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12:00 AM
What makes me think there's probably no other nice explanation is that the answer is $\sum_{k=0}^n \binom{n}{k}\sin\frac{k\pi}{n}=\left(2\cos\frac{\pi}{2n}\right)^n$
 
We get a heart-shaped thing
 
Hello ! what is the principle difference between a regulated and a continuous function?
 
What's the name
 
Cardioid?
 
Yes, that
Which is also the rolling curve on the inside of a circle of radius 2, yeah?
 
12:02 AM
Yep:
aww
 
*outside, then
(Clicking on it works)
 
Yeah.
 
$$(1+e^{i\theta})^2=1+2e^{i\theta}+e^{2i\theta}$$
 
I remember it took me quite a while to actually derive that.
(The rolling thing)
 
And I guess that's not related to the original question
but it's kinda interesting
@Semiclassical Isn't it encapsulated by the above equation?
The center of the outer circle follows $1+2e^{i\theta}$
 
12:05 AM
Eh, I mean in the sense of showing that the angle of rotation of the ball matches up in the right way to the distance travelled by the ball.
 
Ah, OK
Well you could just figure out that it's twice as fast by actually rolling something and seeing how often it returns to its original orientation
and noting that the two circles have the same circumference
 
It does work once you actually think through the angles right.
But it took me a while to see that through correctly.
 
Angles?
 
Sure. There's two.
 
Soft question:
Most of the times, combinatorics seem a bit "informal". Is "graduate-level combinatorics" informal too?
 
12:10 AM
In what contexts is it seen informal?
 
For example: "We have $n$ way to count $example$"
 
I mean those claims would still be justified rigorously
 
"We count case $case$ exactly $k$ times, therefore all the ways are exactly $n/k$"
 
Take some basic example, say, how many possible combinations are there of heads and tails if we flip a coin $n$ times?
 
Does the order matter? lol
 
12:12 AM
Let's say it does
 
@LucasHenrique the combinatorics we had was pretty official
 
Well, $2^n$
 
Yeah, that's something you'd prove outright, by induction, perhaps
It's not really an informal argument
Like you might understand many things using words
But it's all rigorous
 
Like, recurrence?
 
I don't know anything about recurrence
 
12:16 AM
One of the more formal aspects to combinatorics is generating functions. Looots of stuff there.
e.g. $F(x)=\frac{1}{1-x-x^2}=1+1x+2x^2+3x^3+5x^4+8x^5+\cdots$ is the generating function for the Fibonacci numbers
 
Oh I think Laci does a bit with those in his class
So yeah that'll be neat
 
@LucasHenrique they tell you $\binom{x}{k} = \frac{\prod_{i=0}^{k-1}(x-i)}{k!} \in \mathbb{Q}[x]$, what the Vandermonds identity is, and you have to show that $\frac{1}{(1-x)^{m+1}} = \sum_{n=0}^{\infty} \binom{n+m}{m}x^n$ for example.
 
My preferred way to understand that is that $(1-x-x^2)F(x)=1$
Or, more formally: multiplying by $1-x-x^2$ annihilates every coefficient of $F(x)$ except the first.
 
@LucasHenrique and then, formal series etc
 
@Danu I guess it's standard for Sp(n)Sp(1) in GL(4,R) to act via Sp(n) on the left of H^n (=R^4n) and via Sp(1) on the right, so it's Sp(n)xSp(1) mod +/-(1,1). On the interpretation that U(2n) acts from the left on C^2n=R^4n, that should make Sp(n) a subgroup of U(2n) and Sp(1) intersect it in +/-1, in which case I'd expect the intersection of Sp(n)Sp(1) and U(2n) to just be Sp(n)...
 
12:21 AM
The reason I like that is because the $n$th coefficient of that product would be $f_n-f_{n-1}-f_{n-2}$, which vanishes for every integer except $n=0$. (I'm taking $f_{-n}=0$.)
 
@Daminark Weren't you the one with the link to the thing that made you realize you know nothing about combinatorics
 
Nope, I think that was Alessandro
 
To be fair, I don't really know any combinatorics yet either, which I hope to rectify soon
 
Generating functions really are my favorite part of combinatorics.
My favorite bit is probably the linkage to complex analysis.
 
12:24 AM
92
Q: Important formulas in Combinatorics

Gil KalaiMotivation: The poster for the conference celebrating Noga Alon's 60th birthday, fifteen formulas describing some of Alon's work are presented. (See this post, for the poster, and cash prizes offered for identifying the formulas.) This demonstrates that sometimes (but certainly not always) a maj...

I know absolutely nothing of anything listed there ^
 
I recognize some of the formulas referenced in the OP.
 
@Semi I sorta view this the way I view complex analysis in number theory
 
And the "All=exp(connected)" is actually a relevant idea in physics :)
 
I appreciate that there's a connection going on, and want to see how deep of a statement it is in complex analysis
 
@Daminark It's a fairly simple statement, actually.
 
12:26 AM
But I also kinda want for there to be a counting/elementary argument
I mean like, for a given problem
 
Well, the complex-analysis bit isn't intended for that kind of thing. More for asymptotic analysis of counting problems.
 
Like alright, we can use complex analysis to solve it, is it perhaps equivalent to some problem within complex analysis, or whatever analysis is used?
 
Suppose I've got a function $F(x)=\sum_{n=0}^\infty f_n x^n$, with $\{f_n\}$ being the sequence of interest.
The most obvious property of the $\{f_n\}$ is that they should be positive reals, since otherwise they won't be much good for counting things.
 
…reals?
 
That makes sense, what's wrong?
 
12:30 AM
Yeah. If $F(x)$ is an ordinary generating function, they'll be integers.
 
@Daminark As opposed to integers
 
If it's an exponential GF, though, they'll be integers divided by n!.
 
Oh I thought you were surprised at restricting vs complex stuff. And it makes sense that things might not always be integral. Irrational surprises me
 
Yeah, you'd never run into irrational coefficients in a counting generating function. But you could also have a probability generating function, where the $n$th coefficient would be the probability of the $n$th event.
So "nonnegative real" is the only constraint I really care about. (I don't care if a particular instance is zero.)
Aanyways. $F(x)$ is defined as a power series, so it's got some radius of convergence; I'll assume it's not zero.
Hence $F(x)$ defines an analytic function in that circle of convergence.
And then the Cauchy integral formula tells you:
 
What an assumption we've got going on there...
:P
 
12:36 AM
Don't sell it short. It wouldn't be true if we had $f_n=n!$.
 
Oh no I didn't mean this sarcastically
 
Ah.
Factorial growth of combinatoric sequences is actually an issue in physics, because it means that the generating functions are at best asymptotic series.
Anyways.
 
I meant it in the, I know it's often fair and you have something nifty coming up but whoa we've just pressed some buttons
 
$$f_n=\frac{1}{2\pi i}\oint_C f(z)\frac{dz}{z^{n+1}}$$
 
Anyway, continue
Hmm... That is actually convenient now that I think about it
 
12:38 AM
where $C$ is the circle of convergence. (I guess I should've said "analytic on the disk of convergence")
 
@Semiclassical What's an asymptotic series
 
Wait the disk? I know you've got a ball in which you converge but how do you know about the boundary?
 
In mathematics, an asymptotic expansion, asymptotic series or Poincaré expansion (after Henri Poincaré) is a formal series of functions which has the property that truncating the series after a finite number of terms provides an approximation to a given function as the argument of the function tends towards a particular, often infinite, point. Deep investigations by Dingle reveal that the divergent part of an asymptotic expansion is latently meaningful, i.e. contains information about the exact value of the expanded function. The most common type of asymptotic expansion is a power series in either...
@Daminark Hrm. That's a good point.
 
Or do you just choose some proper disk?
 
I guess I'll just say it's some circle which lies inside the disk of convergence.
The precise radius doesn't matter because yay contour integrals.
 
12:40 AM
True dat
 
Anyways. What this has done is translate the problem from finding coefficients of some function $F(x)$ to computing contour integrals.
And that's a great thing because one then has methods of estimating such integrals.
The main work is usually done by the steepest-descent method.
 
Yeah, it's neat.
 
This is actually nifty
 
I vaguely remember this sort of thing from generatingfunctionology
 
12:45 AM
Yeah.
The other reference I have for this stuff is a book by Flajolet and Sedgwick. (And hey, the pdf is on the author's website: algo.inria.fr/flajolet/Publications/book.pdf)
 
I have a small question. If $t(x) = \sum_{k=-n}^{n}c_k e^{ikx}, c_k \in \mathbb{C}$ a trigonometric polynomial, then $$ \int_{-\pi}^{\pi}t(x)e^{-ikx}\mathrm{d}x=\int_{-\pi}^{\pi}\sum_{j=-n}^{n}c_je^{i‌​(j-k)x}\mathrm{d}x=2\pi c_k$$. What happens here? cannot follow.
 
I like it when books are freely available
 
@Kirill There's two steps there. Which one aren't you sure about?
 
'cause otherwise I have to get them for free in questionably legal ways and then feel bad
 
This is so true @Akiva
 
12:47 AM
@Kirill $\int \sum=\sum\int$ and $\int_{-\pi}^{\pi} e^{i kx}dx=2\pi \delta_k$
 
The second step is the interesting one
$\delta_k=\begin{cases}1,&k=0\\0,&k\ne0\end{cases}$
in case he wasn't familiar with that
 
Oh lol yeah I was confused
Okay yeah that makes sense
 
And the picture you should have for that is the point going around the origin and stuff
and I'm already bored of that topic
 
Kek
 
@Semiclassical both: we put the $t(x)$ in in the first step, so (without integral) $(\sum_{k=-n}^{n}c_ke^{ikx}) \cdot e^{-ikx}$. Why does it equal to $\sum_{j=-n}^{n}c_je^{i(j-k)x}$? I do not get this formal correct.
 
12:52 AM
Warning: You're using $k$ in two ways at once.
 
@Semiclassical should I substitute $j$ instead of $k$ from the beginning?
 
One is the $k$ in the exponential you're integrating against, the other is the index of summation.
 
Can we prove that, for any finite set of primes $p_i$, there's a nontrivial automorphism of $\Bbb R[\sqrt{p_1},\sqrt{p_2},\dots,\sqrt{p_k}]$ that fixes $\sqrt{p_1}$… and can we do that without proving that they're linearly independent?
 
Depends what you mean by that.
 
I guess we could take that any pair are linearly independent as a lemma
 
12:53 AM
You had better not relabel $e^{-ikx}$ to $e^{-i jx}$. That'll change the question being asked.
 
and then build the automorphism inductively or something
 
But you should relabel the index of summation to something else, and $j$ is as good a choice as any.
 
for example, $(e^{2ix}+3e^{-ix})e^{-5ix}=e^{(2-5)ix}+3e^{(-1-5)ix}$
 
If so, we get a fairly quick proof of the linear independence of the roots of primes.
 
because $(a+b)c=ac+bc$ and $a^ra^s=a^{r+s}$
 
12:54 AM
And by roots of primes I really mean roots of squarefree stuff, it should be the same
 
@Semiclassical so, put $j$ in the polynomail as a summation index first?
 
Fun fact, $1$ is both squarefree and square
 
Right. That gives you the first equality.
@akiva huzzah definitions :P
For the second equality, it helps to do a step which they don't do explicitly and interchange the sum/integral
which gives $$\sum_{j=-n}^{n}c_j \left(\int_{-\pi}^{\pi}e^{i‌​(j-k)x}\mathrm{d}x\right)=2\pi c_k$$ as what we want to show
Do you know anything about that integral in the middle?
 
@Semiclassical ok, thats clear with the first one. I think it is $2\pi$ for $j=k$ and 0 in other cases. Does it right?
 
Right. Easy enough to show that by considering the cases of $j=k$ and $j\neq k$ separately.
That kills every term in the sum except the $j=k$ case, and leaves $2\pi c_k$ as the result.
 
12:59 AM
It's a fancy way to say the average of the points on a circle is the center
 
Pretty much.
 
@Semiclassical that is why you use Kroenecker-Delta. I have to right your equation down. Integral of the sum is the sum of integrals - clear, but not explicitly
 
which is clearly true by rotational symmetry
 
@Kirill I'm a little paranoid about that interchange, tbh. I forget under what conditions it's justified to do that.
 
@Semiclassical surely, not allways
 
1:01 AM
But this is something which a good text would explain.
 
Speaking of, $\int_{-\pi}^\pi e^{e^{i\theta}}d\theta=2\pi$ for trivial reasons
 
Wait is this about when you can interchange infinite sums with integrals?
 
Yeah.
 
I'm on my phone rn so I can't read the stuff easily
 
@AkivaWeinberger could you please enlarge the expression?
 
1:03 AM
@Semiclassical You had a finite sum though, no?
 
Oh, true.
 
Ah, well you can do it if you know the partial sums are all bounded in absolute value by an L^1 function
 
$$\int_{-\pi}^\pi e^{e^{i\theta}}d\theta=2\pi$$
 
Yeah, that makes it easy. There's only a finite number of terms, so the interchange is fine.
 
I, on the other hand, do not
@Daminark glares
 
1:04 AM
@AkivaWeinberger I still do not see what stands in the power of $e$...:)
 
What? @Akiva
 
$\exp(\exp(i\theta))$
 
@Kirill Akiva's integrand is exp(exp(i*theta)), which is different from what you're talking about
 
@Kirill $e$ to the power of $e^{i\theta}$
@Daminark You can render MathJax on mobile
 
?!?
 
1:05 AM
I am currently on mobile and am rendering MathJax
 
ok thanks
 
Dayum
 
I almost never go on chat from my computer
I'm essentially always on mobile
 
Do you just go on the bookmarks bar as normal?
 
I am able to mo
@Daminark Essentially yeah
 
1:07 AM
Hmm, it's not working, at least not for chrome
 
I'm using Chrome on iPhone
 
Maybe the LaTeX renderer is outdated?
 
No?
Oh wait maybe
Probably not
 
@AkivaWeinberger you mean you wrote $\textt{\begin{cases}...}$ for me from a mobile?
 
I've never changed it yet, maybe there's been an update since like, January
 
1:08 AM
But you can just go to the link and re-copy it
@Kirill Yeah, so?
 
@AkivaWeinberger I just wanted to express my respect for this.
 
feels respected
 
@AkivaWeinberger my phone has a bad character, I need a non-polynomial time to write more than 70 symbols.
 
What if I get addicted to respect?
Does that mean I have to do respectful stuff? Oh god
 
Addictions are not respected usually so don't worry about it
 
1:13 AM
I let power, which is a street drug by the way, go to my head
 
that's no a math conversation
 
@Twink that is both general discussion and questions alike.
@Semiclassical I have written it down, twice. Now I have it, thank you.
 
OK so take a Hamel basis of $\Bbb R$
 
@amWhy Yes. I have multiple accounts now.
 
bill wurtz voice That's math
 
1:17 AM
this chatroom has changed so much since @anon changed his account
pedro and ted don't come here anymore
 
Ted does.
 
but not as before
 
none of the regulars I was familiar with when I started in chat are here anymore
they left before you even came around. things change.
 
I know
:(
@anon give me your whatsapp
 
no
 
1:20 AM
why not? :(
 
not interested
 
you'll never be my friend? :(
you're mean
anyway... people used to talk about math here
now there are many dummies
and trolls
 
a contradiction itself
 
1:25 AM
@Twink yeah, math please
 
Well that happened
So, does the quotient space way of proving that operators on C^n are diagonalizable also extend to the more general case when the minimal polynomial factors completely?
 
1:52 AM
So I found this problem
Consider a finite set $E$. The complement map $f:\mathcal P(E)\to\mathcal P(E)$ is a bijection that satisfies $f(V)\cap V=\emptyset$ for all $V\in\mathcal P(E)$
Can we generalize this to arbitrary downwards-closed families of sets?
Let $A$ be a family of subsets of $E$ such that if $V\in A$ and $U\subset V$ then $U\in A$. Can we find an $f:A\to A$ such that $f(V)\cap V=\emptyset$ for all $V\in A$?
 
Hmm, so our issue is that we don't have a global maximum set
Well, can we partition this set into chains?
 
Hm, let's try $\{1,2\}$, $\{2,3\}$, $\{1\},\{2\},\{3\}$, and $\emptyset$
 
LOL @anon do you remember this? xD math.meta.stackexchange.com/questions/10880/…
those were good times
 
So $\{1,2\}\leftrightarrow\{3\}$, $~\{2,3\}\leftrightarrow\{1\}$, $~\{2\}\leftrightarrow\emptyset$ works
So the things that were in $\{1,2\}$ or $\{2,3\}$ but not both got complemented in $\{1,2,3\}=\{1,2\}\cup\{2,3\}$
and the things that were in both got complemented in $\{2\}=\{1,2\}\cap\{2,3\}$
Hmm
And $\{1,2\}$ and $\{2,3\}$ were our only maximal elements
Hm, I wonder what would happen if we took the downwards closure of $\{1,2\}$, $\{2,3\}$, and $\{3,4\}$
Well we can map $\{2,3\}\leftrightarrow\emptyset$, and then almost anything else after that is allowed
Maybe the most symmetric one is $\{2,3\}\leftrightarrow\emptyset$, $~\{1,2\}\leftrightarrow\{3,4\}$, $~\{1\}\leftrightarrow\{2\}$, $~\{3\}\leftrightarrow\{4\}$
@Daminark Any idea?
 
2:14 AM
I'm thinking about it, it's trickier than I thought since we don't have much structure
My concern is if a family has an odd number of sets
Or wait am I being dumb?
I'm thinking about it, it's trickier than I thought since we don't have much structure
 
@Daminark If it does you could map the empty set to itself… that said, I don't think it could ever happen unless $A=\{\emptyset\}$
You could do inclusion-exclusion on the power sets of each thing in $A$, or something
Why is the empty set disjoint from itself that's so stupid
 
Even worse, every set has a subset from which it's disjoint
 
Is $\mathcal P(A\cap B)=\mathcal P(A)\cap\mathcal P(B)$?
Yeah, yeah?
 
Should be, yeah
 
Hm so PIE does prove that every downwards-closed family has an even number of elements
Maybe we could generalize it to solve the problem, or something
 
2:33 AM
PIE? And lol maybe
 
Principle of inclusion-exclusion
 
Oh, I've never seen that abbreviation before
 
Oh, I have an idea
Separate it into three groups
Those with $1$ in them,
those not in the first group that stay in $A$ when you union $\{1\}$ to them,
and the rest
Call these $X$, $Y$, and $Z$
$Y$ and $Z$ are downwards closed, right?
Oh wait, $Z$ isn't downwards closed
Never mind I need to think more
@Daminark I'm wrong. $A$ need not be even
-_-
$\{1\},\{2\},\emptyset$
 
Rip
 
2:48 AM
I mean still, map empty-set to itself should take care of that
 
So this won't break anything yet
 
No
Correct
 
Ugh, algebra problem. Can't find the 'correct' technique to advance
 
Add 0 to both sides
 
2:49 AM
@AkivaWeinberger Wao
 
$a, b, c \in \Bbb R ^{+}; \quad abc(a+b+c) = 3$. Prove that $(a+b)(b+c)(c+a) \geq 8$
 
First thing I'd take note of is that $2^3 = 8$
 
you're thinking $a=b=c=1$?
 
That's $(\frac3{abc}-c)(\frac3{abc}-b)(\frac3{abc}-a)$
No idea if that's helpful
 
the problem is equivallently written as $e_1 e_3 = 3 \implies e_1 e_2 - e_3 \geq 8$
using symmetric polynomials
LOL
 
2:52 AM
@Semiclassical I mean, I just jumped to the first connection I had between the numbers
 
Right.
 
It's possible you can think of this as a function of 3 variables and use Lagrange multipliers to show that a min is at (1,1,1) or something
But like do we really want to do that?
That sounds overkill
 
well, it's an olympiad problem
we usually do not use calculus or superior level stuff
 
If we expand we get $a^2b+a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2$
Modulo the possibility that I totally fucked that up which isn't impossible
I'm bad at bookkeeping
 
Cauchy–Schwartz deals with sums of products, right? Not products of sums?
 
2:58 AM
Usually I look at Cauchy-Schwarz as the $\|\cdot\|_2$ case of Holder
Holder is $\|fg\|_1 \le \|f\|_p\|g\|_q$
 
Integrals of products then
@Daminark That was probably a typo but I like that Holder is money
 
(You see why I like Holder, it's just a nice clean statement)
 
you guys are using techniques that i've never seen before
 
We haven't actually done anything
 
Yeah this is calculus stuff we're talking about here, much as we haven't used it
 
3:00 AM
We've just confirmed that both Cauchy–Schwartz and Holder are useless :P
(in all likelihood)
 
Hmm
This might be AM-GM
Actually this probably is AM-GM
What's $(a+b+c)^3$?
It's $a^3 + b^3 + c^3 + (a+b+c)(ab + ac + bc)$
So we know that $a^3 + b^3 + c^3 + \frac{3}{abc}(ab + ac + bc) \ge 27abc$
 
Oh, god, good luck
In any case, I'm horrible at this sort of thing
so don't expect much from me here
 
So $a^3 + b^3 + c^3 + \frac{3}{a} + \frac{3}{b} + \frac{3}{c} \ge 27abc$
Yeah merp
Oh wait hold on my shtick might've been wrong
Yeah it was totally wrong, fuck
Wait I'll do it more carefully
 
Have funz
Have funds
Good night
 
Oh wait no it's easy
Wait wait
@Akiva
$(a+b+c)^3 = a^3 + b^3 + c^3 - 3(a+c)(a+b)(b+c)$
And that's at least $27abc$
 
3:11 AM
OK?
 
This has the expression we want in it
 
The inequality ends up on the wrong side, though, no?
 
Oh wait whoops should've ping'd @Lucas
 
$(a+b)(a+c)(b+c)\le\frac13(a^3+b^3+c^3)-9abc$
is what you get from that I think
We want to prove
$(a+b)(a+c)(b+c)\ge8$
The inequality is facing the wrong way
 
Merp
Oh hmm easier thing
Triple AM-GM
$a+b \ge 2\sqrt{ab}$
$a+c \ge 2\sqrt{ac}$
$b+c \ge 2\sqrt{bc}$
So their product is at least $8abc$
So now we want that $abc = \frac{3}{a+b+c} \ge 1$
Or at least that if that's not true you can easily verify that the other product is at least 8
 
3:27 AM
Last night dream, there's this fractal looking thing:
Where every point in the black space you see here, when zoomed in, will give you the same figure show here, no matter how far you zoom
In addition, the pattern you see here is not really just the blue dots, but that it is a grid with a nonlinear scale and where the horizontal gridlines intersect the vertical ones are given by the blue dots
That is, it actually look something like this, and every point in the black space, when zoomed in looks just like this grid shown here
In other words, if we consider each quadrant separately, each line is actually some set that looks like this: $\text{some countably infinite sequence bounded from below} \times [0,\infty)$
More accurately, the set $X$ has the following iterated construction: Let $s_0$ be a countably infinite sequence with infimum at $0$, then:

\begin{align}
X_0 & = s_0\\
X_n & = X_0 \times [0,\infty)\\
X & = \lim_{n\to \infty} X_n
\end{align}
typo: $X_{n+1} = X_n \times [0, \infty)$
So for example, we can have points at ..., 1/100,1/90,1/80,1/70,1/60,1/50,1/40,1/30,1/20,1/10,1,10,20, .... If we zoom into the interval (1/100,1/90), then we might have something like 1/100 + (..., 1/100,1/90,1/80,1/70,1/60,1/50,1/40,1/30,1/20,1/10,1,10,20, ....), and so on
and it does not matter which interval you choose (including intervals like (a,b) such that $\lim_{a\to b}$), how far you zoom, it will have this exact same sturcture
It is possible this set might be isomorphic to the reals as the above feature will mean effectively the infimum can be any real number, thus making the set complete
Alternately, I think this set can be written as $\{x+\{\frac{1}{y}\},x\in \Bbb{R},y \in \Bbb{N}\}$
 
4:40 AM
I'm wondering if there's a clean way to prove that holomorphic functions are $C^1$
Oh apparently there's this nifty thing called Weyl's lemma which should do it for you
I mean, maybe you don't even need all that, just that harmonic functions (as opposed to weakly harmonic functions) are smooth
 

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