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I am facing trouble in the following question

coefficient of $x^{2012}$ in $$\frac{1+x}{(1+x^2)(1-x)}$$

i broke it down as $$(1+x){(1+x^2)}^{-1}{(1-x)}^{-1}$$ and then the coefficient of $x^{2012}$ in this expression is equal to

coefficient of $x^{2012}$ in ${(1+x^2)}^{-1}$+coefficient of $x^{2012}$ in ${(1-x)}^{-1}$+1 $\cdot$ coefficient of $x^{2011}$ in ${(1+x^2)}^{-1}$+1 $\cdot$ coefficient of $x^{2011}$ in ${(1-x)}^{-1}$.

I couldnot proceed after this.please help me in this regard.thanks.

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  • $\begingroup$ I presume you mean the coefficient of the Binomial Series? If so, what are the series of each of the three functions that are multiplied? $\endgroup$ – Brevan Ellefsen Oct 27 '16 at 23:05
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$$\frac{1+x}{(1+x^2)(1-x)}=\frac{x}{1+x^2}+\frac{1}{1-x}$$ use $${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}$$ and let $x\rightarrow -x^2$ $${\frac {1}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n}$$ multiply by $x$ $${\frac {x}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n+1}$$

the second series does not contain power $2012$ because $$2n+1=2012$$ so the coefficient of $x^{2012}$ is 1

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I would rewrite this as $\frac{(1 + x)(1 + x)}{(1 + x^2)(1 - x^2)} = \frac{1 + 2x + x^2}{1 - x^4}$. Then the series for this is simply $(1 + 2x + x^2)\sum\limits_{n=0}^{\infty} x^{4n}$. Then the coefficient of $x^{2012}$ is clearly 1.

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  • $\begingroup$ That's a smart move... $\endgroup$ – imranfat Oct 28 '16 at 0:46

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