How does $\frac4{1-x^3}=\sum_{n\ge 0}4x^{3n}$ equal $4x^0+0x^1+0x^2+4x^3+0x^4+0x^5+4x^6+0x^7+0x^8+\ldots$?

Question

I've been searching through the internet and through SE to find something to help me understand generating functions, but I haven't found anything that would solve my problem with them.

I understand that

$$\frac1{1-x}=\sum_{n\ge 0}x^n\;,\tag{1}$$

gives the sequence $(1, 1, 1, 1,...) $ because $$\frac1{1-x}=\sum_{n\ge 0}x^n\;,\tag{1}$$

is just another way of writing the sum $1+x+x^2+x^3+x^4+...$ and the coefficients of each term are 1, and thus the sequence is $1, 1, 1, 1,...$.

What I don't get is how does

$$\begin{align*} \frac4{1-x^3}&=\sum_{n\ge 0}4x^{3n} \end{align*}\tag{3}$$ equal

$$4x^0+0x^1+0x^2+4x^3+0x^4+0x^5+4x^6+0x^7+0x^8+\ldots$$

and thus the sequence $(4, 0, 0, 4, 0, 0, 4, 0, 0,...)$.

I would say that $$\begin{align*} \frac4{1-x^3}&=\sum_{n\ge 0}4x^{3n} \end{align*}\tag{3}$$

is equal to $$4x^{3\times0}+4x^{3\times1}+4x^{3\times2}+4x^{3\times3}...=4+4x^3+4x^6+4x^9\ldots$$

There is something that I'm completely not understanding and I would like to know what that something is.

Answer 1

$$4x^0+0x^1+0x^2+4x^3+0x^4+0x^5+4x^6+0x^7+0x^8+4x^9\ldots$$

and

$$4+4x^3+4x^6+4x^9\ldots$$

are the same polynomial. Each has a $4$ term, each has a $4x^3$ term, et cetera.

So both can be represented by the sequence $(4, 0, 0, 4, 0, 0, 4, ...)$

The first term of the sequence is the coefficient of the constant term, the second term is the coefficient of the linear term, et cetera.

Answer 2

You are correct, $$ \frac{4}{1-x^3} = \sum_{n=0}^\infty 4x^{3n} = 4 + 4x^3 + 4x^6 + 4x^9 + \dots $$ But the formula you doubt is also right, since all other terms are zero.

Answer 3

You are correct in your thinking, but are missing a key point. $$4x^{3 \times 0} + 4x^{3 \times 1} + 4x^{3 \times 2} + 4x^{3 \times 3}... = 4x^{3 \times 0} + 0x^1+0x^2+ 4x^{3 \times 1} +0x^4+0x^5+ 4x^{3 \times 2} + 0x^7+0x^8+4x^{3 \times 3}$$ This is because adding 0 to an equation does not change the sum.

Answer 4

Take $\dfrac1{1-x}=1+x+x^2+x^3+x^4+...$, multiply by $4$ and replace $x$ by $x^3$ to get $\dfrac4{1-x^3}=4+4x^3+4x^6+4x^9+4x^{12}+...$. Is this unclear ?