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Forum Tester

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:26 pm

This thread primarily is intended for testing integrity & beta testing of new features......

$\begin{aligned} \Updelta p_\lambda&={\frac{1}{2\Uppsi \epsilon \sqrt{( 2\pi \alpha\beta)^2+\pi(4-8\bar{Q})\alpha\beta+1} \sqrt{ 2\pi \epsilon ^2\alpha\beta- \epsilon ( 2\pi \alpha\beta+1)+\bar{Q}}}}\\ & \quad \times \left[\left\{ \epsilon \left(- 2\pi \epsilon \alpha\beta + 2\pi \alpha\beta-2\bar{Q}+ \epsilon \sqrt{( 2\pi \alpha\beta)^2+\pi (4-8\bar{Q})\alpha\beta+1}\right.\right.\right.\\ &\quad-\left.\left.\left.\sqrt{( 2\pi \alpha\beta)^2+\pi(4-8\bar{Q}) \alpha\beta+1}+ \epsilon +1\right)\pi\right\}\right.\\ &\quad+\left.\left\{ \epsilon \left(- 2\pi \epsilon \alpha\beta+ \epsilon \left( 2\pi \alpha\beta+\sqrt{( 2\pi \alpha\beta)^2+ \pi(4-8\bar{Q})\alpha\beta+1}-1\right)\right.\right.\right.\\ &\quad+\left.\left.\left.2\bar{Q}-\sqrt{( 2\pi \alpha\beta)^2+ \pi(4-8\bar{Q})\alpha\beta+1}-1\right)\pi\right\} \right], \end{aligned} $


Last edited by Admin on Sun Oct 30, 2011 10:41 pm; edited 1 time in total

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:27 pm

This thread primarily is intended for testing integrity & beta testing of new features......

$\begin{aligned} \bar{Q}&={\frac{1}{3 \epsilon ^4+24 \epsilon ^2+8}}\left[\vphantom{\left( {\frac{1}{2}}+\alpha \beta \pi \right)^{\frac{1}{2}}}15 { \epsilon }^{3}+8 \epsilon \pi \alpha \beta \left( 4-3 { \epsilon }^{2}-{ \epsilon }^{4} \right)+20 \epsilon \right.\\ & \quad \left.+4\left(\vphantom{\left( {\frac{1}{2}}+\alpha \beta \pi \right)^{\frac{1}{2}}}4 { \epsilon }^{2}{\pi }^{2}{\alpha}^{2}\beta\left( 4-3 { \epsilon }^{2}-{ \epsilon }^{4} \right) ^{2}\right.\right.\\ & \quad \left.\left.+15 \alpha \left( {\frac{4}{3}}+{ \epsilon }^{2} \right){ \epsilon }^{2}\pi \beta \left( 4-3 { \epsilon } ^{2}-{ \epsilon }^{4} \right)-6 \left( \epsilon +1 \right)\left( \epsilon -1 \right)\left(\vphantom{\left( {\frac{1}{2}}+\alpha \beta \pi \right)^{\frac{1}{2}}} { \epsilon }^{8}{\alpha}^{2}{\beta}^{2}{\pi }^{2}\right.\right.\right.\\ & \quad \left.\left.\left.+ \left( {\frac{15}{2}} {\alpha}^ {2}{\beta}^{2}{\pi }^{2}+{\frac{1}{8}}-2 \alpha \beta \pi \right) { \epsilon } ^{6}+\left( -{ \frac{11}{6}} {\alpha}^{2}{\beta}^{2}{\pi }^{2}-{ \frac{33}{2}} \alpha \beta \pi -{\frac{3}{8}}\right ){ \epsilon }^{4}\right.\right.\right.\\ & \quad \left.\left.\left.+\left( -{\frac{16}{3}} {\alpha}^{2}{\beta}^{2}{\pi }^{2}+{ \frac{7}{12}}- {\frac {28}{3}} \alpha \beta \pi \right) { \epsilon }^{2}-{\frac{4}{3}} \left( {\frac{1}{2}}+\alpha \beta \pi \right) ^{2}\right)\right)^{\frac{1}{2}}\right] \epsilon . \end{aligned} $


Last edited by Admin on Sun Oct 30, 2011 10:42 pm; edited 1 time in total

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:27 pm

This thread primarily is intended for testing integrity & beta testing of new features......

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:27 pm

This thread primarily is intended for testing integrity & beta testing of new features......

$\Pi(0,0)=T_{\rm c}^{\,0}\;g^{2}(\alpha)\;\omega_{\rm c}^{-2\alpha}\sum_{n} \int\frac{d^{d}k}{( 2\pi )^{d}}\;\frac{1} {\big( \epsilon _{k}^{2}+\omega_{n}^{2}\big)^{1-\alpha}} $


Last edited by Admin on Sun Oct 30, 2011 10:46 pm; edited 1 time in total

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:28 pm

This thread primarily is intended for testing integrity & beta testing of new features......

$f_{\vec k}\left(\theta,\varphi\right)=-\lambda\frac{m}{ 2\pi \hbar^2} \\ \left(\left\vert V\left(\vec r\right)\right\vert^{\frac{1}{2}}\varphi_f\left(\vec r\right) \epsilon _V\left(\vec r\right),\left\vert V\left(\vec r\right)\right\vert^{\frac{1}{2}}\psi_{\vec k}^+\left(\vec r\right)\right)\:. $


Last edited by Admin on Sun Oct 30, 2011 10:45 pm; edited 1 time in total

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:28 pm

This thread primarily is intended for testing integrity & beta testing of new features......

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:29 pm

This thread primarily is intended for testing integrity & beta testing of new features......

$g( \epsilon ) = \frac{N} {\sigma \sqrt{ 2\pi }}\exp \left (- \frac{{ \epsilon }^{2}} {2{\sigma }^{2}}\right ),$


Last edited by Admin on Sun Oct 30, 2011 10:45 pm; edited 1 time in total

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:29 pm

This thread primarily is intended for testing integrity & beta testing of new features......

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:29 pm

This thread primarily is intended for testing integrity & beta testing of new features......

$\phi^{({\rm scatt})}_{\alpha}(r,\theta)= \sqrt{\frac{2}{\pi ( \epsilon _{0})^{1/2}k_{0}r}} T_{\alpha}(\theta){\rm exp}\left[ -i \sqrt{ \epsilon _{0}} k_{0}r+ i\pi/4 \right], $


Last edited by Admin on Sun Oct 30, 2011 10:44 pm; edited 1 time in total

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:29 pm

This thread primarily is intended for testing integrity & beta testing of new features......

$\begin{aligned} &\left[1+\frac{m\hbar}{2\mu}k_x^2\left(\frac{\Uptheta(\hbar^2k_x^2-2mE)}{\sqrt{\hbar^2k_x^2-2mE}}+\hbox{i}\frac {\Uptheta(2mE-\hbar^2k_x^2)}{\sqrt{2mE-\hbar^2k_x^2}} \right)\right]f(E,k_x,k_y^{\prime})\\ &=-\frac{m}{\mu}\frac{k_x^2}{ 2\pi } \frac{{\hbox{exp}}({\hbox{i}}k^{\prime}_yy_0)}{ k^{\prime2}_y+k_x^2-(2mE/\hbar^2)-\hbox{i} \epsilon }. \end{aligned} $


Last edited by Admin on Sun Oct 30, 2011 10:44 pm; edited 1 time in total

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:30 pm

This thread primarily is intended for testing integrity & beta testing of new features......
$
\begin{aligned} q_n&=\left({\frac{ \epsilon _\ast- \epsilon } { \epsilon _\ast+ \epsilon }}\right)^n q,\quad Q_n={\frac{2 \epsilon } { \epsilon _\ast+ \epsilon }} \left({\frac{ \epsilon _\ast- \epsilon } { \epsilon _\ast+ \epsilon }}\right)^n q,\\ |z|\le L/2:\quad \Upphi=&{\frac{q} {4\pi \epsilon _\ast}} \sum_{n=-\infty}^\infty \left({\frac{ \epsilon _\ast- \epsilon } { \epsilon _\ast+ \epsilon }}\right)^{|n|} {\frac{1} {\sqrt{r^2+(z-nL)^2}}},\\ z>L/2:\quad \Upphi=&{\frac{q}{ 2\pi ( \epsilon _\ast+ \epsilon )}} \sum_{n=0}^\infty\left({\frac{ \epsilon _\ast- \epsilon } { \epsilon _\ast+ \epsilon }}\right)^n {\frac{1}{\sqrt{r^2+(z+nL)^2}}}. \end{aligned}$


Last edited by Admin on Sun Oct 30, 2011 10:43 pm; edited 2 times in total

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:31 pm

This thread primarily is intended for testing integrity & beta testing of new features......

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:32 pm

This thread primarily is intended for testing integrity & beta testing of new features......

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Re: Forum Tester

Post by Admin on Sun Oct 30, 2011 10:32 pm

This thread primarily is intended for testing integrity & beta testing of new features......

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Re: Forum Tester

Post by Admin on Mon Oct 31, 2011 4:22 am

This thread primarily is intended for testing integrity & beta testing of new features......

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Re: Forum Tester

Post by Admin on Mon Oct 31, 2011 4:26 am

This thread primarily is intended for testing integrity & beta testing of new features......

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Re: Forum Tester

Post by Admin on Mon Oct 31, 2011 4:29 am

$\begin{aligned} &\left(1+\frac{m}{\mu}\frac{k_x^2} { 2\pi }\int\frac{{\hbox{d}}\kappa}{\kappa^2+k_x^2-(2mE/\hbar^2)-\hbox{i} \epsilon } \right)f(E,k_x,k_y^{\prime})\\ &=-\frac{m}{\mu}\frac{k_x^2}{ 2\pi }\frac{{\hbox{exp}}({\hbox{i}}k^{\prime}_yy_0)}{k^{\prime}_y+k_x^2-(2mE/\hbar^2)-{\hbox{i}} \epsilon }. \end{aligned}$

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Re: Forum Tester

Post by Admin on Sat Nov 05, 2011 6:52 pm

Sample Code:
Code:
function selectCode(a)
    {
      // Get ID of code block
      var e = a.parentNode.parentNode.getElementsByTagName('CODE')[0];
   
      // Not IE
      if (window.getSelection)
      {
          var s = window.getSelection();
          // Safari
          if (s.setBaseAndExtent)
          {
            s.setBaseAndExtent(e, 0, e, e.innerText.length - 1);
          }
          // Firefox and Opera
          else
          {
            // workaround for bug # 42885
            if (window.opera && e.innerHTML.substring(e.innerHTML.length - 4) == '<BR>')
            {
                e.innerHTML = e.innerHTML + ' ';
            }
   
            var r = document.createRange();
            r.selectNodeContents(e);
            s.removeAllRanges();
            s.addRange(r);
          }
      }
      // Some older browsers
      else if (document.getSelection)
      {
          var s = document.getSelection();
          var r = document.createRange();
          r.selectNodeContents(e);
          s.removeAllRanges();
          s.addRange(r);
      }
      // IE
      else if (document.selection)
      {
          var r = document.body.createTextRange();
          r.moveToElementText(e);
          r.select();
      }
    }
    if(text){}else{ var text = 'Selecionar todos';}
    jQuery(document).ready(function(){
      jQuery("dl.codebox dt").not("dl.spoiler > dt").html('Code: <a href="#" onclick="selectCode(this); return false;" title="Select all the content" class="code-a"> Select Content </a>');
    });

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Re: Forum Tester

Post by Admin on Sun Nov 13, 2011 8:22 pm

a test message as always

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Re: Forum Tester

Post by Admin on Fri Nov 25, 2011 10:29 am



Last edited by Admin on Mon Dec 05, 2011 12:05 pm; edited 1 time in total

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Re: Forum Tester

Post by Admin on Fri Nov 25, 2011 10:45 am


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Re: Forum Tester

Post by Admin on Tue Nov 29, 2011 1:07 am


$\begin{aligned} \langle k_y|G(E,k_x)|y^{\prime}\rangle=&\frac{1} {\sqrt{ 2\pi }}\int\!{\hbox{d}}k_y^{\prime}\hbox{exp}(-\hbox{i}k^{\prime}_yy^{\prime}) \langle k_y|G(E,k_x)|k_y^{\prime}\rangle\\ = &\frac{1}{\sqrt{ 2\pi }}\frac{1}{k^2_y+k_x^2-(2mE/\hbar^2)-\hbox{i} \epsilon } \left[\hbox{exp}(-\hbox{i}k_yy^{\prime})\right.\\ -&\left.\frac{\hbar k_x^2\ell\Uptheta(\hbar^2k_x^2-2mE)}{ \sqrt{\hbar^2k_x^2-2mE} +\hbar k_x^2\ell}\hbox{exp}\!\left(-\hbox{i}k_y y_0-\sqrt{\hbar^2k_x^2-2mE}\frac{|y^{\prime}-y_0|}{\hbar}\right)\right.\\ -&\left.\hbox{i}\frac{\hbar k_x^2\ell\Uptheta(2mE-\hbar^2k_x^2)}{ \sqrt{2mE-\hbar^2k_x^2}+\hbox{i}\hbar k_x^2\ell} \hbox{exp}\!\left(-\hbox{i}k_y y_0+\hbox{i}\sqrt{2mE-\hbar^2k_x^2} \frac{|y^{\prime}-y_0|}{\hbar}\right)\right]. \end{aligned} $

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Re: Forum Tester

Post by Admin on Fri Dec 02, 2011 10:41 am



[doc]src="http://www.scribd.com/embeds/2942002/content?start_page=1&view_mode=list&access_key=key-sxlnrfvqu5s22kzi5xb"[doc]



sdfad


Last edited by Admin on Mon Dec 05, 2011 10:11 pm; edited 5 times in total

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Re: Forum Tester

Post by Admin on Sat Dec 03, 2011 11:15 am

  • kadhadaskdjkadhadaskdjkadhadaskdjkadhadaskdj

[docis]documentId=110829214737-4e3a7b11d0e747048caf9e223bd33f1a[docis]
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Re: Forum Tester

Post by Admin on Mon Dec 05, 2011 10:35 pm

$\frac{1}{\rho }\frac{{{\text{d}}\rho }}{{{\text{d}}z}} + \frac{1}{w}\frac{{{\text{d}}w}}{{{\text{d}}z}} = - \frac{1}{A}\frac{{{\text{d}}A}}{{{\text{d}}z}},$
[doc]src="http://www.scribd.com/embeds/27438612/content?start_page=1&view_mode=list&access_key=key-2ezi6trbyg08mivwsqdb"[doc]

$z\frac{{{\rm d}^{2} w}}{{{\rm d}z^{2} }} + {\left({1 + 2b - Pr - z} \right)}\frac{{{\rm d}w}}{{{\rm d}z}} - (b - 2)w = 0$

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Re: Forum Tester

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