12:03 AM
Hey guys, I think my statistics grading system is bugged, but I'm not entirely sure. I just need some re-assurance for my answer. I entered $9$ for the variance of the Exponential Distribution and it's not accepting it.
https://i.gyazo.com/9544a70d7b7e003abc48e412ac533a44.png
Help is appreciated.

I'm assuming this is an inner regular measure. If the only closed subsets of $E_j$ have zero measure, then by definition of inner regularity, $E_j$ also has zero measure, so $\mu^*(E_j-F_j) \leq \mu^*(E_j) = 0$ @Leaky

@MatheiBoulomenos hmm...
alright

Hello guys
I got a question
if G acts on the set S , it also acts on the set of subsets of S
we can restrict the action to subsets of S with fixed number of elements
What does this mean ?

You can define an action on the power set of $S$ via $gX = \{gx | x \in X\}$ for any $g \in G$ and $X \subset S$

@MatheiBoulomenos @LeakyNun help :)
hmm this was the intro of sylow theorem

12:10 AM
@MatheiBoulomenos I don't think that is what it means

I might add that to avoid comfusion
so from what i understood

I do think that this is what it means

G is a group acting on a set S
G can aslo act on the subsets of set
so assuming that the order of the subsets are the same

@MatheiBoulomenos well he said "restrict"
which means changing the domain

Yes, but there is an induced action from the action on S

12:12 AM
gU wont change the cardinality of the subset U

That's correct

Restriction here means, instead of looking at the whole power set, you note that the image of a k-element set is a k-element set

@Daminark hmm

he meant "corestrict"

oh nvm, I completely missed a part
the part where it says "it also acts on the set of subsets of S"
sorry :P

12:13 AM
np leaky :D

Yes, that was what I was talking about

@MatheiBoulomenos He comeant corestriction

Anyway :)
So bascilly what the action does now

I actually think "corestriction" is a useful word. It's a common thing you do

taking the subsets and shuffling them
is that right?

12:13 AM
yes

the subsets of S are the "elements " here

What does corestriction entail? I thought you were just memeing

Okay nice I dont know why this is usefull right now but at least i got what it means :D
thanks guys :D

I'm thinking if you want to "restrict the codomain", I tend to call it "corestriction", as I reserve "restriction" for restricting the domain, but I guess that's non-standard

@MatheiBoulomenos how does one make sense of the stabilizer of U , subset of S ?

12:15 AM
I was thinking this

Codomain?
Oh like target space

Like, "corestricting" a function $f : A \to B$ to $f : A \to f(A)\subset B$

Yeah sure that works, I guess

is gU = U for u in U ?

@Kasmir yes, that's right, that's the definition of the stabilizer in this case

12:17 AM
hmm
so we multiply by g
and we need to get back our whole set U back
one problem here is that hmm
it can be that many elements of G , sends a subset to itself
I mean do we have a partition on G here?
some elements gU= U, stablizer, other element g'U = X , X subset of S

Well, the cosets $G/ \operatorname{Stab}(U)$ partition $G$, yes. The bijection that you worked so hard on between the orbit and the cosets of the stabilizer still applies
if $G$ acts on the set of its subgroups by conjugation, what will the stabilizer of a subgroup $H$ be?

hmm let me try to make sense of this :D
hmm
do you mean g H g^-1 = H
very good question ._.

Yes, so what is the subgroup of all such $g$ called?

normalizer of H ?

Right!

12:24 AM
:DDDD
Yeeey :D

That's important in applying this stuff to Sylow theory

Hmm but in this case
the subgroups (sets) are not disjoint
they have at least 1 in them all
on my first example, we had subsets of S only

@MatheiBoulomenos how do you derive that from $\forall A \subset \Bbb R^n : \mu^\star(E \cap A) + \mu^\star(E^C \cap A) = \mu(A)$?

You should forget that they are subsets and think of them as points. The partition of $\mathcal P(S)$ that you get is made of subsets that consists of points, but in this case these points are themselves subsets of some set $S$. But this is not important, it's only confusing.
@LeakyNun what exactly?

@MatheiBoulomenos that there's a closed subset such that see above
32 mins ago, by Leaky Nun
"We can find a closed subset $F_j$ of $E_j$ with $\mu^*(E_j-F_j) \le \varepsilon / 2^j$"

12:30 AM
@MatheiBoulomenos okay ill continue with your second example, of subgroups =P

You haven't told me what $\mu^*$ is, so I assumed it was an inner regular measure

I mean more like continue with sylow :D

@MatheiBoulomenos just any outer measure
let's say it's the Lebesgue outer measure

I read somewhere that $C(\{1,...,n\})$ is $*$-isomorphic to the $n \times n$ diagonal matrices. What does $C(\{1,...,n\})$ denote?
Does it denote all continuous functions from $\{1,...,n\}$ to $\Bbb{R}$?

Here's a cute problem I found: If $H$ and $K$ are normal subgroups which intersect trivially, then each element of one commutes with each element of the other

12:43 AM
That's a pretty important lemma

@user193319 I don't think so? $\{1,\ldots,n\}$ is usually considered to by default to have the discrete topology so every map is continuous. That'd mean this set is $\mathbb{R}^n$, which doesn't sound terribly isomorphic to $M_n(\mathbb{R})$
What's this * operation?

"Isomorphic as a C^* algebra" presumably
algebras of functions are commutative so that's certainly not a matrix algebra

I was thinking $f \mapsto diag(f(1),...,f(n))$ would be the isomorphism.

Both dami and I missed "diagonal matrices"
You are correct
and this does refer to continuous functions

hi mike
long time no see

12:47 AM
hi

@MikeMiller Okay! Very good! And the $*$ operation on $C(\{1,...,n\})$ would be $f(x)^* = \overline{f(x)}$, and on the $n \times n$ diagonals it would just be the conjugate transpose?

i keep writing backwards lower case deltas
$\delta$

sup dudes

Yo

12:52 AM
yo

playing some zelda
proving some limit theorem thing

which zelda

oot

The limit theorem zelda!

one of my top 4 zeldas

12:53 AM
yeah the one with the epsilons and the deltas
what if the plural of zelda is zeldae
or zeldi?

The triforce is an artifact that allows you to make every series converge absolutely on demand

1:16 AM

I 100% agree

Huh, never heard of that

@MikeMiller Here is a quote that confuses me: "Every automorphism on $C(\{1,...,n\})$ is a composition with a homeomorphism on $\{1,...,n\}$" My question is, a composition with a homeomorphism and what else? Also, do they mean a homeomorphism from $\{1,...,n\}$ to $\{1,...,n\}$?

Morning

1:35 AM
Anyone here familiar with backpropagation?
Maybe it is not necessary to be familiar with it to answer this question.
Suppose we have functions f, g and h.
g is defined in terms of f.
And we want to find the derivative of h with respect to f.
Now, there's must be a property in calculus which relates this derivative to an expression involving a sum of the partial derivatives of g with respect to f.
Which rule of calculus is it?
I know it's the chain rule, I'm just trying to understand where that sum comes from.

4 hours later…
5:32 AM
[Random]
Some thoughts about the ordinal collapse function as an operator:

5:54 AM

6:44 AM
\begin{align}
C^{}_{\alpha}(A) & = \langle A,+,\omega_{\cdot}\psi_{\cdot}(\{\forall\eta \in A: \eta < \alpha\})\rangle\\
C^0_{\alpha}(S_0) = S_0 & =\{0,1\}\\
C^{n^+}_{\alpha}(S_0) & = C^{}_{\alpha}(C^{n}_{\alpha}(S_0))\\
C_{\alpha}^{\omega}(S_0) & = \bigcup_{n <\omega}C^{n}_{\alpha}(S_0)\\
\psi_{n} (\alpha) & = \gamma = \limsup (C_{\alpha}^{\omega}(S_0)) > \omega_n \wedge \forall \eta < \gamma : \eta 2 < \gamma
\end{align}
The original non operator form by simpleart:
We can make this larger as follows:
\begin{align}
C^{}_{\alpha}(A) & = \langle A,+,\omega_{\cdot}\psi_{\cdot}(\{\forall\eta \in A: \eta < \alpha\})\rangle\\
C^0_{\alpha}(S_0) = S_0 & =\{0,1\}\\
C^{n^+}_{\alpha}(S_0) & = C^{}_{\alpha}(C^{n}_{\alpha}(S_0))\\
C_{\alpha}^{\omega}(S_0) & = \bigcup_{n <\omega}C^{n}_{\alpha}(S_0)\\
\psi_{n} (\alpha) & = \gamma = \limsup (C_{\alpha}^{\omega}(S_0)) > \omega_n \wedge \forall \eta < \gamma : \eta^2 < \gamma
\end{align}
Or even larger:
\begin{align}
C^{}_{\alpha}(A) & = \langle A,+,\omega_{\cdot}\psi_{\cdot}(\{\forall\eta \in A: \eta < \alpha\})\rangle\\
C^0_{\alpha}(S_0) = S_0 & =\{0,1\}\\
C^{n^+}_{\alpha}(S_0) & = C^{}_{\alpha}(C^{n}_{\alpha}(S_0))\\
C_{\alpha}^{\omega}(S_0) & = \bigcup_{n <\omega}C^{n}_{\alpha}(S_0)\\
\psi_{n} (\alpha) & = \gamma = \limsup (C_{\alpha}^{\omega}(S_0)) > \omega_n \wedge \forall \eta < \gamma : \eta 2 < \gamma \wedge 2^{\eta} < \gamma
\end{align}
Therefore simpleart's $\psi_n$ function is a weakely inaccessible cardinal
while the $\psi_n$ in the bottommost block of the above message denotes a strongly inaccessible cardinal
Therefore, it seems that:
8

Ruby, 349 bytes, fψ0(ψ9(9))(9) where f is the fast growing hierarchy and ψ is an ordinal collapsing function described below. Try it online! def f(b,n=0)n<1?(b.class!=Array):(n.times{n+=b==1?n:f(g(n,b),n)};n)end;def g(n,r)a,b=r;f(r)?(r>1?r-1:n):r.size>2?h(n,a,b):f(b)?(b>1?[a,b-1]:b>0?a:[a,n]):...

To get a number bigger than TREE(3), somehow a weakly inaccessible cardinal is needed
As for how to get larger than TREE(3) predicatively, my guess is that the fast growing hierarchy is perhaps the only way (to be checked)
Now, the following is suspected to be highly circular:

7:04 AM
\begin{align}
C^{}_{\alpha}(A) & = \langle A,+,\omega_{\cdot}\psi_{\cdot}(\{\forall\eta \in A: \eta < \alpha\})\rangle\\
C^0_{\alpha}(S_0) = S_0 & =\{0,1\}\\
C^{n^+}_{\alpha}(S_0) & = C^{}_{\alpha}(C^{n}_{\alpha}(S_0))\\
C_{\alpha}^{\omega}(S_0) & = \bigcup_{n <\omega}C^{n}_{\alpha}(S_0)\\
\psi_{n} (\alpha) & = \gamma = \limsup (C_{\alpha}^{\omega}(S_0)) > \omega_n \wedge \forall \eta < \gamma : \eta 2 < \gamma \wedge \eta^2 < \gamma \wedge 2^\eta < \gamma \wedge \omega_\eta < \gamma \wedge 2^\eta < \gamma \wedge \beth_\eta < \gamma \wedge \beth_\eta < \gamma \wedge \text{$\gamma$ is a Sup
Anyway, that's enough rambling about the ordinal collapse function for now.
Now to reprove that there cannot be a largest ordinal (because I failed to find that proof in the transcript)
Let the foundation be some generalisation of ZF with proper classes: Let $\Bbb{On}$ be the class of ordinals
Suppose there exists a proper class $\infty$ such that it is a fixed point under all possible nondecreasing map (including the powerset operation. We insert an extra axiom in this set theory so that the only object which violates cantors theorm is $\infty$, in particular $|\mathcal{P}(\infty)|=|\infty|$). We want to show that the resulting construction is inconsistent unless the proper class of ordinals form a cyclic ordering (and thus violating the axiom of regularity)
Recall that in ordinal multiplication, $\alpha0=0\alpha=0$ because the cartesian product is empty if either input sets is an empty set
Now by the definition $\alpha \infty =\infty$ for any $\alpha \in \Bbb{On}$. what about $\infty \alpha$? If $\infty \alpha > \alpha$ by the nondecreasing property of ordinals, then this contradicts the fact that $\infty$ is the maximum. If $\infty \alpha =\infty$ we have another problem:
Note that for the second condition: $\infty \alpha = \alpha \infty$ for all $\alpha \in \Bbb{On}$. This means, $\infty$ act as a (two-sided) absorber. But we knew in semigroup theory if there is a two sided absorber, then it must be unique
Therefore $\alpha\infty = \infty = 0\infty = 0$ and thus we have $\infty = 0$

7:22 AM
@Secret aren't you getting tired from all that typing?

this contradicts with the extra axiom that is added because $\mathcal{P}(0)=\{0\}=1$. Therefore the extra axiom is inconsistent thus we discard it and go back to ZF extended with proper classes
Jenna: Not when it is on the main computer
(cont.) Now let's check the resulting structure we end up with:
0,1,2,....$\omega$,...,$\omega_{\omega}$,...,$\beth_{\beth_{\omega}}$,..........‌​....,0
This violates the axiom of regularity, hence it is inconsistent in ZF
and therefore there cannot be a largest infinity in ZF
QED

Isn't $0\infty$ undefined?

Uh, in the above proof, $\infty$ is not the infinity in analysis, but it is assumed to be an ordinal which is a fixed point of any nondecreasing map
the proof want to show that a largest infinite ordinal cannot exist in ZF even if we don't care about the powerset axiom
Now whether the existence of $\infty$ is consistent for ZF-Regularity I have no idea, cause I am not good at working with non well founded set theories

I have no idea what "nondecreasing map" and "powerset axiom" and "ZF" mean.

A nondecreasing map is a function $f$, such that $x < y$ means $f(x) \leq f(y)$.

Powerset axiom is an axiom in some set theory which said that given a set S, the set of all subset of S, which is denoted as $\mathcal{P}(S)$ exists.

ZF is Zermelo–Fraenkel set theory, which is tied to the foundation of mathematics and trying to express every mathematical objects as sets

7:40 AM
I'll just assume you know what you're doing

Some update: It is known that $\alpha^0=1$ because the empty cartesian product is a singleton because the only function that maps from the empty set to itself is the empty function. This means at the present definition, even under a non well founded version of ZF the existence of $\infty$ is still inconsistent because then $\alpha^0=0\neq 1$. Even if we ignore that and redefine $\alpha^0=0$, it will not help because then ordinal exponentiation will be broken as the inductive case means
$\alpha^{\beta}=0$ for all $\beta \in \Bbb{On}$
Now, whether having "the largest infinity" (technically incorrect since we have shown above that it will lead to $\Bbb{On}$ taking a cyclic ordering and thus became a non well founded set) by defining it to be fixed under any nondecreasing definable map except exponentiation by zero is consistent needs to be checked more closely before this question is to be asked on main

7:58 AM
@Secret In ZF-P you should be able to get $[0,\omega_1)$ as ordinals but no further

For starters, we can still have cantor's theorem being valid since $2^0=1$ and we can let $0$ to be at least a strongly inaccessible cardinal so that no power set of any ordinal larger than it in the ordering can reach it.
@AlessandroCodenotti Ah right (because $\omega_1$ existence need powerset), but now I am wondering ZF-Regularity
it seems it is consistent and construcible that the ordinals becomes a cyclic ordering with 0 being both a finite ordinal and strongly inaccessible
wait a sec..., I think in doing so I will violate the property of ordinals being heredity transitive sets (because the emptyset has no elements, let alone being transitive). or the definition of emptyset (because the emptyset has no elements)
unless making 0 strongly inaccessible (thus unreachable by any powersets) somehow helps evades the problem

Ordinals needs to be well ordered by $\in$
And stuff violating regularity most likely isn't

37 mins ago, by Secret
0,1,2,....$\omega$,...,$\omega_{\omega}$,...,$\beth_{\beth_{\omega}}$,..........‌​....,0
right, so this construction will not be ordinals by definition, but just some cyclically ordered set

1 hour later…
9:21 AM
Is it possible to define transfinite induction without using heredity transitive sets. That is, does there exists a structure with an ordering isomorphic to the ordinals but without the property of being heredity transitive?
One thing I never understood about "Ordered under $\in$" is why we are doing that besides the reason that the will ensure every number can be written as a set and limit ordinals can be wrote in terms of unions
I am fine having all my structures not ordered under $\in$ if there is somehow a way to reproduce the well ordering of the ordinals, but throwing away the requirement of "under $\in$". Probably set theory just cannot do that job for me. I need something more expressive such as type theory

1 hour later…
10:35 AM
in Mathworks, 32 secs ago, by Secret
I always thought a well ordered collection will mean everything is laid out in discrete chunks on the table and I can easily pick any element I want because they are all uniquely labelled somewhat by the ordering
life will be much easier if I can say $\omega +1 = \omega^+$
and not $\omega+1 = \{0,1,2,3,4,5,...\} \cup \{\omega\}$

10:55 AM
Rant: Why should I care about all the other elements when the element I am interested in is just one of them and possibly its nearest neighbours

@Secret can't you?
@JennaSloan of course he does

@Secret As @LeakyNun said, you trivially can. I think you still don't get the point that if all you want is well-orderings and don't care at all about comparing them, then you have very trivial operations. + is just concatenation.
You just have to deal with the fact that there are uncountably many well-orderings isomorphic to the naturals.

In such framework, those uncountably many well orderings are isomorphic, but cannot be shown to be order isomorphic to the naturals?

You asked this before and I already said they can be shown.

@user21820 what do you mean?

11:03 AM
ah ok, I think I got my brain tied into knots again

And I would prefer if you kept all your thoughts in one place. I don't like having two separate chats with overlapping discussions. =)
@LeakyNun Go to Mathworks and I'll show you.

12:05 PM
Hi!!

To create a team of 5 players out of 11 players there are $\binom{11}{5}$ ways, right?

If we have 7 starters, 15 main dish and 6 dessert, how many different
3 course menus can we create? In this case we cannot use the binomial coefficient, can we?

user84215
Hi.

We have 7 posibilities for the starter, 15 posibilities for the main dish and 6 possibilities fir te dessert. Since they are independet, the number of combinations is equal to the product $7\cdot 15\cdot 6$. Is this correct?

Marystar: seems so

Ok, thanks!

@Secret btw, MathJax doesn't work on codegolf.SE.
Oh hello @Mr.Xcoder

12:21 PM
Hello @SimplyBeautifulArt
I have this problem I don't know how to approach well... Calculate $\big(\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}+...+‌​\frac{\sqrt{10}-\sqrt{9}}{\sqrt{90}}\big):\frac{\sqrt{10}-\sqrt{9}}{\sqrt{10}}$. At first glance, I thought I should write the formula for the general term of that sum, but I didn't really make ends meet.

So you are looking at: $$\sum_{n=1}^9\frac{\sqrt{n+1}-\sqrt n}{ \sqrt{n+1}}$$?

@SimplyBeautifulArt Yes (we didn't learn about sigma-summation yet)

What do you mean by "calculate" it?

Find its numerical value.
Any hints on how I should approach this?

Well, if you want a numeric value, you could always binomial expand.

12:33 PM
@SimplyBeautifulArt The final result of that sum only would be $$\frac{270\sqrt{14}-10\sqrt{105}-12\sqrt{70}-15\sqrt{42}-18\sqrt{35}- 20\sqrt{21}-60\sqrt{3}-70\sqrt{7}-105}{30\sqrt{14}}$$ :P

Ew :P
Also, use some spacing in it to avoid things like that $\sqrt{21\quad}$ from occurring.

I might want to multiply top and bottom by $\sqrt{n+1}$ to get rid of the squareroot denominator
$$\sum_{n=1}^91-\frac{\sqrt n}{ \sqrt{n+1}}$$ looks easier

@Secret Yes I tried, that would get me $$\frac{n+1-\sqrt{n(n+1)}}{n+1}$$ wouldn't it?

$$\sum_{n=1}^9 1-\frac{\sqrt {n(n+1)}}{n+1}$$

Which means $1-\frac{\sqrt{n(n+1)}}{n+1}$

12:39 PM
ok nvm that is not very simplified

Amplifying with the conjugate would give us $$1-\frac{(n-1)\sqrt{n(n+1)}}{n^2-1}$$which wouldn't help at all...

The expression $\sqrt{n+1}-\sqrt{n}$ looks familar, but I don't remember where I have seen that before

$$\sqrt n=\sqrt{n+1}-\frac1{2\sqrt{n+1}} +\mathcal O((n+1)^{-3/2})$$

I still have no idea what to do next but the first term is $1-\frac{\sqrt{1}}{\sqrt{2}}$
@SimplyBeautifulArt @Secret Wait we were all wrong
16 mins ago, by Mr. Xcoder
I have this problem I don't know how to approach well... Calculate $\big(\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}+...+‌​\frac{\sqrt{10}-\sqrt{9}}{\sqrt{90}}\big):\frac{\sqrt{10}-\sqrt{9}}{\sqrt{10}}$. At first glance, I thought I should write the formula for the general term of that sum, but I didn't really make ends meet.
Look at the denominators...

facepalm.
That certainly helps
$$\sum_{n=1}^9\frac{\sqrt{n+1}-\sqrt n}{\sqrt{n(n+1)}}=\sum_{n=1}^9\frac1{\sqrt n}-\frac1{\sqrt{n+1}} =1-\frac1{\sqrt{10}}$$

12:44 PM
Umm, let's see. The general term is $$\frac{\sqrt{n+1}-\sqrt{1}}{\sqrt{n(n+1)}}$$...
@SimplyBeautifulArt I don't understand the transition between the last two steps.

simpleart: Ah, partial fractions

Let me try to write it myself first

@Mr.Xcoder I simplified the first fraction, then it telescopes.

teleswhat? Sorry don't understand half the terms you use :P
@SimplyBeautifulArt Wait, you're not a mod anymore?

@Mr.Xcoder I was never a mod lol

12:46 PM
$$=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}$$
@SimplyBeautifulArt ... facedesk right

@Mr.Xcoder $$(a-b)+(b-c)+ (c-d)+\dots +(y-z) = a-z$$

Oh yeah facepalm, telescopic summation...

@Mr.Xcoder I ran in the election though

(I am dumb, very dumb)

66

The 2017 elections are over. By analyzing the OpaSTV file with the voting data, the result of the 2017 moderator elections are as follows (with no particular order): In the name of all the users on Mathematics SE, congratulations on your victory! You can view a summary report of the election...

2

12:49 PM
$$=\sqrt{\frac{n+1}{n(n+1)}}-\sqrt{\frac{n}{n(n+1)}}$$
$$=\sqrt{\frac{1}{n}}-\sqrt{\frac{1}{n+1}}$$
@SimplyBeautifulArt Am I right so far ^?

Now, the sum is $$\sqrt{\frac{1}{1}}-\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}}-\sqrt{\frac{1}{3}}+..‌​.+\sqrt{\frac{1}{9}}-\sqrt{\frac{1}{10}}$$, right?
$$=\sqrt{\frac{1}{1}}-\sqrt{\frac{1}{10}}=1-\sqrt{\frac{1}{10}}$$
And that gives $$\big(1-\sqrt{\frac{1}{10}}\big):\frac{\sqrt{10}-\sqrt{9}}{\sqrt{10}}$$

Yup
Woohoo, time to golf my code a bit more.

@SimplyBeautifulArt What code? Maybe I can help.

12:54 PM
@Mr.Xcoder You can use \left( ... \right) or \big( ... \big) or \bigg( ... \bigg)
8

Ruby, 349 bytes, fψ0(ψ9(9))(9) where f is the fast growing hierarchy and ψ is an ordinal collapsing function described below. Try it online! def f(b,n=0)n<1?(b.class!=Array):(n.times{n+=b==1?n:f(g(n,b),n)};n)end;def g(n,r)a,b=r;f(r)?(r>1?r-1:n):r.size>2?h(n,a,b):f(b)?(b>1?[a,b-1]:b>0?a:[a,n]):...

@SimplyBeautifulArt I use \big(...\big)

Personally I prefer \left(...\right), since it auto-fits the right size.

No I cannot help, it's in Ruby :'-(
Too late to edit now

I'm just going to make two functions into one twice

Yuppe my final result is correct according to the book, $2\sqrt{10}+7$. Thanks a lot @Sim

1:01 PM
No problem @Mr.Xcoder

The decimal value of the above however, is very interesting.
Namely $13.32456$ (notice the digits after the dot)
They do however have a weirder path if we do more precise calculations ($13.3245553203$)

:P Define pattern

@SimplyBeautifulArt ಠ_ಠ

[Random]

I could be lacking rigour by defining pattern as the deltas between consecutive elements (digits)

1:08 PM
Prove that there are no real transcendentals that is a sum of a real transendental and a complex algebraic number:
Proof:
Let $r,s \in \Bbb{R\backslash Q}$ and $z \in \Bbb{C}$. Suppose $r=s+z$. Then:
$0=\text{Im}(r) = \text{Im} (s) + \text{Im} (z) = \text{Im} (z)$
$r=\text{Re}(r) = \text{Re}(s) + \text{Re}(z) = s + \text{Re}(z)$

@Mr.Xcoder Woo hoo, I cut down about 3 bytes X'D

But we require $z$ to be a complex number with both nonzero real and imaginary parts, therefore contradiction

@SimplyBeautifulArt woah... :P

Oh nvm

1:13 PM
Only 1 byte

It's not running with all those errors

@Secret wdym?

@Secret It should work theoretically.

.code.tio:1:in f': undefined local variable or method a' for main:Object (NameError)
from .code.tio:1:in block in f'
from .code.tio:1:in times'
from .code.tio:1:in f'
from .code.tio:1:in <main>'

Real time: 0.071 s
User time: 0.053 s
Sys. time: 0.014 s
CPU share: 95.17 %
Exit code: 1

Oh whoops
There we go
It now runs
Without errors (ignore the overflow)
I wonder if anyone else is going to post a solution/program to that question.

1:17 PM
.code.tio:1:in i': stack level too deep (SystemStackError)
from .code.tio:1:in i'
from .code.tio:1:in i'
from .code.tio:1:in i'
from .code.tio:1:in i'
from .code.tio:1:in i'
from .code.tio:1:in i'
from .code.tio:1:in i'
from .code.tio:1:in i'
... 8278 levels...
from .code.tio:1:in block in f'
from .code.tio:1:in times'
from .code.tio:1:in f'
from .code.tio:1:in `<main>'

Real time: 0.860 s
User time: 0.791 s
Sys. time: 0.041 s
CPU share: 96.76 %
Exit code: 1
Wish our universe had at least countable memory

It does have countable memory available?

I mean, countably infinite

:P
@Secret Love how it never escapes the i function
Much recursion.
Actually, it does escape the i function
But it spends the majority of the time going through i.
I hope no-one is bothered by my repeated bumping of the question via edits.

[Random]

1:39 PM
\begin{align}
& 0\\
& 1\\
& 2\\
& 3\\
& ...\\
& (0,1,2,3,...)\\
& (0,1,2,3,...,0)\\
& (0,1,2,3,...,0,1)\\
& (0,1,2,3,...,0,1,2)\\
& ...\\
& (0,1,2,3,...)(0,1,2,3,...)\\
& (0,1,2,3,...)(0,1,2,3,...)(0,1,2,3,...)\\
& ...\\
& (\omega,\omega 2,\omega 3,\omega 4,...)\\
& (\omega,\omega 2,\omega 3,\omega 4,...,\omega)\\
& (\omega,\omega 2,\omega 3,\omega 4,...,\omega,\omega 2)\\
& ...\\
& (\omega,\omega 2,\omega 3,\omega 4,...)(\omega,\omega 2,\omega 3,\omega 4,...)\\
& (\omega,\omega 2,\omega 3,\omega 4,...)(\omega,\omega 2,\omega 3,\omega 4,...)(\omega,\omega 2,\omega 3,\omega 4,...)\\

Can any one able to create a tag GATE, NBHM, ISI, GRE, TIFR_GS, etc
otherwise questions get repeated
so many duplicates
or Simply a Tag 'Indian Exam'

not only gre
required Indian entrance examination tag

I don't see why we should have those tags

so many duplicated questions.

1:44 PM
Why should we have a tag dedicated to India's entrance exams?
@ManeeshNarayanan Why should that have anything to do with the tags?

or simply Math Exam

Similar questions will have similar tags.
Also, we have a GMAT tag.

It is also an entrance exam.

No, I will not, unless you explain to me why we need these tags.

Note that $C(\{1,...,n\})$ denote all of the continuous functions from $\{1,...,n\}$ to $\Bbb{C}$. Here is a quote that confuses me: "Every automorphism on $C(\{1,...,n\})$ is a composition with a homeomorphism on $\{1,...,n\}$" My question is, a composition with a homeomorphism and what else? Also, do they mean a homeomorphism from $\{1,...,n\}$ to $\{1,...,n\}$?

1:47 PM
ok.

1:57 PM
A tag of Competitive exam would be a nice one I think which could comprise of a set of questions dedicated to competitive exams also the approach to such questions will be a kind of different! just a thought

@BAYMAX Similar to the (contest math) tag.

bleh, Fourier series convergence. too slooow

I say Fourier series, but what I really mean is the basis $\{f_n\}_{n=1,3,5,\cdots}$ where $f_n(x)=\sqrt{\frac{2}{L}}\cos\frac{n\pi x}{L}$ on $x\in [-L/2,L/2]$
so it's a bit more restricted.
And I'm interested in representing the function $\sqrt{\frac{2}{a}}\cos\frac{\pi x}{a}$ for $x\in[-a/2,a/2]$ (and zero everywhere else) with $a\ll L$
which, unfortunately for convergence, is 1) not smooth at $x=\pm a/2$ and 2) localized to a small region within the domain. not good for convergence

@ManeeshNarayanan You can find a few related discussion about similar tags here: chat.stackexchange.com/transcript/message/40678621#40678621

2:13 PM
(I think that if I make $a$ twice as small while leaving $L$ fixed then the number of terms required for a decent approximation doubles?)

2:42 PM
Anyone interested in taking a shot at answering this question? math.stackexchange.com/questions/2482821/automorphisms-of-c-1-n

so i m good then contest math is good too
@SimplyBeautifulArt :)

@BAYMAX =P

@Semiclassical Yeah convergence of Fourier series becomes faster when you have more and more regularity I think.

yeah, and less localization

3:16 PM
Hi
Any can help me?
I have created an equation for successions, for find a desired number of succession
for example, i want the nineth number of the succession
I made 4 equations, each for successions with different operations, + - * /
but my problem is do the equations generally, only reemplazing the pattern of the succession

hi

and not, specific equations for each operator
Hi

what is the property called where if real $a > 0$ that there exists a real $x$ such that $0 < x < a$

@MeowMix linear order of reals is dense
aka take $x=a/2$

should i state that in my proof? or can i just let it be obvious
the case here is where i have something along the lines of $L < \varepsilon$ such that $L > 0$ and i need to say there exists an $\varepsilon$ where this inequality doesnt hold

3:59 PM
> The quaternion multiplicative inverse f 2 ( q ) = q − 1 {\displaystyle f_{2}(q)=q^{-1}} {\displaystyle f_{2}(q)=q^{-1}} is another fundamental function, but it raises difficult questions such as “What should f 2 ( 0 ) {\displaystyle f_{2}(0)} {\displaystyle f_{2}(0)} be?” and “What q {\displaystyle q} q solves the equation f 2 ( q ) = 0 {\displaystyle f_{2}(q)=0} {\displaystyle f_{2}(q)=0}?”
What? 0 cannot have a multiplicative inverse in quaternions

4:35 PM
[Random]
Some inaccessible ordinal
Let $\alpha$ be (insert suitable description) inaccessible ordinal defined by:
$\alpha := \forall \eta < \alpha, \bigcup_{\eta < \alpha}\eta < \alpha$
Let $\beta$ be (insert suitable description) inaccessible ordinal defined by:
$\beta := \forall \eta < \beta, \bigcup_{\eta < \beta}\beta < \beta$
In words, $\alpha$ is an ordinal not reachable from any supremum of ordinals less than it, and $\beta$ is an ordinal that is of of confinality at least $|\beta|$ and its fundemental sequence is a constant sequence consists of itself
In particular, $\beta$ has an unusual property that even if the sequence has length $|\beta|$ its union will be less than $\beta$ if there is no $\beta$ in the sequence
However, since an ordinal cannot contain itself, thus the fundemental sequence must consists of ordinals less than $\beta$, this makes $\beta$ a kind of inaccessible ordinal

2

Which of the following matrices are non-singular? $I + A$ where $A$ not equal to $0$ is a skew-symmetric real $n\times n$ matrix, $n\geq 2$. Every skew-symmetric non-zero real $5 \times 5$ matrix. Every skew-symmetric non-zero real $2 \times 2$ matrix.

which means what I just said about $\beta$ is nonsense and the only sensible thing here is $\alpha$
Gah, I really need to focus more on programming...

From where did %Ax=x$come from 4:51 PM 0 If I am not mistake,$C(\{1,...,n\})$denotes all of the continuous functions from$\{1,...,n\}$to$\Bbb{C}$. I read somewhere that every automorphism on$C(\{1,...,n\})$is a composition with a homeomorphism on$\{1,...,n\}$. My question is, a composition with a homeomorphism and what else? Als... Anyone interested in taking a shot at answering this question? @ManeeshNarayanan observe that det(A) = 0 not given det(A) is not granted$A$is skew symmetric det(A)=0 is not granted @ManeeshNarayanan do you know what "observe" means? it means it is not granted but you can deduce it 4:55 PM I was asking 1. Explanation given by David Giardo Sir In the answer of 1. it is given$Ax=x$I was wrong there sorry,$det(A) = (-1)^d det(A)$where$d $is order of matrix$A$so if$d$is odd then$det(A) = 0\$