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12:08 AM
how can we show that the standard basis is actually a basis for $\mathbb{R}^n$?
 
12:35 AM
Show that it is a linearly independent generating set
 
what do you mean by "generating set"? like this one en.wikipedia.org/wiki/Generating_set_of_a_group?
 
1:11 AM
The linear span of the vectors is the entire space
 
Hmm, has the "hot meta posts" sidebar always been so ... yellow?
 
1:48 AM
how do I get to display the latex in this chat properly?
 
2:25 AM
@SubhasisBiswas Follow the instructions here: tinyurl.com/cfqcvpc
 
0
Q: Riemann Zeta function with the prime counting function in place of $n$

UltradarkInterested 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
 
2:54 AM
wow, this is $nice$
Thank you rithaniel $\infty$
 
$$n^o\;\;p^ro^bl^em$$
 
3:49 AM
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

JonathanConsider 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...

It's a really simple problem,
but I can't get past what I've already figured out
 
What's simple to you is too hard for me.
what are the prerequisites for understanding this problem?
 
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.
 
4:38 AM
@HenningMakholm I don't think so, but mainly it looks larger(?)
 
5:00 AM
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
 
5:13 AM
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?
 
5:36 AM
9
Q: View of the sky from inside a black hole

user76284Consider 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...

Anyone know if there's a closed form for the integral at the bottom?
 
6:11 AM
@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
 
6:36 AM
@Ultradark That is a cool question and potentially important as well. However, cannot help with that since I suck at prime numbers
 
7:04 AM
@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.
 
Morning all
 
Morning, Meg
 
How's it going?
 
It's going okay. Finishing up my homework for the week before finals.
 
Nice one :)
 
7:15 AM
How about you?
 
just sitting at work preparing for the coming day
 
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).
 
@Rithaniel all products are finite
 
Really? Isn't there a concept of an infinite product, similar to an infinite sum?
 
no, there are no infinite sums
those are called series and are a thing from calculus which should not be confused with sums
 
7:25 AM
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.
 
no, the first is a sequence
the second is a series
 
Ah, right.
 
while it behaves much like a sum, there are tons of pitfalls if you think of it as one.
 
So, what is the multiplicative counterpart to a series, then?
 
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
 
7:27 AM
Ah, okay, well that at least clarifies it for me.
Just unfortunately naming schemes, like homeomorphism versus homomorphism.
 
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
 
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.
 
the reals first, definitely. Or maybe even "reality" first, which the reals model quite well (in 1 dimension)
 
True, though the ability to infinitely zoom in does somewhat find itself at odds with physics and the concept of fundamental particles.
 
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
 
7:33 AM
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.
 
8:11 AM
@TobiasKildetoft ah, right, of course, uncountable algebraically closed fields are classified by characteristic and cardinality
 
right
which is very weird
I mean, try to identify $\mathbb{C}(x)$ inside $\mathbb{C}$.
 
Or as a model theorist would say, $\mathsf{ACF}_p$ is uncountably categorical for every $p$
 
28
Q: Intuitive explanation for how could there be "more" irrational numbers than rational?

OviI'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, ...

Rithaniel: I kinda find a way to visualise that with those processing plots I made last year (typing...)
 
 
1 hour later…
9:21 AM
i thought id try another forum, since the first one on the google search results didn't send me a confirmation email
 
10:20 AM
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?
 
10:46 AM
0
Q: How to solve this Matrix equation?

Rajesh Dachiraju$\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 ...

 
11:45 AM
$\mathbf{U} = (\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})^{-1} \mathbf{X}^HL$
$\mathbf{U}^H\mathbf{U} = \mathbf{U}^H(\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})^{-1} \mathbf{X}^HL$
$\rho = \mathbf{U}^H(\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})^{-1} \mathbf{X}^HL$
$\frac{\partial\rho}{\partial \lambda} = \frac{\partial}{\partial \lambda}(\mathbf{U}^H(\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})^{-1} \mathbf{X}^HL) = 0$
$\frac{\partial\rho}{\partial \lambda} = \frac{\partial}{\partial \lambda}((\mathbf{X}^H \mathbf{X} + \mathbf{I} + \mathbf{\lambda}\mathbf{F})(\mathbf{U}^H)^{-1} \mathbf{X}^HL) = 0$
$\frac{\partial\rho}{\partial \lambda} = \frac{\partial}{\partial \lambda}(\mathbf{X}^H \mathbf{X}(\mathbf{U}^H)^{-1} \mathbf{X}^HL + \mathbf{I}(\mathbf{U}^H)^{-1} \mathbf{X}^HL + \mathbf{\lambda}\mathbf{F}(\mathbf{U}^H)^{-1} \mathbf{X}^HL) = 0$
$\frac{\partial\rho}{\partial \lambda} = \frac{\partial}{\partial \lambda}(\mathbf{X}^H \mathbf{X}(\mathbf{U}^H)^{-1} \mathbf{X}^HL) + \frac{\partial}{\partial \lambda} ((\mathbf{U}^H)^{-1} \mathbf{X}^HL) + \frac{\partial}{\partial \lambda}(\mathbf{\lambda}\mathbf{F}(\mathbf{U}^H)^{-1} \mathbf{X}^HL) = 0$
 
 
3 hours later…
2:42 PM
If $X$ is an $n$-dimensional simplicial complex and $A$ is a $n$-dimensional subcomplex, does it follow that $X = A$?
 
@user193319 $X=\{0, 1\}$, $n=0$, $A=\{0\}$
 
No. Consider a triangle vs a triangle with a line attached to it.
 
sniped :P
 
Or @Leaky's example works too.
 
Okay. Say $A$ is a subcomplex of $X$. Can we say that its dimension is at most $n$?
 
2:53 PM
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!
 
3:05 PM
@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
 
Thank you
 
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?
 
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
 
Can anyone help me out with my combinatorics problem?
I'm having a really hard time progressing onit :/
 
3:23 PM
@LeakyNun Thank you so much.
 
@user193319 it's right below
PP149-150
the paragraph starting with "The Mayer–Vietoris sequence is easy to derive from..."
 
Okay. I see definitions for $\varphi$ and $\psi$. But what about the capitalized versions?
 
that's just the induced maps from the chain maps
 
3:56 PM
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.
$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\lim_{n\to \infty}(n+1)\frac{g_{n+1}(t)}{g_{n}(t)}$
 
Hii @LeakyNun
 
4:26 PM
@Leaky is it obvious that nilpotent groups are not closed under ultraproducts?
 
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?
 
 
1 hour later…
6:04 PM
@AlessandroCodenotti idk maybe take groups that need n steps to be nilpotent
 
6:16 PM
I don't see a simple way to write "takes n steps" in the language of groups though
 
7:13 PM
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?
 
abelian variety crypto
 
7:27 PM
@ÍgjøgnumMeg Do this follow the same type of group operation?
 
erm an abelian variety is an algebraic group so .. group cryptography
otherwise I literally have no idea
 
8:16 PM
@AlessandroCodenotti you don't need to write anything in any language when constructing ultraproduct right
 
$\Psi(s)= \sum_{n=1}^{\infty} \frac{\lambda_n}{n^s}= \sum_{n=2}^{\infty} \frac{1}{\pi(n)^s}$ Does the sequence $\lambda_n$ contain even numbers or $1$
 
@LeakyNun no, but to prove that the ultraproduct isn't nilpotent I was thinking of invoking Los's theorem
 
8:43 PM
oh right
 
9:06 PM
@Secret is there a way to compute the arithmetic progression $\lambda_n$ with a computer program?
 
 
1 hour later…
10:06 PM
@ÍgjøgnumMeg @LeakyNun hi guys! Any idea on my (diffuse) question?
 
10:32 PM
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
can anyone prove this statement?
 
10:56 PM
Primality doesn't look like it's relevant here---this should hold for any odd number $x$ and its digit reverse $y$, provided its digit reverse is also odd.
 
my whole life is a lie: I thought the Fourier series of a continuous periodic function converges pointwise everywhere
 
For example, 537 and 735 are obviously not prime, but 636---the midpoint between them---is a palindrome, and its digit sum is the same as that of 537.
Indeed, if both $x$ and $y$ are even, this also works. But it fails for numbers where one of $x,y$ is even and the other is odd, because then their average isn't an integer.
 
@LeakyNun counterexample?
 
@Semiclassical there's no easy one... I'm still searching the literatures
I've been told that here is an example
now $\displaystyle \sum_{n=1}^{\infty}\frac{\sin\left(2^{n^3}x\right)}{n^2}$ seems like an example as well
I mean isn't the above series itself a Fourier series...
 
11:20 PM
Is the sum of two or more parabolas still a parabola?
I guess so
$x^2 + 3x^2 = (4x^2)$
 
@LeakyNun why not? Seems like it’s straightforwardly a Fourier series for a function periodic on the interval [0,pi]
It’s a lacunary Fourier series tho
 
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall
stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert.
Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
 
$x^2-x^2=0$ and the latter is usually not considered a parabola as far as I know
 
@Semiclassical ^
 
Can’t help but think of Hermite’s line
“I turn with terror and horror from this lamentable scourge of functions with no derivatives.”
 
11:29 PM
Wow, it was @Fargle. Howdy, @Semiclassic, @Leaky, @Thorgott.
 
Hi @ted
 
Is $f(y) = \sum_i y_i^2$ a parabola?
 
@TedShifrin hey... you... must know this... oder
 
The graph of that function is a paraboloid in high dimensions, @nbro.
@Leaky: Aber ich weiß gar nichts.
 
@TedShifrin So it has just one optimum?
 
11:30 PM
Right, the origin.
 
Saw a phrase a few days ago that I hadn’t before : the Weil-Petersson metric
 
@TedShifrin sad
 
@Semiclassic: I berember that vaguely from years ago.
 
@TedShifrin Thanks!
 
@nbro so long as one isn’t doing something silly like f(x,y,z)=x^2+y^2
 
11:32 PM
@Semiclassical In that case, it might not be a paraboloid anymore
 
Right
 
@Semiclassic: That doesn't fit the formula given.
 
could you tell me what the series is used for?
 
@nbro: That would be a parabolic cylinder, of course.
 
the one with $\sum \sin(2^{n^3}x)/n^2$
 
11:33 PM
Creating monsters @LeakyNun
 
Of course I can't tell you that @Leaky. Weird counterexamples are rarely useful other than as oddities. :P
 
I mean, what did they say regarding that series
 
Does anyone ever "use" the Weierstrass nowhere-differentiable continuous function?
 
As I said, tho, that’s an example of a lacunary Fourier series, and lacunary series are weird
 
also, I mean, for my whole life... I believed that continuous functions converge to their own Fourier series...
 
11:35 PM
How does $f(x) = (x - c)^2$ change w.r.t $f(x) = x^2$, where c is a constant?
 
Continuous functions on a finite interval or periodic continuous functions on all of $\Bbb R$?
 
It is a shifted version
But how?
 
You can figure that out, @nbro.
 
Yeah
 
The key, I think, is that the fact that the series is lacunary means that the Fourier coefficients decay very slowly
 
11:35 PM
I think it is shifted to the left
 
@TedShifrin periodic continuous function
 
Depends whether $c>0$ or $c<0$ ?
 
$f:[-\pi,\pi]\to \Bbb R$, $f(-\pi)=f(\pi)$
 
You mean pointwise convergence? Of course not :) You need differentiability hypotheses for that.
You only get $L^2$ convergence.
 
And iirc statements about convergence of Fourier series typically hinge on the rate at which the Fourier coefficients decay?
 
11:37 PM
Which is based on $C^k$ for various $k$.
 
@TedShifrin but do you have a counter-example?
13 mins ago, by Leaky Nun
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
I hope he had cited the example... lol
 
I don't remember if I know an explicit example. I can look back at the notes when I taught this, but that was 30+ years ago.
I just reordered Polya-Szegö. It might have an example, too.
 
who should I ask?
 
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!!
He refers to Zygmund for the proof.
 
The wiki page on lacunary Fourier series references theorems of Kolmogorov and Zygmund: en.m.wikipedia.org/wiki/Lacunary_function
 

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