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12:00 AM
Any fun integrals today?
 
well not really I have been struggling with a ramsey theory thing for the last few days would be nice to do integrals instead
 
What's Ramsey Theory?
(Pardon my ignorance)
 
ssorry was brushing my teeth
 
np
 
the thing I want to understand is basically if you are coloring in a cube there'l always going ot be an arithmetic progression inside it
if we make the dimension of the cube high enough with respect to number of colors and width in the cube
 
12:05 AM
@cassandra0 Progression?
 
like
a straight line which all the same color
 
What do you mean by colouring a cube? The faces?
 
sorry I explained it really badly, but this will help me be much clearer in my own head if I get this right
 
Ok. Glad to help, hahaha!
 
so imagine a 5 dimensional cube divided into n*n*n*n*n subcubes
and we color each one in one of 3 colors say
 
12:10 AM
Ok
 
then there should be a straight line or diagonal line or something, inside it, which is all the same color
 
A diagonal of coloured cubes, I presume?
 
@aDangerousIdea I can't make myself study though, I mostly dont even care to or enjoy doing it so I do something else to take my mind off doing poorly in school
 
@Jordan What do you like to do?
 
@Argon I am exceptionally lazy, but mostly video games, TV, reading and of course wasting a lot of time on reddit
 
12:14 AM
@Jordan Haha reddit
 
Yeah, it is an easy way to spend time doing nothing
 
> In other words, the higher-dimensional, multi-player, n-in-a-row generalization of game of tic-tac-toe cannot end in a draw, no matter how large n is, no matter how many people c are playing, and no matter which player plays each turn, provided only that it is played on a board of sufficiently high dimension H.
that's actually quite a good way to say it
any number of players, any width of board... there is a dimension high enough that the game will end in someone winning
 
Cool! I sorta understand it :)
 
simply because any board filled in with that many colors will have a monochromatic line in it
the proof is a really hard to follow double induction
 
Hmmm... Could you have a factional dimension?
 
12:21 AM
if something has fractional dimension it's a fractal isn't it
 
No idea. Just throwing it out there!
 
@cassandra0 I don't think so
At least, not necesarilly.
 
Hi @Peter!
 
Maybe?
 
@PeterTamaroff, why not
 
12:23 AM
Cantor's set has "Hausdorff" dimension $$\log 2/\log 3$$ IIRC
 
that's a fractal!
 
user19161
How about Pedro's set?
 
leo
hello there!
 
Hi
 
user19161
Hey @amwhy have you had dinner?
 
user19161
12:36 AM
@Argon Awaiting your integral of the day.
 
leo
@PeterTamaroff hola, como va todo?
 
Argon's Integral of the Day: $$\int_0^\infty \frac{\sin\alpha x}{1-e^{\beta x}}\, dx=\frac1{2\alpha}-\frac{\pi}{2\beta}\;
\text{coth}\left(\frac{\alpha \pi}{\beta}\right)$$
@WillHunting
 
I like the calculations about holomorphic functions which have number theory consequences
 
user19161
@Argon Have you done your homework?
 
leo
have you seen parametrizations $\delta$, $\gamma$ of the unit circle in $\Bbb R^2$ (or part of it) so that $\delta\circ\gamma^{-1}$ is not of class $C^\infty$?
 
12:42 AM
@WillHunting Yep
How about you?
 
user19161
@Argon I have no homework! Anyway, maybe I should take a break from SE now...
 
@WillHunting Why?
 
user19161
@Argon Just a thought. I like to utter random thoughts to Aaron.
 
@WillHunting I see. You are a banana.
 
user19161
@Argon Haha, and you are an apple.
 
12:45 AM
@WillHunting Why apple?
 
user19161
@Argon A for Aaron, A for apple. QED.
 
@WillHunting And B for [CENSORED]. QED
 
user19161
@Argon WTF! Shouldn't I be a jackfruit?
 
@WillHunting J for Jambul
 
user19161
@Argon HAHAHAHA, OK I shall be jambul from now, not a banana!
 
12:47 AM
@WillHunting Hahahaha!
 
user19161
@Argon Hmm I think I will send you an email in a minute.
 
@WillHunting Okayz
 
@Argon, do you know any nice number theory?
 
@cassandra0 Not really. My math is sadly quite limited.
Prime number theorem is cool!
 
user19161
@Argon What's prime number theory?
 
1:00 AM
prime number theorem is really hard, but it's related to the zeta function you can do a lot of analysis of it
 
@WillHunting Theorem, sorry
 
something like the zeta(1 + it) I think has no zeros
 
$$\pi(x) \sim \frac{x}{\log x}$$
Which is awesome
 
I actually have to learn the prime number theorem
I don't really understand it
 
It's really deep
 
1:05 AM
RH is just a little side remark that Riemann made while proving PNT.
 
@WillHunting Your favourite hypothesis!
 
user19161
@Argon Yes, we will work on it one day. You do the proof and I do the book cover.
 
XD
 
there's a combinatorics bit due to Chebychev where you prove that c_1 log(x)/x < pi(x) < c_2 log(x)/x
that's quite simple actually I understand that
 
Then, I guess, you must show that $c_1, c_2 \to 1$
 
1:08 AM
oh no these are fixed constants
 
1 and 1?
 
haha
 
:)
 
I didn't actually calculate them I should do that
 
Bleh, I need to learn more math
 
1:13 AM
what does $\sim$ mean?
 
Asymptotic
 
$f(n) \sim g(n)$ means $f(n)/g(n) \to 1$
limit as n to infty
 
$$f(x) \sim g(x) \iff \lim_{x \to \infty} \frac{f(x)}{g(x)} \to 1$$
 
thanks
I'm just trying to understand why $c_1 \log(x)/x < \pi(x) < c_2 \log(x)/x$ doesn't give that
I mean suppose $c_1$ is like $10^{-32}$ and $c_2$ is $2^7$
I will calculate them at some point from the proofs I've got but that'll do for now
 
Divide all by $\frac{\log x}{x}$
Then $$c_1 < \frac{x\pi(x)}{\log x}< c_2$$
Right?
 
1:15 AM
I see
 
And as $x\to\infty$, $c_1 < 1 < c_2$ by PNT
 
so what PNT is telling us is that $\pi(x)$ actually converges TOWARDS $y = \frac{\log(x)}{x}$. it's not just $O(\frac{\log(x)}{x})$
is that right?
I think it's right..
 
Hmmm....
 
like maybe primes 4k+1 converge towards the line $y = \frac{1}{2}\frac{\log(x)}{x}$
seems that
$\pi(x)$ converges towards $y = \frac{\log(x)}{x} + c$
 
No need for $c$, as it becomes insignificant, no?
 
1:23 AM
it would be in $\pi(x)/y(x)$
become insignificant there
 
@leo Ah! Me perdi tu mensaje.
 
but I think $\pi(x)/y(x) \to 1$ implies only than $\pi(x)$ tends towards $y(x)$ plus some constant
 
@cassandra0 What are you discussing?
 
@PeterTamaroff, jus trying to understand the meaning of the result of prime number theorem
 
Consider the following $$\lim_{x\to\infty}\sum_{n=1}^x \frac{1}{x} - \log x = \gamma $$
 
1:26 AM
it says $
\pi(x)\sim\frac{x}{\ln x}.\! $ which means that as $x \to \infty$ we have $\frac{\pi(x) \ln(x)}{x} \to 1$
 
@cassandra0 " The prime number theorem gives a general description of how the primes are distributed amongst the positive integers."
 
But still $$\sum_{n=1}^x \frac{1}{x} \sim \log x$$
 
@Argon That is awfull notation!
Hehehehehe
 
oh my gosh this is so confusing
 
@PeterTamaroff Oh noes! :)
 
1:28 AM
why bloody way up does the fraction go
 
?
 
@Argon, what you've said I understand
 
Ok
 
@Argon Not necesarily. We can say that when $x\to a$ with $a\in \Bbb R$
For example, $\sin x \sim x$ when $x\to 0$
 
@PeterTamaroff Not necessarily what?
Oh, yes
 
1:30 AM
Or more nicely $\sin x= x+o(x)$ when $x\ to 0$
 
You are right, the limit must be given somewhere
 
see n/log(n) seems to be a constant away ?
 
@cassandra0 If you really want to understand the prime number theorem, you should study some analytic number theory first
 
@cassandra0 There exists no constant $C$ such that $\pi(x)+C = \frac{x}{\log x}$
 
@Argon I think that is courtesy of Chebyshev
 
1:33 AM
@Argon, I know but isn't there a constant $C$ such that $\pi(x) + C$ converges towards $\frac{x}{\log x}$? In the sense that $|(\pi(x) + C) - \frac{x}{\log x}| \to 0$?
I just don't really understadn what $\pi(x) \sim \frac{x}{\log(x)}$ tells us
 
Do you know little o?
 
yes
 
$f\sim g \implies f(x) = g(x)+o(g(x))$
 
I see
so if we have $\varepsilon > 0$ there's some $x$ such that $|f(x)-g(x)| < \varepsilon |g(x)|$
but $g(x)$ grows so the bound could be as large as anything
 
Yes
 
1:40 AM
so why is PNT actually stronger than Chebychevs result?
 
Not sure. I really don't know enough number theory to give a good answer.
 
$$c_1 \log(x)/x < \pi(x) < c_2 \log(x)/x$$
this is chebychevs result
@PeterTamaroff, any views on this
 
@cassandra0 Yes.
 
Yay!
Pedro to the rescue.
 
user19161
The jambul listens.
 
1:42 AM
:)
Hi @Will !
 
I actually asked about it
Jambul?
 
HAHAHAHA!
 
this is very useful to me but it's not the same result of chebychev
 
user19161
@PeterTamaroff I am Jambul and Aaron is Apple. QED.
 
user19161
Hey @amwhy we both used ellipses on the comments to the question, haha.
 
1:47 AM
@WillHunting We both have quadrilaterals as our avatars.
 
@WillHunting My use of ellipses is contagious!
 
@amWhy Are you Charlie's replacement?
 
Hahahahaha!
 
@PeterTamaroff How so?
 
hi @amWhy
 
1:49 AM
@amWhy Well...
 
do you do number theory?
 
@cassandra0 Hey!
 
Charlie is in this room!
 
That is completely inappropriate!
 
@cassandra0 hi cassandra!
 
1:49 AM
sorry I just wanted to know
 
@PeterTamaroff Geez
:)
 
@cassandra0 I'm just punning around.
I have to prove some stuff.
Let's get serious.
 
@Peter I like your avatar
 
I don't what happened to Charlie?
I'm clearly missing something...
 
@Argon It is nice, right? Plushies!"
 
1:51 AM
@PeterTamaroff Haha! They are awesome
Hi @amWhy, by the way!
 
Hi, @Argon, how goes it?
 
@amWhy Good, good
You?
 
@Argon Pretty good =)
 
user19161
Good Will Hunting is not good...
 
@amWhy How do you define a dense subset of $\Bbb R$?
 
1:54 AM
Good @Will Hunting
 
@WillHunting Why not?
 
user19161
@PeterTamaroff Don't you know?
 
analytic number theory
 
I'm thinking: $A$ is dense in $\Bbb R$ if for every $x\in\Bbb R$ there exists $\{a_n\}\subset A$ such that $\lim a_n=x$
 
user19161
@PeterTamaroff A subset is dense in a topological space if its closure is the space.
 
1:56 AM
@WillHunting No topological considerations, man,
But yes, then I'm right.
and $a_n\neq x$ right?
 
user19161
@PeterTamaroff Yes, you are right. That is so in a metric space.
 
user19161
@PeterTamaroff Wrong. They could all be $x$.
 
user19161
So obviously R is dense in R.
 
@PeterTamaroff pretty much topologically: "E" dense subset of \Bbb R if every point of \Bbb R is a limit point of E, or a point of E, or both.
 
user19161
One can just take (x,x,x,...) converges to x.
 
1:58 AM
How would I go about finding the general (non-recursive) forumula for $$t_n = 2 t_{n-1} + (-1)^n$$
 
@WillHunting That would allow isolated points.
 
@WillHunting You mean XXX
 
@Argon Dude.
 
@PeterTamaroff Oui?
 
@Argon Analyze odd and even parts.
Profit.
@amWhy Right
 
1:59 AM
@PeterTamaroff Parts?
 
user19161
@PeterTamaroff How does that even come into the picture? We are just talking about dense!
 
@Argon subsequences
 
user19161
@PeterTamaroff In what way?
 
@WillHunting Limit points vs. isolated points.
Its right there, black and white,.
 
@PeterTamaroff Can you explain what you mean?
 
1:59 AM
hi
 

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