Exercise 4.4.18* ($a(-b)^{n-1} U_n(a', -1) = U_{2n}(a,b)$, where $a' = -2 -a^2/b$)

Question

Put $a' = -2 -a^2/b$. Show that $a(-b)^{n-1} U_n(a', -1) = U_{2n}(a,b)$, and that $(-b)^n V_n(a', -1) = V_{2n}(a,b)$.

Richard Ganaye · Accepted Answer

\begin{proof}
Here we assume that $a 
e 0, b 
e 0$, and $D= a^2 +4b 
e 0$ (the definition 4.11 suppose that $D 
e 0$).

By Problem 17, with $r = -b$, we obtain
\begin{align}
(-b)^{n-1}U_n(a', -1) &= (-b)^{n-1} U_n\left( - 2 - \frac{a^2}{b}, -1ight)=U_n(a^2 + 2b, -b^2).
\end{align}
(The definition p.201 of $U_n(a', -1)$ presupposes that $a'$ is an integer.)

Let $\lambda, \mu$ be the (distinct) roots of $p(x) = x^2 - ax -b = (x- \lambda)(x- \mu).$
Then
$$\lambda + \mu = a, \qquad \lambda \mu = -b.$$
Thus
\begin{align*}
&\lambda^2 + \mu^2 = (\lambda + \mu)^2 - 2 \lambda \mu= a^2 + 2b,\
&\lambda^2 \mu^2 = b^2.
\end{align*}
Therefore $\lambda^2, \mu^2$ are the roots of
\begin{align}
q(x) &= (x - \lambda^2)(x - \mu^2) = x^2 -(a^2+2b) x + b^2,
\end{align}
thus
\begin{align*}
U_n(a^2 + 2b, -b^2) &= \frac{\lambda^{2n} - \mu^{2n}}{\lambda^2 - \mu^2}\
&= \frac{1}{a}\  \frac{\lambda^{2n} - \mu^{2n}}{\lambda - \mu}.
\end{align*}
By (1), 
\begin{align*}
a(-b)^{n-1} U_n(a',1) &= aU_n(a^2 + 2b, -b^2)\
&= \frac{\lambda^{2n} - \mu^{2n}}{\lambda - \mu}\
&= U_{2n}(a,b).
\end{align*}

Similarly, by Problem 17 and (2),
\begin{align*}
(-b)^n V_n(a',1) &= V_n(a^2 + 2b, -b^2)\
&= (\lambda^2)^n + (\mu^2)^n\
&= V_{2n}(a,b).
\end{align*}

In conclusion, if $a 
e 0, b 
e 0$, and $D= a^2 +4b 
e 0$, then
\begin{align*}
U_{2n}(a,b) &= a(-b)^{n-1} U_n(a',1) \
V_{2n}(a,b) &= (-b)^n V_n(a', -1),
\end{align*}
where $a'= - 2 - \frac{a^2}{b}$ is an integer.
\end{proof}
Note: An hypothetic patient reader (not me) can consider the exceptional cases $D = 0$ (or $a = 0$) by taking the definition (4.10) rather than (4.11).

Exercise 4.4.18* ($a(-b)^{n-1} U_n(a', -1) = U_{2n}(a,b)$, where $a' = -2 -a^2/b$)

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Exercise 4.4.18* ($a(-b)^{n-1} U_n(a', -1) = U_{2n}(a,b)$, where $a' = -2 -a^2/b$)

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