Its sad to think about all the math that can't be known, there are thousands of papers out there, and tons produced every year, there is no way to keep up.
@ᴇʏᴇs @Incurrence I feel completely broken now. I think I am at rock bottom. This is very good. It means I am about to start a brand new life. I think I will change my reading list too. It will no longer be the 12 holy books I mentioned. Maybe I will change my life plans too.
The people who win such prizes often are the least expected. =)
When I think of how bad this life has been, I then think it could be worse in the next one, so I try to cherish this one and work towards a better next one.
I think I am lucky I am not born in North Korea.
I think I am lucky I don't have ALS.
Actually, do you guys think this site should even exist?
This is a 1978 Putnam problem. As usual, you are farming reputation by posting puzzles without source, this time also with non-original solution. — pizza41 secs ago
It could also be he's in a problem-solving class and the questions were presented to him without a source so maybe he didn't know they're previous contest problems
He was satisfied with our response, so he left, if he wasn't he would have argued probably or given more examples
@DiscipleofBarney Not those two lines, but if other lines went around the same number of singularities they would have the same homotopical equivalence class
So homotopical equivalence is how many 'genus' they pass around
@Incurrence There is no such thing as genus of paths. The relation you're looking for is: the paths are homotopic (either basepoint-preserving homotopic, or freely homotopic). In any case, I don't know if they are or not, since I can't figure out what your green curve does.
Homotopy classes of loops on surfaces are more complicated than just "how many holes they pass around". The fundamental group of the genus g surface is isomorphic to $\langle a_1, b_1, \dots, a_g, b_g | [a_1, b_1] [a_2, b_2] \cdots [a_g, b_g]\rangle$, where that's a group presentation.
This is a very hard group to get a handle of in practice; there's still plenty of research done on, e.g., its automorphism group.
(The torus's fundamental group is abelian, isomorphic to $\Bbb Z \oplus \Bbb Z$, where the isomorphism is given by (number of times one goes around the 'interior loop', number of times one goes around the 'exterior loop'), counted with sign.)
This is a 1978 Putnam problem. As usual, you are farming reputation by posting puzzles without source, this time also with non-original solution. — pizza41 secs ago
First we arrange (without affecting the area) the pieces of the octagon like this:
Now we can see that resulting octagon is actually a square without four small triangles around its vertexes, like shown here
Let's calculate area of the octagon
$$A = (4+2\sqrt{2})^2 - 4\cdot(\sqrt{2}\cdot\s...
@DiscipleofBarney Well for my test, I want to pass 80, but I wasn't sure if the only criteria is that they be closed and negative, so I may not reach 80 today
If $\omega$ is a invariant form (from the antipodal map on the $S^{n}$) and $\omega=d \tau$, for some form $\tau$ on $S^{n}$ then $\omega=d(\tau+i^{*}\tau)/2$. Let $H^{*}(S^{n})^{I}$ the de Rham cohomology of invariant forms on sphere. How to show that the natural map $H^{*}(S^{n})->H^{*}(S^{n})^...
@WillHunting If I have a locally compact Abelian group $G$, I can see that for $\chi\in \hat{G}$, $\hat{f}(\chi) =\sum_{g\in G} f(g)\overline{\chi(g)}$. But in the case when it's not Abelian, $\hat{G}$ is not a group, and I am having trouble understanding the key difference between characters in the Abelian case and representations with their corresponding traces in the non-Abelian case.
oh my god !! it will be 3 only since we have 2^p + 1 = 2 + 1 (mod p) so 3 = 0 (mod p) and 3 = 3 (mod p) since p is prime and 3 is prime then the only number must be 3
I am too tired this week I had 4 exams last few days and presentation today and just working on my last number theory assignment that I have to submit on tuesday then
and then project for number theory on thursday then semester is done :D
First we arrange (without affecting the area) the pieces of the octagon like this:
Now we can see that resulting octagon is actually a square without four small triangles around its vertexes, like shown here
Let's calculate area of the octagon
$$A = (4+2\sqrt{2})^2 - 4\cdot(\sqrt{2}\cdot\s...
