Mathematics - 2022-04-24

Ted Shifrin

00:02

Would a toddler catch all the filth?

I presume munchkin was the toddler, so ... yes.

leslie townes

most of it, yes

Ted Shifrin

@copper Well, I'm over 100K. When does my gift magically arrive?

3

copper.hat

03:07

@TedShifrin Hi Ted, it took months before mine arrived.

A nice mug and a flimsy t-shirt.

leslie townes

03:23

they probably wanna wait and see if this over 100k thing is real, or just the result of a targeted campaign of upvoting by ted sockpuppets.

Ted Shifrin

I’m fake through and through, just like you!

leslie townes

cool moment earlier, ted. we checked the mail and found evidence that a bird had been plucked. we guessed it was a dove. we hunted around the area until we found the scene of where a hawk got it. definitely a dove.

we spent so much time wandering around outside looking at the ground that my wife came out and asked us what we were doing. she was less interested in the dove. didn't even want to see what was left of it.

Ted Shifrin

Munchkin sleuthed?

leslie townes

we sleuthed together. a crow helped us find it. he flew off with a tiny piece of it. we went to where he had been.

Ted Shifrin

No, seeing mauled remains isn’t so exciting.

leslie townes

03:34

hawk kills are pretty clean, it was almost entirely plucked feathers. only a tiny bit of blood, no gore.

we did find about half of a wing and half of a tail

Ted Shifrin

Poor dovey.

leslie townes

i left out the sad part. halfway through our hunt, i noticed that a dove was watching us from the neighbor's roof.

Ted Shifrin

Oh oh … either lover or mother.

leslie townes

yep.

Ted Shifrin

Thereby hangs a tail.

leslie townes

03:37

my daughter's surprisingly non sentimental about this stuff. was really fascinated with it

Ted Shifrin

She should spin quite the story.

porridgemathematics

04:49

does anyone understand the 'integration by part' step here: imgur.com/a/uRukYMX ? We can just assume that $X$ is $\mathbb{C}^n$ and we are integrating with respect to the euclidean volume form

and everything concerned is smooth

Koro

Hi @porridgemathematics!

porridgemathematics

hi @Koro :)

im familiar with 'integration by parts' for differentiable manifolds, using stokes theorem, i figure there is a way to use this here directly that is pretty obvious, but im not able to grok

so we want to write our integrand as $\alpha \wedge d(\beta)$ where $d(\beta)$ is some multiple of the volume form and $\alpha$ is smooth, and then use the usual integration by parts formula, but im not able to resolve what candidates $\alpha,\beta$ should be

uh, actually sorry, $X$ is some compact quotient of $\mathbb{C}^n$, say a complex torus, but we can just defer to euclidean coordinates is what im trying to say

so the volume form just descends

Koro

porridgemathematics: I have a small question on polynomials. I think my proof for that is correct. Can you please take a look at that? Thanks. math.stackexchange.com/questions/4433577/…

porridgemathematics

@Koro im kinda swamped right now, my question is actually for an essay i need to write which i cant really write properly right now because im struggling with basic misunderstandings of things, ill look into it when i have the time if thats okay !

Koro

I plan to place a bounty on the question but I can't do that before 7 to 8 hrs from now.

porridgemathematics

04:54

(the question will not help me 'solve the essay' if anyone is concerned, this is from lecture notes from a course im doing, and you need to understand the lecture notes to write said essay)

Koro

@porridgemathematics that's fine whenever you have time. thanks.

Ted Shifrin

@porridge Take $\bar\partial f\wedge \star v$. $v$ is a (0,1)~form, so type considerations should do it — $\bar\partial$ will agree with $d$.

your “we want to write” sentence above makes no sense.

porridgemathematics

05:11

i meant we want to show that the integrand is one of the summands in $d(\alpha \wedge \beta)$, where $\alpha \wedge \beta$ is an $(2n)-1$ form, and then use stokes theorem, isnt that what integration by parts means in this context?

whats $ \star v$?

Ted Shifrin

Hodge star

porridgemathematics

hm, it seems we haven't covered that so far, in any case ill look it up , thanks for your help

Akiva Weinberger

Say you have the eight points $(\pm1,\pm1,\pm1)$ forming a cube, and you rotate one point 180 degrees around the other

(I.e. around the line through the other and the origin)

I think you get $(\frac53,\frac13,\frac13)$ (up to sign)

I thought I had a question when I started writing this but I don't

leslie townes

i had a dream last night but i forget what it was

Akiva Weinberger

I think lines through the origin form a quandle with the operation of rotating one 180 degrees around the other

Koro

05:40

@leslietownes that's very common.

porridgemathematics

@TedShifrin could we also proceed like this? $\overline{\partial}(\overline{\partial_i}f \overline{v_i} dV) = 0$, hence $-\overline{\partial_i}f \overline{v_i} dV + \overline{\partial}v_i \wedge (f dV_{i}) = 0$, where $dV_{i}$ omits the $\overline{dz^i}$ part, which gives $\overline{\partial_i}f \overline{v_i} dV = \overline{\partial_{i}}\overline{v_i} f dV$?

i should have parenthesized the $(\overline{\partial_i}f)$ in $\overline{\partial_i}f \overline{v_i}$

Koro

06:05

I got answer to my question. :)

My proof is correct.

leslie townes

you've even begun using bill's favorite font

:)

Koro

yes :D.

leslie townes

have you begun using colors?

Koro

yes, i have :)

2

A: Questions about proof regarding the ring $\Bbb Z[x]/(2x^2-4,4x-5)$

It seems they skipped explicit mention of an intermediate step that applies the below extension of a common ring isomorphism theorem (proved e.g. in this answer) If $\varphi: R \to S$ is a surjective ring hom then $\frac{R/ \ker\varphi}{(I+\ker\varphi)/ \ker\varphi} \cong \frac{S}{\varphi(I)}$ f...

leslie townes

well, you could certainly choose worse role models.

i know bill's style has resulted in conflicts with various folks at times, but his answers are generally of very high quality.

Koro

06:10

Leslie: after looking at this answer and the links therein, I could understand my errors in applying the third isomorphism theorem.

leslie townes

some of his answers are better than what you find in the best textbooks.

Koro

The answer has been very helpful to me (and many more answers of his ) :)

I wrote an answer to this post: math.stackexchange.com/questions/4434058/…

But I have not posted it yet as I don't know if it's psq or not

leslie townes

sometimes he'll respond to a question by generalizing it and giving a proof that is simpler and more understandable than the non-generalized version. that is pretty rare.

i don't know that "psq" has a rigorous definition. it looks like one to me. the stuff besides the problem statement is indistinguishable from a rote recitation of stuff that you might see around the relevant definitions. (or a concealed textbook hint.) so, maybe some risk of the question eventually being deleted.

Koro

@leslietownes yes, like finding x in $ax=b$ in some $Z_p$ :).

By the third isomorphism theorem, $$\rm \mathbb{R}[X,Y]/(X^2+Y^2-1,X-2)\cong\frac{\mathbb R[X,Y]/(X-2)}{(X^2+Y^2-1,X-2)/(X-2)}\tag 1$$

Note the homomorphism $\rm \phi: R[X,Y]\to R[Y]$ defined by $\rm \phi (f(X,Y))=f(2,Y)$ has $\rm \ker \phi =(X-2)$.

$\rm (X^2+Y^2-1,X-2)=(X^2+Y^2-1)+(X-2)=(X^2+Y^2-1)+\ker \phi$

With this $(1)$ becomes: $$\rm \mathbb{R}[X,Y]/(X^2+Y^2-1,X-2)\cong \frac{\mathbb R[X,Y]/\ker \phi}{((X^2+Y^2-1)+\ker \phi)/\ker \phi}\cong \color{blue}{\frac{\mathbb R[Y]}{(Y^2+3)}}\tag 2$$, where the second isomorphism is by [this version of the third isomorphism theorem][1].

The above is the answer I wrote. I'm glad I was able to write it.

leslie townes

oh, i see he responded in the comments. i'd consider posting that as an answer. the comments do show that he did work (although it would have been better had he put that into edits on the question).

sometimes when i respond to a PSQ that has become something more than that via comments, i encourage the OP to edit the original question, and/or i refer to the OP's comments at the beginning of my answer.

the idea being that whoever is on a PSQ purge might see that and not delete.

Koro

06:20

yeah, but now it seems that the OP has found their answer.

leslie townes

sheesh. then he should answer his own question. what's with these people.

porridgemathematics

@TedShifrin nvm, there is an error in what i wrote, ignore please

leslie townes

07:32

@slate, your name was a very good opener on wordle the other day. got it in 2.

Slate

I'm glad I could help!

one potato two potato

11:39

If $R$ is an associative $k$ algebra and $I\subset R$ is an ideal and $\phi:R\to R$ is a $k$-linear endomorphism then $\phi(I)\subset I$?

$k$ is a field

Slereah

11:53

In proofs regarding the curvature 2-form, there is usually the use of fundamental vector fields as stand ins for vertical vector fields in general, but as far as I can tell, this is not generally true

why does this work in such circumstances?

For instance in Kobayashi :

The reason for $A^*(\omega(B^*))$ being zero I've seen is that $\omega(B^*)$ is a constant Lie algebra-valued function, but this relies on $B^*$ being a fundamental vector field

and since that reasoning relies on the vector field acting as a derivative, it can't just be approximating it at a single point

and according to Mr. Cap, they are very much not the entirity of the space of sections of the vertical bundle :

Since $V(P)\cong P\times\mathfrak g$, the space of vertical vector fields is isomorphic to $C^{\infty}(P,\mathfrak g)$. The fundamental vector fields (parametrized by elements of $\mathfrak g$ that you are describing are just a very special set of vertical vector fields. — Andreas Cap Dec 18, 2015 at 15:13

Slereah

12:30

Some other paper says they do it "without loss of generality" but do not explain further

Mats Granvik

13:23

$1=1$
$1=(1+1)-1$
$1=(1+1+1)-(1+1)$
$1=(1+1+1+1)-(1+1+1)$
$1=(1+1+1+1+...+1)-(1+1+1+...+1)$
$1=\underbrace{1}_n-\underbrace{1}_{n-1}$

$$T = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 &\cdots \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \vdots&&&&&&&\ddots \end{pmatrix}$$

$$T(n, 1) = 1$$
$$T(n, k) = \left[n >= k \right]\left(\sum_{i=1}^{n-1} T(n - i, k - 1) - \sum_{i=1}^{n-1} T(n - i, k)\right)$$

(*Mathematica start*)Clear[t];
nn = 8;
t[n_, 1] = 1;
t[n_, k_] :=
t[n, k] =
If[n >= k, 1,
0]*(Sum[t[n - i, k - 1], {i, 1, n - 1}] -
Sum[t[n - i, k], {i, 1, n - 1}])
TableForm[T = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]]
Total[Transpose[T]]
(*Mathematica end*)

user193319

13:49

0

Q: Reflection Group of Type $D_n$

Here is the description of the reflection group of type $D_n$ in Humphreys' book Reflection Groups and Coxeter Groups: ($D_n$, $n \ge 4)$ Let $V = \Bbb{R}^n$, and define $\Phi$ to be the set of vectors of squared length $2$ in the standard lattice. So $\Phi$ consists of the $2n(n-1)$ roots $\pm ...

porridgemathematics

15:30

hi, does anyone know if this is true? If $(V,h)$ is a holomorphic vector bundle over a complex $X$, and we take the hermitian connection $D$, i.e. the unique complex,metric connection on $V$, then defining the dual/conjugate connections on the dual/conjugate bundle in the natural way, would the curvature tensor of $(\overline{V})^{\ast}$ and $V$ coincide?

i.e. would its components be the same with respect to the same basis (taking the associated basis)

Ted Shifrin

15:48

@porridge $\bar V$ is not a holomorphic bundle, so theory goes out the window. Better stick just to the dual.

porridgemathematics

@TedShifrin right, but given we extend $D$ to $\overline{V}$ by defining $D_{v}(\overline{\sigma}) = \overline{D_{\overline{v}}(\sigma)}$, can't we get a connection on $\overline{V}$ viewing it as just a complex vector bundle?

and then compute the curvature of this connection?

defining it this way would make it kind of an 'antiholomorphic' connection if you will, i.e $(1,0)$-tangent vectors will kill antiholomorphic sections of $\overline{V}$

and then extend to the tensor-algebra of $V, \overline{V}, V^{\ast}, \overline{V}^{\ast}$ by requiring commutation with contractions, and the product rule with respect to $\otimes$

imbAF

16:40

In the big O notation, when we want to find k,c, depending on the k value, c varies, so which value of k should we consider, the lowest? meaning $k >= 1$

robjohn

17:06

@MatsGranvik Did you have a question?

Mats Granvik

@robjohn Not asking questions and merely saying things has become a thing of the past.

It is actually forbidden not to ask questions. One must have questions.

Jeopardy theme song [10 hours]

►

robjohn

No, just wondering why you posted. It was confusing without any explanation.

Mats Granvik

@robjohn It is meant to be a degenerate case of something more general.

(*Mathematica start*)
Clear[t];
nn = 8;
t[n_, 1] = If[n >= 1, 0, 0];
t[n_, k_] :=
t[n, k] =
If[n >= k, (Sum[t[n - i, k - 1], {i, 1, k - 1}] -
Sum[t[n - i, k], {i, 1, k - 1}]), 1]
TableForm[T = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]]

(*Mathematica start*)
Clear[t];
nn = 8;
t[n_, 1] = If[n >= 1, 0, 0];
t[n_, k_] :=
t[n, k] =
If[n >= k, (Product[t[n - i, k - 1], {i, 1, k - 1}] -
Product[t[n - i, k], {i, 1, k - 1}]), 1]
TableForm[T = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]]

Is there a name for when replacing a sum with a product gives the same answer? @robjohn

And sorry for me being a jerk.

robjohn

Not sure if there is one

Mats Granvik

ok

user10478

18:08

Is the action of switching from a sample mean to a population mean (adding 1 to the denominator when the formula is in a particular common form) ultimately the same action as applying Laplace's rule of succession?

Mats Granvik

All instances of mathematics are degenerate cases of one grand formula. Unfortunately no-one knows what that formula is.

I would have deleted the above, but it is too late now.

user193319

19:04

1

Q: Reflection Group of Type $D_n$

Here is the description of the reflection group of type $D_n$ in Humphreys' book Reflection Groups and Coxeter Groups: ($D_n$, $n \ge 4)$ Let $V = \Bbb{R}^n$, and define $\Phi$ to be the set of vectors of squared length $2$ in the standard lattice. So $\Phi$ consists of the $2n(n-1)$ roots $\pm ...

group-theory reflection root-systems coxeter-groups reflection-group

rb3652

19:31

Hi all, I have a question.

user8469759

19:44

can anyone if have time have a look at this question? cs.stackexchange.com/questions/150971/…

and tell me whether or not the proof I gave is correct

I can migrate the question to mathexchange but I thought it was more suited for computer science

Ted Shifrin

20:17

@porridgemathematics Antiholomorphic bundles on a complex manifold are not well-defined.

Look at transition functions and figure out where you use holomorphic in the usual case.

rb3652

Is there any way to compare a vector and a scalar?

Ted Shifrin

Compare? What does that even mean?

Mats Granvik

Is there a way to compare a vector and a matrix?

Ted Shifrin

There's not even a way to compare two vectors (other than checking whether they're equal).

Mats Granvik

Most of the Riemann hypothesis is like that.

Or proving the irrationality of Euler Gamma. (Comparing a vector to a matrix.)

user10478

20:40

I am confused over the Wikipedia article that discusses why the principle of indifference is often arbitrary (en.wikipedia.org/wiki/Principle_of_indifference). Why does it single out continuous variables. Is it not the case that it is equally arbitrary to make assumptions about discrete quantities competing for the uniform distribution assumptions, i.e., whether square graph paper with an integer number of squares per side between 3 and 5 should be assumed to have ~4 squares per side

or ~17 squares total?

leslie townes

yes. i think the underlying point is just that a choice of what you are indifferent to often has consequences within a model. you can't arbitrarily assign assumed distributions to related quantities.

probability wikipedia is not that great.

user10478

Okie, thanks

leslie townes

it's a good question. i guess whoever wrote that portion of the entry would say that the entry only says it's 'especially' common with continuous variables, not limited to them. that might be true. when things are geometric or pictorial sometimes people get ahead of themselves.

the entry on the bertrand paradox is pretty good.

well, some of it is good. a lot of prob/stat wiki is a disordered brain dump of multiple contributors at different times.

user10478

I was looking at the part at the bottom of the section, "The fundamental hypothesis of statistical physics, that any two microstates of a system with the same total energy are equally probable at equilibrium, is in a sense an example of the principle of indifference. However, when the microstates are described by continuous variables (such as positions and momenta), an additional physical basis is needed in order to explain under which parameterization the probability density will be uniform."

I don't know what the physics words mean, but it sounds like the article is going beyond "especially" there.

leslie townes

yeah, and the section title 'application to continuous variables' is ill chosen.

probably 20 different monkeys at 20 different typewriters at 20 different times.

user10478

20:54

Perhaps even 400 different times!

leslie townes

given what wikipedia is, it's a miracle that so much of it is as good as it is.

they keep rejecting my entry on Lesliecoin as insufficiently notable, and for advertising cryptocurrency. i hope elon musk buys them out and allows free speech on that platform.

user10478

I don't actually know what the mechanism of consensus is on wikipedia.

leslie townes

i don't either.

imbAF

anyone knows what's this identity called and how it's derived : $a^{\frac 1 {log_n(b)}}=n^{log_b(a)}$

Ted Shifrin

21:10

Musk speech, you mean.

leslie townes

a^(1/log_n(b)) = e^(ln(a)/log_n(b)) = e^(ln(a)/ln(b) * ln(n)) = e^(ln(n) * ln(a)/ln(b)) = n^(ln(a)/ln(b)) = n^(log_b(a)). most of this is elementary properties of exponents. a key identity is the change of base identity for logarithms, log_c(x) = log_d(x)/log_d(c), used twice in the above, first with c = n and d = e and x = b, and then again with c = b and d = e and x = a.

ted: yes.

imbAF

thx

@leslietownes you have any idea regarding O notation, but in the CS context ?

leslie townes

i remember the day i learned the chase of base formula for logs. it was homework in my precalculus class in 11th grade. first teacher i'd ever had who actually had a degree in math. cool guy.

imbaf: i have ideas regarding O notation although i plead the fifth as to cs context.

imbAF

ah

One additional question regarding logarithms

How could I, change the following expression to a simpler one.

I am not sure, whether what I'll get, represents the simplest expression or it can be simplified even further

$2^{\sum_{i=1}^n k log(i)}$

I assume k is a constant. Tho nothing, is said about it

leslie townes

for each k and i, we have that k log i is log(i^k). and the sum (on i) of log(i^k) is the log of [the product of i^k (over those i)]. does that help?

imbAF

21:25

$2^{k log(\Pi i)}$

would that be simpler ?

ah ok

so we got the same

thx

leslie townes

yeah.

imbAF

I would ask you about this

1

Q: Finding the constants in Landau notation

I am trying to find the constants $n_0$ and $c$ to show that some given functions belong to the $O(\cdot)$ equivalence class. But, while it seems easy, I am not sure whether I am allowed to do what I will showcase below, or rather, what decides which constants I should take into consideration. Fo...

landau-notation

It's about the $\Omega$ notation

Ted Shifrin

Sometimes it would be considerate to be unlazy and type actual mathematics.

leslie townes

one way of approaching this would be to first try proving that log^8(n) is Omega(log(n)), and then proving that log(n) is Omega(n^(2/3))). you could chase through proofs of both of those things to find c's and n_0's.

imbAF

It's that the commentary, that I did there is important, and I didn't want to write a whole text here

but can't I choose an arbitrary value of n_0 or c?

leslie townes

21:29

i further note that the language "the" constants c, n_0 suggests that only one c and n_0 will do, which is not the case.

imbAF

In CS for the $\Omega$ notation the following commentary is valid: when the device for f(n) fast enough is and n is large enough then f(n) is smaller then g(n)

leslie townes

maybe you could choose an arbitrary value of one of them as long as the other one is appropriately chosen. thinking about this kind of stuff is harder than what the question seems to be asking, however, which is only to identify one pair of c and n_0 that will work.

imbAF

by smaller, we mean that for the same task, less time is needed

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