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  "body": "\n## Definition\n\nLet $n \\ge 1$ and let $P_0, P_1, \\dots, P_n$ be points of $\\R^d$, the control points.\nThe *Bézier curve* of degree $n$ with these control points is the map $B \\colon [0, 1] \\to \\R^d$,\n$$B(t) = \\sum_{k=0}^{n} \\binom{n}{k} (1 - t)^{n-k}\\, t^k\\, P_k.$$\n\n## Remarks\n\nThe coefficients $\\binom{n}{k}(1-t)^{n-k} t^k$ are the [[Bernstein basis polynomials|0C4VWBFY73]] of degree $n$; they are nonnegative on $[0, 1]$ and sum to $1$.\nSo $B(t)$ lies in the [[convex hull|0FQSDC2EQ2]] of the control points, $B(0) = P_0$, $B(1) = P_n$, $B'(0) = n(P_1 - P_0)$ and $B'(1) = n(P_n - P_{n-1})$.\nExample: $n = 2$ with $P_0 = (0, 0)$, $P_1 = (1, 2)$, $P_2 = (2, 0)$ gives $B(t) = (2t, 4t - 4t^2)$, an arc of the parabola $y = 2x - x^2$.\nNon-example: no arc of a circle of positive length is a Bézier curve.\nIndeed, a polynomial map $t \\mapsto (x(t), y(t))$ with $x^2 + y^2 = 1$ must be constant.\nIf $x$ and $y$ had top degree $m \\ge 1$, the coefficient of $t^{2m}$ in $x^2 + y^2$ would be a positive sum of squares.\nA rational Bézier curve adds positive weights, $B(t) = \\sum_k w_k b_{k,n}(t) P_k / \\sum_k w_k b_{k,n}(t)$, and does represent circular arcs.\n",
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