« first day (2314 days earlier)   
00:00 - 08:0008:00 - 15:00

8:00 AM
@Brody oh yeah I realize this would require too much geographical knowledge to guess the answer
I'm from India
@Adeek, the Galois group of $\mathbb{Q}(\zeta_n) \cong \mathbb{Z}_n^\times$, with $\zeta_n$ an $n$th root of unity. When $n$ is prime, $\mathbb{Z}_n^\times$ is cyclic.
hmm let us see @KajHansen
@SoumyoB hmm, the more you know (geography-wise)
oh yeah
oh yeah @KajHansen very nice
I haven't looked at the galois group of the cyclotomic extensions
by the way, does a continuous linear functional $T$ separating two subsets $E$ and $F$ of the domain Banach space with complex numbers as the codomain mean that there exists a $\gamma \in \mathbb{R}$ such that $Re (Tx) < \gamma \leq Re (Ty)$ for all $x \in E$ and $y \in F$?
8:02 AM
but yeah I can believe that.
I will just compute it for extra practice.
where $Re (y)$ means the real part of $y$ for $y\in \mathbb{C}$
hey how did they determine this result @KajHansen ?
@Adeek, let $n$ be prime. Then $(x^n-1)/(x-1)$ is irreducible. The Galois group of an irreducible polynomial acts transitively on its roots. Thus, for any $k$, we have an automorphism determined by $\zeta \mapsto \zeta^k$
Compose this automorphism with itself a bunch of times
I see.
Well, note that $\zeta \mapsto \zeta^k$ will send $\zeta^m \mapsto \zeta^{mk}$
8:09 AM
$mk$ gets reduced modulo $p$
So there's a natural isomorphism between that and $\mathbb{Z}_n^\times$
I see that is awesome
I liked it too
@KajHansen here is a combinatorial argument btw.
How can I show that submanifolds of complex tori have non-negative Kodaira dimension?
8:18 AM
Consider the maps as you said before $\omega \mapsto \omega^k$ for $k = 1,..,4$. Then Since $[Q(\omega) : \mathbb{Q}] = 4$ the above permutations of roots determine our galois group completely. Consider $\sigma_3 : \sigma_3(\omega) = \omega^3$. Then $\sigma_3^2(\omega) = (w^3)^3 = \omega^9 = \omega^2 \implies |\omega| > 2 \implies |\sigma_3| = 4.$
There are only two subgroups of order 4 either $Z_4$ or $Z_2 \times Z_2$ it can't be $Z_2 \times Z_2$ so must be $Z_4$.
Yeah, that's a good argument
There's no algorithm in general to find the generator for $\mathbb{Z}_n^\times$, but it's easy enough for small $n$.
hm what about an example where the ground field has characteristic p.
that has aut group isomorphic to $\mathbb{Z_4}$
oh I think that is easy
the galois group of $F_{2^2}$ is isomorphic to $\mathbb{Z}_4$
Finite field galois groups are always cyclic
It is generated by the frebenius endomorphism
8:26 AM
it it $F_{2^4}$
not $F_{2^2}$
The former
Hi, I saw a continued fraction formula for a tangent function like this:

tan(x) = x/(1-(x2/(3-(x2/(5-(x2/(7-...)))))))

I got this from:

Can someone explain how this function really works? When I tried substituting x with 3.14159, I got a negative result instead of 0.
@morbidCode the exponents are wrong
8:41 AM
@DHMO or I read it wrong. I am using a screen reader so I can't be sure. Can you look at the link where I got it from? It's on exercise 1.39 mitpress.mit.edu/sicp/chapter1/node21.html
@morbidCode I looked.
It is supposed to be tan(x) = x/(1-(x^2/(3-(x^3/(5-(x^4/(7-...)))))))
i.e. it should be 3-(x^3 instead of 3-(x^2
@DHMO ah I see. Thanks!
alright nights @KajHansen
Hello chat.
9:04 AM
@DHMO I am an absolute beginner on math notations. Now that you explained the correct formula, I'm just curious. Could it be the reason for the exponent 2 all over the formula in the book is because it assumes x gets updated every term? For example 3-x^2 where x is the result of x^2 from 1-x^2? Because (x*x)x? I'm not so sure, because your formula is certainly clearer. Or maybe the book's formula is really just wrong?
@morbidCode I don't understand your question
Your link agrees with what I just wrote
in General topology, 22 mins ago, by Secret
I think using this I have figured it out. My analysis is as follows:
in General topology, 2 mins ago, by Secret
Suppose there is a set $X=\{a,b,c\}$ and a topology $\tau_x=\{\emptyset,\{a\},\{b\},\{a,b\},\{a,b,c\}\}=\{\emptyset,A,B,C,D\}$. From this topology and the notion of closeness given in the wikipedia link we can deduce the following:
1. no points are close to the empty set
2. $a\in A$ thus only a is close to A
3. $b \in B$ thus only b is close to B
4. $a,b, \in C$ thus both a and b are close to C
5. $a,b,c \in D$ thus all a,b,c are close to D.

Why the closure under intersections is necessary in the axiom:
9:25 AM
> A chain is only as strong as its weakest link.
Can this be demonstrated using mathematics?
anyone know probability?
Is an antichain as strong as its weakest antilink?
9:41 AM
Q: Essence of definition of normal numbers

sashaConsidering only real numbers in $(0,1)$ Hardy and Wright define simply normal numbers as A number is simply normal in base $r$ ( $r >=2 $ ) if $\; \lim_{n \to \infty }\frac{n_b}{n}=\frac{1}{r}$, where $n_b$ reprsents number of occurrences of digit $b$ ( $0 \le b < r$ ) in the first $n$ digi...

any help is appreciated.
10:20 AM
wooo finally got my probability proof problem done
it was nuts
10:42 AM
in General topology, 6 mins ago, by Secret
However suppose I ask what is the complement of {a} in $\tau_2$, then I get {b,c} which is in $\tau_2$, thus {a} is closed. But in $\tau_2$ there is no notion of "only a is close to the set {a}" because {a} is not in the topology. Therefore, what does it mean when it is found that {a} is closed in $\tau_2$?
by "what does it mean" I mean what information can we learn from knowing that {a} is closed in the topology $\tau_2$?
@Secret why is that notion invalid?
in other words, who said {a} must be in the topology?
Nobody said {a} must be in a topology. But based on the discussion and the notion of closeness, the open sets in a topology tell us which points or bunch of points are close/near to what sets
so if {a} is not present in the topology, then we cannot really say anything about whether a is close to the set {a} in this topology?
but from the definition of closed topological sets, we found the complement of {a} is {b,c} which is in the topology hence an open set. That means b or c are close to the set {b,c} in this topology
Therefore by definition of closed topological set, {a} is a closed set in this topology
so that implies somehow there's a notion of "a but no other points or bunch of points is close to {a}"?
11:11 AM
Is the infinite cylinder with shrinking radius $\bigcup_{t \in \mathbb{R}} \{t\} \times \min(1,\frac{1}{|t|}) \cdot S^1$ a smooth submanifold of $\mathbb{R}^3$?
Or let's say we smooth the ugly parts at $t=1$ and $t=-1$ with a bump function.
11:59 AM
I can't believe I just read a proof of the fact that $C[0,1]$ is not dense in $\mathbb{L}^\infty [0,1]$
how can this even be true
$C[0,1]$ is complete, hence it is a closed subspace of $\mathbb{L}^\infty [0,1]$, hence it cannot be dense in the latter
how can the same proof not apply to any other $\mathbb{L}^p$ space
Hi people! I have a short question on commutative algebra and in particular von Neumann regular rings. Anyone interested?
I just bought 3 false nail box sets
12:26 PM
Hi what are the $O(n,m)$ and $SO(n,m)$ groups called? As in what is their name
would one say Lorentz group for all of them?
(special) Indefinite orthogonal group of signature n,m?
I'm thinking big O notation but that could be different
Lorentz group is the case O(3,1)
I posted my question here: math.stackexchange.com/questions/2043206
I hope I haven't committed any faux pas.
there should be a guideline on how to master proofs. x.x
12:46 PM
@usukidoll Step 1: do 'em a lot.
shouldn't step 1 be understand the definitions and theorems first?
@TobiasShuxueLaoshi why should A' be larger than A?
@usukidoll I'd call that step 0.
1:03 PM
Hi nerds
How to find derivative of a function also limit of any function when there is something like $sum_{k=1}^{n}$ in it?
Hi all
@Ramanujan Could you give an example?
Q: How to find $\lim_{n \to \infty } \frac{ \sum_{k=1}^{n}(1-(1-2^{-k})^n) }{\ln n}$

maxkorHow to find $\lim_{n \to \infty } \frac{ \sum_{k=1}^{n}(1-(1-2^{-k})^n) }{\ln n}$.Is there a simple way?

It's thought,lets see a simpler example
@DHMO hi
ambiguous question is ambiguous
@DHMO so we can't find derivative of a function including sigma?
@Ramanujan i meant, specify the summation
1:13 PM
Q: Interchanging the order of differentiation and summation

FroskoyCan the order of a differentiation and summation be interchanged, and if so, what is the basis of the justification for this? e.g. is $\frac{\mathrm{d}}{\mathrm{d}x}\sum_{n=1}^{\infty}f_n(x)$ equal to $\sum_{n=1}^{\infty}\frac{\mathrm{d}}{\mathrm{d}x}f_n(x)$ and how can it be proven? My intuiti...

$y'=x^2 + x + 1/6$
@Ramanujan no.
y'''' =0
All wrong.
If your sum is finite, the derivative of the sum is the sum of the derivative
1:16 PM
@Ramanujan you're confusing $\sum\limits_{i=1}^x i^2$ with $\sum\limits_{i=1}^n x^2$
that's why I told you to specify the summation
It seems to me that you don't even know what $\sum\limits_a^b c$ means @Ramanujan
$\sum\limits_a^b c$ values of c are from a to be getting added
@Ramanujan then what is $\sum\limits_{i=1}^n i^2$?
$1^2 +2^2 +3^2 +4^2 + \cdots n^2$
and what is its value?
1:22 PM
now, what is $\sum\limits_{i=1}^n n^2$?
It will be same?
of course not
$$\sum_{i=1}^n n^2 = \underbrace{n^2 + n^2 + \dots + n^2}_{n\text{ times}}$$
$n^2 \huge?$
1:25 PM
@Ramanujan no
@DHMO then?
$n^2 + n^2 = 2n^2$
$n^2 + n^2 + n^2 = 3n^2$
$\underbrace{n^2 + n^2 + \dots + n^2}_{n\text{ times}} = \dots$ ?
@Astyx $n^3$?
$\sum\limits_{i=1}^n x^2$= $x^3$?
1:29 PM
12 mins ago, by DHMO
It seems to me that you don't even know what $\sum\limits_a^b c$ means @Ramanujan
Don't confuse me any more,tell me what's wrong with me
@AliCaglayan Because we're essentially adding stuff that weren't there. But of course my intuition must be playing tricks with me. There must be some cancellation going on.
well the ideal generated by those elements is prety big
It seems to almost span A[X]
so I wouldn't think it is a surprise that $\psi$ is an epimorphism
@DHMO will it be right for sum of squares of x natural numbers
1:40 PM
@Ramanujan then it is defined only when x is an integer
which makes differentiation pointless
@DHMO but it is dy/dx
@Ramanujan the domain of the function is not continuous
so the function itself cannot be continuous
so you cannot take its derivative
@DHMO x is not a variable?!?
1:43 PM
@Ramanujan x is a variable
but it can only take integer values
Yes it can be continuous
@Astyx what is the sum of the first 2.5 squares?
That doesn't make sense
That doesn't mean it's not continuous
1:45 PM
What is $\sqrt{-3.5}$ ? It doesn't exist. However $\sqrt\cdot$ is continuous over its domain
@Astyx how does that matter?
its domain is $\Bbb Z$
And ..?
it is a discrete set in $\Bbb R$
And ..?
1:47 PM
so it cannot be continuous
correct me
Q: use of $\sum $ for uncountable indexing set

BeltrameI was wondering whether it makes sense to use the $\sum $ notation for uncountable indexing sets. For example it seems to me it would not make sense to say $$ \sum_{a \in A} a \quad \text{where A is some uncountable indexing set e.g. some $A \subset \mathbb{R}$ } $$ Would it be better to avoid ...

To mess with things further..., check above MSE
What's with all the stars ?
@Astyx it's called starred
1:50 PM
Continuous only means the inverse image of any open set is an open set. On $\Bbb Z$ with the usual topology any function is continuous since $\{n\}$ is open
@Astyx I'm not using that definition of continuous
is the usual topology of $\mathbb{Z}$ discrete topology?
Not sure about terminology @Secret
@DHMO then you're not using the right word
o wait, it cannot be discrete else {n} will be closed since its complement is open
@Astyx I'm using the definition that $f:S\to T$ is continuous if $\forall s \in S: \lim\limits_{x\to s}f(x) = f(s)$
@Secret what is {n}?
1:54 PM
all singletons in Z
e.g. {1}, {2} etc.
@Secret and what's the problem?
limits in $\Bbb Z$ ..?
then what's its derivative?
It doesn't have any
why not?
1:56 PM
(If I understood correctly) since it is given in the usual topology that {n} is open, it means its complement is not in the topology as otherwise {n} will be closed
@Secret what's wrong if {n} is open and closed?
Limits allow you to get arbitrarily close to a value. This is not something you can do in $\Bbb Z$
a continuous function must map open sets to open sets
@AliCaglayan well, if the topology is $\Bbb R$ then you're right
(except for infinities)
1:57 PM
but we're talking about $\Bbb Z$
anything else I am still trying to get my head aorund of (and details have already posted in general topology room)
@Secret and every set in the topology of $\Bbb Z$ is open
@DHMO I didn't mention $\Bbb R$ anywhere
@AliCaglayan you can still apply the epsilon-delta definition of limit in $\Bbb Z$
Oh really? Can you show me
1:58 PM
@AliCaglayan can you give me an example?
Well you can
Just take $\delta\lt1$ and you get anything you want
So it's continuous
but why doesn't it have a derivative?
e.g. I don't recall whether {m,n} for integers n,m is open in the usual topology of $\mathbb{Z}$
2:00 PM
Being continuous doesn't mean it has a derivatiave
@Secret every subset of $\Bbb Z$ is open in the usual topology of $\Bbb Z$
because derivatives are not defined in $\Bbb Z$ i'd say
@AliCaglayan and i'm asking why this specific case does not have a derivative
@Astyx why not?
I've only heard of them in metric vector spaces
isn't $\Bbb Z$ a metric space?
2:01 PM
but not a vector space
If that's the case, then the usual topology is the discrete topology. Hmm, I guess I need to think again later...
@Secret yes it is
To define derivatives you need (or at least that's what I was taught) something which tends to 0 without being 0
This doesn't exist in $\Bbb Z$
For obvious reasons
@Astyx "tends to 0" is ambiguous
so {n} is open under the discrete topology, and also closed because its complement is a subset of $\mathbb{Z}$ which is open in this topology. Hence all subsets of $\mathbb{Z}$ are clopen in this topology and thus disconnected?
2:05 PM
I know
But I recon you can understand what I meant
@Secret All sets in a discrete topology is clopen
And I don't know what disconnected means
@Secret except for singletons
@Astyx maybe
2:08 PM
@Astyx why are singletons {n} not closed. If we take the complement we get $\mathb{Z}-{n}$ and this is a subet of $\mathbb{Z}$ thus open?
@Secret singletons are closed
@Secret I meant they are not disconnected
I see
In the discrete topology the singletons are the connected components
2:09 PM
spaces with this property are called totally disconnected
@Astyx where are you from?
What country ?
2:10 PM
Why ?
And you ?
because spelling "disconnected" as "disconected" reminds me of Spanish (disconectado)
I think French still has the double n
@Astyx je ne prefere pas le dire
Typos happen
(maybe too often)
@Alessandro. Btw I have a conceptual question about closed sets in a topology (I am using the open set convention for topology)
3 hours ago, by Secret
3 hours ago, by Secret
in General topology, 6 mins ago, by Secret
However suppose I ask what is the complement of {a} in $\tau_2$, then I get {b,c} which is in $\tau_2$, thus {a} is closed. But in $\tau_2$ there is no notion of "only a is close to the set {a}" because {a} is not in the topology. Therefore, what does it mean when it is found that {a} is closed in $\tau_2$?
3 hours ago, by Secret
by "what does it mean" I mean what information can we learn from knowing that {a} is closed in the topology $\tau_2$?
I'm not sure what you mean, there are some useful properties of closed sets
such as containing all their limit points
NB By A close to B, I mean in the following sense in the wikipedia article on "closeness"
00:00 - 08:0008:00 - 15:00

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