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Persia Quote \| Reply \| This message was updated on 4/10/2007 5:44:48 AM by Persia	Series posted on: 4/10/2007 5:33:12 AM Hi and thank you for providing such a opportunity. Question: I know this question makes the use of e^j0 = cos0 + jsin0 (0= theater) but I cant quite undrestand how.
davo Quote \| Reply \|	Series replied on: 4/14/2007 9:53:17 AM First write 2/2-e^jx in a+bj form like 2/2-e^jx=2/2-cosx - jsinx ( since e^jx =cosx + jsinx ) = (2/2-cosx - jsinx) (2-cosx+jsinx)/(2-cosx+jsinx) and simplfing the denominator, you will get the answer (if I am not wrong in calculation) 2(2-cosx)/(5-2cosx) + j(2sinx)/(5-2cosx) --1 but the sum of e^jnx/2^n = the sum of cosnx/2^n +jthe sum of sinnx/2^n----2 (the reason is clear from euler's law and applying the sum rule for series) By equating (1) and (2) you will get your answer as 2sinx/5-2cosx =the sum of sinnx/2^n note:the sum of e^jnx/2^n = 2/2-e^jx (given)
Persia Quote \| Reply \|	Series replied on: 4/22/2007 9:52:32 AM Thank you, it really helped.
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