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I'm reading generatingfunctionology, and the author came up with this generating function:

$$ A(x) = \frac{x}{(1-x)(1-2x)} = x\{\frac{2}{1-2x} - \frac{1}{1-x}\} $$

Then he expands it. Which I thought was supposed to be:

$$ \sum_{n \ge 0} (x\{\frac{2}{1-2x} - \frac{1}{1-x}\})^n $$

But judging from the result:

$$ \{2x + 2^2x^2 + 2^3x^3 + ...\} - \{x + x^2 + x^3 + ...\} $$

He seems to be doing:

$$ \sum_{n \ge 0} (x\frac{2}{1-2x})^n - \sum_{n \ge 0} (x\frac{1}{1-x})^n $$ Which means I completely misunderstand the idea behind this process. So why is this the expansion?

Stanislav Bashkyrtsev
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    This is not a step of finding a generating function for $A$ (which is presented as a function of real numbers, not a sequence anyway). It's writing a sum which is exactly equal to $A(x)$ when the sum converges. – aschepler Dec 20 '22 at 14:50
  • Aaah.. That's what I didn't get! The author didn't even mention anything about geometric series, so this must've been very obvious.. for others :) – Stanislav Bashkyrtsev Dec 20 '22 at 14:57

1 Answers1

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The power series for $\frac{1}{1-x}$ (e.g., see geometric series) is $$\frac{1}{1-x} = 1 + x + x^2 + \cdots.$$ Similarly $$\frac{1}{1-2x} = 1 + (2x) + (2x)^2 + \cdots.$$

angryavian
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    And the point of using partial fractions on $A(x)$ is to get the function into a form in which this can be used. If there were higher powers in the denominator eg terms like $\frac {1}{(1-x)^{2}}$ related elementary results would come into play. [note mainly for OPs benefit] – Mark Bennet Dec 20 '22 at 14:59
  • Combined with what @aschepler left as a comment, I now have a full picture. Thanks everyone! – Stanislav Bashkyrtsev Dec 20 '22 at 15:38