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  "body": "\n## Description\n\nThe Bernoulli polynomials are the polynomials $B_n(x) \\in \\Q[x]$, $n \\ge 0$, given by the generating function\n$$\\frac{t \\, e^{x t}}{e^t - 1} = \\sum_{n=0}^{\\infty} B_n(x) \\frac{t^n}{n!}.$$\nEquivalently, $B_n(x) = \\sum_{k=0}^{n} \\binom{n}{k} B_k \\, x^{n-k}$, where $B_k$ are the [[Bernoulli numbers|0KC00K7KY4]] with $B_1 = -\\frac{1}{2}$.\n\n## Remarks\n\nThe first values are $B_0(x) = 1$, $B_1(x) = x - \\frac{1}{2}$, $B_2(x) = x^2 - x + \\frac{1}{6}$ and $B_3(x) = x^3 - \\frac{3}{2} x^2 + \\frac{1}{2} x$.\nEach $B_n$ is monic of degree $n$, and $B_n(0) = B_n$.\nThey satisfy $B_n'(x) = n B_{n-1}(x)$ and $B_n(x + 1) - B_n(x) = n x^{n-1}$ for $n \\ge 1$.\nSumming the second identity gives the sums of powers: for $p \\ge 0$ and $m \\ge 1$,\n$$\\sum_{k=0}^{m-1} k^p = \\frac{B_{p+1}(m) - B_{p+1}(0)}{p + 1}.$$\nCheck with $p = 1$, $m = 4$: $\\frac{B_2(4) - B_2(0)}{2} = \\frac{12}{2} = 6 = 0 + 1 + 2 + 3$.\nThey also satisfy $B_n(1 - x) = (-1)^n B_n(x)$.\n\n## Explanations\n\n### L1 human\n\nThe Bernoulli polynomials are a family of polynomials that turn sums of powers into polynomial formulas.\nTheir values at $0$ are the Bernoulli numbers.\nThe difference $B_n(x + 1) - B_n(x)$ is the simple monomial $n x^{n-1}$, which is what makes the sums telescope.\n",
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