Recent content by cellotim

I have a second-order, nonlinear differential equation: D(y,y',y'',x; y'(x=R)) = 0. Note that there is a "parameter" which is the first derivative of the solution evaluated at a particular point x=R. I want to solve it numerically and use only the solution at y(x=R) and y'(x=R). I don't care...

Yes, I see now. There is a difference. The continuity equation for electromagnetism is, \nabla\cdot\vec{J} + \partial \rho/\partial t = 0 or \partial_\mu J^{\mu} = 0 where J^{\mu} = (-\rho,\vec{v}) is underspecified. But the equation \partial_{\mu} T^{\mu\nu} = 0 has four equations, not...

Yes, I agree. I think it's a good example of a basic mistake and one that's easy to make given the way the geodesic equation is set up. It's a good discussion point for logical fallacies. In this case, it's a classic fallacy of reversed implication, i.e. since constant velocity implies...

I was looking for references on the quadrupolar formula and found this http://adsabs.harvard.edu/full/1992Ap&SS.194..159Y". I was so shocked to find that it had actually been published (although it was nearly 20 years ago) that I had to post this warning that there is a fundamental flaw in the...

In your last line, you take tan10x out of the limit without evaluating it. Put it back in and turn it into cot(10x) to put it on the bottom, then apply L'Hospital's. Make sure to apply exp at the end to get the final answer.

Well, consider the sets of integers (1,...,n) (n+1,...,2n) and so on (and the negatives likewise). Now, these partition the integers. Take n consecutive integers anywhere in the integers and for none of them to divide n exactly you would have to fit them inside one of these sets. That's...

It needs to be diagonal. It remains to prove that that diagonal matrix must be the identity.

The M's are not necessary as far as I can see only that L and L' are lower-triangular and U and U' are upper-triangular, and I mean preserve upper-triangularity of U' to U.

Like before, use the rows that have only a single 1. These allow you to change any value in the column without affecting the rest of the row.

Let me try to reiterate it. Say we have a neighborhood of f(x), V. Then we can find a neighborhood of x, U in X such that f(U) is in V. In the same way, if we have a neighborhood of g(f(x)) W in Z, we can find a neighborhood of f(x) T in Y, such that g(T) is in W. So, we should be able to...

Suppose we have two different LU decompositions, A = LU and A=L'U'. Because A is non-singular L, U, L' and U' are all non-singular and invertible. This implies that U = L^{-1}L'U'. Now you should be able to show that I = L^{-1}L' in order to preserve upper-triangularity.

I see. In your original question, A's first row was unchanged. In that case, just use the new C row to get the result.

Forget about A and B. You can do the operations on C alone. Look at rows 7 and 8.

You type them \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array} where x=tex. Use cccc for a 4x4 matrix instead of ccc.

Use the tex delimiters. Just click on the matrix in my post, you should get a popup with the source code.