Mathematics

12:01 AM

@Eran Cayley's theroem

Every group acts transitively on itself by left multiplication

12:23 AM

@orbit-stabilizer I'm not the best at the dihedral group either, and the only practice I know of is to just find a book with good problems and work through it

Adeek

12:48 AM

Hey there is something stupid that I am stuck on

orbit-stabilizer

@Daminark sounds good

Adeek

I am just trying to understand why is this goes from $C^{2}(X)$ to $C^{0}(X)$?

Daminark

Each of the derivations knocks down a dimension

So if you compose two it'll knock down two dimensions, an then you take the difference as usual

user193319

Claim: If $F \subseteq \Bbb{R}$ is such that $m(\Bbb{R}-F) = c < \infty$, then $F$ is uncountable and unbounded. Proof: If $F$ were countable, then $c = m(\Bbb{R}-F)=m(\Bbb{R}) - m(F) = \Bbb{R}$, which is a contradiction. Again, if $F$ were bounded, then $m(F) < \infty$, then $c = m(\Bbb{R} - F)$ implies $m(\Bbb{R}) = c + m(F) < \infty$, which is another contradiction. Hence, $F$ is uncountable and unbounded. Does this sound right?

Oh, by the way, $m()$ denote the Lebesgue measure.

Adeek

1:04 AM

yeah but we are almost knowing all dimensions ?

@Daminark can yeah I agree with this but here it goes from $C^2$ to $C^0$

Daminark

I mean maybe it's that you're dealing with surfaces or something? Like in principle it'd be l to l-2

Adeek

yeah

I need to stop focusing on such minor details

it is stupid

I think we are picking here a specific l

being l = 2

Mathematica

1:24 AM

hi need help solving (z+i)^5 = 1

orbit-stabilizer

@Daminark Let $G$ act transitively on $X$. Let $N\triangleleft G$. Show that any two $N$-orbits in $X$ have the same cardinality.

Nvm, found an answer: math.stackexchange.com/questions/533199/…

Ah, but what if G is infinite...

Bvss12

1:39 AM

I have a serious problem trying to prove $\bigcap_{i=1}^kker(f_i)\subset ker(f)\iff f\in Span(f_1,...,f_k) $ Can someone help me :( ?

berrygreen

@Mathematica, $e^{5\log(z+i)}=1$, then use the complex definition of the log function

Kevin Driscoll

1:57 AM

@Bvss12 It isnt clear what your notation means. What are $f$ and $f_i$? Im also not sure what $Span(f_1, . . ., f_k)$ means; usually I only see span with elements of a vector space, but your $f$ look like they are maps..

Faust

2:09 AM

Hows everyones exams going?

Kevin Driscoll

Not goin yet here

Bvss12

@KevinDriscoll math.stackexchange.com/questions/2546630/… Thanks Kevin for answering.

I asked the question and found a pretty good proof in this link

But that's part is not too clear for me...

Daminark

@orbit-stabilizer never think about infinite groups, you lose cardinality arguments

orbit-stabilizer

@Daminark lol

10 Replies

2:41 AM

How do I represent a function in in the form of a/(1-bx) as a power series?

nvm, I think I got it... Would it be a/(1-bx) = Sum[n=0,inf]: a(bx)^n ?

What have I done wrong?

nvm

I miscalculated the interval of convergence

I hope you guys are ok with me using this chat like a rubber ducky

nobody actually online here to stop me ;)

Actually legit need help here: The hell am i doing wrong?

plz halp?

10 Replies

3:38 AM

pls?

Adeek

4:01 AM

@Daminark hey

Bvss12

4:28 AM

Hey

Daminark

4:44 AM

Hey @Adeek!

orbit-stabilizer

he has left us

when's your group theory exam @Daminark

Daminark

Friday

orbit-stabilizer

Next week?

Feeling ready?

Daminark

Yeah, and probably

I was okay on the first midterm, better on the second, so hopefully this will go nicely

orbit-stabilizer

That's great!

Mine's Tuesday and I'm not ready at all

Daminark

4:57 AM

Damn, well good luck

Secret

Associahedra: The Shapes of Multiplication | Infinite Series

►

orbit-stabilizer

Currently learning nilpotence and solvability. Sighhh

Wow that was fast entrance and post @Secret

Secret

Hmm, I knew algebra can be visual, but I never thought associative algebra can be done similarly

orbit-stabilizer

Wait

I think I was just reading her blog earlier today!

Secret

@orbit-stabilizer I used to that. Sometimes, I drop a link to seed a discussion and then disappear, other times I just sit back and watch the discussion unfolds

Silent

4:59 AM

10 hours ago, by Silent

Hey! Let $F$ be an archimedean ordered field. Suppose $\bar F$ be Dedekind completion of $F$. Does $\bar F$ has to be archimedean?

Please have a look at this

orbit-stabilizer

taidanae-bradley-7bou.squarespace.com

@Silent sorry, I don't know Dedekind completion

MatheinBoulomenos

@Silent I think the answer is yes, as $F \subset \bar F$ is dense (right?)

Secret

@orbit-stabilizer That one is a subset of the cartesian product between two shapes. In general, if you do a cartesian product between two 2D objects, you will get a 4D one, so to make a torus, it is a subset of that cartesian product. I forgot the details but I will revise later

Dedekind cuts cannot really create infintesimals from a set that does not have any, do they?

Daminark

Oh no visual stuff I'm outta here bye everyone

Secret

lol

Daminark

5:05 AM

Okay not really but lol

Secret

When I finally get back to algebra, I should explore both symbolic and visual ways to represent them. By that time, I will be ok with infinities enough to visualise most dense structures and infinite groups I hope

orbit-stabilizer

I think topology is pointless.

Secret

Not really if you know what kinds to focus on

On a more joking note, there exists pointless topology

Daminark

@orbit en.wikipedia.org/wiki/Pointless_topology

Dammit Secret

orbit-stabilizer

Twas a joke

Daminark

5:08 AM

I can't have you sniping me now

Also hey @Mathein!

MatheinBoulomenos

Hey @Daminark

Secret

The cohomology discussion in recent weeks confuse me like hell, though,which is why I am kinda less active on main and mostly in my own rooms

Daminark

But yeah I dunno, I've been less than interested in the very fancy stuff in point-set, like blah only works in $T_{screw\ you}$ spaces

Secret

well actually, cohomology is still ok because I have been to some lecture. It is those fundamental groups stuff that really confuse me

Daminark

Fundamental groups are an easier concept than cohomology, I'd say

The idea is that you say two functions $f,g:X\to Y$ are homotopic if there is some $h:X\times I \to Y$ such that $h(x,0) = f(x)$ and $h(x,1)= g(x)$

(Everything here being continuous ofc)

So, one thing you have in topology is called the compact-open topology

And these things called compactly generated spaces

Secret

5:13 AM

so all open sets in those topologies are compact sets?

Daminark

Nope

I don't know much about it

But it's something like

Choose $K\subset X$ compact and $U\subset Y$ open, and let $V(K,U) = \{f:X\to Y |\ f(K)\subset U\}$

And let that be a subbasis for the topology

Or something vaguely to that effect

Perturbative

Aren't those called $K$-spaces or something @Daminark

Also morning everyone!

Daminark

The point is, when you let your spaces be compactly generated, it turns out that currying becomes an isomorphism

$C(X\times Y, Z) \cong C(X,C(Y,Z))$

Perturbative

"currying becomes an isomorphism"

Daminark

That's a homeomorphism of topological spaces

MatheinBoulomenos

5:16 AM

Yeah, that's basically the main reason why one should care about compactly generated spaces

Perturbative

Oh shit currying is actually a legit term

Daminark

Currying means you let $f(x,y) = f(x)(y)$

Perturbative

I thought it was a pun on curries or something

orbit-stabilizer

Currying is used when you look at group actions. You have a homomorphism from G to the symmetric group and you have a map from G x X to X that satisfies a couple properties. The relationship between the two is from currying

Perturbative

Ahhh I see

Daminark

5:20 AM

@Mathein and @Perturbative what's going on with you?

MatheinBoulomenos

@Daminark doing fine, thanks. Could use some more sleep I guess. And yourself?

Perturbative

@Daminark It seems that I've lived up to my stereotype, by being Indian and hungry and mistaking "currying" for a pun on actual curry

Daminark

Doing aight. My finals aren't until Friday so I've got some time to rest

@Perturbative OH LOL

Perturbative

Apart from that, I'm doing aight

Daminark

I mean the food curry is the fake one and currying as above is real ofc

Silent

5:24 AM

@MatheinBoulomenos I searched for that, and I got this question, saying that no non-archimedean field is dedekind complete.

MatheinBoulomenos

@Silent interesting

Silent

Well, in that question, we are starting with non-archimedean field, while I was asking if we can end up with non-archimedean field

MatheinBoulomenos

@Daminark @Perturbative obviously, Haskell Curry was not only an important mathematician and theoretical computer scientist who inspired multiple programming languages, but he also made an important contribution to Asian cuisine

Daminark

Lmao

MatheinBoulomenos

@Silent As I wrote above, I think this should follow more or less directly from the fact that any poset is dense in its Dedekind completion

Captain Bohemian

5:31 AM

i just ate curry 10 hours ago, and am hungry again. I am from Asia. I don't know if curry is only food in Asia.

Narcissus jewel

@CaptainBohemian Curry is a global object.

Captain Bohemian

@MatheinBoulomenos wow, can a mathematician and programmer be interested in cuisine? I thought scientists are generally not interested in cuisine.

MatheinBoulomenos

I was just joking

Captain Bohemian

no wonder

BAYMAX

Hi chat!

If we have any set $A \subset \Bbb{R}$, then any method by which we can say that $A$ has an open interval?
Like if we are given this set----> $A = \{\sum_{i=1}^{\infty}\frac{a_{i}}{5^i}, i =0,1,2,3,4\}$

Daminark

5:43 AM

Oh my lord

Perturbative

@BAYMAX $A$ has an open interval or $A$ is an open interval?

Secret

I wish there's a way to separate the geometric series portion from the arbitrary series $a_i$. If we can do so, then if $a_i$ does not dominate the geometric series, then the fact the partial sums of the geometric series converges to zero means the end where the limit is will become open. My guess that in such condition, A can only be a half open interval

BAYMAX

@Perturbative We need to check if $A$ contains an open interval.

Secret

what are the contents of $a_i$?

BAYMAX

sorry

A typo $a_{i} = 0,1,2,3,4$

Secret

6:00 AM

I think $A$ will be half open of the form $[0,\frac{1}{1-\frac{1}{5}})$. This is because $5^i > a_i$ for all i, thus the asymototic behaviour of the summand, hence the series will be dictated by the geometric series $\sum_{i=1}^{\infty}\frac{1}{5^i}$

desmos.com/calculator/yswbimlxyi

Actually, the open end will be something slightly larger than $\frac{5}{4}$. But as you can see, it closely follows the behaviour of the geometric series

Secret

[Integral Project] Hmm, I might be able to use methods like these to compute Waiting's many series

$$\sum_{i=1}^{\infty} \frac{b_i}{c_i}$$

Secret

Arithmetico–geometric sequence

In mathematics, an arithmetico–geometric sequence is the result of the term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression. Put more plainly, the nth term of an arithmetico–geometric sequence is the product of the nth term of an arithmetic sequence and the nth term of a geometric one. Arithmetico–geometric sequences arise in various applications, such as the computation of expected values in probability theory. For instance, the sequence ...

BAYMAX

hmm

Secret

The series you provide has a closed form. Thus yes, A should be half open with the form $[0,a)$ where $a$ is the infinite sum of the arithmetico geometric series $\frac{i}{5^i}$

DanielSank

6:52 AM

in The h Bar, 1 min ago, by DanielSank

$$\int_0^T dt_1 \int_{t_1}^T dt_2 \int_{t_2}^T dt_3 \cdots \int_{t_{n-1}}^T dt_n$$

Anyone have any idea how to do integrals like that?

Balarka Sen

Have you tried induction, or something of that sort?

DanielSank

@BalarkaSen Yeah but I'm not sure what I expect the result to be so induction is still not solid.

Balarka Sen

Ah

DanielSank

@BalarkaSen Maybe I can figure it out by evaluating the $t_n$ integral...

If the original expression is $I_n$, I get $$I_n = T I_{n-1} - \int_0^T dt_1 \int_{t_1}^T dt_2 \cdots \int_{t_{n-2}}^T dt_{n-1} t_{n-1} \, .$$

I now suspect the result should be $T^n/n!$.

Balarka Sen

Hmm

DanielSank

7:06 AM

...which perhaps we can prove by showing that this nested integral is related to a term in the Taylor series of the exponential function.

Balarka Sen

It seems you get a bad recurrence relation. $I_n = TI_{n-1} - T^2/2 I_{n-2} - \cdots$

DanielSank

Yeah but I bet if we take derivatives w.r.t. $T$ or some nonsense we get something good.

Mary Star

9:05 AM

Suppose we have a set X which is not a linear subspace. Can the span of X then have a basis?

Daminark

The span of $X$ will necessarily have a basis, since the span of any set is a vector space

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$Mathematics$ Mathematics