Since the integers are closed under addition and subtraction, the LHS must evaluate to an integer. But $\frac{1}{2}$ is not an integer. So, either $1 -1 + 1 - 1 + 1 + ... \neq \frac{1}{2}$ or the integers are not closed under addition and subtraction. The latter, however, is impossible, so it mus...
@TedShifrin OK, so I found three papers which are far more accessible than Eliashberg-Mishachev's exposition of h-principle, and does it in a much more concretely. These are Rourke & Sanderson's "Compression theorem I, II, III".
The second paper states and proves the following theorem: Suppose $M^n$ is an embedded submanifold of $Q^q \times \Bbb R^k$ where $q - n \geq 1$ and the embedding is "compressible", i.e., $M$ is nowhere tangent to the $x_0 \times \Bbb R^k$'s or equivalently, the projection map to $Q$ is an immersion $M \to Q$.
Suppose $H$ is the subspace in the Grassmannian $\text{Gr}_n(Q \times \Bbb R^k)$ corresponding to the $n$-plane subbundles of $\pi^*TQ$ (which is rank $q$). $U$ be a neighborhood of $H$ is the Grassmannian. Then there is an isotopy $h$ of $M$ in $Q \times \Bbb R^k$ which takes $M$ to $h(TM) \subset U$.
heres the problem i was thinking about: $p$ racers race $r$ races, and each time they get points according to their ranking (1st place gets 1, 2nd places gets 2, etc. etc.). for what values of $p$ and $r$ (specifically) can all racers get the same point value at the end of the races?
for example, $p = 2$, $r = 2$ works, since the first racer can get 1st then 2nd, for a total of 3, and the second can get 2nd then 1st, for the same total of 3
so i have a bunch of observations
for any $r = 2$ it works, since you can just flip the positions
@AlexW A little bit of this, a little bit of that. Most actively thinking about either dynamics (in particular hyperbolic dynamics) or topology (in particular h principles)
@AlexWertheim It is quite interesting actually. The machinery of h-principles gives you a proof of sphere eversion theorem, which you must know already
@Balarka: you overestimate me, unless I'm being obtuse to obvious sarcasm. I don't even know what the sphere eversion theorem says. (Is this a result of Smale's?)
Kenny: fun example. Determine the Jacobson radical of the ring of 2x2 matrices over your favorite field. Then determine whether such a ring contains nilpotents.
@Daminark Well, the latter part is more something I'll be approaching in January. Right now, just my brother, who has a lot of experience with both his own and many students' resumes/CVs.
(Some further food for thought, @Kenny: what's going wrong? You might be familiar with the result that in a commutative ring, the nilradical is the intersection of prime ideals. But in a noncommutative ring, the nilradical is almost never an ideal, let alone the intersection of prime ideals. This brings up another interesting question: what is the right definition of ''prime ideal'' in a noncommutative ring? Lots of fun to be had here.)
@Kenny: it is true, however, that if $I$ is a (left)-ideal of a ring $R$ which consists of nilpotent elements, then $I \subseteq J(R)$. Pretty amazing! Lam's beautiful book "A First Course in Noncommutative Rings" is a great read, if you find these kinds of results interesting.
In the commutative context, the nilradical is a subset of J(R), right? ('Cuz the former is intersection of all prime ideals, and the latter is intersection of all maximal ideals)
Lol, lots of things Balarka. For my money, the biggest problem is that the set of nilpotent elements of a noncommutative ring $R$ need not form an ideal (in fact, I'm guessing this is almost never the case)
Yep! You could try to fix this by taking the ideal generated by the set of all nilpotent elements, but this could be quite large. (In particular, if you look at the preceding example of 2x2 matrix rings over a field: [0, 0; 1, 0] and [0, 1; 0, 0] are both nilpotent elements, but their sum, [0, 1; 1, 0] is a unit, so any ideal containing even these two elements is the whole ring)
Oh by the way so, I know that it is possible to put a product on a cohomology group and get a ring. How it's done is over my head, but whatever the case, are those non-commutative?
Last night dream involve some kind of strange derivative:
On the blackboard the following is written $$u, \downarrow \!\!u = \frac{\partial \downarrow \!\!u}{\partial x}$$ $$(forgot), \frac{\partial v}{\partial u}$$
$\downarrow\!\! u$ denotes that u is decreasing as the derivative is taken. The equation to be solved seemed to be a cross between proportionality problems in high school and differential equation with variation features
In that maths lesson in the dream, the teacher asked us to start putting down arrows and see how the other functions and variables will respond as a first step towards some unspecified optimisation problem
Based on what is said in the dream, it appears such derivative is basically the derivative of some function u(x) restricted to a domain such that u(x) is decreasing. If a point p is specified, it is basically the set of all derivatives such that u(x) x>p is decreasing
standard stars-and-bars trick: solutions $(a_1,a_2,a_3,a_4)$ to $a_1+a_2+a_3+a_4=18$ satisfying $a_1\ge 7$ correspond to solutions $(b_1,b_2,b_3,b_4)$ satisfying $b_1+b_2+b_3+b_4=11$ with no restrictions, via just subtracting $7$ from $a_1$ to get $b_1$ (and leaving all the other variables the same value)
Similarly, there will be $\binom{4}{2}$-many intersections like $|A_1\cap A_2|$ (picking two indices from $\{1,2,3,4\}$ after all), and $|A_1\cap A_2|$ counts solutions $(a_1,a_2,a_3,a_4)$ with $a_1,a_2\ge7$, so you can subtract $7$ from two variables instead there
nice to see the PIE result come out in that way. though in some sense it's none too surprising: it would seem weird to me if you could to rewrite the result in a simple way that didn't have a combinatorial interpretation
Take any solution $(a_1,a_2,a_3,a_4)$ to the equation $a_1+a_2+a_3+a_4=18$ in which $a_1\ge7$. If you subtract $7$ from $a_1$, now it's any of the nonnegative solutions to $a_1+a_2+a_3+a_4=11$.
This can be done by the useful technique of differentiating under the integral sign.
In fact, this is exercise 10.23 in the second edition of "Mathematical Analysis" by Tom Apostol.
Here is the brief sketch (as laid out in the exercise itself).
Let $$ F(y) = \int\limits_{0}^{\infty} \frac{\sin...
well, you sum $|A_i\cap A_j|$ for each choice of $i,j$ and there are $\binom{4}{2}$ such choices
taking for instance $|A_1\cap A_2|$, that counts solutions $(a_1,a_2,a_3,a_4)$ to $a_1+a_2+a_3+a_4=18$ in which both $a_1,a_2\ge7$. Well, guess what happens when you subtract $7$ from both of them...
@Semiclassical This illustrates why I am bad at physics. I suck at handling special cases that only arises for certain values. Whenever I solve problems, I am often greedy and tried to solve the general case and then enumerate the specific cases from it, but it seems such method does not always work
This can be done by the useful technique of differentiating under the integral sign.
In fact, this is exercise 10.23 in the second edition of "Mathematical Analysis" by Tom Apostol.
Here is the brief sketch (as laid out in the exercise itself).
Let $$ F(y) = \int\limits_{0}^{\infty} \frac{\sin...
