Just had fun with my students going through the eight options of double integral exists, iterated integral exists in one order, iterated integral exists in the other order. :P
Since @Hippa is here, I'll certainly be cranky in moments.
@MikeMiller Well, the idea previously was to have SE deal with it comprehensively, but paternity leave postponed the "dealing with it". So now we're playing whack-a-mole in slo-mo.
$x$ is an invertible element of a ring, @Choups? Normally, one would just say such a thing. In English, we also call such things units. The units in $R$ are often written $R^\times$.
But there is no context for $^{-1}$ unless one says beforehand that it makes sense, @Choups.
@MikeMiller Ah. Yes, that was "helpful, we're aware and keep an eye, but for the moment, we're waiting for SE action". Something has happened since making that page less embarrassing ;) Still a little left to brush away, though.
@DanielF @Mike: I had my first obnoxious comment from Dr. Sonnhard, and first wrote a snippy comment back to him, then decided to modify my answer to make his comment even more irrelevant :P
Hey all, I know that $p_n \sim \sum_{k=1}^n \log p_k$, where $p_k$ is the $k$-th prime, but how about the sign of their difference? Does it change infinitely often or not?
@BalarkaSen It's an ugly verification: if $X$ is a complex and $A$ a subcomplex, then $X^n\times I$ is obtained from $X^n \times \{0\}\cup ((X^{n-1}\cup A^n )\times I)$ by attaching copies of $D^n\times I$ along $\partial D^n\times I\cup D^n\times \{0\}$.
@TedShifrin Note that there are noncompact topological groups that have "nonzero Euler characteristic" in any reasonable sense; e.g. $\Bbb{CP}^\infty$ is homotopy equivalent to a topological group, but has nonzero homology precisely in even dimensions.
I have heard that $S^1, S^3, S^7$ are the only topological groups and there should be canonical action on each of these on $S^3, S^7, S^{15}$ respectively using complex number, quaternions and octonions structure on, respectively.
@MikeMiller I guess it would be painful to smooth it by hand (i.e. express the smooted function), but its existence should be enough. How do I, however, justify that in every interval [n,n+1] the derivative is $1$ at least once (assuming that $f$ smoothed is $\mathcal{C}^1$)?
@PedroTamaroff Finding (or proving the impossibility of doing so) a monotone function such that $f\to_\infty l\in\mathbb{R}$ and $f'$ doesn't converge to $0$
@MikeMiller Ugh some kind of exponential branch then ? (two link two steps smoothly)
@PedroTamaroff It's really stupid that you'll sigh at me for not spelling a bunch of huge names correctly. I have worked a whole week for visualizing the Hopf fibration and it's really disheartening to hear sneerish comments after working that hard. I don't know what a fibration is, OK? All I did was tried to find and visualize an S^1 action on S^3 such that S^3/S^1 is homeomorphic to S^2.
The cost of the levels till the second last is equal to $\sum_{j=0}^i \left(\frac{7}{8}\right)^i n $ and $ i $ can be found from $\frac{n}{2^i}=0$ because it has the smallest possible denominator, right?
Is the cost of the last level equal to $T(1) \cdot 3^{i}$ ? Or am I wrong? @DanielFischer
@evinda I don't know, haven't calculated anything. But you should calculate a few, form a hypothesis about what $T(n)$ is, and set out to prove that by induction.
@DanielFischer Yes, the hypothesis can be found from the recursion tree that I drawed and I found that the cost is: $$T(n)=\sum_{j=0}^i \left( \frac{7}{8}\right)^j n=n \frac{\left( \frac{7}{8} \right)^{\lg n}-1}{-\frac{1}{8}}$$
I was wondering which is the cost of the last level of the recursion tree..
