# Hermitian polynomial

Plots of the first five Hermitian polynomials H n

The Hermitian polynomials (after Charles Hermite ) are polynomials with the following equivalent representations:

${\ displaystyle H_ {n} (x) = (- 1) ^ {n} e ^ {x ^ {2}} {\ frac {\ mathrm {d} ^ {n}} {\ mathrm {d} x ^ {n}}} e ^ {- x ^ {2}} \ ,,}$

or. ${\ displaystyle H_ {n} (x) = e ^ {x ^ {2} / 2} \, \ left (x - {\ frac {\ mathrm {d}} {\ mathrm {d} x}} \ right ) ^ {n} \, e ^ {- x ^ {2} / 2} \ ,.}$

The Hermitian polynomials (with a fixed one ) are solutions of the Hermitian differential equation , a linear differential equation of the second order: ${\ displaystyle n}$

${\ displaystyle H_ {n} '' (x) -2 \, x \ cdot H_ {n} '(x) +2 \, n \ cdot H_ {n} (x) = 0 \ qquad (n = 0, 1,2, \ dots).}$

## Explicit representation

The explicit representation is obtained from the first representation with the formula of Faà di Bruno

${\ displaystyle H_ {n} (x) = (- 1) ^ {n} \ sum _ {k_ {1} + 2k_ {2} = n} {\ frac {n!} {k_ {1}! k_ { 2}!}} (- 1) ^ {k_ {1} + k_ {2}} (2x) ^ {k_ {1}}}$

so

${\ displaystyle H_ {0} (x) = 1}$
${\ displaystyle H_ {1} (x) = 2x}$
${\ displaystyle H_ {2} (x) = (2x) ^ {2} -2 = 4x ^ {2} -2}$
${\ displaystyle H_ {3} (x) = (2x) ^ {3} -6 (2x) = 8x ^ {3} -12x}$
${\ displaystyle H_ {4} (x) = (2x) ^ {4} -12 (2x) ^ {2} + 12 = 16x ^ {4} -48x ^ {2} +12}$

Hermitian polynomials can be calculated using the following recursion formulas : ${\ displaystyle (n \ in \ mathbb {N} _ {0}, H _ {- 1} (x): = 0)}$

${\ displaystyle H_ {n + 1} (x) = 2 \, x \, H_ {n} (x) -2 \, n \, H_ {n-1} (x)}$
${\ displaystyle H_ {n} '(x) = 2 \, n \, H_ {n-1} (x)}$

Since one is multiplied with each iteration step, you can quickly see that a polynomial is of degrees . The coefficient of the highest power is . For even , only even powers of occur, correspondingly for odd, only odd powers, which is mathematically identified by the identity ${\ displaystyle x}$${\ displaystyle H_ {n} (x)}$${\ displaystyle n}$${\ displaystyle x ^ {n}}$${\ displaystyle 2 ^ {n}}$${\ displaystyle n}$${\ displaystyle x}$${\ displaystyle n}$

${\ displaystyle H_ {n} (- x) = (- 1) ^ {n} \ cdot H_ {n} (x)}$

expresses.

The recursive representation of the above Hermitian polynomials can also be written as follows through simple substitution : ${\ displaystyle n '= n + 1}$

${\ displaystyle H_ {n} (x) = 2xH_ {n-1} (x) -2 (n-1) H_ {n-2} (x) \, \, \, \, \, \ quad \ quad (n = 1.2 \ ldots)}$

### Pascal source code

With the help of the known initial conditions and the function values ​​can easily be calculated with the following recursive Pascal function: ${\ displaystyle H_ {0} (x) = 1}$${\ displaystyle H_ {1} (x) = 2x}$

 Function Hermite(n:Byte;x:Extended):Extended;
Function Go(m:Byte; p,q:Extended): Extended;
Begin
If n=m Then Go := p
Else Go := Go(m+1, q, 2*x*q - 2*(m+1)*p)
End;
Begin
Hermite := Go(0, 1, 2*x)
End;


The more general derivation formula can be implemented as follows: ${\ displaystyle H_ {n} ^ {(m)} (x) = 2nH_ {n-1} ^ {(m-1)} (x)}$

 Function HermiteAbleitung(n,m:Byte;x:Extended):Extended;
Begin
If m=0 Then HermiteAbleitung:=Hermite(n,x)
Else
If n<m Then HermiteAbleitung:=0
Else If m=1 Then HermiteAbleitung:=2*n*Hermite(n-1,x)
Else HermiteAbleitung:=2*n*HermiteAbleitung(n-1,m-1,x)
End;


## Orthogonality

The Hermitian polynomials satisfy the orthogonality relation with regard to the weight function${\ displaystyle e ^ {- x ^ {2}}}$

${\ displaystyle \ int \ limits _ {- \ infty} ^ {+ \ infty} e ^ {- x ^ {2}} \ cdot H_ {n} (x) \ cdot H_ {m} (x) \, dx = 2 ^ {n} \ cdot n! \ Cdot {\ sqrt {\ pi}} \ cdot \ delta _ {nm}.}$

This means that certain real functions can be expanded into a series according to the Hermitian polynomials .

## Different representation of Hermitian polynomials

Plots of the first five Hermitian polynomials He n (physicist convention)

Another way of defining Hermitian polynomials (physicists' convention) is

${\ displaystyle He_ {n} (x) = 2 ^ {- n / 2} H_ {n} (x / {\ sqrt {2}}) = (- 1) ^ {n} e ^ {x ^ {2 } / 2} {\ frac {\ mathrm {d} ^ {n}} {\ mathrm {d} x ^ {n}}} e ^ {- x ^ {2} / 2}.}$

They are orthogonal to the weight function${\ displaystyle e ^ {- x ^ {2} / 2}}$

${\ displaystyle \ int \ limits _ {- \ infty} ^ {\ infty} e ^ {- x ^ {2} / 2} \, He_ {n} (x) \, He_ {m} (x) \, dx = {\ sqrt {2 \, \ pi}} \, n! \, \ delta _ {mn}}$

and satisfy the differential equation

${\ displaystyle y '' - x \, y '+ n \, y = 0.}$

You can pass through recursively

${\ displaystyle He_ {n + 1} (x) = x \, He_ {n} (x) -n \, He_ {n-1} (x)}$

determine.

## Binomial theorem

For the Hermitian polynomials a formula applies which has a similar form to the binomial theorem . For is ${\ displaystyle a ^ {2} + b ^ {2} = 1}$

${\ displaystyle H_ {n} (ax + by) = \ sum _ {k = 0} ^ {n} {\ binom {n} {k}} a ^ {k} b ^ {nk} H_ {k} ( x) H_ {nk} (y).}$

## Index with negative value

The derivative of the complementary error function is ${\ displaystyle 1- \ operatorname {erf} (x) = \ operatorname {erfc} (x)}$

${\ displaystyle {\ frac {\ mathrm {d}} {\ mathrm {d} x}} \ operatorname {erfc} (x) = - {\ frac {2} {\ sqrt {\ pi}}} e ^ { -x ^ {2}}}$.

The representation of the Hermitian polynomials can thus also be written as follows:

${\ displaystyle H_ {n} (x) = {\ frac {\ sqrt {\ pi}} {2}} (- 1) ^ {(n + 1)} e ^ {x ^ {2}} {\ frac {\ mathrm {d} ^ {n + 1}} {\ mathrm {d} x ^ {n + 1}}} \ operatorname {erfc} (x)}$,

so that one finds for: ${\ displaystyle n = -1}$

${\ displaystyle H _ {- 1} (x) = {\ frac {\ sqrt {\ pi}} {2}} e ^ {x ^ {2}} \ operatorname {erfc} (x)}$.

The functions of higher indices are calculated as:

${\ displaystyle H_ {n-1} (x) = {\ frac {(-1) ^ {n}} {2 ^ {- n} (- n)!}} {\ frac {\ mathrm {d} ^ {-n}} {\ mathrm {d} x ^ {- n}}} H _ {- 1} (x)}$or recursively with .${\ displaystyle H_ {n-1} (x) = {\ frac {1} {2n}} H_ {n} '(x)}$${\ displaystyle n = (- 1, -2, -3, \ dotsc)}$

The functions obtained in this way, like the polynomials with a positive index, satisfy the Hermitian differential equation.

They are:

${\ displaystyle H _ {- 1} (x) = {\ tfrac {1} {2}} {\ sqrt {\ pi}} e ^ {x ^ {2}} \ operatorname {erfc} (x)}$
${\ displaystyle H _ {- 2} (x) = {\ tfrac {1} {2}} (1-x {\ sqrt {\ pi}} e ^ {x ^ {2}} \ operatorname {erfc} (x ))}$
${\ displaystyle H _ {- 3} (x) = {\ tfrac {1} {8}} (- 2x + (1 + 2x ^ {2}) {\ sqrt {\ pi}} e ^ {x ^ {2} } \ operatorname {erfc} (x))}$
${\ displaystyle \ ldots}$

## Applications

The Hermite polynomials get their importance from their versatile applicability in physics. For example, they are needed to construct the orthonormal solution functions of the quantum mechanical harmonic oscillator . These correspond to the Hermitian functions , which are obtained by multiplying with the Gaussian normal distribution and using suitable normalization.

They are also used in the finite element method as shape functions.

The probability density of the non-central Student's t-distribution can be expressed using Hermitian polynomial functions whose index has negative values.