Hello all, this question really has me and some friends stomped so advice would be appreciated.
Ok so, the relevant (dimensionless) continuity equation I have found to be
$$\frac{\partial\rho}{\partial t} + (1-2\rho)\frac{\partial \rho}{\partial x} = \begin{cases} \beta, \hspace{3mm} x < 0 \\...
If ##n## satisfies ##\sigma(n) \geq 3n## then ##n \in N_3##, so ##N_1 \subseteq N_2 \subseteq N_3 \subseteq ...##. OP denotes ##n## by ##N_3## instead but that would just be confusing since there is clearly more than one such number, and it doesn't make a difference to the problem so...
What's...
I believe the condition stated in words is that the number of positive divisors for ##N_k## is greater than or equal to ##k## times ##N_k##, for example, if ##n## is ##N_3## then it must have ##\sigma(n) \geq 3n##. So basically the ##N_k## are best thought of as classes.
after two simple line integrals we find that
$$ f(x,y) = \arctan{p/q} + \arctan{y/x} + \pi/2$$
if ##x > 0## and
$$f(x,y) = \arctan{p/q} + \arctan{y/x} - \pi/2$$
if ## x < 0 ##.
And then we can just take the limits to find the value of the discontinuity (as given above). But what is the...
I think I figured out my error which was that ## Pr(\cup_i E_i) = 1- (1/N)^N##. This probability is in fact equal to ## 1 - N!/N^N## since there are N! orders of picking all the different colors. Now the expression comes out nicely.
I do believe this was @timetraveller123 's idea as well.
Well ##\cup_{i} E_i## is just the event that at least one color is not used, so its probability is given by ##1- (1/N)^N##. Now if I is a subset of {1,...,N} where ##\left | I \right | = l## then ##Pr(\cap_{i\in I} E_i) = (1-l/N)^N## (I'm guessing this is where I'm making a mistake?). So then we...
Homework Statement
"Consider a system with three states, ##|1\rangle , |2\rangle ,|3\rangle ## with energies ##\hbar \omega_1 , \hbar \omega_2 , \hbar \omega_3 ##. the states are then separated by ##\hbar \omega_3 -\hbar \omega_1 = \hbar \omega_{13}## and ## \hbar \omega_3-\hbar \omega_2= \hbar...
Homework Statement
Have to read a paper and somewhere along the line it claims that for any distinct ## \ket{\phi_{0}}## and ##\ket{\phi_{1}}## we can choose a basis s.t. ## \ket{\phi_{0}}= \cos\frac{\theta}{2}\ket{0} + \sin\frac{\theta}{2}\ket{1}, \hspace{0.5cm} \ket{\phi_{1}}=...
That's just how ##T## is defined. It's a transformation that takes a vector from a 3-D space to a 2-D space. In terms of the actual calculation, you let ##T## operate on a basis vectors of E (which is a basis for ##U##). So the first one is ##\begin{pmatrix} 1\\ 0\\0 \end{pmatrix}## which...
Ok I can see the logic that led you through this attempt but it's not quite correct. First off you have a 3x3 matrix for a transformation that goes from a 3-D space to a 2-D one. This is already troublesome. You want your number of columns to be equal to ##\dim{U}## and the number of rows to be...
I don't know much about the specs of specifics laptops on the market right now but I would say that pentium and celeron are in fact NOT worth considering. Also, you'd be surprised at how much memory some programs use up, and a decent cpu is worthless without enough memory. One last point is to...