# Taylor Series for $\dfrac{x}{x ^ 2 + x + 1}$.

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?

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

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 $$\dfrac{x}{x ^ 2 + x + 1}=x-x^2+x^4-x^5+\dots$$

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

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 $$0,1,-1,0,1,-1,0,1,-1,0,...$$

Hint

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

For $$-1

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

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

• Oct 15, 2019 at 18:37
• For this one, it is not needed. It would be if it had $1/(1-x)^3$. Oct 16, 2019 at 17:28
• @martycohen, Please find the updated post Oct 17, 2019 at 5:31