Nicely spotted, that's a factor 4 wrong right there. Actually there is another error on that line, one which together with the other one sadly makes the argument as it is not work anymore. Perhaps I get back to repairing it tomorrow.
I also wrote down your argument with mod 4 in my own words...
The key here is that ##sin(x)## is symmetric around ##pi/2##. Hence the substitution ##t=pi/2+x## may be of use. You then see that a term in your new expression should disappear.
Thank you for your reply! That's what I'm starting to think as well, that mathematics with physics courses may be the better option. And yes the rigour is something that often bother me during physics classes and I find a spend a lot of time trying to fill in all the steps I don't feel are...
Hello! I'm a European physics student with an interest in what I believe to be mathematical physics. I'm torn between going for a master in physics or one in mathematics as more and more it seems to me that the research areas I'm interest are often conducted in the mathematics department...
It's from the 2 in the effective masses ##\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}(m_e^*m_h^*)^{3/4} =\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}\left(\frac{\hbar^2 \hbar^2}{(2a)(2b)}\right)^{3/4}=\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}\left(\frac{\hbar^2}{2\sqrt{ab}}\right)^{3/2}##...
Homework Statement
An intrinsic semiconductor with a direct gap have valance band ##\epsilon_k = E_v-b|k|^2## and conduction band ##\epsilon_k = =E_c+a|k|^2##, with ##E_v=6.0##eV, ##E_c = 5.5eV##, ##a=5.0eV\cdot Å^2##, and ##b=3.0eV\cdot Å^2##.
Calculate the chemical potential ##\mu## and the...
That argument seems fine as far as I can tell. Nice work! So this shows (0,0) can't be part of either set.
Another thing to note is that if ##S=A'\cup B'## we must have ##A \subseteq A'## or ##B \subseteq B'## (or the other way around) since if ##A'## contains a proper subset ##E\subset A## and...
I'm sure you don't but I'm not sure how everything follows from this. Why are ##A## and ##B## the only candidates? I'm with you that ##A## and ##B## are each a connected set but how does this show that every other choice of separated sets doesn't work?
If I take two sets so that their union is...
As for them being connected I think there's still work to do. @pasmith suggested that I show that ##(0,0)## is a limit point of ##A## which we did. (If he/she meant (0,0)?)
As to turn that into a proof that ##S## is connected seems harder. Obviously we have that ##A,B## aren't separated.
Two...
We have this useful theorem stating that if ##p## is a limit point of ##E\subseteq X## then ##f## is continuoius if and only if ##\lim_{x\to p} =f(p)##.
(For ##f:E \to Y## for metric spaces ##X,Y##, ##p\in E##).
And since the limit doesn't exists the function can't be continuous in ##(0,0)##...
I'm sorry but there was a typo in my initial post. It should be ##\{(x,y): -1 \le x \le 0, y=0 \}## i.e. also equal to zero (otherwise it wouldn't have been connected). I fixed that in the initial post now.
Lets name the set with the ##\sin## ##A## and the other one ##B##.
Well the problem...
Why ##(-1/2,0)##? You don't mean ##(0,0)##?. The first open ball obviously doesn't contain any points in the set for ##\epsilon \le 1/2## at least.
The open ball around ##(0,0)## should contain points in the set. We need points satisfying ##\sqrt{x^2+\sin^2 1/x } < \epsilon## using the...