Is this recurrence relation $g_{n+1}=ig_n-g_{n-1}$ is a trivial?

Question

Let $g_1=i$ and $g_2=-1$, where $i=\sqrt{-1}$, and

$$g_{n+1}=ig_n-g_{n-1}$$

For $n=1,2,3,4, ...$ then $g_n:={i, -1, -2i, 3, 5i, -8, -13i, 21, ...}$ respectively.

Is this recurrence relation is trivial?

Answer 1

HINT :

Dividing the both sides of $$g_{n+1}=ig_n-g_{n-1}$$ by $i^{n+1}$ gives $$\frac{g_{n+1}}{i^{n+1}}=\frac{g_n}{i^n}+\frac{g_{n-1}}{i^{n-1}}$$ which can be written as $$G_{n+1}=G_n+G_{n-1}$$ where $G_n=g_n/i^n$.

Answer 2

Since you have $g_{n+1}=ig_n-g_{n-1}$, if we let : $$u_0=\begin{pmatrix} g_1 \\ g_2 \end{pmatrix}=\begin{pmatrix} i \\ -1 \end{pmatrix}$$ $$A=\begin{pmatrix} 0 & 1 \\ i & -1 \end{pmatrix}$$ Then we have, $\forall n$ : $$u_n=Au_{n-1}=A^nu_0$$ Now you can digonalise $A$ to express an explicit expression of $u_n$ and so of $g_n$.

Answer 3

Using the characteristic equation of $g$, $$z^2 - iz + 1$$ we find that the roots are $\frac{\sqrt{5} - 1}{2i}$ and $-\frac{\sqrt{5} + 1}{2i}$. Then, $$g_1 = i = c_1\Big(\frac{\sqrt{5} - 1}{2i}\Big)^1 + c_2\Big(-\frac{\sqrt{5} + 1}{2i}\Big)^1 = \frac{c_1(\sqrt{5} - 1) - c_2(\sqrt{5} + 1)}{2i} \\ g_2 = -1 = c_1\Big(\frac{\sqrt{5} - 1}{2i}\Big)^2 + c_2\Big(-\frac{\sqrt{5} + 1}{2i}\Big)^2 = \frac{c_1(6 -2\sqrt{5}) + c_2(6 + 2\sqrt{5})}{-4}$$ which can be solved to get something very similar to the Fibonacci equation.

Answer 4

Rewrite $$g_{n+1}=ig_n-g_{n-1}=g_{n+1}=ig_n+i^2g_{n-1}$$ and let $g_n=i^nh_n$. So, $$i^{n+1}h_{n+1}=i^{n+1}h_{n}+i^{n+1}h_{n-1}$$ that is to say $$h_{n+1}=h_{n}+h_{n-1}$$