@TedShifrin We proved it by writing out a big array of exact sequences, then starting a particular place tracing all of the logical implications for how an element in one place is connected to elements in the other places
Well, as you know, I'm very fond of integration over the fiber and crossing with $I$ as adjoint operations ... and that gives the usual chain homotopy for homotopic maps giving cohomologous forms.
LOL, my colleague just responded: "Oh and it gets much worse: I’m pretty much convinced (and Roger Howe agrees) that elementary school multiplication actually involves working in category theory." @Mathei @Balarka
@TedShifrin Well, it's hard to sleep and keep folding and unfolding a piece of low production paper which is supposed to fold into a cube at the same time
"New Math" may have been a failure, but "Newer Math" which I propose now, i.e. teaching category theory and cohomology at elementary school is totally reasonable
(As time goes on I also feel like the WTF reaction to the kempner series isn’t really warranted either. It boils down to a sophisticated version of “if you write down a lot of digits randomly, at least one of them is probably a 9.”)
@Narcissusjewel I leave it as an exercise to the reader to figure out the expectation and standard deviation of this data, and fit it with an appropriate normal distribution.
@MikeMiller I know you are more a topologist, but I can't seem to find a proof that the contact fields on a contact manifold (M,\xi) are in bijective correspondence with the sections of the normal bundle to \xi
@anakhronizein Here is my intuition. The contact fields should be the same thing as $f R_\alpha$ for the Reeb field $R_\alpha$, and such functions in the orientable case correspond to sections of the (trivial) normal bundle.
Applying $\mathcal L_X \alpha = e^g \alpha$ to the Reeb field we see that $R_\alpha(\alpha(X)) + d\alpha(R_\alpha, X) = e^g$; we know that $d\alpha(R_\alpha, - ) = 0$, so this is $R_\alpha(\alpha(X)) = e^g$
I suspect this final $e^g$ didn't depend on the choice of $\alpha$
@AkivaWeinberger $\left(\frac{p+q}{p}\right)^{p/q}$, i.e. $\left(\frac{p+q}{p}\right)^{1/q}$, i.e. $(p+q)^{1/q}$ and $p^{1/q}$, i.e. two perfect $q$-powers just $q$ away, i.e. highly unlikely
Let f be a continuous function that maps the unit interval [0, 1] in R to itself. Assume that f has a derivative f' which is defined and continuous on [0, 1] and that |f'(x)| < 1 for x ∈ [0, 1].
Show that there is a constant M < 1 such that for all x, y in [0, 1], |f(x) − f(y)| ≤ M|x − y|
