In polynomial interpolation of two variables, the Padua points are the first known example (and up to now the only one) of a unisolvent point set (that is, the interpolating polynomial is unique) with minimal growth of their Lebesgue constant, proven to be
O(log2n)
[-1,1] x [-1,1]\subsetR2
We can see the Padua point as a "sampling" of a parametric curve, called generating curve, which is slightly different for each of the four families, so that the points for interpolation degree
n
s
s=\lbrace\xi=(\xi | |
Pad | |
1,\xi |
2)\rbrace=\left\lbrace\gamma
|
\right),k=0,\ldots,n(n+1)\right\rbrace.
Actually, the Padua points lie exactly on the self-intersections of the curve, and on the intersections of the curve with the boundaries of the square
[-1,1]2
s | |
\operatorname{Pad} | |
n |
[-1,1]2
2n-1
The four generating curves are closed parametric curves in the interval
[0,2\pi]
The generating curve of Padua points of the first family is
\gamma1(t)=[-\cos((n+1)t),-\cos(nt)], t\in[0,\pi].
If we sample it as written above, we have:
1=\lbrace\xi=(\mu | |
\operatorname{Pad} | |
j,η |
k),0\lej\len;1\lek\le\lfloor
n | |
2 |
\rfloor+1+\deltaj\rbrace,
\deltaj=0
n
j
\deltaj=1
n
k
with
\mu | ||||
|
\right),
η | ||||
|
\right)&jodd\\ \cos\left(
(2k-1)\pi | |
n+1 |
\right)&jeven. \end{cases}
From this follows that the Padua points of first family will have two vertices on the bottom if
n
n
The generating curve of Padua points of the second family is
\gamma2(t)=[-\cos(nt),-\cos((n+1)t)], t\in[0,\pi],
which leads to have vertices on the left if
n
n
The generating curve of Padua points of the third family is
\gamma3(t)=[\cos((n+1)t),\cos(nt)], t\in[0,\pi],
which leads to have vertices on the top if
n
n
The generating curve of Padua points of the fourth family is
\gamma4(t)=[\cos(nt),\cos((n+1)t)], t\in[0,\pi],
which leads to have vertices on the right if
n
n
Kn(x,y)
x=(x1,x2)
y=(y1,y2)
2([-1,1] | |
\Pi | |
n |
2)
\langlef,g\rangle=
1 | |
\pi2 |
\int | |
[-1,1]2 |
f(x1,x2)g(x1,x
|
defined by
Kn(x,y)=\sum
k | |
j=0 |
\hatTj(x1)\hatTk-j(x2)\hatTj(y1)\hatTk-j(y2)
with
\hatTj
j
\hatT0=T0
\hatTp=\sqrt{2}Tp
Tp( ⋅ )=\cos(p\arccos( ⋅ ))
p
s=\lbrace\xi=(\xi | |
\operatorname{Pad} | |
1,\xi |
2)\rbrace
s=\lbrace1,2,3,4\rbrace
n
f\colon[-1,1]2\toR2
x\in[-1,1]2
s | |
l{L} | |
n |
f(x)=\sum\xi\in\operatorname{Pad
s}f(\xi)L | |
n |
s | |
\xi |
(x)
where
s | |
L | |
\xi |
(x)
s | |
L | |
\xi |
(x)=w\xi(Kn(\xi,x)-Tn(\xii)Tn(xi)), s=1,2,3,4, i=2-(s\mod2).
The weights
w\xi
w\xi=
1 | ⋅ \begin{cases} | |
n(n+1) |
1 | |
2 |
if\xiisavertexpoint\\ 1if\xiisanedgepoint\\ 2if\xiisaninteriorpoint. \end{cases}