Solving recurrence relation of two series with generating functions

Question

I'd appreciate your advice on how to solve recurrence relations of the following kind using generating functions: general solution of:

\begin{cases}a_{n+1} &= 5a_n - 3b_n \\ b_{n+1} &= 4a_n - 2b_n\end{cases}

for $n>0$ and particular solution for $a_0 = 2, b_0 = 1$.

I know how to solve recursions where only one series, f.e. $(a_n)_n$ is involved, but not how to do this for $(a_n)_n$ and $(b_n)_n$ at once.

Answer 1

Akin to the matrix approach, simply eliminate one of the variables from the equation. For example,

$$3b_{n+1}=2a_{n+1}+2a_n$$

which can then be substituted into the definition of $a_{n+1}$ to get

$$a_{n+1}=3a_n-2a_{n-1}$$

which is now linear in one recurrence. Similarly, we have

$$4a_{n+1}=5b_{n+1}-2b_n$$

and hence,

$$b_{n+1}=3b_n-2b_{n-1}$$

Answer 2

Note that $$\begin{pmatrix}a_{n+1}\\ b_{n+1}\end{pmatrix}=\begin{pmatrix}5 & -3\\ 4 & -2\end{pmatrix}\begin{pmatrix}a_{n}\\ b_{n}\end{pmatrix}$$

Diagonalize the square matrix to obtain $$A_{n+1}=P^{-1}DPA_n$$ And then $$A_n=P^{-1}D^nPA_0$$

Answer 3

For this specific one, you have that $a_n=b_n+1$ for all $n$:

Proof by induction:

Base: $a_0=b_0+1$

Step. Assume $a_n=b_n+1$

Then $a_{n+1} = 5a_n-3b_n=5(b_n+1)-3b_n=2b_n+5$

and $b_{n+1}=4a_n-2b_n=4(b_n+1)-2b_n=2b_n+4$

And so indeed $a_{n+1}=b_{n+1}+1$

OK, so then you have:

$a_{n+1}=5a_n-3b_n=5a_n-3(a_n-1)=2a_n+3$

and (as we already saw):

$b_{n+1}=2b_n+4$

And, you said that you can solve those yourself.

Answer 4

For this recurrence, $a_{n+1}−b_{n+1}=(5a_n−3b_n)−(4a_n−2b_n)=a_n−b_n$ for all $n,$ so $a_n-b_n=a_0-b_0$.

Therefore, $a_{n+1}=5a_n−3\color{blue}{b_n}=5a_n−3\color{blue}{(a_n−(a_0−b_0))}=2a_n+3(a_0−b_0). $

Now only one series $(a_n)$ is involved, so you should be able to take it from here.