Bernoulli polynomials

Remarks

The first values are $B_0(x) = 1$, $B_1(x) = x - \frac{1}{2}$, $B_2(x) = x^2 - x + \frac{1}{6}$ and $B_3(x) = x^3 - \frac{3}{2} x^2 + \frac{1}{2} x$. Each $B_n$ is monic of degree $n$, and $B_n(0) = B_n$. They satisfy $B_n'(x) = n B_{n-1}(x)$ and $B_n(x + 1) - B_n(x) = n x^{n-1}$ for $n \ge 1$. Summing the second identity gives the sums of powers: for $p \ge 0$ and $m \ge 1$, $$\sum_{k=0}^{m-1} k^p = \frac{B_{p+1}(m) - B_{p+1}(0)}{p + 1}.$$ Check with $p = 1$, $m = 4$: $\frac{B_2(4) - B_2(0)}{2} = \frac{12}{2} = 6 = 0 + 1 + 2 + 3$. They also satisfy $B_n(1 - x) = (-1)^n B_n(x)$.