I have an impulse train given by1R+1+∑k=1Rcos⁡(2kπxR+1)R+1It seems obvious to me that, for x=0, the function returns 1. This is because cos⁡(0)=1, and we therefore end up with 1R+1+RR+1=R+1R+1=1However indeterminate result at x=0. Usually this means there is a division by 0 somewhere. But I can't see any reason for this function to produce an indeterminate result.

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I have an impulse train given by1R+1+∑k=1Rcos⁡(2kπxR+1)R+1It seems obvious to me that, for x=0, the function returns 1. This is because cos⁡(0)=1, and we therefore end up with 1R+1+RR+1=R+1R+1=1However indeterminate result at x=0. Usually this means there is a division by 0 somewhere. But I can&#039;t see any reason for this function to produce an indeterminate result.

Messiah Haynes · Accepted Answer

The only thing I noticed is that the simplification of the result is not as good as it could be since, using another CAS,  ∑k=1Rcos⁡(2kπxR+1)=12(sin⁡(π(2R+1)xR+1)csc⁡(πxR+1)−1)  making  1R+1+∑k=1Rcos⁡(2kπxR+1)R+1=sin⁡(π(2R+1)xR+1)csc⁡(πxR+1)+12(R+1)  Probably, the indetermination comes from the limit of csc⁡(t)  when  t→0 while, using Taylor expansion  sin⁡(π(2R+1)xR+1)=(π(2R+1))R+1x−(π(2R+1))36(R+1)3x3+O(x5)  csc⁡(πxR+1)=R+1πx+πx6(R+1)+7π3x3360(R+1)3+O(x5)  sin⁡(π(2R+1)xR+1),csc⁡(πxR+1)=(2R+1)−2π2R(2R+1)3(R+1)x2+O(x4)

I have an impulse train given by \frac{1}{R+1}+\frac{\sum_{k=1}^R \cos(\frac{2k \pi x}{R+1})}

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