Mathematics - 2023-05-28

Always for you!

copper.hat

12:28 AM

i like to raise the tone

the orientation is reversed when you live in the gutter

12:45 AM

@CottonHeadedNinnymuggins that looks correct. The other counts things wrong.

I'm gonna get me some Magdonal

ask your doctor if magdonal may be right for you

If he says no I'll just throw an apple at him

that should keep him away

Cotton Headed Ninnymuggins

1:14 AM

@robjohn how should I count them?

@TedShifrin Stars and bars because we want indistinct balls into distinct boxes without an injective or surjective restriction. There are indeed ${10/choose 5}$ different Yahtzee rolls. I’m convinced $6$ of them don’t have repeated outcomes

Cotton Headed Ninnymuggins

@CottonHeadedNinnymuggins The way that I commented was correct: $1-\frac{\binom{6}{5}5!}{6^5}=90.74074\%$

There are $\binom{6}{5}$ ways to choose 5 non-duplicates, and $5!$ ways to arrange them. That divided by $6^5$ possible choices.

There must be a way to do it by counting the distinct rolls in the way I did, right? What went wrong with my way?

1:31 AM

I completely agree with robjohn.

@CottonHeadedNinnymuggins You’re convinced?

Note that $\frac{\binom{6}{5}5!}{6^6}=\frac{6\cdot5\cdot4\cdot3\cdot2}{6^5}=\frac{5\cdot4\cdot3\cdot2}{6^4}$ which agrees with your first approach.

Cotton Headed Ninnymuggins

I think the bars and stripes is wrong. Explain the binomial coefficient.

Dannyu NDos

2:03 AM

Whilst studying homotopy theory, there has been something bothering me.

The Hawaiian earring. Homotopies TO it are quite studied, but what about homotopies FROM it?

The Hawaiian earring is just a coarser version of the wedge of countably infinitely many circles. It should still be a co-H-space?

2:38 AM

@TedShifrin stars and bars is correct when counting the number of different Yahtzee rolls (as in the roll 66655 is the same as 56566). I think I make some kind of assumption when I do $6/252$. It’s true that $1$ roll of the $252$ distinct rolls contains 11111, but the chances of getting 11111 is not $1/252$. All I’m confused about at this point is exactly what assumption was made.

2:52 AM

@CottonHeadedNinnymuggins Right. The various options are not all equally likely.

Koro

6:35 AM

How to prove continuity of $f: C\to C: \sum_{n=1}^\infty \frac{a_n}{3^n}\mapsto \sum_{n=1}^\infty \frac{a_{2n}}{3^n}$, where C is Cantor set?

Take $p\in C$. Take any $\epsilon>0$. We want a $\delta>0$, $|x-p|<\delta\implies |f(x)-f(p)|<\epsilon$. Suppose that $x=\sum_{n=1}^\infty \frac{a_n}{3^n}, p=\sum_{n=1}^\infty \frac{b_n}{3^n}$. $|f(x)-f(p)|\le \sum_{n=1}^\infty \frac{|a_{2n}- b_{2n}|}{3^n}\le \epsilon$.
How to get $\delta$ from here?

Koro

7:06 AM

nvm, I got it :-).

porridgemathematics

8:27 AM

if anyone has time and an interest, could you take a look at my question? math.stackexchange.com/questions/4708018/…

it may have a simple solution, and is on perrons method for the dirichlet problem in two dimensions

Mats Granvik

10:15 AM

These are the first few rows of an infinite matrix T from which EulerGamma can be approximated:

$\gamma =-\lim_{n\to \infty } \, \left(\sum _{k=1}^{n^2} \frac{T(n,k)}{k}\right)$ = 0.5772156649015328606065120900824024310421593359399...

A long standing problem in number theory is to prove that EulerGamma is transcendental.

Mankind is not able to handle simultaneous addition, subtraction, and division. Anything involving three ideas or more is difficult.

11:51 AM

@MatsGranvik Have you seen the double series I derived for $\gamma$? Using the Mathematica code at the bottom, it computes 10000 digits in 16 seconds.

Mats Granvik

12:18 PM

@robjohn That is impressive. No I had not seen that. You don't have any subtraction inside the double sum, if I looked at it right.

That's right. No subtractions

porridgemathematics

12:44 PM

i finally answered my question.. 4 hours later, in literally one line and one of those obvious solutions that went over my head :(

i wish it was more blaring to me that if $\phi$ is superharmonic, the perron solution to the dirichlet problem in a domain with $\phi$ as boundary data is always dominated by $\phi$ itself.. thats all it took

here perron solution meaning the upper envelope of all subharmonic functions dominated by the boundary data

user193319

2:16 PM

Suppose that $H$ is a finite-index subgroup of $G$. How do I show that there exists $m \in \mathbb{N}$ such that $g^m \in H$ for every $g \in G$?

Sourav Ghosh

2:48 PM

Let $G$ be the multiplicative group of $\Bbb{R}\setminus\{0\} $ and $H$ be the subgroup consisting of all positive reals. Choose $-1\in G$ . Does there exists any $m\in\Bbb{N}$ such that $-m>0$ ?

porridgemathematics

@SouravGhosh no, but there does exist $m \in \mathbb{N}$ for which $(-1)^m > 0$

you specified multiplicative group..

4:41 PM

@user193319 Do you perhaps have an important additional hypothesis on $H$?

user193319

5:09 PM

@TedShifrin Hmm..not that I am aware of...why? Is it false as stated...I hope not!

Soumik Mukherjee

6:07 PM

@user193319 Hint: try to show that there exists a normal subgroup $N$ of $G$ of finite index such that $N \subset H$

6:43 PM

Hi :) I heard that pi is almost surely normal. I can't find a reference though. Please would anyone help?

I think I misunderstood something. That's not what was claimed, I think. Never mind :)

Hang on. I was right: reddit.com/r/math/comments/fiv4r/…

But that's not a reference . . .

Q: Is $\pi$ almost surely normal? A reference request.

7:10 PM

0

NB: This is a reference-request question and so there's not much context I can give. It is claimed here that $\pi$ is almost surely normal; that is, the probability that $\pi$ is normal is one. I haven't found a reference for this anywhere. If it is true, please would you provide one? Context: I ...

7:21 PM

@user193319 Soumik has given it away. If you expect $m$ to be the index $[G:H]$, it may well be false.

Amin Idelhaj

7:32 PM

Hello

Q: Is $\pi$ almost surely normal? A reference request.

7:46 PM

My question got a downvote and I don't understand why. Please would someone make some suggestions? I know I can't know for sure why . . .

-1

NB: This is a reference-request question and so there's not much context I can give. It is claimed here that $\pi$ is almost surely normal; that is, the probability that $\pi$ is normal is one. I haven't found a reference for this anywhere. If it is true, please would you provide one? Context: I ...

probability number-theory reference-request pi

8:21 PM

the usual form of the heat equation implies we're dealing with the usual euclidean metric on euclidean space

A: Why does the foliation $\mathcal{F}$ of this Lorentzian manifold also solve the backwards heat equation?

8:44 PM

0

I believe the crux of the matter is this: The heat equation in the above form assumes Euclidean space with the usual Euclidean metric; therefore in some sense restricting $\mathcal{F}$ to the usual Euclidean metric implies a solution to the heat equation. $\mathcal{F}$ is only a Cauchy foliation ...

too long for a comment...

@Shaun probably one of the sources in the wikipedia page will do

for instance, it notes: "The concept of a normal number was introduced by Émile Borel (1909). Using the Borel–Cantelli lemma, he proved that almost all real numbers are normal, establishing the existence of normal numbers."

and since pi is a real number...

@Semiclassical “almost all”?

?

i think the answers do a pretty good job of explaining it. there are various precise senses in which normality is an "almost everywhere" or "generic" kind of property, but these results don't imply that any specific number has a better "chance" of being normal, or anything like that.

i would assume, as a starting point, that 99% of what is written about normality in connection with pi, is written as carefully as 99% of what is written about infinite decimals, i.e., with no care for mathematical particulars and lots of accidentally wrong things.

it's always relative to a set, i guess. like, it's conceivable that there's a subset of the reals (containing pi) whose elements are not almost surely normal

...i guess there's some trivial examples of such. e.g., $\{0.1,\pi\}$

8:55 PM

it's a little odd to ask questions of the form, "why would [some mathematical assertion] be true, keep in mind i'm not the one making it, it's just something i've seen, and i won't tell you where, and maybe there is no where." generally, it doesn't give a lot of context for what the OP might know, or not know, or whether there is any gap between the OP's summary of an assertion and what the assertion actually is.

i've sometimes scolded koro for posting questions like that.

right

i guess there's no clean way to go from "elements of X are almost surely normal" to "this particular element of X is almost surely normal"

it's a bit like the countability of the algebraic numbers showing that there are "more" transcendentals than algebraic numbers, in a sense that is both precise, and also not informative about the status of any particular number.

right

@leslietownes Thank you for the feedback :)

@Semiclassical Thank you. I invite you to type up an answer my question.

@TedShifrin except for a set of measure zero, I believe.

Gwyn

9:31 PM

If I proved that $(\log n)/n^a \to 0$ as $n \to +\infty$ for $a>0$, so there exists $N_\epsilon \in \mathbb{N}$ such that $n \ge N_\epsilon \implies |(\log n)/n^a|<\epsilon$, can I say that $(\log x) /x^a \to 0$ as $x \to +\infty$ with $x\in\mathbb{R}_{>0}$ because $\left|\frac{\log x}{x^a}\right| \le \frac{\log \lceil x \rceil}{\lfloor x\rfloor^a}$

and since $\lceil x\rceil,\lfloor x\rfloor \in \mathbb{N}$ if $x>N_\epsilon$ then $\lceil x \rceil> N_\epsilon$ and $\lfloor x \rfloor\ge N_\epsilon$ and so $\frac{\log \lceil x\rceil}{\lfloor x\rfloor^a}<\epsilon$?

9:51 PM

@robjohn But I don’t see how that helps if one cares whether a particular number is normal.

10:25 PM

@Gwyn how do you prove the inequality? Besides, the usual proof is for $x$, and then one deduces the sequence fact.

I just found out we have an Affine Geometry course in our 4th sem.

I got a headstart on everyone else

10:42 PM

@TedShifrin It doesn't.

A: Is it true that $O(ab)=O(ba)$, where $G$ is a group and $a,b \in G$?

Hi :) Why would someone downvote the following answer of mine?

0

There is no need for assumptions on finiteness. Let $a,b\in G$ for a group $G$ and let $m\in\Bbb N$. We have $$\begin{align} b(ab)^mb^{-1}&=b\underbrace{(ab)\dots(ab)}_{m\text{ times}}b^{-1}\\ &=\underbrace{(ba)\dots(ba)}_{m\text{ times}}(bb^{-1})\\ &=(ba)^m. \end{align}$$ Therefore, if $(ab)^m=...

@robjohn That was why I asked my silly question.

The following comment is made on my answer:

The question is a standard, trivial problem in elementary group theory. It doesn't conform to my standards. Instead of answering it, it would be better to look for a duplicate, or maybe a donation on PayPal. — uniquesolution 8 mins ago

I addressed it in a comment that follows.

This is the oldest version of this question I could find, @uniquesolution, and it is precisely because of a duplicate that I posted this answer here, not there; and I'm guessing you've done this because you're unhappy you answered a poor quality question and I called you out on it. — Shaun 4 mins ago

Yes, a childish war. I’ve been in a few myself.

@Shaun The first part of your comment is okay, but continuing with calling them out on another post is not productive.

10:54 PM

I guess so, @robjohn. I was angry . . .

Why is it not meaningful to square the dirac delta function?

It is best not to engage. Answering their question is okay, but posting in anger often leads to ugliness.

I would comment that a common undergraduate mistake is there: If $a^k=e$, it does not follow that the order of $a$ is $k$. Since you’re writing a very undergraduate answer, saying “it follows that” is glib.

@geocalc33 Because you cannot always multiply distributions

@robjohn That's good advice. Thank you.

Shaun, you saw my response to you? I’m not engaging in the childish war, just the math.

11:07 PM

@TedShifrin Indeed. I'm not engaging in it beyond my comment.

@TedShifrin I see. I wonder if one gains anything useful in the special cases where you can square a distribution...

@Shaun I was referring to my subsequent comment on math.

@geocalc33 When they’re actual functions?

11:26 PM

@Shaun adding to @Ted's comment, it would be better to say that $a^k=e\iff b^k=e$ implies that the orders are equal, since your answer, and symmetry, shows that $(ab)^k=e\iff(ba)^k=e$.

@TedShifrin Oh, sorry; no, I didn't see that. Yes, I skipped that detail. Thank you. I will amend the answer now.

11:46 PM

I have edited it, @TedShifrin and @robjohn. What do you think?

math.stackexchange.com/revisions/4702227/2

Gwyn

@TedShifrin I used that the logarithm is increasing to get $x \le \lceil x \rceil \implies [\log x \le \log \lceil x \rceil x]$ and that, for $x \ge 1$ and $\alpha >0$, we have $\lfloor x \rfloor \le x \le \lceil x \rceil \implies \frac{1}{\lceil x \rceil} \le \frac{1}{x} \le \frac{1}{\lfloor x \rfloor} \implies \frac{1}{\lfloor x \rfloor ^a} \le \frac{1}{x^a} \le \frac{1}{\lceil x \rceil ^a}$

I am noticing now a mistake in this message above, I should estimate $\left| \frac{\log x}{x^a}\right \le \frac{\log \lceil x \rceil x}{\lceil x \rceil ^a}$, and use $x \ge N_\epsilon \implies \lceil x \rceil \ge N_\epsilon$ with $\lceil x \rceil \in \mathbb{N}$ and so $\frac{\log \lceil x \rceil}{\lceil x \rceil ^a}<\epsilon$.

Too much LaTeX mess, I will write again tomorrow...sorry for the mess.