Mathematics - 2019-05-06

user193319

00:13

If $\{x,y\}$ is given the discrete topology, is it true that $H_0(\{x,y\}) \cong \Bbb{Z}^2$?

sccoding

02:14

Hello! How can I find 8 distinct non-negative numbers <= 50 such that all the pairwise differences are distinct?

Also, is there any way to generate these numbers (other than use powers of 2)

Rithaniel

05:52

So, given a semigroup $S$, one can adjoin an identity element to create the monoid $S^1$. Alternatively, you can adjoin an absorbing element to create the semigroup $S^0$. My question is: Is there a specific term for $S^0$ similar to how $S^1$ is a monoid? If not, and you wanted such a term so you didn't have to say "semigroup adjoin an absorber" all the time, what would be an appropriate term?

user131753

06:27

0

Q: A question on Rudin's "proof" of the associativity of the convolution of two measures

Background Let $G$ be a locally compact abelian group and $M(G)$ the Banach space of all complex Radon measures on $G$. The convolution of two complex Radon measures $\mu, \lambda\in M(G)$ is defined by specifying a linear functional on $C_0(G)$ (and invoking Riesz representation theorem) as f...

Leaky Nun

07:23

@user193319 yes

@sccoding what's wrong with powers of 2?

these "x without y" questions are really my pet peeves

Alessandro Codenotti

09:00

@user193319 yes, remember that $H_0$ is essentially counting connected components

Or that homology of a disjoint union splits as a direct sum of the homologies

Leaky Nun

10:52

Borel–Kolmogorov paradox

In probability theory, the Borel–Kolmogorov paradox (sometimes known as Borel's paradox) is a paradox relating to conditional probability with respect to an event of probability zero (also known as a null set). It is named after Émile Borel and Andrey Kolmogorov. == A great circle puzzle == Suppose that a random variable has a uniform distribution on a unit sphere. What is its conditional distribution on a great circle? Because of the symmetry of the sphere, one might expect that the distribution is uniform and independent of the choice of coordinates. However, two analyses give contradictory...

Silent

What is gcd of zero polynomial and $x^5+1$?

Leaky Nun

what's the gcd of zero and anything?

Mathphile

does any truncatable circular palindromic prime exist which has more than one digit?

Silent

@LeakyNun I think its anything.

here $x^5+1$

Leaky Nun

then anything it is

Silent

11:16

ok

:)

Let $f(x)=x^{15}+ax^{10}+bx^5+c$ be irreducible where $a,b,c\in \Bbb Z_5$. Then since derivative of this polynomial is zero, its roots in an extension will be multiple roots. Is this extension $\Bbb Z_5[x]/(x^{15}+ax^{10}+bx^5+c)$

Rithik Kapoor

@Mathphile those are a lot of restrictions for a prime

I doubt if a prime like this exists

i wonder if we can prove this

Mathphile

11:35

@LeakyNun any ideas?

Leaky Nun

@Silent no such polynomial can be irreducible

since finite fields are perfect

@Mathphile no

@Silent but yes that would be the extension

Silent

Thank you.

Mathphile

12:16

seems like 373 is the only truncatable circular palindromic prime exist which has more than one digit

Semiclassical

13:45

hi chat

Rudi_Birnbaum

14:00

Hi all! question question:

I got a famliy of "vector fields" $f: \Bbb R^3 \to \Bbb R^3$ that satisy a certain physical property (that is not that relevant now) However they have all in common that theay are pointwise parallel to $\nabla\rho(\vec{r})$.

for $\rho: \Bbb R \to \Bbb R$.

parallel means now a multiple of any scalar including 0.

The question:

How to formulate that properly, shortly and concisely?

I thought of $f$ is the gradient of an arbitrary scalar function $g$ of $\rho$. Like $f=\nabla g(\rho)$. But I am getting unsure ...

1010011010

Hi all

Rudi_Birnbaum

Who can help?

1010011010

I'd like to make an calculus question as attractive as possible

Besides putting a bounty on it, what else could I do to make a calculus question as attractive a possible?

I could include my own efforts but they're not based on anything. I'm completely stumped for the first time ever in some 10y of calculus

swarnim

math.stackexchange.com/questions/3214943/… can anyone please help?

Mathphile

14:38

why are primes of the form 11.......11 always have a prime number of digits?

Leaky Nun

14:54

@Mathphile because say 1111 is divisible by 101

1111 = 101 x 11

111111 = 10101 x 11

111,111,111 = 1,001,001 x 111

Semiclassical

cute

So if the number of digits is composite, then you can always find a factorization of that kind

Mathphile

ahh

got it

is there any prime of the form 12345678910111213......

Mats Granvik

16:10

1234567891=prime
12345678910111=prime

Mathphile

@MatsGranvik they have to be complete numbers if you know what i mean

according to my search of a large range of numbers by pari i haven't found a number

i doubt that a prime number of this kind exists

Silent

A group action is faithful iff every stabilizer is trivial. Am i correct?

Semiclassical

16:28

@Mathphile there's a bit of a law-of-large-numbers aspect to this: the prime number theorem states that $\pi(n)$$ (number of primes less than n) grows like $n/\log n$, so that the probability of a random number from 1 to n is of order $1/\log n$

So for instance the probability of picking a random number from 1 to 10^100 and getting a prime is 0.004. Not likely at all. Combine that with the fact that not many numbers of the form 1234... seem to be prime in the first place

so you're taking just a few samples and looking for something unlikely. shouldn't be too surprising if you don't see it

(of course, making this kind of argument legit is always hard. absent an explanation like Leaky's before, it may be very difficult to conclusively forbid such numbers. But that doesn't make them likely)

An interesting variation: Suppose you start with 2, i.e. consider 234...

In that case both 2 and 23 are primes. Are there any other such examples in this sequence?

My intuition would be: If you start with some positive integer and consider the sequence q,q(q+1),q(q+1)(q+2),... (in your sense), then any such sequence only contains a finite number of primes

oh hey, 23456789 is prime. neat

Mathphile

16:47

yup that definitely is neat

Mathphile

17:23

are there an infinite number of primes of the form 111...11?

Alessandro Codenotti

@Mathphile open

user76284

17:37

3

A: Model of Robinson Arithmetic but not Peano Arithmetic

See John Burgess, Fixing Frege (2005), page 56 : None of the usual associative, commutative, or distributive laws for addition and multiplication can be proved in $\mathsf Q$ [i.e. Robinson arithmetic] nor can even the law $Sx \ne x$. As to this last point, a natural model of $\mathsf Q$ is p...

Are 2 rogue elements necessary? Isn't $\mathbb{N} \cup \{\aleph_0\}$ a model of Q?

Silent

17:59

Please verify this:

2 hours ago, by Silent

A group action is faithful iff every stabilizer is trivial. Am i correct?

Secret

so many open questions in number theory, all because the knapsack problem is so hard to solve and that we have no computable formula of arbitrary primes

Konformist Liberal

18:19

Hi, possibly silly quick question: Is Ho(Top) an abelian category?

S. Crim

18:40

Anyone good with martingales that could help me out with a quick question?

Ultradark

19:40

scientificamerican.com/article/…

Rudi_Birnbaum

20:00

Hi all, is here anyone familiar with vector calculus?

Ted Shifrin

@Rudi: I'm supposed to be more than familiar with it. What's the question?

Rudi_Birnbaum

is that right?

Ted Shifrin

That requires a long time to sort through.

Rudi_Birnbaum

f and rho are scalar functions

B a constant vector

Ted Shifrin

So it's a bunch of product rule.

Rudi_Birnbaum

20:03

Well if you look into it it should be actually pretty simple

yeah CAB BAC stuff

with and without grad

first step is trivial ok?

I just move rho out of the way.

the second step is

from here en.wikipedia.org/wiki/Vector_calculus_identities#Curl

Ted Shifrin

So it looks right, but I'm not checking every line. Do you have a specific question? I don't have the patience to work out everything.

Rudi_Birnbaum

No, it was just that I needed help to check if its correct.

Thanks anyway!

Ted Shifrin

Try a specific example with a nonzero answer.

Rudi_Birnbaum

Well the most important question is if f(x) can be anything else than a constant if the whole thing should vanish

Konformist Liberal

Hi, (possibly silly) quick question: Is Ho(Top) an abelian category?

Rudi_Birnbaum

20:09

I have tested c * grad(rho(r)

thats correct

No the big question is can f be anything else than a constant. And I suppose its no. That would be if that is all correct.

Ted Shifrin

I don't see why it's relevant that these are functions of $\mathbf r$.

The formula should hold for arbitrary functions of $(x,y,z)$.

Rudi_Birnbaum

OK!

Ted Shifrin

Oh, I guess that's what you mean.

Rudi_Birnbaum

I mean its just from the physcis behind

Ted Shifrin

I was thinking functions of $\|\mathbf r\|$.

Your $\mathbf r$ is just the position vector. So this is a totally general formula.

So take $f=x$, $\rho=y$, and $\mathbf B = \mathbf k$. Does it check?

Rudi_Birnbaum

20:12

yes!

test

Ted Shifrin

OK, done.

Rudi_Birnbaum

Internet problem ...

yes i think its right

agree?

Ted Shifrin

Yeah.

Rudi_Birnbaum

cool

then if B can be any constant vector and the whole beast shoud be 0

f(r)=const.

@Ted?

Ted Shifrin

Say what?

You're asking if the thing is $0$, must $f$ be constant?

Daminark

20:20

Hey there

Rudi_Birnbaum

yes?

Ted Shifrin

Obviously, if $f$ and $\rho$ are constant multiples of one another, you always get $0$. And we can figure out other examples.

heya Demonark.

Rudi_Birnbaum

@TedShifrin If you say you could sign the correctness of the eq. with your good name I'll put you on the acknowledgment of the paper

Ted Shifrin

No, that's OK.

Rudi_Birnbaum

OK.

thank you! (its physically quit meaningful)

Ted Shifrin

20:23

You saw my denial that $f$ has to be constant, right?

Rudi_Birnbaum

Oh, example?

Ted Shifrin

Did you see what I wrote?

Rudi_Birnbaum

@TedShifrin here?

Ted Shifrin

Yes.

Rudi_Birnbaum

so where are they?

the examples

Ted Shifrin

20:25

Come on.

Take $f=\rho$ to be an arbitrary function.

You can make up more interesting examples where $\nabla f\times \nabla\rho$ is orthogonal to a given constant $\mathbf B$.

Rudi_Birnbaum

B shall have any orientation.

So it should work for B=z

Ted Shifrin

You mean it has to be $0$ for all vectors $\mathbf B$?

Or some fixed one?

Rudi_Birnbaum

and at the same time for B=x,...

@TedShifrin yes that

Ted Shifrin

Well, if $\mathbf v\cdot\mathbf B=0$ for all $\mathbf B$, then $\mathbf v = 0$ of course.

But I still gave you lots of counterexamples.

Rudi_Birnbaum

can we charcaterize the set of all $f$?

s.t. the expression is = 0 for any B?

Ted Shifrin

20:30

Simply $\nabla f\times\nabla\rho = 0$. Depends on $\rho$.

Rudi_Birnbaum

parallel

I had thought about $\nabla f(\rho)$

Ted Shifrin

What does that mean?

Rudi_Birnbaum

But it seems thats not all

for any scalar $f$

$\nabla f(\rho) = f'(\rho)\nabla \rho$

Ted Shifrin

Oh, you mean any function $f$ that can be written as $F(\rho(\mathbf r))$? Yes, that will do it.

Rudi_Birnbaum

but we loose those which have vorticity

...

Ted Shifrin

20:33

Huh? How? These can have vorticity. It's just parallel to $\nabla \rho$.

Rudi_Birnbaum

ahh - hmm

right, they can't!

Ted Shifrin

Huh?

Rudi_Birnbaum

I understood what you mean.

Ted Shifrin

OK.

Rudi_Birnbaum

OK so $\nabla F(\rho(\mathbf r))$ is fine and "any field parallel to grad rho" is also fine?

to say I mean

Ted Shifrin

20:37

So the question remaining is: If the gradients are parallel, must $f$ be $F(\rho(\mathbf r))$? The answer actually is yes.

Rudi_Birnbaum

@TedShifrin cool!

Ted Shifrin

Assuming $\nabla \rho$ is everywhere nonzero, anyway. There could be crazy stuff otherwise.

Rudi_Birnbaum

Well, the point is, it is zero for some very few points.

Ted Shifrin

Well, if the points are isolated, then what I said should still be fine by continuity.

Rudi_Birnbaum

ok!

\rho is an electron density

in a molecule

Ted Shifrin

20:39

So it has singularities ...

Rudi_Birnbaum

so its quite well behaved

it has some cusps

at each nucleus

but its finite everywhere

Ted Shifrin

So what I said may have global issues, but it's certainly locally correct everywhere.

Rudi_Birnbaum

and continuous

and almost everywhere smooth

Ted Shifrin

Hmm ... Well, "almost everywhere" can wreak havoc.

Rudi_Birnbaum

at the nuclei it makes "cusps" of finite derivative.

Ted Shifrin

20:41

Well, I've given you stuff that's correct except for crazy winding situations.

Rudi_Birnbaum

everywhere positive and integrates to 1

i see!

thanks a lot!

Well one more tiny thing :-D

Ted Shifrin

I need to leave in a moment, but yes?

Rudi_Birnbaum

when are the few zeros of $\nabla \rho$ identical to the zeros in $({\bf B} \times \nabla F(\rho))\rho$?

For invertible/unique $F$?

($\rho$ is zero free)

Ted Shifrin

But remember we said that $\nabla f$ is a multiple of $\nabla\rho$, so $\nabla f$ will vanish whenever $\nabla \rho$ does.

OK, I need to leave for now.

Rudi_Birnbaum

21:00

yes.

thank you @Ted you were of great help!

GFauxPas

Having a hard time understanding how to write this integral in terms of a unit normal vector: let $f, g$ be smooth functions on a compact domain $M \in \mathbb R^n$, let $\mu = \star 1$ the volume form on $M$.

part 1, done: show $df(\star dg) = \nabla f \cdot \nabla g \mu$

part 2, done: show $d \star dg = (\Delta g) \mu$

part 3, done: show $d(f(\star dg)) = (\nabla f \cdot \nabla g + f \Delta g) \mu$

part 4, I need a hint:

kyle campbell

For anyone familiar with statistics: if you set an arbitrary level of significance α to be some value and subsequently you do an experiment and find that your p-value is significantly less than α, is it true that if your p-value is much smaller than α then the experiment has a greater significance than an experiment where the p-value is closer to α?

GFauxPas

let $n$ be the outward pointing unit normal vector field on $\partial M$. write $\frac {\partial g}{\partial n} = \nabla g \cdot n$, show:

$\displaystyle \int_{\partial M} f(\star dg) = \int_{\partial M} f \frac{\partial g}{\partial n} \mu_{\partial M}$

$\mu_{\partial M}$ is the volume form on $\partial M$

I'm not familiar with the properties of $\nabla g \cdot n$

:(

Ultradark

21:22

$\nabla g \cdot n= \nabla \cdot (gn)-g (\nabla \cdot n)$

@GFauxPas

GFauxPas

thanks, that's a start

hm

the introduction of $n$ still is throwing a wrench in my brain, it seems to come out of nowhere

Semiclassical

21:53

@GFauxPas Maybe take consider the example of g(x,y,z)=x^2+y^2. (The lack of z on the RHS is intended)

GFauxPas

Found the theorem I needed to go forward

Semiclassical

Cool

GFauxPas

$\star dx_i = n \mu$

But let me see what you just said

okay, so, $dg = 2xdx + 2ydy, \star dg = 2x dydz - 2ydxdz$?

Arrow

What are some nice geometric examples of closed maps with non-compact fiber? I don't mean things like a constant map from an uncountable discrete space, but something more pictorial.

Mathphile

22:09

0

Q: Circular Happy Palindromic Primes

$(1)$ A circular prime is a prime number with the property that the number generated at each intermediate step when cyclically permuting its (base 10) digits will be prime. For example, 1193 is a circular prime, since 1931, 9311 and 3119 all are also prime. (2) A happy number is defined by the ...

elementary-number-theory prime-numbers conjectures palindrome

any ideas?

Semiclassical

@GFauxPas yeah. Not sure what the volume form on the cylinder is but it should be simple enough

And it should agree with the formula

GFauxPas

so did we have a consensus on whether or not to write $x_i dx^i$ or $x^i dx_i$ for $x_1 dx_1 + x_2 dx_2 + \ldots + x_n dx_n$?

Similarly

If I have a vector (c1,c2,...,cn) wrt a basis (b1,b2,...,bn)

How do I write $c_i b_i$, does it matter ?

What's the geometric interpretation of $\nabla f \cdot n$?

Ted Shifrin

23:13

Neither. You need a summation symbol.

So the area 2-form on the cylinder is $a\,d\theta\wedge dz$, where the radius is $a$. It's not automatically just $\star dg$.

Unless $dg$ is a norm-$1$ $1$-form.

In general, if you have a hypersurface with unit normal $n$, then the "area" $(n-1)$-form is going to be $\sum (-1)^{i-1} n^i dx^1\wedge\dots\wedge\widehat{dx^i}\wedge\dots\wedge dx^n$.

Adam

23:40

They announced on the local news program on the television that online trolls can be convicted and sentenced to up to 5 years imprisonment here pretty scary stuff, I mean ill assume im allowed to keep cyber bullying famous puppets

if someone unimportant makes fun of someone declared to be important I doubt they will swat me and have me in max security for amusing myself on my own profiles

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