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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.

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    $\begingroup$ 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) $\endgroup$ – Michael Joyce Apr 2 '14 at 19:12
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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 $f(x)$.

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  • $\begingroup$ Do you mean I should write $$x^2 +\sum_{n=0}^\infty a_{n+3} x^n \cdot x^3$$? $\endgroup$ – Paze Apr 2 '14 at 19:11
  • $\begingroup$ Yeah that looks better to me. $\endgroup$ – Alex J Best Apr 2 '14 at 19:13
  • $\begingroup$ 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? $\endgroup$ – Paze Apr 2 '14 at 19:25
  • $\begingroup$ 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. $\endgroup$ – Alex J Best Apr 2 '14 at 19:32
  • $\begingroup$ I'm still stuck trying to represent it as $f(x)$. Can you hold my hand a bit on this one? $\endgroup$ – Paze Apr 2 '14 at 22:35
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You have to get rid of the first terms, and add in the new ones: $$ f(z) - a_0 - a_1 z - a_2 z^2 + b_0 + b_1 z + b_2 z^2 $$

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