Hi. I have started with Counting and was trying to understand the concept of choosing. If I flip a fair coin 50 times, I have 2^50 outcomes. Now if I want to know how many ways can I get 5 heads and 45 tails?
I came up with 2^50_C_5 ways of selecting 5 element subsets, and multiply that by 50! because there are 50 places to put those coins
I don't think so. One correct way to do it would be as follows : for the first heads you have 50C1 ways, for the second you have 49C1, ... for the 5th you have 46C1, so overall you have 50*49*48*47*46/2^50 chances
I am thinking of writing another blog post, this time about some of the strengths of higher representation theory and categorification, highlighted by some examples. Not sure how much interest there would be though.
it was an exercise that said "Which of the following belong to the projective plane: (i) a parallelogram (ii) an isosceles triangle (iii) a triangle and its medians"
yes but doesn't the triangle belong to the projective plane in general? why wouldn't an isosceles triangle? projective geometry only uses the straightedge
@Hippalectryon Ahh, I have never studied projective stuff like that. I don't really see the point of doing these things the way they were introduced so long ago
@user19405892 The idea is that in the projective plane, we can no longer really measure lengths, so a triangle being isosceles does not make sense
"the geometry of the straigghtedge at first seems to have very little in connection with the familiar derivation of the name geometry as 'earth measurement'"
Can someone explain why each $x\in\mathbb Z$ has a multiplicative inverse $x^{-1}$?
The division is not defined in the ring. So for example what is the multiplicative inverse of $5$ such that $5*5^{-1}=1$?
Is the fact that $\mathbb Z$ is a ring explained with rational integers? Even though we cannot divide, we can define inverse elements with rational integers (https://en.wikipedia.org/wiki/Integer)?
Hi. I'm trying to prove that $\displaystyle\lim_{(x,y) \to (0,0)}xy=0$. So I need to show that for every $\varepsilon>0$ there is a $\delta>0$ such that $|xy|< \varepsilon$ whenever $\sqrt{x^2+y^2}<\delta$. Can anyone please give me a hint as to what to do next?
@MikeMiller: in the image I sent you, I sometimes use different notation for the same thing ($\| \cdot \|_{H^k} = \| \cdot \|_k$ for example). I was just too lazy yet to go over the document and make all notation "well-defined"
I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not a UFD.
I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to concl...
@MikeMiller: so what I wrote down for it makes sense? and the step that fails in general is switching order of limit and sum to make the sequence Cauchy?
Hrm, the step that fails in general should be a uniformity problem, maybe
In that case the sum is a finite sum but doesn't have bounds independent of the function (you're summing over charts covering the support of $f$, but there can be arbitrarily many of them)
@DanielFischer In this answer (which is sort of related to what we were talking about yesterday), you state that the integral over the large semicircle vanishes because $|\sin (\mu z)|$ grows exponentially as $\text{Im} (z) \to \infty$. Are you implicitly using the proof of Jordan's lemma here?
@Huy: You're summing over a locally finite cover, presumably, and your input is not a function, but a compactly supported function. A compactly supported function hits only finitely many elements in a locally finite cover.
@DanielFischer Hello, a consequence of Morera's theorem is that when a sequence of holomorphic function $f_n$ converges uniformly to $f$ then $f$ is holomorphic then. So when we are ( juste as an example) on the open disc $D(0,1)$ we can write Cauchy's formula for $f$ around a disc $D(0,r)$ for $r<1$, but if we have only uniformly convergence on the circle $bD(0,1)$, it doesn't follow necessarily that $f$ is holomorphic on the hole disc right ?
@JeSuis We're assuming that the $f_n$ are continuous on the closed disk and holomorphic in the interior, aren't we? Use the maximum modulus principle on $f_n - f_m$.
@DanielFischer badly stated sorry, I take a sequence of holomorphic function $(f_n):D\to \Bbb{C}$ where $D$ is the open unit disc, we assume that $f_n$ converges uniformly to $g$ on any circle $\Gamma_r$ centered at $0$ of radius $r<1$. Does it follows that we have uniformly convergence on the closed disc $\overline{D}(0,r)$?
@DanielFischer I just saw it yesterday, so let's see if I can do it :D. As $f_n$ is holomorphic on $D(0,r)$ and continuous on $\overline{D(0,r)}$, and non constant (assuming that the constant case is trivial). Then for $f_n-f_m$ we have for all $z\in D(0,r) \Vert (f_n-f_m)(z)\Vert\le \sup_{\Gamma_r}\Vert f_n-f_m\Vert$ and now we the uniform convergence right ?
@DanielFischer Is the fact that $\left| \sin (\mu z) \right|$ remains bounded on the portions of the contour near the real axis as $R \to \infty$ enough to conclude that the integral vanishes along those arcs as $R \to \infty$? What's bothering me is the fact that $\frac{z}{z^{2}+a^{2}} \sim \frac{1}{z}$ as $|z| \to \infty.$ What's happening to the length of those arcs as $R \to \infty$?
It's the canonical example of a symplectic manifold, but you're thinking too rigidly. We're on a very small symplectic manifold - a small open subset of the usual phase space $\Bbb R^2$ of a particle on a line.
Classical dynamics usually moves outside a disc, methinks.
well, if you start with a harmonic oscillator potential $H=x^2+p^2$, then the phase space is foliated into concentric circles by different choices of energy $H=E$. so if you set a maximum positive energy $E_{max}$, then that'd give you $D^2$.
and more generally it'll be diffeomorphic to $D^2$ if you pick a potential $V(x)$ with a single global minimum and then pick energy to be sufficiently close to that minimum.
not convinced that's the right way to think of that, mind
So the homomorphism has to be zero, since it's a map to an abelian group. But that makes one wonder "Can I prove it's zero without invoking big machines?"
admittedly, the change of perspective more for my own curiosity than to supply any proof.
ahah, found it. in classical mechanics, it's called a canonical transformation.
though, actually, i can't remember if that's the right scope. i.e. every canonical transformation counts as a symplectomorphism, but i don't remember if the reverse is true
Well, I guess simple implies perfect or cyclic of prime order. But it should come as no surprise that diffeomorphism groups are not cyclic of prime order...
I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not a UFD.
I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to concl...
