I'm having trouble finding this sum

[Jncik]

Homework Statement

find

\sum_{-N1}^{+N1}e^{-j\omega n}

Homework Equations

The Attempt at a Solution

Let \lambda = e^{-j\omega}

we have

\sum_{-N1}^{+N1}\lambda ^{n} = \sum_{-N1}^{-1}\lambda ^{n} + \sum_{0}^{+N1} \lambda ^{n}

for the first i have

S = \lambda ^{-N1} + \lambda ^{-N1+1} + \lambda ^{-N1+2} + ... + \lambda ^{-2} + \lambda ^{-1}

-\lambda S = -\lambda ^{-N1+1} - \lambda ^{-N1+2} - \lambda ^{-N1+3} - ... - \lambda ^{-1} - \lambda ^{0}

hence

S = \frac{\lambda ^{-N1} - 1}{1-\lambda }

for the second i have

S2 = \lambda^{0} + \lambda^{1} + ... + \lambda^{N1-1} + \lambda^{N1}
-\lambda S2 = -\lambda^{1} - \lambda^{2} - ... - \lambda^{N1} - \lambda^{N1+1}

hence

S2 = \frac{1 - \lambda^{N1+1}}{1-\lambda}

so the sum is

\frac{1-\lambda^{N1+1} + \lambda^{-N1} - 1}{1-\lambda} = \frac{\lambda^{-N1} - \lambda^{N1+1}}{1-\lambda} = \frac{e^{j \omega N1} - e^{-j \omega N1}e^{-j\omega}}{1-e^{-j\omega}}

but the book says \frac{sin\omega(N1 + \frac{1}{2})}{sin(\frac{\omega}{2})}

 

[Dick]

Multiply numerator and denominator by e^(j*w/2). Now remember e^(jx)-e^(-jx)=2jsin(x).

 

[Jncik]

thanks a lot :)