Recent content by neutrino2063

Ah, it is... thanks, it's fixed now alpha should be (L/2)*k2

I need to somehow simplify: \frac{1}{B^2}=\frac{1+\cos{2\alpha}}{2k_1}+\frac{\sin{2\alpha}}{2k_2}+\frac{\alpha}{k_2} to: B=\sqrt{\frac{2}{L}}\sqrt{\frac{\beta}{1+\beta}} Where: \alpha=\frac{L}{2}k_2 and \beta=\frac{L}{2}k_1 And \beta is also defined transcendentally...

So I have this equation: T=(cos(k2*a)^2+(1/4)*(r^\ast+\frac{1}{r^\ast})*sin(k2*a)*sin(k2^\ast*a))^{-1} where r=k2/k1; r^\ast=\frac{k2^\ast}{k1}; k2^\ast is the complex conjugate of k2 also k2=\frac{\sqrt{2*m*(E-V)}}{\hbar} and k1=\frac{\sqrt{2*m*E}}{\hbar} m,\hbar,a are all...

I am using separation of variables and superposition to solve: u_{xx}+u_{yy}=0; for (x,y) \in (0,L) X (0,H) u(0,y)=f(y); u(L,y)=0; u(x,0)=g(x); u(x,H)=0 Is it correct to assume that I can write my solution as: u=u_1+u_2 Where: u_1 is the solution with BC u(0,y)=0...

Why is \nabla\cdot\vec{A}=\frac{1}{r}\frac{\partial}{\partial r}(rA_{r})+\frac{1}{r}\frac{\partial}{\partial \theta}(A_{\theta}) Where \vec{A}=A_{r}\hat{r}+A_{\theta}\hat{\theta} And \nabla=\hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta} Instead of...