Hmmm, I think I'll give this limit to some students $$\lim_{n\to\infty}\left(\frac{1+\sqrt[\large 2^2]{2!}+\sqrt[\large 3^2]{3!}+\cdots +\sqrt[\large n^2]{n!}}{n}\right)^\sqrt{n}$$
@Vader you already have the tools right in front of you! if u=log(x) and du=1/(ln10 * x) dx then can you rewrite log(x)/x dx in terms of u and du? it's very simple algebra!
you know log(x) is u, and you know dx/x is ln(10)*du, so log(x)/x dx = (ln10) udu
Can anyone give me a hint towards the following problem - in a polynomial ring R[X], is it possible for 1+X to be a product of two non-units? I suspect the answer is no, but I can't see a way of doing this (earlier in the question we had to show that 1+rX is unit iff r is nilpotent, but I don't see how this helps)
@robjohn does it mean that if we have family of continuous linear functionals say $f_1 , ..... , f_n$ , then they are independent if $a_1 f_1 +...... +a_n f_n = 0$ means $a_1=...= a_n=0$ for every x? , where x is the element of the space where $f_i's $ take the value.
if we have $$V(0,y) = \sum ^{\infty}_{n=1} C_n \sin(n\pi y/a) = V_0 (y)$$ we multiply by $\sin (n' \pi y/a)$ on each side and integrating with respect to $y$
I already proved the group of rigid motions of the tet is iso to $A_4$, it was cool. Now, I was given the lattice of subgroups of $A_4$ and I have to prove that lattice is correct. So I want to prove the following: if $\tau,\sigma$ are a two cycle and a three cycle in $A_4$ they generate the whole group.
I can easily find out what the subgroups of order 2,3 are.
All I have to show is there is a unique subgroup of order 4, which I could do using Sylow, since it is normal and a Sylow 2-subgroup, and that any other subgroup is that or either all A_4, @AlexanderGruber
@Astrum So, the general idea is this. Solutions of Laplace's equation are by definition 'harmonic functions' and so the sines and cosines of integer frequencies form an orthonormal basis for these solutions
So we can replace ANY solution and ANY appropriate boundary condition by its Fourier series
and then instead of solving a Differential Equation, we need only to solve an algebraic equation for the Fourier coefficients
@Astrum Okay. Do you find you're getting stucking on doing the integrals/algebra required to find the fourier coefficients, or is it the process of finding them that's messing you up?
@Astrum okay no problem. If you continue having issues maybe I can connet what you're doing here with basic linear algebra in finite dimensions and it might be more clear
I proved quite a lot of stuff just now! @DanielFischer
Quite a lot of work on A_4 lately.
Well, dunno if a lot but significant.
Proved iso to motions of a tet, found lattice of subgroups, proved twas solvable, as is S_4, and found the unique group of order 4 is normal and iso to V_4
@PedroTamaroff Okay, take an element $c$ of order $4$ of $S_4$. Consider its square. Try writing it as the square of two other elements of order $4$, not being $c^{-1}$ either, and fail. Deduce there is no such subgroup ;)
If I have a function which depends explicitly on some variable $x$ and a parameter $\beta$ will and asymptotic expansions of the form $f(x,\beta) = \sum_{n=1}^{\infty} \frac{c_n(x)}{\beta^n}, \beta \to \infty$ necessarily be unique?
@KevinDriscoll well, here it is: $$V_1 (x,y) = \frac{8 V_0}{\pi} \sum _{n=1,3,5...} \frac{\sin ^3 (n\pi /4)}{n} e^{-n\pi x/a}$$ is for the first potential, the one from 0 ---> a/2
@Astrum Fromt he way you described how you did the problem, you found one solution for the region where the BC is V(0) = V0 and another one for the region where V(0)=-V0
@PedroTamaroff Because an element of order $4$ belongs to some Sylow-2 subgroup. Since all such are isomorphic, every Sylow-2 subgroup contains elements of order $4$. The Klein group has none, so you need some.
Call the first $V_1(x,y)$ and the second $V_2(x,y)$
@Astrum Since $V_1$ is valid in a domain containing 0, you should impose the boundary condition that $V_1(x,0)=0$ (but do NOT do the same thing for $V_2$ because that solution isn't valid near 0
@Astrum Then finally you will find that you have 1 free coefficient still available after applying the boundary condition appropriate for $V_1$ and $V_2$
Conceptually you'll have to imagine that we ahve a tiny tiny piece of insulation at $x=0$ so that the two strips of metal at differnt voltages don't actually touch
otherwise current would flow and ruin our boundary conditions
If you're doing basically ANY kind of physics research understanding how to solve differential equations (both numerically and analytically when possible) is VERY important
you are basically doing it all the time
So this particular example fo how to solve Laplace's equation with constant boundary conditions isn't very practical int erms of real experiments and cutting-edge theory
But the concepts of how to solve ODEs using boundary conditions and symmetry and Fourier analysis is SUPER important
@Mike It depends. I don't know what kind of general requirements they might have. But the work itself is going to be focused kind of grunt work. Something not too conceptually difficult that you can learn in your time on the job without a lot of specifici background.
