Recent content by Fightfish

Not sure what it is that you want to achieve but here is a set of pictures of representative states of a Ising system at different temperatures that I've done in the past: You can either start with the uniformly aligned state each time or use the final output system state of the lower...

I usually prefer the alternative - to heat the system up instead. So instead of starting from a random state, I start with the zero temperature limit of perfectly aligned spins. This avoids the problem of getting trapped in local minima.

Could I check what your initial state was? If you started with an arbitrary random state and immediately put it at such a low temperature, it is highly likely for the system to become stuck in a local equilibrium. Most of the Monte Carlo moves will be rejected because the extremely low...

The main reason for the slow speed is because f[x_]:= (some stuff here) is a functional definition, so that whenever ##f(\alpha)## is called for a particular argument ##\alpha##, the sequence of instructions given in (some stuff here) will be evaluated from scratch. So, if (some stuff here)...

Equation (15.7) gives you the instantaneous velocity as a function of time, while Equation (15.8) gives you the expression for the maximum speed of the oscillation. What is the maximum value of ##\sin \omega t##?

Well I wouldn't recommend this approach. You should instead consider the product-to-sum identities in order to simplify the product of trigonometric terms into a simple sum of individual sine and cosine terms. For example, ##2 \cos \theta \cos \phi = \cos (\theta + \phi) + \cos (\theta - \phi)##.

Your integration from lines 3 to 4 is not correct: \int \frac{1}{1 + x^{-2}} \mathrm{d} x \neq \tan^{-1} \left(\frac{1}{x}\right). My suggestion is to instead split the original expression as \frac{x^4 + x^2 + 1}{2(1 + x^{2})} = \frac{x^4 + x^2}{2(1 + x^{2})} + \frac{1}{2(1 + x^{2})}.

When you did the triangle construction in your diagram, you treated ##x## there as a length, which only takes on positive values, but ignored its position relative to where the origin was defined. So, the x-coordinate of the point should in fact be the negative of the ##x## in your derivation.

No there is nothing wrong with the labeling of the coordinates in the diagram. Recall that ##\cos \theta## is negative for ##\pi/2 < \theta < 3\pi/2## (or what you may know as the second and third quadrants).

Or more precisely, E(t) = \left [ E_{0} e^{-\gamma t} e^{i \omega_{0} t} + \mathrm{c.c.} \right] \theta (t), where ##\theta(t)## is the Heaviside step function.

Note that: (1) There is a typo in the expression for ##\alpha(\omega)## - it should be ##E_{0}## instead of ##E## (2) ##E(t)## is not given by the first expression for all time - the excitation only happens at ##t = 0##.

It looks like a J in Fraktur script to me (i.e. the mathfrak command in LaTeX). You can find a reference list of characters in the various scripts here: https://tex.stackexchange.com/questions/58098/what-are-all-the-font-styles-i-can-use-in-math-mode

Ah okay, I was wondering whether you meant that or to look for the repetition cycles (the latter of which I actually tried, and its, well, not that bad I guess).

While studying the repetition period of the powers is a possible method, it can still get rather tedious (especially if you need to do it by hand in a time limited situation). Another method is to try considering ##19 = (20-1)## and performing a binomial expansion of the expression. The modulo...

Have you heard of the power mean inequalities? The hint is to make use of the fact that the arithmetic mean is always greater than or equal to the harmonic mean. (and yes I'm not giving any explicit expressions here because I want you look them up and try to understand them - the Art of Problem...