I'm reading a book on time series and in the book there is a chapter on difference equations and one part has got the following derivations which is unclear to me:
Now suppose that the sequence satisfies $$u_n-\alpha_1u_{n-1}-\alpha_2u_{n-2}=0, \,\,\,\,\,\,\,\,\alpha_2\ne0,\,\,\,\,\,\,\,\,n=2,3,\ldots\tag{3.31}$$ This equation is a homogeneous difference equation of order $2$. The corresponding polynomial is $$\alpha(z)=1-\alpha_1z-\alpha_2z^2,$$ which has two roots, say, $z_1$ and $z_2$; that is, $\alpha(z_1)=\alpha(z_2)=0$. We will consider two cases. First suppose $z_1\ne z_2$. Then the general solution to $\text{(3.31)}$ is $$u_n=c_1z_1^{-n}+c_2z_2^{-n},\tag{3.32}$$ where $c_1$ and $c_2$ depend on the initial conditions. The claim that is a solution can be verified by direct substitution of $\text{(3.32)}$ into $\text{3.31}$: $$\begin{align}(c_1 & z_1^{-n}+c_2z_2^{-n})-\alpha_1\left(c_1z_1^{-(n-1)}+c_2z_2^{-(n-1)}\right)-\alpha_2\left(c_1z_1^{-(n-2)}+c_2z_2^{-(n-2)}\right)\\ &=c_1z_1^{-n}(1-\alpha_1z_1-\alpha_2z_1^2)+c_2z_2^{-n}(1-\alpha_1z_2-\alpha_2z_2^2)\\&=c_1z_1^{-n}\alpha(z_1)+c_2z_2^{-n}\alpha(z_2)=0.\end{align}$$ Given two initial conditions $u_0$ and $u_1$, we may solve for $c_1$ and $c_2$: $$u_0=c_1+c_2\,\,\,\,\,\,\,\,\,\,\,\,\text{and}\,\,\,\,\,\,\,\,\,\,\,\,u_1=c_1z_1^{-1}+c_2z_2^{-1}$$ where $z_1$ and $z_2$ can be solved for in terms of $\alpha_1$ and $\alpha_2$ using the quadratic formula, for example.
When the roots are equal, $z_1=z_2(=z_0)$, a general solution to $\text{(3.31)}$ is $$u_n=z_0^{-n}(c_1+c_2n).\tag{3.33}$$ This claim can also be verified by direct substitution of $\text{(3.33)}$ into $\text{(3.31)}$: $$\begin{align}z_0^{-n} & (c_1+c_2n)-\alpha_1\left(z_0^{-(n-1)}[c_1+c_2(n-1)]\right)-\alpha_2\left(z_0^{-(n-2)}[c_1+c_2(n-2)]\right)\\ &= z_0^{-n}(c_1+c_2n)(1-\alpha_1z_0-\alpha_2z_0^2)+c_2z_0^{-n+1}(\alpha_1+2\alpha_2z_0)\\ &=c_2z_0^{-n+1}(\alpha_1+2\alpha_2z_0). \end{align}$$ To show that $(\alpha_1+2\alpha_2 z_0)=0$, write $1- \alpha_1 z- \alpha_2 z^2=(1 - z_0^{-1} z)^2$, and take derivatives with respect to $z$ on both sides of the equation to obtain $(\alpha_1 + 2 \alpha_2 z)=2 z_0^{-1} (1 - z_0^{-1} z)$. Thus, $(\alpha_1 + 2 \alpha_2 z_0)=2 z_0^{-1} (1 - z_0^{-1} z_0)=0$, as was to be shown. Finally, given two initial conditions, $u_0$ and $u_1$, we can solve for $c_1$ and $c_2$:
The bold region part is unclear for me...
why does $1 - \alpha_1z - \alpha_2z^2 = (1 - z_0^{-1}z)^2 = (1 - z_0^{-1}z)(1 - z_0^{-1}z) = 1 - \displaystyle{\frac{2z}{z_0} + \left(\frac{z}{z_0}\right)^2}$...
Can someone clarify this for me x) Thank you!
