4-5

a) $z+zF(z)+z^2F(z)=z+\sum_{i=2}^\infty F_{i-1}z^i+\sum_{i=3}^\infty F_{i-2}z^i$
$\ \ = 0+z+z^2+\sum_{i=3}^\infty F_{i}z^i=F(z)$

b) 太显然了。。。移项后配方即可
c) 根据级数展开$\frac{1}{1-az}=\sum_{i=0}^\infty az^i$即可
d) $F_i=\frac{1}{\sqrt5}(\phi^i-\hat{\phi}^i)$
$\because|\hat{\phi}^i|<1\therefore \frac{1}{\sqrt5}|\hat{\phi}^i|<0.5$
$\therefore F_i是\frac{1}{\sqrt5}\phi^i的最近的整数$
e) $F_{i+2}=\frac{1}{\sqrt5}(\frac{3+\sqrt5}{2}\phi^i-\frac{3-\sqrt5}{2}\hat{\phi}^i)=\frac{1}{\sqrt5}(\sqrt5\phi^i+\frac{3-\sqrt5}{2}(\phi^i-\hat{\phi}^i))\geq \phi^i$