Solve the reccurence relation $a_n = a_{n-2}, a_0=a_1 = 1$

Question

This is my first time working through this type of problem and I am looking to see if I have worked it out correctly, thanks!

Solve the reccurence relation $$a_n = a_{n-2}, a_0=a_1 = 1$$

First we get the characteristic equation:

$$x^n = x^{n-2}$$

Dividing by the smallest we get,

$$x^2=1$$ $$x = \pm 1$$

Using the auxiliary equation: general solution

$$a_n = A_1x^{n+1} + A_2x^{n+1}$$

Using conditions in the equation and our solution for $x$ gives,

$$a_0 = 1 = A_1(1) + A_2(-1) = A_1 - A_2$$ $$a_1 = 1 = A_1(1)^2 = A_2(-1)^2 = A_1 + A_2$$

Solving for $A_1, A_2$ I got $A_1 = 1, A_2 = 0$

Therefore plugging into the general solution I got,

$$a_n = 1^{n+1}$$

Answer 1

$$a_{2n}=a_{2n-2}=a_0=1$$

$$a_{2n+1}=a_{2n-1}=a_1=1$$

thus

$$\forall n\in \Bbb N \;a_n=1$$

Answer 2

Use the method of generating functions (overkill but why not). Set $a_n=0$ for $n<0$ and extend the recurrence relation to $$ a_n=a_{n-2}+\delta_{n,1}+\delta_{n,0}\quad (n\geq 0)\tag{1} $$ so that it is valid for all $n\geq 0$ where $\delta$ is the Kronecker Delta. Let $A(x)=\sum_{n=0}^\infty a_n x^n$ and multiply both sides of (1) by $x^n$ and sum on $n$ to get that $$ A(x)=x^2A(x)+x+1\iff A(x)(1-x^2)=1+x. $$ In particular $$ A(x)=\frac{1+x}{(1-x)(1+x)}=\frac{1}{1-x}=\sum_{n=0}^\infty x^n $$ whence $$ a_n=1\quad (n\geq0). $$

Answer 3

We transform $a_{n} = a_{n-2}$ to $a_{n+2} = a_n$ and $a_0=a_1=0$.

Using generating functions (with $A(x):=\sum_{n \geq 0} a_{n} x^n$) this leads to

$\begin{eqnarray*}\sum_{n \geq 0} a_{n+2} x^n &=& \sum_{n \geq 0} a_{n} x^n \\ \Leftrightarrow \frac{A(x)}{x^2} - \frac{1}{x^2}-\frac{1}{x} &=& A(x) \\ \Leftrightarrow A(x) = \frac{1}{1-x} &=& \sum_{n \geq 0} x^n \end{eqnarray*}$

If we look now at the coefficient of $[x^n]$ with see that it is $1$ and therefore $a_n = 1$.