Mathematics - 2023-08-19

PM 2Ring

03:48

@ThomasFinley Yes, we want to help people to learn. But Stack Overflow / Stack Exchange was not created to be a help desk. Stack Exchange sites are knowledge repositories. Please see math.meta.stackexchange.com/q/21893/207316 & meta.stackexchange.com/q/256189/334566

5

The chat rooms are a little different to the main sites, and each chat room has its own culture. But most of the regulars in chat aren't interested in delivering answers to random questions on a silver platter. We tend to follow the philosophy of "teach a man to fish". We want to assist you in the process of understanding, and figuring out the answer yourself.

5

Thomas Finley

04:15

@PM2Ring You got it wrong, I presume. I solved the problem. I had an answer. I posted the pictures. Then, I needed someone to verify the legitimacy of it. I asked a person to read it and enquired whether he is interested or not and in the former case, I suggested reposting the pictures of the solution again. But then, another member felt it incorrect because he thought it would be a spamming. But I gave him to understand my reasons and I feel we both have resolved our issues amicably.

And I feel, we both choose to forget 'bout it and blaming things as a sine-qua-non miscommunication maybe, because, I was a bit hasty than usual.

Thomas Finley

05:15

I think the set of all odd functions from F1 to F2 is not a subspace.

Oh no! My bad, it is.

Ignore this.

one potato two potato

06:11

I'll just gonna take a conformal field theory course. The actual title is 'Analysis of Random conformal fields' and it seems the theory is governed by probabilistic methods. It's a good chance to learn probabilistic methods

one potato two potato

10:06

henryseg.github.io/cohomology_fractals

arxiv.org/abs/2002.00239 for explanations

Jakobian

10:46

@onepotatotwopotato I'm going to learn conformal field theory out of spite now

I'm joking, but that'd be funny

one potato two potato

I'm not particularly interested in CFT but rather in learning or seeing the usage of probability methods.

(which is not in graph theory)

one potato two potato

11:00

sometimes people say stochastic methods not probabilistic methods. I don't know the difference.

People are tempted to say stochastic than probabilistic at some point.

Jakobian

@onepotatotwopotato there's no difference I'm pretty sure

one potato two potato

stochastic is easier to pronounce. Probabbblllistic

Xander Henderson

11:48

@onepotatotwopotato It's good to take classes in things you think don't interest you.

Sometimes, you learn that you like it.

Jakobian

12:16

**Theorem.** The following are equivalent (in ZFC):
\begin{enumerate}
\item Continuum hypothesis
\item There is a unique hyperreal field of size continuum
\item There is a unique hyperreal field of siz continuum of the form $C(\mathbb{N})/M$ for some maximal ideal $M$
\end{enumerate}

I guess this isn't supported by mathjax

one potato two potato

can't believe the statement then because it's not technically texed.

Jakobian

$$\begin{align*}\text{CH}\& \iff \text{There is a unique hyperreal field of size continuum} \\ \&\iff \text{There is a unique hyperreal field of size continuum of the form $C(\mathbb{N})/M$ for a maximal ideal $M$ \end{align*}$$

this doesn't compile either? ...

$$\text{CH} \iff \text{There is a unique hyperreal field of size continuum} \\ \iff \text{There is a unique hyperreal field of size continuum of the form }C(\mathbb{N})/M\text{ for a maximal ideal }M $$

!!!

$$\iff\text{There is a unique real-closed }\eta_1\text{-field of size continuum}$$

(continuum is the smallest size such fields can be, so the theorem postulates existence of smallest such field)

sorry for spamming, I was excited

robjohn

12:47

@Jakobian I don't believe that the enumerate environment is supported inside a math environment.

Jakobian

Oh you're right

Ajay

Can someone check to see if my question makes sense and is answerable?

robjohn

@Jakobian Is this what you were trying to get?

$$\begin{align}\text{CH}& \iff \text{There is a unique hyperreal field of size continuum} \\ &\iff \text{There is a unique hyperreal field of size continuum of the form $C(\mathbb{N})/M$ for a maximal ideal} M\end{align}$$

Ajay

0

Q: Illustrative examples of the fixed points

As per the title, I was hoping to find clear illustrations of fixed point in digital images. Here is one example I have, Example 1: Starting with the base image in Figure 1, we append two copies of the image, and attach them at the base vertices of the original image. Notice that Figure 1 and F...

I'm worried it might not be clear.

Soumik Mukherjee

If $f: \Bbb{R} \to \Bbb{R}$ be defined by $f(x)=\frac{3x^2+1}{x^2+3}$

Xander Henderson

12:59

@Ajay That worry is well founded. I don't understand the question.

Soumik Mukherjee

what is the maximum time given to edit message?

If $f: \Bbb{R} \to \Bbb{R}$ be defined by $f(x)=\frac{3x^2+1}{x^2+3}$
$f^{\circ n}=f^{\circ n-1}\circ f$ then what will be $\lim_{n \to \infty} f^{\circ n}(x)$ for $x=2$ and $x=\frac12$

The Empty String Photographer

@SoumikMukherjee $\arccos(-0.41614683654) \text{ minutes}$

Soumik Mukherjee

any idea on this about how to proceed?

@TheEmptyStringPhotographer ooh

The Empty String Photographer

@SoumikMukherjee (Use that) ---> (🖥️)

sunny

13:16

Related to this answer, if $f(x)=\sum_{n=0}^{\infty}a_nx^n$ has nonzero radius of convergence $R$ and $a_0\neq0$, then how come $\sum_{n=1}^{\infty} |a_n| |x|^n $ is continuous in $|x|<R$? Moreover, why can we pick a radius such that $\sum_{n=1}^{\infty} |a_n| |x|^n \le 1$?

Xander Henderson

@SoumikMukherjee The map has only one fixed point: $f(1) = 1$. So if the limit exists, it must be $1$. The difficulty is in showing that the limit exists.

robjohn

@SoumikMukherjee $1$, I think

Since $1\lt f(x)\lt x$ for $x\gt1$ and $1\gt f(x)\gt x$ for $x\lt1$. In fact, $f(x)-x=\frac{(1-x)^3}{x^2+3}$

Xander Henderson

My first instinct would be to make an appeal to the contraction mapping principle. It should be sufficient to show that $|f'(x)| < 1$ for $x \in (1/2-\varepsilon, 2+\varepsilon)$.

@robjohn Assuming $x > 0$. But yeah, that's a nicer argument.

Ajay

@XanderHenderson Oh well, I suppose I'll keep it up until the close votes take it down. I can't think of a better way of phrasing it right now. Sob!

Xander Henderson

@Ajay So you aren't even going to try to explain what you mean?

Ajay

13:26

@XanderHenderson I'm still trying to edit, I haven't given up yet.

sunny

@sunny of course, $f(x)=\sum_{n=0}^{\infty}a_nx^n$ is continuous in $|x|<R$, but $\sum_{n=1}^{\infty} |a_n| |x|^n$ is a different function.

Soumik Mukherjee

@XanderHenderson @robjohn yes, the answer is $1$, thanks for the help

robjohn

@sunny Note that for $|x|\lt R$, the convergence of each is absolute

sunny

@robjohn hmm yes, $f(x)$ converges absolutely in $|x|<R$, but how is this related to continuity of $\sum_{n=1}^{\infty} |a_n| |x|^n$?

Xander Henderson

@sunny Think about it for a hot second. What do you know about this series?

You know that $\sum a_n x^n$ converges absolutely on a disk with radius $R$, yes?

sunny

13:40

yes

Xander Henderson

And does that converge to a continuous function?

sunny

yes

Xander Henderson

So in what ways could $\sum |a_n x^n|$ fail to converge to a continuous function?

What kind of discontinuities could exist, and where?

sunny

hmm interesting question, let me ponder

Xander Henderson

Another thing to think about: what can you say about $\sum |a_n| y^n$?

For which $y$ does that converge? And for those $y$, does it converge to a continuous function?

Ajay

13:44

@XanderHenderson Which part of my question was not clear to you?

Xander Henderson

@Ajay I genuinely have no idea what you are trying to ask.

What do you mean by a "fixed point"?

What is a "fixed point of a digital image"?

Ajay

After some research into the topic, it became clear that in digital images the fixed point does not refer to an actual point on the Cartesian plane, rather it refers to an image which is an element of the sequence.

It's what i'm trying to show with the triangle example.

Xander Henderson

Give definitions.

What is a "fixed point"?

Ajay

A point which satisfies f(x) = x

Xander Henderson

Okay, so what kind of $x$ are you dealing with, and what is $f$?

Jakobian

13:53

the $x$ on the keyboard, and $f$ is a letter

Xander Henderson

@Jakobian Not helpful. :/

Jakobian

I know I didn't read the convo

yet

Lord Pooty

Xander Henderson

Yes, I've seen your pictures. But that doesn't answer my question.

Lord Pooty

I'm not sure how else to explain it.

Xander Henderson

13:57

Well, if you can't explain it, how can anyone help you?

Lord Pooty

I simply append two copies of P_0, and attach them at the base vertices of the original image. And I repeat this

Jakobian

@robjohn yeah, thanks!

Lord Pooty

I need to go away and think for a bit. You made a very good observation @XanderHenderson about f and x. Thanks :)

Xander Henderson

It seems to me that what you are actually doing is gluing together three copies of $P_{n}$ in order to obtain $P_{n+1}$. In this case, the map seems to be a map of sets, and the fixed point is an unbounded set made up of a bunch of right triangles.

But, again, your question is extremely unclear. It is hard to help when you don't grok the question.

Jakobian

@sunny remember all the exercises you used to do about uniform convergence?

sunny

14:06

@Jakobian yes :)

Jakobian

what was the main method to prove a series $\sum_n f_n$ is continuous in a given domain $D$?

sunny

to show it converges uniformly on a subinterval of $D$

Jakobian

yeah. You can apply this here with no problem

sunny

yeah, I have realized the following:

Jakobian

@sunny the second part, Joel Cohen explains in his answer

see the part in brackets

sunny

14:18

if $f(x)=\sum_{n=0}^{\infty}a_nx^n$ converges in $|x|<R$, then we know 1) it converges absolutely in $|x|<R$ and 2) it's continuous in $|x|<R$ and 3) it's uniformly convergent on $|x|\leq \rho$ where $0\leq\rho<R$ (let's call this a theorem).

I was wondering whether $g(x)=\sum_{n=1}^{\infty}|a_nx^n|$ is also continuous in $|x|<R$. For sure, $h(x)=\sum_{n=0}^{\infty}|a_nx^n|$ is continuous $|x|<R$ by the theorem, since we know it's a power series that converges on $|x|<R$ according to the theorem. And hence $g(x)=h(x)-1$ is also continuous, since $a_0=1$.

@Jakobian when Joel states that $\sum_{n=1}^{\infty}|a_nx^n| \leq 1$ for all $|x| \le \rho$, is this the definition of continuity with $<$ replaced by $\leq$?

Jakobian

14:46

@sunny I might be a bit pedantic here, but $h(x) = \sum |a_n|\cdot |x|^n$ is a power series in $|x|$ with the same radius of convergence

@sunny what does it matter

If $\sum ... < 1$ for $|x| < \delta$ then you can just take $|x|\leq \delta/2$

Xander Henderson

@Jakobian inorite?

1 hour ago, by Xander Henderson

Another thing to think about: what can you say about $\sum |a_n| y^n$?

geocalc33

15:19

What are the drawbacks of a Fourier transform defined on a compact subset of $\Bbb R^n?$

Xander Henderson

You gotta be much more specific than that...

geocalc33

15:49

For a Schwartz function $f$ on $\mathbb R$, let $\Gamma(f,s)=\int_{\mathbb R^\times} |x|^sf(x){dx\over |x|}$. The usual Gamma factor is obtained by taking a Gaussian. The local functional equation (proven by changing variables in the defining integrals, in the range $0<{\rm Re}(s)<1$), is
$$\Gamma(f,s)\cdot \Gamma(\hat{g},1-s)= \Gamma(\hat{f},1-s)\cdot \Gamma(g,s)$$
for any two Schwartz functions $f,g$. And Riemann's argument proves
$$
\Gamma(f,s)\cdot \zeta(s) = \Gamma(\hat{f},1-s)\cdot \zeta(1-s)

How does the analysis differ?

Xander Henderson

What, precisely, do you mean by "Schwartz on $\Omega$"? Smooth and vanishes on the boundary?

geocalc33

mathworld.wolfram.com/SchwartzFunction.html

I'm using this definition

Xander Henderson

@geocalc33 That's the definition on $\mathbb{R}^n$.

geocalc33

smooth yes...vanishes on the boundary, not necessarily

Xander Henderson

How are you defining such a function on $\Omega$?

If I recall correctly (and it's been a while for me), the Fourier transform of a compactly supported function is no longer compactly supported. So I don't see any reason to expect that the Fourier transform would be nicely behaved on a subset $\Omega \subseteq \mathbb{R}^n$.

Also, what does it mean to "take a Gaussian"?

geocalc33

16:00

The Gaussian function is self dual under the Fourier transform so things work out nicely and you get the "usual" Gamma factor of $\pi^{-s/2}\Gamma(s/2)$

Xander Henderson

@geocalc33 "The" Gaussian function?

If $f$ is a Gaussian function, then $\mathscr{F}(f)$ is also a Gaussian function.

But there are lots of Gaussian functions.

geocalc33

@XanderHenderson Yes, exactly. Yes, the compactly supported function is no longer compactly supported under the Fourier transform..It is still in the Schwartz space however

Xander Henderson

@geocalc33 Not if it doesn't vanish at the boundary...

Jakobian

@XanderHenderson sounds about right, after the Fourier transform it should be more akin to the exponential

Xander Henderson

But if it doesn't vanish at the boundary, then it wasn't really Schwartz to begin with...

geocalc33

16:04

@XanderHenderson It tends to zero at the boundary

Xander Henderson

7 mins ago, by geocalc33

smooth yes...vanishes on the boundary, not necessarily

Which is it?

I'm confused...

geocalc33

It hits a singularity at the boundary but tends to zero

Xander Henderson

What?

That doesn't make sense.

What do you mean by "it hits a singularity"?

geocalc33

oh okay i got it - I defined the interval as an open interval $(0,1)$

so you can say that yes it goes to zero as $x \to 1$ lets say

Singularity meaning function is not defined at $x=1$

Xander Henderson

@geocalc33 It wasn't defined at $x=1$ in the first place. But if it goes to zero, that means that it can be continuously extended to $x=1$ by zero.

And if it is supposed to be Schwartz, then we would expect that continuous extension to satisfy $f^{(n)}(1) = 0$ for all $n \in \mathbb{N}$.

But I'm not really sure that it makes sense to think about the Fourier transform on $(0,1)$, unless you are thinking about periodic functions on $\mathbb{R}$. The Fourier transform, in generality, should be thought of as an operator on locally compact abelian groups (something something), which takes functions on that group to functions on the dual group.

Soumik Mukherjee

16:11

@XanderHenderson I went to search 'inorite' on dictionary thinking it was an actual word 💀

Xander Henderson

$\mathbb{R}^n$ is self-dual; the dual of $S^1$ (where $S^1$ can be thought of as an interval with endpoints identified, hence periodic functions) is $\mathbb{Z}$; etc. Google "Pontryagin dual".

Anywho, I have errands to run.

geocalc33

@XanderHenderson Okay

copper.hat

Curious why someone would downvote a 7 year old innocuous question (not answer).

one potato two potato

16:31

because it can be done and nobody can stop it

Shaun

Hi :) I made a case to undelete and reopen something:

0

A: Requests for Reopen & Undeletion Votes (volume 01/2022 - today)

Please undelete and reopen https://math.stackexchange.com/q/4754844/104041 It was deleted, presumably, because some users do not like the fact that $$x+1=x-2$$ has $x=\infty$ as a solution; a controversial opinion. The comments say the question is nonsense. It is not. It is also claimed that it i...

Shaun

16:45

I don't get the downvotes on it, unless it's the same crowd that deleted the original question . . .

I think I made a strong case.

Ted Shifrin

@copper.hat Is anything truly innocuous?

geocalc33

I don't get the 2 downvotes on my question. Now it's only sitting at +12

ridiculous

one potato two potato

17:04

downvote downvote downvote

geocalc33

I don't understand the upvotes either - my question could have been asked by a clever highschooler

Shaun

I don't see a link to your question, @geocalc33. Perhaps I could give some feedback on it . . .

$Mathematics$ Helpful Commentary

Do you want constructive feedback for your Mathematics Stack E...

geocalc33

@Shaun Okay I will link it it 30 seconds

I wrote a good poem recently too

Fourier transform.
Bounded domain.
Restriction to the domain $I=(0,1).$
Fourier transformed.

robjohn

@geocalc33 welcome to the Dumbfounded By Downvote club.

4

@TedShifrin does being innoculated count?

Ted Shifrin

@robjohn only if your eyes are wide open.

ZaWarudo

17:17

Is this proof that the graph of the inverse function can be obtained by the graph of the function by exchanging $x$ with $y$ correct? Proof: by definition, $(x,y) \in \text{graph}(f) \iff y=f(x)$. Since $f$ is invertible, we have $y=f(x) \iff x=f^{-1}(y)$. But $(y,x) \in \text{graph}(f^{-1}) \iff x=f^{-1}(y)$. Hence, $(x,y) \in \text{graph}(f) \iff (y,x) \in \text{graph}(f^{-1})$; thus, we can obtain the graph of $f^{-1}$ by exchanging $x$ and $y$ in the graph of $f$.

robjohn

@TedShifrin dilation innoculation?

Ted Shifrin

I was thinking of ocu with eye, like binocular or monocle.:)

@ZaWarudo Overly wordy, but sure.

Amira

hi can you please help me ?

in nonhomogenous heat equation :$u_{t}-\Delta u=f(x,t)$ in U smooth open of $R^{n}$ and f $\in L^{2}((0,T),L^{2](U))$i find that u $\in C((0,T},L^{2}(U)) \cap L^{2}((0,T),H_{0}^{1}(U))$ and i wanted to prove uniqueness but in the hint thay say that $u_{t}\in L^{2}((0,T),H_{0}^{1}(U)\cap H^{2}(U))$ i can prove that but my question is why it's a a condition to prove uniqueness? thanks

冥王 Hades

17:40

I haven’t slept even a tiny bit in over 30 hours

any longer and my brain will ascend to a higher plane of existence

Koro

Every great circle on a 2-sphere is a geodesic. But why is every geodesic on a 2-sphere is a circle?

Ted Shifrin

More likely, lower.

Semiclassical

@冥王Hades go sleep

Koro

17:57

if I could show that every such geodesic is locally an arc of a circle then I'm done.

(i.e., for some point p on the geodesic, there exists an open set whose intersection with the geodesic is an arc of a circle.)

Ted Shifrin

Uniqueness of geodesics?

Koro

This is because of the existence & uniqueness of a maximal geodesic.

or probably my question is wrong. Should I have said every maximal geodesic is a great circle?

Ted Shifrin

I’m saying that’s a proof. Considering the acceleration vector gives an elementary proof. But I’m leaving now.

Koro

ok. My intended statement was: Every maximal geodesic on a circle is either a point or a great circle.

I made mistake in stating it earlier.

yeah, I'll try proving this.

Xander Henderson

@copper.hat Did it get bumped recently?

I will sometimes downvote an old question which gets bumped to the top of the list in order to make it eligible for Roomba deletion.

Koro

18:08

hmm I see now how it follows from the existence and uniqueness of maximal geodesics.

copper.hat

@XanderHenderson I don't think so. math.stackexchange.com/questions/1504483/…

Its part of my colonial prerogative to establish northern hemisphere dominance

albeit i do not have colonial roots

nor an interest in northern hemisphere dominance

Xander Henderson

Huh...

More on topic at History of Science and Mathematics, but an interesting question.

No idea why it would have gotten a downvote.

copper.hat

yep, that was mentioned.

i don't care about it other that my never ending quest to vaguely understand humans

and still puzzled how serial voting removal was applied to a different se/so site on which i have only two answers.

again, i give a s* about the rep, but am still curios

then again, apparently there is a business in selling rep... an alternative route to the orange mse jumpsuit

@XanderHenderson thanks, i assume that upvote was you.

no need of course.

Xander Henderson

For the record, I think it is an interesting question, and the answer which references Stoke's theorem is, I think, an interesting answer. Upvoted both.

copper.hat

yep, i upvoted the Stokes one a long time aog.

@TedShifrin there is a secret message to the new world order hidden in the angle question.

John Bull

18:22

Hello, can someone help me regarding functions?

if someone can please help me with this question:
https://math.stackexchange.com/questions/4755220/calculating-fnj-from-fj-frac-a-j-i-a-ij-b-a-ij-j

Koro

why is a great circle a maximal geodesic on S^2?

I know that the great circle is a geodesic.

Xander Henderson

In other news... it took almost 2 hours to get a flu shot this morning. Mostly because my brother's insurance is all f'd up. And it shouldn't be.

Stupid social security administration... :(

copper.hat

are you concerned about getting the flu or potentially spreading it?

Koro

why is the green curve not a maximal geodesic on the 2-sphere??

Green curve is the union of a great circle (colored in yellow) and a hair protruding from the circle.

Xander Henderson

18:41

It isn't even a geodesic, no?

Koro

I'm trying to understand why not.

Geodesic to an n-surface S at p in Tp S (the tangent space to S at p) is a curve a: I--> S such that a(0)= p and the acceleration vector a(double dot) (t) is orthogonal to T{a(t)} S for every t in I.

geocalc33

19:06

analytic continuation is the mistress I once knew

but I lost her. Now my heart yearns once more. And I realize I must rotate the wick of the candle and light the flame. Continue our love.

How am I this good at poetry

Xander Henderson

@geocalc33 Dunning-Kreuger?

geocalc33

@XanderHenderson I feel like I am really good at poetry though. I don't think it's Dunning-Kreuger effect

Xander Henderson

@geocalc33 Uh huh.

geocalc33

You'll be pleased and glad to hear I have a potentially illuminating approach to my question. Get this: Use analytic continuation to extend the PDE's to a quote on quote larger domain. I mean what could go wrong right?

Xander Henderson

"quote on quote"?

geocalc33

19:20

I am drawing inspiration from physics but I insist that I don't care about it.

Yes quote on quote

Xander Henderson

@geocalc33 Nonsense.

I believe you mean "quote-unquote".

geocalc33

@XanderHenderson I don't know what you mean but it's fine :)

oh

okay I gotcha

copper.hat

Dunning Kruger is my middle name.

geocalc33

I just learned something new in english

Xander Henderson

@copper.hat That's two middle names. You've failed again.

geocalc33

19:22

boom roasted.

copper.hat

No, I am absolutely right

Xander Henderson

merriam-webster.com/dictionary/quote%2C%20unquote

copper.hat

I mean, everything I say is a lie.

geocalc33

McManus is my middle name

Xander Henderson

@copper.hat But what about that that thing you just said!?

copper.hat

19:23

To answer that I must direct you to my middle name.

Xander Henderson

I see.

geocalc33

That's pretty neat -that that's your middle name @copper.hat

copper.hat

Edmund Gerard Thomas

Seriously.

Two of my brothers wanted at least of one of their children to have Danger as a middle name.

Strangely, their spouses nixed that idea.

Xander Henderson

I wanted to have twins.

I was going to name one of them "Control" and the other "Treatment'.

geocalc33

I am going to name my child Jergen Jongledence

copper.hat

19:27

:-). I know a family who have a Bill and William (not an aggregated family).

geocalc33

@XanderHenderson that is deeply disturbing

copper.hat

My cultural insensitivity & lack of awareness resulting in my bursting out laughing because I thought it was a joke.

Maybe Null as a middle name?

Xander Henderson

@geocalc33 Aw... I thought it was a good joke.

copper.hat

Or $H_0$?

Xander Henderson

@copper.hat How much would it suck for a second child to be named $H_a$?

copper.hat

19:29

:-)

Xander Henderson

"Look, $H_a$, you're the backup plan. If something happens to your older brother, we'll still have you!"

2

copper.hat

:-). I have encountered some strange names over the decades, but am unable to recall them now of course.

The usual ones with Vietnamese names in an English speaking country.

I often wonder what Irish names sound like to other cultures.

geocalc33

the last name Butkus

I would have a hard time not laughing during introductions

Jakobian

@XanderHenderson Hausdorff measure?

copper.hat

I met a George Harrison recently.

Surprisingly he was aware of someone else of the same name. I had to tell him of course.

Xander Henderson

19:34

@Jakobian Alternative hypothesis. In contrast to $H_0$, the null hypothesis.

Hausdorff measure is usually $\mathscr{H}^s$ or $\mathcal{H}^s$.

geocalc33

I met a guy named Joseph R. Biden

I also met a professor at some college in new york (state)

he gave a talk about knots

copper.hat

that is knot true

geocalc33

I asked a question at the end (when it opened up for questions)

bwDraco

I dug out an fun book titled The Number Devil... and this quote caught my eye:

Xander Henderson

@bwDraco I have that book.

bwDraco

19:39

> ...we can't even prove with any certainty that no perfect solution exists. That would be something at least. It would allow us to stop looking for one. Besides, proving that no proof exists is a proof in itself of sorts.

(with respect to the shortest path problem)

I am in awe at how this book covers the halting problem, of all things.

(page 229)

Jakobian

shortest path as in on graphs?

bwDraco

Yes

Jakobian

people solve these with a probabilistic algorithm iirc

bwDraco

Start from page 226.

Jakobian

this is also called the traveling salesman problem I think

bwDraco

19:44

> proving that no proof exists is a proof in itself of sorts

Is this indeed referring to the halting problem?

geocalc33

could refer to any problem

Jakobian

for example, people were wondering about continuum hypothesis

Continuum hypothesis is a statement that cardinality of $\mathbb{R}$, continuum, which is $2^{\aleph_0}$, is equal to $\aleph_1$

bwDraco

NP completeness!

Jakobian

people proved that in axioms of ZFC, which is what mathematicians use, you cannot derive CH (short for continuum hypothesis)

so this is a proof that a proof doesn't exist, to be exact, a proof of CH doesn't exist in ZFC

There is also the statement of Godel's incompleteness theorem, which says that if we have a particular kind of theory, then there exist statements which one cannot prove

Gödel's incompleteness theorems

Gödel's incompleteness theorems are two theorems of mathematical logic that are concerned with the limits of provability in formal axiomatic theories. These results, published by Kurt Gödel in 1931, are important both in mathematical logic and in the philosophy of mathematics. The theorems are widely, but not universally, interpreted as showing that Hilbert's program to find a complete and consistent set of axioms for all mathematics is impossible. The first incompleteness theorem states that no consistent system of axioms whose theorems can be listed by an effective procedure (i.e., an algorithm...

for most mathematicians, it's of value to know what can and what cannot be proved

bwDraco

Well, not shortest path problem. Traveling salesman problem.

Jakobian

19:51

oh, are those actually different?

ah yeah they are. Sorry

Ted Shifrin

@Koro Points are not considered.

Are you using local length minimization or velocity covariant constant as the definition? Either way, your green squiggle fails.

bwDraco

$P \stackrel{?}{=} NP$

...wow.

Ted Shifrin

@copper Huh? What angle question?

bwDraco

19 mins ago, by bwDraco

> ...we can't even prove with any certainty that no perfect solution exists. That would be something at least. It would allow us to stop looking for one. Besides, proving that no proof exists is a proof in itself of sorts.

(with respect to the traveling salesman problem)

I'm utterly confused here... but it looks like they're talking about the fact that the traveling salesman problem (in decision form) is NP-complete.

Koro

20:07

1

Q: Every geodesic $f$ on the unit n-sphere is of the form $f(t)= \cos (at) e_1+\sin (at) e_2$ for some orthogonal vectors $e_1,e_2\in \mathbb R^{n+1}$

I know that if $f(t)= \cos (at)e_1+ \sin(at)e_2$, then $\ddot f(t)=-a^2 f(t)= \pm N(f(t))\perp T_{f(t)}S$, where $N(p)= \frac{\nabla f(p)}{\|\nabla f(p)\|}$ is the unit normal vector, $S= g^{-1}(1), g(x_1,x_2,...,x_{n+1})= \sum_{i=1}^{n+1} x_i^2$ and $T_{f(t)}S=$ the tangent space to $S$ at $f(t)...

multivariable-calculus differential-geometry vectors

@TedShifrin that acceleration at t is orthogonal to the tangent space at a(t).

copper.hat

@TedShifrin An oldie, math.stackexchange.com/questions/1504483/…

Ted Shifrin

@Koro So it’s easy vector calculus to see this forces a great circle.

bwDraco

I wrote some annotations in the margins of the book years back... and it brings up stuff like the Fermat polygonal number theorem (page 98).

My mind is spinning.

Shaun

21:28

Why was this closed as lacking details/clarity!?

Ted Shifrin

Lack of effort?

robjohn

22:42

Perhaps lack of details or clarity?

robjohn

22:57

There are some confusing facts stated before the question: "it seems that all groups of order $112=24\times7$ have an element of order $14$. Could you please help me about the proof of it?" So it is a PSQ along with a lack of clarity formed by the confusing prolog.

sunny

If someone has some time to spare, I would be a happy camper if this answer, specifically to question 3, could be checked, i.e. compare it with my comment and tell me if the answer is right or my comment. Maybe we're both wrong?

For instance, the answer claims $|f(x)|=\sum_{n=0}^{\infty}\left|a_{n}\right||x|^{n}$ and at the end, that $\delta <0$. I think this can not be right.

Ted Shifrin

LOL.

robjohn

@TedShifrin yes?

Xander Henderson

@sunny It seems like you have asked at least three questions in that post. Moreover, it seems like most of those questions are questions which you have asked in here over the last couple of days. It kind of feels like you have ignored what has been said here, posted a question on the main site, ignored the answer given, and are asking us to go over the same material. Again.

I'm not sure that I have the energy to engage again.

Ted Shifrin

The beginning “obviously “ is garbage. I do not have the energy to read all that.

robjohn

23:05

Ok. So you were just agreeing with my critique.

Ted Shifrin

@robjohn Of course $\delta$, like her good friend $\epsilon$, cannot be negative.

No, we’re on different questions,

robjohn

ok

Ted Shifrin

I was answering sunny, sorta.

Details and clarity is the button I push for no effort.

$Mathematics$ Helpful Commentary

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