I'm just finishing the leftovers of an apricot tart I made when I had friends over the other night. Mostly just apricots, but a French short pastry and a bit of sugar ... :(
> Call a real number repetitive if for every k you can find a string of k digits that appears more than once in its decimal expansion. The problem is to prove that if a real number is repetitive then so is its square.
I want to say that when a certain cohomology class vanishes on the total space of a bundle, I get a specific representative of a cohomology class of degree something less on the base space.
Also I think I've heard the phrase "What goes up must come down," as well as "What goes down must come up," used way too often at my school when talking about our elevators
If some polynomial of curvature vanishes on the frame bundle, I get an actual form that it's d of that lives downstairs. Like $\Omega = d\omega_{12}$ for surfaces ... but if curvature vanishes, you can make a global $\omega_{12}$ on the manifold.
It's a more sophisticated version of that. I'll think a bit.
Anyhow, in that situation, I would call $\omega_{12}$ the transgression. Chern's proof of Gauss-Bonnet was a far more complicated version of this.
My question was independent of that. I have a phobia of dying elevators. Even in Evans Hall at Berkeley I often walked up and down either 7, 8, 9, or 10 floors because someone got stuck a few times.
It used to drive me nuts how many people (particularly college students) would insist on taking the elevator so as not to walk one flight of stairs ... at UGA.
Let $r(a,b)$ be a rational number depending on $a,b$ and nonnegative. For every $b$ there is an $a$ such that $r(a,b)$ is not $0$.
Let $C(a,b)$ be a squarefree positive integer depending on $b$ and different for every $b$.
Consider for positive integers $I,J$ :
$$ S_J = \sum_{j=1}^J \sum_{i=1}...
But like, there was this one problem in high school which was multiple choice, each of them was a 4-way inequality, and I spent a full minute trying to figure out why all 4 answer choices were the same
Anyone have any idea why $Sp(n)\cdot Sp(1)/(U(2n)\cap Sp(n)\cdot Sp(1))\cong Sp(1)/U(1)$? Or why $U(2n)\cap Sp(n)\cdot Sp(1)\cong Sp(n)\cdot U(1)$? Perhaps... the magician @arctictern? :P
@LucasHenrique Pell's theorem says that you have infinitely many solutions to $x^2-ny^2=1$, right? The $\Bbb Z[\sqrt n]$ stuff tells you that if you have one, you have infinitely many. But you still need to show that at least one exists in the first place.
> Call a real number repetitive if for every k you can find a string of k digits that appears more than once in its decimal expansion. The problem is to prove that if a real number is repetitive then so is its square.
Sure, @Blue. It's a curve that hits each curve in the family tangentially. For example, suppose I take the family of lines $(\cos t)x + (\sin t) y = 1$ as $t$ varies from $0$ to $2\pi$. Draw those. Can you guess the envelope?
@AkivaWeinberger well it is quite simple. I am nowhere near equipped to solve that, but here is a sketch of a proof. First off, the numbers will be rationals that are not integers. Then you just have to prove that all rationals that are not integers are repetitive. With those two tools in hand, the proof is trivial.
the non-integer rationals are closed with regards to squaring.
in order to have a series that repeats for all n, you must have a series you can arbitrarily grow. This implies that the series is periodic at some point
therefore => rational
Anonymous
I guessed it from the graph (Yup, I "saw" it :) ). But how would I derive the equation mathematically? @TedShifrin
@Blue: This is actually an exercise in my multivariable math book. If the family of curves is given by $f(x,y,t)=0$, you solve $f(x,y,t)=0$, $\partial f/\partial t (x,y,t)= 0$.
@TedShifrin I'm a bit new to multivariable calculus. Could you please explain what $\partial f/\partial t (x,y,t)= 0$ means? Are you taking partial derivative w.r.t each of the variables x,y, t separately ?
That's not trivial, @Blue. You need the multivariable chain rule to prove it, and you need to understand the geometry of the gradient vector. So maybe wait ...
Anonymous
8:57 PM
@TedShifrin Okay. I should try to cover up multi-variable calculus fully first before delving in differential equations maybe.