Complex Summation: Understanding Discrete Time Function

[WolfOfTheSteps]

This is not really a homework problem, but I'm studying a text, and I came across this:

http://img198.imageshack.us/img198/4586/sumh.jpg

I know how to get that fraction with the exponents in it (using a summation formula). But for the life of me, I can't figure out how to manipulate that fraction to give the final result.

For example, if I put k=0 into that fraction, I get 0/0, not 5. I tried a bunch of manipulation of the fraction to get sines and cosines, and make the denominator real, but I still can't get a closed form solution that gives the final result.

What am I missing?

Thanks!

UPDATE: Forgot to mention, this is a discrete time function. k is always an integer.

 

[WolfOfTheSteps]

I totally forgot about that factoring trick:

<br /> 1-e^{-j x} = e^{-j x/2}(e^{j x/2}-e^{-j x/2}) = e^{-j x/2}jsin(x/2)<br />

That's all I needed!

UPDATE:

Wait, I was wrong.

Even with that factoring trick, I get:

<br /> e^{-j(\pi k-\pi k/5)}*sin(\pi k)/sin(\pi k/5)<br />

But this is still 0/0 for k=0. How do I get the real result?? I'm so frustrated with this!

 

[n!kofeyn]

No matter what k is, the top is 0. If k is not 0,+/-5,+/-10,..., then the bottom is not 0, so that is how they get 0 for otherwise. For k=0,+/-5,+/-10,..., you need to find the limit as k approaches those values, because 0/0 has no meaning. Use l'Hospital's rule to evaluate your function at those points.

 

[WolfOfTheSteps]

No matter what k is, the top is 0. If k is not 0,+/-5,+/-10,..., then the bottom is not 0, so that is how they get 0 for otherwise. For k=0,+/-5,+/-10,..., you need to find the limit as k approaches those values, because 0/0 has no meaning. Use l'Hospital's rule to evaluate your function at those points.

Thanks, that makes sense.

Also, instead of going to L'Hospital's rule, I could just go back to the summation for k=0,+-5, etc and show that it is a summation of ones... while the fraction would prove the "0 otherwise" for the other k values. This would work too.

Thanks a lot!

 

[n!kofeyn]

No problem. Yea, you're right about the summation giving you the five.