Mathematics - 2022-12-24

Ted Shifrin

00:18

Happy holidays, all.

Paramanand Singh

00:44

@TedShifrin the question has some close votes as well and it may get closed for lack of context. Some users (myself excluded) also down vote answers to such questions. That maybe the case here.

@robjohn let's also get that by a set theorist (Asaf) : math.stackexchange.com/a/3731233/72031 (this includes my comment as well)

Ted Shifrin

01:12

I refuse to be so pedantic as to insist that the rationals aren’t a subset of the reals and the reals aren’t a subset of the complexes. Sue me.

leslie townes

the next thing ted will try to tell us is that an integer isn't an equivalence class of pairs of natural numbers, or that the natural numbers don't include 0.

i think i agree with you, ted. i'm not sure i would call the set theoretic distinctions 'pedantic,' as much as i would say, there is no reason to presume that everything mathematical "really" starts with some kind of set theory that we either choose to ignore (by identifying different sets or choosing particular embeddings) or not. many mathematical conversations just don't start with, or even involve, set theory.

lawvere theory, on the other hand...

Derivative

01:28

Am I going nuts? If $A$ is a PID then $A[x]$ is a PID? I thought the integers provided a counterexample

PM 2Ring

@leslietownes Lawvere & Surely was a great sitcom.

Ted Shifrin

@Derivative UFD, not PID

Paramanand Singh

@Derivative I think that works for UFDs and not PID's

Derivative

my professor thinks otherwise

Ted Shifrin

he misspoke

Your book did not.

Derivative

01:34

he put this on the test and marked it wrong

god damn

and I emailed him about it and he's insisting on it

Ted Shifrin

Then give him the counterexample, with proof.

Derivative

this is gonna go over well

I'm going to try anyway

Ted Shifrin

This is in (pretty much) every text.

Be polite.

If a student showed me I’d made an error, I was usually very grateful (and usually impressed).

Oliver Díaz

@Koro: Every vector space $V$ over a field $F$ has a Hamel basis, that is, there is a set $B\subset V$ such that $B$ is linearly independent (any finite linear combination $a_1v_1+\ldots + a_Nv_n=0$ with $a_j\in F$ and $v_j\in B$ yields $a_j=0$ for all $1\leq j\leq N$) and for any $x\in V$, there is a finite set $\beta_1,\ldots, \beta_N\in F$ and $u_1,\ldots,u_N\in V$ such that $x=\beta_1u_1+\ldots +\beta_N u_N$.

Derivative

that has been my experience with my professors as well

but this guy...

Ted Shifrin

01:37

This was a T/F question?

Derivative

yes

he's sort of nuts. This is the first class he's teaching, and he's going to fail most of the class

Oliver Díaz

@Koro: All this depends on the Axiom of Choice (or equivalences of it, for example Zorn's lemma).

Derivative

there are 21 students enrolled and only 12 showed up for the final

Ted Shifrin

Never taught before?! Unusual to start at this level.

Derivative

he was a TA before at most

Ted Shifrin

01:38

If he continues to insist, you might go see the department head.

He’s a postdoc?

Derivative

no, postdocs don't teach in my country

Ted Shifrin

What’s his field?

Derivative

in my university*

he's an algebraist!! He has publications on algebraic number theory!

shocking right

well I sort of don't want to cause trouble, because I'm gonna need to ask the department head a lot of other things in the future

Ted Shifrin

Oh, this is ridiculous. Are you sure you’re telling us the complete question?

leslie townes

@PM2Ring :D :D :D

Derivative

01:41

"Se A é domínio de ideais principais, então A[T] é domínio de ideais principais." T/F?

you get the pciture

Oliver Díaz

@Koro: This posting may be of interest to you

Ted Shifrin

Crazy. Be polite but persistent (with proof).

leslie townes

that's really something else. i had a near-incompetent postdoc as an undergrad, but they weren't wrong-bad, just, bad-bad.

Derivative

I'm going to take classes in another university and I'm gonna have to ask the department to approve that the classes count towards my degree

so I didn't want to be the person to do this

Ted Shifrin

Don’t be scared to be right. Just be polite.

The department head should not be happy that less than half the class showed up to the final.

BTW, this was my job (associate head) for 8 years. Ask for approval before, not after, you take the class.

PM 2Ring

01:45

For rather large values of half. ;) 12/21 turned up.

Derivative

no, no the process is different here

you can consult before, but the final decision is only made after taking the class

different country

Ted Shifrin

I understand, but do consult.

Derivative

I did

Ted Shifrin

OK

Derivative

thanks though

Ted Shifrin

01:47

Can you prove $\Bbb Z[T]$ is not a PID?

shintuku

02:23

do we get interesting qualitative differences in cyclic groups as we consider higher and higher orders of them?

leslie townes

shin: the cycles get longer

shintuku

which, of course, means the group tends to be healthier

Ted Shifrin

Not really.

Amin Idelhaj

02:47

I heard there's a party in here, how are you guys doing?

Ted Shifrin

03:36

Heya, Demonark.

robjohn

06:46

@TedShifrin I am not so pedantic, either; for the most part, I consider $\mathbb{Q}\subset\mathbb{R}$ and $\mathbb{R}\subset\mathbb{C}$. However, I don't deny that there is a place for such pedantry, and so I brought up that $x\mapsto x^2$ on $\mathbb{R}$ might not be considered the same as $x\mapsto x^2+0i$ in every sense.

Yai0Phah

@robjohn I think that, unless in the context of ZFC or set-theoretic constructions, one does not distinguish them. Subset really means a chosen embedding in normal math.

It is an arbitrary isomorphic thing, but a chosen thing.

robjohn

As I said, I am not so pedantic, and to keep from going mad, I consider $\mathbb{R}\subset\mathbb{C}$.

we often mod by isomorphisms

Yai0Phah

Complex numbers have a nontrivial Galois group over rational numbers.

We only fix an embedding from R to C.

robjohn

07:01

$i$ is a solution to $z^2+1=0$, but which one?

Yai0Phah

Sometimes there are real mathematical consequences. If I remember correctly, Serre has an example of an algebraic variety X over a number field K such that the topological space of its analytification varies when we conjugate K in C.

leslie townes

uh oh, now i don't know how i'll sleep at night.

Yai0Phah

Maybe the homotopy type also varies.

leslie townes

maybe we should go back to saying there aren't any solutions

Magnus Alexander

10:36

Can anyone give me a hint? $\sum\limits_{n=2}^{\infty} \dfrac{1}{n^{\alpha}-1}$ converges just as like as $\sum\limits_{n=2}^{\infty} \dfrac{1}{n^{\alpha}}$ ?

BAYMAX

2

Q: A pen-and-paper proof for a matrix implication.

Suppose $A = \begin{bmatrix} x & 1\\ y & 0\end{bmatrix}, B = \begin{bmatrix} z & 1\\ w & 0\end{bmatrix}$, for $x,y,z,w \in \Bbb{R}$. I have observed by considering many examples of $x,y,z,w$ that: If all the eigen values of $A^2B$ and $AB^2$ are less than one in absolute value $\implies$ $\det(A...

If you have any ideas!

Jakobian

11:13

@robjohn how I see it is that rationals, real numbers, all of those further constructions are as follows

after already having constructed the rationals, we forget about the original natural numbers, and instead use the copy of them

or you could always redefine complex numbers so that $\mathbb{R}$ is a subset of $\mathbb{C}$ in the literal way

then there's no problems in its definition

cristallo

13:34

1

Q: How do I prove that this map on the quotient space is well defined?

Let $K$ be a field and $X\subset K^n$. Then define $$\mathcal{O}(X):=K[X_1,...,X_n]\big/I(X)$$ where $I(X):=\{f\in K[X_1,...,X_n]: f(x)=0~~\forall x\in X\}$. I need to show that $$\phi:\mathcal{O}(X)\rightarrow Map(X,K); ~~~~~\bar f\mapsto \left(\phi(\bar f):x\mapsto f(x)\right)$$ is well define...

abstract-algebra algebraic-geometry solution-verification commutative-algebra quotient-spaces

can someone help me here?

Adil Mohammed

13:50

Why is slope of the sphere $x^2+y^2+z^2=1$ at the point (2/3, 1/3, 2/3) equal to $\frac{\partial y}{\partial z}$

when i visualise the partial differentiation of z wrt y, i am picturing a line thats parallel to yz plane.... hows that the tangent of the line?

oh wait that part is mentioned somewhere else in my textbook, brb

Magnus Alexander

14:19

Hello, does anyone know how to reduce the power in the differential equation $y''=xyy'$?

Simd

15:26

I have a basic embarrassing question. Take this equality...

the left part seems to be the product of two terms. The first is a function of x and y and the second is a function of t and x

but the righthand part is a functjon of x and y only

what simple thing am I missing?

Ted Shifrin

It is not a product. It’s stupid engineering notation for iterated (double) integrals.

Simd

@TedShifrin oh!!

I would never have guessed that

@TedShifrin how did you know this?

AMDG

15:51

There's something particularly nice I realized about differentiation of the exponential functions. Tell me something: how do most people learn (in public school) what the derivative of the circular functions are? I assume if anything it's just rote memorization as usual and they just give a formula. Is that correct?

shintuku

yeah, the implicit function formula

ah nvm, you said circular function, as in the trig ones?

AMDG

Yep

shintuku

it consists in learning soh cah toa and how to compute trig functions with the calculator

and their derivatives are just given

AMDG

Cringe

shintuku

sometimes they make nice little pictures for the intuition of their derivatives

i was shown that traveling slope .gif for the intuition of the derivative

Feynman_00

15:56

I guess using the definition of the derivative and $\lim_{x\to0} \sin x/x=1$ limit (?)

AMDG

Well, there's an equally nice and intuitive algebraic understanding of the derivatives of sine and cosine: $$\frac{d^n}{dt^n} \cos(t) = \Re(\frac{d^n}{dt^n} e^{it})$$ and of course sine has it's analog in there.

Where $i=\sqrt{-1}$

Bruh, just show them this. It hits like a train once you realize it.

(oh and $t\in\Bbb{R}$)

Anyways, just thought I'd mention it. It came to me like a week ago or so.

shintuku

cool stuff!

AMDG

:D

Also, if complex-valued functions are "4D", then why not just provide a representation as an animation with three dimensions and one dimension of time? Make a gif :P

Nevertheless I think just showing the two planes for the domain and codomain is simpler and more intuitive. It's translating points on the complex plane, not necessarily building a 4D structure per se even if it can be reified as that.

I mean, technically you can just present real-valued functions the same way as one plane showing the set of points $(x, 0)$ and the other showing $(0, y)$ or $(x, y)$. The way I see it, when not isolating domain or codomain, the image of the relation is a distinct shape unique to that relation.

Joe Shmo

17:24

math.stackexchange.com/a/2278507/270202 : "the singularities at $\cos z = 0$ are removable..." why?

robjohn

18:16

@JoeShmo Since $|f(z)|\le M|\cos(z)|$, $\left|\frac{f(z)}{\cos(z)}\right|\le M$. If $f(z)$ is bounded in a neighborhood of $z_0$, then any singularity at $z_0$ is removable.

Ted Shifrin

Because $f$ must vanish there too. Or read the comments. It’s officially the Riemann extension thm.

Hi @robjohn. Happy holidays!

Joe Shmo

@robjohn oh I see. Just because..

Goku カカロット

Anyone enjoying holidays? I'm not

robjohn

@TedShifrin To you, too! It will be 82° here on Sunday

Ted Shifrin

While our friends eastward freeze badly!

Joe Shmo

18:18

It's COLD

Happy holidays to all!

shintuku

holidays, i.e., unlimited math time

robjohn

@JoeShmo No, that is part of Riemann's Theorem: if $f$ is holomorphic away from $z_0$ and bounded in a neighborhood of $z_0$, then any singularity at $z_0$ is removable.

Joe Shmo

ah, I guess what I'm saying is (and to paraphrase Ted) that if $f$ didn't vanish at $z_n = (2n+1) \pi$, then the quotient $\frac {f(z_n)} {\cos z_n}$ cannot be bounded (in modulus)

robjohn

It is true that when an analytic function vanishes, it vanishes to order $1$, but that is also something that needs to be proven

Joe Shmo

It can vanish to order $\ge 1$ as far as I'm concerned, but I need order at least $1$. Am I doing something stupid?

meaning, can I get away without Riemann extension theorem here?

also, while I have you, there's the following claim here: $\frac 1 3 < |\cos z| < \frac 5 3$ on $|z| = 1$. The upper bound I have from the Taylor expansion of $\cos z$ (achieved at $\pm i$). How to get the lower bound?

Ted Shifrin

18:32

You’re going to reprove it. One line proof.

Joe Shmo

I'm going to reprove what?

Ted Shifrin

Why not the lower from Taylor?

Joe Shmo

one sec ^

also, reprove what? the upper bound? there's a better proof?

Ted Shifrin

@Simd It's standard engineering/physics notation. I do not like it, but they use it. I write $\int_a^b \int_{f(x)}^{g(x)} h(x,y)\,dy\,dx$.

lorenzo

Hello and happy holidays. If someone has some time to spare I would like to have some feedback on a proof of mine ( math.stackexchange.com/q/4604900 ). Thanks

Ted Shifrin

18:40

no, reprove Riemann extension.

Joe Shmo

👍

Joe Shmo

18:52

OK, yeah, I can get this wolframalpha.com/… as the lower bound from Taylor, but I wouldn't know that it's $> \frac 1 3$ without wolfram

I mean I could probably go from there by induction, but that leaves a lot to be desired

Feynman_00

I don't know if this is true but...

Joe Shmo

there's gotta be a complex trig identity here that does the job (evident by the $\cos,\ \cosh$ dance on wolfram)

Feynman_00

Joe Shmo

This cat belongs to Schrodinger, Feynman. A physicist's favorite animal. Indeed, a proof is simultaneously true and false, until the cat calls it one way or the other.

The things you do to this animal in the name of science, of course.. are repulsive.

Feynman_00

@TedShifrin I agree with you about the latter being a better notation, although I should mention I've only used the notation proposed by OP in Real Analysis only and the one you proposed in Physics sometimes

A notation we use more frequently in physics is $\int_a^b dx\int_{\alpha(x)}^{\beta(x)}dy f(x,y)$

@JoeShmo Schrödinger hated cats unfortunately (seriously)

Joe Shmo

19:00

I'd never have guessed.

I myself am a dog person.

Feynman_00

And so was E. Schrödinger :P

Here something about it

Joe Shmo

19:17

^not by induction, but I'm getting the bound I need, via left methods (and "preknowing" that $\cos 1 \approx 0.54 < 0.55$)

Derivative

20:12

yikes

Yai0Phah

In fact, I don't know how to construct natural numbers.

shintuku

@Yai0Phah $n = \{ n-1 , \cdots, 1, 0 \}$

where 0 is the empty set

have fun

Yai0Phah

What axiom are you using?

Derivative

Look into the Peano axioms

Yai0Phah

I am not talking about Peano axioms. I am talking about constructing natural numbers in set theory.

shintuku

20:17

yeah they're usually constructed with peano axioms

Yai0Phah

No, Peano axioms are an axiomatization of natural numbers.

shintuku

whoops maybe, in any case, @Yai0Phah the construction i'm using is $x^+ = x \cup \{ x \}$, where $x^+$ is the natural number after $x$ and $0$ is given by the empty set

it's the construction that starts at page 38 of Goldrei's Classic Set Theory

Yai0Phah

OK, that is precisely axiom of infinity.

I have never learned set theory.

To see that the induction axiom holds for this construction, do we also need the axiom of replacement?

Ted Shifrin

20:39

@JoeShmo Review what you know about alternating series. Doesn't that do it?

Joe Shmo

well that's how I prove it essentially

you end up with $|\cos z |\ge 1 - \sum_{n=0}^\infty \frac 1 {(2n+2)!}$, the latter term you could write as $\cos 1 - \epsilon$, $\epsilon < \cos 1$ here

Ted Shifrin

It's something most calculus students learn. If you have $\sum (-1)^n a_n$ with $a_n>0$ and decreasing, then the error between a partial sum $\sum_{n=1}^k a_n$ and the series is simply $a_{k+1}$.

You get a bounding interval by going back to the preceding negative term.

Joe Shmo

lost you. where do you apply this ^^ fact?

Ted Shifrin

For example, if you have $a=1-1/3+1/5-1/7+1/9 - \dots$, then $2/3 < a < 2/3+1/5$.

Joe Shmo

oh..

dur

Ted Shifrin

20:52

I didn't pay attention to your particulars, but I assume that's all that's going on.

Joe Shmo

yeah..

it's not really that the error term of the partial sums is $a_{k+1}$ though

Koro

I have a small confusion in the following statement:X and Y are topological p: X-->Y is a quotient map is equivalent to the fact that p is continuous and maps saturated open sets to open sets.

Joe Shmo

it's just true that $a > 1 - \frac 1 3 = \frac 2 3$

Koro

Is the statement true?

Joe Shmo

since the rest of the terms are positive

Koro

20:54

I think that we must add surjectivity of p as well.

Ted Shifrin

@JoeShmo I thought you had the cos Taylor series. What do you mean, the rest of the terms are positive?

Joe Shmo

in your example

Ted Shifrin

My example was meant to be permanently alternating!

Joe Shmo

$a = (1 - 1/3) + (1/5 - 1/7) + \ldots$

Ted Shifrin

Notice the $-$ before the $\dots$.

Joe Shmo

20:57

yeah I know

Ted Shifrin

Oh, yeah, you can look at it that way. Or you can think about the even partial sums increasing and the odd partial sums decreasing (or vice versa)>

Joe Shmo

yeah no, but my logic doesn't work in the complex case though

which is probably the trouble I ran into

does yours (?) For $|\cos z| = |1 - z^2/2 + z^4/4! - z^6/6! + \ldots$| the above reasoning (at least what I said) doesn't help me

I'm lying. it does help me, eventually.

Ted Shifrin

21:21

I was doing it with $|z|$ inserted.

So then you need to use triangle inequality (both usual and reverse).

Joe Shmo

yeah

I'm not quite doing it like you suggest, but same same

basically, just to put this to rest:
$$|\cos z| = |1 - \frac {z^2}{2!} + \frac {z^4}{4!} - \frac {z^6}{6!} + \frac {z^8}{8!} - \ldots| =\\
|1 - |\frac {z^2} {2!}|\cdot|1 - \frac {z^2} {4 \cdot 3}| - |\frac {z^6} {6!}|\cdot|1 - \frac {z^2} {8 \cdot 7}| - \ldots| \ge\\
|1 - \frac{1}{2!} - \frac{1}{6!} - \frac{1}{10!} - \ldots| \ge\\
1 - \frac 1 2 - \frac 1 6 > \frac 1 3$$

no thats not true.... ^ whatever, I'll get back to it

Ted Shifrin

Wow. This is definitely more complicated than I was thinking. Let me ponder with pencil.

Joe Shmo

I'm getting it to work with the $\cos 1$ business

but nothing straight forward

Ted Shifrin

Is their particular lower bound, as opposed to say, $1/2$, which is easy as I was describing earlier, actually important?

Joe Shmo

1/3

no its meant as a prequel to the following Rouche question

wolfram says that it's just under 1/2

1/3 is actually a very loose bound

basically the third line up there can be written as $1 - \cos 1 + \text{extra}$, and that's as good as I'm able to get it for now

Ted Shifrin

21:35

So, upper bounds are trivial with just inserting $|z|=1$ and using triangle inequality. The lower bound is a bit more subtle.

Joe Shmo

yes

Ted Shifrin

Here's what I'm thinking. Write $|\cos z| = |1-A| \ge 1-|A|$, where $A=z^2/2-z^4/4!+z^6/6! - \dots$.

Joe Shmo

yeah but now the triangle inequality isn't working in your favor

Ted Shifrin

Now we want the largest $|A|$ can be, so that will just be $1/2 + 1/4! + 1/6! + \dots$ for a definite bound.

Yes, it is.

Unless I'm being stooopid.

So that sum is just $\cosh 1 - 1$, which is a bit bigger than $1/2$.

Joe Shmo

that's what I have more or less though

Ted Shifrin

21:40

So $1-|A|\ge 2-\cosh 1 \approx .457$. Way better than $1/3$.

Joe Shmo

i.e., you have to know what $\cos, \cosh$ is at $1$

yeah yeah I have that bound

Ted Shifrin

I think that's allowed, but I still don't know why we care about this.

Joe Shmo

with cosine, not cosh

Ted Shifrin

It should be with $\cosh$, unless my logic is flawed.

Joe Shmo

I fanagled it so it looks like cos(1)

Ted Shifrin

21:42

I'm not sure what value of $z$ on the unit circle actually maximizes $|A|$. It's not obvious to me right this second.

Joe Shmo

yeah, its not as obvious as the upper bound, and I don't love that

I wouldn't know on the quals that cos(1) ~= .55

Ted Shifrin

On a qual no one would ask for that. It's not unreasonable to ask you to get to $2-\cosh 1$ as an upper bound

What I did is just one or two straightforward lines, no trickery.

Joe Shmo

I agree that $2 - \cosh 1$ is a perfectly valid upper bound

but the question explicitly asks for 1/3

I know -- I had this bound all along (again, with cos, not cosh)

Ted Shifrin

Was this an actual qual question with the 1/3?

Joe Shmo

yes

Ted Shifrin

21:50

Bad question.

Move on.

Joe Shmo

fwiw, how it works with $\cos 1$ is for example
$$
|1 - \frac{1}{2!} - \frac{1}{6!} - \frac{1}{10!} - \ldots| =\\
|\cos 1 - (\frac 1 {4!} + \frac {1}{8!} + \ldots)| \ge\\
\cos 1 - \frac 2 {4!} > \frac 1 3
$$
since, say,
$$
|\frac 1 {4!} + \frac {1}{8!} + \ldots| \le\\
\frac 1 {4!} (1 + \frac 1 2 + \frac 1 4 + \ldots) =\\
\frac 1 {4!} \cdot 2
$$

Ted Shifrin

No, your signs aren't right for $\cos 1$.

Joe Shmo

where

you add and subtract

that's what the second line is

I think it's good (?) everything permutes obviously because the whole thing is absolutely bounded by $e$

Ted Shifrin

22:07

Oh, I see.

Joe Shmo

anyway, I've been looking for something better than this ^ since the morning, and I'm coming up short. wasted the entire day on this nonsense

Ted Shifrin

Yeah, stop wasting time on it.

People do often write bad qual questions. Things happen. We changed the structure of qual committees at UGA in an effort to minimize that. When the "non-expert" did his/her job (and actually tried to work the exam and made the "experts" change things to make it doable/correct, the system works. Sadly, typically the non-expert is way too lazy to bother (or too ignorant). But when I was the "non-expert" in both real analysis and algebra, I actually did spend the time/effort to improve the exams.

Not that the students excelled for the most part, anyhow.

Joe Shmo

what was the derivative criteria for the order of poles

Ted Shifrin

I never use/teach that.

Just think about the Laurent series.

Joe Shmo

you have a pole of order $n$ if $f^{(n)}(z)|_{z=z_0} \ne 0$, for the least such $n$?

Ted Shifrin

22:19

That's clearly nonsense.

Joe Shmo

k let me write out the laurent series

oh there are a few good questions here

Ted Shifrin

You can deduce the criterion (which has limits in it, not derivatives) from the Laurent series, immediately.

Joe Shmo

well that I know

the smallest negative index $m$ appearing in the laurent series

is the order of the pole $z_0$, the laurent series being expanded about the pole $z_0$

Ted Shifrin

Right.

Joe Shmo

ok, are singularities at $\infty$ special?

theyre always isolated or something (why do I care?)

Ted Shifrin

22:24

No, they aren't always isolated!

What about $1/\sin(z)$?

Joe Shmo

$1/ \sin z$ has a simple pole at $2 n \pi$, no?

$n \in \mathbb{Z}$

evident from the taylor expansion

Ted Shifrin

$n\pi$, yes, but you were asking about $\infty$

Joe Shmo

yeah, excuse me

let me write it out. at $\infty$ means expand in $1/z$, take $z \to 0$

Ted Shifrin

So it'll look like an essential singularity. But is it?

Goku カカロット

Turkish Kebab, anyone?

Joe Shmo

22:32

yeah I have no idea what's happening

Ted Shifrin

Sure, as long as it isn't dried out.

Joe Shmo

things on the denominator are acting up

Ted Shifrin

Well, if you have a valid Laurent series expanded about $z_0$, then my definition of essential singularity is that there are infinitely many terms with negative exponent. But what does valid mean?

$\sum \dfrac1{n!z^n}$ is an example of such a Laurent series. But what do I mean by valid?

Goku カカロット

@TedShifrin it is dried out

Ted Shifrin

Aww :(

That's sadly too typical.

Goku カカロット

22:36

Don't worry you can just put it in some water and then eat like. Like dunking oreos in milk

Ted Shifrin

Yuck.

And I don't like oreos, either.

Goku カカロット

Ever had a chicken flavored KitKat?

Joe Shmo

oh, Ted.. $1/6! + 1/10! + \ldots \le 2/6!$ whence the sum up there should be $> 1/2 - 2/6! > 1/3$

no cosines, no nonsense

that's what they wanted probably

Ted Shifrin

OK, and why is $1/10! + 1/16! + \dots$ obviously $<1/6!$?

Joe Shmo

for the same reason that $1/4! + 1/8! + \ldots < 2/4!$

Ted Shifrin

22:39

Did you tell me that reason?

Joe Shmo

yuh

Ted Shifrin

Anyhow, you changed the subject.

Joe Shmo

yes, yes. it just hit me like lightning when I wasn't thinking about it

back to our biz: I have no clue what a "valid" laurent series is

Ted Shifrin

It has to converge on an annulus.

Joe Shmo

ok

and what does the choice of annulus decide?

Ted Shifrin

22:43

Well, is the function we're talking about holomorphic on $0<|z|<r$ for some $r$?

Joe Shmo

your function should be, yes

Ted Shifrin

What about $1/\sin(1/z)$?

And why should mine be?

Joe Shmo

for example, $1/\sin(z)$ should be holomorphic on $0 < |z| < \pi$

Ted Shifrin

But we're not looking at that.

But I agree that that is true.

Joe Shmo

ok

dont know whats going on with $1/\sin(1/z)$

isn't that conformal mapping stuff?

well, ok, $\sin(1/z)$ has an essential singularity at $z=0$

Ted Shifrin

22:56

No, it does not. Because by definition an essential singularity must be isolated.

Joe Shmo

oh?

meaning, it's (laurent) expansion about $0$ is invalid?

CroCo

@Yai0Phah I don't understand what you mean by "a relative concept"? The convexity definition is clear.

Ted Shifrin

Right, because there are holes in the annulus accumulating at $0$.

Joe Shmo

I'm losing you. what holes? Every annulus has a hole

Ted Shifrin

You did not read what I said.

Where is $\sin(1/z)=0$?

Joe Shmo

23:04

nowhere. only tends to $0$ as $z \to \infty$

Ted Shifrin

Say what?

Joe Shmo

oh, sorry

at $z=\frac 1 {n \pi}$

Ted Shifrin

Uh huh, which limit to ... ?

Joe Shmo

0

Ted Shifrin

Precisely.

Joe Shmo

23:09

I see

so the answer to your question is no

Ted Shifrin

Whatever the question was, I've forgotten. But probably so. What about the other series?

Joe Shmo

every annulus $0 < |z| < r$ contains a singularity of $1/\sin(1/z)$

Ted Shifrin

Right. So not isolated singularity.

Joe Shmo

what is an isolated singularity?

one that's isolated from all the other singularities?

Ted Shifrin

Hasn't your course/text covered all this?

Joe Shmo

23:11

long, long ago

Ted Shifrin

But, yes, obviously so.

Joe Shmo

and maybe, who remembers anymore

ok, so all of that gives me what?

there is no valid laurent expansion of $1/\sin(1/z)$?

Ted Shifrin

I was warning you that not every Laurent series with infinitely many negative powers represents an essential singularity.

To talk about a Laurent expansion, the function must be holomorphic on some annulus $r<|z|<R$.

Time to reread some basics.

Joe Shmo

ok, what do I read?

Ted Shifrin

I am actually disappearing to the kitchen.

Your favorite textbook on complex analysis would be good.

Joe Shmo

23:14

jeremy orloff doesn't have notes on singularity classifications

Yai0Phah

@CroCo You don't say that a set $E$ is convex. Rather, you say a subset $E$ of a real affine space $X$ is convex.

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