Christoffel–Darboux formula

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).

Christoffel–Darboux formula—if a sequence of polynomials ${\displaystyle f_{0},f_{1},\dots }$ are of degrees ${\displaystyle 0,1,\dots }$, and orthogonal with respect to a probability measure ${\displaystyle \mu }$, then

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

where ${\displaystyle h_{j}=\|f_{j}\|_{\mu }^{2}}$ are the squared norms, and ${\displaystyle k_{j}}$ are the leading coefficients.

There is also a "confluent form" of this identity by taking ${\displaystyle y\to x}$ limit:

Christoffel–Darboux formula, confluent form—$${\displaystyle \sum _{j=0}^{n}{\frac {f_{j}^{2}(x)}{h_{j}}}={\frac {k_{n}}{h_{n}k_{n+1}}}\left[f_{n+1}'(x)f_{n}(x)-f_{n}'(x)f_{n+1}(x)\right].}$$

Proof

Lemma—Let ${\displaystyle p_{n}}$ be a sequence of polynomials orthonormal with respect to a probability measure ${\displaystyle \mu }$, such that ${\displaystyle p_{n}}$ has degree ${\displaystyle n}$, and define$${\displaystyle a_{n}=\langle xp_{n},p_{n+1}\rangle ,\qquad b_{n}=\langle xp_{n},p_{n}\rangle ,\qquad n\geq 0}$$(they are called the "Jacobi parameters"), then we have the three-term recurrence^[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, ${\displaystyle \langle xp_{n},p_{k}\rangle =\langle p_{n},xp_{k}\rangle }$, so if ${\displaystyle k\leq n-2}$, then ${\displaystyle xp_{k}}$ is a linear combination of ${\displaystyle p_{0},...,p_{n-1}}$, and thus ${\displaystyle \langle xp_{n},p_{k}\rangle =0}$. So, to construct ${\displaystyle p_{n+1}}$, it suffices to perform Gram-Schmidt process on ${\displaystyle xp_{n}}$ using ${\displaystyle p_{n+1},p_{n},p_{n-1}}$, which yields the desired recurrence.

Proof of Christoffel–Darboux formula

For any sequence of nonzero constants ${\displaystyle c_{0},c_{1},\dots }$, we can change each ${\displaystyle f_{k}}$ to ${\displaystyle c_{k}f_{k}}$, and both sides of the equation would remain unchanged. Thus WLOG, scale each ${\displaystyle f_{n}}$ to ${\displaystyle p_{n}}$.

Since ${\displaystyle {\frac {k_{n+1}}{k_{n}}}xp_{n}-p_{n+1}}$ is a degree ${\displaystyle n}$ polynomial, it is perpendicular to ${\displaystyle p_{n+1}}$, and so ${\displaystyle \langle {\frac {k_{n+1}}{k_{n}}}xp_{n},p_{n+1}\rangle =\langle p_{n+1},p_{n+1}\rangle =1}$. Thus, ${\textstyle a_{n}={\frac {k_{n}}{k_{n+1}}}}$.

Base case:

$${\displaystyle {\frac {k_{0}}{k_{1}}}{\frac {p_{0}(y)p_{1}(x)-p_{1}(y)p_{0}(x)}{x-y}}={\frac {k_{0}}{k_{1}}}{\frac {p_{0}(y){\frac {x-b_{0}}{a_{0}}}-p_{0}(x){\frac {y-b_{0}}{a_{0}}}}{x-y}}=1}$$

Induction:

$${\displaystyle {\begin{aligned}&(x-y)\sum _{j=0}^{n}p_{j}(x)p_{j}(y)=(x-y)\sum _{j=0}^{n-1}p_{j}(x)p_{j}(y)+p_{n}(x)p_{n}(y)(x-y)\\&{\stackrel {\text{ (IH) }}{=}}{\frac {k_{n-1}}{k_{n}}}\left[p_{n}(x)p_{n-1}(y)-p_{n}(y)p_{n-1}(x)\right]+(x-y)p_{n}(x)p_{n}(y)\end{aligned}}}$$

By the three-term recurrence,

$${\displaystyle {\begin{aligned}&xp_{n}(x)=a_{n}p_{n+1}(x)+b_{n}p_{n}(x)+a_{n-1}p_{n-1}(x)\\&yp_{n}(y)=a_{n}p_{n+1}(y)+b_{n}p_{n}(y)+a_{n-1}p_{n-1}(y)\end{aligned}}}$$

Multiply the first by ${\textstyle p_{n}(y)}$ and the second by ${\textstyle p_{n}(x)}$ and subtract:

$${\displaystyle (x-y)p_{n}(x)p_{n}(y)=a_{n}\left[p_{n}(y)p_{n+1}(x)-p_{n}(x)p_{n+1}(y)\right]+a_{n-1}\left[p_{n}(y)p_{n-1}(x)-p_{n}(x)p_{n-1}(y)\right]}$$

Now substitute in ${\textstyle a_{n}={\frac {k_{n}}{k_{n+1}}},\quad a_{n-1}={\frac {k_{n-1}}{k_{n}}}}$ and simplify.

Specific cases

Hermite

The Hermite polynomials are orthogonal with respect to the gaussian distribution.

The ${\displaystyle H}$ polynomials are orthogonal with respect to ${\displaystyle {\frac {1}{\sqrt {\pi }}}e^{-x^{2}}}$, and with ${\displaystyle k_{n}=2^{n}}$.$${\displaystyle \sum _{k=0}^{n}{\frac {H_{k}(x)H_{k}(y)}{k!2^{k}}}={\frac {1}{n!2^{n+1}}}\,{\frac {H_{n}(y)H_{n+1}(x)-H_{n}(x)H_{n+1}(y)}{x-y}}.}$$The ${\displaystyle He}$ polynomials are orthogonal with respect to ${\displaystyle {\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}x^{2}}}$, and with ${\displaystyle k_{n}=1}$.$${\displaystyle \sum _{k=0}^{n}{\frac {He_{k}(x)He_{k}(y)}{k!}}={\frac {1}{n!}}\,{\frac {He_{n}(y)He_{n+1}(x)-He_{n}(x)He_{n+1}(y)}{x-y}}.}$$

Laguerre

The Laguerre polynomials ${\displaystyle L_{n}}$ are orthonormal with respect to the exponential distribution ${\displaystyle e^{-x},\quad x\in (0,\infty )}$, with ${\displaystyle k_{n}=(-1)^{n}/n!}$, so$${\displaystyle \sum _{k=0}^{n}L_{k}(x)L_{k}(y)={\frac {n+1}{x-y}}\left[L_{n}(x)L_{n+1}(y)-L_{n}(y)L_{n+1}(x)\right]}$$

Legendre

Associated Legendre polynomials:

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

Christoffel–Darboux kernel

The summation involved in the Christoffel–Darboux formula is invariant by scaling the polynomials with nonzero constants. Thus, each probability distribution ${\displaystyle \mu }$ defines a series of functions$${\displaystyle K_{n}(x,y):=\sum _{j=0}^{n}f_{j}(x)f_{j}(y)/h_{j},\quad n=0,1,\dots }$$which are called the Christoffel–Darboux kernels. By the orthogonality, the kernel satisfies $${\displaystyle \int f(y)K_{n}(x,y)\mathrm {d} \mu (y)={\begin{cases}f(x),&f\in \operatorname {Span} \left(p_{0},p_{1},\ldots ,p_{n}\right)\\0,&\int _{a}^{b}f(x)p_{\ell }(x)\mathrm {d} \mu (x)=0(\ell =0,1,\ldots ,n)\end{cases}}}$$In other words, the kernel is an integral operator that orthogonally projects each polynomial to the space of polynomials of degree up to ${\displaystyle n}$.

See also

• Turán's inequalities
• Sturm Chain