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For the given generating function $A(z)=1/(1-x^3)$ what will be the numeric function.. I can find out for $A(z)=1/(1-x^2)$ but in the first case we get two factor in the denominator $(1-x)$ and $(x^2+x+1)$, now how do we find the numeric function for $1/(x^2+x+1)$?

Teddy38
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shadow kh
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  • What do you mean by numeric function? Given any $x$ you can compute it. Are you looking for the terms of the Taylor series at zero? Are you looking for the partial fraction decomposition? What is the numeric function for $\frac 1{1-x^2}$? That example might help understand what you are looking for? – Ross Millikan Nov 10 '17 at 17:13

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We don't need to factorize the denominator. Let's recall the geometric series expansion. \begin{align*} \frac{1}{1-y}=1+y+y^2+y^3+\cdots\qquad\qquad |y|<1 \end{align*}

We obtain with $y=x^3$ \begin{align*} \frac{1}{1-x^3}=1+x^3+x^6+\cdots\qquad\qquad |x|<1 \end{align*} and the coefficients are $1$ if the powers of $x$ are congruent $0\,\textrm{mod}(3)$ and $0$ otherwise.

Markus Scheuer
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