Recent content by physicsjock

http://img96.imageshack.us/img96/8606/qqqqbj.jpg so what I'm thinking is that you let x_n+3 + x_n+2<=x_n+2 + x_n+1 <= x_2 + x_1 then show that the first term can eventually go to the last term squeezing the term into the middle but the problem with this is i don't see where...

thanks hallsofivy, yea i used that general solution to apply variation of parameters, well i used y1=e^(2x), y2 = e^(-2x) y3=x, y4 =1, Ill do what you said and try somthing of the form you said, I'm just not sure what this means "And that should tell you that what you give...

hey, i have this 4th order ode question that I've been working on, the homogeneous solution was easy enough by finding the particular solution has become a bit annoying, the ode is y'''' - 4y'' = 5x2 - e2x I have gotten the particular solution using variation of parameters...

hey, If you have say, cos(x+β) where β is the phase and it fluctuates randomly (not just small fluctuations large ones) between 0 and 2∏ the average value of cos(x+β) would still be 0 right? thanks

Hey I've been trying to show that \frac{1}{\sqrt{1+u^2 -2xu}} is a generating function of the polynomials, in other words that \frac{1}{\sqrt{1+u^2 -2xu}}=\sum\limits_{n=0}^{\infty }{{{P}_{n}}(x){{u}^{n}}} My class was told to do this by first finding the binomial series of...

hey guys, my lecturer skipped the proof to show that \frac{1}{\sqrt{1+u^2 -2xu}} is a generating function of the polynomials, he told us that we should do it as an exercise by first finding the binomial series of \frac{1}{\sqrt{1-s}} then insert s = -u2 + 2xu he then said to expand...

oh cool thanks! so the total wave function is going to be, \psi_{T} =\frac{1}{\sqrt{2}} \psi_{\alpha}(x_{1}) \psi_{\beta}(x_{2})+\psi_{\beta}(x_{1}) \psi_{\alpha}(x_{2}) where α and β are sets of quantum numbers (same of different) so the only difference between \psi_{\alpha} and...

So for even functions there is a probability peak in the centre of the step, and for odd functions the probability is zero in the centre of the step, is that right? So the reason the correction is smaller for odd functions is because there is less probability of the particle being in the...

hey, say you have a infinite potential well of length L, in the middle of the well a potential step of potential V and length x. Inside the well is a particle of mass m. why are the first order energy corrections large for even eigenstates compared to odd ones? also, say well...

Hey, I had this question I thought i answered but now I'm questioning if the group operation is supposed to be addition and not multiplication. The question \varphi :\mathbb{Z}\to {{\mathbb{Z}}_{n}}\,\,\,\,\,where\,\,\varphi (a)=\text{Remainder}\left( \frac{a}{n} \right) When I...

Thanks for all your answers guys so to get a^kb^ja^mb^n = a^{k-m}b^{j+n} b^ja^m = a^{-m}b^{j} Thanks heaps!

I tried it again today with a fresh start and I ended up getting the same thing, The thing that makes me sus about it is the +/- inside the square root

How can you assume a^jb^ka^mb^n=a^ja^mb^kb^n ? The dihedral group doesn't commute so you can't assume that ^ can you? I tried using the group property ba=a-1b but it just made it uglier I might be able to instead say b^na^m is also an element of the domain so \theta...

Hey I've been working on this question, How that the following is a homomorphism \theta :{{D}_{2n}}\to {{D}_{2n}}\,\,\,givenby\,\,\,\theta ({{a}^{j}}{{b}^{k}})={{b}^{k}}\,\,\, \theta ({{a}^{j}}{{b}^{k}})\theta ({{a}^{m}}{{b}^{n}})={{b}^{k}}{{b}^{n}} \theta...

Hey HallsofIvy Thanks for replying, Are you sure its not: \pm dy \sqrt{y^2- 2c}= dx since you take the dx to the right side, and the inverted root onto the left? using the initial condition I get c=0 since \begin{align} & y'(0)=\pm \frac{1}{\sqrt{{{y}^{2}}(0)-2c}}=\pm...