is composed of elementary particles, every rep (assuming you're working in a nice enough setting) is a sum of irreps

My entire knowledge of physics is contained in chapters 1, 3, and 7-12 of this book. Which is not to say that I have mastered that material. Only that I have spent some time with it.

I've also skimmed chapter 14, but not thought hard about any of it.

on that analogy one might be tempted to say stupid stuff like "non-modular rep theory of finite groups is all about figuring out elementary particles, because for a given set of elementary particles, there's only one way of putting them together", "modular rep theory of p-groups is the opposite: there's only one elementary particle, but lots of non-trivial ways you can put it together with itself"

my advisor had a short set of notes entitled 'does quantum field theory exist?' broadly speaking, his answer was 'no, or at least not mathematically in many of the contexts where people are using it.' he posed it as a kind of challenge to the future to figure it out. i declined the challenge.

I think there are versions of QFT like topological QFT or algebraic QFT that are mathematically sound. But that doesn't cover how a lot of physicists work with them

i suspect many physically phenomenon are modeled simplistically and that as one approaches 'reality', many things break down and need a new approach.

the problem is that if the barrier to entry is high, it will not get enough eyeballs to sort things out quickly.

is there any way to get my mathjax to make a tiny bit of space between the norm symbols in $||T||||\vec x||$

Use \|?

e.g. \|T\| \|\vec{x}\| is typeset as $\|T\| \|\vec{x}\|$. (assuming you have chatjax running)

ah very nice thank you

boo, \vec :(

@shintuku If the space in $\|T\|\|\vec{x}\|$ is not big enough, you can add a \, between to get $\|T\|\,\|\vec{x}\|$

i sometimes would break up stuff like that with a \cdot

or if there were more than two norms floating around, decorate the norms to indicate what norms they were

However, other than spacing, \| works with contextual sizing as in $\left\|\int_{\mathbb{R}}\vec{x}\,f\!\left(\vec{x}\right)\mathrm{d}x\right\|=1$

and the mathrm to boot. pure evil.

$\|T\|\hphantom{abcdefghijklmnopqrstuvqxyz} \|\vec{x}\|$

@leslietownes that is the way it is in most books, and if you need to use $d$ as a variable...

$\int f(d)\mathrm{d}d$

in some books. i've certainly noticed it becoming more common as america, and more generally the world, loses its way.

welcome back @Ted

@LukasHeger precisely!

@LukasHeger $\int f(d)\,\mathrm{d}d$, duh.

$\int_{\partial D} \mathrm{d} d \, \delta(d)$

silly question, but does anybody remember how wiki gets the upper bound $\sup_{\text{arc}} \frac{1}{z^2 +1} \le \frac{1}{a^2 - 1}$ in en.wikipedia.org/wiki/…

you need to get some $\mathfrak{D}$s and $\mathcal{D}$s and $\Delta$s in there

$\int_{\partial D}\frac{\partial}{\partial d_1} \frac{\partial}{d_2}d(d_1,d_2)\,\mathrm{d}d_1\,\mathrm{d}d_2$

and don't forget $\mathscr{D}$

oh, is it just that $a$ is a point on the arc?

@MatheusSousa The probability of getting $b$ burnt bulbs is $\frac{\binom{5}{b}\binom{7}{6-b}}{\binom{12}{6}}$

Now you need to compute the expectation of $b$ and $b^2$

Ah, ok

so, it's a hypergeometric problem

but one thing

Do you know how do that?

is hypergeometric distribution a special case for negative binomial distribution?

where n is the number of lamp get in the sample, N is the total of lamps and K is the total of burnt lamps accordingly with N.

it's possible, and perhaps even advisable, to compute these things from the definition in this one case, before seeing how it fits with a general formula

yeah, its true

@MatheusSousa what does that formula give you?

@shintuku And use /,

@TedShifrin do you mean \,?

@robjohn yes, stupid ipad typo.

@TedShifrin mobile sucks

Ah, I’m always late.

You and the pooch recovered?

@TedShifrin me: most of the way; pooch: still working on it.

Good and somewhat good!

I wish we could get her much better, soon.

Though, I see small improvements.

Another vet to consult?

We have two, the Urgent Care vet we first saw, and our regular vet.

we are waiting on a fecal test result. I think they should push rather than wait the 3-6 days for processing.

If you pay extra sometimes they do.

It has been 6 days, but they probably don't count weekends.

I'm sure I could bribe someone, but our vet has no control of how long the lab takes. I would need to find someone there to bribe.

Yup

My son was the one who got the sample and it was minimal. I asked them to make sure it was enough. I am going to be really upset if the result comes back on Friday and they say the sample was not enough.

Heads will roll

This could indicate that a particular antibiotic is needed, and it has been a week so far. If they say we need to wait another week, I will make sure that it gets done in a day.

Good luck! 💙

The FedEx courier: what's in the package? Me: important shit.

Biohazard.

@robjohn Flagged for abusive language. :P

@TedShifrin Thanks! We're keeping our fingers crossed.

@XanderHenderson you mods are so touchy!

@robjohn Good luck.

@copper.hat Thanks. We are trying. So is she.

@copper Your son having a blast at college?

@TedShifrin Not really sure, but we Facetimed over the weekend which was a tonic for me :-).

Good.

While checking in to her return flight to London tomorrow on Virgin, my daughter discovered she was rescheduled to Saturday! I am annoyed but secretly relish the idea of a few extra days. Selfish, I know...

My son went to sign up for clubs, but was told that because of his AP credits that he is considered a sophomore for which the club rush (?) was the previous day.

That is disappointing.

That’s crap.

someone should be able to do something about that.

Yes

I am not sure where to start, I suspect he does not realize that something could be done.

@copper.hat yeah, when you are that age, without the proper experience, you often just give up. Tell him to find out who to talk to.

club rush? isn't that the kind of ape-the-east-coast, frat-house, follow-the-leader regimentation we created the UC system to escape from?

@leslietownes I think that is the term he used, I doubt he is frat house material :-).

I have no contacts in UCSC, I know the chap who was the chancellor last century, but suspect those contacts are too remote in time & hierarchy.

it would disappoint me to learn that they even had frats there. when i was at berkeley, you joined clubs by attending their meetings. they did have an event at the beginning of the year with a lot of student organizations in one place, but it was not the exclusive route to joining. or even a route to joining.

My kids get annoyed when I push to make things happen, so I need to tread carefully.

(Which is understandable.)

i guess space could be a premium with clubs that require the use of school facilities. there are only so many places you can do judo, or whatever.

I loved judo. Usually got trounced, but it was great exercise.

i suggest engaging in light cyberbullying of official UCSC social media accounts via a sock puppet. you can do this with or without his approval.

I might try the provost for a start.

oh, definitely.

my daughter's covid test came back negative, so weirdly, i can feel kind of OK, or maybe even happy, about hearing her coughing all the time.

Is she in good spirits?

yeah, more or less.

last night she said "i'm happy i have a lot of parents." i asked how many she had, and she counted one, two, three, pointing at me, the cat, and my wife.

she gets her best ideas from the cat.

And her claws!

Generally if they are eating & in good spirits I am not concerned. I had a close call (need a less melodramatic description) with my son & pneumonia when he was much younger. I keep a $5 pule oximeter handy to check oxygen saturation.

we have one of those too. sometimes i check mine just to feel good at something. 99% spo2! i can't be stopped!

i have a vector field from $\mathbb{Z}^2 \to \mathbb{R}^2$ i'd like to approximate with a differential equation

is this possible?

Consistently below 95% without extenuating circumstances (race, for example) is a cause for concern.

when i bought it (and why i bought it), i was in the low 90s for a few days. last fall.

now it's just nothing but 99%. A+ every time.

:-).

If you want a true differential equation, you need a vector field on $\Bbb R^2$. How will you interpolate?

right, i'm expecting that although my dataset is in $\mathbb{Z}^2$, the corresponding differential equation would be $\mathbb{R}^2 \to \mathbb{R}^2$

i'm just trying to figure out where i could begin learning how to do this stuff

you have a lot of blanks to fill in. maybe you have some model for whatever is generating the data that can help. it feels generally underdetermined to me. there is no general theory that just fills this in.

I know no remotely standard way to proceed, other than modeling it discretely, not continuously.

for $z \in \mathbb{Z}$ and $r \in \mathbb{R}$ with $z < r < z +1$ and a vector field given by $f:\mathbb{R}^2 \to \mathbb{R}^2$ modeling acceleration as a function of direction, couldn't we estimate $f(r)$ with the information given by acceleration at $f(z)$?

and acceleration at $f(z+1)$. i'm using a physics example but warning i know no physics and i'm trying to make a model of consumption in economics

heh, the above is actually $f:\mathbb{R} \to \mathbb{R}^2$, so let's say acceleration with respect to time, my bad, but yeah, wouldn't it be possible with info from $f(z), f(z+1)$?

nvm, the above makes no sense, i will work on this question more

Vague question: is there some form of product on tensors that does not change the rank of the original tensors?

@gian Nope.

@TedShifrin

Why is this?

What do you call a product of tensors? I presume that by rank you mean the total “order” covariant + contravariant.

@LukasHeger sorry, i don't think i quite understand... i'm guessing what this means is that a collection of morphisms $\{\operatorname{Spec}(B_i)\rightarrow\operatorname{Spec}(A)\}$ constitute a faithfully flat covering when the images of the $\operatorname{Spec}(B_i)$ cover $\operatorname{Spec}(A)$, and the corresponding $\{A\rightarrow B_i\}$ are faithfully flat?

@Ted Shifrin yes that's what I mean by rank. I was just curious if such a construction can be made, since the only thing I've seen is the tensor product which does change the tensor rank.