Homework - Resolve the recurrence relation

Question

What's the closed formula of this recurrence relation? $$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$

Answer 1

A generating function approach would make this straightforward:

$$G(x) = \sum_{n=0}^\infty a_n x^n = 1+2x+\sum_{n=2}^\infty a_n x^n$$ $$=1+2x+\sum_{n=2}^\infty (a_{n-1} + 2a_{n-2} + 2^n)x^n$$ $$=1+2x+\sum_{n=1}^\infty a_n x^{n+1} + 2 \sum_{n=0}^\infty a_n x^{n+2} + \sum_{n=2}^\infty 2^n x^n$$ $$=1+2x+x(G(x)-1)+2x^2G(x)+\frac{2^2x^2}{1-2x}$$

Thus $$G(x)(1-x-2x^2) = 1+x+\frac{2^2x^2}{1-2x}$$ and $$G(x) = \frac{1+x}{1-x-2x^2} + \frac{2^2x^2}{(1-2x)(1-x-2x^2)} = \frac{1+x}{(1+x)(1-2x)} + \frac{2^2x^2}{(1-2x)(1+x)(1-2x)}$$ $$= \frac{1}{(1-2x)} + \frac{2^2x^2}{(1-2x)^2(1+x)}$$

Now you can use partial fractions, and then expand in terms of geometric series to find the taylor coefficients of $G(x)$ which are the terms $a_n$.

Answer 2

This is nonhomogeneneous difference equations. First solve the homogeneous equation $$ a_n - a_{n-1} - 2a_{n-2} = 0 \quad (1) $$ Let $a_n = r^n$, so that $a_{n-1} = r^{n-1}$ and $r_{n-2} = r^{n-2} = 0$. Replacing in (1), we have $$ r^n - r^{n-1} - 2r^{n-2} = 0 \quad r^2 - r - 2 = 0 \quad \Rightarrow \ r_1 = -1, \ r_2 = 2 $$ $$ a_{nh} = C_1(-1)^n + C_22^n $$ Particular solution: Let $a_{np} = a_n = An2^n$. The presence of n is due to the fact that $2$ is a root of the equation (1). So, $a_{n-1} = A(n-1)2^{n-1}$ and $a_{n-2} = A(n-2)2^{n-2}$. Substituting in the given equation, we have $$ An2^n + (An - A)2^{n-1} - 2A(n-2)2^{n-2} = 2^n \quad \Rightarrow \quad A = \frac{2}{3} $$ Thus, $a_n = a_{nh} + a_{np} = C_1(-1)^n + C_22^n + \frac{2n2^n}{3}$. But, $a_0 = a(0) = 1$ and $a_1 = a(1) = 2$. We have the system $$ \begin{cases} 1 = a_0 = C_1 + C_2\\ 2 = a_1 = -C_1 + C_2 + \frac{8}{3} \end{cases} \quad \Rightarrow \quad C_1 = 8/9, \quad C_2 = 1/9 $$ Thus, $$ a_n = \frac{8(-1)^n}{9} + \frac{1}{9}2^n + \frac{2n2^n}{3} $$

Answer 3

The best course of action for simple inhomogeneous recurrences is to make use of a smart "change of variables" (read: substitute with another recurrence relation) to turn it into a homogeneous recurrence. A nice observation here is to notice that $2^n$ is itself a recurrence relation, namely:

$$y_n=2y_{n-1},\,\,(y_0=1)$$ Let's use this to our advantage:

$$\begin{align} a_n=a_{n-1}+2a_{n-2}+2^n &\Leftrightarrow 2^n=a_n-a_{n-1}-2a_{n-2} \\ &\Leftrightarrow 2\cdot2^{n-1}=a_n-a_{n-1}-2a_{n-2} \\ &\Leftrightarrow 2\cdot (a_{n-1}-a_{n-2}-2a_{n-3})=a_n-a_{n-1}-2a_{n-2} \\ &\Leftrightarrow a_n-3a_{n-1}+4a_{n-3}=0 \end{align}$$

Bam, homogeneous recurrence. Can apply the characteristic equation and finish it by yourself?

Edit: The characteristic equation for the last recurrence is $x^3-3x+4=0$, whose roots are $2,2$ and $-1$. It is well-known then, that given these roots, the solution is of the form: $$a_n=(An+B)2^n+C(-1)^n$$

For some constants $A,B,C$. To find out what the constants are, we plug in values of $n$ for which the value of $a_n$ is known. The values of $a_0,a_1$ were given, but we need three to solve a linear system with three variables. We can easily calculate $a_2$: $$a_2=a_1+2a_0+2^2=8$$

Now we solve the system:

$$\begin{cases} (A\cdot0+B)2^0+C(-1)^0=1 \\ (A\cdot1+B)2^1+C(-1)^1=2 \Rightarrow \\ (A\cdot2+B)2^2+C(-1)^2=8 \end{cases}$$

$$\Rightarrow \begin{cases} B+C=1 \\ 2(A+B)-C=2 \\ 4(2A+B)+C=8 \end{cases}$$

Solving this will yield $A=\frac{2}{3}$,$B=\frac{5}{9}$ and $C=\frac{4}{9}$. Therefore, the closed form is: $$a_n=\frac{(6n+5)2^n+4(-1)^n}{9}$$