# Ordinary Generating functions for $b_n$

Problem

Let $f(x)$ be a ordinary generating function for the sequence $\{\ a_0, a_1, a_2... \}\$ Find the ordinary generating function for $b_0 = b_1 = 0, b_2 = 1$ $b_n = a_n$ for $n \geq 3$.

Also find the generating function for $b_n = 0$ for even $n$, $b_n = a_n$ for odd $n$.

My attempts

For the first problem, $A(x) = 0+0+1x^2+a_3x^3+a_4x^4...$

So $$g(x) = x^2 + \sum_{n=3}^\infty a_n x^n$$

$$\Rightarrow x^2 + \sum_{n=0}^\infty a_n x^{n+3}$$ $$\Rightarrow x^2 + \sum_{n=0}^\infty a_n x^n \cdot x^{-3}$$ $$\Rightarrow x^2 + f(x) \cdot x^{-3}$$

For the next problem I've got the sequence to be: $$\{\ 0,a_1 , 0 , a_2 , 0, a_3\ldots \}\$$

So $$A(x) = \{\ a_1 x + a_3 x^3 + a_5 x^5 \ldots \}\$$

Which implies $$g(x) = \sum_{n=0}^\infty a_n x^n - \sum_{n=0}^\infty a_{2n} x^n$$

$$g(x) = f(x) - \sum_{n=0}^\infty a_{2n} x^n$$

But I can't get further. I hope my progress so far is correct in these.

Thank you.

• A word of advice: any time you write an identity of power series, it is a good idea to check and make sure the coefficients of $x^n$ are the same on both sides for small values of $n$. You'll likely spot your errors (pointed out by Alex) that way. (+1 for showing the work on your problem) Apr 2, 2014 at 19:12

So for the first one there are a couple of issues, when you go from summing from $n = 3$ to $n=0$ you add 3 on to the power, but not the index of the coefficient. Also when you factor out the $x^3$ it turns into $x^{-3}$. Fix these things and see how far you can get.

For the second I'd recommend looking at $f(-x)$ and comparing that to $f(x)$.

• Do you mean I should write $$x^2 +\sum_{n=0}^\infty a_{n+3} x^n \cdot x^3$$?
– Paze
Apr 2, 2014 at 19:11
• Yeah that looks better to me. Apr 2, 2014 at 19:13
• I can't get further with that one. I'm having troubles representing it as $f(x)$ now. I looked into $f(-x)$. How does $$f(x)+2 f(-x)$$ sound for an answer for the second problem?
– Paze
Apr 2, 2014 at 19:25
• That's definitely the right idea, as Micheal suggested, try it on a small example though, e.g. $f(x) = 1$. Yeah that first one is a little hard to represent with just $f(x)$ like terms, however using terms like $f(0)$ I think it should be possible though. Apr 2, 2014 at 19:32
• I'm still stuck trying to represent it as $f(x)$. Can you hold my hand a bit on this one?
– Paze
Apr 2, 2014 at 22:35

You have to get rid of the first terms, and add in the new ones: $$f(z) - a_0 - a_1 z - a_2 z^2 + b_0 + b_1 z + b_2 z^2$$