For example in my textbook it says that the sequence for the generating function
$$f(x) = (2x-3)^3$$
is
$$-27,54,-36,8,0,0,0,\dots$$
I can see that plugging $0$ into the function gives me $-27$, but this doesn't hold true for $1$ and $54$. How do I get this sequence or the correct sequence for another generating function like
$$f(x) = \frac{x^3}{1-x^2}\;?$$
On
You are looking for a sequence $a_n$ with $$f(x) = \sum_{k=0}^{\infty}a_nx^k$$
Now $f(x) = (2x-3)^3 = -27x^0+54x^1-36x^2+8x^3+0x^4+0x^5+\dots$
so the generated sequence is $\{-27,54,-36,8,0,0,\dots\}$
On
In general, if a function can be represented as an infinite series $$f(x)=\sum_{n=0}^\infty a_n x^n = a_0+a_1x+a_2x^2+\cdots,$$ then it is said to be a generating function for the sequence $$(a_n) = a_0, a_1, a_2, \dots$$
For your first example, we can write $$f(x)=(2x-3)^3 = -27 + 54x - 36x^2 + 8x^3+0.x^4+0.x^5+\cdots$$ and so it generates the sequence $$-27, 54, -36, 8, 0, 0, 0, \dots$$
On
Let $f(x)=(2x-3)^3$; then after multiplying out we have $f(x)=-27+54x-36x^2+8x^3$, so if we write $f$ as a power series, $$f(x)=\sum_{k\ge 0}a_kx^k\;,$$ we must have $a_0=-27,a_1=54,a_2=-36,a_3=8$, and $a_k=0$ for $k\ge 4$.
For the second problem you have $$f(x)=\frac{x^3}{1-x^2}\;.$$ Deal with the denominator first. You should know the simple generating function that gives the sum of a geometric series:
$$\frac1{1-x}=\sum_{k\ge 0}x^k\;.$$ Replace $x$ by $x^2$, and you have
$$\frac1{1-x^2}=\sum_{k\ge 0}\left(x^2\right)^k=\sum_{k\ge 0}x^{2k}\;.$$
Finally, multiply by $x^3$:
$$\frac{x^3}{1-x^2}=x^3\sum_{k\ge 0}x^{2k}=\sum_{k\ge 0}x^{2k+3}\;.$$
Think of this now as $\sum_{k\ge 0}a_kx^k$; what are the coefficients $a_k$? It may help to write out a few terms of this sum:
$$\frac{x^3}{1-x^2}=x^3+x^5+x^7+x^9+\ldots\;.$$
Clearly $a_k=0$ when $k$ is even and when $k=1$, and $a_k=1$ when $k$ is an odd number greater than $1$. Thus, the associated sequence is $$\langle 0,0,0,1,0,1,0,1,0,\dots\rangle\;.$$
For the first example, these are the cofficents of the polynomial.
For the second example, $1/(1-x^2)=1+x^2+x^4+.....$, therefore $x^3/(1-x^2)=x^3+x^5+x^7+....$, thus the generating sequence is : 0,0,0,1,0,1,0,1,0,...