Haialarmz6

2022-01-29

I have an impulse train given by
$\frac{1}{R+1}+\frac{\sum _{k=1}^{R}\mathrm{cos}\left(\frac{2k\pi x}{R+1}\right)}{R+1}$
It seems obvious to me that, for x=0, the function returns 1. This is because $\mathrm{cos}\left(0\right)=1$, and we therefore end up with $\frac{1}{R+1}+\frac{R}{R+1}=\frac{R+1}{R+1}=1$
However 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.

Messiah Haynes

The only thing I noticed is that the simplification of the result is not as good as it could be since, using another CAS,
$\sum _{k=1}^{R}\mathrm{cos}\left(\frac{2k\pi x}{R+1}\right)=\frac{1}{2}\left(\mathrm{sin}\left(\frac{\pi \left(2R+1\right)x}{R+1}\right)\mathrm{csc}\left(\frac{\pi x}{R+1}\right)-1\right)$
making
$\frac{1}{R+1}+\frac{\sum _{k=1}^{R}\mathrm{cos}\left(\frac{2k\pi x}{R+1}\right)}{R+1}=\frac{\mathrm{sin}\left(\frac{\pi \left(2R+1\right)x}{R+1}\right)\mathrm{csc}\left(\frac{\pi x}{R+1}\right)+1}{2\left(R+1\right)}$
Probably, the indetermination comes from the limit of while, using Taylor expansion
$\mathrm{sin}\left(\frac{\pi \left(2R+1\right)x}{R+1}\right)=\frac{\left(\pi \left(2R+1\right)\right)}{R+1}x-\frac{{\left(\pi \left(2R+1\right)\right)}^{3}}{6{\left(R+1\right)}^{3}}{x}^{3}+O\left({x}^{5}\right)$
$\mathrm{csc}\left(\frac{\pi x}{R+1}\right)=\frac{R+1}{\pi x}+\frac{\pi x}{6\left(R+1\right)}+\frac{7{\pi }^{3}{x}^{3}}{360{\left(R+1\right)}^{3}}+O\left({x}^{5}\right)$
$\mathrm{sin}\left(\frac{\pi \left(2R+1\right)x}{R+1}\right),\mathrm{csc}\left(\frac{\pi x}{R+1}\right)=\left(2R+1\right)-\frac{2{\pi }^{2}R\left(2R+1\right)}{3\left(R+1\right)}{x}^{2}+O\left({x}^{4}\right)$

Do you have a similar question?