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  "body": "\n## Definition\n\nThe *Euler polynomials* $E_k(x)$, for $k \\ge 0$, are defined by the generating function\n$$\\frac{2 e^{xt}}{e^t + 1} = \\sum_{k \\ge 0} E_k(x) \\frac{t^k}{k!}.$$\nThe *periodic Euler functions* $\\widetilde{E}_k$ are the functions on $\\R$ with $\\widetilde{E}_k(x) = E_k(x)$ for $0 \\le x \u003c 1$ and $\\widetilde{E}_k(x + 1) = -\\widetilde{E}_k(x)$ for all $x$.\n\nLet $m \\ge 1$ and $n \\ge 1$ be integers, let $0 \\le h \\le 1$, and let $f$ be a [[smooth function|0BE3QDH73A]] on an open interval that contains $[0, n]$.\n*Euler–Boole summation* is the identity\n$$\\sum_{j=0}^{n-1} (-1)^j f(j + h) = \\frac12 \\sum_{k=0}^{m-1} \\frac{E_k(h)}{k!} \\left( (-1)^{n-1} f^{(k)}(n) + f^{(k)}(0) \\right) + \\frac{1}{2 (m-1)!} \\int_0^n f^{(m)}(x) \\, \\widetilde{E}_{m-1}(h - x) \\, dx,$$\nand the finite sum over $k$ is the Euler–Boole expansion of the alternating sum to order $m$.\n\n## Remarks\n\nThe first Euler polynomials are $E_0(x) = 1$ and $E_1(x) = x - \\frac12$.\nThey satisfy $E_k(x + 1) + E_k(x) = 2x^k$, which is the source of the formula.\n\nExample: take $f(x) = x$, $h = 0$, $n = 2$ and $m = 2$.\nThe left side is $0 - 1 = -1$.\nThe term $k = 0$ gives $\\frac12 (-f(2) + f(0)) = -1$, the term $k = 1$ gives $\\frac12 \\cdot (-\\frac12) \\cdot (-f'(2) + f'(0)) = 0$, and the integral vanishes because $f'' = 0$.\nBoth sides equal $-1$.\n\nNon-example: the identity is for alternating sums.\nFor the plain sum $\\sum_{j=0}^{n-1} f(j)$ the right side is wrong in general, and the Euler–Maclaurin formula applies instead.\n",
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