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Christoffel–Darboux formula

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In mathematics, the Christoffel–Darboux formula or Christoffel–Darboux theorem is an identity for a sequence of orthogonal polynomials, introduced by Elwin Bruno Christoffel (1858) and Jean Gaston Darboux (1878). It states that

j = 0 n f j ( x ) f j ( y ) h j = k n h n k n + 1 f n ( y ) f n + 1 ( x ) f n + 1 ( y ) f n ( x ) x y {\displaystyle \sum _{j=0}^{n}{\frac {f_{j}(x)f_{j}(y)}{h_{j}}}={\frac {k_{n}}{h_{n}k_{n+1}}}{\frac {f_{n}(y)f_{n+1}(x)-f_{n+1}(y)f_{n}(x)}{x-y}}}

where fj(x) is the jth term of a set of orthogonal polynomials of squared norm hj and leading coefficient kj.

There is also a "confluent form" of this identity by taking y x {\displaystyle y\to x} limit: j = 0 n f j 2 ( x ) h j = k n h n k n + 1 [ f n + 1 ( x ) f n ( x ) f n ( x ) f n + 1 ( x ) ] . {\displaystyle \sum _{j=0}^{n}{\frac {f_{j}^{2}(x)}{h_{j}}}={\frac {k_{n}}{h_{n}k_{n+1}}}\left.}

Proof

Let p n {\displaystyle p_{n}} be a sequence of polynomials orthonormal with respect to a probability measure μ {\displaystyle \mu } , and define a n = x p n , p n + 1 , b n = x p n , p n , n 0 {\displaystyle a_{n}=\langle xp_{n},p_{n+1}\rangle ,\qquad b_{n}=\langle xp_{n},p_{n}\rangle ,\qquad n\geq 0} (they are called the "Jacobi parameters"), then we have the three-term recurrence p 0 ( x ) = 1 , p 1 ( x ) = x b 0 a 0 , x p n ( x ) = a n p n + 1 ( x ) + b n p n ( x ) + a n 1 p n 1 ( x ) , n 1 {\displaystyle {\begin{array}{l l}{p_{0}(x)=1,\qquad p_{1}(x)={\frac {x-b_{0}}{a_{0}}},}\\{xp_{n}(x)=a_{n}p_{n+1}(x)+b_{n}p_{n}(x)+a_{n-1}p_{n-1}(x),\qquad n\geq 1}\end{array}}}

Proof: By definition, x p n , p k = p n , x p k {\displaystyle \langle xp_{n},p_{k}\rangle =\langle p_{n},xp_{k}\rangle } , so if k n 2 {\displaystyle k\leq n-2} , then x p k {\displaystyle xp_{k}} is a linear combination of p 0 , . . . , p n 1 {\displaystyle p_{0},...,p_{n-1}} , and thus x p n , p k = 0 {\displaystyle \langle xp_{n},p_{k}\rangle =0} . So, to construct p n + 1 {\displaystyle p_{n+1}} , it suffices to perform Gram-Schmidt process on x p n {\displaystyle xp_{n}} using p n , p n 1 {\displaystyle p_{n},p_{n-1}} , which yields the desired recurrence.


Proof of Christoffel–Darboux formula:

Since both sides are unchanged by multiplying with a constant, we can scale each f n {\displaystyle f_{n}} to p n {\displaystyle p_{n}} .

Since k n + 1 k n x p n p n + 1 {\displaystyle {\frac {k_{n+1}}{k_{n}}}xp_{n}-p_{n+1}} is a degree n {\displaystyle n} polynomial, it is perpendicular to p n + 1 {\displaystyle p_{n+1}} , and so k n + 1 k n x p n , p n + 1 = p n + 1 , p n + 1 = 1 {\displaystyle \langle {\frac {k_{n+1}}{k_{n}}}xp_{n},p_{n+1}\rangle =\langle p_{n+1},p_{n+1}\rangle =1} . Now the Christoffel-Darboux formula is proved by induction, using the three-term recurrence.

Specific cases

Hermite polynomials:

k = 0 n H k ( x ) H k ( y ) k ! 2 k = 1 n ! 2 n + 1 H n ( y ) H n + 1 ( x ) H n ( x ) H n + 1 ( y ) x y . {\displaystyle \sum _{k=0}^{n}{\frac {H_{k}(x)H_{k}(y)}{k!2^{k}}}={\frac {1}{n!2^{n+1}}}\,{\frac {H_{n}(y)H_{n+1}(x)-H_{n}(x)H_{n+1}(y)}{x-y}}.} k = 0 n H e k ( x ) H e k ( y ) k ! = 1 n ! H e n ( y ) H e n + 1 ( x ) H e n ( x ) H e n + 1 ( y ) x y . {\displaystyle \sum _{k=0}^{n}{\frac {He_{k}(x)He_{k}(y)}{k!}}={\frac {1}{n!}}\,{\frac {He_{n}(y)He_{n+1}(x)-He_{n}(x)He_{n+1}(y)}{x-y}}.}

Associated Legendre polynomials:

( μ μ ) l = m L ( 2 l + 1 ) ( l m ) ! ( l + m ) ! P l m ( μ ) P l m ( μ ) = ( L m + 1 ) ! ( L + m ) ! [ P L + 1 m ( μ ) P L m ( μ ) P L m ( μ ) P L + 1 m ( μ ) ] . {\displaystyle {\begin{aligned}(\mu -\mu ')\sum _{l=m}^{L}\,(2l+1){\frac {(l-m)!}{(l+m)!}}\,P_{lm}(\mu )P_{lm}(\mu ')=\qquad \qquad \qquad \qquad \qquad \\{\frac {(L-m+1)!}{(L+m)!}}{\big }.\end{aligned}}}

See also

References

  1. Świderski, Grzegorz; Trojan, Bartosz (2021-08-01). "Asymptotic Behaviour of Christoffel–Darboux Kernel Via Three-Term Recurrence Relation I". Constructive Approximation. 54 (1): 49–116. arXiv:1909.09107. doi:10.1007/s00365-020-09519-w. ISSN 1432-0940. S2CID 202677666.


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