Find generating functions given a sequence, and vice versa?

Question

Given the sequence (0, 0, 0, 6, -6, 6, -6, 6...), how do I find the generating function?

Additionally, given a function like $ f(x)= \frac{(x^3)}{1-x^2} + 3x^7 - 11$ , how do I get the sequence of coefficients like the above? I am aware of the generalized binomial theorem for the form $ f(x) = (1+x)^n $ but I am not sure how to apply that to more complex functions.

Answer 1

Note that \begin{eqnarray*} \frac{1}{1-x^2} =1 +x^2+x^4+x^6+ \cdots \end{eqnarray*} Your first sequence (assuming the indexing starts at $0$) is \begin{eqnarray*} (0,0,0,6,-6,6,-6,\cdots) = 6x^3(1 +x^2+x^4+x^6+ \cdots )-6x^4(1 +x^2+x^4+x^6+ \cdots ) \\ =\frac{6x^3(1-x)}{1-x^2}= \frac{6x^3}{1+x}. \end{eqnarray*} Your second function $f(x)=\frac{x^3}{1-x^2} +\color{red}{3x^7}\color{blue}{-11}$ gives the sequence $(\color{blue}{-11},0,0,1,0,1,0,\color{red}{4},0,1,0,1,0,\cdots)$.

Answer 2

For a given sequence $a_n$ the generating function is given by

$$f(x)=\sum_{n=0}^{\infty}a_nx^n.$$

Sometimes you can find a closed form fir this.

Having the generating function like the example you gave, you will have to turn all functions to Taylor series representations. Add all coefficients and you have your $a_n$. E.g. using geometric series

$$f(x)=-11+3x^7+ x^3/(1-x^2)=-11+3x^7+x^3(1+x^2+x^4+...)$$ $$=-11x^3+x^5+4x^7+x^{9}+x^{11}+x^{13}+...$$