7. [40] Let $q(x)=q^{1}(x)=2 x^{2}+2 x-1$, and let $q^{n}(x)=q\left(q^{n-1}(x)\right)$ for $n>1$. How many negative real roots does $q^{2016}(x)$ have?
Answer: $\frac{2^{2017}+1}{3}$
Define $g(x)=2 x^{2}-1$, so that $q(x)=-\frac{1}{2}+g\left(x+\frac{1}{2}\right)$. Thus
$$
q^{N}(x)=0 \Longleftrightarrow \frac{1}{2}=g^{N}\left(x+\frac{1}{2}\right)
$$
where $N=2016$.
But, viewed as function $g:[-1,1] \rightarrow[-1,1]$ we have that $g(x)=\cos (2 \arccos (x))$. Thus, the equation $q^{N}(x)=0$ is equivalent to
$$
\cos \left(2^{2016} \arccos \left(x+\frac{1}{2}\right)\right)=\frac{1}{2} \text {. }
$$

Thus, the solutions for $x$ are
$$
x=-\frac{1}{2}+\cos \left(\frac{\pi / 3+2 \pi n}{2^{2016}}\right) \quad n=0,1, \ldots, 2^{2016}-1 .
$$

So, the roots are negative for the values of $n$ such that
$$
\frac{1}{3} \pi<\frac{\pi / 3+2 \pi n}{2^{2016}}<\frac{5}{3} \pi
$$
which is to say
$$
\frac{1}{6}\left(2^{2016}-1\right)<n<\frac{1}{6}\left(5 \cdot 2^{2016}-1\right) .
$$

The number of values of $n$ that fall in this range is $\frac{1}{6}\left(5 \cdot 2^{2016}-2\right)-\frac{1}{6}\left(2^{2016}+2\right)+1=\frac{1}{6}\left(4 \cdot 2^{2016}+2\right)=$ $\frac{1}{3}\left(2^{2017}+1\right)$.