solving recursion relation but need to prove that x_n is real

Question

Solve this recursion $x_n = 6x_{n−1}−25x_{n−2}, x_0 = 0, x_1 = 1$. Even though the characteristic equation will have complex roots, the x_n will be real, Verify this. I solved this equation but having problem verifying that X_n is real even though it has complex roots, any suggestions on how to do that?

Answer 1

You actually do not even need the characteristic polynomial to show that, given your initial conditions, every $x_n$ will be a real number. This can be proven using induction

For our base case we say that $x_0, x_1$ are real numbers. In your case they are.

Now we can prove by induction. Assume that $x_{n-1}, x_{n-2}$ are real numbers. From your relation we know that $$ x_n = 6x_{n-1} - 25x_{n-2} $$

but the product of real numbers is a real number, and the sum/difference of real numbers is a real number, and since $x_{n-1}, x_{n-2}$ are real, then $x_n$ must be real.

By our base case we know that for all $n$, $x_n$ is real

Answer 2

To prove $x_n$ is real by mathematical induction, first note for the base case that $x_0$ and $x_1$ are real.

Now assume $x_k$ is real for all $k\le n-1$.

Then $x_n=6x_{n-1}-25x_{n-2}$ is real because $x_{n-1}$ and $x_{n-2}$ are (by the assumption).

We have shown $x_k$ is real for $k=0$ and $k=1$

and that if $x_k$ is real for all $k\le n-1$ then $x_n$ is real.

It follows by induction that $x_n$ is real for all $n\in \mathbb N$.

Alternatively, you may find that $x_n=\dfrac{(3+4i)^n-(3-4i)^n}{8i}.$

That is, $x_n=\dfrac{z^n-\overline z^n}{8i}=\dfrac{z^n-\overline {z^n}}{8i}=\dfrac {w-\overline w}{8i},$ where $z=3+4i$ and $w=z^n$.

But if $w=a+bi$ with $a,b\in\mathbb R$, then $\dfrac {w-\overline w}{2i}=b$ is real, so $x_n=\dfrac b4$ is too.