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Forum Tester

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Re: Forum Tester

Post by Admin on Tue Dec 06, 2011 1:55 am

This is & am talking & about

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Re: Forum Tester

Post by Admin on Tue Dec 06, 2011 5:19 am

  • kadhadaskdjkadhadaskdjkadhadaskdjkadhadaskdj

[doc]src="http://www.scribd.com/embeds/27438612/content?start_page=1&view_mode=list&access_key=key-2ezi6trbyg08mivwsqdb"[doc]
[docis]documentId=110829214737-4e3a7b11d0e747048caf9e223bd33f1a[docis]


$\frac{1}{\rho }\frac{{{\text{d}}\rho }}{{{\text{d}}z}} + \frac{1}{w}\frac{{{\text{d}}w}}{{{\text{d}}z}} = - \frac{1}{A}\frac{{{\text{d}}A}}{{{\text{d}}z}},$


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

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Re: Forum Tester

Post by Admin on Tue Dec 06, 2011 5:49 am

$\frac{1}{\rho }\frac{{{\text{d}}\rho }}{{{\text{d}}z}} + \frac{1}{w}\frac{{{\text{d}}w}}{{{\text{d}}z}} = - \frac{1}{A}\frac{{{\text{d}}A}}{{{\text{d}}z}},$


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

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Re: Forum Tester

Post by Admin on Tue Dec 06, 2011 4:20 pm

Admin wrote: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}} $
No way No way
Admin wrote: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} $


No way No way

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Re: Forum Tester

Post by Admin on Thu Dec 08, 2011 12:05 pm

this is the answer Pakistan

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Re: Forum Tester

Post by Admin on Fri Feb 03, 2012 11:47 pm

657468165677435

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Re: Forum Tester

Post by Admin on Fri Feb 03, 2012 11:51 pm

$\frac{{{d^3}}}{{d{x^3}}}\arctan x = 2\frac{{3{x^2} - 1}}{{{{(1 + {x^2})}^3}}}$


Last edited by Admin on Sat Sep 15, 2012 3:29 am; edited 1 time in total

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Re: Forum Tester

Post by Admin on Fri Feb 03, 2012 11:53 pm

13543546343541346546

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Re: Forum Tester

Post by Admin on Fri Feb 03, 2012 11:56 pm

78888888888888888888888888646546

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Re: Forum Tester

Post by Admin on Sat Feb 04, 2012 10:45 am

54654513211111111111111


Last edited by Admin on Sat Feb 04, 2012 10:51 am; edited 7 times in total

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Re: Forum Tester

Post by Admin on Sun Feb 05, 2012 11:00 pm

Admin wrote:54654513211111111111111

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Re: Forum Tester

Post by Admin on Sun Feb 05, 2012 11:14 pm

Admin wrote:657468165677435

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Re: Forum Tester

Post by Admin on Mon Feb 13, 2012 8:32 pm

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

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Re: Forum Tester

Post by Admin on Fri Feb 17, 2012 7:05 am

Horizontal Example:

Heading 1 Heading 2
content content
contentcontent
Vertical Example:

Heading 1 content
Heading 2 content
Heading 3content

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Re: Forum Tester

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