Mathematics - 2019-04-26

johnny09

00:08

how can we show that the standard basis is actually a basis for $\mathbb{R}^n$?

Thorgott

00:35

Show that it is a linearly independent generating set

johnny09

what do you mean by "generating set"? like this one en.wikipedia.org/wiki/Generating_set_of_a_group?

Thorgott

01:11

The linear span of the vectors is the entire space

Henning Makholm

Hmm, has the "hot meta posts" sidebar always been so ... yellow?

Subhasis Biswas

01:48

how do I get to display the latex in this chat properly?

Rithaniel

02:25

@SubhasisBiswas Follow the instructions here: tinyurl.com/cfqcvpc

Ultradark

0

Q: Riemann Zeta function with the prime counting function in place of $n$

Interested in the following function: $$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$ where $\pi(n)$ is the prime counting function. When $s=2$ the sum becomes the following: $$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...

I like how the asker (me) asked such a great question

Subhasis Biswas

02:54

wow, this is $nice$

Thank you rithaniel $\infty$

Rithaniel

$$n^o\;\;p^ro^bl^em$$

Subhasis Biswas

xD

Jonathan

03:49

hey everyone, i'm having a really hard time with a particular combinatorics problem

and any help would be really appreciated.

0

Q: # of possible binary string arrangements given constraint

Consider a random binary string where each bit can be set to 1 with probability $p$. Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...

probability combinatorics

It's a really simple problem,

but I can't get past what I've already figured out

Subhasis Biswas

What's simple to you is too hard for me.

what are the prerequisites for understanding this problem?

Jonathan

Being able to take a function and seeing how to pass in parameters to get desired value.

the function Z already provides answers.

Extracting the right one is the difficult part, by passing in x and y.

I'd be willing to offer a bounty as soon as I can,

if someone answers it.

Alex Clark

04:38

@HenningMakholm I don't think so, but mainly it looks larger(?)

Alessandro Codenotti

05:00

Uhm I got a notification about the "new mobile chat" with which you can edit and star messages, but I was always able to and nothing changed for me

Silent

05:13

The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.

Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?

user76284

05:36

9

Q: View of the sky from inside a black hole

Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$). Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...

general-relativity black-holes astrophysics curvature geodesics

Anyone know if there's a closed form for the integral at the bottom?

Alessandro Codenotti

06:11

@Silent you don't get a unique one

Take any trascendental extension of $\Bbb C$, such as $\Bbb C(x)$, and take its algebraic closure

Secret

06:36

@Ultradark That is a cool question and potentially important as well. However, cannot help with that since I suck at prime numbers

Tobias Kildetoft

07:04

@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.

Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.

ÍgjøgnumMeg

Morning all

Rithaniel

Morning, Meg

ÍgjøgnumMeg

How's it going?

Rithaniel

It's going okay. Finishing up my homework for the week before finals.

ÍgjøgnumMeg

Nice one :)

Rithaniel

07:15

How about you?

ÍgjøgnumMeg

just sitting at work preparing for the coming day

Rithaniel

Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).

Tobias Kildetoft

@Rithaniel all products are finite

Rithaniel

Really? Isn't there a concept of an infinite product, similar to an infinite sum?

Tobias Kildetoft

no, there are no infinite sums

those are called series and are a thing from calculus which should not be confused with sums

Rithaniel

07:25

My understanding that $\{\frac{1}{2^n}\}_{n\in\mathbb{N}_0}$ is an infinite series, while $\sum_{n=0}^{\infty}\frac{1}{2^n}$ is an infinite sum.

Tobias Kildetoft

no, the first is a sequence

the second is a series

Rithaniel

Ah, right.

Tobias Kildetoft

while it behaves much like a sum, there are tons of pitfalls if you think of it as one.

Rithaniel

So, what is the multiplicative counterpart to a series, then?

Tobias Kildetoft

Just think of how much structure you need to even make sense of it. And even then, not all series converge (i.e. "make sense")

it is (unfortunately) just called an infinite product

but once again, you need much more structure to make sense of it

and even then they might not all make sense

Rithaniel

07:27

Ah, okay, well that at least clarifies it for me.

Just unfortunately naming schemes, like homeomorphism versus homomorphism.

Tobias Kildetoft

It is actually a bit of a miracle that the structures we are most used to (i.e. the reals) have enough structure to make sense of all of these amazing things from calculus

and of course schemes being a mathematical thing makes that sentence even more true

Rithaniel

Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.

Tobias Kildetoft

the reals first, definitely. Or maybe even "reality" first, which the reals model quite well (in 1 dimension)

Rithaniel

True, though the ability to infinitely zoom in does somewhat find itself at odds with physics and the concept of fundamental particles.

Tobias Kildetoft

sure, but the mathematical study of the reals is older than knowledge about fundamental particles and quantum mechanics

And at most scales needed, it works great

Rithaniel

07:33

I still am kind of amazed that any two rationals can be arbitrarily close to one another yet there will always be more reals between them than there are rationals.

Also, indeed it does.

Alessandro Codenotti

08:11

@TobiasKildetoft ah, right, of course, uncountable algebraically closed fields are classified by characteristic and cardinality

Tobias Kildetoft

right

which is very weird

I mean, try to identify $\mathbb{C}(x)$ inside $\mathbb{C}$.

Alessandro Codenotti

Or as a model theorist would say, $\mathsf{ACF}_p$ is uncountably categorical for every $p$

Secret

28

Q: Intuitive explanation for how could there be "more" irrational numbers than rational?

I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...

elementary-set-theory cardinals

Rithaniel: I kinda find a way to visualise that with those processing plots I made last year (typing...)

Adam

09:21

i thought id try another forum, since the first one on the google search results didn't send me a confirmation email

Reety

10:20

Hey I'm a little confused how in fourier analysis the function sin((nπx)/L) are like the basis vectors in algebra, can anyone care to explain?

Rajesh Dachiraju

10:46

0

Q: How to solve this Matrix equation?

$\mathbf{U} = (\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})^{-1} \mathbf{X}^HL $ $\rho(\lambda) = \mathbf{U}^H\mathbf{U}$. Find $\lambda \in (0,\infty)$ such that $$\frac{\partial{\rho}}{\partial{\lambda}} = 0$$ $X$ is $n\times m$ semi-orthogonal matrix $F$ is $m\times ...

linear-algebra real-analysis matrix-equations

Secret

11:45

$\mathbf{U} = (\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})^{-1} \mathbf{X}^HL$

$\mathbf{U}^H\mathbf{U} = \mathbf{U}^H(\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})^{-1} \mathbf{X}^HL$

$\rho = \mathbf{U}^H(\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})^{-1} \mathbf{X}^HL$

$\frac{\partial\rho}{\partial \lambda} = \frac{\partial}{\partial \lambda}(\mathbf{U}^H(\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})^{-1} \mathbf{X}^HL) = 0$

$\frac{\partial\rho}{\partial \lambda} = \frac{\partial}{\partial \lambda}((\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})(\mathbf{U}^H)^{-1} \mathbf{X}^HL) = 0$

$\frac{\partial\rho}{\partial \lambda} = \frac{\partial}{\partial \lambda}(\mathbf{X}^H \mathbf{X}(\mathbf{U}^H)^{-1} \mathbf{X}^HL + \mathbf{I}(\mathbf{U}^H)^{-1} \mathbf{X}^HL + \mathbf{\lambda}\mathbf{F}(\mathbf{U}^H)^{-1} \mathbf{X}^HL) = 0$

$\frac{\partial\rho}{\partial \lambda} = \frac{\partial}{\partial \lambda}(\mathbf{X}^H \mathbf{X}(\mathbf{U}^H)^{-1} \mathbf{X}^HL) + \frac{\partial}{\partial \lambda} ((\mathbf{U}^H)^{-1} \mathbf{X}^HL) + \frac{\partial}{\partial \lambda}(\mathbf{\lambda}\mathbf{F}(\mathbf{U}^H)^{-1} \mathbf{X}^HL) = 0$

user193319

14:42

If $X$ is an $n$-dimensional simplicial complex and $A$ is a $n$-dimensional subcomplex, does it follow that $X = A$?

Leaky Nun

@user193319 $X=\{0, 1\}$, $n=0$, $A=\{0\}$

Fargle

No. Consider a triangle vs a triangle with a line attached to it.

Leaky Nun

sniped :P

Fargle

Or @Leaky's example works too.

user193319

Okay. Say $A$ is a subcomplex of $X$. Can we say that its dimension is at most $n$?

Silent

14:53

I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.

Please explain this comment!

Leaky Nun

15:05

@user193319 sure

@Silent if a power series converges for some $z_0$, then it converges for all $z$ with $|z|<z_0$

that's the whole reason why "radius of convergence" is a thing

Silent

Thank you

user193319

On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?

Leaky Nun

suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable

@Silent

Jonathan

Can anyone help me out with my combinatorics problem?

I'm having a really hard time progressing onit :/

Silent

15:23

@LeakyNun Thank you so much.

Leaky Nun

@user193319 it's right below

PP149-150

the paragraph starting with "The Mayer–Vietoris sequence is easy to derive from..."

user193319

Okay. I see definitions for $\varphi$ and $\psi$. But what about the capitalized versions?

Leaky Nun

that's just the induced maps from the chain maps

N. Maneesh

15:56

Let $g : [0,\frac{
1}
{2}
] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{
1}
{2}
] → \mathbb R$ by $g_1 = g$
and
$g_{n+1}(t) = \int_0^t
g_n(s) ds,$
for all $n ≥ 1.$ Show that
$lim_{n→∞}
n!g_n(t) = 0,$
for all $t ∈ [0,\frac{1}{2}]$
.

Can you give some hint?

My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$

If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.

$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\lim_{n\to \infty}(n+1)\frac{g_{n+1}(t)}{g_{n}(t)}$

user8718165

Hii @LeakyNun

Alessandro Codenotti

16:26

@Leaky is it obvious that nilpotent groups are not closed under ultraproducts?

Rudi_Birnbaum

Hi all!

I have a question

I have a bilinear functional that is bounded from below

I try to approximate the minimum by a ansatz-function that is a linear combination

of any independent functions of the proper function space

I now obtain an expression that is bilinear in the coeffcients

using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)

I get a set of $n$ equations with the $n$ the number of coefficients

a set of n linear homogeneus equations in the $n$ coefficients

Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists

This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.

Avoiding the neccessity to solve for the coefficients.

I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.

I wonder if there is something deeper in the background, or so to say a more very general principle.

Has anyone an idea what I mean?

Leaky Nun

18:04

@AlessandroCodenotti idk maybe take groups that need n steps to be nilpotent

Alessandro Codenotti

18:16

I don't see a simple way to write "takes n steps" in the language of groups though

tarit goswami

19:13

Can we construct any other curve like elliptic curve and define a group operation on the points on that curve to make a new Cryptographic scheme?

ÍgjøgnumMeg

abelian variety crypto

tarit goswami

19:27

@ÍgjøgnumMeg Do this follow the same type of group operation?

ÍgjøgnumMeg

erm an abelian variety is an algebraic group so .. group cryptography

otherwise I literally have no idea

Leaky Nun

20:16

@AlessandroCodenotti you don't need to write anything in any language when constructing ultraproduct right

Ultradark

$\Psi(s)= \sum_{n=1}^{\infty} \frac{\lambda_n}{n^s}= \sum_{n=2}^{\infty} \frac{1}{\pi(n)^s}$ Does the sequence $\lambda_n$ contain even numbers or $1$

Alessandro Codenotti

@LeakyNun no, but to prove that the ultraproduct isn't nilpotent I was thinking of invoking Los's theorem

Leaky Nun