Editing Bernoulli polynomials (section)

==Differences and derivatives==

The Bernoulli and Euler polynomials obey many relations from [[umbral calculus]]:
<math display="block">\begin{align}
\Delta B_n(x) &= B_n(x+1)-B_n(x)=nx^{n-1}, \\[3mu]
\Delta E_n(x) &= E_n(x+1)-E_n(x)=2(x^n-E_n(x)).
\end{align}</math>
({{math|Δ}} is the [[forward difference operator]]). Also,
<math display="block"> E_n(x+1) + E_n(x) = 2x^n.</math>
These [[polynomial sequence]]s are [[Appell sequence]]s:
<math display="block">\begin{align}
B_n'(x) &= n B_{n-1}(x), \\[3mu]
E_n'(x) &= n E_{n-1}(x).
\end{align}</math>

===Translations===

<math display="block">\begin{align}
B_n(x+y) &= \sum_{k=0}^n {n \choose k} B_k(x) y^{n-k} \\[3mu]
E_n(x+y) &= \sum_{k=0}^n {n \choose k} E_k(x) y^{n-k}
\end{align}</math>
These identities are also equivalent to saying that these polynomial sequences are [[Appell sequence]]s. ([[Hermite polynomials]] are another example.)

===Symmetries===

<math display="block">\begin{align}
B_n(1-x) &= \left(-1\right)^n B_n(x), && n \ge 0, 
\text{ and in particular for } n \ne 1,~B_n(0) = B_n(1)\\[3mu]
E_n(1-x) &= \left(-1\right)^n E_n(x) \\[1ex]
\left(-1\right)^n B_n(-x) &= B_n(x) + nx^{n-1} \\[3mu]
\left(-1\right)^n E_n(-x) &= -E_n(x) + 2x^n \\[1ex]
B_n\bigl(\tfrac12\bigr) &= \left(\frac{1}{2^{n-1}}-1\right) B_n, && n \geq 0\text{ from the multiplication theorems below.}
\end{align} </math>
[[Zhi-Wei Sun]] and Hao Pan <ref>{{cite journal |author1=Zhi-Wei Sun |author2=Hao Pan |journal=Acta Arithmetica |volume=125 |year=2006 |pages=21–39 |title=Identities concerning Bernoulli and Euler polynomials  |issue=1 |arxiv=math/0409035 |doi=10.4064/aa125-1-3|bibcode=2006AcAri.125...21S |s2cid=10841415 }}</ref> established the following surprising symmetry relation: If {{math|1= ''r'' + ''s'' + ''t'' = ''n''}} and {{math|1= ''x'' + ''y'' + ''z'' = 1}}, then
<math display="block">r[s,t;x,y]_n+s[t,r;y,z]_n+t[r,s;z,x]_n=0,</math>
where
<math display="block">[s,t;x,y]_n=\sum_{k=0}^n(-1)^k{s \choose k}{t\choose {n-k}} B_{n-k}(x)B_k(y).</math>