like right for a sphere no matter how it's embedded you expect the heat after a long time to uniformize out the the initial average of the heat distro right
it doesnt matter how the sphere is stuck up in space, it's mean curvature can go crazy but the distribution doesnt care (if it's like properly insulated or w.e.)
OK, so this comes from taking the defining equation $de_i = \sum \omega_{ij}e_j$ and differentiating it by the product rule, DogAteMy. Give it a try. [$d$ of LHS is $0$ of course.]
ricci flow is also all about the heat equation lol
but idk if there's any physics beyond some differential geometers being like "hey heat equation smooths shit out what if we apply heat equation to geometric quantities and see what happens"
yeah idk enough about geometric flows yet, only a tiny tiny bit
the idea for ricci flow is that Ricci curvature is kind of like the second derivative of the metric that looks most like $\Delta$ kind of so $g_{t} = - \text{Ric}(g)$ is kind of like the heat equation kind of
the problem with this one @Semi ( i guess it's not a problem, it's why it's so fruitful to study) is that this thing can get crazy singularities because deforming along mean curvature isnt very nice
Sorry about the longwinded route, DogAteMy. There's a learning curve to this method, but it's extremely powerful. See section 3.3 of my diff geo notes if you ever get interested.
@TedShifrin i was eventually able to show it was unsolvable in the way it was stated and the prof fixed it, actually was the third problem on the same assignment i was able to show was either incorrect or unsolvable
@Faust: I think it's OK for a prof to screw up occasionally in class or on homework, but it shouldn't be that often. I'm fond of "Prove or give a counterexample" questions, where there's room for error. :P
@TedShifrin yeah he made up a nice question but it wasnt actually quite right, so he changed it so show why it wasnt true and added a second part but made anther mistake in his additional part, i was fine with the mistake question the unsolvable one sucked cause i spent like 8hrs on it before finally being able to show it was not provable.
@TedShifrin the definition that I like is that $\alpha_1 \equiv \alpha_2$ iff there exists $\beta$ cycle in $\mathbb{P}^1 \times X$ such that if we restrict $\beta$ on the fibers $\{t_0\} \times X$ and $\{t_1\} \times X$ we will get $\alpha_1$ and $\alpha_2$.
Karim: That's the primary notion. Two divisors $D$, $D'$ are linearly equivalent if there is a holomorphic mapping $f\colon X\to \Bbb P^1$ with $D=f^{-1}(0)$ and $D' = f^{-1}(\infty)$. (Of course, you could make those any two values you want.)
So Balarka's cobordism comment is (topologically speaking) to take a curve from $0$ to $\infty$ and take the preimage. What does that give you "upstairs"?
It's called a "pair of pants." Draw a pair of pants. The waist is joined by the pair of pants to the two cuffs at the bottom of the legs. However, you can't have a circle homologous to a disjoint union of two circles. (Number of connected components can't change!)
yeah for example for rational equivalence it is also if we take a circle let us flip the pair of pants upside down. Then, if we look from bottom to top we see that bottom circle is rationally equivalent to the top one
I meant we were drawing a smooth picture of a surface in $\Bbb R^3$, with cobordant $1$-manifolds. ... Algebraic cycles are complex (algebraic) subvarieties, not real ones.
You left off a final slash before }, @Zophikel.
Oh, and you have \big\ in front of nothing. It needs to be before the final }.
I go to the casino with $50. I decide I'm going to go to the roulette table and repeatedly bet $10 on black until I run out of money. Each bet, if I win, pays my original bet back plus $10 (1 to 1 payout).
Let's assume that the house edge on a $10 bet is exactly 50 cents.
What is the expected value of the total number of bets I make before running out of money?
if i have to integrate 1 /(4x^2+1) i know this is can be tan^-1 but if i just apply u substitution and get a natural log of of u where u = 4x^2 +1 is this also correct? I'm a bit confused since i can get two different results
@TannerSwett You should go from the definition of expectation: E[number of bets]=1*(probability of going out on first bet)+2*(probability of going out on second bet)+3*(probability of going out on third bet)+...
You may be able to estimate the expected value by some other method, but there's no way around finding the probabilities if you want to actually compute it
