Polynomials and Infinite products

Polynomials are undoubtedly a beautiful representation of abstraction by humans. Here we restrict our explorations in polynomials that carry integral powers/coefficients. Here are few simpler polynomials and their factorizations.

Some examples

$1. \text{ } f(x) = 1 + x + x^2 + x^3 = (1 + x)*(1 + x^2) \\ 2. \text{ } f(x) = 1 + 2x + 3x^2 + 6x^3 + 5x^4 + 10x^5 = (1 + 2x)(1 + 3x^2 + 5x^4) \\ 3. \text{ } f(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 = (1 + x)(1 + x^2)(1 + x^4)$

For now, leaving all the rules of convergence, that play a tricky role of deciding the expansions, we take it for granted that some kind of convergence rule would hold and go on to investigate on the patterns of expansions in these polynomials.

Standard ones

Let us consider the standard ones, that comes from the binomial expansion of the simplest of the polynomial of degree 1.

$f(x, n) = (1 + x)^n = \sum_{0\le r \le n }C(n, r)x^{n-r}$

With some small change, let us replace one of them with $(1-x)$ to see how the coefficients changes:

$f(x, n) = (1 + x)^{n-1}(1-x) = 1 + \sum_{1\le r \le {n-1}}$C(n,r+1)-C(n,r)$x^{n-r} -x^n$

Let us investigate on the following pattern.

Let $p + q = n$ . Then, the coefficients of the following polynomial could be found.

Some Generalization

$f(x,n) = (1 + x)^p(1-x)^q \\ \\ \text{Coefficient of $x^k$ is given by, } \sum_{0\le i \le p, 0\le j \le q \\i+j = p+q-k}C(p, i)C(q, j)(-1)^j$