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  "body": "\n## Example\n\nLet $n \\ge 1$ and consider the curve $\\gamma \\colon [0, \\pi] \\to [-1, 1]^2$,\n$$\\gamma(t) = \\bigl(-\\cos((n + 1) t), \\ -\\cos(n t)\\bigr).$$\nThe *Padua points* of degree $n$ are the points $\\gamma\\bigl(k\\pi / (n(n + 1))\\bigr)$ for $k = 0, 1, \\dots, n(n + 1)$.\nThese are the self-intersections of $\\gamma$ together with its points on the boundary of the square.\nFor $n \\ge 2$ several of these parameter values give the same point, and there are exactly $(n + 1)(n + 2)/2$ distinct Padua points.\n\nThis number is the dimension of the space of [[polynomials in two variables|07GNX8GCYP]] of total degree at most $n$.\nThe Padua points are a [[unisolvent point set|5276MDSDQV]] for this space.\nSo each choice of values at the Padua points is taken by exactly one polynomial of total degree at most $n$.\n\nCheck for $n = 1$: the parameters $0$, $\\pi/2$, $\\pi$ give the points $(-1, -1)$, $(1, 0)$, $(-1, 1)$.\nThey are not collinear, so they are unisolvent for the polynomials $a + bx + cy$.\n\nThe [[Lebesgue constant|7V7KWNK9HN]] of interpolation at the Padua points is $O(\\log^2 n)$ as $n \\to \\infty$.\n\n## Explanations\n\n### L1 human\n\nIn one variable, Chebyshev nodes are a near-ideal choice of points for polynomial interpolation on an interval.\nIn the square, finding a good set of points that also gives a unique interpolating polynomial is much harder.\nThe Padua points solve this: they are the crossing points of a Lissajous-type curve that sweeps through the square, plus its touching points on the edges.\nInterpolation at these points is unique and numerically well behaved.\n",
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