Mathematics - 2023-07-25

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robjohn

00:44

@XanderHenderson pancakes cooked in an oven?

Usually, I see them cooked on a griddle, or in a pan.

Not covered, however.

冥王 Hades

I’m surprised I haven’t developed diabetes or some kind of medical problem associated with eating too much

shintuku

aeporreca.org/blog/semiring-of-dynamical-systems

coooooooooooooooooooooooooooooooooooooooooooooool

Xander Henderson

01:00

@robjohn I do them on the stove.

In a covered cast iron skillet.

At least, that is how the Japanese style of "fluffy pancake" is made.

Though, like I said, they are really more like mini souffles.

They are very yummy.

冥王 Hades

01:27

Today I learned that donuts can be filled with cheese too

the inventor must be a genius

Ted Shifrin

Blech.

冥王 Hades

Cloox.

Xander Henderson

01:43

@冥王Hades What kind of cheese?

Cream cheese? ricotta? quark?

Jakobian

@冥王Hades you also learned that you can be filled with cheese by eating those

Xander Henderson

Sweetened? or savory?

I've made donuts that I have then filled with filling made of quark, honey, and very finely diced dried apricots.

They were great.

Ted Shifrin

Apricots sound fantabulous. Cheese, not so much, unless it's slightly sweetened ricotta.

D.C. the III

Then we're not talking about donuts anymore are we?.............Today I've observed Xander is the Master Baker to Ted's Master Chef

Ted Shifrin

@Xander might take offense at that.

Xander Henderson

01:54

@TedShifrin Quark is kind of like ricotta. And, like I said, I added honey. :D

Jakobian

typical donut in here doesn't have a hole and they are usually filled with some filling

cheese is one common choice

nothing out of the ordinary

Xander Henderson

@Jakobian What kind of cheese?

Jakobian

I'm more of an eater

white cheese of some kind

Xander Henderson

@Jakobian ... :/

Jakobian

is what was in all of mine

Xander Henderson

01:58

there are so many cheeses that are "white"

like, most of them, maybe

Jakobian

cottage cheese maybe

most recipes I see seem to mention either cottage cheese or plainly white cheese

Xander Henderson

Cottage cheese could work.

It is the right thing for blintzes.

Jakobian

they recommend it to be semi-fat

Xander Henderson

FULL FAT OR DEATH!

Jakobian

or even lean cheese

user image

D.C. the III

02:03

If one is having a pastry you're not eating it to be health conscious.................

Jakobian

donuts in Poland look like this and they are usually filled with jam, cheese, or other things

@D.C.theIII that's why the recipes are probably saying it because it tastes better if it's semi-fat

D.C. the III

I couldn't imagine that being the case. the full fat will just give so much more richness and flavour.....but I'm not a baker....

Jakobian

marmalade is another common choice for a filling of our donuts

Ted Shifrin

02:56

Better than the way too sweet everything in the US.

1 hour later…

MathCrackExchange

04:17

Damn those look tasty

3 hours later…

I am kind of a language dev

07:03

@Jakobian fill it with peanut butter! 🥜🧈

2 hours later…

Unknown x

09:21

user image

How does $f(0)=3$ in the converse argument?

I was stucked at $f(0)=3$
$|f(0)|\leq 3$ can be proved using Cauchy-schwartz inequality. I am unable to prove $f(0)=3$.I even substitute the values $T^\star(0)=AT(0)$,$N^\star(0)=AN(0)$ and $B^\star(0)=AB(0)$ in the $f(0)$ expression. I was unable to reduce the expression. Could you give me hint?

Unknown x

09:36

If $\tilde{T}(0)={T}^\star(0)$,$\tilde{N}(0)={N}^\star(0)$ and $\tilde{B}(0)={B}^\star(0)$. I can say $f(0)=3$. Here all the unit vectors are different. How did the author got $f(0)=3$?

PNDas

10:11

If $E|X|<\infty$ then you can write the characteristic function $\varphi_X(t)\sim 1+itEX$. Can I make the hypothesis weaker? e.g. $E|X|=\infty$?

Jakobian

10:48

@PNDas wdym

If it's infinite then E[X] doesn't exist

If $\varphi_X$ is differentiable then $-i\varphi_X'(0)$ serves as a generalization of $E[X]$ and then the formula holds, I suppose

Jakobian

11:12

1

A: Differentiability of characteristic functions and moments of the corresponding measure

Zygmund's example is the discrete random variable $X$ where $\mathbb{P}(X=\pm n) = \frac{C}{n^2\log(n)}$ for integer $n \geq 2$. $C$ is a unique constant that makes this a probability distribution. You can calculate that $X$ does not have a finite first moment because $$\mathbb{E}(|X|)= \sum_{n=2...

one potato two potato

12:02

what is the image of the horizontal lines on the upper half plane of $\Bbb C$ under mapping $z\mapsto z^{1/3}$?

one potato two potato

12:15

nvm

Mats Granvik

@robjohn Do you know how to program the Euler Maclaurin summation formula for a general function in Mathematica?

5

Q: Euler-Maclaurin summation

I want to compute asymptotic approximations to partial sum of harmonic series in Mathematica, using Euler-Maclaurin summation formula. f[x_] := 1/x em[n_] := Integrate[f[x], {x, 1, n}] + (f[1] + f[n])/2 + Sum[BernoulliB[2k]/(2k)! ((D[ f[x], {x, 2 k - 1}] /. x -> n) - D[f[x], {x, 2 k ...

calculus-and-analysis summation

Never mind. I discovered the error in my assumption about the output from my program with polynomials. The output does not converge to a constant.

sunny

12:48

I'm looking to compute the sup-norm of $f_n(x)=\frac{2x}{1+nx^2}$ over $[0,\infty)$. The derivative of $f_n(x)$ is $\frac{2(1-nx^2)}{(1+nx^2)^2}$. Setting it to zero, yields $x=\pm \frac{1}{\sqrt{n}}$, so $x=\frac{1}{\sqrt{n}}$ is a critical point of interest, where we have $f_n\left(\frac{1}{\sqrt{n}}\right)=\frac{1}{\sqrt{n}}$. At $x=0$, we have $f_n(0)=0$. In the limit that $x$ tends to infinity, we have that $f_n(x)$ tends to $0$. Is $x=\frac{1}{\sqrt{n}}$ necessarily a maximum? Why?

According to the extreme value theorem, a continuous function on a compact set attains a maximum and minimum on that set, however, here we are dealing with $[0,\infty)$, so there's some uncertainty regarding whether or not the function attains a maximum/minimum or not.

123

13:11

Hello Everyone...

Mats Granvik

Hi

Jakobian

13:39

@Thorgott does $|\text{Gal}(K/F)| = [K:F]$ when $[K:F]$ is infinite?

Jakobian

14:08

5

A: Degree of extension of fixed field by infinite set of automorphisms.

No. If $E$ is the algebraic closure of the field $\Bbb{F}_p$, then we can think of $E$ as the nested union $$K_0\subset K_1\subset K_2\subset \cdots,$$ where the field $K_\ell$ is th unique (up to isomorphism) field of $p^{\ell!}$ elements. Then $E=\bigcup_{i=0}^\infty K_i$ is a countably infinit...

it doesn't, it can even be that the left side is greater than the right side

Thorgott

14:21

@Jakobian is K/F a Galois extension or are you just writing Gal(K/F) for the automorphism group of an arbitrary field extension

Jakobian

14:34

It's the group of $F$-automorphisms of $K$ where $K/F$ is any field extension, yes

sunny

@sunny could we argue like this? Surely, $f_n\left(\frac{1}{\sqrt{n}}\right)=\frac{1}{\sqrt{n}}$ is not a minimum, since $f_n(0)=0$. Neither is it a saddle point, for if it where, then there'd be some $a$ such that $f_n(a)>f_n\left(\frac{1}{\sqrt{n}}\right)$. Since $f_n(x)$ approaches $0$ as $x\to\infty$, there must be another critical point $c$ in addition to $x=\frac{1}{\sqrt{n}}$, but there is none. So $x=\frac{1}{\sqrt{n}}$ has to be the maximum. Grateful for any response on this.

Thorgott

yeah, then no

you can also have the LHS trivial with the RHS arbitrarily large

by taking $I$ to be your favorite index set and considering $F=\mathbb{F}_p(x_i,\,i\in I)$ and $K=\mathbb{F}_p(x_i^{1/p},\,i\in I)$

geocalc33

14:51

wish my knowledge on differential 2 forms was exact

Shaun

15:04

Hi :) Please may I have some feedback on the following?

-1

Q: Two exercises by Robinson on supersolvable groups seem to contradict.

This is concerning (part of) Exercise 5.4.5 and Exercise 5.4.6 of Robinson's, "A Course in the Theory of Groups (Second Edition)". I have done the first one; the second might take me a while. The Details: The dihedral group is given by the presentation $$D_{2n}\cong \langle r,s\mid r^n, s^2, srs=...

group-theory direct-product solvable-groups

The downvote baffles me.

Ted Shifrin

@Unknownx Your first sentence is correct. Why are all the unit vectors different? You're missing the whole point of the construction.

@sunny Once and for all, you should prove the following. If $f$ is positive on an open interval (for example, $(a,b)$) and approaches $0$ at the boundary, then $f$ has a global maximum. You want to choose an appropriate closed interval $[c,d]\subset (a,b)$ and apply the extreme value theorem on $[c,d]$. If you've chosen the subinterval correctly, it will follow immediately that that is the maximum point for all of $(a,b)$.

Jakobian

@sunny you can argue based on the sign of the derivative of $f_n$ in $[0, 1/\sqrt{n}]$ and in $[1/\sqrt{n}, \infty)$. It tells you $f_n$ needs to increase, then decrease. So $1/\sqrt{n}$ is a maximum.

Ted Shifrin

15:20

But, of course, as Jakobian points out, a simple graphing exercise from beginning calculus is sufficient in concrete examples.

sunny

15:42

I see, thanks for the response. I liked my saddle-point argumentation :) but I guess yours are cleaner

Ted Shifrin

Ordinarily, we don't use the "saddle point" terminology for functions of a single variable.

You mean a critical point that is not a local extremum. I agree that for multivariable functions we call that a saddle point.

sunny

Yes.

1 hour later…

PNDas

17:10

@Jakobian Why $E|X|=\infty$ imply $EX$ doesn't exist?

robjohn

17:29

@MatsGranvik I have a Mathematica function that I wrote for that. It may be similar to yours. I will look at your question in a while.

Ted Shifrin

@robjohn @Xander, et al. Isn't this totally bizarre?

Xander Henderson

@TedShifrin Yup. Weird.

Ted Shifrin

17:44

After he was so grievously insulted, I checked, and he has posted some true mathematics answers in the ten years of membership. So what the ... ?

Mats Granvik

17:59

@robjohn I have not written other code for the Euler Maclaurin summation formula than this one for the Riemann zeta function: pastebin.com/QebeXVXD And that also i copied from the french site numbers.free about the Riemann zeta function.

Jakobian

@PNDas from definition

Mats Granvik

Today I asked bing chat GPT to write a program for computing the Euler Maclaurin summation formula as a Mathematica program. It did write something but the program failed. It was not accurate.

Jakobian

ChatGPT is not intelligent on its own so that makes sense

we had this type of discussion before

@PNDas what's your definition of expectation of a random variable?

copper.hat

@PNDas the expectation (or integral) is not defined unless $|X|$ is integrable. You can make special cases, such as $X \ge 0$, but in general, absolute integrability is required to avoid infinities of opposite directions.

Jakobian

expectation is based on Lebesgue theory of integral and there we can define $\mathbb{E}X$ if $\mathbb{E}X_-$ or $\mathbb{E}X_+$ is finite where $X_- = \max(0, -X)$ and $X_+ = \max(0, X)$

copper.hat

18:10

Or what @Jakobian said...

Jakobian

but in general when $\mathbb{E}|X|$ is infinite then even if one of those quantities is finite then $\mathbb{E}X$ won't be, so we can say it doesn't exist (well technically it's infinite if one of those is finite, but still...)

I'm not really excluding that people are using other types of integration in probability when necessary, though I never really heard of anything like that.

excluding stochastic integrals

Soumik Mukherjee

Weierstraß polynomials. To get at the deeper properties of the ring $\mathcal{O}_n$, we have to study the local structure of holomorphic functions more carefully. We will proceed by induction on $n \geq 0$, by using the inclusions of rings $\mathcal{O}_{n−1} \subseteq \mathcal{O}_{n−1}[z_n] \subseteq \mathcal{O}_n$.

I have some doubts here, we are considering the set of germs at the origin in $\Bbb{C}^n$. So we can have $\mathcal{O}_n$, but how are we defining $\mathcal{O}_{n-1}$ at the origin in $\Bbb{C}^n$

MathCrackExchange

18:35

@SoumikMukherjee what's definition of $\mathcal{O}_k$?

You are studying some deep stuff, man! :)

Ted Shifrin

You are talking about germs of holo fns in $z_1,\dots,z_{n-1}$.

MathCrackExchange

You are free diving in the deep water, while I'm on the beach.

Unknown x

@TedShifrin Conversaly, We assume $\kappa=kappa^\star$ and $\tau=tau^\star$. We need to prove corresponding curves $C$ and $C^\star$ are congruent. Right?

Soumik Mukherjee

@TedShifrin Yes

MathCrackExchange

@SoumikMukherjee what topic? Complex algebraic geometry?

Soumik Mukherjee

18:39

@MathCrackExchange No, I am just at the beginning, probably that's why I am getting stuck so often

@MathCrackExchange Yes

MathCrackExchange

Thx. I got to go, got to work etc. -_-

Ted Shifrin

@SoumikMukherjee Yes? I answered your question.

Unknown x

Our aim is to find a map in such a way that $C^\star$ is the image of $C$.

map should be rigid.

Ted Shifrin

@Unknownx You’ve been on this for weeks.

Unknown x

@TedShifrin sorry. I was engaged in other duties :(

Ted Shifrin

18:42

That’s what the orthogonal matrix lining up the Frenet frames at time $0$ does.

Unknown x

Then why did you construct $\tilde{\alpha}$?

Ted Shifrin

That lines the curves up at time $0$, and then you show they agree for all time.

Soumik Mukherjee

@TedShifrin I am a little confused, what is the ambient space where we are defining these germs, for $\mathcal{O}_n$, this is germs of holo fns in $n$ variables, for $\mathcal{O}_{n-1}$, this is germs of holo fns in $n-1$ variables. When we are writing them as one is subset of other then what is the underlying space and points we are working at?

Unknown x

@TedShifrin Yes. Now I understood the importance of $\tilde{\alpha}$.

Soumik Mukherjee

Oh I get it, functions in $n-1$ variable can be considered as functions in $n$ variable with the last variable ignored

Ted Shifrin

18:50

@SoumikMukherjee You’re working inductively/recursively. Just like you can think of a polynomial in $x,y$ as a poly in $y$ with coefficients that are polynomials in $x$.

Soumik Mukherjee

Yes yes, Thank you :)

Unknown x

But in the definition of $f(s), $ $f(0)=\tilde{T}(0)\cdot{T}^\star(0)+\tilde{N}(0)\cdot{N}^\star(0)+ \tilde{B}(0)\cdot{B}^\star(0)$. Right?

I agree $\tilde{T}(0)\cdot \tilde{T}(0)=1$ and ${T}^\star(0)\cdot {T}^\star(0)=1.$

similarly for $B$ and $N$ as well.

Ted Shifrin

So?

Unknown x

Is $\tilde{T}(0)\cdot{T}^\star(0)=1$?

Ted Shifrin

Of course.

Unknown x

18:57

How?

Ted Shifrin

How was the tilde curve defined?

Unknown x

$\tilde{\alpha}=\Psi \alpha$

Right?

Ted Shifrin

And how was $\Psi$ defined?

Unknown x

$\Psi(x)=Ax+b$

where $b=\alpha^\star(0)=\alpha(0)$

$A$ is obtained from the Orthogonal transformation of basis as you mentioned earlier.

Ted Shifrin

I mentioned, but you have not understood it, apparently.

Unknown x

19:05

We now define a rigid motion $Ψ$ as follows. Let
$b = α^∗(0) − α(0)$, and let $A$ be the unique orthogonal matrix so that $AT(0) = T^∗(0), AN(0) =
N^∗(0),$ and $AB(0) = B^∗(0).$. This is your statement in the beginning.

Could you tell me where I miss the idea?

Atleast give me hints.

$\tilde{T}(0)\cdot{T}^\star(0)=(A\alpha'(0)+(\alpha^\star(0) − \alpha(0)))\cdot A \alpha'(0)$. Am I correct?

@TedShifrin Where?

Jakobian

19:39

@Thorgott I have two propositions and I'm wondering how close they are too each other. The first one says that a finite extension $K/F$ is Galois iff $[K:F] = |\text{Gal}(K/F)|$. The second says that if $F$ is a fixed field of a finite group $G$ of automorphisms of $K$, then $|G| = [K:F]$.

Of course, the second one helps to get the first one, but I'm wondering if I need to care about the second one

Ted Shifrin

@Unknownx By construction, $\tilde T(0) = T^*(0)$, etc.

PNDas

For example, similar to the example you gave, if you consider $P(X=(-1)^nn)=\frac{C}{n^2\log n}$, $n\geq2$. Then $E|X|=\infty$ but $EX<\infty$.

Jakobian

@PNDas huh?

No, $EX$ doesn't exist

PNDas

$EX=C\sum \frac{(-1)^n}{n\log n}$ right?

Jakobian

no it's undefined

Unknown x

19:43

@TedShifrin Is it from $AT(0) = T^\star(0)$?

PNDas

Then we are considering different definitions

Jakobian

Where did you find your definition?

Ted Shifrin

@Unknownx Yes.

PNDas

Okay what you are saying is correct. I'm wrong here.

Unknown x

@TedShifrin But $\tilde{\alpha}=\Psi\alpha$. Right?

PNDas

19:46

The quantity on the right is not $EX$. Expectation is not defined.

Ted Shifrin

Yes. Translation does not affect derivatives.

Jakobian

Since you're in $\mathbb{R}$ we could define something like the "principal value" $p.v. E[X] = \lim_{n\to \infty} E[X1_{[-n, n]}]$ and that would make sense I suppose

but it does use that we are in $\mathbb{R}$ pretty explicitly

Writing $\mathbb{E}[X]$ in this case is strange since e.g. it doesn't give you how you should order your summation, defining it in this case is problematic, for Lebesgue integrals what we care is absolute convergence

Unknown x

@TedShifrin Now I understood. Thank you very much.

Ted Shifrin

You’re welcome.

PNDas

Yeah the rhs is $\lim E[|X|1_{|X|\leq n}]$.

For some reason, I convinced my self that this is $EX$.

Jakobian

19:50

This would probably give you $-i\varphi_X'(0)$ in this case which I said to you before that we could consider as some kind of "generalized mean"

Thorgott

20:09

@Jakobian They're more or less equivalent

MathCrackExchange

20:52

@Shaun check your e-mail :>

MathCrackExchange

21:08

@Shaun I'm blown away by Teams. It looks really nice and a great alternative to a whole independent website

It's just what we need too to ask for help, much better than chat room

And also to post study group times etc

Shaun

@MathCrackExchange Hi. Yeah, I agree.

MathCrackExchange

What it needs is for the Admins (us) to create tags at some point

I guess we'll create them as we go along

Shaun

@MathCrackExchange That sounds about right.

MathCrackExchange

So I'll be there, and I'll fill in some Q&A's when I study. I think each time you have a question that you have to research, we should put it in Teams, that way future readers will be accelerated in a way

So maybe I won't require book purchase, but just say that it's highly recommended and that pirate copies are not allowed

Shaun

I'm still at that conference, by the way. I have to get up early tomorrow. The afternoon is left free, so they've crammed in a bunch of lectures in the morning. I should try & sleep soon.

@MathCrackExchange Good idea.

MathCrackExchange

21:13

Feel free to edit the site with content whenever you're free. It's yours too, you're an Admin

Okay, sleep well, friend!

2 hours later…

robjohn

23:43

@TedShifrin The answer is definitely strange.

The question is simply asking why $\int_{[0,1]}1\,\mathrm{d}t=\int_0^11\,\mathrm{d}t=1$ instead of $\int_{[0,1]}1\,\mathrm{d}t=\int_1^01\,\mathrm{d}t=-1$.

which could be confusing when one starts path integration.

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