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I have task: Let $f(x)=\frac{x}{1-2x^2}$. Appoint $[x^n]f(x)$.

It's from example exam from Discrete mathematics. In task there isn't anymore information. I don't know what means $[x^n]f(x)$. Anybody can help?

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    $\begingroup$ perhaps it means finding the coefficient of $x^n$ in the series expansion of $f(x)$. $\endgroup$ – Anurag A Jun 24 '14 at 16:07
  • $\begingroup$ I found in lecturer notes something like this $[n]={1,2,...,n}$ but i don't know how use it for this task. $\endgroup$ – warpath Jun 24 '14 at 16:10
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    $\begingroup$ Anurag A is right about the meaning of the symbol. To solve the problem, write down the expansion of $\frac{1}{1-2x^2}$ using the familiar $\frac{1}{1-t}=1+t+t^2+\cdots$, multiply by $x$, and write down the coefficient of $x^n$. $\endgroup$ – André Nicolas Jun 24 '14 at 16:22
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First, the series representation of the expression $\frac{1}{1 - 2x^2}$ is $$\sum_{i = 0}^{\infty} (2x^2)^i \,\,.$$ It follows that the series expansion for $f(x)$ (which is $x$ multiplied by the above expression) is given by $$\sum_{i = 0}^{\infty} x(2x^2)^i = \sum_{i = 0}^{\infty} 2^i x^{2i + 1} \,\,.$$ Notice that each power of $x$ in this series is odd, so we assume the arbitrary $n$ given in the statement of the question is odd, in other words $n \in \{2m + 1 \mid m \in \mathbb{Z}\}$. Now, if we set $n = 2i + 1$ for $0 \leq i \in \mathbb{Z}$, we obtain $i = \frac{n-1}{2}$ for any odd $n \geq 1$ and so we conclude that $$[x^n]f(x) = 2^{\frac{n-1}{2}} \,\,.$$

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