Hello Everybody,
I am a new member to this forum, and my name is Andy Charles. Mostly I always stick to various forums to get informations about various life experiences and to get updated. I was searching for a good forum and finally decided to make my favorite. I found this forum very informative and the members are very helping and also there are many interesting topics are discussed here.
I hope I can enjoy my stay here.
Thanks a lot
Andy Charles
Newbie From CA!!
Moderator:kamrul2010
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- Posts:4
- Joined:Thu Nov 03, 2011 8:20 pm
- Location:CA
- Contact:
- nafistiham
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- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
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Re: Newbie From CA!!
$welcome$ $to$ $our$ $group$......
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
- Contact:
Re: Newbie From CA!!
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: Newbie From CA!!
This is funnier.
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
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Re: Newbie From CA!!
*Mahi* wrote:This is funnier.
woow, it is really ammazing .thanks...
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.