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I'm trying to calculate the taylor series of $\dfrac{x}{x ^ 2 + x + 1}$.

Algebraic manipulation didn't get me anywhere, since the roots of $ x ^ 2 + x + 1 $ are complex.

Integrate or derive made the problem worse

Any tips on how to proceed?

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    $\begingroup$ Complex roots shouldn't faze one. But your function equals $\frac{x-x^2}{1-x^3}$. $\endgroup$ Oct 15, 2019 at 18:26
  • $\begingroup$ @gmn_1450 Observe that $(1+x+x^{2})^{-1}=1-(x+x^2)+(x+x^2)^{2}+...$, if $|x+x^{2}|<1$ $\endgroup$ Oct 15, 2019 at 18:26
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    $\begingroup$ Partial fractions. $\endgroup$ Oct 15, 2019 at 18:33
  • $\begingroup$ Taylor series can be centered lots of places. Do you mean a Taylor series centered at $0$? $\endgroup$ Oct 15, 2019 at 19:38

4 Answers 4

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There seem to be many ways to go about this, so here is one: put $\dfrac{x}{x ^ 2 + x + 1}=\sum_{n=0}^{\infty} a_nx^n$, then $$x=\sum_{n=0}^{\infty} a_nx^n(x^2+x+1)=a_0+(a_1+a_0)x+\sum_{n=2}^{\infty}(a_n+a_{n-1}+a_{n-2})x^n,$$ and by comparing the coefficients we get $a_0=0$, $a_1=1$, and $a_n+a_{n-1}+a_{n-2}=0$ for $n \geq 2$. You can see the coefficients repeat ($a_2=-1,a_3=0,a_4=1,\dots$), so we have $a_{3k}=0, a_{3k+1}=1, a_{3k+2}=-1$, or in other words $$ \dfrac{x}{x ^ 2 + x + 1}=x-x^2+x^4-x^5+\dots $$

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$\begin{array}\\ \dfrac{x}{x ^ 2 + x + 1} &=\dfrac{x(1-x)}{1-x^3}\\ &=(x-x^2)\sum_{k=0}^{\infty} x^{3k}\\ &=\sum_{k=0}^{\infty} x^{3k+1}-\sum_{k=0}^{\infty} x^{3k+2}\\ \end{array} $

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Suppose we have the recurrence $$q_{n+2}=aq_{n+1}+bq_n\qquad n\ge0$$ where $q_0,q_1$ are given. We then define the generating function for this sequence, $$q(x)=\sum_{n\ge0}q_nx^n.$$ We see that $$q_{n+2}x^{n+2}=ax\cdot q_{n+1}x^{n+1}+bx^2\cdot q_nx^n,$$ so that, upon applying $\sum_{n\ge0}$ to both sides, $$q(x)-q_1x-q_0=ax(q(x)-q_0)+bx^2q(x)$$ and $$q(x)=\frac{(q_1-aq_0)x+q_0}{1-ax-bx^2}.$$ We apply this to the problem in question ($a=b=-1$), and get $$\frac{x}{x^2+x+1}=\sum_{n\ge0}q_nx^n,$$ where $$q_{n+2}+q_{n+1}+q_{n}=0\qquad q_0=0,q_1=1.$$ The first few values for $q_n$ are $0,1,-1,0,1,-1,0,1,-1,0,...$

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Hint

$$(1-x)(1+x+x^2)=?$$

For $-1<x<1,$

$$(1+x+x^2)^{-1}=(1-x)(1-x^3)^{-1}=?$$

Using Binomial Series/ Infinite Geometric Series, $$\dfrac x{1+x+x^2}=x(1-x)(1-x^3)^{-1}=(x-x^2)\sum_{r=0}^\infty(x^3)^r=\sum_{r=0}^\infty(x^{3r+1}- x^{3r+2})$$

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