Editing Hermite polynomials (section)

==Definition==
Like the other [[classical orthogonal polynomials]], the Hermite polynomials can be defined from several different starting points. Noting from the outset that there are two different standardizations in common use, one convenient method is as follows:

* The '''"probabilist's Hermite polynomials"''' are given by <math display="block">\operatorname{He}_n(x) = (-1)^n e^{\frac{x^2}{2}}\frac{d^n}{dx^n}e^{-\frac{x^2}{2}},</math>
* while the '''"physicist's Hermite polynomials"''' are given by <math display="block">H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}.</math>

These equations have the form of a [[Rodrigues' formula]] and can also be written as,
<math display="block">\operatorname{He}_n(x) = \left(x - \frac{d}{dx} \right)^n \cdot 1, \quad H_n(x) = \left(2x - \frac{d}{dx} \right)^n \cdot 1.</math>

The two definitions are not exactly identical; each is a rescaling of the other:
<math display="block">H_n(x)=2^\frac{n}{2} \operatorname{He}_n\left(\sqrt{2} \,x\right), \quad \operatorname{He}_n(x)=2^{-\frac{n}{2}} H_n\left(\frac {x}{\sqrt 2} \right).</math>

These are Hermite polynomial sequences of different variances; see the material on variances below.

The notation {{mvar|He}} and {{mvar|H}} is that used in the standard references.<ref>{{harvs|txt|first=Tom H. |last=Koornwinder|first2=Roderick S. C.|last2= Wong|first3=Roelof |last3=Koekoek|first4=René F. |last4=Swarttouw|year=2010}} and [[Abramowitz & Stegun]].</ref>
The polynomials {{mvar|He<sub>n</sub>}} are sometimes denoted by {{mvar|H<sub>n</sub>}}, especially in probability theory, because
<math display="block">\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}</math>
is the [[probability density function]] for the [[normal distribution]] with [[expected value]] 0 and [[standard deviation]] 1.
* The first eleven probabilist's Hermite polynomials are: <math display="block">\begin{align}
\operatorname{He}_0(x) &= 1, \\
\operatorname{He}_1(x) &= x, \\
\operatorname{He}_2(x) &= x^2 - 1, \\
\operatorname{He}_3(x) &= x^3 - 3x, \\
\operatorname{He}_4(x) &= x^4 - 6x^2 + 3, \\
\operatorname{He}_5(x) &= x^5 - 10x^3 + 15x, \\
\operatorname{He}_6(x) &= x^6 - 15x^4 + 45x^2 -  15, \\
\operatorname{He}_7(x) &= x^7 - 21x^5 + 105x^3 - 105x, \\
\operatorname{He}_8(x) &= x^8 - 28x^6 + 210x^4 - 420x^2 + 105, \\
\operatorname{He}_9(x) &= x^9 - 36x^7 + 378x^5 - 1260x^3 + 945x, \\
\operatorname{He}_{10}(x) &= x^{10} - 45x^8 + 630x^6 - 3150x^4 + 4725x^2 - 945.
\end{align}</math>
* The first eleven physicist's Hermite polynomials are: <math display="block">\begin{align}
H_0(x) &= 1, \\
H_1(x) &= 2x, \\
H_2(x) &= 4x^2 - 2, \\
H_3(x) &= 8x^3 - 12x, \\
H_4(x) &= 16x^4 - 48x^2 + 12, \\
H_5(x) &= 32x^5 - 160x^3 + 120x, \\
H_6(x) &= 64x^6 - 480x^4 + 720x^2 - 120, \\
H_7(x) &= 128x^7 - 1344x^5 + 3360x^3 - 1680x, \\
H_8(x) &= 256x^8 - 3584x^6 + 13440x^4 - 13440x^2 + 1680, \\
H_9(x) &= 512x^9 - 9216x^7 + 48384x^5 - 80640x^3 + 30240x, \\
H_{10}(x) &= 1024x^{10} - 23040x^8 + 161280x^6 - 403200x^4 + 302400x^2 - 30240.
\end{align}</math>
<!--
As an alternative to calculating {{mvar|n}}th-order derivatives of {{math|''e''<sup>−{{sfrac|''x''<sup>2</sup>|2}}</sup>}} and {{math|''e''<sup>−''x''<sup>2</sup></sup>}}, an easier, less computationally-intensive method of sequentially deriving individual terms of the {{mvar|n}}th-order Hermite polynomials is to consider the combination of coefficients in the corresponding terms in the {{math|(''n'' − 1)}}th-order Hermite polynomial. For the probabilist's notation, follow the following rules:
# For the starting point in the sequence, the zeroth-order polynomial ({{math|''He''<sub>0</sub>}}) is equal to 1.
# The first term has a power of {{mvar|x}} equal to the given {{mvar|n}}th-order polynomial being derived, and the coefficient of this term is 1.
# The power of {{mvar|x}} of each successive term is two less than the preceding term.
# The coefficient of each term after the first term is calculated by taking the coefficient of the same-numbered term in the {{math|(''n'' − 1)}}th polynomial, and adding to it the product of the power of {{mvar|x}} and the corresponding coefficient of the immediately preceding term in the {{math|(''n'' − 1)}}th polynomial.
# All even-numbered terms in each polynomial are negative, and all odd-numbered terms are positive.

Thus, for {{math|''He''<sub>6</sub>}}, {{math|1=''n'' = 6}} and hence the first term is {{math|''x''<sup>6</sup>}}, with a coefficient of 1. The second term has a power of {{mvar|x}} equal to {{math|1=6− 2 = 4}}. The coefficient is obtained by taking the coefficient of the second term in {{math|''He''<sub>5</sub>}} (which is 10) and adding to it the product of the power of {{mvar|x}} and its coefficient in the first term of {{math|''He''<sub>5</sub>}} (which are 5 and 1, respectively). Thus, {{math|1=10 + 5 × 1 = 15}}. Make this coefficient negative, since this is an even-numbered term. The third term in {{math|''He''<sub>6</sub>}} has a power of {{mvar|x}} equal to 2 (which is 2less than the power of {{mvar|x}} in the second term), and its coefficient is {{math|1=15 + 3 × 10 = 45}}, where 15 is the coefficient of the third term in {{math|''He''<sub>5</sub>}}; 3 is the power of {{mvar|x}} in the second term of the {{math|''He''<sub>5</sub>}} polynomial; and 10 is the coefficient of the second term in {{math|''H''<sub>5</sub>}}. This coefficient (45) is positive, since this is an odd-numbered term. Finally, the fourth term in {{math|''He''<sub>6</sub>}} is the zeroth power in {{mvar|x}} (which is 2less than the power of {{mvar|x}} in the third term), and its coefficient is {{math|1=0 + 1 × 15 = 15}}, where 0 is the coefficient of the (nonexistent) fourth term in the {{math|''He''<sub>5</sub>}} polynomial; 1 is the power of {{mvar|x}} in the third term of {{math|''He''<sub>5</sub>}}, and 15 is the coefficient of the third term in {{math|''H''<sub>5</sub>}} . This coefficient (15) is made negative, since this is an even-numbered term. Thus, {{math|1=''He''<sub>6</sub>(''x'') = ''x''<sup>6</sup> − 15''x''<sup>4</sup> + 45''x''<sup>2</sup> − 15}}.

For {{math|''He''<sub>7</sub>}}, the first term is {{math|''x''<sup>7</sup>}}; the second coefficient is {{math|1=15 + 6 × 1 = 21}} (negative, i.e. {{math|−21''x''<sup>5</sup>}}). The third coefficient is {{math|1=45 + 4 × 15 = 105}} (positive, i.e. {{math|105''x''<sup>3</sup>}}). The fourth coefficient is {{math|1=15 + 2 × 45 = 105}} (negative, i.e. −105''x''). Thus, {{math|1=''He''<sub>7</sub>(''x'') = ''x''<sup>7</sup> − 21''x''<sup>5</sup> + 105''x''<sup>3</sup> − 105''x''}}.

For the physicist's notation, follow the following rules:
# For the starting point in the sequence, the zeroth-order polynomial ({{math|''H''<sub>0</sub>}}) is equal to 1.
# The first term has a power of {{mvar|x}} equal to the given {{mvar|n}}th-order polynomial being derived, and the coefficient of this term is {{math|2<sup>''n''</sup>}}.
# The power of {{mvar|x}} of each successive term is two less than the preceding term.
# The coefficient of each term after the first term is calculated by taking the coefficient of the same-numbered term in the {{math|(''n'' − 1)}}th polynomial, multiplying it by 2, and then adding to it the product of the power of {{mvar|x}} and the corresponding coefficient of the immediately preceding term in the {{math|(''n'' − 1)}}th polynomial.
# All even-numbered terms in each polynomial are negative, and all odd-numbered terms are positive.

Thus, for {{math|''H''<sub>6</sub>}}, the first coefficient is {{math|1=2<sup>6</sup> = 64}} (i.e. {{math|64''x''<sup>6</sup>}}). The second coefficient is {{math|1=2 × 160 + 5 × 32 = 480}} (negative, i.e. {{math|−480''x''<sup>4</sup>}}). The third coefficient is {{math|1=2 × 120 + 3 × 160 = 720}} (positive, i.e. {{math|720''x''<sup>2</sup>}}). The fourth coefficient is {{math|1=2 × 0 + 1 × 120 = 120}} (negative, i.e., −120). Thus, {{math|1=''H''<sub>6</sub>(x) = 64''x''<sup>6</sup> − 480''x''<sup>4</sup> + 720''x''<sup>2</sup> − 120}}.

For {{math|''H''<sub>7</sub>}}, the first coefficient is 2<sup>7</sup> = 128 (i.e., {{math|128''x''<sup>7</sup>}}). The second coefficient is {{math|1=2 × 480 + 6 × 64 = 1344}} (negative, i.e. {{math|−1344''x''<sup>5</sup>}}). The third coefficient is {{math|1=2 × 720 + 4 × 480 = 3360}} (positive, i.e. {{math|3360''x''<sup>3</sup>}}). The fourth coefficient is {{math|1=2 × 120 + 2 × 720 = 1680}} (negative, i.e. {{math|−1680''x''}}. Thus, {{math|''H''<sub>7</sub>(''x'') = 128''x''<sup>7</sup> − 1344''x''<sup>5</sup> + 3360''x''<sup>3</sup> − 1680''x''}}.

Recognizing that {{math|1=''H''<sub>0</sub> = 1}}, these rules can be followed to sequentially derive all {{mvar|n}}th-order Hermite polynomials from {{math|1=''n'' = 1}} towards infinity, and can be computer-coded relatively easily for practical applications. -->{| class="wikitable"
|+Quick reference table
!
!physicist's
!probabilist's
|-
|symbol
|<math>H_n</math>
|<math>\operatorname{He}_n</math>
|-
|head coefficient
|<math>2^n</math>
|<math>1</math>
|-
|differential operator
|<math>(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}</math>
|<math>(-1)^n e^{\frac{x^2}{2}}\frac{d^n}{dx^n}e^{-\frac{x^2}{2}}</math>
|-
|orthogonal to
|<math>e^{-x^2}</math>
|<math>e^{-\frac 12 x^2}</math>
|-
|inner product
|<math>\int H_m(x) H_n(x) \frac{e^{-x^2}}{\sqrt{\pi}}dx = 2^n  n! \delta_{mn}</math>
|<math>\int \operatorname{He}_m(x) \operatorname{He}_n(x)\, \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}} \,dx = n!\, \delta_{nm},  </math>
|-
|generating function
|<math>e^{2xt - t^2} = \sum_{n=0}^\infty  H_n(x) \frac{t^n}{n!}</math>
|<math>e^{xt - \frac12 t^2} = \sum_{n=0}^\infty \operatorname{He}_n(x) \frac{t^n}{n!} </math>
|-
|Rodrigues' formula
|<math>\left(2x - \frac{d}{dx} \right)^n \cdot 1 </math>
|<math>\left(x - \frac{d}{dx} \right)^n \cdot 1 </math>
|-
|recurrence relation
|<math>H_{n+1}(x) = 2xH_n(x) - 2nH_{n-1}(x)</math>
|<math>\operatorname{He}_{n+1}(x) = x\operatorname{He}_n(x) - n\operatorname{He}_{n-1}(x)</math>
|}
<gallery widths="300" heights="300">
File:Hermite poly.svg|The first six probabilist's Hermite polynomials <math>\operatorname{He}_n(x)</math>
File:Hermite poly phys.svg|The first six physicist's Hermite polynomials <math>H_n(x)</math>
</gallery>