LASD is basically a criminal gang. they do all kinds of bad stuff in communities where people can't afford lawyers. not surprised.

you need more training to TA calculus 1 than you need to work in the sheriff's department. people from other countries are blown away by this.

for more on my wildly anarchistic views, consult my podcast, Leslie Speaks.

i've had uniformly good experience with LAPD and LBPD but i'm white and never needed them for anything serious or had them come in and police something that didn't need policing.

OTOH, who wants to go into Compton to re-enforce the wearing of masks?

i see the point, but compton is actually a very masked area. i think this decision is more about exurbs of santa clarita.

i got my second shot across the street from the compton border. 1000% mask complaince, everything calm and normal. i wish i could say the same about huntington beach.

my wife's dad was born in compton and remembers riding a horse in a dried river there. you see less of that these days.

my daughter was born in torrance next to the mall where the key scenes of 'jackie brown' were filmed. she's going to enjoy that movie later.

the bail bonds office is now a set of condos but the mall is still the same.

Is there a name for the arithmetic or manipulation of mathematical symbols themselves as a metamathematical arithmetic?

die hard is another LA county classic. it accurately depicts the drive from LAX to century city in its opening scenes. nakatomi plaza is still there.

that annoying movie with emma stone and ryan gosling a few years ago had some scenes shot a few blocks from where we used to live. there was a cute apartment building. good location scout.

they also filmed in a dive bar that nobody in that movie would have gone to.

AMDG i can't think of an applicable buzzword but formal methods or formal systems or symbolic manipulation, something like that.

to veer slightly back closer to topic.

Probably symbolic manipulation I guess.

symbolic logic, perhaps

if you're familiar with edsger dijkstra a lot of his work had that flavor and it is all free on the web. symbolic manipulation one step away from meaning.

Like if I wanted to say something like, "The symbolic delta for every integer reciprocal $\frac{1}{n}$ for every integer $n$ is $\frac{0}{1}$."

i never met him but got a lot of useful insights from his printed work. i couldn't tell if it was handwritten or if he had a font made of his handwriting, his handwriting was that consistent.

nice

my handwriting is a tire fire of inconsistency.

See I got the idea earlier that what might be feasible is instead to study patterns in symbols themselves for solving problems as a general algorithm. I mean it's the most obvious thing given that it's what we do as humans already, but it's rarely applied in computing, or at least in what I've attempted.

And then find a way to map symbolic operations onto mathematical operations.

I mean, it's easy for us to see the difference between the first reciprocal and the next, so why not make it easier for computers to aid us better?

Also, I don't know why, but I keep getting all kinds of ideas when I look at NP-hard or NP-complete problems the last few days.

Consider the bin packing problem. It seems that at least for rectangle packing, spirals are closely related to the bin packing problem. Consider just the fibonacci spiral, for instance, and consider the rectangles whose sides are tangent to a point on the fibonacci spiral. That can also be extended to the generalization of the fibonacci numbers. The result is always a perfect rectangle with no wasted space.

Then, Mathologer released this video which I find to be incredibly interesting as something to research. I think I know which area of mathematics I love most. I think it's discrete mathematics.

So is there a variation of pascal's triangle for ratios or integer reciprocals?

i dunno quite what you mean by that. the farey sequence comes to mind as somehow analogous and yet also kind of not.

Hi, I am very new to Sylow's theorem. I think that there's been a typo here in proof 1: it should say that no. of Sylow p subgroups in a group of order pq (p<q and p doesn't divide q-1) divides q and not pq.

proofwiki.org/wiki/Cyclic_Groups_of_Order_p_q

here p and q are primes

I say so because Sylow's theorem says in a group of order $p^nm$ such that P does not divide m, the no. of Sylow p subgroups is 1 mod p and divides m.

So in the proof "divides pq" was not required to be considered and only divides q would have sufficed.

@Leslie any suggestions on this please?

@leslietownes Looks pretty similar to me... upload.wikimedia.org/wikipedia/commons/thumb/b/b9/…

You're right, it isn't quite the same for sure, however, I think it will be useful to look into anyways. I want to study these relationships and whatnot so that I can get a better algorithm and implementation for the division algorithm I rediscovered, or at least assume that I rediscovered.

What I'd like to see is if I can't actually make an O(n) division algorithm for n-digit values for any given base.

If I can find that, then substring searches, for example, can be sped up since one can find the right-most occurrence of a substring in a string using $\lfloor \log_2(\frac{x}{y})\rfloor$ where $x$ is the string and $y$ is the subtring interpreted as two binary integers iff y is a substring of x. In practice, it requires multiplying the result of the check for correctness by the result of the logarithm.

Hence why I was asking earlier for a fast way to multiply by zero or one because the result of any boolean operation is true or false.

divides pq and being relatively prime to p implies divides q. i forget how many number theoretical hoops this requires.

@Wolgwang Anyone?

if i get an implicit function $\phi$ from the function $f$ when $f=0$, is there a way to interpret change in $\phi$ as change in $f$?

@leslietownes if divides q then clearly divides pq also but there is no need to consider that. Right?

I mean we need to consider only these divisors of q that is, 1 and q only. No need to consider 1,p,pq,q

This is so by Sylow's theorem.

Right?

@leslietownes The one thing I hate about mathematics sometimes is the fact that it formally demands too much rigor at times, or so it would seem, for even the simplest and most elementary parts of math. On the contrary, that rigor gives us absolute certainty about our conjectures. I just simply wish that that rigor only demanded as much rigor as how simple or complicated a given question is, particularly in obvious cases. For everything else, rigor is obviously justified.

Like, I don't get the point of trying to rigorously prove a simple, obvious, and intuitive fact like 1 + 1 = 2 through some winding, convoluted path to prove that such an equation is true. It is not justified in itself to prove something so simple to do so; perhaps in other contexts where the certainty of a particular result involving the addition operation is dubious is justified, but not for this.

Anyways, would love to discuss more, but I have to go to sleep. Good night!

Same old Euclidean algorithm/Bezout proof.

Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$. Suppose $f(\vec a) = 0$ and $\frac{\partial f (\vec a)}{\partial x_n} \neq 0$ and using the implicit function theorem let $x_n$ be implicitly defined from $f= 0$ by $x_1,...,x_{n-1} = \underline x$ such that $\phi(\underline x) = x_n$. Notice therefore that $\vec x = \langle \underline x, \phi(\underline x) \rangle$

now, observe this beautiful beautiful graph, the labour of minutes and seconds of Paint:

now... why exactly are we justified in interpreting change in $\phi$ with respect to $x_j$ as a quotient of change in $f$ with respect to $x_j$ and $x_n$?

it bothers me because $f$ is set equal to $0$, otherwise we could easily designate a piece of it and say: hey, this is the part that changes with respect to $x_n$

in the above graph, $e$ subscript is a standard vector, and $\vec b = \langle \underline a + t\vec{e_j}, \phi(\underline a + t\vec{e_j}) \rangle$

there are lots of letters around, you are going to cause yourself harm if you restrict yourself to variants of $x$ :-)

my bad hehe

@shintuku take $f(x,y) = ax+by$ with $a \neq 0$. then $\phi(y) = - {by \over a}$.

note that ${\partial f (x,y) \over x} =a, {\partial f (x,y) \over y} =b$.

work with simple cases first.

when you lift weights you start small and work up.

@leslietownes if $n\mid pq$ and $(n,p)=1$ then $\overbrace{nx+py=1}^\text{Bezout}\implies n\underbrace{\left(qx+\frac{pq}ny\right)}_{\in\mathbb{Z}}=q$

@copper.hat right, that's an example I'm comfortable with

since we isolate either x or y, we can then say that f(x,y) changes as a function of x or y

well, it is a model for more complicated stuff.

right, i was also feeling like i was rushing for no reason hehe

if $f(x+h,y+j) \approx {\partial f(x,y) \over x} h + {\partial f(x,y) \over y} j = 0$ then you see that $h \approx -\left( {\partial f(x,y) \over x} \right)^{-1} {\partial f(x,y) \over y} j$, or ${\partial \phi(y) \over \partial y} = -\left( {\partial f(x,y) \over x} \right)^{-1} {\partial f(x,y) \over y}$.

there's bezout again.

the only magic is in proving the existence of $\phi$. you can use newton's method if you are familiar.

is there a minus? $h \approx -(etc)$

yes. i have been know to make a mistake.

your point makes perfect sense

i guess the work now is to get rid of the approximation

alright, thank you very much! i'll meditate on this a while

the approximation is resolved by a fixed point iteration.

god bless you

Any suggestions on this please?

Here's the background:

@Koro it would be clearer if you said what divides these things.

Hi @robjohn

No. of Sylow p subgroups is the divisor

In android phone, mod symbol in your name in chat looks red(noticed just now) colored but in iphone it's different color.

@Koro the diamond?

$\text{robjohn}\,\raise{1pt}{\tiny\blacklozenge}$

the lozenge.

lozenge is a great word, we should say it more.

so when we need to remove a moderator, do we have to dislozenge them?

(I can't tell if that is meant to be a pun on dislodge, but if so, nice :P)

@robjohn yes

The diamond is grey on my Samsung phone ¯\_(ツ)_/¯

@hyper-neutrino that is precisely what I was thinking of

it's a light blue in chrome.

Red on my OP

@Koro is it a different color than my name?

or hyper's name

I think light blue on my iPhone and chrome/opera also

@robjohn yes

that is odd. Both are blue on my computer

@hyper-neutrino also a different color than the name that precedes it?

weird

it is #4979b9 for me on firefox (light blue)

@robjohn please see that

@robjohn no, it is the same

@Leslie : do you have any suggestions for my question?

on Firefox for MacOS, it is x4979B9

@hyper-neutrino same thing

@robjohn Same on chrome.

it is defined directly in the CSS file as .username.moderator{color:#4979b9} in row 1 column 8930 so I would find it surprising to be different on any normal desktop browser

@hyper-neutrino is that the diamond or the text of the name?

both

next to each chat message: <div class="username moderator owner"><span style="float:right">&nbsp;♦</span>hyper​‑neutrino</div>

when expanding the user profile: <h4 class="username moderator" title="47.2k">hyper-neutrino<span> ♦</span></h4>

why does one of them have the span to the left and using inline CSS to float to the right? who tf knows. chat is just broken like that

i believe it is to prevent the diamond from disappearing if there is not enough space - previously, my name was "HyperNeutrino", so for people with smaller screens or high page zoom, the diamond would cause my name to entirely disappear, but the diamond itself would stay

500% zoom

the diamond would be the ghost of a departed quantity

@robjohn I just want to double check; $f(x)=x^m$ near $0$ is in $o(x^{m+1})$, right? So does $f(hu)=u^{m}f(h)$ hold?

@schn no it is not. take the ratio: $\lim\limits_{x\to0}\frac{x^m}{x^{m+1}}=\infty$

True, it is in $o(x^{m-1})$, sorry.

yes

Hey, I just wanted to ask if someone knows anyone who has knowledge regarding Random Walks in Random Environment, especially Sinai’s Walk?

@robjohn So any function $f(x)\in o(x^{m-1})$ can be written as $f(hu)=u^mf(h)$.

For $x$ near $0$.

@schn no, I never claimed that you could use the same function

That is why I stated it with $g$

:)

and the exponent should be $f(hu)=u^{m-1}g(h)$ for some $g\in o\!\left(x^{m-1}\right)$

But $f(x)=x^m$ is written $f(hu)=u^mf(h)$, specifically with $u^m$ and not $u^{m-1}$?

So? $g(x)=xf(x)$

$f(hu)=u^{m-1}g(h)=u^{m-1}hh^m=u^{m-1}h^{m+1}$.

Also, $g(x)=xf(x)=x^{m+1}\in o(x^{m})$.

@schn which is a subset of $o\!\left(x^{m-1}\right)$

If $f(hu)=u^mf(h)$ then $g(h)=uf(h)$ or $g(u,h)=uf(h)$.

@robjohn I disagree with this statement, since $f(hu)=u^{m-1}g(h)=u^{m-1}hh^m=u^{m-1}h^{m+1}\neq u^m h^m$ for $f(x)=x^m$.

You must have meant $g(x)=uf(x)$ or $g(x,u)=uf(x)$.

Or equation 4 should read for $f(x), g(x) \in o(x^{m-1})$, $f(hu)=u^mg(h)$.

Sorry, $g(h)=uh^m\in o\!\left(h^{m-1}\right)$

Right, so $g(x)=uf(x)$. Thanks for helping me out, @robjohn ! You deserve a heart <3

But more questions might drop in about this. Anyway, it does not seem to be easy to be rigorous about this.

@schn I don't like using little-o. I try to use big-O when I can.

@schn To be even clearer, I sometimes use $[x,y]_\#$ which I use to represent some point in $[x,y]$.

@robjohn How is it related to the little-o or big-O notations?

@robjohn By the way, if $f(x)=x^m\in o\!\left(x^{m-1}\right)$, then you claim $f(hu)=u^{m-1}g(h)$ for $ g\in o\!\left(x^{m-1}\right)$. Why not just make it $u^m$ instead of $u^{m-1}$?

@schn Not sure what you are asking. make what $u^m$? The formula you are quoting says that if $f\in o\!\left(x^{\color{#C00}{m-1}}\right)$ then $f(hu)=u^{\color{#C00}{m-1}}g(h)$.

@robjohn Sorry I meant more in general, if $f(x), g(x)\in o\!\left(x^{m-1}\right)$, does it not hold that $f(hu)=u^mg(h)$?

@schn no, why would it?

With $f(x)=x^m$, it seems like one would get away with $f(hu)=u^mg(h)$.

yes

Do you have a counterexample?

For when $f(x), g(x)\in o\!\left(x^{m-1}\right)$ but $f(hu)=u^mg(h)$ does not hold.

@schn that is easy, because your question is wrong: $f(x)=x^m$ and $g(x)=x^{m+1/2}$

what you want to ask is for $f\in o\!\left(x^{m-1}\right)$ can we always find a $g\in o\!\left(x^{m-1}\right)$ so that...

or the negation

but your question asks about any two functions $f,g\in o\!\left(x^{m-1}\right)$

You're right, that seems easy :)

Now, would you say that one can always find a $g\in o\!\left(x^{m-1}\right)$ given an $f\in o\!\left(x^{m-1}\right)$ so that $f(hu)=u^mg(h)$ holds?

I guess it is just reversing the roles of $f$ and $g$ you previously gave.

So $f(x)=x^{m+1/2}$ and $g(x)=x^m$.

That's a no.

@schn that is one particular $g$ for which it doesn't hold, you need to show that no $g$ holds for your statement.

What we know is that $\lim\limits_{hu\to0}\frac{f(hu)}{(hu)^{m-1}}=0$

Or according to the assumption $\lim\limits_{hu\to0}\frac{f(hu)}{(hu)^{m-1}}=\lim\limits_{hu\to0}\frac{u^mg(h)}{(hu)^{m-1}}=\lim\limits_{hu\to0}\frac{ug(h)}{h^{m-1}}\stackrel{?}{=}0$.

I have integers 3..10. I want some simple function so that f(a) + f(b) < f(a+b) but f(a+1) + f(b) < f(a+b) for any a, b I choose from the range. What the easiest way to do that?

@schn you only need $\lim\limits_{h\to0}\frac{g(h)}{h^{m-1}}=0$

@robjohn why is that the necessary condition from $\lim\limits_{hu\to0}\frac{ug(h)}{h^{m-1}}$?

that is the necessary condition for $g\in o\!\left(h^{m-1}\right)$

I suppose I could define f(x) = x^(1+epsilon) for some epsilon?

@robjohn True.

But according to the assumption it needs to be checked that $\lim\limits_{hu\to0}\frac{ug(h)}{h^{m-1}}=0$, right?

@schn according to the assumption, that is true. it does not need checking

The assumption is $f(hu)=u^mg(h)$, where $f(x)\in o\!\left(x^{m-1}\right)$, so yes, I agree.

Instead one needs to check...

$\lim\limits_{h\to0}\frac{f(hu)}{u^mh^{m-1}}=0$.

Or $\lim\limits_{h\to0}\frac{f(hu)}{h^{m-1}}=0$.

If you know how to put the lid on this short proof that one can not find a function $g(x)\in o\!\left(x^{m-1}\right)$ given a function $f(x)\in o\!\left(x^{m-1}\right)$ such that $f(hu)=u^mg(h)$ holds, let me know @robjohn some time. Thanks for the very generous help though.

The way I see it is that $f(x)\in o\!\left(x^{m-1}\right)\subseteq o\!\left(x^{m}\right)\subseteq ...$ and thus one can pull out at least a factor $h^{m}$ out of $f(hu)$, so $f(hu)=h^{m}p(h,u)$ where $p$ is some arbitrary function. If this is correct, then $\lim\limits_{h\to0}\frac{f(hu)}{h^{m-1}}=0$ does indeed hold and thus $f(hu)=u^mg(h)$.

@schn wait, $o\!\left(x^m\right)\subset o\!\left(x^{m-1}\right)$ for $x$ near $0$

near $\infty$, the reverse holds true.

You are right. So $f(x)\in o\!\left(x^{m-1}\right)\subseteq o\!\left(x^{m-2}\right)\subseteq ...$, but one could still pull out, at most probably, a factor of $h^m$ from $f(hu)$, no?

that is the idea.

Then $\lim\limits_{h\to0}\frac{f(hu)}{h^{m-1}}=0$ and $f(hu)=u^mg(h)$ would both hold.

Since if $\lim\limits_{h\to0}\frac{f(hu)}{h^{m-1}}=0$ holds, then also $\lim\limits_{h\to0}\frac{f(hu)}{u^mh^{m-1}}=0$ and thus $g(h)=\frac{f(hu)}{u^m}$ is proven for any $f(x),g(x)\in o\!\left(x^{m-1}\right)$.

Isn't the bin packing problem just an n-dimensional knapsack problem?

Pain, I don't have mathematica, and wolfram alpha won't give any answer for $\prod_{n=0}^{\infty} \frac{1}{1-n^{-x}}$

A Serre functor $S_D$ is an equivalence $S_D : D \to D$ such that for any two objects $A,B \in D$ (here you need $D$ to be a k-linear triangulated category), $$\eta_{A,B} : \operatorname{Hom}(A,B) \to \operatorname{Hom}(B,S_D(A))^{*}$$ is an isomorphism, which is functorial in $A,B$.

An example of serre functor would be the identity for the triangulated category of vector spaces, and a more nontrivial one serre duality for derived category of coherent sheaves on an integral scheme (I do not remember if this is true for general schemes)

@SayanChattopadhyay So? $\Bbb Z[n]$ is still a spanning class of the triangulated category hoChain

Sure that is true, I checked both the conditions

I think you're going off the rabbithole of definitions without a real understanding for what these things mean.

I was asking about your example of spheres, because I do not know how to check both the conditions there

@BalarkaSen Yeah I feel that too, there's too much of language

Then you should think about that instead of thinking about Serre functors. A healthy practice is to not say a definition you do not understand. It is a basic algebraic topology question. $[X, S^n] = 0$ for all $n$, does that mean $X$ is contractible?

But I do not understand what the stable homotopy category even is, I have never seen a defintion

Then why are you bothering?

Because I was looking for examples, and you told me this works

Never mind

I also told you hoCh works, which is an example you understand.

Yup.

@Sayan: I still find the algebraic topology question intriguing (unlike general nonsense about Serre functors), and here are my thoughts: If $X$ is an acyclic space, $[X, S^n] = 0$ for all $n$. Indeed, consider the skeletal inclusion $S^n \to K(\Bbb Z, n)$, where we build the Eilenberg-Maclane space by adding a 0-cell, then an $n$-cell, then $n+2$ cells and higher. Let $F$ be the homotopy fiber. $[X, K(\Bbb Z,n)] = H^n(X; \Bbb Z) = 0$, so any map $X \to S^n$ can be homotoped to a map $X \to F$.

$F \to S^n \to K(\Bbb Z, n)$ is a fibration, running the homotopy LES gives $\pi_{n+1} K(\Bbb Z, n) \to \pi_n F \to \pi_n S^n \to \pi_n K(\Bbb Z, n)$; the last map is an isomorphism. So $\pi_n F = 0$, and of course $\pi_k F = 0$ for all $k < n$ as well. So $F$ is $n$-connected.

Say $m > n$ is where first $F$ has nontrivial homotopy, $\pi_m F \neq 0$. We can build a $K(\Bbb Z, m)$ out of $F$ by attaching more cells. This gives a map $F \to K(\Bbb Z, m)$, and we can again take homotopy fiber, $F'$. Same argument as above gives that the map $X \to F$ is actually homotopic to $X \to F'$.

This way you can climb the connectivity ladder, and get a map from $X$ to a contractible space, which is contractible. So $[X, S^n] = 0$.

So the claim is NOT true. But this is only unstably so, in the stable homotopy category the question is "Does $[\Sigma^\infty X, \mathbb{S}] = 0$ imply $X$ is contractible?". I do not know the answer.

who cares about maps into spheres

Maybe we should use the Barratt-Priddy-Quillen theorem. Use loopspace-suspension adjunction on $[\Sigma^\infty X, \mathbb{S}]$ and the BPQ isomorphism $\Omega^\infty \mathbb{S} \cong \Bbb Z \times B\Sigma_\infty$.

@Thorgott Mapping spaces $[X, Y]$ are well-studied in algebraic topology.

yes, but why those into spheres specifically

what's good about them

Spheres are the simplest spaces. Shrug. Also, closely related with cohomology as spheres appear as skeleta of $K(\Bbb Z, n)$.

Rationally, mapping into spheres is exactly what you'd want to do, as rational spheres are $K(\Bbb Q, n)$. If $n$ is odd or something, for even there's something else.

I see

math contest: youtube.com/watch?v=ojjzXyQCzso

Is there/should there be a way to retrieve the script of a comment once it's too old to edit?

LaTeX or MathJax script or w/e it's called.

@AMDG It might be better if you start at $n=1$.

@user10478 right click on MathJax, then show math as TeX.

I can't tell if it's working or not.

Sorry to bother you again about this @robjohn (not expecting an answer of course), but why would $\lim_{h\to0}\frac{f(hu)}{(hu)^{m+1}}=\infty$ as stated in your comment mean that a factor of $hu^{m+1}$ can not be pulled out of $f(hu)$, as specified in your comment?

By the way, equation 4 makes sense. So $g(h)$ is a function that possibly has a $u$ in it.

@schn just use the functions we were talking about and plug things in.

@schn If we could pull out a factor, what's left should at least be bounded (if not tend to $0$)

That is what I'm suspecting, but feels like a very daring statement. What $\lim_{hu\to0}\frac{f(hu)}{(hu)^{m+1}}=\infty$ is saying (I changed $h$ to $hu$ beneath $\mathrm{lim}$) is that $f(x)\not\in o(x^{m+1})$.

@schn yes.

it was only specified to be in $o\!\left(x^{m-1}\right)$

it is not even in $o\!\left(x^m\right)$

we are talking about $f(x)=x^m$, correct?

We were before, but now it is $f(x)=\frac{x^m}{\log{x}}\in o(x^m)$.

really weird moment today at a kiddie sports thing. there was a kid named eoin, a kid named rien, and my daughter, all with red hair. half of the field was red hair. not normal for orange county.

@schn okay, then the same is true except that it is in $o(x^m)$

Right.

i half expected someone to set off a flare or for a fight to break out.

@leslietownes for what? Oh, I see

i had a bag full of potatoes just in case.

it was a children's introduction to sport class. ages probably 2-4. not a lot of sport going on. just running around with soccer balls.

rien's mother had very dark hair, i wonder how our recessive genes survived that.

or maybe she colors it.

my wife masquerades as a redhead.

@robjohn But just because $f(x)=\frac{x^m}{\log{x}}\not\in o(x^{m+1})$ does that mean one can not pull out a factor of $hu^{m+1}$?

@schn "Pulling out" the factor leaves one with something not even in $o(1)$. It blows up near $0$.

the withdrawal method is discouraged for other reasons, too. that's my copper hat comment of the day.

catholic roulette

copper if you look above i had to confront a football pitch littered with irish descended children today.

i wanted to riot.

@robjohn In regards to which function are you claiming this?

$\frac{x^m}{\log(x)}\in o\!\left(x^m\right)$ but you cannot pull out a factor of $x^{m+1}$

the same is true if $x^{m+1/2}$

Salutations to the peanut gallery.

@dc3rd been a while

@robjohn And you “proved” it by showing that $\frac{x^m}{\log(x)}\not\in o\!\left(x^{m+1}\right)$ :) I don’t see the connection yet.

@schn what does it mean to you to "pull out" a factor of $x^{m+1}$?

Meaning that you can write f(x)=x^{m+1}g(x)$ for some function $g$.

I've been keeping busy and focused robjohn. Still working away at Ted's course, but keeping distractions to an absolute minimum and staying single tasked focus when working.

and what property do you want $g$ to have?

@dc3rd cool. how is Ted's course going?

@leslietownes the funny (& good) thing is that irish pitches are very much genetically diverse now, even in my last home town. i was helping a friend move a few years ago and one of the lads who was helping was from poland, but spoke fluent irish.

no offence to all parties involved, but i would not consider irish/polish to be diverse in that regard).

i love to hear it although i do recall someone advocating for an irexit recently

get rid of all the EU strangulations

eu overreach

F/G don't care about smaller countries and throw them under the bus.

no equivalent to senate

it's a weird question of institutional design. as you imply, we have almost the opposite problem in our federal system here.

i don't know how to solve it.

@robjohn It should be $o(x^{-1})$ since $f\in o(x^{m})$, right?

@schn okay, but $o(x^{-1})$ might blow up at $0$.

Very good actually very fulfilling.....also has had me questioning how things are taught in the school system. DOing his course has enlightened me to the ideas a lot more and provided purpose to why we learn what we learn. Also his explanations a lot better than explanations I'd recieved in the courses I've taken in the past............

but I'm not sure if this is a result of me just being more mathematically mature after having done Spivak and such or if it is exclusively down to how his text is written and the lectures......probably a bit of both

@robjohn it seems to be the only option though, if $f\in o(x^m)$, no?

good teaching is hard to scale, other than in the rough form of putting a finished product on youtube and anyone who wants can click through.

@schn The only option for what? I am not sure where this is supposed to be going.

i am more cynical than most about the project of teaching people to teach, i think the best you can do is to get people to stay away from a fairly small list of bad habits. often it is good enough but it isn't necessarily good.

@leslietownes it is....it is not easily quantifiable, plus everybody has their own idisyncrasies when it comes to learning....if you are passionate and want to deep dive into a subject

but there are entire departments set up that advocate the contrary, and what do i know.

there's accumulation of facts and learning. do not confuse.

my wife will admit she is not the world's greatest teacher. she can track metrics like, don't give so many low grades, or don't give a test that people can't finish, but the rest of it is kind of unfixable. which is fine.

funny about the idea of "good enough"......was reading a book and it was discussing the idea of our current obsession with perfection and optamization...and then I thought to mathematics and how even in that realm "good enough" still has to be accepted even if not prferred

i was not very good at offering encouragement to precocious students. i focused firmly at the D to B- range of students and probably ignored people that i should not have.

@copper.hat.....very very understated concept....

@schn: I am not sure that a calculus can be made from little-o when combining two variables like this. It often needs to be looked at on a case by case basis. The fact that our classification is based on $\lim\limits_{x\to0}\frac{f(x)}{x^m}=0$, with no detail about how it goes to $0$ often makes a calculus difficult to derive. This is why I avoid little-o when possible.

you did the good thing in my opinion Leslie.....the A range students are motivated and don't need the nudging along and encouragement that the rest do.

my attitude was that the A students don't even need me. which maybe they did. the good news is i am no longer polluting young minds.

for me my fastest learning is with someone else with similar motivation.

i remember spending an awful lot of time on encouraging students whose grades were lower to actually come to office hours. i hated when i'd do office hours and it would just be the A student showing up and asking me about fractals or something. although those were fun conversations.

@leslietownes fun but often off-topic

i'd engage if there was nobody who had 'real' questions but it was frankly a waste of everybody's time.

if often find there are sound bite insights which change my approach in a fruitful way. plus a lot of grind.

the only value i saw in teaching was getting D students to C+ students and after a while the smallness of that goal wore me down. i didn't have a broader view of it

the idea of putting 21-year-old me in front of a classroom is hilarious. i'm surprised the universe allowed it to happen.

is there any geometrical approach to pseudconvex domains?

@leslietownes I certainly pushed C and B students to improve, but I am confident that most of my A students appreciated my helping them be even better. It’s about lots more than grades. But I confess that one reason I retired early was the feeling I was no longer motivating students as much as in the past.

i think i basically ignored A students which is something i wish i could go back and correct. i was focused on keeping people from losing scholarships etc. and kind of brushing off the people who might end up going to grad school in something.

i did write them letters of recommendation.

I worked with any student who made an effort.

i could have used more of those. :)

Mostly students who didn’t care avoided me. When

i am linkedin 'friends' with a number of former students, several of whom went on to very impressive careers although not through the grad school route. one of them may give me a job some day.

i taught courses for which they had no option, I still did my best to get them to be successful.

Very few of mine went to grad school.

Heh, if only I had attended the school you taught at. I might have had interesting classes.

why isn't Collect collecting the terms? It's mathematica

c-130 just lumbered over

they are all different terms from what i can see

oh mathematica was thinking that ax was a variable, not a times x. Nevermind

First lesson in Mathematica. You have variables $ax^3$, etc. You need a space so that $a$ and $x$ are both variables.

I was too late.

I understand there is a new trend on homework. Instead of each student doing the homework, the students from groups of 2 to 4 students and do the homework together.

Is this a good idea? I think not.

i think it is a bad idea.

What it means is that there is less homework for the prof / TA to grade.

have a nice evening

@Bob Depends.

What is the goal of a homework assignment?

this is a good point. if the idea is to simulate a social interaction and problem solving in a workplace type environment, maybe a good structure. if the goal is to assess what anybody in particular is capable of doing, horrible idea.

Personally, I tend to give assignments to students so that they will work through problems so that when I test them using exams, they will, hopefully, have built familiarity.

I don't grade homework for correctness, only completion. I figure that any student who "cheats" is going to fail the exams, anyway, and exams make up the bulk of the available points.

So I don't give two f***s if students collaborate. Indeed, I think that they probably should collaborate.

i had a husband/wife style relationship with my physics lab partner, we worked very well together, but if i had not been able to choose or gotten the wrong group it could have been unfair to me.

it helps if you let students choose the groups and if as xander mentions the work itself does not matter that much except on the experience side.

With respect to written assignments, I tell students that they should collaborate, but that everyone has to turn in their own writing. I give zeros to copy-pasta, but I give students all the opportunities to revise and turn in new work.

in retrospect that was a really weird relationship. each thursday i'd bring her groceries and she'd cook dinner and then we'd do our physics.

These assignments are meant to improve mathematical writing ability, so, again, there is no harm in letting students work together, as long as there is improvement from one iteration to the next.

@leslietownes Heh.

the skill being tested there is simply the ability to recognize how one draft improves upon another. who cares how they develop it.

i didn't like being randomly sorted into groups in high school or college. although there is an aspect of this that is very reflective of 'real life.' but let me choose who i work with and i do OK.

the whole point in education after all is to teach the student and develop their skills - assessments are just to make sure they've reached that point, which is why I've always believed that a marking system where building on your past and resolving your mistakes is a good idea. the point isn't that when you were tested the first time you got it right, but that when you are tested for the last time that you have finally mastered the concepts

at least that's from the perspective of a random barely-even-adult who has been through a whole one year of university :p

when exams in my classes where cumulative i tended to have a policy of letting a grade on a later exam replace a grade on an earlier one if it was better.

the problem with having that kind of policy is some students may see it as a kind of lottery where if they just roll the right dice at the final, all will be forgiven, which is self destructive. but no system is perfect.

@leslietownes I never assign groups. I hated group projects as a high school student and as an undergraduate (I felt like I generally carried my "team", and resented it), and would have preferred to work alone. For the students like myself, I would prefer that they have the option to work alone.

On the other hand, some people really thrive in groups, and I want them to be able to learn in the way that works best for them.

i just emailed my lab partner about the good old days. i saw on social media that like me, she has a child now. almost 25 years ago. goodness. that was a good group.

i have some of our assignments in a box of college stuff in my office. i sent her one of our worst-graded lab reports.

@leslietownes I have a similar policy. I prefer to give a lot of short, low stakes exams, and a relatively high-stakes final. If the final is better than a midterm, I will replace the midterm exam score with the final exam score. If you learn it by the end, more power to you.

I also now work at an institution where I can withdraw a student at any time, up to and including the last day of class. I tell my students that if they don't take the final, I'll give them a W. Most see this as a threat, but some of them recognize that this is a "Get out of F-Jail Free" card.

that's cool. i've never taught at a place that had a withdrawal policy like that. the period was to my mind needlessly short, often before people had taken even one exam.

i tried to schedule my exams so you'd get one before the deadline but it was hard sometimes.

@porridgemathematics Hi, please see the message I've written.

my experience with group stuff is that mostly i ended up doing the work.

they knew i cared, so why bother

also my experience.

having some group stuff is reasonable, but not the sole evaluating criterion

my daughter was in a fairly motivated group, but my son had a different experience

my keychain comes from one of these arrangements. i partnered with someone on an exam in high school with the approximate understanding that i would do everything, and she bought me a keychain. 25 years later here it is, still aggregating my keys.

starting with the common core implementation in middle school

common core is really tough, particularly in math. previous iterations were less sensitive to teacher training, which is abysmal.

@vitamind A pseudoconvex domain is one which admits a plurisubharmonic exhaustion function. How well do you understand plurisubharmonicity?

It is my impression that US math students are falling behind the rest of the world.

we continue to lower standards