Editing Stirling number (section)

==As change of basis coefficients==
Considering the set of [[polynomial]]s in the (indeterminate) variable ''x'' as a vector space,
each of the three sequences
:<math>x^0,x^1,x^2,x^3,\dots \quad (x)_0,(x)_1,(x)_2,\dots \quad x^{(0)},x^{(1)},x^{(2)},\dots</math>
is a [[Basis (linear algebra)|basis]].
That is, every polynomial in ''x'' can be written as a sum <math>a_0 x^{(0)} + a_1 x^{(1)} + \dots + a_n x^{(n)}</math> for some unique coefficients <math>a_i</math> (similarly for the other two bases).
The above relations then express the [[change of basis]] between them, as summarized in the following [[commutative diagram]]:
: {{Dark mode invert|[[File:Stirling numbers as polynomial basis changes.png|center|500x500px|A diagram of how different [[Stirling number|Stirling numbers]] give coefficients for changing one basis of polynomials to another]]}}
The coefficients for the two bottom changes are described by the [[#Lah numbers|Lah numbers]] below.
Since coefficients in any basis are unique, one can define Stirling numbers this way, as the coefficients expressing polynomials of one basis in terms of another, that is, the unique numbers relating <math>x^n</math> with falling and rising factorials as above.

Falling factorials define, up to scaling, the same polynomials as [[Binomial coefficient#Binomial coefficients as polynomials|binomial coefficients]]: <math display="inline">\binom{x}{k} = (x)_k/k!</math>. The changes between the standard basis <math>\textstyle x^0, x^1, x^2, \dots</math> and the basis <math display="inline">\binom{x}{0}, \binom{x}{1}, \binom{x}{2}, \dots</math> are thus described by similar formulas:
: <math>x^n=\sum_{k=0}^n \biggl\{{\!n\! \atop \!k\!}\biggr\} k! \binom{x}{k} \quad \text{and} \quad \binom{x}{n}=\sum_{k=0}^n \frac{s(n,k)}{n!} x^k</math>.

===Example===
Expressing a polynomial in the basis of falling factorials is useful for calculating sums of the polynomial evaluated at consecutive integers.
Indeed, the sum of falling factorials with fixed ''k'' can expressed as another falling factorial (for <math>k\ne-1</math>)

:<math>\sum_{0\leq i < n} (i)_k = \frac{(n)_{k+1}}{k+1} </math>

This can be proved by [[mathematical induction|induction]].

For example, the sum of fourth powers of integers up to ''n'' (this time with ''n'' included), is:

:<math>\begin{align}
\sum_{i=0}^{n} i^4 & = \sum_{i=0}^n \sum_{k=0}^4 \biggl\{{\!4\! \atop \!k\!}\biggr\} (i)_k = \sum_{k=0}^4 \biggl\{{\!4\! \atop \!k\!}\biggr\} \sum_{i=0}^n (i)_k =  \sum_{k=0}^4 \biggl\{{\!4\! \atop \!k\!}\biggr\} \frac{(n{+}1)_{k+1}}{k{+}1} \\[10mu]
& = \biggl\{{\!4\! \atop \!1\!}\biggr\} \frac{(n{+}1)_{2}}2
  + \biggl\{{\!4\! \atop \!2\!}\biggr\} \frac{(n{+}1)_{3}}3
  + \biggl\{{\!4\! \atop \!3\!}\biggr\} \frac{(n{+}1)_{4}}4
  + \biggl\{{\!4\! \atop \!4\!}\biggr\} \frac{(n{+}1)_{5}}5 \\[8mu]
& = \frac12 (n{+}1)_{2} + \frac73 (n{+}1)_{3} + \frac64 (n{+}1)_{4} + \frac15 (n{+}1)_{5}\,.
\end{align}</math>

Here the Stirling numbers can be computed from their definition as the number of partitions of 4 elements into ''k'' non-empty unlabeled subsets.

In contrast, the sum <math display="inline">\sum_{i=0}^n i^k</math> in the standard basis is given by [[Faulhaber's formula]], which in general is more complicated.