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Secret
Secret
1:04 AM
[The Cult of Infinity]
0. Begin by specifying an alphabet $\mathcal{A}$. This give us the symbols to be used. Being human beings, our alphabet is finite despite it can be made unbounded to account for the development of our society
1. Next, define a formal grammar $\mathcal{G}$ which provide the rules in producing new strings and how they are concatenated, truncated etc.
 
Alex
Alex
Who are you communicating this to @Secret?
 
Secret
Secret
No one, cause chat is empty
 
Balarka Sen
Balarka Sen
It is not
There are at least three people here
3 is not equal to 0
ur argoment is invalid
 
Secret
Secret
well a moment ago Alex is not on the side bar, and you are grayed out
And I am a non-person, anyway :P
 
Balarka Sen
Balarka Sen
THE MOMENTS GONE NOW
MWAHAHAH
Get rekt
 
Alex
Alex
1:08 AM
Get rekt!
 
Akiva Weinberger
Akiva Weinberger
Let $N$ be larger than any number that will be defined by mathematicians
 
Balarka Sen
Balarka Sen
Get rekt in turn @Alex
 
Secret
Secret
dies
 
Alex
Alex
get sniped @Alex
 
Akiva Weinberger
Akiva Weinberger
$f$ is continuous iff $f(x+\frac1N)-f(x)$ is smaller than any number that will be defined by mathematicians
 
Secret
Secret
1:09 AM
@AkivaWeinberger That's basically Cantor's Absolute Infinity $\Omega$
 
Akiva Weinberger
Akiva Weinberger
assuming $f$ has been defined by mathematicians
 
Secret
Secret
Well, if $f \in \text{Some Uncountable (Cardinality) Set}$, then it can be indefinable, thus it works :P
 
Akiva Weinberger
Akiva Weinberger
Assume $N/2$ is also larger than anything we'll ever define, as well
and any similarly such defined function on $N$
 
Secret
Secret
Actually, can we ever formally prove the existence of an element of actual infinity from below?
 
Akiva Weinberger
Akiva Weinberger
Apparently Everest pronounced his name /ˈiːvrɪst/ (EEV-rist)
 
Secret
Secret
1:16 AM
Btw if $g(N)$ is indefinably infinite for any relation $g$, then $f(x+\frac{1}{N})$ is also indefinably infinite, and hence $f(x+\frac{1}{N})-f(x)$ is Indefinably infinite-finite=indefinably infinite or undefined. Hence either there are no functions, or there are no continuous $f$ s :PPPP
 
Akiva Weinberger
Akiva Weinberger
I guess I mean specifically the functions $g$ that go to infinity
 
Secret
Secret
hmm...
ok $f(x+\epsilon)-f(x)$ is continuous indeed
though I knew very little how continuity is defined in the hyperreals
$st(f(x+\epsilon))=f(x)$?
 
Secret
Secret
Still, it seems that without any axioms of infinite objects, it seems to be impossible to distinguish between an unbounded collection vs an (actual) infinite object
 
user338510
What is math?
 
Secret
Secret
It's complicated
Modern maths is basically rooted in logic, which itself is disputed
Otherwise, some say it is a collective creative activity on making objects that are not limited to the confines of common sense or reality
Other said it is a systematic study of rules in formal systems
So, use the axiom of finite choice and pick your poision :P
 
user338510
1:27 AM
Math is useless.








(now roast me)
 
Secret
Secret
It is pointless, though :P
17
Q: Useless math that became useful

ppaulojrI'm writing an article on Lychrel numbers and some people pointed out that this is completely useless. My idea is to amend my article with some theories that seemed useless when they are created but found use after some time. I came with some ideas like the Turing machine but I think I'm not gr...

nt.number-theory big-list mathematics-education gm.general-mathematics
$$\int idk = ik + C$$
Done roasting
:P
 
user338510
oof
 
Balarka Sen
Balarka Sen
rekt
 
Abra001
Abra001
pointless topology, not surprising, i see enough people around who think all maths is pointless.
 
Secret
Secret
It is one of the oldest math jokes and the best counter to all arguments of pointless by turning it into a hilary mess
> Q: *stares blankly, jaw slowly unhinging*

A: Exactly.

This enormous project to prove everything—one of the purest mathematical enterprises ever undertaken—didn’t just end with a feeble flicker and a puff of smoke. Far from it.

Sure, it didn’t accomplish its stated goals. But by clarifying (and, at times, revolutionizing) ideas like “proof,” “truth,” and “information,” it did something even better.

It gave us the computer, which in turn gave us… well… the world we know.

Q: So the pure mathematics being done today might, someday, give us a new application as transformative as the compu…
 
user338510
1:34 AM
Math Teachers are Trash.
 
Secret
Secret
But still, what can we gain from studying actual infinity?
Actual infinity
In the philosophy of mathematics, the abstraction of actual infinity involves the acceptance (if the axiom of infinity is included) of infinite entities, such as the set of all natural numbers or an infinite sequence of rational numbers, as given, actual, completed objects. This is contrasted with potential infinity, in which a non-terminating process (such as "add 1 to the previous number") produces a sequence with no last element, and each individual result is finite and is achieved in a finite number of steps. == Anaximander == The ancient Greek term for the potential or improper infinite was...
(Nonconstructivist) Mathematicians are more familiar of them in the form of infinite cardinals
> Potential infinity versus actual infinity
The distinction between potential infinity and actual infinity goes back to Aris- totle. A detailed, nuanced discussion can be found in Books M and N of the Metaphysics [1, 3], which constitute Aristotle’s treatise on the philosophy of mathematics. Aristotle’s position is that, while potential infinities have an ob- jective existence in reality, actual infinities do not. This is in the context of a broader argument against Plato’s theology.
In modern mathematics, the prime example of potential infinity is the nat- ural number sequence 1, 2, 3, . .…
If Proof Theory and the Hilbert Program is what gave us computers and information technology in general
Then the next big breakthrough in mathematics that depends on foundational stuff must have to do with understanding infinity
in particular the open question:
> Does Actual Infinity exists in reality?
> Though the formalist revolution is an undeniable fact of mathematical (and perhaps scientific) history, some questions about it still remain -- e.g., is formalism good or bad? Some scientists and mathematicians have suggested
that mathematics, no longer tied to its origins in physics, is developing into a baroque art form, a thing of great embellishments and few uses; that mathematics has been reduced to a mere game of meaningless marks on paper. Others have argued that mathematics turns out to be useful in surprising and unexpected ways, just because mathematicians have concerned themsel…
 
PVAL-inactive
PVAL-inactive
1:56 AM
This sites awful
 
Secret
Secret
which?
 
Secret
Secret
So the notion of finitisability might be what need to focus on exactly how much maths can be deal with without actual infinity
 
Secret
Secret
2:11 AM
[The Cult of Infinity]
0. Begin by specifying an alphabet $\mathcal{A}$. This give us the symbols to be used. Being human beings, our alphabet is finite despite it can be made unbounded to account for the development of our society
 
Abra001
Abra001
 
Secret
Secret
Lame proof
Still, what happens if we do wrote that in:
Define $\infinity$
 
Leyla Alkan
Leyla Alkan
How is the green line obtained? And also what is the use of $N\gt N_1$ or $n\geq N$ here, since $N$ is appearing nowhere in the proof
 
Secret
Secret
Next multiply it by i to give $i\infty$. Then
Define $i\infty=8$
Therefore we have: $a+i\infty=a+8$
Multiply it by i again to give:
$ia+\infty$
$1q=\infty, 0=8, -8=0, -1(\infty)=0$
If $0=8$ then:
8+x=x for all x
Thus we are now in a $\mathbb{Z}/8$ version of the complex numbers adjoined with infinity
 
Balarka Sen
Balarka Sen
@AkivaWeinberger We need you
Immediately
Very urgently
 
MatheinBoulomenos
MatheinBoulomenos
2:24 AM
@TobiasKildetoft I've worked out the remaining three-dimensional case
 
Secret
Secret
Now to check whether it is consistent by adding the rule: $i\infty=8,i8=\infty$...
 
MatheinBoulomenos
MatheinBoulomenos
@TobiasKildetoft The idea is to show that if $A$ is a three-dimensional algebra such that $A/\operatorname{rad}(A)\cong \Bbb C \times \Bbb C$, then $A$ has a faithful two-dimensional representation
 
Secret
Secret
We knew that $i^2=-1$. Thus $ii8=-8=i\infty=8$. This is ok, because 8=0, thus it is expected that it is its own additive inverse
Now what about: $ii\infty=-\infty=i8=\infty$. This is also ok, thus here $\infty$ is its own additive inverse
 
MatheinBoulomenos
MatheinBoulomenos
@TobiasKildetoft Then we can use this the classification of subalgebras of $\operatorname{M}_{2\times 2}(\Bbb C)$ described in this MO post to see that $A$ is actually isomorphic to the $2\times 2$ upper triangular matrices: mathoverflow.net/a/27379/117693
 
Secret
Secret
But wait: $i0=0$ and $i8=\infty$ and $0=8$. Thus $\infty=0$. Therefore:
By trying to do exactly as that faux proof said we will have: $1/0=0,-10=0,1/0=0$
Thus 1/0 does not exist, as required
 
Abra001
Abra001
2:33 AM
@LeylaAlkan definitions are scrambled and i think a step ((n-N1)eps/2)/n is skipped. it doesn't matter since (n-N1)<n
 
Akiva Weinberger
Akiva Weinberger
@BalarkaSen Wagh
 
Balarka Sen
Balarka Sen
@AkivaWeinberger Does Darude Sandstorm have a secretly encoded message in Morse code???
 
Leyla Alkan
Leyla Alkan
@Abra001 Can you use mathjax please :)
 
Balarka Sen
Balarka Sen
Only someone of high Rick and Morty intelligence can answer this
 
Akiva Weinberger
Akiva Weinberger
.…- ...…- ...…- ...…- ...…- ...…-
is the message
Wait, spaces are as long as dashes
So no spaces
 
Balarka Sen
Balarka Sen
2:39 AM
What does that translate to in plebian
 
Akiva Weinberger
Akiva Weinberger
There's some good information here
Very well-annotated
 
Abra001
Abra001
there is 6 letters 5 identical. no matches found in english ....
 
Balarka Sen
Balarka Sen
Oh god
The first “dun” of this song is meant to communicate the importance of existential nihilism in post modern culture–especially in social areas concerning neo feminism, political apathy, and retardation. With that being said, with each “dun” there is a contrived imagery of a child with downsyndrome in 2008 saying “dun du duuu” with another child mocking him– pounding the side of his hand to his chest with each “dun.”
JESUS
Clearly a reference to Darude’s father. Here Darude expresses his compassion towards those of us who don’t have a father, as seen in “dadadsadada”.
Darude would make a similar line, 5 seconds after this one (“Dddddddd ddadadadadaddadadadadadaadadadadadad”) where he would elaborate on this.
 
Daminark
Daminark
This warms my heart
 
Akiva Weinberger
Akiva Weinberger
> FADE

This line represents Darude’s life after this song.
@Abra001 Waaaaa?
 
Abra001
Abra001
2:52 AM
this lyrics reminds me of crazyfrog (and it has lyrics too).
 
Secret
Secret
@BalarkaSen wait what? Those "dun" actually have meaning? I thought they are just setting up the tension mindsplode
 
Akiva Weinberger
Akiva Weinberger
They don't, the Genius link I posted is a joke
 
Abra001
Abra001
3:08 AM
my kitten just meowed when she saw these lyrics, i think they do mean something.
 
Leyla Alkan
Leyla Alkan
Why do they only look at the limit at one side($r\rightarrow 1^{-}$) Is it because otherwise we wouldn't have $\sum_{n=1}^{\infty}(-r)^n=\frac {-r}{1+r}$ ?
 
Akiva Weinberger
Akiva Weinberger
 
Secret
Secret
Akiva: I predict the cat will end up landing its paws all over the video
 
Akiva Weinberger
Akiva Weinberger
@LeylaAlkan It doesn't even converge for $r>1$
 
Leyla Alkan
Leyla Alkan
@AkivaWeinberger yeap, this is what I tried to mean
 
Akiva Weinberger
Akiva Weinberger
3:20 AM
It doesn't converge when $r$ is bigger than $1$, and it doesn't converge when $r$ is equal to $1$. But it does converge when $r$ is less than $1$, so we can take the limit as $r$ approaches $1$ from below. Intuitively, this gives you what the value "should be" at $r=1$.
Meaning it gives you what $\sum(-1)^n$ "should be".
 
Leyla Alkan
Leyla Alkan
okay, thanks..
 
Drew Brady
Drew Brady
Anyone here w/ knowledge of functional analysis? math.stackexchange.com/questions/2693221/…
 
Akiva Weinberger
Akiva Weinberger
@BalarkaSen BLARK
You know functional analysis I think
 
MatheinBoulomenos
MatheinBoulomenos
@DrewBrady your justifications are fine
 
Drew Brady
Drew Brady
Thanks
 
Leyla Alkan
Leyla Alkan
3:27 AM
Do you also know how is this formula made up @AkivaWeinberger ?
 
MatheinBoulomenos
MatheinBoulomenos
You can rewrite $(-1)^nnr^n$ as $n(-r)^n$, this is the derivative of a geometric series
 
Akiva Weinberger
Akiva Weinberger
@LeylaAlkan Oh there's a nice trick for that
Take $\sum(-1)^nr^n$ and differentiate it with respect to $r$
So $\sum(-1)^nr^n=\frac1{1+r}$ so differentiating gets $\sum(-1)^nnr^{n-1}=-\frac1{(1+r)^2}$
Multiply both sides by $r$ and you get your answer
It's not all that different from the sum in this question
(if you substitute $x=-r$ in there and multiply by $r$ you get your sum)
To be honest, I'm only sharing that because I like the answer that I wrote there
 
Leyla Alkan
Leyla Alkan
Hahaha, it seems that you waited too long for such a question :)
I'll check your answer soon too
 
Drew Brady
Drew Brady
You can also get to it without differentiation
But it is less fun. But you can work backwards from the RHS. For example (1 + r)^{-2} = (\sum_1^\infty (-r)^n)^2
 
Leyla Alkan
Leyla Alkan
3:46 AM
Okay, thanks..
 
Abra001
Abra001
4:24 AM
@AkivaWeinberger a proof of this equality crossed my mind but i strongly think it's tackled atleast in one of these so many answers.
oh yeah i knew, it's 3rd proof written by the op.
also the bounty-elected answer doesn't seem simple to digest from exactly the beforelast edge.
 
Xander Henderson
Xander Henderson
4:41 AM
I don't suppose that there is anyone here with a good knowledge of harm. anal. who might be willing to have a look at this answer and explain what I did wrong? The authors of the text clearly are not expecting that answer...
 
Leyla Alkan
Leyla Alkan
the $N^{th}$ Cesaro sum of the series $\sum_{k=1}^{\infty}c_k$ is $\sigma^N=\frac {s_1+\cdots s_N}{N}$, where $s_n=\sum_{k=1}^{n}c_k$

So $\sigma^n-\frac{n-1}{n}\sigma^{n-1}= \frac {s_n}{n}$ but here it says $\sigma^n-\frac{n-1}{n}\sigma^{n-1}= \frac {c_n}{n}$ . I didnt understand why
By the way are these two notations the same, $\sum_{k=1}^{\infty}c_k$ and $\sum c_n$ ?
By the way I mistakenly typed $\sigma_n$'s as $\sigma^n$'s
 
Tanuj
Tanuj
5:19 AM
Hello guys ! I've got a question to ask
$$\lim_{x \to \infty}\left(\dfrac{2}{\pi}(x+1)\cos^{-1}\left(\dfrac{1}{x}\right)-x\right)$$
My try: Since , $x \to \infty$ , $\dfrac{1}{x} \to 0$ , and therefore , $\cos^{-1}\left(\dfrac{1}{x}\right) \to \dfrac{\pi}{2}$.
This however , gives me the answer as $1$ , which is not correct. Can anyone tell me where I messed up ?
 
Secret
Secret
5:34 AM
You have "$\infty-\infty$" when you try to directly evaluate the limit like that. The xs cannot cancel that way
 
Tanuj
Tanuj
@Secret hmm , can I not directly put the value of the variable when I have $\infty-\infty$ form ?
@Secret if not , then what should I do ?
 
Semiclassical
Semiclassical
One way to get a sense of what's going wrong is to rewrite it as $$x\left[\frac{2}{\pi}(1+x^{-1})\cos^{-1}(x^{-1})-1\right]$$
hmm. That doesn't really help
For reference, the correct answer is apparently $1-2/\pi$ (via mathematica)
 
Tanuj
Tanuj
yea that's the correct answer.
 
Semiclassical
Semiclassical
and we can note that $\lim_{x\to\infty}\frac{2}{\pi}\cos^{-1}(1/x)=1$
 
Tanuj
Tanuj
@Semiclassical I did that
 
Semiclassical
Semiclassical
5:40 AM
So we can remove that from the limit and what remains is $$\lim_{x\to\infty}\left[\frac{2}{\pi} x\cos^{-1}(x^{-1})-x\right]$$
Which evidently should equal $-2/\pi$
 
Secret
Secret
That's a $\infty - \infty$ form. I am not sure how it can be broken down as Lhopital on the arccos looks messy
 
Semiclassical
Semiclassical
Which we can rewrite to $\lim_{x\to\infty}\frac{2}{\pi}\frac{\cos^{-1}(1/x)-\pi/2}{1/x}$
 
Tanuj
Tanuj
@Semiclassical Not sure how .
@Semiclassical from this , how did you get the next step
 
Semiclassical
Semiclassical
Well, at this point, you can note that $t=1/x\to 0$ as $x\to\infty$
so this last limit can be equivalently written as $\lim_{t\to 0}\frac{2}{\pi}\frac{\cos^{-1}(t)-\pi/2}{t}$
 
Tanuj
Tanuj
@Semiclassical ah got it.
 
Semiclassical
Semiclassical
5:45 AM
not so easy to see, I gotta say
Probably the smart thing to do would be to do $t=1/x$ from the start
 
Tanuj
Tanuj
@Semiclassical why can't we directly evaluate the value of that $\cos^{-1}$ thing and why do I have to solve the bracket and then put its value ?
 
Semiclassical
Semiclassical
@Tanuj you mean, back here?
 
Tanuj
Tanuj
@Semiclassical what I'm saying is , when I directly put $\cos^{-1}\left(\dfrac{1}{x}\right)=\dfrac{\pi}{2}$
 
Semiclassical
Semiclassical
So if you could do that, then $\lim_{x\to\infty}\left[\frac{2}{\pi} x\cos^{-1}(x^{-1})-x\right]$ would be 0 not $-2/\pi$.
 
Secret
Secret
I am not even sure how to solve $\lim_{t\to 0}\frac{2}{\pi}\frac{\cos^{-1}(t)-\pi/2}{t}$, the denominator blow up while the numerator does not
 
Tanuj
Tanuj
5:51 AM
@Semiclassical won't it be $1$ ?
 
Semiclassical
Semiclassical
@Tanuj No, I've subtracted out the 1 already
@Secret let $f(t)=\cos^{-1}t$
then you've got $\lim_{t\to 0}\frac{1}{t}[f(t)-f(0)]=f'(0)$
@Secret also, the denominator -> 0 as t->0
 
Tanuj
Tanuj
@Semiclassical When I directly put $\cos^{-1}\left(\dfrac{1}{x}\right)=\dfrac{\pi}{2}$ in $$\lim_{x \to \infty}\left(\dfrac{2}{\pi}(x+1)\cos^{-1}\left(\dfrac{1}{x}\right)-x\right)$$ , I get the incorrect answer as $1$ , right ? So why can I directly put $\cos^{-1}\left(\dfrac{1}{x}\right)=\dfrac{\pi}{2}$ in $\lim_{x\to\infty}\frac{2}{\pi}\cos^{-1}(1/x)$
 
Semiclassical
Semiclassical
that's right. Similarly, if you do the same substitution into $\lim_{x\to\infty}\left[\frac{2}{\pi}x\cos^{-1}\left(\frac{1}{x}\right)-x\right]‌​$, you'll incorrectly conclude that this limit is zero.
the reason I write focus on this is because it makes the issue clearer: You've got $x\left[\frac{2}{\pi}\cos^{-1}\left(\frac{1}{x}\right)-1\right]$. The point is now that while the term in square brackets goes to zero as $x\to\infty$, the $x$ in front blows up.
 
Tanuj
Tanuj
Why can I use the value of $\cos^{-1}$ thing later but not at the start ?
 
Secret
Secret
@Semiclassical Ah I see, the derivative is hiding in plain sight
 
Tanuj
Tanuj
5:59 AM
@Semiclassical okay.
 
Semiclassical
Semiclassical
$$\dfrac{2}{\pi}(x+1)\cos^{-1}\left(\dfrac{1}{x}\right)-x=x\left[\dfrac{2}{\pi} \cos^{-1}\left(\dfrac{1}{x}\right)-1\right]+\dfrac{2}{\pi}\cos^{-1} \left( \dfrac{1}{x} \right)$$
If you blindly plug $x\to\infty$, you'd have $\infty\cdot 0+1$
and while the $1$ is fine, the first product is indeterminate.
so more care than just "plug it in" is needed
 
Tanuj
Tanuj
@Semiclassical Got you !
@Semiclassical Though after the first step , I get something like $$\lim_{x\to\infty}\left[\frac{2}{\pi} x\cos^{-1}(x^{-1})-x\right]$$
What's the next step from here ? It's still $\infty-\infty$ form
 
Semiclassical
Semiclassical
same as I did before.
let $t=1/x$ and recognize what you've got as a difference quotient.
 
PrithiviRaj
PrithiviRaj
$\lim_{x\to\infty}\left[\frac{2}{\pi} x\cos^{-1}(x^{-1})-x\right]=\lim_{t\to 0} \frac{\left[\frac{2}{\pi} \cos^{-1}(t)-1\right]}{t}$
and this is in $\frac{0}{0} $ form
(use L' Hospital) , we'll get the limit as $\frac{-2}{\pi}$
 
Tanuj
Tanuj
6:14 AM
$\dfrac{\dfrac{2}{\pi}\cos^{-1}(t) -1}{t}$ , where $t \to 0$
Oh yea exactly .
@Semiclassical @Secret @PrithiviRaj Thanks a lot guys ! Finally got it down.
 
PrithiviRaj
PrithiviRaj
:)
@LeylaAlkan , Is Abel summable same as Cesaro summable ?
 
Tanuj
Tanuj
Okay guys , I got one more question on limits I can't solve
$$\lim_{x\to 0}\left[\dfrac{a^{\tan x}-a^{\sin x}}{\tan x -\sin x}\right]$$
 
Leyla Alkan
Leyla Alkan
@PrithiviRaj Abel is stronger than Cesaro and Cesaro $\implies $ Abel, so they are not same
 
PrithiviRaj
PrithiviRaj
okay!
 
Tanuj
Tanuj
For this , I basically cannot use L' Hospital as it gets too messy. What should be the line of thinking now ?
 
Secret
Secret
6:31 AM
My first hunch will be the substitution u = sin x
 
Tanuj
Tanuj
@Secret hmm , wouldn't that complicate $\tan x$
 
PrithiviRaj
PrithiviRaj
$\lim_{x\to 0}\left[\dfrac{a^{\tan x}-a^{\sin x}}{\tan x -\sin x}\right]=$\lim_{x\to 0}\left[\dfrac{a^{\tan x- \sin x}-1}{\tan x -\sin x}\right]\cdot a^{sin x}$
 
Secret
Secret
Hmm...
$a^{\tan x}-a^{\sin x}=a^{\sin x}(a^{\tan x - \sin x} - 1)$
 
Tanuj
Tanuj
Hmm
 
PrithiviRaj
PrithiviRaj
let $ \tan x - \sin x = t$ and now ,our limit can be written as $\lim_{x\to 0} a^{\sin x} \cdot \lim_{t\to 0}\left[\dfrac{a^{t}-1}{t}\right]$
 
Tanuj
Tanuj
6:42 AM
I guess there won't be any other way...
@PrithiviRaj yea I got it , thanks :)
 
PrithiviRaj
PrithiviRaj
Can we use two variables at a time in a limit ? , I'm not sure ,( $t$ and $x$)
 
Tanuj
Tanuj
@PrithiviRaj That won't affect the answer . It's just a variable
 
PrithiviRaj
PrithiviRaj
oh , I see . Okay!
and sorry for my MathJax ! I'm a novice.
 
Tanuj
Tanuj
all cool , don't worry
Another question I have to ask.
 
PrithiviRaj
PrithiviRaj
sure
 
Tanuj
Tanuj
6:51 AM
Can't type it (I'm lazy) , so
 
Secret
Secret
I think in a multi variable limit perspective, the change of variable specify the curve which the limit is to be made, so the problem still has only one degrees of freedom and hence the limit should be able to decompose into an iterated limit
 
Tanuj
Tanuj
@Secret cool.
So in this question , do I check for which function , limit does not exist ?
 
Secret
Secret
Rearrange the function so that y is the subject, then it should become clear when taking the squareroot
 
Tanuj
Tanuj
@Secret ok , $y$ replaces $x$ everywhere in the question then , what next ?
 
PrithiviRaj
PrithiviRaj
I think C & D are continuous.
 
Tanuj
Tanuj
6:56 AM
@PrithiviRaj How , what's your line of thinking ?
 
PrithiviRaj
PrithiviRaj
trace those functions and you'll get semicircles!
and for A & B , find limit at $x=o$ .
 
Tanuj
Tanuj
@PrithiviRaj Ah , okay . Hmm , any other approach though ? Like checking their limits ?
 
MatheinBoulomenos
MatheinBoulomenos
@TobiasKildetoft good morning!
 
Tobias Kildetoft
Tobias Kildetoft
@MatheinBoulomenos Good morning
 
Silent
Silent
From baby Rudin: 'If $\{E_n\}$ is a sequence of closed nonempty bounded sets in a complete metric space $X$, if $E_n\supset E_{n+1}$, and if $\lim\limits_{n\to \infty} \text{diam} E_n=0$, then $\cap_1^{\infty} E_n$ consists of exactly one point.' Can we discard 'bounded' from the hypothesis, because $\lim\limits_{n\to \infty} \text{diam}\, E_n=0$ implies bounded, doesn't it?
 
Tanuj
Tanuj
7:04 AM
@PrithiviRaj any idea on how to proceed now ?
 
MatheinBoulomenos
MatheinBoulomenos
@TobiasKildetoft does it seem right that any indecomposable 3-d algebra with 1-dimensional radical is isomorphic to the upper triangular matrices?
 
Tanuj
Tanuj
@Secret What to do next ?
 
Silent
Silent
@@TobiasKildetoft, will you please look at my above question?
 
Tobias Kildetoft
Tobias Kildetoft
@MatheinBoulomenos It does not sound unreasonable at least. I did not check your argument properly yet
 
PrithiviRaj
PrithiviRaj
just check continuity at $x=0$ for A ,B . and you'll get the answer.
@Tanuj
 
Tanuj
Tanuj
7:07 AM
@PrithiviRaj Okay , why not at $x=2$ ? Also , what about the other two options ? Where do I check the limit for them ?
@PrithiviRaj But the graphical approach is far too easy ! I just didn't think about it.Thanks.
 
PrithiviRaj
PrithiviRaj
we don't need to make it complicated !
:)
 
Tanuj
Tanuj
yea
 
Kasmir Khaan
Kasmir Khaan
7:20 AM
Hi yall :D
 
Silent
Silent
Let $f$ be a uniformly continuous function from a subset $E$ of a metric space $X$ to metric space $Y$ and $E$ is bounded in $X$. Is $f(E)$ a bounded subset of $Y$?
 
user8663905
user8663905
Hi, can anyone give me an example of a function such that that function applied to the intersection of several subsets does not equal the intersection of that function applied to each subset individually?
Or, in other words: f(\bigcap_{k\in I}^{ }S_{k}) \neq \bigcap_{k\in I}^{ } f(S_{k})
Hmmm, is that not proper Latex formatting?
Oh, wait, forgot the $
 
abenthy
abenthy
If $g \colon \mathbb{R}P^n \to \mathbb{R}P^\infty$ is injective, does the induced map on cohomolog $g^* \colon H^1(\mathbb{R}P^\infty) \to H^1(\mathbb{R}P^n)$ map a generator to a generator?
 
user8663905
user8663905
$f(\bigcap_{k\in I}^{ }S_{k}) \neq \bigcap_{k\in I}^{ } f(S_{k})
Well, it's not meant to necessarily be an infinite set of subsets.
Just a the intersection of an arbitrary set of subsets.
(I'm still trying to figure out what's wrong with my latex
$$f(\bigcap_{k\in I}^{ }S_{k}) \neq \bigcap_{k\in I}^{ } f(S_{k})$$
That worked.
Didn't intend to center it, so apologies for that.
I'm looking for an example of a function for f such that that statement is true. I know that it must not be injective, but I'm not entirely sure why not
 
Tobias Kildetoft
Tobias Kildetoft
8:12 AM
@MatheinBoulomenos I looked at your argument for dimension $3$ now and it looks good
Do you have any idea for whether one might be able to construct a positive basis for that algebra? And same question for the full algebra of $2\times 2$ matrices (since this would be a start on the general case of semisimple ones).
 
MatheinBoulomenos
MatheinBoulomenos
@TobiasKildetoft hmm, I'll think about it
I have to go now
thanks for your feedback
 
Tobias Kildetoft
Tobias Kildetoft
8:30 AM
Does anyone have a good suggestion for a feed reader to use with arXiv feeds?
 
 
1 hour later…
Abcd
Abcd
9:52 AM
How to find common normal to two curves?
 
MatheinBoulomenos
MatheinBoulomenos
10:21 AM
@TobiasKildetoft a positive basis for the 2x2 upper triangular matrices is $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$
that means every $3$-dimensional algebra has a positive basis
 
Tobias Kildetoft
Tobias Kildetoft
@MatheinBoulomenos Nice. And even one with the nice property we were considering that allowed for direct products (maybe we should give that property some name)
 
MatheinBoulomenos
MatheinBoulomenos
strongly positive?
idk
"really nice basis"
 
Tobias Kildetoft
Tobias Kildetoft
strongly positive is not a bad choice. Could also call it $1$-avoiding
 
MatheinBoulomenos
MatheinBoulomenos
$1$-avoiding is more descriptive
 
Daminark
Daminark
Hey everyone!
 
MatheinBoulomenos
MatheinBoulomenos
10:27 AM
Hey @Daminark
 
Tobias Kildetoft
Tobias Kildetoft
@Daminark Hi
Similar properties for other elements of the basis might also be of interest at some point, so one could have $x$-avoiding for some basis element $x$ (maybe with a left- and a right version)
So this property is actually fairly strong, since then the span of the basis elements other than $1$ becomes a 2-sided ideal.
 
Drew Brady
Drew Brady
Anyone here able to help me with this problem: imgur.com/a/hUlw5
Functional analysis
I figured out part (a) I think, I am working on (b)
 
Tobias Kildetoft
Tobias Kildetoft
So in particular, there is no way for the algebra of $2\times 2$ matrices to have a positive basis with this property
 
MatheinBoulomenos
MatheinBoulomenos
right
 
abenthy
abenthy
If I pick a linear injection $R^n \to R^\infty$, it gives a map $RP^n \to RP^\infty$. Is it clear that this map induces a nontrivial map in cohomology $H^1(RP^\infty;Z_2) \to H^1(RP^n;Z_2)$?
 
Drew Brady
Drew Brady
10:34 AM
For part (a) isn't it just that in a separable Hilbert space bounded seq => has weakly convergent subseq => image has norm convergent subsequence by assumption. ?
 
Daminark
Daminark
Yeah that's it
 
Drew Brady
Drew Brady
cool. any idea how to prove the hint? I'm probably being stupid
 
Daminark
Daminark
In fact you can generalize this to saying that if $X$ is reflexive and $T:X\to Y$ is completely continuous (weak convergent => strong convergent), then $T$ is compact. This is by Eberlein-Smulyan
 
Drew Brady
Drew Brady
first of all $TTx -> 0$ in norm means $\langle TT^\astx, TT^\ast x\rangle \to 0$ right?
 
Daminark
Daminark
Yeah
 
Drew Brady
Drew Brady
10:37 AM
The point is to conclude that $\langle T* x_n , T*x_n \rangle \to 0$
like the definition of adjoint means that $\langle T^\ast x_n, T^\ast x_n \rangle = \langle x_n, TT^\ast x_n \rangle$
Daminark?
 
Daminark
Daminark
Still here, you're doing fine
 
Drew Brady
Drew Brady
yeah. I just don't really know what to do at this point htough
 
Daminark
Daminark
Well I'll phrase things slightly differently
Let's say $\{x_n\}$ is a bounded sequence in $H$. $T^*$ is a bounded linear map, $T$ is compact, so $TT^*x_n$ has a convergent subsequence $TT^*x_{n_k}$
 
Drew Brady
Drew Brady
by convergent you mean in the norm
 
Daminark
Daminark
Now $\|T^*x_{n_k} - T^*x_{n_j}\|^2 = \langle T^*x_{n_k} - T^*x_{n_j}, T^*x_{n_k} - T^*x_{n_j}\rangle = \langle TT^*(x_{n_k} - x_{n_j}), x_{n_k} - x_{n_j} \rangle$
Yeah I mean in norm here. But anyway, take this last inner product and push it through Cauchy-Schwarz
That's bounded by $K\|TT^*(x_{n_k} - x_{n_j})\|$ where $K = 2\sup_n \|x_n\|$
So that goes to 0, meaning the $T^*x_{n_k}$ form a Cauchy sequence. And so you're done
 
Drew Brady
Drew Brady
10:49 AM
oh. nice!
I keep forgetting to use completeness.
 
Daminark
Daminark
Turns out this works in Banach spaces as well
 
Drew Brady
Drew Brady
yeah
 
Daminark
Daminark
The more general proof is a bit more work, uses Arzela-Ascoli
 
Drew Brady
Drew Brady
Ok one other question Daminark
It's this problem: imgur.com/a/JqRke
I think I can do (a)
Basically what I did was I said if x is in ker(A) then 0 = BAx. Since BA = I_1 - E_1, this means I_1x = E_1x.
So the identity map restricted to ker(A) is a compact operator
Hence the unit ball (in ker(A)) is compact set
=> finite dimensional
 
Daminark
Daminark
Yup
 
Drew Brady
Drew Brady
11:00 AM
Ok, so part (b) is trickier lol
We want show that if y_n = Ax_n -> y then y = Ax for some x
Like if x_n are bounded I can do it.
Because then we can find a subsequence x_n_k such that E_1 x_{n_k} converges.
Then BAx_{n_k} + E_1{x_n_k} -> By + x*
(x* being the limit of E_1 x_{n_k}
And so y = lim_k A x_{n_k} = A (lim_k x_{n_k}) = A(By + x*), as desired
but like idk what to do if x_n are not bounded...is there a nice way to reduce?
 
Daminark
Daminark
Do pay attention to the hint
When I was reviewing for my functional exam there was a problem that said that if $T:X\to X$ is a compact operator and $X$ is infinite-dimensional, then you can find some sequence $x_n$ on the unit sphere such that $Tx_n \to 0$. The proof was that if you couldn't, this would amount to having $\|Tx\| \ge c\|x\|$ for some $c > 0$
This implied that $T$ was injective, and thus isomorphic to its image. But this inequality also gives closed image since if you have some Cauchy sequence in the image $y_n = Tx_n$, you have $\|x_n - x_m\| \le \frac{1}{c}\|y_n - y_m\|$, so the $x_n$ are Cauchy and thus converge
 
MatheinBoulomenos
MatheinBoulomenos
@TobiasKildetoft after trying random stuff for a while, I can't find a positive basis for $M_{2\times2}(\Bbb C)$. Maybe we should try to write a program that tries random stuff? Or try to prove that there is no positive basis?
 
Daminark
Daminark
Point is, that type of inequality is very useful. I'm about to go to sleep now but yeah hopefully that helps
 
Drew Brady
Drew Brady
hm. okay.
 
Akiva Weinberger
Akiva Weinberger
11:16 AM
@XanderHenderson It never before occurred to me that the abbreviation of harmonic analysis is "harm anal"
 
quallenjäger
quallenjäger
:D:D:D:D:D:D
@AkivaWeinberger Good one.
 
Drew Brady
Drew Brady
Anyone can help me with part(b) of this problem? imgur.com/a/JqRke
it is functional analysis
 
Secret
Secret
11:56 AM
@Tanuj Sorry I was away inside my work when you send me that message. Seems PrithvRaj got you covered. Anyway, if you want to do it purely algebraically, you actually have to check that all boundary points are continuous, which here are the 3 points -2,0,2. But again, graphs of squareroots are relatively easy to deal with (those equations will give you semicircular arcs) so go with the easy root unless the exercise said no to that
@alisha, who send you here. There are no records of your summoning into the math chat
 
Alisha
Alisha
@Secret Why?
 
Tanuj
Tanuj
@Secret Hakuna Matata :)
@Secret lol
 
Secret
Secret
//help
 
Alisha
Alisha
###################### Help ######################

==================== Commands
//WhoMade      | Gets the user ID of the user who created a command
//alive        | Used to check if the bot is working
//remhome      | Removes a home room - Admins only
//kill         | They must be disposed of!
//ban          | Bans a user from using the bot. Only usable by hardcoded bot admins
//promote      | Changes a users rank.
//help         | Lists all the commands the bot has
//time         | What time is it?
 
Secret
Secret
ugh whatever, @PrincessLuna will deal with this
 
Shobhit
Shobhit
12:01 PM
//time
 
Alisha
Alisha
@Shobhit Fri, 16 March 13:01:33.0445 2018 +01 CET (+0100)
 
Shobhit
Shobhit
damn
 
Tanuj
Tanuj
//time
 
Alisha
Alisha
@Tanuj Fri, 16 March 13:02:11.0955 2018 +01 CET (+0100)
 
Secret
Secret
in Sandbox, 2 days ago, by Alisha
@PrincessLuna My friend is is in charge...
 
Tanuj
Tanuj
12:02 PM
incorrect ! haha got you !
 
Secret
Secret
anyway, better not to mess with bots here, we don't want him to end up like pseudohuman being retracted from the network
 
Drew Brady
Drew Brady
anyone able to help w/ functional ?
 
Secret
Secret
I don't think any functional analysis guys are around atm
3
Q: What is the interface between functional analysis and algebraic geometry?

gradstudentThis is a very open ended curiosity of mine and I would be grateful to hear any comments in this direction. In particular I am interested in functional analysis/algebraic geometry books/papers references which show this bridge from functional analysis into algebraic geometry. I am not sure ...

ag.algebraic-geometry fa.functional-analysis big-list big-picture reading-list
and therefore, I don't think Mike (currently the only person closest to analysis that is on) can help with hilbert space stuff
I don't recall if alessandro does functional analysis
//unsummon 36
 
Alisha
Alisha
@Secret 2 more votes required
 
Secret
Secret
bah, looks like only the bot owners can deal with this

 Modern Abstract Analysis

For functional analysis, measure theory, and related areas. M...
You can also post the question there to get more relevant people to answer
 
Secret
Secret
12:41 PM
O they dried out...
 
Balarka Sen
Balarka Sen
@AkivaWeinberger Very little
 
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