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I'm working as an assistant in an algebra course. We're looking at homigenous linear recurrence relations of second order.

For real valued initial conditions, is it possible for the constants in the solutions of the homogenous linear recurrence relation of second order to take non-real values?

I think it shouldn't be possible. Working a little bit with relations with complex eigenvalues, I get $$A cos(n \alpha + \beta) = -Bsin(n \alpha + \beta) , \forall n \in \mathbb N_0$$ Where $A$, $B$, $\alpha$ and $\beta$ are real-valued constants

I want to prove that this can only happen when A = B = 0 (in case it's true of course)

Edit: Alpha and Beta are not given, they could take any value.

Francisco José Letterio
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1 Answers1

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If the recurrence is

$x_{n+2} =ax_{n+1}+bx_n $, the characteristic equation is $r^2-ar-b = 0$.

The roots are $r_1, r_2 =\dfrac{a\pm\sqrt{a^2+4b}}{2} $, so if $a^2+4b < 0$ the roots are complex.

If the solution is $x_n =ur_1^n+vr_2^n $, then $x_0 = u+v$ and $x_1 = ur_1+vr_2$.

Solving these, $x_1 =ur_1+vr_2 =ur_1+(x_0-u)r_2 =u(r_1-r2)+x_0r_2 $ so $u =\dfrac{x_1-x_0r_2}{r_1-r_2} $ and $v =x_0-u =\dfrac{x_0(r_1-r_2)-x_1+x_0r_2}{r_1-r_2} =\dfrac{x_0r_1-x_1}{r_1-r_2} $.

Note that $r_1-r_2 =\sqrt{a^2+4b} =c $ and $r_1r_2 =-b $ so $u =\dfrac{x_1-x_0r_2}{c} $ and $v =x_0-u =\dfrac{x_0r_1-x_1}{c} $.

The solution is then

$\begin{array}\\ x_n &=\dfrac{x_1-x_0r_2}{r_1-r_2}r_1^n+\dfrac{x_0r_1-x_1}{r_1-r_2}r_2^n\\ &=\dfrac{x_1r1^n-x_0r_2r_1^n+x_0r_1r_2^n-x_1r_2^n}{r_1-r_2}\\ &=\dfrac{x_1r_1^n-x_0r_2r_1^n+x_0r_1r_2^n-x_1r_2^n}{c}\\ &=\dfrac{x_1(r_1^n-r_2^n)+x_0(r_1r_2^n-r_2r_1^n)}{c}\\ &=\dfrac{x_1(r_1^n-r_2^n)+x_0r_1r_2(r_2^{n-1}-r_1^{n-1})}{c}\\ \end{array} $

so that if $\dfrac{r_1^n-r_2^n}{c} $ is real, the solutions are real.

$r_1^n, r_2^n =(a\pm c)^n/2^n$ so

$\begin{array}\\ r_1^n-r_2^n &=\dfrac1{2^n}\sum_{k=0}^n\binom{n}{k}(a^kc^{n-k}-(-1)^{n-k}a^kc^{n-k})\\ &=\dfrac1{2^n}\sum_{k=0}^n\binom{n}{k}(a^{n-k}c^{k}-(-1)^{k}a^{n-k}c^{k})\\ &=\dfrac1{2^n}\sum_{k=0}^{(n-1)/2}2\binom{n}{2k+1}a^{n-2k-1}c^{2k+1}\\ \text{so}\\ \dfrac{r_1^n-r_2^n}{c} &=\dfrac1{2^{n-1}}\sum_{k=0}^{(n-1)/2}\binom{n}{2k+1}a^{n-2k-1}c^{2k}\\ \end{array} $

and this is real whether $c$ is real or imaginary.

marty cohen
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