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I recently derived the poles of the $z$-transform of the Fibonacci sequence to be the golden ratio by doing the following:

$u(k+2)=u(k+1)+u(k); u(0)=u(1)=1$

$z^2U(z)-z^2u(0)-zu(1) = zU(z)-zu(0)+U(z)$

$U(z)(z^2-z-1) = z^2$

$U(z) = \displaystyle\frac{z^2}{z^2-z-1}$

poles: $(z^2-z-1) = 0\rightarrow z = \displaystyle\frac{1\pm\sqrt{5}}{2}$

I could see that they turned out to be the golden ratio when I did this but I'm wondering why that is. What about z-transforms and poles cause this outcome? Why does this result make sense?

QuantumSuperfield
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driventoasterstrudel
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1 Answers1

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Let us start with something a bit more general and consider the sequence $$ u(n) = \alpha z_1^{n}+\alpha z_2^{n} $$ where $\alpha=(z_2-1)/(z_2-z_1)$, $\beta=(1-z_1)/(z_2-z_1)$, $z_1>1$, and $z_2<-1$.

From this expression, we obtain $$ \begin{array}{rcl} L(n)&:=&\dfrac{u(n+1)}{u(n)}\\ &=&\dfrac{\alpha z_1^{n+1}+\alpha z_2^{n+1}}{\alpha z_1^{n}+\alpha z_2^{n}}\\ % &=&\dfrac{(z_2-1)z_1^{n+1}+(1-z_1)z_2^{n+1}}{(z_2-1)z_1^{n }+(1-z_1)z_2^{n}}\\ % &=&\dfrac{z_1[(z_2-1)z_1^{n}+(1-z_1)z_2^{n}]+(1-z_1)z_2^{n+1}-z_1(1-z_2)z_2^n}{(z_2-1)z_1^{n }+(1-z_1)z_2^{n}}\\ % &=&z_1+\dfrac{(1-z_1)z_2^{n+1}-z_1(1-z_2)z_2^n}{(z_2-1)z_1^{n }+(1-z_1)z_2^{n}}\\ % &=&z_1+\dfrac{z_2^n(z_2-z_1)}{(z_2-1)z_1^{n }+(1-z_1)z_2^{n}}\\ % &=&z_1+\dfrac{(z_2-z_1)}{(z_2-1)\left(\dfrac{z_1}{z_2}\right)^n+(1-z_1)}. % \end{array} $$ Since $z_1>z_2$, we have that $(z_1/z_2)^n\to\infty$ as $n\to\infty$, which implies that $L(n)\to z_1$ as $n\to\infty$.

If we pick $z_1=(1+\sqrt{5})/2$, and $z_2=(1+\sqrt{5})/2$, then the sequence is the Fibonacci sequence and we have that

$$ \lim_{n\to\infty} \dfrac{u(n+1)}{u(n)}=z_1=\dfrac{1+\sqrt{5}}{2}. $$

KBS
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