Mathematics - 2023-01-25

@TedShifrin yeah, the picture displayed in chat. Like how I'm God Cloth Seiya

12:20 AM

@PM2Ring There is this, too

@robjohn You have thereby worked challenge problems in three of my books!

@TedShifrin In that one answer? I feel accomplished!

For extra credit, without any calculations, use a bit of projective geometry to deduce that the middle part is a subset of a hyperboloid of one sheet.

@robjohn Indeed! I'd forgotten about that excellent answer, but I upvoted it last time you linked it. :)

@TedShifrin Does it have to do with the rulings?

12:30 AM

> Projecting this line onto a plane containing the axis of the cylinder and rotating the cylinder yields the line

I'm curious to learn of a simpler way than that ^ which shows you get a hyperboloid of 1 sheet.

It's nice to play with a pair of parallel circles linked by strings, so you can rotate them back & forth, seeing the cylinder transform into the hyperboloid.

I guess its even better if you use stiff wire instead of string, so you can see that the lines remain straight.

@robjohn Yessir.

Quite different from a cylinder in an important way!

The cylinder is the degenerate case, where the two sets of rulings merge into one set.

Indeed.

12:52 AM

sagecell.sagemath.org/…

2

That picture appears in two of my books. I drew mine with Mathematica, of course.

I'm wondering how an educator could take advantage of the interactive aspect of it.

It's a very satisfying structure. I'm sure it's been impressing people since basket weaving was invented. :)

Water tower structural stability!

Nice application 👍

1:03 AM

Also cooling towers. And rocket exhausts.

@user4539917 what’s the largest acute angle possible between the rulings?

hmmm...

Maybe an interactive application …

Have you decided about teaching the epsilon class yet professor? @TedShifrin

@user4539917 SageMath can easily be embedded into a Web page. Some educators make extensive use of it, eg math.gordon.edu/ntic/ntic/frontmatter-1.html

1:16 AM

That's a nice way to build a course/textbook @PM2Ring

@user4539917 By inaction I have de facto decided.

:(

@user4539917 Hint: In my program, set m=24, delta=6

2:05 AM

@TedShifrin two rulings at every point?

2:19 AM

@robjohn Right. Now what are the (projective) options for a doubly ruled surface?

2:45 AM

@TedShifrin Hyperbolic paraboloid and hyperboloid of one sheet

@robjohn does your logic teaching program have any graphics other than truth tables/charts.

3:01 AM

@user4539917 It has a derivation module (proofs), Invalidity (proving statements incorrect), Parsing (building a tree from logical sentences), Truth Tables, Recognizing Rules, Symbolization (translating English to symbols)

Not sure that there are any fancy graphics

I guess it depends on what kinds of graphics you're talking about

Any judicious use of colour.

@user4539917 errors are indicated in red, answers are marked in red and green in the selection list for incorrect and correct.

but not really much color

3:29 AM

What is a smooth curve?

The book first defines: two parametrizations $z:[a,b]\to \mathbb C$ and $\bar z: [c,d]\to \mathbb C$ to be equivalent if there exists a continuously differentiable bijection $T:[c,d]\to [a,b]$, with $T'(x)>0$ for every x in [c,d].

So here what I did was: consider S:= the set of all parametrizations from closed intervals in R to $\mathbb C$. Then I defined $\sim$ as follows: $z_1\sim z_2$ iff $z_1$ and $z_2$ are equivalent (as per the above given definition).

the definition should relate z and (oof on this notational choice) bar z to T in some way.

oh sorry, I forgot to add that one. we have $\bar z(y)= z(T(y))$ for every y in [c,d].

I showed that $\sim $ is an equivalence relation on S.

Then the book says: the family of all parametrizations that are equivalent to z(t) determines a smooth curve $\gamma \subset \mathbb C$, namely the image of [a,b].

I don't understand this part.

Is it true that : take z(t) in S, then $[z(t)]\in S/\sim$ is a curve?

$S/\sim$ is the set of all equivalence classes of $\sim$.

Or

smooth curve is not a map, rather an image.

completion of the third line above: namely the image of [a,b] under z with the orientation given by z as t travels from a to b.

that came out in bits and pieces but that's what it sounds like to me. the 'smooth curve' associated to the set of parametrizations is just the image of any one parametrization (as a point set), perhaps regarded with the orientation, perhaps not.

3:46 AM

@robjohn Projectively those are the same. Of course, there’s also the plane. There are fancy proofs and easier ones.

@leslietownes yes. Image of any element of [z(t)] is the same.

so it makes sense to speak of 'the image of [z(t)]'.

I understood it now. :-)

I have one question though: do you have any objections to the fact that S is a set?

i don't know what a "parametrization" is (the above only defines equivalence of them), but assuming that has a sensible definition, no, i have no objection to S being a set

oh I didn't define parametrization. A parametrized curve is a function from z from closed interval in R to C.

The book also does not seem to define 'parametrization' but defines parametrized curve.

I suppose both are supposed to mean the same thing.

@DLeftAdjointtoU Hi!!

feels like somewhere there could be differentiability hypotheses, or the author at least intended to include them, in the definition of 'parameterized curve'. otherwise the resulting [implicit?] definition of 'smooth curve' does not resemble what one would think.

that was added after defining the parametrized curve. The addition was called 'regularity conditions', which are as follows:

z(t) is continuous and differentiable on [a,b] with z'(t) non zero on [a,b].

I don't understand however why z'(a), z'(b) were required to be non zero.

3:59 AM

maybe for no reason, or maybe to avoid goofy edge cases. having a nonzero derivative is good because it gives you a well defined tangent direction.

but yeah, S is totally a set. you could realize it as a subset of the set of ordered pairs (A,B), where A is an element of {[a,b]: a, b in R, a < b} representing the choice of domain, and B is the subset of R x C representing the graph of the curve (plus whatever regularity hypotheses you want to tack on).

the pair set axiom and whatever axioms you use to encode a function as a set would be enough, if you accept that R and C are sets.

@leslietownes I recently heard of 'categories and functors'. I don't yet know what they are. But that made me doubt the setness of S.

4:37 AM

it helps that the domains of things in S are all subsets of a single common set (R) and the codomains are all, if not the same, at least subsets of a single common set (C). this isn't strictly necessary for set-ness, but it sure helps.

something like "all maps from one smooth manifold to another", okay, that could be trouble, or at least you might have to make a bunch of arbitrary choices if you wanted to realize that as a set. choices likely having little to do with whatever it was that you would want to study with that collection, but necessary to make sense of that thing as a set.

4:56 AM

Would that make the empty set the only set without any subsets.

Isn’t it itself a subset?

only one way for it not to be, which doesn't seem to get very far

5:22 AM

$\emptyset \subset \emptyset$ it just goes on forever

is there no end to nothingness

@robjohn :-) my granny used to churn butter from milk & salt. its quite a bit of work.

I wonder how much stuff would break in math if you just added an (extremely unjustifiable) axiom that stops that chain at some arbitrary (possible huge) finite number. Like "You can only have a sequence of inclusions of length $10^{100}$" or something.

nothing, because of transitivity.

That would disqualify sets with more elements than that.

@copper.hat Yeah, I suppose so. The other option is, depending on how you implement it, it contradicting transitivity and resulting in an inconsistent system of axioms.

So, either nothing breaks, or everything breaks

5:42 AM

@Rithaniel No more divergent series

No more series, period.

shintuku

no more sets because they are subsets of themselves

Yeah, if you try to get something like it to work, it just results in ultrafinitism

are you series?

(That is a terrible pun)

6:05 AM

:-) my strength

6:39 AM

💪🏋️‍♀️ (-:

Market is in red today.

Suppose that f is holomorphic in an open set $\omega$ and given that Re(f) is constant, how do I show that f is constant?

The problem is that $\omega$ is not given to be connected.

well, you do need to assume that omega is connected, or the result is not true.

for omega connected you can prove it in any number of ways, but it sounds like you know that.

Ohh

Leslie, it's exercise no. 13 in Stein and Shakarchi's complex analysis book

Part b) is to prove the same conclusion if Im(f) is constant.

part(c) is to prove the same conclusion if |f| is constant.

But the connectedness is not given. If it were given, then the result would follow by Cauchy Riemann equations.

6:59 AM

OK? you found a minor error in the book?

No, I thought that the result was true /more general.

But it appears that we do need connectedness.

11 mins ago, by leslie townes

well, you do need to assume that omega is connected, or the result is not true.

if omega is a union of disjoint nonempty open sets A and B, you can exhibit counterexamples to each of (a), (b), (c), e.g. for (a) consider f defined to be 0 on A and i on B

definitely a goof on their part

yeah, like in real case we could say: f being 1 on [0,1] and 0 on [2,3].

@leslietownes I had seen that already :D.

weird that a popular text would have such an error so early on, maybe it's in the errata or has been fixed in a new printing

user 726941

an early error usually drives away readers

(or at least me :)

7:37 AM

koro: CR equations are probably the closest path to the result from the definitions, but you should think about other proofs you might be able to give, using theorems about analytic functions.

9:45 AM

@Koro I think the reason is basically because your 'curve' $C$ here is supposed to mean a connected one dimensional manifold with boundary, and it turns out that for such objects, (and you can show this basically by definition), they always have a single coordinate chart, namely, a global parameterization $\gamma(t) : [0,1] \rightarrow C$

for any (lets stick to connected) smooth manifold $M$ of any dimension with or without boundary, a local parameterization is just a local chart

well, its the inverse of one

so given your $\gamma$ here really ought to be a chart map, and you are viewing $C$ as embedded in the plane or some euclidean space, you really want $\gamma'$ to have nonzero derivatives, including up to the endpoints.

in non-smooth category of course, i.e. not even $C^1$ or differentiable or something, you have a general concept of something called a rectificable curve

and you can actually do basically all of complex analysis with these things

but they are pretty badly behaved

(can be)

these are the images of maps of bounded variation from some closed interval of the real line into $\mathbb{C}$

and it makes sense to define line-integrals in terms of the push-forward measures they induce on their image w.r.t the lebesgue measure

and this is actually what you are doing anyway

what you end up getting here is their absolutely continuous part of the stieltjes measures they induce, in general $\gamma'$ is a.e. defined for these maps because it turns out bounded variation functions are always the difference of monotone functions

(here of course we are saying the real/imaginary parts are.. )

@Koro also for the other proofs, think about the open mapping theorem, if Re(f) or Im(f) or |f| or arg(f) even lol, are constant and your f lives on a domain (connected open set), thats bad news

Q: Show that $\lim_{r\to 1^-}\sum_{n=1}^\infty r^n a_n=\sum_{n=1}^\infty a_n$ if $\sum_{n=1}^\infty a_n$ converges.

11:19 AM

0

Given that $\sum_{n=1}^\infty a_n$ converges, it is to be shown that $\lim_{r\to 1^-}\sum_{n=1}^\infty r^n a_n=\sum_{n=1}^\infty a_n$. I tried to use the following version of integration by parts: $\sum_{n=M}^Na_nb_n=a_NS_N-a_MS_{M-1}-\sum_{n=M}^{N-1}(a_{n+1}-a_n)S_n,$ where $S_n=\sum_{i=1}^n b_i...

Does that not follow from Abel’s Theorem?

that's what (Abel theorem) is to be proven :)

The solution is available in theorem 8.2 of Rudin's PMA.

But I want to do it the way Stein Shakarchi suggested in one of the exercises.

11:49 AM

@robjohn it seems that we have to use that only.

and probably that's what Stein and Shakarchi also meant.

But I think that then there was no point in giving the hint that: use summation formula as it is not needed per se.

We have $\sum_{n=1}^Nr^n a_n =\sum_{n=1}^Nr^n (s_n -s_{n-1})=r^Ns_N+(1-r)\sum_{n=1}^{N-1}r^n s_n$

Taking $N\to \infty$, we get: $\sum_{n=1}^\infty r^n a_n =(1-r) \sum_{n=1}^\infty r^n s_n$ (assumption: 0<r<1)

From here: using $\epsilon, \delta$ definition of limit, the result follows.

The point is nowhere did I use the summation formula.

I have deleted the post.

12:19 PM

I dug up an old proof I wrote 10 years ago:

$$
\begin{align}
\sum_{n=0}^\infty a_n(1-z^n)
&=(1-z)\sum_{n=0}^\infty\sum_{k=0}^{n-1}a_nz^k\\
&=(1-z)\sum_{k=0}^\infty\sum_{n=k+1}^\infty a_nz^k\\
&=(1-z)\sum_{k=0}^\infty t_{k+1}z^k\\
&=(1-z)\sum_{k=0}^{N-1}t_{k+1}z^k+(1-z)\sum_{k=N}^\infty t_{k+1}z^k\\
\left|\,\sum_{n=0}^\infty a_n(1-z^n)\,\right|
&\le|1-z|\sum_{k=1}^N|t_k|+\frac{|1-z|}{1-|z|}\sup_{k\gt N}|t_k|
\end{align}
$$

If $z\in\mathbb{R}$ and $0\le z\lt1$, then $\frac{|1-z|}{1-|z|}=1$

darkexodus

Guys, what is wrong with option A?

it's the union, not the intersection.

darkexodus

I know its union. During test I couldn't find a counter which is covered by option A that should not be in P. Can you drop a hint?

That's because there are not any. It is the other way around. There are things in P that are not covered by option A.

For example $(0,3)\in P$, but $0\not\equiv3\pmod{12}$

darkexodus

12:35 PM

Thanks for the help! Don't know why I couldn't come up with a counter. I tried with some random values but it didn't help.

12:54 PM

@Koro does the exercise not specify "domain"?

that's how complex analysis textbooks usually work around stuff like this

I never read the book in question, to be clear

Bob Langefeld

1:09 PM

Question / conjecture in graph theory and combinatorics. In a graph, define 2nd order friends, as the friends of your friends, that don't intersect with your friends.

Now if for every vértice, we have that if selecting randomly a neighbor twice in a row

On average is 1st order friend and not a 2nd order friend

Then and only then we have that everybody has more friends than second order friends

Any thoughts?

@Thorgott still as a general rule of thumb, at least for a first course in complex analysis, its probably best to stick to thinking about only domains. apparently what was at issue was that the author said 'open set' and probably (i want to say almost certainly, if he wrote a whole textbook on the subject) meant 'domain'

also Im not going to bother with reading milnors proof of what we were talking about yesterday lol, i will just accept it and move on, it seems way too pedantic to worry about for me at least rn

not that the proof is unimportant, im sure if i understood it completely it would be useful, i just keep getting sidetracked by things like this

like, I ended up trying to get through a proof of seeley's extension theorem after I was annoyed by how awkwardly manifolds with corners/boundary's seem to be defined in the books I otherwise find really clear so far

i keep getting stuck in this loop of getting sidetracked by these annoying details when I should be working on pretty much unrelated things lol, idk if this is common for other students

i also wanted to , after koro posted about abels theorem, reduce that down to an application of a generalized form of the DCT on the right measure space lol.. im still convinced this is doable

(and the generalization of abels theorem to sectorial limits)

1:26 PM

yeah, that is very fair

its like, productive procrastination..

I never heard of Seeley's extension theorem lol

neither had I haha

smoothing corners is a small technicality, but it's not as bad

yeah, once my OCD issues with manifolds with boundary charts are resolved, I'm okay with manifolds with corners. I just delete them, and prove stokes on manifolds with corners via proving first: $\int_{M} \omega = \int_{M \setminus C} \omega$ ($\omega$ compactly supported, top degree), and $C$ the set of corner points

1:31 PM

yeah, same difference

then $M \setminus C$ is a manifold with boundary, (with the induced atlas from $M$, forgot to say $M$ is oriented), and I apply stokes as usual

these technical details of differential topology (gluing along boundaries, smoothing corners, etc.) are all explained well in Wall's book, but it still is something that's only worth worrying about if you have to

walls book, will add it to the list - and yeah i certainly dont have to, haha

these ad-hoc tricks usually suffice for what I need

2:05 PM

@Thorgott the domain was given as open set in complex space.

an error it is then

yeah

3:07 PM

When talking about inflection points, is it common to classify them as inflection point with positive slope and inflection point with negative slope?

Xander Henderson

3:23 PM

@ILikeMathematics I have never done so, but I don't see why you couldn't.

It's a property of the second derivative (changing from $f''>0$ to $f''<0$ or vice versa) so I don't know why you'd care about the first derivative, but do what you want

Oh, you know what, it can tell you if the downhill flow gets "stuck" or not

This feels like dynamics / catastrophe theory

4:11 PM

@AkivaWeinberger With "gets stuck", do you mean if there is an edge at the point of inflection?

amWhy

@XanderHenderson Since you've time, could you begin to address my question in the mods' office, at least to start?

^^^ Not an emergency, and if you've got to leave math.se altogether, to, say, teach, or prepare, or grade, I understand. Just let me know.

4:30 PM

@ILikeMathematics that is not a common classification

Thanks, Xander

Alright

Xander Henderson

@ILikeMathematics I didn't do anything...

You responded

@XanderHenderson Well, you said you've never classified them like that, so that implies that it's not very common and should answer my question

Xander Henderson

@ILikeMathematics Well, it isn't common in introductory calculus, which does not go deep; nor is it common in fractal geometry, where "differentiable" is not common, let alone twice differentiable.

So I am a biased source. :P

4:35 PM

Alright

5:04 PM

@ILikeMathematics Like if you load up desmos.com and compare x^3+x and x^3-x, the latter has a 'cup' where water can collect

so if a drop of water is flowing downhill it will stop in the middle instead of going off to minus infinity

user250478

5:24 PM

Hello to the chat, I wondered (as curiosity) if there are variants with fractals of optical illusions like than the Kanizsa's triangle or for the Ebbinghaus illusion. Wikipedia has the articles, in particular, Optical illusion and for Koch snowflake and it is well known the Sierpinski triangle.

5:42 PM

Up at almost 3 AM

hello!

'Sup

@robjohn After my comment before this comment of yours, I was going to comment-'I know that you will do it without using N that I used above, and that you'll directly handle infinity instead of handling N. And my prediction was correct. :-)

Not much.

How bout you?

Just contemplating existence itself nothing much

5:44 PM

I want to know if there is any result that talks about the following:-

Hm. I’m contemplating what would happen if stands really existed.

Would they be visible, invisible, or would they be the “missing texture”?

Suppose that f is defined on a set $\Omega\subset C$ that contains an open set U. Suppose that f is complex differentiable at some u in U, can it be said that f' is complex differentiable at u too?

I’m gonna learn how to render MathJax in my head. God help me.

@parz you mean in chat?

Yes.

Because mobile can’t render it.

5:47 PM

@AkivaWeinberger Yeah, thanks

@parz In the chat description, you'll find a link there after 'LATEX in chat'. If you follow the instructions there, you'll render mathjax.

I’m too lazy to click on links.

There is only 1- the minimum possible no. of links. :)

Rendering MathJax in my head is a much faster alternative.

Obviously.

there is that.

5:50 PM

@parz stands from JoJo?

Yes.

Probably invisible

I know that the conclusion I stated above holds if f is complex differentiable on an open set containing closure of a disk.

(i.e., the conlusion holds 'locally')

But I wonder if it holds 'pointwise'.

5:51 PM

Koffka ring illusion

Quite beautiful.

@user250478 Doesn't answer your question but you might find it interesting^

Three shades of grey throughout but at the end it looks like four

I’m going to code golf language. Woohoo.

TRANSLATION: v><>puttlang!!

Suppose that f is defined on a set $\Omega\subset C$ that contains an open set U. Suppose that f is complex differentiable at some u in U, can it be said that f' is complex differentiable at u too?
I know that the conclusion I stated above holds if f is complex differentiable on an open set containing closure of a disk.
(i.e., the conlusion holds 'locally')

Jakobian

6:22 PM

Is $C(X, \mathbb{R})$ separable where $X$ is locally compact separable metric when equipped with compact-open topology?

koro: yes, you would need some hypothesis like that to be sure that f' is even defined other than at u (i.e., to know that f' is something that can be differentiated at u). e.g. f(z) = |z|^2 is differentiable at 0 but at no other point.

thanks a lot, Leslie.

8:03 PM

@leslie Is Munchkin staying healthy and out of trouble for a change?

she has a lingering cough, but who doesn't this time of year.

yesterday one of the younger kids poured sand from the playground on her head, so munchkin pushed her into a chain link fence.

we tried to explain at home that this wasn't setting a good example for the younger kid, who had just turned 3 and maybe didn't know how to behave. munchkin responded, "she wasn't setting a very good example."

Catching a cold is so much fun

I love not being able to breathe without effort

Munchkin is like Tromp: She can do no wrong, and it's always everyone else's fault. Narcissist down pat.

I've no cough, but I've had a runny nose for weeks.

8:18 PM

Kyoto is really cold during this time

Well, it is winter ?

Yes.

$-4^\circ$C

Is that unusual for Tokyo? It's sort of usual for the northeastern US (where it gets colder indeed).

I don't know about Tokyo but its pretty cold for Kyoto

Oops. My typo.

8:25 PM

I was stupid enough to travel barefooted in that kind of cold with basically no warm clothes, just for fun

Hence, my cold

That sounds even stupider than I was at your age.

I've done lots of stupid stuff

Probably why I have negative friends

I'm not sure about that, unless your stupidity negatively impacted them.

Well, I mostly helped them with homework. Particularly mathematics homework.

But that's all, other than that, none

8:47 PM

@Koro You can replace the infinite sums with sums to $M$ and not change the inequalities, then let $M\to\infty$.

9:52 PM

Does anyone else like the Trash chatroom? Just me?

UnderMathUate

11:16 PM

I'm struggling with this proof right now (it's from my Real Analysis course)

Let $S$ be an ordered set. Let $B \subset S$ be bounded (above and below). Let $A \subset B$ be a nonempty subset. Suppose all the infimums and supremums exist. Show that $$inf(B) \leq inf(A) \leq sup(A) \leq sup(B_)$$

I have an idea of why this works. I drew a picture to illustrate the situation.

I'm just not sure how to start :/

What’s an obvious upper bound of $A$?

UnderMathUate

An upper bound of B?