How do we meaningfully interpret $\sum_{k = 0}^n x^k = \frac{1 - x^{n+ 1}}{1 - x}$ in the context of combinatorics?

Question

Consider the generating function $$\sum_{k = 0}^n x^k = 1 + x^2 + x^3 + \; ... + \; x^n \tag{1}$$

The coefficient of $x^k$ in $(1)$ represents the number of ways to get $k$. For example, for $n = 6$, $a_{k >0}$ could represent the number of ways to get $k$ in a single dice roll. This provides a direct way to relate the generating function to a combinatorial problem. Furthermore, it preserves the addition and multiplication properties (for more dice roll) in combinatorics.

We can now algebraically demonstrate that $$(1) = \frac{1 - x^{n + 1}}{1 - x} \tag{2}$$

This representation seems to distance itself away from the combinatorial equivalence of coefficients and solutions. The involvement of the $(1 - x)$ term in the denominator seems to suggest that some kind of division is going on.

Is there a way to meaningfully interpret $(2)$ such that we preserve the combinatorial representation that we started with?

Answer 1

Since $1-x^{n+1}$ has a negative coefficient, it is not possible to interpret it as a generating function whose coefficients directly count things. It is however possible to rearrange the formula:

$$ (1+x+\cdots+x^n)\times\frac{1}{1-x^{n+1}}=\frac{1}{1-x} $$

which, as power series, is

$$ (1+x+\cdots+x^n)(1+x^{n+1}+x^{2(n+1)}+\cdots)=1+x+x^2+\cdots. $$

For the first generating function, the coefficient of $x^k$ counts how many numbers there are in $\{0,\cdots,n\}$ which are equal to $k$. For the second generating function, the coefficient of $x^k$ counts how many multiples of $n+1$ are equal to $k$. Therefore, the coefficient of $x^k$ in their product ought to count how many pairs $(a,b)$ there are with $a\in\{0,\cdots,n\}$, $b\in\{0,n+1,2(n+1),\cdots\}$ and $a+b=k$. Which is $1$ for each $k$.

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There is a way to interpret negative coefficients combinatorially: they amount to uncounting. So whoever is in charge to counting things is told certain rules about how they add or subtract to the tally depending on which object they're looking at. As a result, there can be negative tallies. (You can interpret inclusion-exclusion this way, since overcounts need to be corrected with uncounting to get correct counts.)

Here, $1-x^{n+1}$ counts $0$ once and $n+1$ minus one times. As a result, the coefficient of $x^k$ in the expansion of $(1-x^{n+1})(1+x+x^2+\cdots)$ ought to tally ordered pairs $(a,b)$ with $a+b=k$ once when $a=0$, minus once when $a=n+1$, and zero times when $a\ne0,n+1$. For any $k\ge n+1$, it is counted and uncounted exactly once each, which cancel each other out, whereas for $k<n+1$ the tally is just one. Thus the result of the expansion is $1+x+\cdots+x^n$.

Answer 2

We need to get some new interpretation first, let $x^n$ be understood as a 'list of n elements' , so for instance, $x^0$ is a list of no elements. Now, let $L$ be all finite lists, we have:

$$L = x^0 + xL \tag{1}$$

What does addition mean here? You can understand it as an 'or'. Let me say the above equation in English :

"All finite list is empty or has a first entry"

Here

1. $L$-> finite list

2. is -> $=$

3. empty list = $x^0$

4. has a first entry--> an x followed by another list.

Now let's rearrange (1):

$$ L-xL = x^0$$

$$ L = \frac{1}{1-x} = 1+x+x^2....$$

But what does that mean? You could read it as saying, any finite list contains either no items or one item or two items or three or.... and so on.

Take from Professor Ghrist's original calculus course. I've yet to find where he got the idea from but nonetheless I thought ti was profound. You can also check out this question I posted.