Mathematics - 2013-06-16

leo

00:03

I think $Y$ must be Hausdorff, because then compact sets in $Y$ are also closed

user1

00:38

@leo I do think $\lambda$ is continuous without any extra conditions on $Y$.

leo

@user1 how?

lets denote $V(A,B)=\{f\in C(X,Y) : f(A)\subseteq B\}$

user1

@leo If $U$ is open in $Y$, note that $F^{-1}$ of $U$ is open in $X\times[0,1]$, where I use $F$ to mean the homotopy.

anon

@leo You're already using the letter $F$. How about $V(A,B)$ instead (as seen on the wikipedia article on compact-open topology).

@PeterTamaroff hello. brb tho.

user1

@leo If $f_t$ maps a compact set $K$ into $U$, then that means $K\times{t}\subseteq F^{-1}(U)$.

leo

the set $\varphi=\{V(A,B):\text{$A$ is compact in $X$ and $B$ is open in $Y$}\}$. This set is a subbase for the compact open topology, so it's enough to show that the inverse image of any of those $V(A,B)\in \varphi$ by $\lambda$ is open on $[0,1]$. I'm stuck o that

@anon thanks

user1

00:44

@leo Let $O\times V$ be an open box around $K\times{t}$ in the preimage of $U$.

leo

okay

user1

I think that the set of all $t$ in $V$ satisfy $f_t\in V(K,U)$

I did not use compactness of $K$, though, which worries me slightly.

leo

@user1 I started that way. I as trying to proof that those $t$ are exactly $v$, which seemed weird to me as well

@user1 With Hausdorffness of $Y$, it follows easy that the complement of those $t$ is closed

I'll think

I hope

user1

01:04

@leo I was wrong to assume that $K\times \{t\}$ fits neatly into an open box. What you have to do is construct a family of open boxes, one for each $\langle k,t\rangle$ and then use the compactness of $K\times \{t\}\subseteq K\times[0,1]$ to show that there is a finite subfamily.

leo

@user1 On it. I'll let you know how it goes

user1

@leo :)

@leo I actually have to go, but I would still like to know if this works.

leo

@user1 I'll tell you later, thanks :-)

Peter Tamaroff

01:39

@anon Are you back?

anon

yes

Peter Tamaroff

Yay

The problem is as follows. Suppose that $f(z)=z+a_2z^2+a_3z^3+\cdots$. We can write $$\frac{zf'(z)}{f(z)}=A(z)=1+A_1z+A_2z^2+\cdots$$
Show that if $|A_n|\leq 2$ then $$|a_n|\leq n$ for each $n$.

It is what I showed you before.

The best I could do is $$n|a_n|\leq 2\sum_{k=1}^n |a_k|$$

But that recurrence gives me a bigger bound, I think.

@anon

anon

01:54

dunno

Peter Tamaroff

@anon Bahoolas.

I bet Brian Scott can figure this one out.

leo

02:15

@user1 The key observation is $$\lambda^{-1}(V(K,U))=\{t\in I:K\times \{t\}\subseteq F^{-1}(U)\}$$

Ethan

04:19

If $c_q(n)$ is a ramanujan sum
Then, $$\frac{(1-x^a)^m}{a}\sum_{t=0}^{a-1}\frac{c_a(a-t)}{(1-e^{2\pi i t/a})^m}=\sum_{\gcd(a ,\text{ S})=1}x^{\text{S}}$$
Where the rhs ranges over all $m$ tuples, $\text{S}$ whose sum is co-prime to $a$ in $\mathbb{(\frac{Z}{Za})^m}$

Ethan

05:03

$$0<\sum_{n\leq x}\ln(\frac{\gcd(a,n)}{a})<\ln(a)$$

$\forall a$

If $\chi$ is the non principal character modulo $4$, and
$$\sum_{n=1}^\infty\frac{q(n)}{n^s}=\frac{L(s,\chi)\zeta(s)}{\zeta(2s)}$$
And $\Pi(x)$ is the number of primes less then or equal to $x$ of the form $n^2+1$,
$$\Pi(x)=\frac{4\sqrt{x}}{\ln(x)}\sum_{n\leq \sqrt{x}}\frac{q(n)\mu(n)\ln(n)}{n}+\text{somthing small}$$

$$\sum_{n\leq x
}_{n\equiv b\text{ mod }a}d(n)=\frac{\phi(a)}{a^2}x\ln(x)+\frac{\phi(a)}{a^2}x(2\gamma-1+2\sum_{p\mid a}\frac{\ln(p)}{p-1})+O(x^{1/2})$$

$\forall a,b \mid \gcd(a,b)=1, 1\leq b\leq a$

Ethan

10:08

$$\int_{0}^\infty\frac{\sqrt{2}}{\sqrt{e^t-1}}\frac{\sqrt{2}}{\sqrt{e^{t/2}+1}}\frac{\sqrt{2}}{\sqrt{e^{t/4}+1}}\frac{\sqrt{2}}{\sqrt{e^{t/8}+1}}\frac{\sqrt{2}}{\sqrt{e^{t/16}+1}}....dt=\sqrt{\frac{\pi}{2}}\zeta(\frac{3}{2})$$
$$\sum_{n=-\infty}^\infty\frac{2^n}{q^{2^n}+1}=\frac{1}{\ln(q)}$$

Julian Kuelshammer

12:04

This "emptying-the-unanswered-queue" business makes me reaching my daily-vote limit each day. Today 11h before the end of the day.

skullpatrol

It is a noble ~~cause~~ crusade :-)

Julian Kuelshammer

:-)

skullpatrol

As Feynman said "Know how to solve every problem that has been solved."

Julian Kuelshammer

There should be some more "crusaders". At the moment we are making some progress, but very slowly. Although sometimes it just needs checking a simple linear algebra answer and then upvote.

skullpatrol

12:46

Hi @amWhy

amWhy

@skullpatrol Hello, skull :)

skullpatrol

@amWhy How are you?

amWhy

@skullpatrol good...still half-asleep, but coming to... How are you?

skullpatrol

@amWhy Fine thanks.

user1

@leo I agree; it contains the problem in a space we know a lot more about (when comparing $K\times I\subseteq X\times I$ with $Y$).

skullpatrol

12:57

@amWhy did you see anon's answer?

amWhy

@skullpatrol where?

skullpatrol

@amWhy here

amWhy

It's saying pretty much what's already been said.

skullpatrol

@amWhy why was it made into a community wiki?

amWhy

if you exclude $x = y$, then the first method can be employed...but excluding $a = b$ means you omit a possible scenario. The second method doesn't exclude a = b.

anon chose to post as "CW".

skullpatrol

13:05

@amWhy There is no y?

amWhy

I meant if you exclude $a = b$...sorry.

skullpatrol

np

@amWhy Thanks for all the help, I appreciate it :-)

amWhy

@skullpatrol You're welcome! It's always a pleasure, especially when the "asker" is a "bud"...

skullpatrol

:D

amWhy

@skullpatrol I like your $\ddot \smile$

skullpatrol

13:13

@amWhy Thanks

DanZimm

13:45

How come I feel so inadequate when I look at math graduate programs?

skullpatrol

13:58

Maybe because you're not a math graduate :P

DanZimm

:P

skullpatrol

Just my opinion, please don't be offended...

PooyaM

Hi there!

skullpatrol

Hi

PooyaM

@skullpatrol What is your major?

skullpatrol

14:05

@PooyaM I don't have one.

@PooyaM What is yours?

PooyaM

Electrical Engineering, I was looking for a Mathematics major

@skullpatrol Electrical Engineering, I was looking for a Mathematics major

DanZimm

@skullpatrol no, I am nowhere near a math graduate - still in undergrad

Julian Kuelshammer

hi

DanZimm

hello

Julian Kuelshammer

@PooyaM you need one for the US or not country-specific?

PooyaM

14:12

@JulianKuelshammer Actually I want to talk to a Math major...

@JulianKuelshammer here

@JulianKuelshammer Seems like you are a Post-Doc in Math!!!

Julian Kuelshammer

@PooyaM yep

BenjaLim

@JulianKuelshammer Massive fail

PooyaM

@JulianKuelshammer Great...I'm looking for double majoring with Mathematics...What is your opinion?

Julian Kuelshammer

@PooyaM That's always difficult to tell when you don't have that much information. That's why I asked which country you are in. It depends for example on how much the overlap is between mathematics and electrical engineering. And of course also on how you are currently doing with your math courses. From my perspective (teaching some of the engineers) what they do in their math courses is often quite different from what a mathematician does.

PooyaM

14:30

@JulianKuelshammer I agree with you about the difference of teaching mathematics in engineering schools and math schools...Actually, I've taken some online mathematics courses like Intro. of Algorithms(71/100), Linear and Discrete Optimization(Didn't finished it because of university midterms but till week 5/6 I completed 100/100 and I barely found complications throughout the course) Cryptography(which is little far from math ,97/100). I live in Iran, so I chose Electrical engineering to

@JulianKuelshammer get a good job. The other problem is if I wanted to pursue my interest, there were too many and too diverse. Even in Mathematics, I can't choose an area to do research in. Optimization, Number theory,Graph theory, Complexity theory and AI. I think that is the reason why I didn't pursue Math or Physics major at the first place!

Julian Kuelshammer

@PooyaM What about talking to some math professor at your local university? They probably know better, what is possible, or maybe they also know people who already did what you are asking about.

nitrous2

Should I delete my response

1

Q: How to calc the square root of a number without calculator?

How can I find the square root of a number without using a calculator?

PooyaM

@JulianKuelshammer I talked to three professors in out Math dept. They told me that if you don't want to be a professor, don't double major with math. They told me you should choose between a well paid job(which is earned by pursing Engineering) and a low paid stressful job a.k.a being a professor in math

@JulianKuelshammer I just don't know what to do. I find courses like Electrical Machines and Microelectronic circuits too naive. I see no deep idea in engineering courses but simple,linear,approximate yet working and robust

Julian Kuelshammer

14:46

@PooyaM To be honest I know nothing about the job situation in Iran. In Europe there are plenty of jobs for mathematicians outside University, like working for insurances, for banks, as a consultant, or also in the physics/engineering direction. Here, I think most people would see it as an advantage to have a degree in math, but such things differ between countries. I can understand that you are lacking deep ideas (maybe also reasoning) in engineering courses from what I'm teaching engineers.

PooyaM

@JulianKuelshammer For example in Digital Logic Design we learned " mealy and moore algorithm " . But by learning I mean, I know how to do it just like the computer does! I don't know how that algorithm is designed. I think these have something to do with Markov Processes. And every time I read some articles in Wikipedia discussing these ideas, my heart starts beating and I think I'm wasting my time in linear time independent systems!

@JulianKuelshammer And here also there is job outside of univ. for math majors. But I don't think a bank clerk thinks about proving whether P=NP

Julian Kuelshammer

@PooyaM Usually not, that's right. :-)

PooyaM

@JulianKuelshammer I think,the only way I have is choosing a single area and do research in it. And I should self study the prerequisites. But I loved to be like Von Neumann and Shannon

@JulianKuelshammer and Gauss and Poincare and Decartes and Leibniz :( :( :( :(

Julian Kuelshammer

@PooyaM then you have to do a lot of research :-) Your professors might be right: Then it is a stressful and (at least in comparison) low-paid job. You have to make the decision. It's not obvious in the beginning how far one can get.

PooyaM

@JulianKuelshammer Thank you for talking with me :)

robjohn

15:38

@nitrous2 I assumed that the OP was asking for a numerical method to get square roots, as one would get using a calculator, but not using a calculator. What you wrote is correct, just not what the OP was looking for (imo).

robjohn

15:55

It certainly is quiet right now

Peter Tamaroff

16:38

@robjohn scream

robjohn

@PeterTamaroff why?

Peter Tamaroff

@robjohn Sometimes there is no reason.

Or maybe there is, but one knows not why.

robjohn

okay

Peter Tamaroff

@robjohn I have a problem.

robjohn

yes?

Peter Tamaroff

16:40

@robjohn For a few days now.

Let $\langle a_n\rangle$ be a sequence of complex numbers, and $\langle p_n\rangle$ a sequence of non negative real numbers. Let $A(z)$ and $P(z)$ be their related powerseries. We write $A(z)\ll P(z)$ if $|a_n|\leq p_n$ for each $n$.

I have proven that $A\ll P$ and $B\ll Q$ implies $A+B\ll P+Q$ and $A\cdot B\ll P\cdot Q$.

Now, the problem is as follows:

Suppose $f(z)=z+a_2z^2+a_3z^3+\cdots$ (yes, $a_0=0,a_1=1$). Suppose further that $$\frac{zf'(z)}{f(z)}\ll \frac{1+z}{1-z}$$ Show that $|a_n|\leq n$ for each $n$.

Note that if we write $A(z)=zf'(z)/f(z)$, then $A(0)=1$. So we can write $$A(z)=1+\sum_{k\geq 1}A_kz^k$$ The above means that $|A_k|\leq 2$, since $$\frac{1+z}{1-z}=1+2z+2z^2+2z^3+\cdots$$

The best I could do is deduce $$n|a_n|\leq 2\sum_{k=1}^n |a_k|$$

@robjohn =)

robjohn

Nothing comes to mind immediately. I will think about it.

Peter Tamaroff

16:56

@robjohn Thank you.

AlanH

hello y'all

robjohn

$$
1+2a_2z+3a_3z^2+4a_4z^3+\dots\ll1+(a_2+2)z+(a_3+2a_2+2)z^2+(a_4+2a_3+2a_2+2)z^3+\dots
$$

simple algebra and induction after that, it seems

AlanH

17:13

@robjohn When I use the bookmark start Chatjax, it doesn't work. I have to click on render Chatjax every time instead. Do you know why this could be? (Does it matter where I keep the bookmark, say in my add-on bar for Firefox instead?)

robjohn

17:31

@AlanH If the renderMathJax bookmark works, then if the start ChatJax bookmark is installed in the same place, it should work.

The only thing they do differently is start ChatJax starts a loop to rerender when needed

Peter Tamaroff

@robjohn You cannot have complex numbers on a series that is a majorant, can you? A majorant is supposed to be of positive real numbers. I mean, on the RHS of the $\ll$

I mean, the algebra of $\ll$ is not the same of $\leqslant$.

AlanH

I'm reading Ahlfors and he says different regions may have the same closure. Can someone provide an example of this? I don't quite see it

Peter Tamaroff

@AlanH Take the open disk $D=\{|z|<1\}$

And then take $D\cup \{(1,0)\}$

Or $D\cup A$ where $A\subseteq \partial D$

More dramatically $\overline{\Bbb Q}=\overline{\Bbb R}=\Bbb R$ in the metric space $\Bbb R$ with the usual topology.

AlanH

So $D$ union any piece of the unit circle has the same closure as D alone, is what you're saying?

Peter Tamaroff

@AlanH Yes.

Because in fact $\overline{D}=D\cup\partial D$.

AlanH

17:46

@PeterTamaroff ah. thanks.

robjohn

So use the same idea, but write
$$
\frac{f'(z)}{f(z)/z}=1+b_2z+b_3z^2+b_4z^3+\dots
$$
then
$$
1+2a_2z+3a_3z^2+4a_4z^3+\dots=1+(a_2+b_2)z+(a_3+a_2b_2+b_3)z^2+(a_4+a_3b_2+a_2b_3+b_4)z^3+\dots
$$
subtract to get
$$
a_2z+2a_3z^2+3a_4z^3+\dots=b_2z+(a_2b_2+b_3)z^2+(a_3b_2+a_2b_3+b_4)z^3+\dots
$$
Then use that $|b_k|\le2$ to get
$$
a_2z+2a_3z^2+3a_4z^3+\dots\ll 2z+2(|a_2|+1)z^2+2(|a_3|+|a_2|+1)z^3+\dots
$$
then it is algebra.

Peter Tamaroff

18:08

@robjohn Yes, that is exactly what I got, sire.

I think?

I must have slipped somewhere. You're getting $(n-1)|a_n|\leq 2\sum_{k=1}^{n-1}|a_k|$

robjohn

@PeterTamaroff Then you use that $2(1+2+3+\dots+n)=(n+1)n$ and cancel

Peter Tamaroff

@robjohn Heh, I wasn't too far away. Damn me.

My induction hypothesis is $|a_k|\leq k$ for $k=1,\dots,n-1$. Done.

I will rewrite this to get it neat and tidy.

@robjohn Thank you dearly.

robjohn

@PeterTamaroff no problem

amWhy

Hi, @robjohn

Hi, @Peter

robjohn

what's up?

amWhy

18:17

@robjohn you asking me? Not much...just saying "hello"...

robjohn

@amWhy okay. Hello

Peter Tamaroff

@robjohn What are you up to this Sunday?

AlanH

Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is separable. *Hint* Fix $\delta > 0$, and pick $x_1 \in X$. Having chosen $x_1, \dots, x_j \in X$, if possible, so that $d(x_i, x_{j+1}) \geq \delta$ for $ i=1,2\cdots,j $. Show that this process must stop after a finite number of steps, and that $X$ can therefore be covered by finitely many neighborhoods of radius $ \delta $. Take $ \delta = 1/n (n = 1,2,\cdots) $, and consider the centers of the corresponding neighborhoods.

Peter Tamaroff

@AlanH That is from Rudin

AlanH

Yeah

@PeterTamaroff Are you able to view the whole thing? I copied it from texshop

Peter Tamaroff

18:30

@AlanH Let me try and explain.

Fix $\epsilon >0$, and $x_1\in X$ arbitrary. First, I want to show that for this given $\epsilon$ we can find a finite set of points $\{x_1,x_2,\dots,x_p\}$ such that the $p$ open balls $B(x_i,\epsilon)$ cover $X$.

AlanH

@PeterTamaroff You're just showing the dense part right? Or what part are you explaining?

Peter Tamaroff

Now, if $X\subset B(x_1,\epsilon)$, we're done. If not, there exists $x_1\neq x_2$ such that $d(x_1,x_2)\geq \epsilon$. If $X\subset B(x_1,\epsilon)\cup B(x_2,\epsilon)$ we're done. If not, there exists $x_3\neq x_2,x_1$ such that $d(x_1,x_3),d(x_2,x_3)\geq \epsilon$.

Now, if this process didn't terminate, we would find a sequence $\{x_i\}$ in $X$ such that $d(x_i,x_j)\geq \epsilon$ if $i\neq j$, which is absurd, since this would have no convergent subsequence, hence no limit point.-

AlanH

yes, exactly

Peter Tamaroff

Now, we have proven in particular that for each $n\in\Bbb N$ there exists an integer $p_n$ such that $$\{x_{n,1},\dots,x_{n,p_n}\}$$ is a $1/n$-net for $X$; that is, the set of open balls with radius $1/n$ and centers at the $x_{n,i}$ cover $X$.

Now consider the countable set $$\{x_{n,i}:n\in \Bbb N\; , 1\leq i\leq p_n\}$$

Show this is dense in $X$.

@AlanH I mean that $$\bigcup_{i=1}^{p_n} B(x_{n,i},1/n)\supset X$$

AlanH

@PeterTamaroff right, so I thought it was just dense by the fact that we have an open cover. Every point in X is either a limit point or is one of those centers

Peter Tamaroff

18:39

@AlanH This last part shouldn't be much of a fuzz. Fix $\epsilon >0$, and take $x\in X$, and let $K=\{x_{n,i}:n\in \Bbb N\; , 1\leq i\leq p_n\}$. We want to show $B(x,\epsilon)\cap K\neq \varnothing$.

Now, take $N$ such that $1/N<\epsilon$.

AlanH

okay okay stop

why is it necessary to do that?

that's where I was confused. isn't our construction already enough to say its dense by open covers?

Peter Tamaroff

@AlanH That is what dense means: $K$ is dense in $X$ iff for each $\epsilon >0$ and $x\in X$, $K\cap B(x,\epsilon)\neq \varnothing$, that is $\overline K=X$.

AlanH

I was using the definition that every element of $X$ is either a member of the set (in this case the centers), or is a limit point of that set. equivalent, no?

Peter Tamaroff

Having taken such $N$, note that $$X\subseteq \bigcup_{i=1}^{p_N}B(x_{N,i},1/N)$$ But then $x$ must be in some of said balls, so $d(x_{N,j},x)<1/N<\epsilon$ for some $1\leq j\leq p_N$, so $x_{N,j}\in B(x,\epsilon)\cap K$.

@AlanH Sure.

AlanH

@PeterTamaroff or maybe you should just finish explaining and i'll ask my questions afterwards

Peter Tamaroff

18:45

@AlanH I'm done.

You can read it now, think about it. It is a nice exercise.

AlanH

@PeterTamaroff So are there other points in $X$ besides the centers?

Peter Tamaroff

@AlanH You mean in $K$; yes?

robjohn

@PeterTamaroff just doing some things around the house. Probably go out for an early dinner.

Peter Tamaroff

@robjohn Do you have Father's Day there?

robjohn

@PeterTamaroff I think it is a pretty universal holiday

Peter Tamaroff

18:49

@robjohn Heh, didn't know.

Mother's Day is on a different day there.

robjohn

@PeterTamaroff really?

when is it there?

Peter Tamaroff

@robjohn October 20th

AlanH

@PeterTamaroff Why do you say $K$? Earlier you stated to fix $\epsilon$ and take $x\in X$.

robjohn

@PeterTamaroff That is quite a ways off. Is it always that date or is it always a Sunday?

Peter Tamaroff

@AlanH Could you repeat your question?

BenjaLim

18:53

5am here in australia

AlanH

@PeterTamaroff K is just the set of centers, right? Is there anything else in K?

Peter Tamaroff

@AlanH Nothing else. The centers.

They are the points that make the $1/n$-nets for $X$ for each $n=1,2,\dots$.

AlanH

@PeterTamaroff Then what is X comprised of?

Peter Tamaroff

@AlanH Isn't $X$ your space to begin with?

You said "Let $X$ be a metric space..." ¬¬

AlanH

@PeterTamaroff yeah

Peter Tamaroff

18:56

@AlanH So...?

@BenjaLim 4pm here!

AlanH

@PeterTamaroff I don't know, I"m getting really confused. I've been on this problem for two days now. bah.

Peter Tamaroff

@AlanH OK, let's start over.

AlanH

@PeterTamaroff no no, its just the dense part I'm having trouble with

Peter Tamaroff

@AlanH OK. The two important things are: for each $\epsilon$, we can make $1/n<\epsilon$ for some $n$. For each $n$, $$\bigcup_{i=1}^{p_n}B(x_{n,i},1/n)$$ covers $X$.

So, pick $x\in X$ in your space, and $\epsilon >0$.

To show $K$ is dense, we ought to show $B(x,\epsilon)\cap K\neq \varnothing$ for each $x\in X$ and $\epsilon >0$. This is saying that $X\subseteq \overline{K}$. Since trivially $\overline K\subseteq X$, we get that $\overline K=X$, that is, $K$ is dense.

But given $\epsilon >0$; we can pick $n_0$ such that $1/n_0<\epsilon$. Since the balls corresponding to said $n_0$ cover $X$, $d(x,x_{n_0,j})<1/n_0<\epsilon$ for some $j$ (that is $x$ must be in some of the balls, here I'm saying it is in the ball with center in $x_{n_0,j}$), which means that $x_{n_0,j}\in B(x,\epsilon)\cap K$. Do you see?

skullpatrol

19:17

Hi @MarianoSuárez-Alvarez

skullpatrol

19:28

@MarianoSuárez-Alvarez how are you?

skullpatrol

19:47

Hi @Argon

Hi @keri welcome to the chat.

keri

hello

skullpatrol

@keri how are you?

keri

awesome, just spending the evening with graph theory :)

and u?

skullpatrol

@keri Fine thanks :-)

keri

good :D

skullpatrol

19:53

:D

keri

why is it so quiet?

skullpatrol

Sometimes it gets that way, while other times you can't get a word in edge-wise...

...it really depends on who is talking.

keri

So let's throw a math party :D

skullpatrol

We have a "Party Zone" for that :D

$Mathematics$ 0

keri

I mean something more math than party ;)

skullpatrol

20:03

Oh... icic

;-)

@keri Gotta run, see ya later.

keri

bye

skullpatrol

:D

Argon

@skullpatrol Hi Skull

skullpatrol

@Argon Hi/bye

4 mins ago, by skullpatrol

@keri Gotta run, see ya later.

Argon

@skullpatrol Bye

skullpatrol

20:11

:D

Peter Tamaroff

20:44

@anon

anon

yes

Peter Tamaroff

Remember my problem the other day, about $\ll$?

anon

the one starred on the side?

Peter Tamaroff

@anon Oh, yes. Didn't see it was starred.

I have my screen with too much zoom.,

Well, I should be getting that $$n|a_{n+1}|\leq 2\sum_{k=1}^n |a_k|$$ Assuming $|a_k|\leq k$ for $1,\dots,n$ one gets $$n|a_{n+1}|\leq 2\sum_{k=1}^n k\implies |a_{n+1}|\leq n+1$$ so the assertion is proven.

anon

pretty sure you wrote $n|a_n|\le $ yesterday (otherwise I could have told you that!)

Peter Tamaroff

20:48

@anon Sorry?

anon

19 hours ago, by Peter Tamaroff

The best I could do is $$n|a_n|\leq 2\sum_{k=1}^n |a_k|$$

Peter Tamaroff

Oh, well, I keep failing in my manipulations, yes.

anon

I attempted the induction idea using what you wrote yesterday, which had $a_n$ on the left instead of $a_{n+1}$.

Peter Tamaroff

@anon But it fails, doesn't it?

Dominic Michaelis

does one know a category theoretical proof of sylow group theorems ?

anon

20:51

@PeterTamaroff you just proved that $|a_k|\le k$ for $k=1,\cdots,n$ implies $|a_{n+1}|\le n+1$, did you not?

or by "I should be getting that [blah]" are you saying that you haven't got [blah] but you think you need to get there (which sounds reasonable)?

Peter Tamaroff

@anon Yes, that is what I'm saying. I should be getting that, but I am not. =)

I must be overlooking something in my manipulations. Gurrdarmn.

Does anyone know why movies look so much homemade in a "HD" television (compared to the more unrealistic one has in "older" TVs)? The image has a faster dynamic, for example, makes it look strange.

@anon Why are there so many groups of order $2^n$?

anon

@PeterTamaroff framerate + motion interpolation

Peter Tamaroff

@anon So, how does one fix that? Watching newer movies? Even "The Dark Night Rises" looked silly, or "The Avengers", though this one not so much.

anon

@PeterTamaroff that isn't entirely clear to me. it follows intuitively from the conjectre that the number if iso classes of groups of order p^n grows as p^O(f(n)), and the fact that f(n) is independent of p is the more fundamental fact in need of explanation

I do not have an hdtv, so I do not share your 1st world problems :-)

Peter Tamaroff

@anon What is $f(n)$?

anon

21:03

in the context of my response it doesn't matter what f(n) is, but it happens to be more precisely put as (2/27)n^3+O(n^(8/3))

Peter Tamaroff

@anon I hadn't changed my TV set since 2007/8. =P

Dominic Michaelis

@PeterTamaroff @PeterTamaroff I always thought it is "The Dark Knight Rises"

Peter Tamaroff

@DominicMichaelis \MEFACEPALMS

I surely meant that.

Dominic Michaelis

Finally i found a mistake :D

Peter Tamaroff

OK, this is probably the tenth time I will do this calculation.

Dominic Michaelis

21:16

so the 10 th pairwise different answer ?

Peter Tamaroff

@DominicMichaelis No, I am doing something wrongly and I cannot see where.

Dominic Michaelis

It is to late for me to be a help

Bageer

Finally, I can vote people down! (starting power trip) I will probably have to wait till school starts again before I get to put it to good use through.

Peter Tamaroff

FUCK YEAH. Did it right this time. @anon I got it.

I have to get me a treat now.

mick

22:03

@keri define math party with ZFC

what is a math party ?

hey guys

http://en.wikipedia.org/wiki/Yitang_Zhang
Is this for real ? Are we closer to the proof of prime twins ? I think he is wrong.

@PeterTamaroff hi

@robjohn hi

@anon hi

Peter Tamaroff

@anon Let $B_n$ denote the possible ways of writing $n$ as a sum of $1,2,3,4$. Then $$\sum_{j\geq 0}B_j\zeta^j=\prod_{k=1}^4(1-\zeta^k)^{-1}$$ Yes?

Note that order counts.

@mick On what grounds?

anon

@PeterTamaroff yes, but order does not count

Peter Tamaroff

@anon Oh, then I'm wrong?

mick

@PeterTamaroff because most sieve methods assume some structure such as multiplicative functions or are probabilistic in nature. Further Arxiv is full of false proofs of the twin prime conjecture based on sieves.

Peter Tamaroff

@anon "Two sums consisting of the same terms but in different order are regarded as different."

AlanH

22:22

@PeterTamaroff @PeterTamaroff So we know that there exists $x$ such that $d(x,x_{n_0,j}) < 1/n_0 < \epsilon$ because the union of open balls with radius $1/n_0$ covers $X$. Is this correct? I think the whole time I was struggling with why such an $x$ existed.

@anon greetings

Peter Tamaroff

@AlanH Wait!

AlanH

waiting...

Peter Tamaroff

We picked an $x\in X$ arbitrarilty. What we want to show is that $x_{n_0,j}$ exists and is in $B(x,\epsilon)\cap K$!

AlanH

@PeterTamaroff okay, so the question is flipped, how do we know there exists such an $x_{n_0,j}$ that is within an epsilon distance?

mick

Maybe I need lessons in sieve theory @PeterTamaroff

Peter Tamaroff

22:24

:9918755 Because we chose $n_0$ so that $1/n_0<\epsilon$

And because for any $n$, $$X\subseteq \bigcup_{i=1}^{p_n}B(x_{n,i},1/n)$$

AlanH

@PeterTamaroff Okay, now i believe it. Thanks for taking the time to help.

Shall I repay you in starred messages or by voting up some answers? The choice is yours

Peter Tamaroff

@AlanH I don't care about starring.

robjohn

@mick hey there... what's up?

well, off to get Father's Day dinner. BBL

mick

BBL ?

Bon appetit mon ami !

Peter Tamaroff

@mick "Big Baked Lasagna"

mick

22:37

aha

im looking for introductions into sieve theory for dummies like me :)

@PeterTamaroff @robjohn @anon

Peter Tamaroff

@mick No clue, Mick.

mick

something easier than Terry tao's writings plz

sieve theory is so different from other math or it seems ...

:/

Peter Tamaroff

@anon I am doing some combinatorics.

anon

23:14

@PeterTamaroff I am eating some caramel delights.

Peter Tamaroff

@anon I need an "obviousmeter".

$$(1+x)(1+x^2)(1+x^3)\cdots =\frac1{(1-x)(1-x^3)(1-x^5)\cdots}$$

One can see that if you multiply by $1-x$ on the LHS, all the terms of the form $(1+x^{2^k})$ die.

Then, upon multiplication by $1-x^3$, all terms of the form $1+x^{3\cdot 2^k}$ die.

And so forth.

anon

mhmm

Peter Tamaroff

So, this is saying "every integer can be written in the form $\text{odd}\cdot \text{pureblood even}$" @anon

anon

yes

Peter Tamaroff

@anon What kind of treat are you having?

anon

23:23

these mofos

Peter Tamaroff

@anon I am speechless.

@anon cf. 0:30

@anon What does that have?

$Mathematics$ 0

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$Mathematics$ Mathematics