How to prove that this finite sum is true

Question

Does anyone know how I could show that

$$ \sum_{n = -N_1}^{N_1}{e^{-i\omega n}} = \frac{\sin( \omega ( N_1 - \frac{1}{ 2} ))} {\sin( \frac{\omega}{2} ) } $$

Solution: $$ = \sum_{n=0}^{N_1} ( e^{i \omega n} + e^{-i \omega n} ) - 1 $$

$$ = \sum_{n=0}^{N_1} ( e^{i \omega n} + e^{-i \omega n} ) - 1 $$

$$ = \frac{1}{1 - e^{i \omega}} + \frac{1}{1 - e^{-i\omega} } - 1 $$

$$ = \frac{1 - e^{i \omega} + 1 - e^{-i \omega} }{ ( 1 - e^{i \omega} )( 1 - e^{-i \omega} )} - 1 $$

$$ = \frac{2 - e^{i \omega}- e^{-i \omega} }{ ( 1 - e^{i \omega} )( 1 - e^{-i \omega} )} - 1 $$

Really don't know how to get the $\sin$ function from here.

When I try to use geometric series it starts getting even more complex,as shown in the question.I know in order to have sin I need to have 2i somewhere, thats why I introduced it. — KillaKem, Apr 16 '13 at 16:35
the summation formula you have is of infinite Geometric Series, but in your case the number of terms is not necessarily infinite — lab bhattacharjee, Apr 16 '13 at 16:45

lab bhattacharjee · Accepted Answer · 2013-04-16T16:34:47.303

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HINT: It is a Geometric Series and using Euler's formula $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$

$$\sum_{n = -N_1}^{N_1}{e^{-i\omega n}} $$ $$= e^{i\omega N_1}\cdot\frac{(e^{-iw})^{2N_1+2}-1}{e^{-iw}-1}$$

$$=\frac{e^{\frac{(2N_1-1)iw}2}-e^{-\frac{(2N_1-1)iw}2}}{e^{\frac{iw}2}-e^{-\frac{iw}2}}$$

$$=\frac{2i\sin \frac{(2N_1-1)w}2}{2i\sin \frac w2}$$

edited Apr 16 '13 at 16:34

answered Apr 16 '13 at 15:54

lab bhattacharjee

274,582

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And use $i$, not $j$. Because it's an $I$maginary number, not $J$imaginary number. :) – bob.sacamento Apr 16 '13 at 16:04
@bob.sacamento, have you seen the notation in the question? – lab bhattacharjee Apr 16 '13 at 16:06
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Well, of course I was hoping KillaKem would see my comment too. Just a joke. Engineers always use $j$. I love to rag them about it. – bob.sacamento Apr 16 '13 at 16:09
Just after reading @bob.sacamento's comment I realized $,j=i=\sqrt{-1},$ . I know some people somewhere write $,j,$ , but I think the overwhelming majority of mathematicians in the world and in this sector of the galaxy write $,i,$ . People should be aware of this, imo. – DonAntonio Apr 16 '13 at 16:09
Changed from j to i. – KillaKem Apr 16 '13 at 16:21
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Rule of thumb: Electrical engineers who work with circuits, Smith charts, and the like use $j$; all other sentient carbon-based life forms use $i$. The reason for this is that they (the EE's) discovered current before imaginary numbers. – Ron Gordon Apr 16 '13 at 16:38
Which identity did you use to move from the summation to that expression just below it? – KillaKem Apr 16 '13 at 16:40
@KillaKem, so did I. Please find the edited answer. – lab bhattacharjee Apr 16 '13 at 16:41
@KillaKem, http://en.wikipedia.org/wiki/Geometric_series – lab bhattacharjee Apr 16 '13 at 16:42
Seriously guys, I didn't expect anybody to change anything. It was just a joke. Even though it should, of course, be $i$ instead of $j$. – bob.sacamento Apr 16 '13 at 16:49

How to prove that this finite sum is true

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