Suppose have a quantum-mechanical wavefunction in position space $\psi(x)$. I want to Fourier transform this to momentum space.
I'm just interested in playing around with various $\psi(x)$, so I'll take it as being known symbolically. I can then in principal have MMA do this either using the usual Integrate command or using FourierTransform.
...however, this seems like overkill. I really don't care about the Fourier transform $\tilde{\psi}(k)$ is symbolic, so I should just do it numerically.
@TobiasKildetoft Do we verify that $2^{6} \neq 1 \mod{13}$ and $2^{2} \neq 1 \mod{13}$ and so we deduce that 2 is a generator of $\mathbb{Z}_{13}^{\star}$ ?
that seems like it should be the best option, but the connection between DFT and (an approximation of) the usual Fourier transform is not entirely obvious
@Semiclassical How about writing a very simple program to do the integral numerically if all you're going to be doing is 1D integrals you can do that very quickly. And you wont have to bother with all the weird cases NFourierTransform is trying to take care of and check against
In general if the order of the group will be $n=p_1^{q_1} \cdots p_2{q_r}$ we will look at the values $g^{\frac{n}{p_i}}$ for $i=1, \dots, r$, right? @TobiasKildetoft
@KevinDriscoll I suspect the real 'best way' is to figure out how DFT relates to the Fourier transform, since the former takes advantage of the efficiency of FFT
@LandonZeKepitelOfGreytBritn ew
i guess it teaches you to be critical about what the text says
> It is true, however, that a countable union of countable sets need not be comparable with ℵ1 without choice. In fact, we can have a non-well-orderable set that can be written as a countable union of sets of size 2.
@Semiclassical I don't think one should buy/download an engineering book in order to call into question everything which is stated in it. It's only chapter 3, imagine what type of errors I might encounter once I reach eg chapter 17...
> Interestingly enough, Dedekind-finiteness can be graded into various level of finiteness, so some sets are more finite than others. For example, it is possible for a Dedekind-finite set to be mapped onto N, which in some way makes it "less finite" than sets which cannot be mapped onto N.
@Semiclassical I don't know a lot about DFT, but I am kinda dubious about it. The local density approximation is not systematically-improvable. It works, but it takes a lot fo outside work to put into it to get something reasonable.
$\frac{Y}{R} = G(1-HY)$ I don't wee why that's nonsense at least now one part is correct... ie the fraction but the other side of the equations is far from being what is needed
His personal animus toward me had nothing to do with "geometric approach." It had to do with his mental state and the fact that I corrected errors he made in answers on main, and that infuriated him.
BTW, @Balarka, this question should remind you of our discussions in here. I'm still not sure what the question means. And I'm not sure Moishe Cohen is right, either.
I just recently proved that if $J \subseteq M_n(R)$ is a left ideal, where $R$ is some ring, then $J^T := \{A^T \mid A \in J \}$ is a right ideal, and vice-versa. This got me wondering: is there some necessary and sufficient conditions involving a nonempty subset $J$ being an ideal and $J^T = J$,...
Holy shit, past me was a genius. I had forgotten that I did it, but past me left present me a to-do list of things that need to be finished in this paper.
This may be the nicest thing past-me has ever done specifically for present-me
For example, my aunt did not develop symptoms at the usual time. But well into her 30s has a psychotic break due to extreme emotional and physical trauma. It took both.
I just recently proved that if $J \subseteq M_n(R)$ is a left ideal, where $R$ is some ring, then $J^T := \{A^T \mid A \in J \}$ is a right ideal, and vice-versa. This got me wondering: is there some necessary and sufficient conditions involving a nonempty subset $J$ being an ideal and $J^T = J$,...
So, here’s a riddle / problem for this week (stolen from 538.com)
“You, a very wealthy aristocrat, own 25 horses. You’re bored one day (you’re bored everyday) and decide to amuse yourself by identifying the three fastest horses in your stable.However, your personal racetrack is severely lacking in capacity, and you can only race five horses at a time. What is the minimum number of races you’ll need to organize to identify your three fastest horses?”
(Source above should have been FiveThirtyEight.com)
Now that can't be right. We live in a deterministic universe. So if you just know the initial state f the universe, then the answer is 0 races. You can just derive which hours is the fastest form basic physical laws.
there is no maximum, race the same 5 horses over and over, keep going until you run out of money and have to auction the horses the money, squat in your own home as the bank takes it away from you
Thats right. So that want an algorithm that gets the right answer every time and they want to know what is the minimum number of scheduled races that that algorithm must run
@Semiclassical I don't see why minimum is a problem. It would make sense, clearly you have some dumb upper bound, which is to choose 5 horses, race them, then take the winner of that, race them, etc
@Daminark I take @Semiclassical 's objection to be that the minimum number of races required to determine the 3 fastest is not well defined, by itself. Because if for the first race you pick 5 horses randomly, you can get lucky and put all 3 fastest in that race. Then you take the horse who finished 3rd and run it in 5 more races, it wins every time. That's 6 races and you're done
But obviously that only works if you get lucky to start out with
