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I've been getting the characteristic equation from relations of the form

$$U_n=3U_{n-1}-U_{n-3}$$

Thanks to this question I made before: How to get the characteristic equation?

Now, I recently have been asked to get such equation from this:

$$f(n) = \begin{cases}5 & \text{ if n = 0} \\ f(n-1) + 3 + 2n + 2^n \end{cases}$$

I tried to apply the same idea. I took:

$$F_n=F_{n-1} + 3 + 2n + 2^n$$

Changed the subscripts to exponents...

$$F^n=F^{n-1} + 3 + 2n + 2^n$$

Replaced $F$ by the desired variable...

$$x^n=x^{n-1} + 3 + 2n + 2^n$$

Divided by the smallest exponent (I guess it is $n-1$):

$$x=1+ 3^{-x+1} + 2n^{-x+1} + 2^{n-x+1}$$

And this doesn't look right. As you can see, the terms $3$, $2n$ and $2^n$ messed up the whole thing.

What did I do wrong, and how can I get the characteristic equation from this?

Community
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Saturn
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    There are a number of posts on MSE already dealing with this question, try typing "non-homogeneous recurrence" in the search box. – David Mar 27 '14 at 01:54
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    @David: Thanks, I didn't know that's how it's called haha. I will take a look at them. – Saturn Mar 27 '14 at 01:56

2 Answers2

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Use the machinery of generating functions. Define $F(z) = \sum_{n \ge 0} f(n) z^n$, and write your recurrence as: $$ f(n + 1) = f(n) + 5 + 2 n + 2^{n + 1} $$ Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize: \begin{align} \sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\ \sum_{n \ge 0} z^n &= \frac{1}{1 - z} \\ \sum_{n \ge 0} n z^n &= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\ &= \frac{z}{(1 - z)^2} \\ \sum_{n \ge 0} 2^n z^n &= \frac{1}{1 - 2 z} \end{align} Pulling all together: $$ \frac{F(z) - 5}{z} = F(z) + \frac{5}{1 - z} + \frac{2 z}{(1 - z)^2} + \frac{2}{1 - 2 z} $$ Solving for $F(z)$, as partial fractions: $$ F(z) = \frac{5 - 13 z + 8 z^2 - 2 z^3}{1 - 5 z + 9 z^2 - 7 z^3 + 2 z^4} = \frac{2}{1 - 2 z} + \frac{1}{(1 - z)^2} + \frac{2}{(1 - z)^3} $$ Using the (generalized) binomial theorem: \begin{align} f(n) &= 2 \cdot 2^n + \binom{-2}{n} (-1)^n + 2 \cdot \binom{-3}{n} (-1)^n \\ &= 2^{n + 2} + \binom{n + 2 - 1}{1} + 2 \cdot \binom{n + 3 - 1}{2} \\ &= 2^{n + 1} + n + 1 + \frac{(n + 2) (n + 1)}{2} \\ &= 2^{n + 1} + \frac{n^2 + 5 n + 4}{2} \end{align}

Maxima's help with the algebra is gratefully acknowledged.

vonbrand
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The characteristic equation gets you solutions of a homogeneous equation, in this case $f(n) = f(n+1)$ (but that's too trivial, those solutions are constant). You then have to add particular solutions of the non-homogeneous equation. In this case look for a polynomial of degree $2$ solving $f(n) = f(n+1) + 3 + 2n$ and $2^n c$ solving $f(n) = f(n+1) + 2^n$.

Robert Israel
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