Non-homogenous recurrence relation

Question

I need to solve the recurrence relation $A(n)=2A(n-2)+ 2^{n-2}$. I tried writing out equations up to the $A(2)$ and multiplying by powers of two and adding all the equations together then all the terms cancelled but after that I couldn't find the sum of the powers of two. I forgot to mention that the base is $A(2)=2$ and n is even

Answer 1

Let us compute some values of A(n):

$n=2, A(2)=2$

$n=4, A(4)=2A(2)+2^2=8=2^3$

$n=6, A(6)=2A(4)+2^4=32=2^5$

$n=8, A(8)=2A(6)+2^6=128=2^7$

You should see the pattern, It seems that $A(n)=2^{n-1}$. This is true and can be easily shown by induction.

Answer 2

You do this much the same way as you might solve a differential equation: first find the general solution to the homogeneous part of the equation, and then find a particular solution that you can add to that general solution.

The homogeneous equation here is:

$$ A(n)=2A(n-2) $$

Because $n$ is even, the general solution to this is easily seen to be:

$$ A_C(n)=B.2^{n/2} $$

where $B$ is an arbitrary constant.

We now seek a particular solution. One thing we could try would be just setting:

$$ A_P(n)=2^{n-2} $$

But then $2A_P(n-2)+2^{n-2}=2^{n-1}+2^{n-2}\neq2^{n-2}$.

That didn't work. We somehow need to find some way to make the two terms on the left combine into one term. If we try

$$ A_P(n)=2^{n-1} $$

then we get:

$$ 2A_P(n-2)+2^{n-2}=2^{n-2}+2^{n-2}=2^{n-1}=A_P(n) $$

Thus, $A_P(n)=2^{n-1}$ is a particular solution to the equation. The general solution then is:

$$ A(n)=A_C(n)+A_P(n)=B.2^{n/2}+2^{n-1} $$

Since $A(2)=2$, we have

$$ 2=B.2^1+2^1=2B+2 $$

So $B=0$, and the solution is $A(n)=2^{n-1}$.

Answer 3

It never hurts in such problems to start by gathering some numerical data:

$$\begin{array}{rcc} n:&2&4&6&8&10&12\\ A(n):&2&8&32&128&512&2048\\ A(n):&2^1&2^3&2^5&2^7&2^9&2^{11} \end{array}$$

There’s a very obvious pattern here, that leads to the conjecture that $A(n)=2^{\text{what function of }n?}$ if $n$ is even. Complete the conjecture correctly, and you should have little difficulty using mathematical induction to prove that it’s correct.

Alternatively, let $B(n)=A(2n)$; then $$B(n)=2B(n-1)+2^{2(n-1)}=2B(n-1)+4^{n-1}\;,$$

and ‘unwrap’ the recurrence:

$$\begin{align*} B(n)&=2B(n-1)+4^{n-1}\\ &=2\Big(2B(n-2)+4^{n-2}\Big)+4^{n-1}\\ &=2^2B(n-2)+2\cdot 4^{n-2}+4^{n-1}\\ &=2^2\Big(B(n-3)+4^{n-3}\Big)+2\cdot 4^{n-2}+4^{n-1}\\ &=2^3B(n-3)+2^2\cdot4^{n-3}+2\cdot 4^{n-2}+4^{n-1}\\ &\;\vdots\\ &=2^kB(n-k)+2^{k-1}\cdot 4^{n-k}+2^{k-2}\cdot 4^{n-k+1}+\ldots+2\cdot 4^{n-2}+4^{n-1}\\ &\;\vdots\\ &=2^{n-1}B(1)+\sum_{i=1}^{n-1}2^{i-1}\cdot4^{n-i}\\ &=2^n+\sum_{i=1}^{n-1}2^{i-1}\cdot4^{n-i}\\ &=2^n+\sum_{i=1}^{n-1}2^{i-1}\cdot2^{2n-2i}\\ &=2^n+\sum_{i=1}^{n-1}2^{2n-i-1}\\ &=2^n+\sum_{i=n}^{2n-2}2^i\\ &=2^n+2^n\sum_{i=0}^{n-2}2^i\\ &=2^n\left(1+\sum_{i=0}^{n-2}2^i\right)\\ &=\;? \end{align*}$$

Answer 4

Generating functions to the rescue. Define $g(z) = \sum_{n \ge 0} A(n) z^n$, write: $$ A(n + 2) + 2 A(n) + 2^n $$ Multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums to get: $$ \frac{g(z) -A(0) - A(1) z}{z^2} = 2 A(z) + \frac{1}{1 - 2 z} $$ As partial fractions: $$ A(z) = \frac{1 - 2 A(0) - 2 (A(1) - 1) z}{2 (1 - 2 z^2)} + \frac{1}{2 (1 - 2 z)} $$ You could now split the first term into partial fractions (but with irrational coefficients) or just take: \begin{align} \frac{1}{1 - 2 z^2} &= \sum_{n \ge 0} 2^n z^{2 n} \\ \frac{z}{1 - 2 z^2} &= \sum_{n \ge 0} 2^n z^{2 n + 1} \end{align} i.e., split into even and odd cases. The second term is just a geometric series.