Exercise 4.4.17 ($U_n(ar, br^2) = U_n(a,b) r^{n-1}$)

Question

Show that $U_n(ar, br^2) = U_n(a,b) r^{n-1}$ for $n \geq 1$, and that $V_n(ar, br^2) = V_n(a,b)r^n$ for $n\geq 0$.

Richard Ganaye · Accepted Answer

ewcommand{\N}{\mathbb{N}}

\paragraph{ Ex 4.4.17} 
{\it Show that $U_n(ar, br^2) = U_n(a,b) r^{n-1}$ for $n \geq 1$, and that $V_n(ar, br^2) = V_n(a,b)r^n$ for $n\geq 0$.
}

\begin{proof} The Lucas's sequences $U_n(a,b)$ and $V_n(a,b)$ are defined by
\begin{align}
U_n(a,b) &= \frac{\lambda^n - \mu^{n}}{\lambda - \mu},\qquad V_n(a,b) = \lambda^n + \mu^n,
\end{align}
where $\lambda, \mu$ are the distinct roots of the polynomial $p(x) = x^2 - a x - b = (x- \lambda)(x- \mu)$.

Since $U_n(0,0)$ is not defined, we assume that $r 
e 0$ ($r \in \mathbb{R}$).

We define $q(x) = x^2 -ar x - br^2$. Then
\begin{align*}
q(x) &= x^2 -ra x - b r^2\
&= r^2 \left[ \left(\frac{x}{r} ight)^2 - a \left(\frac{x}{r} ight) - b ight]\
&= r^2 p\left(\frac{x}{r}ight)\
&= r^2 \left( \frac{x}{r} - \lambda ight) \left( \frac{x}{r} - \mu ight)\
&= (x - \lambda r) (x - \mu r).
\end{align*}
The roots of $q(x)$ are $\lambda r, \mu r$, and are distinct. Therefore, by (1), for all $n \in \N$,
\begin{align*}
U_n(ar,br^2) &= \frac{(\lambda r)^n - (\lambda r)\mu^{n}}{\lambda r  - \mu r}\
&=r^{n-1}  \frac{\lambda^n - \mu^{n}}{\lambda - \mu}\
&=  U_n(a,b)r^{n-1}.
\end{align*}
and
\begin{align*}
V_n(ar, br^2) &= (\lambda r) ^n + (\mu r)^n\
&=V_n(a,b) r^n.
\end{align*}
In conclusion, if $n \in \N$, and $r \in \mathbb{R}^*$,
$$U_n(ar, br^2) = U_n(a,b) r^{n-1},\qquad V_n(ar, br^2) = V_n(a,b)r^n.$$
\end{proof}

Exercise 4.4.17 ($U_n(ar, br^2) = U_n(a,b) r^{n-1}$)

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Exercise 4.4.17 ($U_n(ar, br^2) = U_n(a,b) r^{n-1}$)

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