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  "body": "\n## Definition\n\nLet $f \\colon \\R \\times \\R^n \\to \\R^n$, and consider the [[initial value problem|0FXDW992DS]] $y' = f(t, y)$, $y(t_0) = y_0$, with step size $h \u003e 0$ and $t_j = t_0 + j h$.\nLet $k \\ge 1$, and suppose that $y_{j-k+1}, \\ldots, y_j$ are known.\nLet $P$ be the [[interpolation polynomial|SS1YMS2NPC]] of degree at most $k$ with $P(t_{j+1-i}) = y_{j+1-i}$ for $i = 0, \\ldots, k$.\nThe *$k$-step backward differentiation formula* determines $y_{j+1}$ by the condition\n$$P'(t_{j+1}) = f(t_{j+1}, y_{j+1}).$$\nWith the backward difference operator $\\nabla y_{j+1} = y_{j+1} - y_j$, this condition reads\n$$\\sum_{i=1}^{k} \\frac{1}{i} \\nabla^i y_{j+1} = h f(t_{j+1}, y_{j+1}).$$\n\n## Remarks\n\nIt is an implicit [[linear multistep method|BX4D85A79P]] in which only $\\beta_k \\ne 0$.\nExample: for $k = 1$ the formula is $y_{j+1} - y_j = h f(t_{j+1}, y_{j+1})$, the backward Euler method.\nFor $k = 2$, $\\nabla y_{j+1} + \\frac12 \\nabla^2 y_{j+1} = \\frac32 y_{j+1} - 2 y_j + \\frac12 y_{j-1}$, so the formula is $y_{j+1} - \\frac43 y_j + \\frac13 y_{j-1} = \\frac23 h f(t_{j+1}, y_{j+1})$.\nNon-example: the Adams–Bashforth formula $y_{j+1} - y_j = h \\left( \\frac32 f_j - \\frac12 f_{j-1} \\right)$ is a linear multistep method but not a BDF, since it uses $f$ at old points and not at $t_{j+1}$.\nThe $k$-step formula has order $k$; it is zero-stable for $1 \\le k \\le 6$ and not zero-stable for $k \\ge 7$; the formulas with $k = 1, 2$ are A-stable.\n",
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