2

I used partial fractions but the obtained formula is only correct for the first two elements.

$\dfrac{x}{(1+x+x^2)}=\dfrac{x}{(1+a_1x)(1+a_2x)}=\dfrac{A_1}{(1+a_1x)+A_2(1+a_2x)}$

$x=\dfrac{-1 \pm\sqrt3i}{2}$

Then let $a_1=\dfrac{-1+\sqrt3i}{2}$ and $a_2=\dfrac{-1-\sqrt3i}{2}$

and then calculated $A_1$ and $A_2$ to obtain the formula

$a_n=\dfrac{(-1)^n}{\sqrt3i} \cdot \left(\left(\dfrac{-1-\sqrt3i}{2}\right)^n-\left(\dfrac{-1+\sqrt3i}{2}\right)^n\right)$

The first three elements become $0$, $1$, $1$, but the answer from the Taylor series is $0$, $1$, $-1$.

Ken
  • 3,751
jjstef
  • 21
  • It should be $0,1,-1$. To simplify calculations, let $\zeta=\dfrac{-1+\sqrt3i}2$. Note that $\zeta^2=\dfrac{-1-\sqrt3i}2$, $\zeta^3=1$, and $\zeta+\zeta^2=-1$. – Akiva Weinberger Jun 05 '15 at 02:25
  • $\zeta$ is called a third root of unity, because $\zeta^3=1$. ("Unity" meaning "one.") There are three third roots of unity; the other two are $\zeta^2$ and $1$. In the complex plane, these form the vertices of an equilateral triangle. – Akiva Weinberger Jun 05 '15 at 02:27

1 Answers1

1

Partial fractions are not the most efficient way here.

Our expression is $\frac{x(1-x)}{1-x^3}$. Expand $\frac{1}{1-x^3}$ using $\frac{1}{1-t}=1+t+t^2+t^3+t^4+\cdots$.

André Nicolas
  • 507,029
  • Didn't see initial simplification of the function denominator. Thank you. Power series obtained gives coefficients 0, 1, -1, 0, 1, -1, ... which agree from Taylor series expansion. If the denominator of a function cannot be factored, should partial fractions ever be considered. Also, the formula I developed using partial fractions appears to incorrectly predict a +1 for the 3rd coefficient. Can't find the error. – jjstef Jun 05 '15 at 16:41
  • This was a very special case, where the denominator can be made very nice by multiplying by a suitable factor. The idea will work for denominators $1+x+\cdots+x^k$, and close relatives, but not in general. In the comments there is a suggestion for how to do the computation more smoothly. That will help to avoid mechanical errors. – André Nicolas Jun 05 '15 at 16:52