Mathematics - 2013-10-02 (page 2 of 3)

Nick

09:00

It holds because I derived it. Maybe I wasn't clear about that part

a*ar*ar^2...

Tobias Kildetoft

yeah, I meant the full thing you want to show, not just the product

Nick

=> a^n * r ^ ( 1+ 2 + 3 ... n-1 )

O_O ohh, right

Ok, it just seems to prove true whenever I change the value of n

Tobias Kildetoft

@Nick I mean write up the full set of terms for some small $n$ and see how they happen to match up on both sides

Nick

O_O I think I found the way!

I don't need to prove, what I thought I needed to prove.

$P^2 = (S / R)^n$ can be easily proved!

Wow, that was a heck of a lot of thought for tht.

@Tobias: I'm sorry I made you scream

$S/R = a{ ( r^n - 1) / (r - 1)} / [1/a ( { ( r^{n - 1}) / (r - 1) } * r^{(1 - n)} ) ] = a^2 * r^{(n-1)}$

So,

$(S/R)^n = a^2n * r^{n(n-1)}$

Thus,

$P^2 = (S/R)^n = a^{2n} * r^{n(n-1)}$

Hence proved!

@Tobias: Did that make sense?

ah, I suck at latex. but I hope you got the gist of it

Tobias Kildetoft

09:22

looks good

For some reason I had not even considered using that simpler form for the sums

Nick

@Tobias: The reason I couldn't prove this was because I had no idea how to really find $R$. Once I did that (thanks to you actually). It began to make sense

so, uh, thank you again for the help. :D

I really appreciate it.

Tobias Kildetoft

no problem

Nick

I have to go now. I have start on trigonometry. Bye! See you later

Oh wait, my funny proof for the day

Mats Granvik

11:37

log(a) ? log(b) = a + b

Tobias Kildetoft

@MatsGranvik are you asking what should go at the ?

Chris's sis

Greetings

@robjohn are you around?

Karl Kronenfeld

@Nick What if you are dividing by zero at the end?

somaye

12:21

hi

@JayeshBadwaik how are you?

Jayesh Badwaik

12:32

Hi, I'm good.

What about you?

Mats Granvik

12:50

@TobiasKildetoft yes, that is my question. We/I know that log(a)+log(b) = a * b. Of course exp(log(a)) + exp(log(b)) = a + b but I am wondering if there is another way. But I might not be able to use it, if there is an answer, because I am adding and subtracting infinity.

Tobias Kildetoft

@MatsGranvik I don't think there is any other operation that $a\cdot b = e^a + e^b$ that will work here.

Mats Granvik

Ok

user87637

13:20

@JayeshBadwaik I have reverted to my original 9 holy books, lol.

robjohn

13:40

@Chris'ssis Off to the park. Back in about 45 min

Chris's sis

@robjohn ok. I wanted to show you a proof. (then later)

robjohn

14:12

@Chris'ssis I am back

Not much went on while I was away, I see :-)

Chris's sis

@robjohn ok :-)

what'sup

hi

Chris's sis

@robjohn sent

@what'sup hi

what'sup

how are you ?

hi @robjohn

sorry it's the internet

robjohn

@what'sup Hello

@Chris'ssis got it

Chris's sis

14:20

@robjohn ok

what'sup

@robjohn how are you?

@Chris'ssis any challenges today ???

Chris's sis

@what'sup my greatest challenge to me is myself. It's hard to defeat myself, and it's every day a try. :D

what'sup

math challenge ??

Chris's sis

@what'sup I didn't post things today. (not yet) :-)

what'sup

OK

Chris's sis

14:27

@what'sup trying to compute this without pen and paper might be sort of fun $$\lim_{n\to\infty}n 2^n \int_1^n \frac{1}{(1+x^2)^n} \ dx $$

what'sup

14:52

have you tried to use DCT ? @Chris'ssis

Don Larynx

hi everyone :)

what'sup

hi @DonLarynx

Chris's sis

@what'sup does it help? No, I didn't try, but only high school tools.

Don Larynx

in a dihedral group, $s=r^i$ never holds for any $i$. What about $i=\frac{n}{2}$?

what'sup

OK

Basant Sharma

15:02

how can i contact a particular user ?

@arthur fisher...u there?

Arthur Fischer

@BasantSharma Someone rang?

(Well, almost rang.)

Basant Sharma

yes yes

i needed some help

Arthur Fischer

@BasantSharma With what?

Basant Sharma

it is regarding one answer you posted

Arthur Fischer

@BasantSharma Okay..... (I'm not sure I'm going to like this.)

Basant Sharma

15:17

haha don't worry it's just a clarification

0

Q: Constructing a local nested base at a point

I am trying to prove the following: "Let $X$ be a first countable space and $x$ a member of $X$. Prove that there is a local nested basis $\{S_n\}_{n=1}^\infty$ at $x$." Since $X$ is first countable there is a countable local base $\mathcal{B}_x$ at $x$. Constructing a nested sequence of sub...

general-topology

my question is why the set Si is a finite intersection of open neighbourhoods of x, and therefore Si is itself an open neighbourhood of x

is it necessary that it must be a finite intersection?

Arthur Fischer

@BasantSharma According to the definition $S_i = U_1 \cap \cdots \cap U_i$, so that seems like a finite intersections of the $U_j$'s.

Each $U_j$ is open, so $S_i$ is also open.

Basant Sharma

the symbols are all messed up

i cant understand anything

:(

ok ok i got it...that's exactly my question...why does it neeed to be a finite intersection

Arthur Fischer

@BasantSharma Are you using robjohn's chatJax?

Basant Sharma

it could be an arbitrary intersection as well

Arthur Fischer

@BasantSharma Do you believe that for each natural number $i$ the set $\{ 1 , \ldots , i \}$ is finite?

Basant Sharma

15:24

yes

Arthur Fischer

Then there is nothing else. We define for each natural number $i$ the set $S_i$ to be the intersection $U_1 \cap \cdots \cap U_i$; an intersection of finitely many of the $U_j$'s.

(Of course, $S_i$ might equal some other intersection, but we don't really care about that.)

Mats Granvik

LambertW

Basant Sharma

now i get it

at least beginning to understand what's going on

@ArthurFischer Thanks a lot

Arthur Fischer

@BasantSharma No problem. Anything else?

Basant Sharma

@ArthurFischer that would be all...thanks for your time...i will bug you again if i get stuck :P

Arthur Fischer

15:32

@BasantSharma Cheers!

Basant Sharma

@ArthurFischer cheers ! :)

Mats Granvik

Booyakasha

robjohn

Bernoulli says that for $k\le n$
$$
\begin{align}
1-\frac1{4n^2\log\left(1+\dfrac{k^2}{n^2}\right)}\quad\text{increases to}\quad1-\frac1{4k^2}
\end{align}
$$
Thus, for all $k$,
$$
1-\frac1{k^2}\le\left\{\begin{array}{}
1-\frac1{4n^2\log\left(1+\dfrac{k^2}{n^2}\right)}&\text{when }k\le n\\
1&\text{when }k\gt n
\end{array}\right.
$$

Thus, we can use dominated convergence on the product

Chris's sis

@robjohn hehe, yes!

Mats Granvik

I need a faster computer.

Chris's sis

15:46

@robjohn also for the problem in that proof it's enough to make use of the dominated convergence for series (it's my second solution). It's amazingly powerful.

@robjohn I think above there is a small typo in the second accolade.

robjohn

@Chris'ssis where?

Chris's sis

@robjohn Ah, sorry. I misread something there. It's OK.

@MatsGranvik why? It's about Maple and Mathematica tasks?

Mats Granvik

@Chris'ssis Yes, Mathematica. I have a zeta function limit integral which is slow.

Jack M

16:05

does anyone have any thoughts on this problem? math.stackexchange.com/questions/92270/…

I came across it last night and I was really surprised to find the solution is that complicated

Mats Granvik

16:30

Clear[x, n, k, s, a1, nn, b1, mm]
mm = 3;
b1 = Expand[
Sum[Exp[Limit[
1/(s - 1)*
Sum[(1 - If[Mod[k, n] == 0, n, 0])/(k)^(s - 1), {k, 1, mm*n}],
s -> 1]]*(-x)^n, {n, 0, 32}]];
a1 = 1 + Integrate[b1, x];
x = N[(1 - 11/8)/Exp[1], 30];
Print["here"]
N[2*Pi*Exp[1]*a1, 30]
N[2*Pi*Exp[1]*x/LambertW[x], 30]

Clear[x, n, k, s, nn]
x = N[(2 - 11/8)/Exp[1], 30];
Print["here"]
N[2*Pi*Exp[1]*a1, 30]
N[2*Pi*Exp[1]*x/LambertW[x], 30]

Varying variable mm still lets the approximation of the zeta zero stay close to LambertW approximation

My hope is that letting mm--> go to infinity one could get rid of -11/8, but might not work. I don't know how to efficiently compute and integrate the power series for large n.

mm=1 is equal to LambertW

*x/Lambert(x)

Should have been Zeta[s] instead of 1/(s-1)

Mats Granvik

16:54

stay close is an exaggeration

Lord_Farin

17:40

Today is my first anniversary on MSE. Yay.

Jack M

can anyone think of any semester projects for a math student?

I'm in second year

right now I was thinking either elementary galois theory, geometric topology (proving facts about graphs, things like or euler's formula) or non standard infinite series (ramanujan sums etc)

Lord_Farin

@JackM Those could all work. Which area(s) of mathematics are you interested in/have you taken courses in?

Jack M

I've taken courses in linear algebra, multivariable calc, analysis, and basic topology

Lord_Farin

Which of those did you find most enjoyable?

Jack M

tough question

I dislike multivariable calc, but only because I'm struggling with it a bit

I tend to enjoy mathematics that I can relate to the real world

although I guess the definition of "real world" is subjective

for instance, that's what attracted me to galois theory (unsolvability of polynomaisl in radicals, impossibility of geometric constructions)

both of which are theorems which make statements about reality, not just abstract mathematical constructs

again it's subjective, and I regularly find abstract mathematics interesting as well

Lord_Farin

17:59

I see. Well, Galois theory does require quite a footing in abstract algebra, which comes in handy at a lot of places in maths. So it could provide you with a track and a clear goal. Along the way, you'd pick up a nice bunch of mathematics that'll come in useful many times.

Jack M

I thought about functional analysis, because I like the idea of applying topology to sets of functions

but I don't know anything about functional analysis, so I don't know what direction the project would go in

I would need a specific theorem or question to direct the project

Lord_Farin

Functional analysis will require a solid footing in analysis, linear algebra to yield some intuition (L.A. is basically finite-dimensional F.A.) and comfortability with topology.

Jack M

those sound like reasonable pre-reqs for me

then again, how solid a footing in analysis?

my footing is solid, but I'm not sure if it's far enough up the mountain

Lord_Farin

So it's challenging, but rewarding. It's been a while since I was involved in F.A., so I'm not really qualified to set a target for your endeavour that I can confidently call "feasible".

With solid footing, I mean: Fluency in $\varepsilon$-$\delta$ arguments.

Jack M

I'm not even certain functional analysis is the correct term

basically, I've seen people talk about function sets as topological spaces, and I find that interesting

Lord_Farin

18:07

F.A. is very much about that.

Jack M

I'll add it to my list

with a question mark

Lord_Farin

Basically, I'd suggest: pick a nice theorem that you think is really cool, approach a professor and ask how long it'd take to learn that, and what you'd need to know. (And other nice theorems that you could and/or would learn, etc. etc.. The spinoff questions can be endless.)

Jack M

that's how I was picturing it

Lord_Farin

@JackM It's a sound approach. Just accrue some ideas, and go for it. :)

Pedro Tamaroff

18:48

@Lord_Farin Le Lawdy is around.

Chris's sis

@robjohn I think there is something unclear to me as regards my second approach with DCT. I'm writing the proof right now and maybe you might take a look at it.

Shiester

Are you guys good at calculating probability?

There's this real qt3.14 grill i work with and i think she's into me. she walks by me at lunch every day, looks at me and says hello, and continues walking off. she could easily take a shorter route to the lounge. However, I'm not exactly in the best shape, im sort of tall, im big into video games and I don't have a huge house. I also like axe body spray. what are the odds I might be able to go the distance with her?

Pedro Tamaroff

@Shiester qt3.14 LOL.

robjohn

19:04

@Chris'ssis look at the image you sent?

@PedroTamaroff does that mean she is approximately cute?

Pedro Tamaroff

@robjohn PUNPNUPNUPUNUPNU

cyberskull

qt$\pi$

Chris's sis

@robjohn I didn't send anything yet but I'd like to send it in a few minutes. :-)

robjohn

@cyberskull interminably cute?

Pedro Tamaroff

@cyberskull Her beauty is transcendental?

cyberskull

19:07

:D

robjohn

@PedroTamaroff she's cute, but I can't say why, rationally.

Pedro Tamaroff

@robjohn LULZ.

cyberskull

19:24

qt$\pi$ certification

;-)

Chris's sis

19:39

@robjohn sent it. Pls take a look over it when you have some time.

robjohn

@Chris'ssis got it

Chris's sis

@robjohn ok :-)

robjohn

@Chris'ssis That is pretty much how the argument goes in the Dominated convergence argument I gave for the product

Chris's sis

@robjohn is my way completely correct, nothing missing?

robjohn

@Chris'ssis It looks okay. I don't see anything missing, but some people are more pedantic than others.

Chris's sis

19:47

@robjohn thank you very much.

@robjohn That's true.

robjohn

@Chris'ssis You said that something had been unclear. Did you clear that up?

Lord_Farin

@PedroTamaroff Hello there, Pedro. It seems you've changed your name. :)

De-anglicised, from the looks of it.

Chris's sis

@robjohn that point was related to the fact that some authors would also have treated in my proof the case $k>n$, but I guess this is related to the "pendatic" word you talked about above. :-)

Lord_Farin

Hello there @Chris'ssis. It's been a while.

Chris's sis

@Lord_Farin Hello!!! Long time I didn't see you around. Where have you been? :-))))

Pedro Tamaroff

19:55

@Lord_Farin Yeah.

Lord_Farin

@Chris'ssis Where have I been. Good question. For the most part, I just redistributed my time; MSE chat was among the losers...

Chris's sis

I need to use a larger keybord since I barely manage to write things with this small one.

Arkamis

@cyberskull Skully

Chris's sis

@Lord_Farin That's not bad. Did you study/prepare for your thesis?

cyberskull

@Arkamis Hi, how are you?

Arkamis

19:56

Exhausted

Lord_Farin

@Chris'ssis In fact, I've graduated at the end of August.

Mostly, I spent time catching up on ProofWiki business.

Pedro Tamaroff

@Lord_Farin Congratulaciones.

Arkamis

Quarter-point of the season already

Lord_Farin

@PedroTamaroff Thanks. :)

Chris's sis

@Lord_Farin $\Large \text{Congratulations!}$

Lord_Farin

19:58

@Chris'ssis Thank you, thank you. :)

cyberskull

@Arkamis What is goin on with them giants?

Pedro Tamaroff

Wonder if @TedShifrin is related to her.

Arkamis

@cyberskull I don't know, but I love it.

Lord_Farin

(Question to self: Hm, what'd be the response if I told it was with distinction?)

Pedro Tamaroff

@Lord_Farin Honours?

Chris's sis

19:59

@Lord_Farin I envy you (in a positive sense, of course). :D

jinawee

Can Laplace equation solutions have inflection points?

Lord_Farin

@PedroTamaroff Not exactly from my knowledge of what "Honours" means. It just means I got quite high grades during my MSc.

Pedro Tamaroff

@Lord_Farin Ah, cool!

@Lord_Farin What is your thesis about?

Lord_Farin

@PedroTamaroff Building up logic in a category-theoretic framework.

Pedro Tamaroff

@Lord_Farin Could you explain in a softened version?

Lord_Farin

20:08

@PedroTamaroff Lots of the operations in logic (conjunction, disjunction, quantification) can be described using category-theoretic methods, applied to suitable categories. My thesis builds up this theory basically from scratch, and goes all the way to describing some analogues of model-theoretic results, as well as proving some category-theoretic results using the framework in a non-obvious way.

Pedro Tamaroff

@Lord_Farin reads

@Lord_Farin Ahm OK.

Lord_Farin

@PedroTamaroff How about you? Are you writing a thesis? Have you written one? What is/was it about?

Pedro Tamaroff

@Lord_Farin I'm a freshman, dawg.

Lord_Farin

@PedroTamaroff Sorry. I've been away. Memory is leaky.

Chris's sis

@robjohn I think we might modify DCT for series to make it clearer. Let me change it a bit.

Lord_Farin

20:17

@JasperLoy Are you the (in)famous user that this room has shouted for for many, many weeks?

robjohn

@Chris'ssis Series are just integrals over the integers with discrete measure

Lord_Farin

@robjohn +1 to measure theory. :)

user87637

@Lord_Farin Well, I am the real JL who always pisses off people, yes.

Chris's sis

@robjohn right.

Lord_Farin

@JasperLoy A pleasure to meet you.

user87637

20:18

@Lord_Farin We have spoken before, actually, I think.

Chris's sis

@JasperLoy you're a pleasant person to me. :D

user87637

Also, beware of imposters. They pretend to be me by using a blue square, mostly.

user87637

In fact, you can't be sure I am who I claim I am...

Pedro Tamaroff

@JasperLoy Of course one can.

user87637

The imposters will also say "I am a banana!"

Lord_Farin

20:20

@JasperLoy Oh well, that holds true for any user who has at one point had their account deleted, I suppose.

@JasperLoy This is useful information only if you don't say that.

user87637

@Lord_Farin You have a very secretive name, lol.

Lord_Farin

@JasperLoy Is that so? Nonetheless, people around MSE (and a few other sites, like ProofWiki) seem to "know" the person behind the name a bit.

But alas, I like my internet anonymity; hence I won't disclose my real name.

user87637

@Chris'ssis I remember you said your ex had a wedding.

user87637

@Lord_Farin Hehe, I vaguely recall, perhaps wrongly, that it was disclosed by someone else indirectly in this room.

Chris's sis

@JasperLoy yeah ...

user87637

20:24

@Chris'ssis You can forget about him then. But isn't it too early to get married at your age?

Lord_Farin

@JasperLoy Or you could just be fishing for information with that sentence... :)

Chris's sis

@JasperLoy never too early :-)

user87637

@Chris'ssis Yeah, considering how one can have kids at sixteen.

user87637

@Lord_Farin Haha, I think you are C******, but never mind.

Chris's sis

@robjohn we might use $\lim_{n\to\infty} \sum_{k=1}^{N(n)}$ instead of using $\infty$ at the top of summations. Thus we avoid any possible confusion as regards inequalities involving $k$'s. Then we only consider $k \le N(n)$. Am I wrong?

@JasperLoy life is full of mystery.

user87637

20:27

@Chris'ssis You misspelled mystery, lol.

robjohn

@Chris'ssis I try to extend the sum to $\infty$

user87637

@robjohn Yes. I am still thinking if the math world would be better with you, lol.

user87637

Anyway, to those wondering how I am now, I am still very crazy. I hope to recover by the end of next year.

user87637

@cyberskull I think maybe I made Charlie upset when I said I did not wanna talk to her anymore.

cyberskull

Orly?

robjohn

20:33

Here, hear! I agree :-)

Lord_Farin

@robjohn Good to hear that.

user87637

I miss Charlie Brown and Snoopy.

Chris's sis

@robjohn OK. I only wanted to be more specific as regards the foregoing theorem.

user87637

My favourite American drama seems to be The Wonder Years, but I only watched the first five episodes.

robjohn

@Chris'ssis You really need to include all the integers in the domain of summation, or things get confusing.

user87637

20:36

Wait, I see the starred message. Is there a Klein-Smith theorem? I only know about the Krull-Schmidt theorem.

user87637

It is interesting to know that there is the BCH code in coding theory and the BCH theorem in algebra, where the BCH stands for six different names.

Chris's sis

@robjohn Sure. But isn't my way of presenting things clear enough? For instance, we may consider that $N(n)=n^2$, and then, let's say, we have to consider $1 \le k\le n^2$.

Lord_Farin

@JasperLoy What about Cantor-Bernstein-Schröder vs. Cauchy-Bunyakovsky-Schwarz?

user87637

@Lord_Farin Ah, I did not think of that.

Lord_Farin

@JasperLoy Yeah, Bunyakovsky is a tad long, so many leave him out.

robjohn

20:40

@Lord_Farin I'd only heard Cauchy-Schwarz until I started on MSE.

Never heard of Bunny-kovsky :-)

user87637

@robjohn You see the longer version in the books where the authors are very particular about including as many names as possible.

Chris's sis

@robjohn In the textbooks here you find it as Cauchy-Bunyakovsky-Schwarz.

Arkamis

@Lord_Farin Have you ever done any game theory work?

Lord_Farin

@Arkamis Nothing beyond rock-paper-scissors.

Arkamis

hm

robjohn

20:42

@Lord_Farin lizard-spock?

user87637

Grothendieck generalised Riemann-Roch to Grothendieck-Riemann-Roch and Chern generalised Gauss-Bonnet to Chern-Gauss-Bonnet, both of which are connected by the Atiyah-Singer index theorem, fascinating.

Arkamis

I'm looking into some relationships between category theory and game theory, and I find a lot of passing references to vague ideas, but nothing concrete, and I think there is a lot of applicability there.

Lord_Farin

@robjohn In my first year of BSc. I wrote a piece on a variant with like 100 options. I've since forgotten everything about it, though (except writing it, obviously...).

user87637

@Lord_Farin Oh, you need to write such things in the first year of undergrad?

Lord_Farin

@JasperLoy Well, it wasn't very deep. More of a cursory glance.

user87637

20:44

@Lord_Farin What is variant and options, something in finance?

Pedro Tamaroff

@JasperLoy The amount of surnames is too damn high!

user87637

@PedroTamaroff Wait for the Tamaroff-Loy theorem, hehe.

Lord_Farin

@JasperLoy Probably, but that's not what I meant.

user87637

@Lord_Farin Oh, now I understand!

Lord_Farin

@JasperLoy I'll probably have to disclose my identity to be able to join in...

A small price, if it were to happen.

user87637

20:47

@Lord_Farin Yes, you are Farin, lol.

Lord_Farin

@JasperLoy Tamaroff-Loy-Farin :P

"The Tamaroff-Loy-Farin Theorem establishes a closed form for the number of upvotes an MSE answer gets, as a function of time, based on the triviality of the subject matter and the depth of the answer."

3

Chris's sis

21:01

@Lord_Farin Doesn't this one look like a rare gem? $$\sum_{k=2}^{\infty}\frac{3^k-1}{4^k } \zeta(k+1)$$

(maybe it's time for some glasses here)

Lord_Farin

@Chris'ssis Hm, your edit makes it look a little more mundane.

Chris's sis

@Lord_Farin now it's just fine. :-)

Lord_Farin

@Chris'ssis It's still nice, but now I can at least think of ways to approach it.

Danny

21:18

chirs

chriz

Pedro Tamaroff

@Lord_Farin It's nice my name goes first. =D

We should start naming theorems cyclically to avoid that.

Lord_Farin

@PedroTamaroff I sense trouble with the linear nature of paper.

masfenix

Hey guys, was wondering if you can validate my reasoning to a question. I want to show that Schwartz functions are not dense in $L^\infty$. a simple counter example is any constant function $f$, say $f(x) =1 \forall x$. Clearly no Schwartz functions can be used to approximate this. but how do I turn it into a formal proof?

Right now my reasoning is that $|g - f| < \epsilon$ is impossible because $g \in S$ will vanish to zero outside the compact set.

Chris's sis

I'm amazed how beautiful some things are.

masfenix

21:38

anyone have an idea on how I can formalize the proof?

Eric Stucky

21:52

3

Q: Surprising limit (probability of no two coinciding pairs)

I stumbled upon this question by random chance. The motivation is kind of long, the question is pretty short; if you're just here for the limits, feel free to skip to the break. I'm taking five courses this semester, all of which meet on two days, and I happened to notice that all of my days see...

probability combinatorics limit recreational-mathematics

Sorry, I was testing how that feature works :S

Pedro Tamaroff

@EricStucky Use [text](link)

Eric Stucky

like this ?

yes, like that.

Thank you :)

Pedro Tamaroff

22:08

@anon Well, I got to give an answer with that construction I told you some days ago.

Filtration, you called it?

Twink

9

Q: The fastest trajectory that a particle should follow through two different mediums

Hello, after 3 failed attempts to solve this problem, I decided to start a bounty for this question. Please, I need a complete answer with an interpreation of the final result. Thank you in advance. Problem: A particle travels at speed $v_a$ in a medium $A$ and at speed $v_b$ in an medium $B$. ...

calculus optimization

anon

@PedroTamaroff yes

Pedro Tamaroff

@anon Guess you can give a cool answer there =)

anon

some general comments. "indicator" is not entirely ubiquitous in your usage, so I wouldn't ask anybody to "recall" it. it probably deserves to be pointed out that $|a|_H$ is the order of $aH$ in the quotient $G/H$. Also $G'$ is often used to denote the derived subgroup, so we want to avoid the prime. A more philosophical point is that this "adjoin an element repeatedly to create a tower" construction is ubiquitous in all of algebra, between fields, rings, groups, modules, monoids and friends

note one doesn't even need the indicator at all to define filtrations $\langle a\rangle\subset\langle a,b\rangle\subset\langle a,b,c\rangle\subset\cdots$, the indicator is useful in determining the successive indices (more informatively, the factors of the series as cyclic groups)

as a historical tangent, filtrations where the composition factors are cyclic (as here, in which case the group is called solvable) is important in galois theory: this tells us when all the algebraic elements in a field extension (equivalently, roots of a polynomial) can be expressed in terms of nested radicals or not

Danny

dr.twink

masfenix

22:31

Does anyone know why the $L^p$ continuity of translations is defined $1 \leq p < \infty$?

Pedro Tamaroff

@anon Let me correct all that. Thanks.

masfenix

The continuity of translation says that define $f_t = f(x - t)$ then $|| f_t - f ||_p < \epsilon $

Pedro Tamaroff

@anon Right.

@anon Heh, that I should read about.

Would you think it is wise to adjoin those comments to my answer?

anon

sure

I'll go ahead

Pedro Tamaroff

@anon You'll do it?

Cool.

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$Mathematics$ Mathematics