# Recurrent equation system

How to solve this recurrent equation system?

$$\begin{cases}a_{n+1}=a_n+2b_n\\ b_{n+1}=2a_n+b_n\end{cases}$$

The possible solutions should be \begin{cases}a_n=\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\ b_n=\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{cases}

• What is $a_0$ and $b_0$? Looks like an induction question to me. – max_zorn Jan 8 '18 at 16:09

Let $X_n =(a_n,b_n)$ with $X_0=(1,0)=(a_0,b_0)$

\begin{cases}a_{n+1}=a_n+2b_n\\ b_{n+1}=2a_n+b_n\end{cases} Then it is equivalent $$X_{n+1} = AX_n \Longleftrightarrow X_n =A^nX_0$$

where, $$A= \begin{pmatrix}1&2\\ 2&1\end{pmatrix}$$

Then prove by induction that,

$$A^n= \begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n &\frac{1}{2}3^n-\frac{1}{2}(-1)^n\\\frac{1}{2}3^n-\frac{1}{2}(-1)^n&\frac{1}{2}3^n+\frac{1}{2}(-1)^n\end{pmatrix}$$ Therefore you will get the desired result. $$X_n= \begin{pmatrix}a_n\\b_n\end{pmatrix}= \begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n &\frac{1}{2}3^n-\frac{1}{2}(-1)^n\\\frac{1}{2}3^n-\frac{1}{2}(-1)^n&\frac{1}{2}3^n+\frac{1}{2}(-1)^n\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\=\begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\ \frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{pmatrix}$$

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

So, $a_{n+1}-2a_n-3a_{n-1}=0$.

The roots of $x^2-2x-3=0$ are $-1$ and $3$.

$a_n=\alpha (-1)^n+\beta(3)^n$ for some $\alpha$ and $\beta$.

$2b_n=a_{n+1}-a_n=\alpha(-1)^{n+1}+\beta(3)^{n+1}-\alpha (-1)^n-\beta(3)^n=-2\alpha(-1)^n+2\beta (3)^n$

$b_n=-\alpha(-1)^n+\beta (3)^n$

Write your system in vectors and matrices:

$$\Big({a \atop b}\Big)_{n+1} = \Big({ 1 \; 2 \atop 2 \; 1}\Big) \Big({a \atop b}\Big)_{n}$$

Then use eigenvalues (3 and -1) and eigenvectors (1,1) and (1,-1) of the matrix. This gives you that the solutions must be of the form (with the eigenvalues)

$${a_n=A 3^n+ A(-1)^n}\\ {b_n=B 3^n- B (-1)^n}$$

The constants $A$ and $B$ must be determined by initial conditions (not given in the question). You get $a_1 = 2A$ and $b_1 = 4 B$.

Assume we know $a_1$ and $b_1$.

First, $a_{n+1}+b_{n+1}=3(a_n+b_n) = \cdots = 3^n(a_1+b_1)$.

And secondly, $a_{n+1}-b_{n+1}=-(a_n - b_n) = \cdots = (-1)^n (a_1 - b_1)$.

And

$$a_{n+1} = \frac{(a_{n+1}+b_{n+1})+(a_{n+1}-b_{n+1})}{2}$$

$$b_{n+1} = \frac{(a_{n+1}+b_{n+1})-(a_{n+1}-b_{n+1})}{2}$$

Done!

• then false>>>>>>> – user518372 Jan 9 '18 at 15:22

Let $a_0=A, b_0=B$ Add your two equations:

$$a_{n+1}+b_{n+1}=3(a_n+b_n)=3^{n+1}(A+B)$$

Subtract the second equation from the first

$$a_{n+1}-b_{n+1}=-(a_n-b_n)=(-1)^{n+1}(A-B)$$

Since you don't give initial values, you can't specify a particular solution from the data provided.