(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> | + | == Opening Challenge == |
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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: | ||
| − | <math> \left( {e^{j 2 \pi }}\right) ^t = 1^t = 1 </math> | + | <math> \left( {e^{j 2 \pi }}\right) ^t = 1^t = 1 </math> |
| + | * Hmmm... but then <math>1^{\frac{1}{2}}=1</math>, however we know that <math>1^{\frac{1}{2}}=\sqrt{1}=\pm1</math>. -pm | ||
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A great source would be this web site: | A great source would be this web site: | ||
| − | http://www00.wolframalpha.com/input/?i=(e^(2+i+pi+))^t | + | http://www00.wolframalpha.com/input/?i=(e^(2+i+pi+))^t --- Adam Frey |
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| + | [https://kiwi.ecn.purdue.edu/rhea/index.php/Midterm_Cheat_Sheet] <= returns to Midterm Cheat Sheet. | ||
Latest revision as of 09:21, 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 $
- Hmmm... but then $ 1^{\frac{1}{2}}=1 $, however we know that $ 1^{\frac{1}{2}}=\sqrt{1}=\pm1 $. -pm
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.
