Recent content by Schwarzschild90

Hi Ray. Yes and I've developed my solution much more. It's still not quite correct though. Will someone help me?

Homework Statement 2. and 3. Relevant equations and the attempt at a solution We find the L^2 projection as such: <b_j , e_j > , where e_j is orthonormal basis j. Now set b_j = < x^2 , e_j > for 1 \leq j \leq 3 .

We have that O(j) for k = 1.. M-1 is a column matrix, but does this imply that O(j) is a row vector with matrix elements (k,j)=(1,1), (1,2), (1,3), (1,4)?

I read it through, but we haven't worked that much in-depth with eigenvalues and eigenvectors with respect to the lattice Lalplacian or used applied linear algebra sufficiently for me to easily understand that. So it's slightly above my mathematical skills, but I'll talk it through with my...

Okay, I knew that definition of the lattice Laplacian. It's what we used in the course, but it was not defined as such. Right. Next step is solving the characteristic equation for the eigenvalues of the system.

Right. How is the lattice Laplacian commonly defined?

Homework Statement Homework Equations The lattice laplacian is defined as \Delta^2 = \frac{T}{\tau} , where T is the transition matrix \left[ \begin{array}{cccc} -2 & 1 & 0 & 0 \\ 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & -2 \end{array} \right] and \tau is a time constant, which is...

I see your point. For a 4 x 4 matrix, I would expect a 4/5% probability of winning if I have 4 coins and my opponent has one?

I thought about that, but decided to take a leap of faith and take tau = 1. With tau = 1, then a 100 x 100 matrix presents me with 99,01% chance of ultimately ruining my opponent, as expected. I present a plot of U(l) versus position l.

But does r > 0 not imply that tau is likewise greater? Does it hold for a 100 x 100 matrix as well? A plot of the ultimate probabilities show a linear relationship between matrix elements and ultimate probabilities. If only one is the correct value to pick for tau, then it should not appear...

But you determine the first column of u, while the professor said that only the first row is of interest.

I figured as much, but I wanted to know for sure. I set tau=1 for simplicity, as I thought it simplified the expression, but I also see there's a rigorous mathematical reason to do so.

Correct. I am given a specific matrix. What do you mean by no choice for tau? (I might have difficulties putting your explanation into context, as I am not familiar with your notation) I determined the ultimate probability by taking tau = 1 and computing the inverse of T, like so: U=-(inv(T)/tau).

Supposedly it's just the rate the ultimate probability is reached with. The ultimate probability is inversely proportional to tau, so large values of tau might correspond to years. I've emailed the professor to clarify things.

I want to expound on the ultimate probability as determined by matlab. Taking tau = 1, we get the following expression for the ultimate probability, taking l = 1. (So the first position in the n_0 column vector is = 1.) U(l) = - \frac{T^{-1}}{\tau} n_0 = \left[\begin{array}{cc} 0.8 & 0.6...