## Introducing LaTeX Math Typesetting

$$mc^2(\frac{hc}{\lambda_0}-\frac{hc}{\lambda_1})=\frac{hc}{\lambda_0}\frac{hc}{\lambda_1}(1-cos\theta)$$
$$hmc^3(\frac{1}{\lambda_0}-\frac{1}{\lambda_1})=\frac{h^2c^2}{\lambda_0\lambda_1}(1-cos\theta)$$
$$\lambda_0\lambda_1(\frac{1}{\lambda_0}-\frac{1}{\lambda_1})=\frac{h^2c^2}{hmc^3}(1-cos\theta)$$
$$\lambda_1-\lambda_0=\frac{h}{mc}(1-cos\theta)$$

$$E_0=hf_0$$ $$f_0=\frac{c}{\lambda_0}$$
$$E_1=hf_1$$ $$f_1=\frac{c}{\lambda_1}$$

 Quote by HalfManHalfAmazing $$\oint E\cdot dA = \frac{Q_{enc}}{\epsilon_0}$$ Yahoo! Finally produced Gauss' Law in latex!
This would be better :

$\oint_\mathcal{S} \mathbf{E} \cdot d\mathbf{A} = \frac{Q_\textrm{enc}}{\epsilon_0}$

 Hi all, Here is my question: In the $$E>U_0$$ potential barrier, there should be no reflected wave when the incident wave is at one of the transmisson resonances. Assuming that a beam of particles is incident at the first transmission resonance, $$E=U_0+(\frac{\pi^2 h^2}{2mL^2})$$, combine the continuity conditions to show that B=0. Here are the continuity conditions: 1st $$A+B=C+D$$ 2nd $$k(A- B)=k^{'}(C-D)$$ 3rd $$Ce^{ik^{'}L}+De^{-ik^{'}L}=Fe^{ikL}$$ 4th $$k^{'}(Ce^{ik^{'}L}-De^{-ik^{'}L})=kFe^{ikL}$$ A couple more equations that we already know are $$k=\sqrt{\frac{2mE}{h^2}}$$ and $$k^'=\sqrt{\frac{2m(E-U_0)}{h^2}$$ Here is my attempted solution: I divided the 4th equation by K and then set equation 3 and 4 equal to each other. I then used the new equation to solve for C in terms of D giving me $$C=De^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}$$ where I substituted $$\frac{\pi}{L}$$ in for k'. I substituted this result into the first equation to now give me $$A+B={De^{-2i \pi}\frac{k^{'}+k}{k^{'}-K} + D$$. Solving for D gives me $$\frac{A+B}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K} + 1} = D$$ Now, plugging in our solutions for D and C into the 2nd equation $$\frac{(A+B)e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}+1}-\frac{A+B}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}+1}=\frac{k}{k^{'}}(A-B})$$ At this point, it seems impossible to simplify the equation to a point where it is obvious that B = 0. Am I on the right track or is there an easier way?
 $$17(1-\frac{1}{17^2})^\frac{1}{2}$$ Does anyone know of a website that allows me to use tex code and generate latex graphics, that isn't necessarily part of a forum like this one? BTW, the guy who coded LaTeX into this forum is a bloody wizard. EDIT: God! LaTeX is amazing! I've been checking the examples in this thread... what an invention!
 Found one here: http://at.org/~cola/tex2img/index.php (sorry for double post, I can't find the EDIT button...)
 $$\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$\frac{x-vt}{\sqrt{2}}$$ $$\sqrt{2}$$
 Recognitions: Gold Member Science Advisor Staff Emeritus Test... $A = A'$ - Warren
 Recognitions: Gold Member Science Advisor Staff Emeritus Test $E = mc^{10}$ ???