Line 1: Line 1:
 +
== Opening Challenge ==
 +
 +
 
<math> e^{j 2 \pi t}  = \left( e^{j 2 \pi} \right)^t= \left( cos{2 \pi } + j sin{2 \pi } \right)^t= 1^t =1 </math>    This is incorrect (as one sentence). But,   
 
<math> e^{j 2 \pi t}  = \left( e^{j 2 \pi} \right)^t= \left( cos{2 \pi } + j sin{2 \pi } \right)^t= 1^t =1 </math>    This is incorrect (as one sentence). But,   
  

Revision as of 06:34, 23 July 2009

Opening Challenge

$ e^{j 2 \pi t} = \left( e^{j 2 \pi} \right)^t= \left( cos{2 \pi } + j sin{2 \pi } \right)^t= 1^t =1 $ This is incorrect (as one sentence). But,

It is correct that:

$ \left( {e^{j 2 \pi }}\right) ^t = 1^t = 1 $


However

$ \left( e^{j 2 \pi t} \right) = \left( cos{2 \pi t} + j sin{2 \pi t } \right) $

So there error would be that (2 Pi t) is the theta, so it must stay in the theta when converted into sin and cos.

Therefore the "t" may not be separated into the exponent in that case.

But if "t" starts off in the exponent, then the result will equal 1.


A great source would be this web site:

http://www00.wolframalpha.com/input/?i=(e^(2+i+pi+))^t --- Adam Frey


[1] <= returns to Midterm Cheat Sheet.

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett