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Use generating functions to solve the recurrence relation

$$ a_{n} = 3a_{n−1} + 2 $$

with initial condition $a_{0} = 1$.

If I can bring it to $ a_{n}=k a_{n-1} $ I can solve it easily. Thank you

Mhenni Benghorbal
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The Answer
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5 Answers5

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Here is a start

$$ \sum_{n=0}^{\infty}a_{n+1}x^n = 3 \sum_{n=0}^{\infty} a_{n}x^n + 2\sum_{n=0}^{\infty} x^n. $$

Now, recall this

$$ \sum_{n=0}^{\infty} x^n =\frac{1}{1-x}.$$

I think you can finish it now. Just follow the technique in this problem.

Community
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Mhenni Benghorbal
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Suppose $$ \begin{align} f(x) &=\sum_{n=0}^\infty a_nx^n\\ &=1+\sum_{n=1}^\infty(3a_{n-1}+2)x^n\\ &=1+x\sum_{n=0}^\infty(3a_n+2)x^n\\ &=1+3xf(x)+\frac{2x}{1-x}\\ (1-3x)f(x)&=\frac{1+x}{1-x}\\ f(x)&=\frac{1+x}{(1-x)(1-3x)}\\ &=\frac2{1-3x}-\frac1{1-x}\\ &=\sum_{n=0}^\infty(2\cdot3^n-1)x^n \end{align} $$ Equating coefficients yields $$ a_n=2\cdot3^n-1 $$

robjohn
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$A_n = 2*3^{n} - 1 $. This should do it

imranfat
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    actually, I believe that $a_n=2\cdot3^n-1$. Yours doesn't match for $n=0$. In any case, it would be nice to provide some explanation. – robjohn May 22 '13 at 16:57
  • When I put this equation in my TI, it provides the values 1,5,17,53,161 for the values x=0,1,2,3,4... That corresponds with A0=1, A1=3(1)+2=5, A3=3(5)+2=17,etc At first I typed it wrong – imranfat May 22 '13 at 17:41
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Let $h_n=a_n3^{-n}$. What is $h_n-h_{n-1}$? Now telescope. This gives that $$h_{n}-h_0=2\sum_{k=1}^n\frac{1}{3^k}$$

One can look at generating functions, but it proves much more tortuous. In general, a recurrence of the form $$x_{n+1}=ax_n+b$$ can be reduced by $y_n=x_n a^{-n}$ by $$y_{n+1}=y_n+\frac{b}{a^{n+1}}$$ and upon telescoping to $$y_{n+1}-y_0=b\sum_{k=0}^n \frac{1}{a^{k+1}}$$ that is $$x_{n+1}=b\sum_{k=0}^n a^{n-k}+a^{n+1}x_0$$

$$x_{n+1}=b\sum_{k=0}^n a^{k}+a^{n+1}x_0$$

Pedro
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You want to solve $a_{n} = 3a_{n−1} + 2$ and you said that you could do it if you could bring it to the form $a_{n} = 3a_{n−1}$.

Here is one way to do that:

Let $a_n = b_n+c$, where $c$ is a constant to be determined. We want to choose $c$ so that $b_n = 3 b_{n-1}$.

Substituting the expression for $a$, $b_n+c = 3(b_{n-1}+c)+2 = 3 b_{n-1} + 3c+2 $, or $b_n = 3 b_{n-1}+2c+2$.

To make $b_n = 3 b_{n-1}$, we need $2c+2 = 0$, so $c = -1$. Then $b_n = 3 b_{n-1}$, so $b_n = 3^n b_0$.

Writing $b_n$ in terms of $a_n$, $a_n-c = 3^n(a_0-c)$ or $a_n = 3^n a_0 - 3^n c + c = 3^n a_0-c(3^n-1) = 3^n a_0 + 3^n - 1 $ (since $c = -1$).

This idea of modifying a recurrence so that it becomes easier to solve is a very general problem solving technique. It looks like this:

  1. You have a hard problem.
  2. Convert it to an easier problem.
  3. Solve the easier problem.
  4. Convert that solution to a solution of the hard problem.

Think of multiplying by using logs, for example.

In your case, your problem was converted to $b_n = 3 b_{n-1}$ and, when the solution to this was found, the $b$'s were converted back to the original $a$'s.

marty cohen
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