Recent content by cup

I had the same issue a while ago when I installed a new graphics card. At first, my computer acted like yours except that it kept restarting (instantly) over and over. My power supply was too weak. After upgrading the power supply my computer acted exactly like yours. I then proceeded...

Justin, In the finite case, can you state clearly what \mathbf{B} and \mathbf{B}_{dip} are chosen to be, and why this should represent the desired physical situation as R \to \infty?

Sorry about that. Well, please show me the finite derivation you speak of. This is what I get when I attempt it: \left\{ \begin{array}{c c} \mathbf{B}_{dip} = \frac{\mu_0}{4 \pi r^3} [ 3(\mathbf{m} \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{m} ] & r > R \\ \mathbf{B}_{dip} =...

A formula such as U_{em} = \frac{1}{2} \int (\epsilon_0 E^2 + \frac{1}{\mu_0} B^2) d^3r comes with certain requirements on the objects represented by the symbols in the formula. The integrand has to be a function. If you go ahead and (implicitly) assume that it's a function. This...

To clarify: you should look around to find out if there is any formalism for squaring the dirac delta function. If there isn't then the starting point of the derivation is unfounded.

I didn't read trough this thread very carefully, but are you sure that you are allowed to do the manipulations U = \frac{1}{2\mu_0} \int (\mathbf{B} + \mathbf{B}_{dip})^2 d^3r U = \frac{1}{2\mu_0} \int (B^2 + B^2_{dip} + 2 \mathbf{B} \cdot \mathbf{B}_{dip})d^3r On...

Reading more advanced calculus is a fine way to become "mathematically mature". Then, after reading advanced calculus (theorem-proof based), you are probably ready to read a book on Real Analysis. The difference between "real analysis" and "advanced calculus" is fuzzy, so you will be well...

As far as I know (still a student), (Abstract) Algebra is very applicable. And not just to other kinds of mathematics. Mathematicians, "pure" and "applied", "should" have some basic knowledge of algebraic structures and techniques. I think even some engineers study abstract algebra. It's the...

Never mind. I found my own error. I was thinking M_{\pi (i)} / M_{\pi (i-1)} instead of M_{\pi (i)} / M_{\pi (i) - 1} which is what the theorem says. Sorry about that. Cool theorem, though. :smile:

Hello friends. I am working trough "Abstract Algebra" by Dummit & Foote. I recently got to section 3.4, on composition series and "the Hölder program". The Jordan-Hölder theorem states: Let G be a finite, non-trivial group. Then: 1) G has a composition series. 2) If \{ 1 \} = N_0 \leq N_1...

I find that it's easier to do these kinds of calculations on a computer, using Tex. As long as you use a "vertical" coding style (one term per line only), it won't get messy. The ability to copy and paste really helps.

Example: If z = \frac{a-bi}{c+id}e^{-ig} then z^* = \frac{a+bi}{c-id}e^{+ig}

(In the following, z* denotes the complex conjugate of z.) The first thing to do is to make the denominator real (you will see why this makes sense): U \frac{ \left( e^{a(H-y)} e^{i[a(H-y) +nt]} - e^{-a(H-y)} e^{-i[a(H-y) +nt]} \right) \left( e^{aH} e^{iaH} - e^{-aH}...

Assuming that U, a, H are all real, it is very simple to take the complex conjugate: Just change the signs of all the i's.

Answer: xn = 11n yn = 13n-2 zn = 6-19n where n is any integer. Correct?