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08:01
Hmm, I'll work it out
What goes wrong with sending a sequence $(a_n)$ in the sequence $(a_n/n)$ if it's not in $c_0$?
... I think that just works
But yeah let's see if Ted's thing works as well
If you have $(a_n)$ and $(b_n)$, then you let $c_n = a_n - b_n$, so $\|c_n\|_{\infty} < \epsilon$
Continuity at points of $c_0$ is the only issue.
I think.
Neat. There is actually an example of an everywhere smooth, nowhere analytic function.
08:08
Hmm
@Tobias good lord
It's like $\sum_{n = 0}^\infty \exp(-2^n) \cos(2^n x)$.
Or something
@BalarkaSen Yeah, something coming from a cumulative distribution of some kind
It's a modification of Weierstrass's function
Which is like $\sum_{n = 0}^\infty 2^{-n} \cos(2^n x)$?
Maybe more complicated
Ah, right, if you pick a sequence in $\ell^{\infty}\setminus c_0$ converging to an element of $c_0$ there are issues. The ones agreeing with $1/n$ for the first $j$ terms and being constantly $1/j$ afterward for example seems problematic
Why a problem?
08:16
So let $x_j$ be the element that does that
Right whoops
But yeah so $f(x_j) = (\frac{1}{j}, \frac{1}{2j},\ldots)$
So the $f(x_j) \to 0$ while $x_j \to (1,\frac{1}{2},\ldots)$
Huh?
You using my f or A's?
Yeah
Which function?
08:20
$(\limsup a_n)/n$
Yeah
The point is that if you let $x_j = (1,\frac{1}{2},\ldots,\frac{1}{j},\frac{1}{j},\ldots)$, then by your $f$, we have $f(x_j) = (\frac{1}{j},\frac{1}{2j},\ldots)$
Yeah, I see.
So maybe I need not limsup but norm!
That fixes this issue, I think.
But the idea of sending the sequence $(a_n)$ to the sequence $(\frac{a_n}{n})$ for $x\notin c_0$ that Alessandro posed feels like it works
08:24
Hullo
Interesting question, but it's past my bedtime.
Yeah same I've got a midterm tomorrow
I dunno why I'm still awake
Night, back to this tomorrow.
@Daminark hm, no wait, where does $f(x_j)$ go with my function? I think that's still a problem
@Daminark from my point of view you said the same thing yesterday :P
Bye @Ted
Technically my midterm is today
Even on my side, it's 2:30 AM
But it's not really tomorrow until I wake up or the sun rises
@Alessandro I think you're still okay though, right?
Or wait hmm
:thonk:
08:29
Bye @Ted. I am very glad that people are thinking about this
$f(x_j)$ I think will converge to $(\frac{1}{n^2})$
So that's still bad
I guess the question is whether we should be trying to prove that no such retraction exists
I'm starting to think that's the case
Apparently there exists some Lipchitz continuous function, a friend of mine said so
Do try the norm idea to see if it works, though at this point I also gotta sleep
(Tbh the fact that you can have Banach spaces with non-complemented closed subspaces is probably good reason to believe the Normie Wildberger had a point...)
With that, I bid you nerds farewell!
Sure. Thanks for all the help
08:49
I guess norm idea won't work. If we let $x_j = (1/j,\frac{1}{2},\ldots,\frac{1}{j},\frac{1}{j},\ldots)$, then we have $f(x_j) = (\frac{1}{j},\frac{1}{2j},\ldots)$
0
Q: GATE $2018$ - Probability question : occurrence of green and red faces when die is rolled $7$ times

Mithlesh Upadhyay A die is colored green on $4$ sides and red on $2$ sides. if the die is rolled $7$ times. Then which of following option is correct regarding occurrence of green and red faces? (A) $2$ green and $5$ red (B) $3$ green and $4$ red (C) $4$ green and $3$ red (D) $5$ green and $...

Some exam questions are strange. I wonder if we can write a proof checker to see whether the question is actually made correctly...
That will really help the exam administration process since it will reduce the chance of having questions being reviewed and then found wrong
I might have to think and discuss in logic room later on what it means for a question to be sound if the answer is not known in advance
On mobile. Check soon
Please do!
09:02
Sunn O))) aren't really my thing
I haven't listened to any of their solo work
I plan to, at some point
I just love Scott Walker
They make this weird drone doom metal thing
Yeah. I can catch the heavy droning in this album too
If you want another interesting post rock album try "not for want of trying" by maybeshewill by the way
Thanks, I shall check this out!
09:08
Listened to a good traditional heavy metal album called "Satan's Hallow" by Satan's Hallow. But they almost copied a Rainbow's song then stopped listening
But I loved the female vocalist's voice.
This shows how when given a sufficiently destructive force, even some of the most numerous species (practically speaking almost infinite like) will plummet to zero
2
Sapiens
Meanwhile, mathematics structures where finite objects can reach infinite objects and vise versa is not very common without some mapping
The most well known example where such occurs philosophically speaking, is calculus
 
3 hours later…
12:01
(Again not directed toward anyone) I guess either someone created a sock or someone is star spamming
Hi @AkivaWeinberger
long time no see !
Hi
Fun fact, a ton of people is around a dozen people
2
Good morning everyone! Where can I ask a logic puzzle? One of these popular puzzles with true and false statements. I am completely puzzled myself :) Thank you!
@Jason There is a puzzling SE site. It might be on topic there
www.puzzling.stackexchange.com
12:18
thank you guys!
@Jason Also try Rayomond Smulliyan (or something similar named guy)'s "What's the name of this book" (or some similar name, i forgot)
Gotch @Jason Check out this book:
12:53
Abcd: To answer your question, it is similar to this:
The Game is a mental game where the objective is to avoid thinking about The Game itself. Thinking about The Game constitutes a loss, which must be announced each time it occurs. It is impossible to win most versions of The Game. Depending on the variation of The Game, the whole world, or all those aware of the game, are playing it all the time. Tactics have been developed to increase the number of people aware of The Game and thereby increase the number of losses. Though the origins of The Game are unknown, a game featuring ironic processing was played by Leo Tolstoy in 1840. == Gameplay == There...
In short, similar to how it is impossible to win most of the game, it is almost impossible to stop The Plan
On a more iluminati-ish analogy: The Plan is what makes the chat flow
(Fun fact: The chemistry incident have somehow managed to cause the topology group to stop reducing weirdness (probably because with a better grasp of manifolds, I can contribute to the discussion). Currently, the weirdness reduction is only concentrated for a couple of users. You know when they happen when you saw messages that are somewhat out of context and incomprehensible)
In other news, I cannot think of any maths to discuss about yet
13:18
Barlarka:
Scott walker album:
Lullaby = sinister sounding
Fetish = strange
Bull = reminds of rave and rock
Herod = mix of the criminal underworld and chanting
Brando = of the 5 I like this one the most. There seemed to be a drama weaved into it since it started out lively, then go mysterious, and then the lyrics suggest some kind of struggle, then becomes hopeful, and then becomes dwelling and then more power struggle
Music wise, Brando also has the most variation
13:46
Hi,
$a$ is an integer, such write in basis $b$ (is palindrom) with only $\{0,1\}$ and the size is $n<b$.
Is it true that $a^2$ is a palindrom in basis $b$ ?
I cannot think of any non number theoric ways to answer that question (and I don't know enough number theory to figure out what tools I need to answer it too)
But from a purely algebraic point of view it will probably had something to do with the following matrix:
I think it's true, and there are a big trick elementary behind
Let $d_i$ be the digits of $b$. Treat $b$ as a vector, compute its outer product with itself, and analyse the symmetry of the resulting matrix
e.g. if b =1001, the matrix will be:
a is the number and b the basis
a is palindrom
in basis b
ah sorry, then typo, switch all instance of b above to a
Regardless of basis, the following must be true:
$1 \cdot 1 =1$
$a \cdot 0 = 0$
Thus for $a = 1001_b$ we should get:
$$[a] \otimes [a] = \begin{pmatrix} 1 & 0 & 0 & 1\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\1 & 0 & 0 & 1\end{pmatrix}$$
and $a^2$ is then given by summing all the entries of the matrix
We can now observe that, for $a$ with $n$ instances of $1$s, $a^2$ will have $n^2$ instance of $1$s. Thus the algebraic way boils down to whether $n^2$ is palidrome in base $b$. That I am not very sure because I am terrible with carryovers
13:56
Doesn't this follow from just writing up the entries in the square of a polynomial and noting that the coefficients stay below $b$, so that we really can treat everything as polynomials
I think so, we are basically expressing $a$ as a polynomial of powers of $b$ and thus its digits becomes the coefficients
[hs]C'est pas bien Gérard d'espionner pour avoir la réponse ![\hs]
@Secret : the answer is not trivial, yes ?
I am not sure whether there is an efficient way to check whether $n^2$ is palidrome in $b$ though other than writing it in base $b$ by expanding it in terms of polynomials involving $b$ and then check that the nth and n-k th matches for each k < n/2
uh wait a sec... I forgot something...
$a=1001_b = 1b^0+0b+0b^2+1b^3$
$a^2 = \sum_{ij}([a] \otimes [a]) =$
$$\begin{pmatrix}1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\1 & 0 & 0 & 1 \\\end{pmatrix}$$ and the position of each entry is:
\begin{pmatrix}b^0 & b^1 & b^2 & b^3 \\ b^1 & b^2 & b^3 & b^4 \\b^2 & b^3 & b^4 & b^5 \\b^3 & b^4 & b^5 & b^6 \end{pmatrix}
so reading out the results, only the off diagonal entries that are symmetric about the diagonal add together
14:10
@Secret : do you want the answer ?
and we get $a^2 = 10(1+1)001_b$
where stuff in brackets will depend on $b$
If $b=2$, carryover occurs
and we get $a^2 = 110001_2$ which is not palidrome. Otherwise $a^2 = 102001_b$ which is also not palidrome
@Dattier sure
(cannot handle the general case)
ok, I give you the trick
14:14
@AkivaWeinberger Your average person weighs >150lbs???
lemma : if P,Q polynôme palindrome, then P*Q is polynome palindrome
@Secret
why this lemma is true
@Secret if you don't know there are a second elemantary lemma behind
lemma 2 : P polynom palidrom iff $X^{deg(P)}P(1/X)=P(X)$
@Secret : ok ?
@anakhronizein I looked it up - average worldwide for adults is like 135lb, but in North America it's like 175lb
Also, the word "around" gives me some wiggle room :P
(I am at the worldwide average)
@Secret : ok ?
thinking...
Maybe I should find the averages by gender
14:24
You should include children. ;)
the average person has 1 breast
Is $\arctan(-x)= \arctan(x)$?
$X^{deg(P)+deg(Q)}P(1/X)Q(1/X)=P(X)Q(X)$ how is this palidrome?
Sorry, haven't studied inverse trig but there's a question in my current chapter complex numbers which involves this part^ of inverse trig
@Secret MAGIC!
14:27
@Gérard : je n'aime pas la magie
@Slereah Beautiful
je préfère la cuisine$
The average person has $2-\epsilon$ arms
@Secret Musically I like Brando the most. Herod is a lyrical masterpiece of the album. I also like Lullaby
What if there's a person with two million arms throwing off the average
14:29
@Secret : do you agree with lemma 2 ?
@Slereah I think we'd know
@Dattier I tried to prove it by brute forcing the multiplication and all I get is a mess. I don't know of clever ways to deal with it since my knowledge on polynomial ideals is almost nonexistent
In generall, this is why I don't like multiplying two polynomials
too much nesting to track where things go
@AkivaWeinberger That is only if by "average" you mean "mean"
I would argue that the average person has two arms
since the mode is a much better measure of central tendency in this situation :P
@BalarkaSen For me, Herod sounds really good in representing a criminal underworld movie scene. It has the dense mysterious impression in it
I don't know about that. It's telling a biblical story
14:33
Ah that explains why the lyrics sounds like chanting
(Of the infantile genocide as described in the old testament)
Yeah
Does anyone notice that MSE is about to cross 900k questions in 4 more questions! Woohoo! :D
2 more to go
@Secret : reprise that slowly is not difficult
just 1 more....
and it's there!!!!
bye (@Gérard maintenant tu connais la solution)
Falcon Heavy launch scheduled for 1:30pm EST tomorrow
@Gérard : pourquoi voulais-tu absolument connaître la réponse ?
@GaurangTandon how do you know it is 900k?
O wow the n cancels
14:46
@Secret : ok ?
P(1/X) reverses the powers of n, thus literally flip the polynomial left to right. Since P is palidrome, "filling the vacancy" by multiplying by $X^n$ gives the original polynomial, as required
@anakhronizein check the count on this page math.stackexchange.com/questions (make sure that ignored questions on the questions page are grayed out instead of hidden in your user preferences)
@Secret : exact
Oh I see now, thanks!
@anakhronizein You're welcome
14:48
@Secret : so ok ?
Now: If $P,Q$ palidrome then by the rules of degrees we get $deg(PQ) = deg(P)+deg(Q)$. Suppose this new polynomial is $R$, we want to know whether it is palidrome. Thus we compute $P(1/X)Q(1/X)=R(1/X)$ which reverses the ordering of the powers around. Now we multiply both sides by $X^{deg(P)+deg(Q)}$ to get:
$$X^{deg(P)+deg(Q)}P(1/X)Q(1/X) = X^{deg(P)+deg(Q)}R(1/X)$$
@Secret : exact
Now since P,Q is palirome, by lemma 2 the LHS is P(X)Q(X)=R(X). Now using lemma 2 again, we see the equation is that of a palidrome polynomial equation, therefore $R$ must be palidrome
@Secret : exact
that's prove the lemma1
Now back to the original question: $a$ can be expanded as a polynomial P(b) with deg(n) and n < b. We knew that P(b) is palidrome and that all coefficients are 0 or 1. Thus when $a^2=P(b)P(b)$ is computed by lemma 2 we knew that the product is palidrome. In addition, the coefficients being 0 or 1 means their products lies in the ideal {0,1} and thus no carryovers can occur, hence $a^2$ is always palidrome
14:56
@Secret : exact
hmm, this substitution $$P(X) := P(1/X)$$is useful. I should keep that in mind in the future...
that's classic in cryptographie
@Secret : the answer is elementary but not trivial, ok ?
ok I guess....
thanks, because @Gérard pretend the contrary
@Gérard is a french agent of DGSE in cryptographie departement
@Astyx : salut, alors cela se passe bien à l'X ?
tiens @Gérard un que tu vas kiffer
Find $p$ prime number with : $P(x)=x^{\frac{p-1}{2}} \mod p$, $P(1)=P(2)=P(3)=...=P(22)$, and with :
$p\neq 11013658661829071$ and $p \neq 197521$.
15:14
those are some really weird numbers to be not equal to...
11013658661829071 is a solution
197521 too
@Gérard vous payez combien à la DGSE ?
@Gérard pour toi c'est gratuit lol
il suffit que tu me le demandes poliment
Oui plutôt bien
Jamais je n'aurais imaginé faire ce que je fais maintenant
Mais je n'ai aucun regrets
Et tu fais quoi ?
@Astyx
Je suis en gendarmerie
service informatique ?
15:25
Entre autres oui
tu as quels grades ?
Aspirant
Comme tous les X en première année
tu n'es pas officier ?
Si, Aspirant c'est le premier grade d'officier
ah
tu connais Gérard commandant du service crypto-math à la DGSE
15:27
Non vu que je ne suis pas à la DGSE
Mais il doit y avoir des X qui le connaissent
Et pourquoi ne pas être aller à la DGSE jouer aux espions lol
Pour être franc j'avais la flemme de remplir le dossier
Ahahahaha
Et il parait que c'est pas aussi excitant que ça en a l'air
(Après rien ne dit que j'aurais été pris)
je ne sais pas, mais je travaille avec la DGSE et toi aussi, d'ailleurs
lol
et c'est vrai
tu sais pourquoi ?
15:30
Mmm ça dépend de ta définition de "travailler avec"
Mais dis moi
bravo : on fournit des données à ses agences
de notre plein gré
en venant sur le net ou encore plus en venant sur des sites comme celui-ci
Certes
Tu sais à quoi sert se site pour la DGSE
?
MSE ?
0
Q: Ideal Generated by Nilpotent Elements is a Nilpotent Ideal

user193319 Prove that if $R$ is a commutative ring and $N=(a_1,...,a_m)$ where each $a_i$ is nilpotent., then $N$ is a nilpotent ideal. So we need to find a $k \in \Bbb{N}$ such that $N^k = (0)$. Working through small values $m$ and $k$, I can see that $k = 1 + \max \{a_1,...,a_m\}$ seems to work. E.g...

15:32
oui
@Astyx
Pas spécifiquement
Ah trouvé de nouvelles idées (cryptanalyse), car les matheux sont malades du classique
Ah oui
Enfin c'est pas juste MSE j'imagine
tout à fait
Mais MSE reste la plus grosse plate frome du genre
Plutôt toute la doc sur internet
C'est la plus userfriendly si tu veux
15:34
la plus vivante
Mais sinon t'as beaucoup plus d'information sur Arxiv etc je pense
mais non, c'est pas vivant, les maths c'est avant tout de l'apprentissage
dans MSE on voit la pensée se déployé
dans Arxiv on a gommer les indices qui permettent de rendre la solution intelligible
@user193319 Try $km$
(assuming your logic is true, I haven't studied that in a long while)
pour ne pas changer les habitudes, voilà une énigme
Find $p$ prime number with : $P(x)=x^{\frac{p-1}{2}} \mod p$, $P(1)=P(2)=P(3)=...=P(22)$, and with :
$p\neq 11013658661829071$ and $p \neq 197521$.
Je t'avoue que juste maintenant, je n'ai pas du tout le temps
15:42
@Astyx : ok, d'autres ici auront peut-être le temps
Sur ce j'y vais
À plus tard
Au revoir
Find $p$ prime number with : $P(x)=x^{\frac{p-1}{2}} \mod p$, $P(1)=P(2)=P(3)=...=P(22)$, and with :
$p\neq 11013658661829071$ and $p \neq 197521$.
Any of you here familiar with do Carmo's Riemannian Geometry?
(By Manfredo Perdigão do Carmo, of Brazil)
I, and many other, are
(I are?)
You are the twenty-first letter of the alphabet
I just got the book, and I'm gonna study it with my teacher for the semester
Is it good?
15:50
I like it.
Arright
That's good
I think it's the standard Riemannian geometry text
I have no idea if I have the prereqs
If you learn it count me in
@BalarkaSen Oh, like Hatcher for AlgTop?
15:51
Something like that
@BalarkaSen Would be hard seeing as you're far away?
Or I guess like, talk about stuff with him and then talk about stuff with you separately
Incidentally: Do I capitalize the "do"?
I know like the first 7 chapters and I haven't done tons of exercises. My Riemannian geometry is weak
@AkivaWeinberger That'd be fun, yep
Stronger than mine
@BalarkaSen there's a few vying for that position, but it's one of the more popular ones
@AkivaWeinberger do Carmo is fantastic
That's good to hear
I remember hearing people being unsure of Rudin, which is what I was doing before this
15:55
I've never used baby nor papa Rudin, but "functional analysis" and "fourier analysis on groups" are quite nice.
I assume I'm holding a translation
Yep- "Translated by Francis Flaherty"
Second edition?
good
> "Translator's note: It is intended that this translation follow the original Portuguese closely."
Duh
15:57
hullo guys
What, otherwise would I just think you made shit up?
@AkivaWeinberger I would be temped to fix proofs.
He actually did, apparently
"Minor corrections"
He presumably means symbols, etc.
But some of the proofs are confusing, admittedly.
And the definition of parametrized surfaces.
No complete rewriting of proofs, then
15:59
Probably not
Anyway, you should read it
the exercises are really good too
Arright, will do
i dont like the riemannian category too much

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