Write down the number of equations and the number of variables. If there are more variables than equations, then there are infinitely many real solutions and I would be surprised if you can't find an integer solution.
I don't know for certain, though, because I have no interest in trying to examine that more closely.
@AliCaglayan I am just checking tensor product properties for modules. Just to verify my understanding I checked many of them there is something I am stuck on.
Consider the map from $(M \times N) \otimes P \rightarrow (M \otimes P) \times (N \otimes P)$ given by $(x,y) \otimes p \mapsto ((x \otimes p), (y \otimes p)$. This map is well defined since the map $((x,y),p) \mapsto ((x \otimes p), (y \otimes p))$ is bilinear.
I'm a physicist. I'm reasonably good at math. I would like to understand stochastic processes better than I do now. In particular, I'd like to be able to see the links between diffusion-type equations, stochastic differential equations, and Langevin type equations.
I'd really like to find a resource that explains these things in enough depth that I believe the results, and such that I can apply them to real problems. At the same time, I don't really need to see all the epsilon-delta proofs.
Consider the problem $U_{n+1} = U_n + h \Lambda U_n$. $\Lambda$ is diagonizable and is eigen-decomposed to $\Lambda = VDV^{-1}$, where D and V $\in R^{d\times d}$, D is diag. How to proceed the induction to show that the Forward Euler method applied to this problem gives $U_n = V(I + hD)^n V^{-1} y_o$?
@ZachHauk Perhaps. But comparison is unnecessary as regards my observation that it's surprising that you've heard of algebraic geometry by middle school.
A drunk man stands with a cliff one step to his left.
He takes steps randomly left and right.
Each step has probability $p$ of going left and probability $q=1-p$ of going right.
Each step is the same size.
If allowed to randomly step indefinitely, what is the probability that the drunk falls off...
(If you try a coordinate transformation which involves stretching instead of just rotating, you'll see that the transformation is not quite what you expect)
@Semiclassical Right, I shouldn't have mentioned covectors. Sorry.
Not sure you saw what I said after that, but you might check out the second-to-last chapter in Altland & Simon's text on condensed matter field theory. @DanielSank
(Second-to-last chapter in the edition I know is classical nonequillibrium theory, which is what you'd want. Last chapter is quantum nonequillibrium theory, stuff like Keldysh.
@A---B For what it's worth, if you go to the physics site's chat room and ask 0celo7, he will give you a good answer. In fact, I will ping him now with a link to your question.
@ZachHauk It involved talking to the administration, I don't remember exactly. The fact that I took the AP Calculus BC test right before high school and got a 5 (out of 5) probably helped.
I noticed just now that in a chapter I never read before, there's a discussion of dissipative quantum tunneling. I am rather interested in reading that as it's directly relevant to my day job!
Okay, why don't you use the matrix transformation of a rotation? i.e $$\left( {\begin{array}{cc} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{array} } \right)$$
@Semiclassical Neat. I actually could stand to learn about boson coherent state path integration. I saw it in a course and I understood why it was nice, but I can't reconstruct it now.
@DanielSank actually, that riemann sum comes from consider the integral $$\int_{-3}^{-1} \dfrac{1}{x^2} dx$$ Easy one with hardest sum I've never seen :P
because that would show that the set of all orthogonal vectors to some $\vec{v}$ is a subspace spanned by the most possible linearly independent vectors whose dot product with $\vec{v}$ is 0
@Semiclassical Just as an aside, in quantum computing, it turns out to be very important to understand the two level Landau-Zener problem with arbitrary trajectories, i.e. not just the linear one that people usually talk about.
But, anyways, there are models where the scattering amplitudes can be done exactly, and quite recently it's been discovered that this collection of models is actually larger than people realized.