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theorem

By: anta40 on Apr 11th, 2012  |  syntax: Latex  |  size: 1.18 KB  |  hits: 37  |  expires: Never
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  1. \documentclass[a4paper]{article}
  2. \usepackage{amsmath}
  3.  
  4. \begin{document}
  5.  
  6. Suppose you want to publish something that is as simple as
  7. \begin{align}
  8.     1 + 1 = 2
  9. \end{align}
  10.  
  11. This is not very impressive. If you want your article to be accepted by IEEE reviewers,
  12. you have to be more abstract. So, you could complicate the left hand side of the expression
  13. by using
  14.  
  15. \begin{align*}
  16. 1 = ln(e) \; and \; 1 = sin^2 x + cos^2 x
  17. \end{align*}
  18.  
  19. The right hand side can be stated as
  20. \begin{align*}
  21.    2 = \sum_{n=0}^{\infty}\frac{1}{2^n}
  22. \end{align*}
  23.  
  24. Therefore, Eq.(1) can be expressed more ``scientifically`` as:
  25. \begin{align}
  26.    ln(e)+(sin^2 x + cos^2 x) = \sum_{n=0}^{\infty}\frac{1}{x^n}
  27. \end{align}
  28.  
  29. which is far more impressive. However, you should not stop here. The expression
  30. can be further complicated by using
  31. \begin{align*}
  32.    1 = cosh(y)\sqrt{1 - tanh^2 y} \; and \; e = \lim_{z \rightarrow 0}{(1 + \frac{1}{z})}^z
  33. \end{align*}
  34.  
  35. Eq. (2) may therefore be written as
  36. \begin{align}
  37.    ln \left[ \lim_{z \rightarrow 0} (1 + \frac{1}{z})^z \right ] + (sin^2 x + cos^2 x) = \sum_{n=0}^{\infty} \frac{cosh(y \sqrt{1 - tanh^2 y})}{2^n}
  38. \end{align}    
  39.  
  40.  
  41. \end{document}