/sci/ - Science & Math

>>10422106
ok, I think I get what your problem is.

the thing is, while intereference is also happening with only one slit, the resulting pattern is continuous. it is still a refraction pattern tho, but because it originates from the same wave there is no discrete destructive interference. there is destructive interference, but it is continuous, so you dont see exactly where it happens and where it doesnt happen.

now with two slits on the other hand even tho they share the same original wave, the different locations of the slits guarantees that the refraction patterns won't be the exact same for both slits, they will be VERY similar but not perfectly. so now between those two waves, exiting the slits, destructive interference does not happen continuosly but discretely.

I hope this makes sense and maybe pic related help a little so get the mind juices flowing

since there are undeniable parallels:
whats /g/s opinion of /sci/?

do you go there? why? why not?

I was doing some exercises in Ahlfors and came across the classic problem of computing [math]i^i[/math]. Clearly, they want you to express [math]i=\exp\left(\frac{1}{2}\pi i+2\pi n i\right)[/math] so that you can distribute the exponent through this and write:
[math]i^i=\exp\left(-(2n+\frac{1}{2})\pi\right)[/math]

But this got me thinking, what is stopping you from doing this iteratively to an arbitrary complex number [math]z=e^{i\theta}[/math]? So a first order approximation might look something like:
[math]z=(e^{1+2\pi n_1i})^{i\theta}=e^{i\theta}e^{-2\pi n_1}[/math]
Continuing to perform this expansion on [math]e^{i\theta}[/math] will just output more terms that look like [math]e^{-2\pi n_1}[/math], but if we conduct the expansion on one of these terms then we will find:
[math]z=e^{i\theta}(e^{1+2\pi i n_2})^{-2\pi n_1}=e^{i\theta-4\pi ^2 n_1n_2 i}e^{-2\pi n_1}[/math]

How does this get resolved? Do we have to conclude that [math](e^a)^b\ne e^{ab}[/math] in general?