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12:22 AM
What’s the order type of the set of all integer polynomials, where $f \leq g \leftrightarrow \exists x \in \mathbb{Z} : \forall y \in \mathbb{Z} : f(y) \leq g(y)$?
Are the set of integer polynomials with non-negative coefficients order-isomorphic to the set of ordinals below $\omega^\omega$?
 
12:52 AM
@Ultradark Darboux theorem
 
1:03 AM
Also, what’s the relation between addition and multiplication of these polynomials and Hessenberg addition and multiplication of their corresponding ordinals?
 
1:20 AM
5
Q: Orthogonal complement to subspace of $n \times n$ matrices with trace = $0$

sawgholFor an inner product space $W = \{A \in M_{n \times n}(\mathbb{R}) \mid tr(A)=0\}$, with inner product $\langle A,B\rangle:=tr(A^TB)=\sum_{i=1}^n\sum_{j=1}^na_{ij}b_{ij}$, find $W^{\perp}$. I know that: $dim(W) = n^2 -1$, and that $W \oplus W^{\perp} = M$ But I'm not sure of how to proce...

How did sawghol in his answer conclude that $C=I$?
 
 
2 hours later…
3:24 AM
@N.Maneesh it's easy to see this directly
you can just write out a basis here
 
 
1 hour later…
4:26 AM
1
Q: Can you work as a math lecturer for free?

Shuheng ZhengIf someone has a PhD in some math or math related area (say CS theory or electrical engineering), is it possible for this person to teach an undergrad course, like say Intro Complex Analysis, at an university without asking for re-enumeration? The motivation for not asking for re-enumeration is t...

 
Does anyone know what functions of the form $x^{3x^2+x^x}+5x^{1+x}+x^2$ are called?
 
complicated :P
/s
variables with polynomial exponents?
 
Polynomial towers maybe
12
Q: Every non-increasing sequence of polynomial towers stabilizes -- Finitary proof

Vladimir ReshetnikovIn this question we are concerned only with positive integers $\mathbb N$ and other finitary objects that can be encoded using integers. A term function means a total computable function $\mathbb N^n\to\mathbb N$ with one or multiple arguments. Letters $m,n,k$ range over $\mathbb N$. Letters $f,...

 
5:25 AM
Can someone give hint on this one, only a pointer: to show that if $d| r$ then $2^d - 1 | 2^r - 1$. Here $a | b$ means $b$ is divisible by $a$
 
@jeea hello!
1
Q: Prove that if $d$ divides $n$, then $2^d -1$ divides $2^n -1$

flamingohatsProve that if $d$ divides $n$, then $2^d -1$ divides $2^n -1$. Use the identity $x^k -1 = (x-1)*(x^{k-1} + x^{k-2} + \cdots + x +1)$

Just google something similar to your question and you will find this ^^^^^^
 
ohh yes thanks, @manooooh
 
 
1 hour later…
6:38 AM
@Ultradark I don't know French.
Anyway, I have a very general question. I'm looking for some examples in math where suppose something (anything) that depends on a parameter $n$ has different behaviors/values/limits/etc. when $n$ is a large but finite number, and when $n \to \infty$.
Any cute examples?
Some behavior/property/etc. (anything) changes in a non-trivial manner when the limit $n \to \infty$ is taken.
I realize it's a vague question.
 
7:17 AM
Morning all
 
7:49 AM
@Avantgarde It's probably hard to do this in the way that you'd want, because limiting behavior/asymptotic behavior pretty much exactly deal with "n large enough". I guess a cheeky example might be $(-1)^n$---for all finite $n$, this is meaningful, but the limit doesn't exist.
 
8:40 AM
@UserX H is of order 6. Z4 x Z6/H is of order 24/6 = 4, no?
@Avantgarde There are notions of scaled limits in probability, where you can view continuous random variables as limit (in distribution) of the form $\lim_{n \to \infty} n^{-1/\alpha} X_n$ where $\alpha$ is some real constant and $X_n$ are discrete distributions.
As a simple example, consider $X_n \sim \text{Geo}(\lambda/n)$ distributed geometrically, so $\Bbb P(X_n = m)$ is the probability that a coin with success probability $\lambda/n$ returns head after $m$ tosses.
Then $X_n/n$ converges in distribution to the exponential distribution $\text{Exp}(\lambda)$ which is a continuous distribution, very different as a random variable than the $X_n$'s.
One way to interpret it as "you have a coin with very low success probability, of order 1/n, but since that'd take very long time to return a head usually, of order n, you scale it by n"
Then there's the classical example of the central limit theorem, where no matter whatever random variable $X_0$ you start with, as long as it has finite second moment, taking iid copies $X_1, X_2, X_3, \cdots \sim X_0$ and taking the scaled limit $\lim_{n \to \infty} n^{-1/2}(X_1 + \cdots + X_n)$ one always ends up (in dsitrbution) at a normal distribution.
This is a sort of "second order ergodic theorem", which sort of says that this scaled mixing process converges to the normal distribution (in an appropriate space of distributions) regardless of the initial condition
As another example, random walks on $\Bbb Z$ are mathematically very easy to describe as just the discrete-time process $(S_n)_{n \geq 1}$ where $S_n = \sum_{1 \leq i \leq n} X_i$, $X_i$ being iid Bernoulli variables, taking $1$ or $-1$ with probability $1/2$
One can obtained Brownian motions on $\Bbb R$, which are usually complicated to describe properly, as scaled limits of random walks on $\Bbb Z$. Namely, take $B_t = \lim_{n \to \infty} n^{-1/2} S_{\lfloor nt \rfloor}$ (this exists pointwise for $t$ by CLT). Then $(B_t)_{t \in [0, 1]}$ is in fact a Brownian motion on $[0, 1]$.
 
 
1 hour later…
10:07 AM
One of the most useful thing about matrix is how it parallelise a lot of things:
For example, this matrix is actually a brute force approach on solving the following operator equation:
$$133540 (a) 8740 (b) 12600 (c) 7600 = 143860 (d) 8740 (e) 12600 (f) 7600$$
This is useful when you have some assignment where you knew the numbers involved, and yet tried so hard with the theory and unable to get the solution
3
Q: Solving equations where the solution is an operator

Luka HorvatOk, so here's some context. Solving regular equations we might have something like this: $2 + x = 5$, solving for $x$ we get 3. We might even have an equation like $x + y = 5$ where there are multiple solutions. But what's in common with all these equations is that the process, or the algorith...

These equations are part of a more general class of functional equations. In the context of Banach spaces, where the unknown operators are bounded operators, things can get very complicated:
 
10:24 AM
(actually forget what I said about bounded operators)
3
Q: How to solve differential equation with operator as coefficient?

Chetan WaghelaI have been working on a problem in Quantum Mechanics and I have encountered a equation as given below. $$\frac{d\hat A(t)}{dt} = \hat F(t)\hat A(t)$$ Where ^ denotes it is an operator How will this differential equation be solved? Will the usual rules for linear homogeneous first order diffe...

 
 
1 hour later…
11:43 AM
well Homological algebra is a bit intense
 
12:16 PM
Morning, math chat
 
Hullo
 
Hello
Suppose X is a convergent square matrix with real elements, Xt is its transpose, and M is some other real matrix with same dimensions. Does a sum from i = 1 to infinity where the general term is of the form X^i M Xt^i have a name or might there be some related results? I think this sum should converge for arbitrary real M, any chance the result could be expressed in closed form in terms of M?
 
12:40 PM
Let $\Gamma$ be some group, and let $\ell^2(\Gamma) := \{f : \Gamma \to \Bbb{C} \mid \sum_{\gamma \in \Gamma} |f(\gamma)|^2 < \infty\}$. Given $\gamma \in \Gamma$, define $u_\gamma : \ell^2(\Gamma) \to \ell^2(\Gamma)$ by $u_\gamma(f)(\eta) := f(\gamma^{-1} \eta)$. I am reading an expository paper in which in claims that $u_\gamma$ is a unitary operator and that $\gamma \mapsto u_\gamma$ defines a unitary representation (called the left regular representation).
However, this seems to be a mistake. Shouldn't $u_\gamma$ actually be $u_\gamma (f)(\eta) := f(\gamma \eta)$?
Defined the former way, I wasn't able to show that $\gamma \mapsto u_\gamma$ is a homomorphism.
 
12:56 PM
Nevermind. I'll ask on main.
0
Q: Left Regular Representation of a Group

user193319Let $\Gamma$ be some group, and let $$\ell^2(\Gamma) := \{f : \Gamma \to \Bbb{C} \mid \sum_{\gamma \in \Gamma} |f(\gamma)|^2 < \infty\}.$$ Given $\gamma \in \Gamma$, define $u_\gamma : \ell^2(\Gamma) \to \ell^2(\Gamma)$ by $u_\gamma(f)(\eta) := f(\gamma^{-1} \eta)$. I am reading an expository pap...

 
 
2 hours later…
2:39 PM
@user193319 How does it make any difference? Their $u_\gamma$ would be your $u_{\gamma^{-1}}$
 
Can we say that a function, which is continuous at all points in which it exists, is continuous even thought it doesn't exist at all points in the real line?
Like the square root function
 
3:03 PM
@Simone Let me break it down for you.
Take a function $f$. In the definition of $f$, there are two sets: the domain and the codomain.
Sometimes we write it as $f\colon A\to B$, where $A$ is the domain, and $B$ is the codomain.
A function $f\colon A\to B$ is continuous if it is continuous at x for all x in A.
The square root function is continuous on its domain $[0,\infty)$.
But it is not defined elsewhere.
So to answer your question, for a function from R to R to be continuous, it must be both defined AND continuous at each x in R. However, you can have functions from a subset of R to R which are continuous, defined similarly as I mentioned above.
Not every continuous function is a function R-->R
Hopefully that settles it for you.
 
If I have two commuting, selfadjoint operators, is <x,Ax><x,Bx>=<x,ABx>?
 
@JannikPitt no, take both to be the identity
 
maybe if x is a unit vector?
that would be the case I need
 
3:20 PM
I think the case of A, B being projection operators onto disjoint subspaces is going to be a problem?
Since then AB annihilates everything but the inner products on the left need not vanish
 
@Semiclassical not disjoint, you mean with intersection {0}
yeah there's definitely an issue
 
@anakhro so I'd say it is true that the square root function is continuous.
Correct?
 
@Simone if you want to be really careful then say continuous on its domain
so there's no possible confusion
 
That's what I'd say aswell, but that's not what I was asked XD
 
3:40 PM
@RyanUnger ya, I was being careless
 
4:01 PM
Does anyone know what techniques might be useful for dealing with/simplifying a recursively-defined coordinate replacement with an integral? Specifically, if it helps, f(x)=x + integral(sin(f(x)/f(x)^2) from 1 to x, where f(x) k=0 is f(x)=x? Using Weierstrass t-replacement I end up with f(x)=f(x), which is correct but not helpful. And I don't know what to do with the endless stream of gamma functions I get when I do a definite integral and induction.
 
4:39 PM
@Simone Define "the square root function", please.
 
@anakhro $f(x)=\sqrt{x}$
 
@Simone that's only half of the definition. What is the domain and codomain?
 
the real numbers
 
No, that's not even a function.
 
-1 is a real number
 
4:48 PM
the domain is [0,\infty)
the function is defined on the reals
 
Please be more careful with your words.
 
?
 
The square root function with domain [0,\infty) is not "defined on the reals".
It is defined on the subset [0,\infty) of the reals.
 
ok
so is it continuous?
 
Is what continuous?
 
4:50 PM
Part of learning math is learning how to be pedantic. Both -1 and 1 have the property that $x^2=1$. So, you have to restrict the codomain for the square root to be a function.
Otherwise it is a relation (if I am using that term correctly).
 
I don't have to restrict the codomain,
it is R
So: is the square root function continuous?
 
What is "the square root function"?
 
You've heard of the "vertical line test," no? Where a function must return a unique output for any particular input?
 
are you taking the piss?
 
4:55 PM
Would you also agree that $\sqrt{1}=\pm 1$?
 
@Simone no, you are failing to be unambiguous.
 
$\sqrt{x}$ is the square root function
 
@Simone no, that's notation that is ambiguous.
 
$\sqrt{x}$ is an operation that returns two values, thus it cannot be a function.
 
That's not a "function".
 
4:57 PM
Well, it returns two values when $x\in\mathbb{R}$
 
I guess I'll ask the question later
 
A function is a single-valued, total relation $R\subseteq A\times B$ for two sets $A,B$.
Until you can give $A,B$ and the relation, we can't actually answer your question.
 
See, to answer accurately, a person really needs to know how the function is defined. A particular definition of the square root function might not actually be continuous depending on that.
 
@Simone the point is that, for example, $2^2 = (-2)^2 = 4$, so $\sqrt{4} = \pm 2$, and you need to choose which of these you mean
 
$\sqrt{x}$ whose domain is [0,\infty) and its image is also [0,\infty)
 
5:01 PM
There ya go
 
the everyday milquetoast square root
fro goodness sake
 
Yes, that's continuous.
 
THANK YOU!!!
 
@Simone pedantry is a good quality in mathematics
Rudeness is not :P
 
it was perfectly clear
 
5:02 PM
@simone Not quite.
 
you're getting me agitated
 
(Us in here have been in the non-milquetoast stuff for a while. We need extra specificity.)
 
I mean.. we don't exist to answer your questions, so the fact that you are getting agitated isn't of any concern to anyone except you
 
sure. Have a nice day
 
5:05 PM
I said thank you
 
Yeah, @ÍgjøgnumMeg, he said it all large in caps, all happily, almost with a great big hug!
 
Don't be antagonistic.
(to everyone present, that is.)
 
@Rithaniel want to fight?
 
Sure, so long as I can pick the nature of the combat.
Who can optimize a puzzle in the video game TIS-100 to the lowest number of nodes.
 
D:
I guess you win this time, RITHANIEL.
 
5:25 PM
There's actually a three time Putnam fellow, JP Grossman, who set most of the records I know of on TIS-100.
 
@TedShifrin so the other day I was asking myself conceptually what is the meaning of polynomial equations.
I think I came with a satisfying answer. It is a way to detect the complexity of the underlying field by seeing how geometrical data are arranged.
I think I am satisfied with this conceptual data. Let me know what you think of this.
 
Problem: Show that every group of order 36 has a nontrivial normal subgroup.
Question: Does this require an application of Sylow's theorem?
Please don't give full solution.
I just want to know if I'm headed in the right direction.
 
It does look like they'd be applicable, @user193319
 
I can show there is a sylow subgroup of order 9. To show that it is normal, I need to show that it is the only one. Currently, I know that the number of sylow subgroups is either 1 or 4.
I can't rule out the case that there are 4.
 
What can you say about the number of elements of a particular order in either case?
 
5:39 PM
I'm not sure what you are asking.
 
what's the question here?
 
So, you're saying that there exists either one or four Sylow 3-subgroups. How many elements are accounted for in those subgroups in either case?
 
Each one has the same number of elements---namely, nine---because they are all conjugate to each other.
 
Indeed, and what are the orders of those elements?
 
3 or 9, I guess.
 
5:42 PM
or 1.
 
True.
What follows though?
 
Because they're subgroups, they must contain the identity. So, there are 8 elements not of order 1 in those subgroups. If there are 4 such subgroups, then you have 32 such elements, no?
 
I'm not sure I understand. Why can't the subgroups have overlap?
 
Well, you know the order of an element of a subgroup must divide the order of the subgroup, right?
 
I'm quite lost by your reasoning.
We have four sylow subgroups.
Each has 8 nonidentity elements.
 
5:46 PM
Ah, I see what you're getting at.
So, we don't know that there isn't a subgroup of order 3 in common between two of the Sylow 3-subgroups.
Well, that's actually a little bit more difficult than I was initially anticipating.
 
The intersection of all $p$-sylow subgroups is a normal $p$-subgroup, called the $p$-core
Either this is nontrivial and you're done, or it is trivial and then you can do an elements counting argument like Rithaniel was suggesting I guess
 
I've not heard of this $p$-core. How do you prove that it's normal?
 
@AlessandroCodenotti What was the actual question ?
 
@Rithaniel Conjugate of a $p$-sylow is a $p$-sylow
@SubhasisBiswas Every group of order 36 has a nontrivial normal subgroup
 
Ah, and since the intersection is the same it has to be normal. That's super easy.
 
5:57 PM
@AlessandroCodenotti is the following formulation correct: Every non-abelian group of order 36 is not simple.
 
It's true if you include abelian things as well.
 
@BalarkaSen umm..isn't that obvious.
 
Sure, I am just not sure why you had to explicitly say "non-abelian" :P
 
@BalarkaSen that part is the most interesting
still trying to get a good grip on sylow
will try this. I am sorry that I failed you on the classification of simple finite groups
 
Okay, so in the case of 36, can you rule out that two 3-subgroups have a non-trivial intersection, but the collection of all four 3-subgroups has a trivial intersection?
 
6:02 PM
hey @BalarkaSen what do you think of my conceptual categorization ?
 
@BalarkaSen, Well, I visited to the TIFR website to check out the selection procedure for Masters in Mathematics. Apparently they want only the EXCEPTIONALLY TALENTED students. I failed even before I applied for the entrance.
 
How do they evaluate the EXCEPTIONALLY TALENTED quality?
 
i don't know. Takes one to know one
 
@Rithaniel I don't think the 3-core argument works, yeah. The intersection of two 3-Sylows has to be of order at least 3. But then it's normalizer contains the 3-Sylows as well, because groups of order 9 are abelian. That should be big enough to carry out the argument.
 
Maybe I should repeat the characterization again.
@BalarkaSen I was thinking the other day what is actually a polynomial equation instead of thinking about it as formal sum. I wanted to understand what it represents.
Given a field k. Polynomial equations gives us a way to detect the complexity of the underlying field by seeing how geometrical data are arranged.
 
6:13 PM
Fields came after polynomial equations lmao
I would say that's a totally wrong way to think about things
Field theory was developed to study polynomials
 
what do you think about polynomial equation when you think about this data ?
do you think of them as formal sums ?
 
I don't know man, they're the most natural thing in the world. Quadratic equations historically came from word problems involving finding out the number of people in a game when you know the number of times they have mutually shook hands, etc
Italian dudes tried to do cubics and quartics as a natural generalization
Then nobody was able to solve the quintics, hence the development of field theory, to prove it cannot be "solved"
 
unsolvability of the quintic
 
They're at the inception of mathematics, asking what they mean philosophically sounds like a wrong question to ask
Polynomials are concrete things. They're not sheaf of schemes over etale sites
 
6:18 PM
Sheaves of sheaves is where the real fun is
 
A sheaf of wheat is the true fun
 
What about near-rings?
 
Yeah I see.
 
Actually is there any construction in which a sheaf on $X$ with values in $\mathrm{Sh}_X$ is used or are sheaves of sheaves luckily not a thing?
 
Or other obscure algebraic structures that might have arisen due to mathematical curiosity instead of real-world investigation
 
6:20 PM
Lol
 
@BalarkaSen But wait, we can go further! Iterate the construction to get $n$-sheaves!
 
They have sheaves with values in $\infty$-categories
Those are called simplicial sheaves or smth
 
Insert the "too busy thinking whether they could to consider whether they should" meme here
 
I can't find sheaf of sheaves in nlab, maybe there's still hope for humanity
 
6:22 PM
I think sheaf is just a way to detect algebraic data locally. I guess a sheaf of a sheaf your putting some kinda of topology on your sheaf and maybe detecting different way of arranging algebraic data over your object ?
 
a sheaf of sheaves of sheaves of rings
 
sheaves are a topic I have yet to touch on.
 
the word sheaf is looking weird
 
sheef
 
schif
 
6:24 PM
@BalarkaSen Let me tell you funny story. My wife always jokes with me about how mathematicians come up with funny words for things.
 
Schüff
 
For instance she was telling what does the word sheaf even represents
I told her it probably represents something in french
 
well a sheaf of wheat is like a bundle of wheat
 
It's uh literally what a sheaf is
 
6:25 PM
I see
 
"sheaf of stalks"
algebraic geometry is agriculture on steroids
 
lmao
brb
 
chat dead?
 
Supposedly the very first numeral systems were developed by Pharaohs to keep track of how much food they had.
 
6:28 PM
@Rithaniel look where we ended up
just wait until something apocalyptic happens and takes us back to the stone age
 
"Aha! Today we have $\aleph_0$ much food"
 
"That's a lotta food"
 
"How much more do we need until we get $\aleph_1$ food? We've been at 0 for what feels like forever!"
 
@Balarka I have a dumb question
What's an inner automorphism of the Calkin algebra exactly?
 
I don't know what the Calkin algebra is
 
6:32 PM
Oh so you take a separable infinite dimensional Hilbert space $H$ and look at $B(H)$, the algebra of bounded operators on $H$
It contains the closed subalgebra $K(H)$ of compact operators, the Calkin algebra is the quotient
 
Ah
So Fredholm guys are the ones invertible in $B(H)/K(H)$, right?
 
the Calkin algebra because all such Hilbert spaces are isomorphic of course
@BalarkaSen yes
Ah so I guess that an inner automorphism is conjugation by a Fredholm operator
 
Yup
 
Nice
Describing the outer automorphisms or even proving they exist was a big problem until it was shown independent of ZFC apparently
 
Whoa, crazy
 
6:35 PM
A professor will come to give a talk on this independence proof in couple of weeks, hence my curiosity
 
@quid I improved the query: data.stackexchange.com/math/query/1058887/… Now it automatically creates the archive.org recovery link for the suspicious deleted questions!
 
I'm not seeing why an operator is Fredholm iff invertible in the Calkin algebra
 
@quid And, it shows only the really, trully suspicious cases, in a much better coprehensible form (f.e. empty columns were removed).
 
@AlessandroCodenotti If $F : H \to H$ is Fredholm, $\ker F \subset H$ is finite-dimensional, so $H = (\ker F)^\perp \oplus \ker F$, where $F$ kills the second factor and is injective on the first factor. $F : (\ker F)^\perp \to \text{range}\, F$ is bijection, so invert it to get $G : \text{range} F \to (\ker F)^\perp$ and I think range is a closed subspace of $H$, so you can extend that to $G : H \to H$.
Uh
I guess you just set $G$ to be zero on $(\text{range} F)^\perp$, that's finite dimensional too (cokernel)
 
Balarka is doing functional analysis now
Scary times
I like how every question with Balarka starts as someone asking him something he knows nothing about but he just figures out the whole theory on the spot
6
 
6:52 PM
Then $GF = I - K_1$ where matrix of $K_1$ is just a bunch of $1$'s on the $\ker F$ block and $FG = I - K_2$ where matrix of $K_2$ is a bunch of $1$'s on the $(\text{range} \, F)^\perp$ block
 
Drive to understand
 
Something like that
@RyanUnger Nah, I have heard people talk about Fredholm theory at different times, I picked up some disjoint things by osmosis
Don't really know it
 
Makes sense
 
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