Editing Stirling number (section)

==Inversion relations and the Stirling transform==

For any pair of sequences, <math>\{f_n\}</math> and <math>\{g_n\}</math>, related by a finite sum Stirling number formula given by

:<math>g_n = \sum_{k=0}^{n} \left\{\begin{matrix} n \\ k \end{matrix} \right\} f_k, </math>

for all integers <math>n \geq 0</math>, we have a corresponding [[generating function transformation#Inversion relations and generating function identities|inversion formula]] for <math>f_n</math> given by

:<math>f_n = \sum_{k=0}^{n} \left[\begin{matrix} n \\ k \end{matrix} \right] (-1)^{n-k} g_k. </math>

The lower indices could be any integer between <math display="inline">0</math> and <math display="inline">n</math>.

These inversion relations between the two sequences translate into functional equations between the sequence [[generating function|exponential generating functions]] given by the [[Stirling transform|Stirling (generating function) transform]] as

:<math>\widehat{G}(z) = \widehat{F}\left(e^z-1\right)</math>

and

:<math>\widehat{F}(z) = \widehat{G}\left(\log(1+z)\right). </math>

For <math>D = d/dx</math>, the [[differential operators]] <math>x^nD^n</math> and <math>(xD)^n</math> are related by the following formulas for all integers <math>n \geq 0</math>:<ref>''Concrete Mathematics'' exercise 13 of section 6. Note that this formula immediately implies the first positive-order Stirling number transformation given in the main article on [[generating function transformation#Derivative transformations|generating function transformations]].</ref>

:<math>
\begin{align}
(xD)^n &= \sum_{k=0}^n S(n, k) x^k D^k \\
x^n D^n &= \sum_{k=0}^n s(n, k) (xD)^k = (xD)_n = xD(xD - 1)\ldots (xD - n + 1)
\end{align}
</math>

Another pair of "''inversion''" relations involving the [[Stirling numbers]] relate the [[finite difference|forward differences]] and the ordinary <math>n^{th}</math> [[derivative]]s of a function, <math>f(x)</math>, which is analytic for all <math>x</math> by the formulas<ref>{{cite journal|last1=Olver|first1=Frank|first2=Daniel|last2=Lozier|first3=Ronald|last3=Boisvert|first4=Charles|last4=Clark|title=NIST Handbook of Mathematical Functions|journal=NIST Handbook of Mathematical Functions|date=2010|url=https://www.nist.gov/publications/nist-handbook-mathematical-functions}} (Section 26.8)</ref>

:<math>\frac{1}{k!} \frac{d^k}{dx^k} f(x) = \sum_{n=k}^{\infty} \frac{s(n, k)}{n!} \Delta^n f(x)</math>

:<math>\frac{1}{k!} \Delta^k f(x) = \sum_{n=k}^{\infty} \frac{S(n, k)}{n!} \frac{d^n}{dx^n} f(x). </math>