(This post is mostly intended for my own reference, as I found myself repeatedly looking up several conversions between polynomial bases on various occasions.)

Let ${\mathrm{Poly}_{\leq n}}$ denote the vector space of polynomials ${P:{\bf R} \rightarrow {\bf R}}$ of one variable ${x}$ with real coefficients of degree at most ${n}$. This is a vector space of dimension ${n+1}$, and the sequence of these spaces form a filtration:

$\displaystyle \mathrm{Poly}_{\leq 0} \subset \mathrm{Poly}_{\leq 1} \subset \mathrm{Poly}_{\leq 2} \subset \dots$

A standard basis for these vector spaces are given by the monomials ${x^0, x^1, x^2, \dots}$: every polynomial ${P(x)}$ in ${\mathrm{Poly}_{\leq n}}$ can be expressed uniquely as a linear combination of the first ${n+1}$ monomials ${x^0, x^1, \dots, x^n}$. More generally, if one has any sequence ${Q_0(x), Q_1(x), Q_2(x)}$ of polynomials, with each ${Q_n}$ of degree exactly ${n}$, then an easy induction shows that ${Q_0(x),\dots,Q_n(x)}$ forms a basis for ${\mathrm{Poly}_{\leq n}}$.

In particular, if we have two such sequences ${Q_0(x), Q_1(x), Q_2(x),\dots}$ and ${R_0(x), R_1(x), R_2(x), \dots}$ of polynomials, with each ${Q_n}$ of degree ${n}$ and each ${R_k}$ of degree ${k}$, then ${Q_n}$ must be expressible uniquely as a linear combination of the polynomials ${R_0,R_1,\dots,R_n}$, thus we have an identity of the form

$\displaystyle Q_n(x) = \sum_{k=0}^n c_{QR}(n,k) R_k(x)$

for some change of basis coefficients ${c_{QR}(n,k) \in {\bf R}}$. These coefficients describe how to convert a polynomial expressed in the ${Q_n}$ basis into a polynomial expressed in the ${R_k}$ basis.

Many standard combinatorial quantities ${c(n,k)}$ involving two natural numbers ${0 \leq k \leq n}$ can be interpreted as such change of basis coefficients. The most familiar example are the binomial coefficients ${\binom{n}{k}}$, which measures the conversion from the shifted monomial basis ${(x+1)^n}$ to the monomial basis ${x^k}$, thanks to (a special case of) the binomial formula:

$\displaystyle (x+1)^n = \sum_{k=0}^n \binom{n}{k} x^k,$

thus for instance

$\displaystyle (x+1)^3 = \binom{3}{0} x^0 + \binom{3}{1} x^1 + \binom{3}{2} x^2 + \binom{3}{3} x^3$

$\displaystyle = 1 + 3x + 3x^2 + x^3.$

More generally, for any shift ${h}$, the conversion from ${(x+h)^n}$ to ${x^k}$ is measured by the coefficients ${h^{n-k} \binom{n}{k}}$, thanks to the general case of the binomial formula.

But there are other bases of interest too. For instance if one uses the falling factorial basis

$\displaystyle (x)_n := x (x-1) \dots (x-n+1)$

then the conversion from falling factorials to monomials is given by the Stirling numbers of the first kind ${s(n,k)}$:

$\displaystyle (x)_n = \sum_{k=0}^n s(n,k) x^k,$

thus for instance

$\displaystyle (x)_3 = s(3,0) x^0 + s(3,1) x^1 + s(3,2) x^2 + s(3,3) x^3$

$\displaystyle = 0 + 2 x - 3x^2 + x^3$

and the conversion back is given by the Stirling numbers of the second kind ${S(n,k)}$:

$\displaystyle x^n = \sum_{k=0}^n S(n,k) (x)_k$

thus for instance

$\displaystyle x^3 = S(3,0) (x)_0 + S(3,1) (x)_1 + S(3,2) (x)_2 + S(3,3) (x)_3$

$\displaystyle = 0 + x + 3 x(x-1) + x(x-1)(x-2).$

If one uses the binomial functions ${\binom{x}{n} = \frac{1}{n!} (x)_n}$ as a basis instead of the falling factorials, one of course can rewrite these conversions as

$\displaystyle \binom{x}{n} = \sum_{k=0}^n \frac{1}{n!} s(n,k) x^k$

and

$\displaystyle x^n = \sum_{k=0}^n k! S(n,k) \binom{x}{k}$

thus for instance

$\displaystyle \binom{x}{3} = 0 + \frac{1}{3} x - \frac{1}{2} x^2 + \frac{1}{6} x^3$

and

$\displaystyle x^3 = 0 + \binom{x}{1} + 6 \binom{x}{2} + 6 \binom{x}{3}.$

As a slight variant, if one instead uses rising factorials

$\displaystyle (x)^n := x (x+1) \dots (x+n-1)$

then the conversion to monomials yields the unsigned Stirling numbers ${|s(n,k)|}$ of the first kind:

$\displaystyle (x)^n = \sum_{k=0}^n |s(n,k)| x^k$

thus for instance

$\displaystyle (x)^3 = 0 + 2x + 3x^2 + x^3.$

One final basis comes from the polylogarithm functions

$\displaystyle \mathrm{Li}_{-n}(x) := \sum_{j=1}^\infty j^n x^j.$

For instance one has

$\displaystyle \mathrm{Li}_1(x) = -\log(1-x)$

$\displaystyle \mathrm{Li}_0(x) = \frac{x}{1-x}$

$\displaystyle \mathrm{Li}_{-1}(x) = \frac{x}{(1-x)^2}$

$\displaystyle \mathrm{Li}_{-2}(x) = \frac{x}{(1-x)^3} (1+x)$

$\displaystyle \mathrm{Li}_{-3}(x) = \frac{x}{(1-x)^4} (1+4x+x^2)$

$\displaystyle \mathrm{Li}_{-4}(x) = \frac{x}{(1-x)^5} (1+11x+11x^2+x^3)$

and more generally one has

$\displaystyle \mathrm{Li}_{-n-1}(x) = \frac{x}{(1-x)^{n+2}} E_n(x)$

for all natural numbers ${n}$ and some polynomial ${E_n}$ of degree ${n}$ (the Eulerian polynomials), which when converted to the monomial basis yields the (shifted) Eulerian numbers

$\displaystyle E_n(x) = \sum_{k=0}^n A(n+1,k) x^k.$

For instance

$\displaystyle E_3(x) = A(4,0) x^0 + A(4,1) x^1 + A(4,2) x^2 + A(4,3) x^3$

$\displaystyle = 1 + 11x + 11x^2 + x^3.$

These particular coefficients also have useful combinatorial interpretations. For instance:

• The binomial coefficient ${\binom{n}{k}}$ is of course the number of ${k}$-element subsets of ${\{1,\dots,n\}}$.
• The unsigned Stirling numbers ${|s(n,k)|}$ of the first kind are the number of permutations of ${\{1,\dots,n\}}$ with exactly ${k}$ cycles. The signed Stirling numbers ${s(n,k)}$ are then given by the formula ${s(n,k) = (-1)^{n-k} |s(n,k)|}$.
• The Stirling numbers ${S(n,k)}$ of the second kind are the number of ways to partition ${\{1,\dots,n\}}$ into ${k}$ non-empty subsets.
• The Eulerian numbers ${A(n,k)}$ are the number of permutations of ${\{1,\dots,n\}}$ with exactly ${k}$ ascents.

These coefficients behave similarly to each other in several ways. For instance, the binomial coefficients ${\binom{n}{k}}$ obey the well known Pascal identity

$\displaystyle \binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$

(with the convention that ${\binom{n}{k}}$ vanishes outside of the range ${0 \leq k \leq n}$). In a similar spirit, the unsigned Stirling numbers ${|s(n,k)|}$ of the first kind obey the identity

$\displaystyle |s(n+1,k)| = n |s(n,k)| + |s(n,k-1)|$

and the signed counterparts ${s(n,k)}$ obey the identity

$\displaystyle s(n+1,k) = -n s(n,k) + s(n,k-1).$

The Stirling numbers of the second kind ${S(n,k)}$ obey the identity

$\displaystyle S(n+1,k) = k S(n,k) + S(n,k-1)$

and the Eulerian numbers ${A(n,k)}$ obey the identity

$\displaystyle A(n+1,k) = (k+1) A(n,k) + (n-k+1) A(n,k-1).$