(New page: <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> It is correct that: <math> \left( {e^{j 2 \pi }}\right) ^t = 1^t ...)
 
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<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>     
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<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, 
  
 
It is correct that:
 
It is correct that:

Revision as of 16:29, 22 July 2009

$ 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

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett