p.s. I used: ##x=\left(\frac{E-E_c}{k_B\:T}\right)## for the integral: ##\int \:g\left(E\right)f\left(E\right)dE \rightarrow \int \:x^{\frac{1}{2}}exp\left(-x\right)dx##
Hello,
I've seen in a few books on solid state physics that one can deduce an expression for average K.E.:
$$<\:K.E.>\:=E_c+3/2\:k_B\:T$$
from the following:
$$<\:K.E.>\:=\:\frac{\int \:\left(E-E_c\right)g\left(E\right)f\left(E\right)dE}{\int \:g\left(E\right)f\left(E\right)dE}$$
I can't...
I get the feeling this isn't the best way to simplify but what I get is as follows:
$$\frac{sin\left(\frac{3\pi }{8}\right)}{sin\left(\frac{3\pi }{4}\right)}=\sqrt{\frac{2+\sqrt{2}}{2}}$$
If you then multiply through by ##2^{\frac{1}{2}}## you get...
I'm assuming the following identity has something to do with it, but can't seem to put it all together:
$$sin\left(x\right)=\frac{e^{ix}-e^{-ix}}{2i}$$
working backwrads from the solution given in the question and putting aside the prefactor of ##2\pi i## that comes from the theorem, that means that the sum of the residues should give ##2^{\frac{1}{2}}sin\left(\frac{3\pi }{8}\right)##
I can't see how to get there from:
$$\frac{e^{\frac{3\pi...
Are you saying:
$$\left(z-e^{\frac{3\pi i}{4}}\right)f\left(z\right)=\left(z-e^{\frac{3\pi i}{4}}\right)\frac{z^{\frac{1}{2}}}{\left(z-e^{\frac{3\pi i}{4}}\right)\left(z-e^{\frac{-3\pi i}{4}}\right)}$$
so that as z goes to z0:
$$\frac{e^{\frac{3\pi i}{8}}}{\left(e^{\frac{3\pi...
Okay cool. Am I right that you get:
$$\cos \left(\frac{\pi }{4}\right)+isin\left(\frac{\pi }{4}\right)=e^{\frac{i\pi }{4}}$$
and
$$\cos \left(\frac{3\pi i}{4}\right)+isin\left(\frac{3\pi i}{4}\right)=e^{\frac{3\pi i}{4}}$$
And if so, is the next step:
$$\left(z-e^{\frac{\pi...
Yes exactly, simple poles. Firstly, I'm just checking to see if the values I have obtained for z0 at the poles are correct, they seem unwieldly. If they are correct, you multiply f(z) by (z-z0) and evaluate the result at =z0.. I've tried to do that but can't see how it would get you to the...
Ah yes, so sorry!! The integral should give:
$$2^{\frac{3}{2}}\pi \:isin\left(\frac{3\pi }{8}\right)$$
I got the numerators and denominators mixed up! Soz
I haven't that's the problem. The first step, in the process I'm familiar with, is to factorise the denominator so that you can find values of z for which their are isolated singularities. I've tried to do that and I've given my results in the original post. Not sure where to go next..
p.s. I'm...
Hello,
Thanks for the responses. The integral is a closed contour integral (with a branch cut along the negative x-axis so the function is not multi-valued) but I couldn't find the correct symbol with the integral sign having a circle. The residue theorem states that in place of calculating the...