I have an impulse train given by
$$\sum_{j=0}^r \frac{\cos \bigl(\frac{2 \pi jx}{r+1} \bigr)}{r+1}$$
Plotted for $r=1$ (blue), $5$ (orange) and $8$ (green), this gives:
Ignore the case of $r=1$, as this is a simple $\cos$ curve.
Where, in terms of $x$, do the maxima and minima occur (in the given ranges), and what are their values in terms of $x$ and $r$?
Differentiation gives me
$$-\sum_{j=0}^r \frac{\frac{2 \pi j}{r+1}\sin \bigl(\frac{2 \pi jx}{r+1} \bigr)}{r+1}=0$$
but I'm not sure how to proceed from there.
Edit:
As per Yves' suggestion below, I tried to figure out the sum, but I hit a wall even before I get to differentiating:
$$\cos x=\frac{e^{ix} + e^{-ix}}{2}$$
Therefore
$$\sum_{j=0}^r \frac{\cos \bigl(\frac{2 \pi jx}{r+1} \bigr)}{r+1}=\frac{1}{2(r+1)} \biggl(\sum_{j=0}^r e^{\frac{i2 \pi j x}{r+1}}+ \sum_{j=0}^r e^{\frac{-i2 \pi j x}{r+1}} \biggr)$$
Next,
$$\sum_{j=0}^{r-1} a^j=\frac{1-a^{r+1}}{1-a}$$
and setting $a=e^{\frac{i2 \pi x}{r+1}}$ (and therefore $a^{r+1}=e^{i2 \pi x}$), we have
$$\frac{1}{2(r+1)} \biggl(\sum_{j=0}^r e^{\frac{i2 \pi j x}{r+1}}+ \sum_{j=0}^r e^{\frac{-i2 \pi j x}{r+1}} \biggr)=\frac{1}{2(r+1)} \biggl(\frac{1-e^{i2 \pi x}}{1-e^{\frac{i2 \pi x}{r+1}}}+\frac{1-e^{-i2 \pi x}}{1-e^{\frac{-i2 \pi x}{r+1}}}\biggr)$$
$$=\frac{1}{2(r+1)} \biggl(\frac{\bigl(1-e^{i2 \pi x}\bigr)\bigl(1-e^{\frac{-i2 \pi x}{r+1}}\bigr)+\bigl(1-e^{\frac{i2 \pi x}{r+1}}\bigr)\bigl(1-e^{-i2 \pi x}\bigr)}{\bigl(1-e^{\frac{i2 \pi x}{r+1}}\bigr)\bigl(1-e^{\frac{-i2 \pi x}{r+1}}\bigr)}\biggr)$$
...but where do I go from there?