Mathematics - 2014-09-06 (page 1 of 3)

Vanio Begic

00:09

Hello people

I have a quick question I need help with

namely how does one go about proving that \lim_{x \to 0} f(x) = \lim_{x \to a} f(x-a)

robjohn

@AlecTeal I've always seen the commutator as a function $[x,y]:G\times G\to G$

Vanio Begic

Anyone?

Alec Teal

00:45

@robjohn I've partially solved it now, thanks for confirming.

The name really threw me off you see. I thought it was supposed to commute.

@robjohn can I ask a soft-question with a one line answer? It's really not worth posting one over.

I'm gonna ask anyway, worst case you stay quiet :P

@robjohn I like to think of quotient groups as "removing a property of a group" for example .... take $G$, an Abelian group, and let $H$ be the subgroup of all elements of $G$ that have finite order.

If you look at $G/H$ in that case, all the elements ('cept the identity) have infinite order.

So @robjohn my question is, how do I work out what properties I'm removing, for example $\mathbb{Z}/<6>$ is of order 6 and are the congruence classes of 6. - what did I remove though?

@robjohn if I have $\mathbb{Z}_6/\{0,3\}$ the resulting 3 elements have order 3, we removed the things of order 2. $\mathbb{Z}_6/\{0,2,4\}$ has 2 elements (of order 2), did we remove the ones of order 3?

Now I'm stuck with $D_{\text{square}}/\{R_0,R_2\}$ I'm not sure what I've taken away, it's certainly not order-2-ness.

This isn't like a book question, I'm "stuck" in the sense of trying to make this "we remove properties" thing more concrete.

robjohn

@AlecTeal: I have to take my dog to the park I may not be able to answer your question, but I will try when I get back.

Alec Teal

Thanks

Balarka Sen

@AlecTeal It's really bad to think of it that way, not to mention if it's even correct.

Might be correct though

$\langle 6 \rangle$ in $\Bbb Z$ is the additive subgroup of $\Bbb Z$ generated by $6$, i.e., multiples of $6$. $\Bbb Z/\langle 6 \rangle$ is the group made out of identifying the multiples of $6$ with the additive identity of $6$.

Alec Teal

01:01

@BalarkaSen there's a lot of truth to it

@BalarkaSen take ANY group you like, if you quotient out the commutator group (everything from the group generated by elements of the form $aba^{-1}b^{-1}$ - which is only the identity when a and b commute) you are left with an Abelian group!

Balarka Sen

@AlecTeal problem is that it doesn't work like that for infinite groups.

Alec Teal

So G/the stuff that doesn't commute = Abelian!

Balarka Sen

abelian groups are always simple enough and easy enough.

Alec Teal

@BalarkaSen a group removing all the finite things has just infinite stuff left (and the identity)

Balarka Sen

@AlecTeal No, no, I mean there exists (infinite) groups $G$ such that $G/H \cong G$ for some subgroup $H$.

That's where your analogy breaks down.

Alec Teal

01:08

.... @BalarkaSen I'm okay with it breaking down eventually.

You're keeping everything that is the identity of the quotient group.

That's a weird way of saying it.

Oh this is a very intuitive way into the first isomorphism theorem....

Ahh!

Balarka Sen

@AlecTeal ?

Alec Teal

@BalarkaSen the "natural morphism" is $f:G\rightarrow G/H$ with $f(g)=Hg$

Balarka Sen

Yes.

Wait, what?

Alec Teal

the kernel of that has the properties "you keep" from choosing H.

Take the commutator one, if the commutator is the identity, then the elements commute

Balarka Sen

You've lost me.

Alec Teal

01:13

Okay @BalarkaSen there's this thing, the commutator, it's an element of a group of the form $aba^{-1}b^{-1}$. If this $=e$ then $ab=ba$

Balarka Sen

I know.

Let's just leave out commutativity and step into the mess.

Alec Teal

So if we let H be the group generated by the commutators (which is normal, trustme)

$G/H$ is Abelian (commutative)

Balarka Sen

What d'you mean by "kernel has the property of "keeping" from choosing H"?

Alec Teal

We keep all the things with commutators = the identity (commutative elements)

Balarka Sen

i thought we were talking of kernels, not commutators?

in general, explain me your statement

Alec Teal

01:15

G/H is isomorphic to .... something - I've seen a property and (evidently) am trying to scrape a concrete sense of it together.

Balarka Sen

I understand the kernel of the map $g \mapsto gH$ as the elements of $G$ which leaves $H$ fixed after multiplication.

Alec Teal

Hmm...

Balarka Sen

@AlecTeal Let $f$ be your map $G \to H$. Think about $G/\text{ker} f$.

Alexander Gruber

01:48

hey, anybody know of a good funding source for open access fees?

i'm accepted to a hybrid open access journal and am trying to make the article available to the general public.

Gustavo Montano

05:15

@r9m. Last episode -> too good.

Hippalectryon

@Chris'ssis Don't worry, i screenshot it as always

r9m

07:51

@GustavoMontano :D Totally !!!!!!!! I got occasional goosebumps =P

Gustavo Montano

08:13

@r9m. The mystery of Sasuke, Obito V.S. Kakashi, Madara's back-stabbing obito, Obito turnaround.

Ugh, it's so good ^_^.

r9m

08:42

@GustavoMontano :-)

Khallil

08:55

All in one episode as well, @r9m & @Gustavo!

Hey, @r9m!

I bought Naruto's 42nd, 43rd and 66th volumes. ^_^

Jiraiya's end, Itachi vs Sasuke, Sasuke learning the truth and Obito's Jinchuuriki transformation.

Gustavo Montano

09:21

@Khallil - volumes? As in manga?

Khallil

Indeed.

Gustavo Montano

Great stuff!

Hmmm, weird question, but - how many episodes would fit in a volume?

Excluding fillers, of course.

Khallil

It depends per volume so I can't put a number on it.

Recently, the studio have been mowing through chapters, so they've decided to make the next two episodes filler.

(Mecha Naruto T_T)

That's volume 42. The J-man! Bawls

Gustavo Montano

What, like - the NEXT 2 weeks will be fillers?

Khallil

Indeed.

It's a shame, but that's how it is.

This website outlines a few of the upcoming episodes until October.

Kasper

10:13

This would be such a usefull tool: authorea.com if it supported amsthm

It is kind of github for latex documents.

Chris's sis

10:33

Greetings

Nick

If
$$y\cdot(3-x) = z\cdot ( 3-y ) = x\cdot(3-z)$$
What's the super quickest way to prove $x = y = z$ ?

Chris's sis

@Hippalectryon by the way, did you finish it?

Nick

(Tell me if you have any mathgic trick, otherwise it's just trivial substitution)

usukidoll

try x = 1 y = 1 z = 1? @Nick

I think if z x y is equal to 0 1 2 3........ we will have everything equal to each other

Vanio Begic

10:51

Hello guys

anybody there

?

90intuition

is an ideal of a subring also an ideal of the ring itself ?

Nick

11:15

@usukidoll: :D as great as guessing values is, I was hoping for some quick algebraic method. The quickest way I've found is equating everything to $a$ and then some kung fu which proves it.

V-Moy

12:04

I'm wondering about the answer of this question, can you give me a list of 10 users who is the best at integral question (integrator)?

Khallil

I will answer your question ... with a question! Are you a girl, @V-Moy?

If so, what does your username stand for?

(My username is my best friend's name!)

V-Moy

Why didn't you see my profile? V for Valentina & as far as I knew, Valentina is a girl's name.

I am not VALENTINO

That's a boy

Khallil

Sorry! I didn't check your profile. I hardly check anybody's profile.

That's a nice name, Valentina.

@V-Moy sounds a bit more masculine than Valentina, which is why I asked!

whs

V-Moy

12:19

@Khallil Hey! You also like anime. Your profile pic in Anime & Manga is Shisui Uchiha. I am a Naruto fan anime & manga but I like Sasuke Uchiha

Khallil

I think Sasuke's my favourite character. His mystery is his main allure!

Shisui's got my favourite technique!

The Kotoamatsukami. ^_^

Nick

Guys, in $\LaTeX$, how do I clean up statements like $$4((S^2)^2 + 2S^2C^2 + (C^2)^2 + 3S^2C^2) $$ The math looks so messy. I need better brackets.

V-Moy

I like Susanoo!

Khallil

\left( and \right) usually make things look neater, @Nick.

The Susanoo is awesome!

Are you up to date with the manga, @V-Moy?

V-Moy

Rinnegan is also awesome. Are you high school student @Khallil?

Nick

12:23

@Khallil: Thank you, one million times.

Khallil

I'll be a university/college student in about a month, @V-Moy! ^_^

No prob, @Nick!

V-Moy

@Khallil Yup! My fav manga Naruto, Bleach, Detective Conan

I also like Hunter X Hunter

Ohhh, you will be a freshman

Khallil

Yep! The Hunter X Hunter anime is one of my favourites. Naruto's my favourite manga. I bought 3 volumes a short while ago and have read 1 and a half volumes. I never read the Itachi vs. Sasuke fight but it's better in the Anime. There's nothing like an animated Kirin!

V-Moy

@Khallil and @Nick Answer my question! give me a list of 10 users who is the best at integral question (integrator) in Math SE.

Khallil

@Chris'ssis is very good at integrating. @robjohn is also fantastic at everything. @DanielFisher is awesome. @r9m is wicked. I've seen a few great posts from @Lucian.

V-Moy

12:27

Itachi is death. Sasuke kills him

Khallil

I can't really think of any others, @V-Moy.

V-Moy

@Khallil Don't tag their names

Khallil

Why not?

The more the merrier. ^_^

V-Moy

@Khallil Who is your fav character in HxH?

Khallil

It's a joint tie between Ging, Hisoka and Killua. Don't spoil anything for me, @V-Moy! I've only just reached Gon's return during the presidential election between Leorio and Pariston.

Killua's Godspeed is amazing.

V-Moy

12:33

I like Hisoka

Khallil

He's such an awesome character. He only looks out for his own entertainment and is ready to turn on everyone!

V-Moy

I won't spoil anything but your chapter is obsolete, too far.

That's weird, the main character is Gon but you didn't like it either

Khallil

Killua's got way more going on. Gon is extremely pure but that makes him less interesting than the other characters who have grim pasts.

V-Moy

@Khallil I also ask a question on Anime & Manga about Sasuke's rinnegan, maybe you can answer it: anime.stackexchange.com/q/8848/4559

Yup! I agree with you, Killua is cooler than Gon.

Khallil

There isn't a solid answer to that question, @V-Moy.

I think it's because he only received half of Hagoromo's power.

Chris's sis

12:41

A question as you might rarely meet on MSE - math.stackexchange.com/questions/921346/a-series-from-dreams

Khallil

Why do you think you dream of math, @Chris'ssis?

Chris's sis

@Khallil That was a series like many others that appeared during my dreams.

Khallil

Yes, but that doesn't answer my question ...

Chris's sis

@Khallil Perhaps it's because I spend a lot of time doing math.

V-Moy

@Khallil But Sasuke looks weird only possess a rinnegan

Khallil

12:44

That makes sense, @Chris'ssis.

V-Moy

@Chris'ssis Your dream like Ramanujan

Khallil

He doesn't in my opinion, @V-Moy. If anything, Kishimoto probably gave Sasuke only one Rinnegan to separate him from the other Rinnegan users. Just like how Sasuke has many outfit changes during the anime. He just enjoys writing Sasuke into the story.

Will Hunting

13:15

@V-Moy math.stackexchange.com/users/50290/ethan this is the incarnation of Ramanujan

V-Moy

@WillHunting Are you Jasper?

Will Hunting

@V-Moy Yes, I am.

V-Moy

I think it would be Cleo

Will Hunting

I am the incarnation of a banana.

V-Moy

Why did you always say banana?

Will Hunting

13:19

I like bananas.

V-Moy

Are you Don King Kong?

Will Hunting

I am Gorilla.

V-Moy

It's Saturday night @WillHunting. Why didn't you hangout?

Will Hunting

@V-Moy Well, I stay at home most of the time nowadays.

V-Moy

If I were 18, I'd hangout with my friends

Will Hunting

13:24

I was in the army when I was 18, lol.

V-Moy

So, you are a SG citizen then. Do you speak Malay?

Will Hunting

Yes. No.

V-Moy

That's weird. Are you Chinese?

Will Hunting

Yes.

V-Moy

The truly Asian! lol

Will Hunting

13:27

I am in this world, but I am not of this world.

V-Moy

Apa kabar @WillHunting?

Will Hunting

@V-Moy Sorry, I do not understand your language.

V-Moy

This is funny, you're a SG, but you don't understand. lol

Alizter

@V-Moy Do not be worried over serial downvotes. The worst you can do is be upset about them. Just ignore them and move on :)

V-Moy

It's long time ago @Alizter. I've tried to forget it

@WillHunting Where did you go? I'm just kidding

Alizter

13:35

@V-Moy The system will try to revert it but there is no point being upset. It happened to me

@WillHunting Gone to see a girl?

V-Moy

@Alizter You too? The downvoters always piss me off

Alizter

@V-Moy That is their aim usually. Don't be pissed off so that they fail their task.

@V-Moy I once got serial downvoted for rejecting a users edit

V-Moy

I'll try but it's hard for me

Alizter

What do you like to do?

V-Moy

@Alizter Hey! You're sixteen!

Alizter

13:38

I am

V-Moy

I like reading manga and watching anime

Alizter

@V-Moy Then do that :P

or go on brilliant.org

V-Moy

@Alizter I am also a user there

Are you also user there?

Alizter

@V-Moy I know I recall seeing some of your problems

V-Moy

Really? What is your user id on brilliant.org

Alizter

13:41

brilliant.org/profile/ali-jxv2u9/feed

Vrouvrou

hi

V-Moy

@Alizter There you are. You're cool. Almost all subjects level 5

Vrouvrou

have you an idea :math.stackexchange.com/questions/920885/…

Alizter

@V-Moy I spent wayy too much time on there

V-Moy

@Vrouvrou Sorry, I don't understand. I am high school student

@Alizter I am rare to visit brilliant

Alizter

13:45

@V-Moy I used to go on it alot but I started school work and stuff

V-Moy

@Alizter I love having fun here

School stuff is always suck!

Alizter

@V-Moy I am gonna solve your another monster integral problem

V-Moy

Okay, then. Write a solution if you can answer it @Alizter

@Alizter Btw, is that your dog in you prof pic?

Alizter

yes

its my grandads

my dog is 30 cm long

he is a pug ^_^

V-Moy

The one in the picture is not you?

I wish I had a puppy

Alizter

13:52

I am in the picture

Chris's sis

I just created something very cute ...

Alizter

@Chris'ssis That one your dreamt about was very fastly converging

Chris's sis

@Alizter Yeah, I know. :-)

V-Moy

I know this is inappropriate question since you're a woman but I'm wondering, how old are you @Chris'ssis?

Chris's sis

@V-Moy I'm young.

Alizter

13:59

@Chris'ssis < 100

V-Moy

@Chris'ssis Around 25-30?

Chris's sis

@V-Moy Why do you ask?

V-Moy

@Alizter Do you know brilliant users that are also MSE users? I only know 3, Calvin, Tunk-Fey & Pranav

@Chris'ssis Just curious

Alizter

@V-Moy Hmm not really sure. I know people who are not too fond of brilliant.org

V-Moy

Who?

Alizter

14:10

can't remember exactly but I have seen a few people look at it and then say they don't like it

Chris's sis

@V-Moy You'll find more about me when I publish my book. ;)

rehband

Mystical Greetings :)

V-Moy

@Chris'ssis What kind of book?

Chris's sis

@V-Moy The book will be mainly a collection of limits, series and integrals created by me, but at the same time it will contain some nice problems (already known) where the elementary solutions are not known yet.

3

V-Moy

@Chris'ssis Let me know then if you have published it. Can't it

Chris's sis

14:19

@V-Moy OK

Alizter

@V-Moy Try and annoy this guy

V-Moy

Not now @Alizter. I have to go. My mom & I will pick me dad. Bye...

robjohn

@r9m: did you see the comment I wrote yesterday about one of your blog posts not being rendered by MathJax because of weird unicode characters inserted?

It looks like 'en-dash' –

park time... bbl

Will Hunting

14:36

@Chris'ssis Include your photo and email too, lol.

Chris's sis

@WillHunting lol :-)))

@robjohn also this one seem to be interesting $$\sum_{k=0}^{n} \frac{H_{k+1} H_{n-k+1}}{(k+1)(n-k+1)}$$

Will Hunting

Today I saw a laptop installed with Windows 8 and Office 2013 at about 400 USD, quite cheap.

Will Hunting

14:57

@Chris'ssis What book did you learn calculus from?

Chris's sis

@WillHunting Well, I read some Romanian books and also some books by foreign authors, but not from cover to cover, only those parts that I needed. Then I read some papers, visited some sites like MSE, had discussons on some topics with persons that were good at math (like robjohn).

@WillHunting I think doing your own research is the best way to learn great things. Well, that "great" is relative.

Will Hunting

@Chris'ssis I have still not found the perfect calculus book. All of them have some problem or other, lol.

Chris's sis

@WillHunting Which one is the calculus book you like the most?

Will Hunting

@Chris'ssis Currently, I intend to use Marsden's Calculus I, II and III.

@Chris'ssis You should say the most instead.

Chris's sis

@WillHunting Yeah, corrected! :D

Will Hunting

15:03

@Chris'ssis You will need someone to correct the English in your book too!

Chris's sis

@WillHunting Never heard of them.

@WillHunting hahaha, good point! ;)

Hippalectryon

15:20

@Chris'ssis (about your last ping) nope, I didnt finish it, I have two exercises sheets to finish

@Chris'ssis And i need your help :c

@Chris'ssis You're gonna say it's trivial, but I can't manage to find the limit of $\sum\limits_{k=0}^n\frac{1}{n^2+k^3}$

Chris's sis

@Hippalectryon What did you try?

Khallil

What does 'the limit' mean in this case, @Hippa?

Chris's sis

@Hippalectryon You can do many things here!

Hippalectryon

@Khallil $\underset{n\rightarrow\infty}{\lim}\sum\limits_{k=0}^n\frac{1}{n^2+k^3}$

@Chris'ssis Well i tried majorating by stuff but it didn't tend to 0

I found the following majorations :

Chris's sis

@Hippalectryon Stop for a second. Have you ever thought to approximate that sum by an integral?

Hippalectryon

15:31

$\ln(1+\sqrt{2})$ and $\int_0^1\frac{\mathrm{d}x}{\sqrt{1+x^3}}$ (is that last one even known ?)

@Chris'ssis ?

Chris's sis

@Hippalectryon Exactly. Did you think to get a rough approximation by using integrals?

$$\int_0^n \frac{1}{n^2+x^3}\ dx $$

Q.E.D. (I let you finish the rest)

Hippalectryon

@Chris'ssis Oh btw

Chris's sis

@Hippalectryon Ooooooo, great job!!! :-)))))

Hippalectryon

@Chris'ssis Wuut ? how do you turn an integral into a sum like that ? Riemann stuff majoration ?

Chris's sis

@Hippalectryon mathworld.wolfram.com/Maclaurin-CauchyTheorem.html

Hippalectryon

15:34

@Chris'ssis -___-

I made a mistake it's $\underset{n\rightarrow\infty}{\lim}\sum\limits_{k=0}^n\frac{1}{\sqrt{n^2+k^3}}$

square root

Khallil

Hey, @Sawarnik.

Hippalectryon

Of course I know the one without the root >:c

Chris's sis

@Hippalectryon lol, you change it now? :D

Sawarnik

Hey.

Hippalectryon

@Chris'ssis Nah i was just like 'hey where is the root gone in his integral' :)

Then i looked at what I had posted above

And uuuh

@Chris'ssis The majorations above are (of course) for the version with the root

Hence the root in the integral btw

@Chris'ssis What should I try ?

Chris's sis

15:40

@Hippalectryon hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

@Hippalectryon AM-GM as a first step ? :-)

robjohn

@Chris'ssis that has Cauchy products written all over it.

Chris's sis

@robjohn Yeah, indeed. :-)

Hippalectryon

@Chris'ssis I was wondering, is the integral i posted defined ?

I mean known

Chris's sis

@Hippalectryon Well, first, let's finish this one.

Hippalectryon

@Chris'ssis Give me some time :c

I'm filling forms at the same time

Chris's sis

15:44

@Hippalectryon $$\frac{1}{n }+\sum\limits_{k=1}^n\frac{1}{\sqrt{n^2+k^3}}\le \frac{1}{n }+\sum\limits_{k=1}^n\frac{1}{\sqrt{2\sqrt{n^2k^3}}}$$

Sawarnik

@Khallil What was that? Hmm, not good Khallil.

Hippalectryon

@Chris'ssis Why the $1/n$ ?

Chris's sis

@Hippalectryon To avoid that $0$ in denominator (see the right side).

Hippalectryon

Uh k

Chris's sis

:17539304 From that point it looks you're done immediately.

Hippalectryon

15:48

@Chris'ssis Thanks :)

Chris's sis

@Hippalectryon :D

Hippalectryon

@Chris'ssis immediatly ?

I still need to show the limit of the sum

Chris's sis

@Hippalectryon Yeap. That part is a piece of cake.

Hippalectryon

@Chris'ssis really? uh ...

Chris's sis

@Hippalectryon Yeap, you can do that part in more ways.

How do you want me to finish it?

Hippalectryon

15:51

Elementary stuff

Chris's sis

@Hippalectryon Apply Cesaro-Stolz theorem

Hippalectryon

WAIIT

Lemme try something first

:C

Chris's sis

@Hippalectryon OK:-)

Hippalectryon

Urm but ...

It equals $1/n+1/\sqrt{2n}\sum_{k=1}^n1/\sqrt[4]{k^3}$

Is there a good formula for $\sum_{k=1}^\infty k^a,a\in\mathbb{R}$ ?

@Chris'ssis Stoltz cesaro ? where do you see that ?

Chris's sis

Stolz–Cesàro theorem

In mathematics, the Stolz–Cesàro theorem, named after mathematicians Otto Stolz and Ernesto Cesàro, is a criterion for proving the convergence of a sequence. Let and be two sequences of real numbers. Assume that is strictly monotone and divergent sequence (i.e. strictly increasing and approaches or strictly decreasing and approaches ) and the following limit exists: Then, the limit also exists and it is equal to ℓ. The general form of the Stolz–Cesàro theorem is the following (see http://www.imomath.com/index.php?options=686): If and are two sequences such that is monotone and unbounded...

Hippalectryon

15:57

@Chris'ssis I know that. Where do you see $a_n$ and $b_n$ ?

Chris's sis

@Hippalectryon $a_n=\sum_{k=1}^n1/\sqrt[4]{k^3}$ and $b_n=\sqrt{n}$

Hippalectryon

And where do you see $a_{n-1}-a_{n}$ ?

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$Mathematics$ Mathematics