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22:00
aight
ill be studying general analysis and measure theory at the same time, itll be a fun summer :D
and by general analysis i mean chapters out of baby rudin
Your spelling could use some study too.
@DanZimm Rudin's Real and Complex Analysis should give you a good grounding in measure theory - but it is not easy reading - there are gentler books
@DanZimm I thought your pic was millenium Falcum
@OldJohn When I went to the analysis professor as a first years, he sent me home with that book.
@JonasTeuwen Good man - gave you a challenge :)
22:04
@OldJohn hrm thank you for the suggestion, I've actually got that one handy too (accidentally got that instead of baby rudin at first, boy was that a mistake)
@Charlie negative, I think it's calculus on a napkin from my first calculus course some odd years ago
@OldJohn They kept doing that. After it was Stein's differentiability properties of functions.
@DanZimm It is a tough book, and I have never finished it, but the first few chapters should cover what you want
@OldJohn Actually he gave me functional analysis, but then 'perhaps this one is good to start with'.
@JonasTeuwen by no means do I ever claim, or do I think I will ever claim that I "know" english :P
@OldJohn aight cool, thanks!
@DanZimm hmmm
22:05
@JonasTeuwen ROFL! I found his FA seriously tough going
and "Fourier Analysis on Groups" completely defeated me
@OldJohn Hah, my advisor gave it for his reading class.
The complete book.
I thought it was actually quite a pleasant read the FA.
Never understood why people thought it was so hard. I just thought it is one of them classicals. Everybody knows it, nobody really tries to read it.
@DanZimm From afar our avatars sure look similar ...
A bit like the guy who gave the Dalai Lama a pocket watch so he could take it apart, then put it back together ... working perfectly :)
@skullpatrol yipyipyip
@Charlie time to change?
22:07
@Charlie - problems with the yip-key again?
@PeterTamaroff your work on Cesaro theorem reminds me of a nice question: Let $a>0,x_0>0$. Consider the recurrence $x_{n+1}=x_n+a^{x_{n}}$ $\forall n \in \mathbb{N}$. Then compute $\lim_{n\to\infty} \displaystyle\frac{x_n}{\ln n}$.
@skullpatrol indeed, but I promise you I didn't steal yours :P
@OldJohn yipyipyip
yipyipyip
@Charlie Yeah - that key is definitely stuck :)
22:08
@skullpatrol it's up to you. would you use the other i made you?
@OldJohn it happens
@Charlie Na...I like it just where it is :D
@skullpatrol really? where is it?
@PeterTamaroff sorry for pinging.
@Charlie where you sent it.
@skullpatrol ah :)
@skullpatrol I want to make another skull
frontal
22:11
@Charlie later...if you have extra time
@skullpatrol yipyipyip
@Charlie yipyipyip
what is an alpha whore, i must go look at this conversation
hahahaha
lmao
22:15
@DanZimm hehe I like things, things like me
@Chris'swisesister Will look into it.
@skullpatrol hahahah yipyipyip
@Charlie it seemed to some up my life pretty well
@DanZimm A couple of other books I found useful for measure theory were Dudley (Real analysis and probability) and Royden (real analysis)
22:16
Aight, do you mind putting that in an answer to my question in the forums (or w.e you call them)
@Charlie :D:D:D
@DanZimm hmmm
@Charlie I definately like things... i hope some things like me
@DanZimm of course
22:17
@OldJohn I prefer Folland above all.
sigh did horrible on my analysis assinment =|
Dudley and Royden have too much air.
i think its safe to say my computer likes me
@JonasTeuwen what do you mean by air/
@Chris'swisesister Oh, but $$=\lim {a^{x_n}}\log\left (\frac{n}{n+1}\right)$$ by Cesaro, yes?
@DanZimm hot air
22:18
@Faust7 oh noes
@Faust7 D:
I have heard good things about Folland - but never actually opened a copy, unfortunately
It's not surprising analysis hates me =)
@Charlie is your ava just a question mark?
@DanZimm The Riddler
22:19
@OldJohn hope it was about Gerald B folland Author of Advanced calculus ? worst textbook i ever owned burned after i finished my final
there are bats in it
looking it up right meow
oh ok
wasnt*
@skullpatrol know a nice name? Wilhelmina
@Charlie For what?
22:21
aight afk for a little
dont say anything funny while i leave :P_
@PeterTamaroff Isn't that similar to having $\sum b_n\to\infty$ and $a_n/b_n\to L$ then $\sum a_n/\sum b_n\to L$?
@DanZimm Right, man, Right, please
@OldJohn Do so. Now!
like moving up to beta whores
@robjohn Yeah, guess so.
22:22
@skullpatrol just a nice name
@JonasTeuwen It will have to wait -got to get ready for a trip to Poland first :)
it's williamina
@skullpatrol not that anyone cares.... (Big Bang Theory)
@robjohn Could you help me count something?
@PeterTamaroff Dracula can help you
22:25
The Count
@OldJohn Yes, so you can wait until you're in the grave 8-(.
That's what most do.
Wait wait wait for the right moment. And BANG! They're dead.
Still waiting.
@JonasTeuwen I'm still waiting for the right moment to stop waiting for the right moment.
@PeterTamaroff your proof looks good, but no better than the other proofs presented.
@PeterTamaroff is it above 4?
@Charlie Aah finally you repeat.. :D
22:27
I feel like I am too young to have some vertebrae fused.
@skullpatrol ah it really makes me laugh
Should I wait, wait, wait 8-)?
@Charlie Yes, it is funny :DDD
@JonasTeuwen No - if you wait too long, you wake up dead :)
@robjohn Yep, sadly.
22:28
@JonasTeuwen do you have disc problems?
@robjohn Yes, at two levels.
@JonasTeuwen get the surgery. There is no reason to live in pain
@skullpatrol the spiders on the wall hahhahahahha
@robjohn Yes, but the recovery time is extremely long and there are a lot of complicating factors that make them rather not do it.
@JonasTeuwen There goes your football scholarship...
22:29
@Charlie :D:D:D:D:D
@OldJohn if you are dead, you cannot wake up...
@robjohn I don't get the joke!
I wanted to ask a professor of neurosurgery yesterday if he also cut open phd students, but I didn't get the chance 8-).
@Charlie yes - it was a (lame) joke
@Charlie As Zombie.
22:30
@Charlie Oh... look who knows so much!
@JonasTeuwen i was going to say that
But you didn't.
Why?
@robjohn I don't know anything, Robs
@JonasTeuwen you posted first
I know less than nothin'
Why were you so slow as you initiated it? 8-).
22:31
@PeterTamaroff things moved while I was clicking.
@skullpatrol Well, at least you know that. It's a start.
@skullpatrol negative knowledge?
@robjohn knowledge about the negative side of life...
@JonasTeuwen i was goig to write "unless you area zombie"
@PeterTamaroff What did you need to count? sheep?
22:32
@robjohn There is something that needs to keep the conservation laws in order 8-).
@skullpatrol I only know I'm getting old
@robjohn Let $X=\{1,2,\dots,100\}$ and define in $P(X)$, $A\sim B\iff A\triangle B$ has at most $2$ elements. I need to find the number of sets $B$ for which $\{1,2\}\sim B$.
I have split this in some cases.
@Charlie Nobody gets young.
@skullpatrol i know
but I'm getting old
that's the point
I am nearly young.
22:35
I even have white hair
Liek a wietch
like rogue
@PeterTamaroff $99\cdot3+\binom{98}{2}+1$
@skullpatrol :-/
@robjohn OK.
How did you do it?
22:41
Just count?
There is also the Erdos trick.
$\{1\}$ and $\{\}$ or one of the $98$ others
$\{2\}$ and $\{\}$ or one of the $98$ others
$\{1,2\}$ and $\{\}$ or one of the 98 others
$\{1,2\}$ and any pair from the $98$ others
$\{\}$
@PeterTamaroff I think that is all
If $B\triangle \{1,2\}$ has zero elements, then $B=\{1,2\}$.
Suppose now it has $1$ element.
Then we can take $B$ to be $\{1,2\}\cup$ all the possible subsets of $\{3,\dots,100\}$, @rob yes?
@skull
I meant all the possible singletons.
23:01
@anon !!!
hello charlie
what's up?
@anon not much, same of always, and you?
pretty good
@anon good!!!
23:03
...
@anon what's the reason?
summer
@anon oh!
ultimate enabler of laziness
23:05
@anon hehehe
I don't know what's good anymore: summer, winter, spring, autmn.. As long as I can breath and sleep, it's ok for me.
the autumn is a pirate
hehe
but it's on the wrong eye
23:09
@anon Could you help me with a counting problem?
BINGO!!!
I have to count the ways to place 83 (indistinguishable) balls into 5 numbered boxes such that there are an even number of balls in the even ones, and an odd number of balls in the odd ones.
Now, let me try first.
I put one ball in each 1,3,5 since I want odd.
Then I just use 80 balls as 40 pairs yes?
@skullpatrol :D dinner!
23:10
So the answer is...
@Charlie later
$$40+5-1\choose 40$$
@anon Yes?
that feel when you finished eating something minutes ago but have no memory of it
@anon Am I left or right?
Answer should be $7\cdot 11\cdot 41\cdot 43$.
sure, that carves out a slice in the sample space
23:13
@anon What?
only 7*11*41*43? I'm skeptical.
$$44!/40!4!$$
@anon Isn't the argument OK?
I put 3 of the 83, left with 80, then put them in pairs of 2, since I want $2k+1$ and $2j$ to appear in the boxes.
I don't know what "use 80 balls as 40 pairs" means exactly
@PeterTamaroff I've counted all those
@PeterTamaroff all possible subsets of size 1
@anon I mean, instead of putting balls separately, put balls in pairs of two, as if we glued them together, since we want to get even and odds.
23:17
oh, I see
For example $$2\cdot 9+1\; \mid \;2\cdot 7\; \mid \;2\cdot 13+1\; \mid \;2\cdot 1\; \mid \;2\cdot 10+1$$
@anon So, yes?
How do you get 40+5-1 choose 40?
@Ethan Have you seen Boya's formula?
what?
no
23:24
I can't see it
show me
I can't find anything named 'boyas' online
$$2\pi {e^{ - \gamma }} = \prod\limits_{n = 2}^\infty \exp \left[\binom{n+1}{2}^{-1}\zeta(n)\right] $$
2
hmm
@anon There we go.
Looks nice
the "exp" notation is ugly
23:26
@Ethan Well, it has $2,\pi,\gamma,e,T_n,\zeta$. Lots of stuff related!
and the fact that the lower bound is $2$ also makes it look kinda ugly
But other then that I like it
@anon Oh noes!
keep your pants on
nvm its ugly
Its not even an infinite product
its an indexed sum
indeed
23:28
It looks like who ever wrote it is hiding somthing why else write it as a product when its better understood as a sum, taking the logarithm of both sides would make it into a nicer form
though I don't know about "ugly"
Also the use of binomial coeiffients is missleading
For such a small binomial you should just write it out
as $\frac{n(n+1)}{2}$
@Ethan You mean "gross"? FACEPALM, man. FACE-FUCKING-PALM.
heh
The entire identity seems to be dressed up I am looking at a lie
I am to harsh
loll
Well first off Taking logarithms eliminates the exponential factor
23:33
May be ethanol is not to hash, butt lather sense the exp is super flous it is a lie fence grows.
The $\ln(2\pi)$ is remminant of a zeta series that appears in the asymptotic expansion of stirlings formulas
and $\frac{1}{n(n+1)}$ is usually used as a telescoping tool
You should end up with terms of the form $\frac{\zeta(n)-1}{n}$
and it is well know $\gamma=\sum_{n=2}^\infty \frac{\zeta(n)-1}{n}$
or somthiung along those lines
eulergamma? someone's been using woflram.
let me checlk
@anon you know what series I am refering to
I am to lazy to derive it
you use the maclaurin series for the logarithm, to get an expansion for the logarithm in terms of harmonic numbers
subtract the logarithm off the first harmonic number let n tend to infinity you get eulers constant
and the other terms are zeta functions
some series of the form $$\sum_{n=2}^\infty \frac{\zeta(n)-1}{n}=+\-\gamma$$
somthing like this
I used a similar technique a while back in an attempt to prove stirlings formula
@PeterTamaroff anyway its not very pretty
@PeterTamaroff this is what google translates for the 3rd line
"I have baptized, how could it be otherwise, as the identity of Boya"
23:42
@Ethan Not bad.
i have baptized?
Anyway the identity is pretty disgusting, as is the author's praise of it
can I quote you on that?
The mathematical constants don't naturally appear in the formula, they are jammed into the formula
the exponential factor wouldn't have come in, unless he exponentiated both sides at the end
@Ethan Baptize means "name".
He also adds the factor of $2$ in at the end so he can re-write the quadratic as a binomial coefficient
23:48
@anon I have worked that $(3,n)=1\implies (5n+3^{n+1},4n-3^n)=1,17$ by simple division rules. Can you do it using congruences?
equals one or 17?
actually you want to subtract 3(4n-3^n), but that's what peter means by "simple division rules"
nvm, i am to lazy to do this, my head hurts anyway
(Euclidean division)
I would not do it with congruences
23:51
@anon Yeah, just mod out by one and another.
Ultimately, one arrives to $d\mid (17n,17 \cdot 3^n)=17(n,3^n)=17$
@Ethan There are too few words and too many formulas there. It is not nice to read.
:9465638 I remember we talked about $M(x^{-1})\log x$ some time ago.
Yes I know
this reminds me of this
@anon The guy is sweating like a fucking pig. My god.
the video was hilarious at the time
"....at the time." Important words! Yes!
Who is Steve Ballmer?

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