# Sum of a finite exponential series

by ElfenKiller
Tags: exponential, finite, series
 P: 9 1. The problem statement, all variables and given/known data Given is $\sum_{n=-N}^{N}e^{-j \omega n} = e^{-j\omega N} \frac{1-e^{-j \omega (2N+1)}}{1 - e^{-j\omega}}$. I do not see how you can rewrite it like that. 2. Relevant equations Sum of a finite geometric series: $\sum_{n=0}^{N}r^n=\frac{1-r^{N+1}}{1-r}$ 3. The attempt at a solution Or is the above result based on this more general equation: $\sum_{n=0}^{N}ar^n=a\frac{1-r^{N+1}}{1-r}$? Although I think the equation in (2) is just this equation for a=1, right? So, I know how to get to the 2nd term in (1), i.e., $\frac{1-e^{-j \omega (2N+1)}}{1 - e^{-j\omega}}$, but I have no idea why it is multiplied by the term $e^{-j\omega N}$.
 PF Gold P: 1,125 Did you notice that the sum you are trying to compute actually starts from n=-N and not n=0? I think you can get the answer you want by making a change of variable and then using the geometric series equation you have identified.
P: 9
 Quote by danago Did you notice that the sum you are trying to compute actually starts from n=-N and not n=0? I think you can get the answer you want by making a change of variable and then using the geometric series equation have identified.
Yes, I've noticed that it starts there. That's why I thought it can be rewritten as $\frac{1−e^{-j\omega(2N+1)}}{1−e^{−jω}}$, but the solution states that this fraction is multiplied by $e^{−jωN}$.

 PF Gold P: 1,125 Sum of a finite exponential series Are you sure that the exponential term in front of the fraction does have a negative sign? I just tried doing the working and ended up with a positive sign, i.e.: $$\sum^{N}_{n=-N} e^{-j\omega n} = e^{j\omega N} \frac{1-e^{-j\omega(2N+1)}}{1-e^{-j\omega}}$$ I did it by making the substitution $\phi=n+N$. I will check my working again. EDIT: I have checked over my working and have convinced myself that the negative should not be there. It is late here so i could easily have made a mistake though
P: 9
 Quote by danago Are you sure that the exponential term in front of the fraction does have a negative sign? I just tried doing the working and ended up with a positive sign, i.e.: $$\sum^{N}_{n=-N} e^{-j\omega n} = e^{j\omega N} \frac{1-e^{-j\omega(2N+1)}}{1-e^{-j\omega}}$$ I did it by making the substitution $\phi=n+N$. I will check my working again. EDIT: I have checked over my working and have convinced myself that the negative should not be there. It is late here so i could easily have made a mistake though
Okay, thank you. For me, it is not about the sign in the exponent. I do not see why we have to multiply by the term in front of the fraction. But I think I rewrote the equation in the wrong way. Can you give me your steps?
 PF Gold P: 1,125 You have transformed the upper and lower limits of the sum, however you have not applied the same transformation to the variable n in the summand. If $\phi=n+N$, then the new limits of the sum will be $\phi=0$ and $\phi=2N$. You must then also replace the 'n' in the summand with $n=\phi-N$. If you do this then you will get the right answer. EDIT: The transformed sum will be: $$\sum^{2N}_{\phi=0} e^{-j\omega (\phi-N)} = e^{j\omega N} \frac{1-e^{-j\omega(2N+1)}}{1-e^{-j\omega}}$$
 PF Gold P: 1,125 Maybe it will be easier to understand if we look at why what you did isn't quite correct. $$\sum^{N}_{n=-N} e^{n} = e^{-N}+e^{-N+1}+...+1+e^1+...+e^{N-1}+e^N$$ $$\sum^{2N}_{n=0} e^{n} = 1+e^{1}+...+e^{2N-1}+e^{2N}$$ See how they are not the same?
 P: 9 Ah, I see the problem now. Thanks!
PF Gold
P: 1,125
 Quote by ElfenKiller Ah, I see the problem now. Thanks!
No problem!

 Related Discussions Differential Equations 1 Calculus & Beyond Homework 0 Calculus & Beyond Homework 10 Calculus 4