# sequences - recurrence relation [closed]

I have to find the expression of $(y_n)$ defined by :

$$y_{n+1}=a y_n+b z_n+c$$

where $(z_n)$ is an arithmetico-geometric sequence :

$$z_{n+1}=d z_n+e$$

and $a,b,c,d,e$ real numbers.

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## closed as too localized by Yemon Choi, Andrés Caicedo, Qiaochu Yuan, Andreas Blass, Bugs BunnySep 25 '12 at 20:31

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It's actually not hard to find the general solution for the $$z_n$$: there is a particular solution, and then you add the general solution of the homogeneous recurrence. So you should try the same strategy for the $$y_n$$. – Charles Matthews Sep 25 '12 at 16:08

Note that the vector $X_n = \pmatrix{y_n\cr z_n\cr}$ satisfies the recurrence $X_{n+1} = A X_n + C$ where $$A = \pmatrix{a & b\cr 0 & d\cr},\ C = \pmatrix{c\cr e\cr}$$ and thus $$X_n = A^n C X_0 + (A - I)^{-1} (A^n - I) C$$ Moreover, $$A^n = \pmatrix{ a^n & b (a^n - d^n)/(a-d)\cr 0 & d^n\cr}$$ This is all assuming $1,a,d$ are distinct. The other cases may be obtained as limits.
One standard way to solve recurrence relations is with generating functions. In this case, let $f$ and $g$ be the ordinary generating functions of the sequences $y$ and $z$. Then the generating function equivalent of your recurrence relations would be $$\frac{g(x)-g(0)}x=d\cdot g(x)+\frac e{1-x}$$ and $$\frac{f(x)-f(0)}x=a\cdot f(x)+b\cdot g(x)+\frac c{1-x}.$$ You can then solve these relations for the generating functions $$g(x)=\left(\frac{e\cdot x}{1-x}+g(0)\right)\cdot \frac 1{1-xd}$$and $$f(x)=\left(x\cdot b\cdot g(x)+\frac{x\cdot c}{1-x}+f(0)\right)\frac 1{1-ax}.$$ Lastly, you need to find the partial fraction decomposition of $f$. Using geometric series and its derivatives, you can then read off the coefficients of the partial fraction decomposition to get an explicit solution for the terms in your sequences.