Mathematics - 2019-03-10

12:31 AM

In mathematics, the universality of zeta-functions is the remarkable ability of the Riemann zeta-function and other, similar, functions, such as the Dirichlet L-functions, to approximate arbitrary non-vanishing holomorphic functions arbitrarily well. The universality of the Riemann zeta function was first proven by Sergei Mikhailovitch Voronin in 1975 and is sometimes known as Voronin's Universality Theorem. == Formal statement == A mathematically precise statement of universality for the Riemann zeta-function ζ(s) follows. Let U be a compact subset of the strip { ...

Secret

4:15 AM

@Ultradark The pattern that governs all scientific breakthrough is the most stubborn one that not even a probability distribution can fit it

Ultradark

4:50 AM

Is it possible to mash the critical strip to the unit square

probably fairly easy to do

$$ \zeta(\sigma+it) \to (0,1) \times (0,i) $$

Joe Shmo

5:07 AM

I have a question here that states that for Newton's method for finding the matrix inverse (iterating $X_{k+1} = X_k(2I - AX_k)$), and the error defined as $\epsilon_k = ||I - AX_k||$, it follows that $\epsilon_{k+1} = \epsilon_k^2$, but I can only show $\epsilon_{k+1} \le \epsilon_k^2$ (which is all that is required to establish a quadratic rate of convergence). I can't quite establish the equality, is it true that they are equal?

Iced Palmer

7:32 AM

Hi -- off topic --. If finding a basis to put an operator's matrix in Jordan form, how do I identify which generalized eigenvectors $v$ will generate the cycles of my basis?

Iced Palmer

9:53 AM

(related to my previous question) Given a formula for a linear operator $T$ and a Jordan canonical form for the matrix of $T$, how can I identify the basis that gives this Jordan matrix? Is there a more civilized approach than searching for cycles of generalized eigenvectors?

Rudi_Birnbaum

Hi guys!

There is this "transcendental number decoding" tool. You enter "3.14.." and it gives you π. Its a webtool from a French(?) mathematician. I can't find the page anymore .. Who knows what I mean?

(read: "transcendental" also "irrational")

something like irrational number decoding

got it again wayback.cecm.sfu.ca/projects/ISC/ISCmain.html

Cool things here btw: plouffe.fr/Simon%20Plouffe.htm

Ooker

10:27 AM

can anyone help me with this meta question?

-1

Q: Are questions about professional mathematician culture and language on-topic?

I want to ask this questions: What points mathematicians need in an explanation about red herring names? This question seems to be off-topic in Math Edu, since the site is more about teachers educating students, not researchers "educating" fellow researchers (naming conventions and culture). May...

Math geek

0

A: How does $\cos\alpha dA=dydz$ come?

Take $\alpha$ for instance. That is defined by the author to be the angle between the x axis and $\mathbf{n}$, so that $\cos\alpha = \mathbf{n}\cdot\mathbf{i}$, that is, the x component of the unit normal. Now $dA$ is the area spanned by two vectors $\mathbf{r_u}du$ and $\mathbf{r_v}dv$. What we ...

Impending Uncertainty

11:03 AM

hello

anybody here ?

user21820

@ImpendingUncertainty Yes, why?

Impending Uncertainty

I have a question

user21820

About...?

Impending Uncertainty

yeah , so i have a radial function ,radially decreasing function in Rn, how can I prove that I can approximate it with a finite linear combinations ( with positive coefficients) of indicator functions of balls centered at 0 ?

user21820

11:22 AM

← 6 messages moved from CRUDE

@ImpendingUncertainty Your question is too imprecise, but just approximate the radial function itself (whose parameter is just the radius), using a monotonic step function. I'll leave the rest to you.

Impending Uncertainty

I did that . Say f(x)=g(|x|) . I tried to approximate g(|x|) with a step function say $$ \sum_{i=1}^{N} a_i \chi_{(c_i,d_i)(|x|) $$ .

If I am not wrong then : $$ \chi_{(c_i,d_i)(|x|)=\chi_{B(0,d_i)}(x)-\chi_{B(0,c_i)} $$ so when I will plug it back to the sum it would give some negative stuff .

Perhaps there is something obvious I dont see

Desura

12:09 PM

Hey, I'm looking at Tic-Tac-Toe on a $n \times n$ board, $n \geq 5$. Assuming we have a pairing strategy on this board, we can show that there is a pairing strategy for the $(n+2) \times (n+2)$ board. I may be blind, but I don't see why a similar argument does not work for the $(n+1) \times (n+1)$ board. Anyone has an idea ? I'm readinf Tic-Tac-Toe Theory from Beck

user193319

12:24 PM

The maps $a,b,$ and $c$ are normal homomorphisms. The blue sequences are exact and represent maps that are given. The problem I am working on asks me to find the red sequences and show that they are exact.

I have found every map and showed that they are all well-defined---except for the map $d : \ker c \to Coker ~a$

I can't figure out how to show this map is well-defined....(give me a minute to type up what I have)

Let $x \in \ker c$. Since $g$ is surjective by exactness, there is some $y \in B$ such that $g(y) = x$. By commutativity, we have $g'(b(y)) = c(g(y)) = c(x) = 0$, so $b(y) \in \ker g' = f'(A')$, so by injectivity there is a unique $z \in A'$ such that $f'(z) = b(y)$. Define $d$ to be $d(x) = f'(z) a(A)$.

I am pretty certain that this is the right way to define $d$; I am just having a tremendous amount of trouble showing that it is well-defined. I could use some help.

MatheinBoulomenos

@user193319 this definition doesn't make sense

you want $d(x)=za(A)$

user193319

Oh, whoops you're right. That was a typo.

So, potential ambiguity arises when we choose $y \in B$ such that $g(y) = x$. To show well-definedness, I need suppose there are two elements $y_1,y_2 \in B$ such that $g(y_i) = x$. This in turn gives rise to $z_1,z_2$. I need to show that $z_1 a(A) = z_2 a(A)$. But this is turning out to be difficult.

MatheinBoulomenos

in this case, you have $g(y_1 y_2^{-1})=1$, so there is some $w \in A$ with $f(w)=y_1 y_2^{-1}$

Then $f'(a(w))=b(f(w))=b(y_1) b(y_2)^{-1}=f'(z_1) f'(z_2)^{-1}$, so by injectivity of $f'$, we get $a(w)=z_1 z_2^{-1}$

which implies $z_1 a(A) = z_2 a(A)$

user193319

12:40 PM

D'oh

Thanks!

Gaurang Tandon

2:09 PM

Hello, could anyone please look at this proof of L(V,W) being a vector space and explain why the associativity of three linear transformations T,U,V is expressed as $T+(U+V)=(T+U)+V$ instead of $T(UV)=(TU)V$? I'd expect "associativity" to imply that you can apply the linear transformations in any order on the given vector, so you can apply UV first and then T, or V first and then TU.

user193319

Because the vector space structure is addition, not multiplication (composition) of operators.

A vector space is an abelian group with respect to its "addition" operation; composition of linear operators is not commutative, so it cannot be the "addition" operation of any vector space structure.

Gaurang Tandon

oh

cool thanks! @user193319

user193319

No problem.

JamalS

3:24 PM

Are “sheaf cohomology” and “étale cohomology” equivalent terms?

loch

4:13 PM

@JamalS etale cohomology is a special case of sheaf cohomology, where you're looking at sheaves defined on the etale topology

Ultradark

4:25 PM

Is a^a always irrational? Where a is a rational number in (0,1)

Ultradark

4:44 PM

Nevermind

It's algebraic

user193319

5:47 PM

@MatheinBoulomenos Okay. I am trying to show exactness at $\ker c$, and I am running into some trouble. Showing exactness at $\ker c$ is equivalent to showing that $g(\ker b ) = \ker d$. Let $x \in g(\ker b)$. Then $x = g(y)$ for some $y \in \ker b$. To calculate $d(x)$, choose $b \in B$ such that $g(b) = x$. Then there is a unique $a' \in A'$ such that $f'(a') = b(y) = 1$, so $a' =1$ by injectivity. Hence $d(x) = a' a(A)=a(A)$, so $x \in \ker d$.

What worries me is that I was able to conclude a' = 1$; that seems too strong of a conclusion. However, I can't spot the error. (by the way, I'm having trouble with the other set inclusion, but let's put that to the side for a moment)

Here's the picture again for reference: i.stack.imgur.com/wddl3.png

MatheinBoulomenos

6:05 PM

@user193319 there is no error

user193319

Really? Hmm...That's surprising. Here's my attempt for the other set inclusion...

MatheinBoulomenos

it's not so surprising because you made a specific choice

you don't show that $a'=1$ fo every possible choice of $b \in B$ such that $g(b)=x=g(y)$, you only showed it when you use $y$ which is special because it's in $\mathrm{ker}(b)$

user193319

Let $x \in \ker d$. Then $d(x) = a(A)$. Choose $y \in B$ such that $g(y) = x$. Then there is a unique $a' \in A'$ such that $f'(a') = b(y)$ and $d(x) = a' a(A)$. This means $a' a(A) = a(A)$, so $a' = a(z)$ for some $z \in A$. Then $f'(a') = b(y)$ becomes $f'(a(z))=b(y)$ and commutativity of the diagram means $b(f(z)) = b(y)$ which implies $b(f(z)y^{-1}) = 1$ which implies $f(z)y^{-1} = w \in \ker b$ or $y = w^{-1}f(z)$. Hence $x = g(y) = g(w^{-1}f(z))$...

The only thing I don't see is why $w^{-1}f(z)$ is in $\ker b$.

Ultradark

6:23 PM

@MatheinBoulomenos does it make sense to define a finite field of rational numbers?

Tobias Kildetoft

@Ultradark no

Ultradark

Why not?

Tobias Kildetoft

because there is no such thing

Ultradark

And the reason is because It would not be closed?

Tobias Kildetoft

that is a fairly meaningless statement on its own.

Dair

6:34 PM

I guess you could have $0, 1 \in \mathbb{Q}$ as opposed to $0, 1 \in \mathbb{Z}$ (and thus different underlying representation) and define $+$ and $\times$ to be isomorphic to $\mathbb{Z_2}$. But I suppose that is against the spirit of the question, unless I am misunderstanding.

Tobias Kildetoft

@Dair those would no longer be rational numbers

Dair

@Tobias Yeah, you're right. nvm. my bad,

since the representation is only "meaningful" when coupled with the appropriate operations...

MatheinBoulomenos

@user193319 so you have shown that $yf(z)^{-1} \in \mathrm{ker}(b)$, this means you're basically done

$g(yf(z)^{-1})=g(y)=x$

you want to show that $y \in \mathrm{ker}(b)$, I don't think that's true in general

user193319

6:53 PM

Dang it...so simple...Thanks for your help.

user193319

7:21 PM

Let $G$ be a non-abelian group of order $p^3$. By a certain theorem, we know there is a subgroup $H$ of order $p^2$. Because $H$ has order $p^2$, it must be abelian. The center $Z(G)$ must have order $p$, for if it had order $p^2$, then $G/Z(G)$ would be cyclic, which would contradict the fact that $G$ is non-abelian. This means that $Z(G) = \langle z \rangle$ for some $z \in G$....

I want to show that $G$ contains an element of order $p$ which is not in $H$. My thought was that $z$ can't be in $H$, but i am not certain of this.

If $z \in H$, then $Z(G) \le H$. Since $H$ is commutative, I thought this might mean $Z(G) = H$ (which would be a contradiction); but I don't think this is necessarily true.

Tobias Kildetoft

7:49 PM

@user193319 The claim is not true unless you assume that $p$ is odd.

user193319

Whoops...I forgot to mention that.

Tobias Kildetoft

$H$ could definitely contain $z$

But note that you can assume that $H$ is cyclic, for a start

user193319

Why is that?

Tobias Kildetoft

because otherwise $H$ would have some extra elements of order $p$ that would work.

user193319

would work? What do you mean by that?

Tobias Kildetoft

7:53 PM

Woops, I missed the requirement on the element

I was thinking it had to not be central

So if all elements not in $H$ have order $p^2$, what happens?

user193319

Then the claim would be true.

Hmm...maybe not...

Tobias Kildetoft

So how many subgroups of index $p$ would this give?

user193319

If $y$ is an element of order $p^2$ not in $H$, the $\langle y \rangle$ contains an element of order $p$; but that element might be in $H$.

How many subgroups? I'm not sure. Does this relate to Sylow's theorem? I can't use that theorem for this problem.

MatheinBoulomenos

$H$ always contains $z$ in fact

Tobias Kildetoft

I was mainly thinking out loud

(and trying to not use the classification of groups of order $p^3$).

user193319

8:02 PM

Ha...Classifying noncommutative groups of order $p^3$ is the problem I am working on.

According to my book there are only two. See page 48 jmilne.org/math/CourseNotes/GT.pdf

Tobias Kildetoft

that is correct

The proof I recall starts with classifying $p$-groups with a cyclic subgroup of index $p$

user193319

I was going on the hint in the back of the book whose link I just posted; but I am open to simpler proofs.

MatheinBoulomenos

can you use group cohomology?

user193319

Haha, probably not.

Joe Shmo

8:28 PM

is anyone well-versed in numerical techniques?

BAYMAX

9:01 PM

any help in proceeding ?

$K(A) = ||A|| ||A^{-1}||$

I am thinking how to implement this?

to obtain the inequality, l ike I am thinking to consider $(A - \hat{A}) x = e$

Spencer1O1

For $x_1,x_2,y_1,y_2,z \in \mathbb{N}$ and $x_1 \neq x_2 \neq y_1 \neq y_2 \neq z$, is $x_1^2 + y_1^2 = z$ and $x_2^2 + y_2^2 = z$ possible?

I have been trying to figure it out.

I just need this piece of information for a proof, but I am having a hard time disproving that it is possible.

BAYMAX

yup

take $x_{1} = 3, x_{2} =4$

and

$y_{1} = 4, y_{2} =3$

$z = 25$

Spencer1O1

But $x_1 \neq y_2$

BAYMAX

hmm

Spencer1O1

And $x_2 \neq y_1$

My intuition is that it isn't possible.

BAYMAX

9:14 PM

sems like possible as i a m thinking that the $(x_{1},y_{1})$, $(x_{2},y_{2})$ could be points on circle

but they also need to be natural numbers

Spencer1O1

If you square a really large power of 2 and square a small power of 2, and add them, other squares of powers of 2 can't add up to it.

The problem that I am trying to solve has to do with spheres, but if you can disprove it with circles than it disproves it with spheres

So in the problem $(x_1,y_1)$ and $(x_2,y_2)$ are points on a circle.

BAYMAX

feels it has connection with number theory let me think

Spencer1O1

It is a number theory problem that I am trying to solve, but I don't know anything that is related to it.

BAYMAX

$33^2 + 4^2 = 1105$

$32^2 + 9^2 = 1105$

Spencer1O1

Cool!

But is it possible to add $x_3$ and $y_3$???

That is the question now...

BAYMAX

9:31 PM

yo

$31^2 + 12^2$

you could do a bit of coding to check it out for other numbers

Ultradark

$\pi \sum_{s=2}^{\infty} \frac{1}{2^s}$

Spencer1O1

Cool, I will

BAYMAX

@Ultradark $\frac{\pi}{2}$, making some statement here or checking for the latex to work :) ?

Spencer1O1

@Ultradark it is $\pi/2$

Ultradark

that's the answer?

Spencer1O1

9:35 PM

yes

Ultradark

how did you get that so fast?

BAYMAX

observe that the terms of the summation follows geometric progression

Spencer1O1

Because the sum of $1/2 + 1/4 + 1/8... = 1$ so $1/4 + 1/8...$ is $1/2$. Then multiply by pi

Ultradark

oh yeah i forgot

thanks

Spencer1O1

@BAYMAX I don't really want to brute force my problem with a computer program, and the reason why is because it would take forever. Here is the whole problem. $a,b,c,d,f,g,h,k,l,m \in \mathbb{N}$ and $a^2+b^2+c^2=m$, $d^2+f^2+g^2=m$, $h^2+k^2+l^2=m$, $a^2+d^2+h^2=m$, $b^2+f^2+k^2=m$, $c^2+g^2+l^2=m$, $a^2+f^2+l^2=m$, and $c^2+f^2+h^2=m$

BAYMAX

9:46 PM

why i see that the last 4 equations kind of violate your conditions

Spencer1O1

It is a 3x3 grid of squares, and the sum of the rows, columns, and diagonals is the same number.

Dair

@Spencer1O1 openproblemgarden.org/op/magic_square_of_squares

BAYMAX

hhe

:)

if ny1 solves it consider me givng mea share!!

Spencer1O1

yes

I would

BAYMAX

:)

Spencer1O1

9:57 PM

50%

BAYMAX

@Dair any help on this?

Dair

@BAYMAX Sorry, I've got nothing.

BAYMAX

no problem

Spencer1O1

I have been working on the magic square of squares for a year...........

Ultradark

10:35 PM

I asked a combinatorics question

I've been asking it for a year

The question asks to find the number of intersections

for example if you take a square grid

and copy it and rotate the copy by $\pi/4$

you get more intersections than the original $n\times n$ grid

but how many more?

is it just $(n^2)^2=n^4$?

PLZHELP

anyone on?

Ultradark

IamonIamonIamon

PLZHELP

I have a very beginner question posted here math.stackexchange.com/questions/3142781/…

can you solve a matrix in 2 directions?

Ultradark

0

Q: Counting the points of intersection

A set of equations is given as follows: $$ x^{\Omega}+y^{\Omega}=1 $$ $$ (1-x)^{\Omega}+y^{\Omega}=1 $$ $$ x^\Omega+(1-y)^\Omega=1 $$ $$ (1-x)^\Omega+(1-y)^\Omega=1. $$ $(\Omega\subset\Bbb Q) =\{1.1,1.2,1.3,...,N\},$ where $N\in\Bbb N$. I was able to calculate the cardinality of $\Omega:$ ...

combinatorics permutations combinations computability

let me know if anything is unclear

@PLZHELP Sorry I don't know how to help you

Joe Shmo

Let $(\lambda, v)$ be an eigen-pair of $A$. Suppose I found an approximate eigenvector to $A$, call it $\hat{v}$, such that $||\hat{v} - v|| = \delta$. suppose the approximate eigenvalue corresponding to $\hat{v}$ is $\hat{\lambda}$. I need to show $(\hat{\lambda} - \lambda) = O(\delta^2)$, but not really sure how

That is, $\hat{\lambda} = \dfrac{\hat{v}^*Av}{\hat{v}^*v}$. Further suppose $A$ is Hermitian.

Seth

11:31 PM

Can someone help me understand how $ B\overline{AB} + \bar A \cdot \overline{A+B} $ can be simplified to $ \bar A $? Seems simple but I'm not figuring it out. (boolean algebra)

I can do:
$ B(\bar A + \bar B) + \bar A \cdot \bar A \cdot \bar B $
$ B\bar A + B\bar B + \bar A \cdot \bar A \cdot \bar B $
$ B\bar A + \bar A \cdot \bar B $
$ B \cdot \bar A \cdot \bar B = \bar A $
But I'm pretty sure I violate order of operations to get from the 2nd last step to the final step, no?

user193319

11:48 PM

@MatheinBoulomenos or @TobiasKildetoft Any tips on how to solve the problem of characterizing all noncommutative groups of order $p^3$? My book gives the following hint: "...there is a normal subgroup $N$ of order $p^2$, which is commutative. Now show that $G$ has an element $c$ of order $p$ not in $N$, and deduce that $G = N \rtimes \langle c \rangle$, etc."

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$Mathematics$ Mathematics