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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.

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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.

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  • $\begingroup$ better $(VU)_{n+1} = \Lambda (VU)_{n}+(V \delta_n)$ $\endgroup$ – G Cab May 25 '19 at 17:29
  • $\begingroup$ @GCab Yes. I agree. Cleaner. $\endgroup$ – Cesareo May 25 '19 at 17:38
  • $\begingroup$ Under which topic/subject could I learn more about this please? $\endgroup$ – NoChance May 25 '19 at 19:56
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    $\begingroup$ @NoChance the theme of linear difference equations feeds on linear algebra. In the non-linear world, each case is an independent universe. There is a book that is old but available on the web and has very illustrative chapters on the equations of differences. A TREATISE ON THE CALCULUS OF FINITE DIFFERENCES. George Boole. archive.org/details/cu31924031240934 $\endgroup$ – Cesareo May 25 '19 at 20:45
  • $\begingroup$ @Cesareo, thank you kindly sir. Such a nice help. $\endgroup$ – NoChance May 25 '19 at 23:03

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