Solving this inhomogenous system of two linear recurrence relations

I have this system of recurrence relations (where $$\forall n\in\Bbb N: b^{(n)}, c^{(n)}\in\Bbb R$$): $$\begin{pmatrix}b^{(n+1)}\\c^{(n+1)}\end{pmatrix} = \begin{pmatrix}1\over2^{n+1}\\0\end{pmatrix} + \begin{pmatrix}\frac34 & \frac14 \\ \frac14 & \frac34\end{pmatrix} \begin{pmatrix}b^{(n)} \\ c^{(n)}\end{pmatrix}.$$

How can I find an explicit formula for all the $$b^{(n)}, c^{(n)}$$?

My question stems from this question. The latter question is trivial to solve if my question from above is answered.

I thought about using some fixed point Theorem, but the calculations seem to get messy.

Hint.

Making

$$M=\frac 14\left( \begin{array}{cc} 3 & 1 \\ 1 & 3 \\ \end{array} \right) = V^{\dagger}\cdot\Lambda\cdot V,\ \ U_n=\left( \begin{array}{c} b_n \\ c_n \\ \end{array} \right)$$

with

$$V = \frac{1}{\sqrt 2}\left( \begin{array}{cc} 1 & 1 \\ -1 & 1 \\ \end{array} \right)$$

and

$$\Lambda = \left( \begin{array}{cc} 1 & 0 \\ 0 & \frac 12 \\ \end{array} \right)$$

Then

$$U_{n+1}=V^{\dagger}\cdot \Lambda\cdot V\cdot U_n + \delta_n$$

This is a linear recurrence with solution

$$U_n = U_n^h+U_n^p\\ U_{n+1}^h - M\cdot U_n^h = 0\\ U_{n+1}^p - M\cdot U_n^p =\delta_n$$

then

$$U_n^h = V^{\dagger}\cdot\Lambda^n\cdot V\cdot C_0$$

now calling $$U_n^p = V^{\dagger}\cdot \Lambda\cdot V\cdot C_n$$ and substituting we have

$$V^{\dagger}\cdot\Lambda^{n+1}\cdot VC_{n+1}=V^{\dagger}\cdot\Lambda^{n+1}\cdot VC_n +\delta_n$$

giving

$$C_{n+1}-C_n = V\Lambda^{-(n+1)}\cdot V^{\dagger}\cdot \delta_n = \frac 14\left( \begin{array}{c} 2+2^{-n} \\ -2+2^{-n} \\ \end{array} \right)$$

etc.

and finally

$$U_n = V^{\dagger}\cdot\Lambda^n\cdot V\cdot\left(C_0+C_n\right)$$

NOTE

According to @G Cab calling $$W = V\cdot U$$ and $$\rho = V\cdot\delta_n$$ we can write also

$$W_{n+1} = \Lambda\cdot W_n + \rho_n$$

which is easier to handle.

• better $(VU)_{n+1} = \Lambda (VU)_{n}+(V \delta_n)$ – G Cab May 25 '19 at 17:29
• @GCab Yes. I agree. Cleaner. – Cesareo May 25 '19 at 17:38