Editing Hermite polynomials (section)

=== Completeness relation ===
The [[Christoffel–Darboux formula]] for Hermite polynomials reads
<math display="block">\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}.</math>

Moreover, the following [[Borel functional calculus#Resolution of the identity|completeness identity]] for the above Hermite functions holds in the sense of [[distribution (mathematics)|distributions]]:
<math display="block">\sum_{n=0}^\infty \psi_n(x) \psi_n(y) = \delta(x - y),</math>
where {{mvar|δ}} is the [[Dirac delta function]], {{math|''ψ''<sub>''n''</sub>}} the Hermite functions, and {{math|''δ''(''x'' − ''y'')}} represents the [[Lebesgue measure]] on the line {{math|1=''y'' = ''x''}} in {{math|'''R'''<sup>2</sup>}}, normalized so that its projection on the horizontal axis is the usual Lebesgue measure.

This distributional identity follows {{harvtxt|Wiener|1958}} by taking {{math|''u'' → 1}} in [[Mehler's formula]], valid when {{math|−1 < ''u'' < 1}}:
<math display="block">E(x, y; u) := \sum_{n=0}^\infty u^n \, \psi_n (x) \, \psi_n (y) = \frac{1}{\sqrt{\pi (1 - u^2)}} \, \exp\left(-\frac{1 - u}{1 + u} \, \frac{(x + y)^2}{4} - \frac{1 + u}{1 - u} \, \frac{(x - y)^2}{4}\right),</math>
which is often stated equivalently as a separable kernel,<ref>{{Citation | last1=Mehler | first1=F. G. | title=Ueber die Entwicklung einer Function von beliebig vielen Variabeln nach Laplaceschen Functionen höherer Ordnung | url=http://resolver.sub.uni-goettingen.de/purl?GDZPPN002152975 | language=de |trans-title=On the development of a function of arbitrarily many variables according to higher-order Laplace functions |id={{ERAM|066.1720cj}} | year=1866 | journal=Journal für die Reine und Angewandte Mathematik | issn=0075-4102 | issue=66 | pages=161–176}}. See p. 174, eq. (18) and p. 173, eq. (13).</ref><ref>{{harvnb|Erdélyi|Magnus|Oberhettinger|Tricomi|1955|page=194}}, 10.13 (22).</ref>
<math display="block">\sum_{n=0}^\infty \frac{H_n(x) H_n(y)}{n!} \left(\frac u 2\right)^n = \frac{1}{\sqrt{1 - u^2}} e^{\frac{2u}{1 + u}xy - \frac{u^2}{1 - u^2}(x - y)^2}.</math>

The function {{math|(''x'', ''y'') → ''E''(''x'', ''y''; ''u'')}} is the bivariate Gaussian probability density on {{math|'''R'''<sup>2</sup>}}, which is, when {{mvar|u}} is close to 1, very concentrated around the line {{math|1=''y'' = ''x''}}, and very spread out on that line. It follows that
<math display="block">\sum_{n=0}^\infty u^n \langle f, \psi_n \rangle \langle \psi_n, g \rangle = \iint E(x, y; u) f(x) \overline{g(y)} \,dx \,dy \to \int f(x) \overline{g(x)} \,dx = \langle f, g \rangle</math>
when {{math|''f''}} and {{math|''g''}} are continuous and compactly supported.

This yields that {{mvar|f}} can be expressed in Hermite functions as the sum of a series of vectors in {{math|''L''<sup>2</sup>('''R''')}}, namely,
<math display="block">f = \sum_{n=0}^\infty \langle f, \psi_n \rangle \psi_n.</math>

In order to prove the above equality for {{math|''E''(''x'',''y'';''u'')}}, the [[Fourier transform]] of [[Gaussian function]]s is used repeatedly:
<math display="block">\rho \sqrt{\pi} e^{-\frac{\rho^2 x^2}{4}} = \int e^{isx - \frac{s^2}{\rho^2}} \,ds \quad \text{for }\rho > 0.</math>

The Hermite polynomial is then represented as
<math display="block"> H_n(x) = (-1)^n e^{x^2} \frac {d^n}{dx^n} \left( \frac {1}{2\sqrt{\pi}} \int e^{isx - \frac{s^2}{4}} \,ds \right) = (-1)^n e^{x^2}\frac{1}{2\sqrt{\pi}} \int (is)^n e^{isx - \frac{s^2}{4}} \,ds.</math>

With this representation for {{math|''H<sub>n</sub>''(''x'')}} and {{math|''H<sub>n</sub>''(''y'')}}, it is evident that
<math display="block">\begin{align}
 E(x, y; u) &= \sum_{n=0}^\infty \frac{u^n}{2^n n! \sqrt{\pi}} \, H_n(x) H_n(y) e^{-\frac{x^2+y^2}{2}} \\
  &= \frac{e^{\frac{x^2+y^2}{2}}}{4\pi\sqrt{\pi}}\iint\left( \sum_{n=0}^\infty \frac{1}{2^n n!} (-ust)^n \right ) e^{isx+ity - \frac{s^2}{4} - \frac{t^2}{4}}\, ds\,dt \\
  & =\frac{e^{\frac{x^2+y^2}{2}}}{4\pi\sqrt{\pi}}\iint e^{-\frac{ust}{2}} \, e^{isx+ity - \frac{s^2}{4} - \frac{t^2}{4}}\, ds\,dt,
\end{align}</math>
and this yields the desired resolution of the identity result, using again the Fourier transform of Gaussian kernels under the substitution
<math display="block">s = \frac{\sigma + \tau}{\sqrt 2}, \quad t = \frac{\sigma - \tau}{\sqrt 2}.</math>