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Bruno J. · #5 June 17th, 2009, 04:33 PM

The operation "reflection about 2" is the map $s: x \mapsto 4-x$

The operation "reflection about 7" is the map $t: x \mapsto 14-x$

Starting with $x=0$ we can obtain the zeroes of $f$ by repeatedly applying $s,t$ in any combination. We have

$t \circ s (x)= 14-(4-x)=10+x$

Moreover $s^2=t^2=1$ , the identity map, so we can consider only alternating sequences of $s,t$ . If we compose $(t \circ s)$ with itself n times then $(t \circ s)^n (x) = 10+(10+(10+...+(10+x)...)= 10n+x$ (where the exponent means composition - careful!)

So now we have three possible types of different sequences of composition:
$(t \circ s)^n\: (x) = 10n+x$
$s\circ (t \circ s)^n\: (x) = 4-(10n+x)$
$(t \circ s)^n \circ s\: (x) = 10n+(4-x)$

Since $x=0$ is our starting point, the zeroes of $f$ are of the form $10n, 10n+4, -10n+4$ for all $n \in \mathbb N$ . It is easy to see that they are almost in bijection with the multiples of 5 on the interval $[-2007,2007]$ . Counting carefully we get $2\Big\lfloor\frac{2007}{5}\Big\rfloor+2=804$ zeroes.