# Hermite Polynomial: 9 Complete Quick Facts

Content

The Hermite polynomial is widely occurred in applications as an orthogonal function. Hermite polynomial is the series solution of Hermite differential equation.

## Hermite’s Equation

The differential equation of second order with specific coefficients as

d2y/dx2 – 2x dy/dx + 2xy = 0

$\frac{d^{2} y}{d x^{2}}-2 x \frac{d y}{d x}+2 n y=0$

is known as Hermite’s equation, by solving this differential equation we will get the polynomial which is Hermite Polynomial.

Let us find the solution of the equation

d2y/dx2 – 2x dy/dx + 2ny = 0

$\frac{d^{2} y}{d x^{2}}-2 x \frac{d y}{d x}+2 n y=0$

with the help of series solution of differential equation

$\begin{array}{l} y=a_{0} x^{m}+a_{1} x^{m+1}+a_{2} x^{m+2}+a_{3} x^{m+3}+\ldots \ldots .+a_{k} x^{m+k} \\ y=\sum_{k=0}^{\infty} a_{k} x^{m+k}\\ \frac{d y}{d x}=\sum a_{k}(m+k) x^{m+k-1}\\ \frac{d^{2} y}{d x^{2}}=\sum_{k=0}^{\infty} a_{k}(m+k)(m+k-1) x^{m+k-2} \end{array}$

now substituting all these values in the Hermite’s equation we have

$\Rightarrow \quad \sum a_{k}(m+k)(m+k-1) x^{m+k-2}-2 x \sum a_{k}(m+k) x^{m+k-1}+2 n \sum a_{k} x^{m+k}=0 \Rightarrow \quad \sum a_{k}(m+k)(m+k-1) x^{m+k-2}-2 \sum a_{k}(m+k) x^{m+k}+2 n \sum a_{k} x^{m+k}=0 \Rightarrow \quad \sum a_{k}(m+k)(m+k-1) x^{m+k-2}-2 \sum a_{k}[(m+k)-n] x^{m+k}=0$

This equation satisfies for the value of k=0 and as we assumed the value of k will not be negative, now for the lowest degree term xm-2 take k=0 in the first equation as the second gives negative value, so the coefficient xm-2 is

a0m (m-1)=0 ⇒ m=0,m=1

as a0 ≠ 0

$a_{0} m(m-1)=0 \Rightarrow m=0, m=1$

$as \quad a_{0} \neq 0$

now in the same way equating the coefficient of xm-1 from the second summation

$a_{1} m(m+1)=0 \Rightarrow\left[\begin{array} { l } { a _ { 1 } \text { may or may not be zero when } m = 0 } \\ { a _ { 1 } = 0 , \text { when } m = 1 } \end{array} \quad \left(\begin{array}{l} m+1 \neq 0 \text { as } \mathrm{m} \text { is } \\ \text { already equal to zero } \end{array}\right)\right.$

and equating the coefficients of xm+k to zero,

ak+2(m+k+2)(m+k+1)-2ak(m+k-n) = 0

$a_{k+2}(m+k+2)(m+k+1)-2 a_{k}(m+k-n)=0$

we can write it as

ak+2 = 2(m+k-n)/(m+k+2)(m+k+1) ak

$a_{k+2}=\frac{2(m+k-n)}{(m+k+2)(m+k+1)} a_{k}$

if m=0

ak+2 = 2(k-n)/(k+2)(k+1) ak

$\quad a_{k+2}=\frac{2(k-n)}{(k+2)(k+1)} a_{k} \quad$

if m=1

ak+2 = 2(k+1-n)/(k+3)(k+2) ak

$a_{k+2}=\frac{2(k+1-n)}{(k+3)(k+2)} a_{k}$

for these two cases now we discuss the cases for k

When $m=0, ak+2= 2(k-n)/ (k+2)(k+1)} ak$

If, $k=0 a2 =-2 n/2 a0=-n a0$

$k=1, a3=2(1-n)/6 a1 =-2(n-1)/3 ! a1$

If $k=2, a4 =2(2-n)/12 a2 =2 (2-n)/12 (-n a0) =22 n(n-2)/4 ! a0$

$When \quad m=0, a_{k+2}=\frac{2(k-n)}{(k+2)(k+1)} a_{k}$

$If \quad k=0, a_{2}=\frac{-2 n}{2} a_{0}=-n a_{0}$

$If \quad k=1, a_{3}=\frac{2(1-n)}{6} a_{1}=-2 \frac{(n-1)}{3 !} a_{1}$

$If \quad k=2, a_{4}=\frac{2(2-n)}{12} a_{2}=2 \frac{(2-n)}{12}\left(-n a_{0}\right)=(2)^{2} \frac{n(n-2)}{4 !} a_{0}$

$If \quad k=3, a_{5}=\frac{2(3-n)}{20} a_{3}=\frac{2(3-n)}{20}\left(-\frac{2(n-1)}{3 !} a_{1}\right)=(2)^{2} \frac{(n-1)(n-3)}{5 !} a_{1}\\ a_{2 r}=\frac{(-2)^{r} n(n-2)(n-4) \ldots \ldots(n-2 r+2)}{(2 r) !} a_{0}\\ a_{2 r+1}=\frac{(-2)^{r}(n-1)(n-3) \ldots \ldots(n-2 r+1)}{(2 r+1) !} a_{1}=0$

so far m=0 we have two conditions when a1=0, then a3=a5=a7=….=a2r+1=0 and when a1 is not zero then

$\begin{array}{c} y=\sum_{k=0}^{\infty} a_{k} x^{k} \\ y=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+\ldots \ldots \ldots \\ =a_{0}+a_{2} x^{2}+a_{4} x^{4}+\ldots . .+a_{1} x+a_{3} x^{3}+a_{5} x^{5} \end{array}$

by following this put the values of a0,a1,a2,a3,a4 and a5 we have

$\begin{array}{l} =a_{0}\left[1-\frac{2 n}{2 !} x^{2}+\frac{2^{2} n(n-2)}{4 !} x^{4}-\ldots+(-1)^{r} \frac{2}{(2 r) !} n(n-2) \ldots(n-2 r+2) x^{2 r}+\ldots\right] \\ +a_{1} x\left[1-\frac{2(n-1)}{3 !} x^{2}+\frac{2^{2}(n-1)(n-3)}{5 !}-\ldots .\right. \\ \left.+(-1)^{r} \frac{2^{r}}{(2 r+1) !}(n-1)(x-3) \ldots(n-2 r+1) x^{2 r}+\ldots\right] \\ =a_{0}\left[1+\sum_{r=1}^{\infty} \frac{(-1)^{r} 2^{r}}{(2 r) !} n(n-2) \ldots(n-2 r+2) x^{2 r}\right] \\ \left.+a_{0}\left[x+\sum_{r=1}^{\infty} \frac{(-1)^{r} 2^{r}}{(2 r+1)}(n-1)(n-3) \ldots(n-2 r+2) x^{2 r+1}\right] \quad \text { (If } a_{1}=a_{0}\right) \end{array}$

and for m=1 a1=0 by putting k=0,1,2,3,….. we get

ak+2 = 2(k+1-n)/(k+3)(k+2)ak

$a_{k+2}=\frac{2(k+1-n)}{(k+3)(k+2)} a_{k}$

$\begin{array}{l} a_{2}=-\frac{2(n-1)}{3 !} a_{0} \\ a_{4}=\frac{2^{2}(n-1)(n-3)}{5 !} a_{0} \\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ a_{2 r}=(-1)^{r} \frac{2^{r}(n-1)(n-3) \ldots(n-2 r+1)}{(2 r+1) !} a_{0} \end{array}$

so the solution will be

$=a_{0} x\left[1-\frac{2(n-1)}{3 !} x^{2}+\frac{2^{2}(n-1)(n-3)}{5 !} x^{4} \cdots+\frac{(-1)^{r} 2^{r}(n-1)(n-3) \ldots(n-2 r+1)}{(2 r+1) !} x^{2 r}+\ldots\right]$

so the complete solution is

$y=A\left[1-\frac{2 n}{2 !} x^{2}+\frac{2^{2} n(n-2)}{4 !} x^{4}-\ldots\right]+B\left[1-\frac{2(n-1)}{3 !} x^{2}+\frac{2^{2}(n-1)(n-3)}{5 !} x^{4} \ldots\right]$

where A and B are the arbitrary constants

## Hermite Polynomial

The Hermite’s equation solution is of the form y(x)=Ay1(x)+By2(x) where y1(x) and y2(x) are the series terms as discussed above,

$y_{1}(x)=1-\frac{2 n}{2 !} x^{2}+2^{2} n \frac{(n-2)}{4 !} x^{4}-\frac{2^{3} n(n-2)(n-4)}{6 !} x^{6}+\cdots$

$y_{2}(x)=x-\frac{2(n-1)}{3 !} x^{3}+\frac{2^{2}(n-1)(n-3)}{5 !} x^{5}-\frac{2^{3}(n-1)(n-3)(n-5)}{7 !} x^{7}+\cdots$

one of these series end if n is non negative integer if n is even y1 terminates otherwise y2 if n is odd, and we can easily verify that for n=0,1,2,3,4…….. these polynomials are

1,x,1-2x2, x-2/3 x3, 1-4x2+4/3x4, x-4/3x3+ 4/15x5

$1, x, 1-2 x^{2}, x-\frac{2}{3} x^{3}, 1-4 x^{2}+\frac{4}{3} x^{4}, x-\frac{4}{3} x^{3}+\frac{4}{15} x^{5}[/lac so we can say here that the solution of Hermite’s equation are constant multiple of these polynomials and the terms containing highest power of x is of the form 2nxn denoted by Hn(x) is known as Hermite polynomial ## Generating function of Hermite polynomial Hermite polynomial usually defined with the help of relation using generating function [latex]\mathrm{e}^{\left(2 x t-t^{2}\right)}=\sum_{n=0}^{\infty} \mathbf{H}_{\mathrm{n}}(\mathbf{x}) \frac{\mathrm{t}^{\mathrm{a}}}{\mathrm{n} !}, \quad$

\begin{aligned} \mathrm{e}^{\left(2 x t-t^{2}\right)}=\mathrm{e}^{2 u t} \mathrm{e}^{-t^{2}} &=\left[\sum_{m=0}^{\infty} \frac{(2 \mathrm{xt})^{\mathrm{m}}}{\mathrm{m} !}\right]\left[\sum_{\mathrm{k}=0}^{\infty} \frac{\left(-\mathrm{t}^{2}\right)^{\mathrm{k}}}{\mathrm{k} !}\right] \\ &=\sum_{\mathrm{n}=0}^{\infty} \sum_{\mathrm{k}=0}^{[\mathrm{n} / 2]} \frac{(-1)^{\mathrm{k}}(2 \mathrm{x})^{\mathrm{n}-2 \mathrm{k}}}{\mathrm{k} !(\mathrm{n}-2 \mathrm{k}) !} \mathrm{t}^{\mathrm{n}} \end{aligned}

[n/2] is the greatest integer less than or equal to n/2 so it follows the value of Hn(x) as

$\mathrm{H}_{\mathrm{n}}(\mathrm{x})=\sum_{\mathrm{k}=0}^{[\mathrm{n} / 2]} \frac{(-1)^{\mathrm{k}} \mathrm{n} !}{\mathrm{k} !(\mathrm{n}-2 \mathrm{k}) !}(2 \mathrm{x})^{\mathrm{n}-2 \mathrm{k}}$

$where \quad \left[\frac{\mathrm{n}}{2}\right]=\left\{\begin{array}{ll}\frac{\mathrm{n}}{2}, & \text { if } \mathrm{n} \text { is even } \\ \frac{\mathrm{n}-1}{2}, & \text { if } \mathrm{n} \text { is odd }\end{array}\right.$

this shows that Hn(x) is a polynomial of degree n in x and

Hn(x) = 2nxn + πn-2 (x)

$\mathrm{H}_{\mathrm{n}}(\mathrm{x})=2^{\mathrm{n}} \mathrm{x}^{\mathrm{n}}+\pi_{\mathrm{n}-2}(\mathrm{x})$

where πn-2 (x) is the polynomial of degree n-2 in x, and it will be even function of x for even value of n and odd function of x for odd value of n, so

Hn(-x) = (-1)n Hn(x)

$\mathrm{H}_{\mathrm{n}}(-\mathrm{x})=(-1)^{\mathrm{n}} \mathrm{H}_{\mathrm{n}}(\mathrm{x})$

some of the starting Hermite polynomials are

H0(x) = 1

H1(x) = 2x

H2(x) = 4x2 – 2

H3(x) = 8x3-12

H4(x) = 16x4 – 48x2+12

H5(x) = 32x2 – 160x3+120x

$\begin{array}{l} \mathrm{H}_{0}(\mathrm{x})=1 \\ \mathrm{H}_{1}(\mathrm{x})=2 \mathrm{x} \\ \mathrm{H}_{2}(\mathrm{x})=4 \mathrm{x}^{2}-2 \\ \mathrm{H}_{3}(\mathrm{x})=8 \mathrm{x}^{3}-12 \\ \mathrm{H}_{4}(\mathrm{x})=16 \mathrm{x}^{4}-48 \mathrm{x}^{2}+12 \\ \mathrm{H}_{5}(\mathrm{x})=32 \mathrm{x}^{5}-160 \mathrm{x}^{3}+120 \mathrm{x} \end{array}$

## Rodrigue Formula of Hermite polynomial | Generating function of Hermite polynomial by Rodrigue Formula

Hermite Polynomial can also be defined with the help of Rodrigue formula using generating function

$\mathrm{H}_{\mathrm{n}}(\mathrm{x})=(-1)^{\mathrm{n}} \mathrm{e}^{\mathrm{x}^{2}} \frac{\mathrm{d}^{\mathrm{n}}}{\mathrm{dx}^{\mathrm{n}}}\left(\mathrm{e}^{-\mathrm{x}^{2}}\right)$

since the relation of generating function

$\mathrm{e}^{2 \mathrm{tx}-\mathrm{t}^{2}}=\mathrm{e}^{\mathrm{x}^{2}-(\mathrm{t}-\mathrm{x})^{2}}=\sum_{\mathrm{n}=0}^{\infty} \frac{\mathrm{H}_{\mathrm{n}}(\mathrm{x})}{\mathrm{n} !} \mathrm{t}^{\mathrm{n}}$

Using the Maclaurin’s theorem, we have

$\left.\frac{\partial^{\mathrm{n}}}{\partial \mathrm{t}^{\mathrm{n}}}\left(\mathrm{e}^{2 \mathrm{tx}-\mathrm{t}^{2}}\right)\right|_{\mathrm{t}=0}=\left.\mathrm{e}^{\mathrm{x}^{2}} \frac{\partial^{\mathrm{n}}}{\partial \mathrm{t}^{\mathrm{n}}}\left(\mathrm{e}^{-(t-\mathrm{x})^{2}}\right)\right|_{\mathrm{t}=0}=\mathrm{H}_{\mathrm{n}}(\mathrm{x})$

or

$\left.\frac{\partial^{\mathrm{n}}}{\partial \mathrm{t}^{\mathrm{n}}}\left[\mathrm{e}^{-(\mathrm{t}-\mathrm{x})^{2}}\right]\right|_{\mathrm{t}=0}=\mathrm{e}^{-\mathrm{x}^{2}} \mathrm{H}_{\mathrm{n}}(\mathrm{x})$

by putting z=x-t and

$\frac{\partial}{\partial \mathrm{t}}=-\frac{\partial}{\partial \mathrm{z}}$

for t=0,so z=x gives

$\begin{array}{l} \left.(-1)^{\mathrm{n}} \frac{\mathrm{d}^{\mathrm{n}}}{\mathrm{d} \mathrm{z}^{\mathrm{n}}}\left(\mathrm{e}^{-z^{2}}\right)\right|_{\mathrm{z}=\mathrm{x}}=(-1)^{\mathrm{n}} \frac{\mathrm{d}^{\mathrm{n}}\left(\mathrm{e}^{-\mathrm{x}^{2}}\right)}{\mathrm{dx}^{\mathrm{n}}}=\mathrm{e}^{-\mathrm{x}^{2}} \mathrm{H}_{\mathrm{n}}(\mathrm{x}) \\ \therefore \mathrm{H}_{\mathrm{n}}(\mathrm{x})=(-1)^{\mathrm{n}} \mathrm{e}^{\mathrm{x}^{2}} \frac{\mathrm{d}^{\mathrm{n}}}{\mathrm{d} \mathrm{x}^{\mathrm{n}}}\left(\mathrm{e}^{-\mathrm{x}^{2}}\right) \end{array}$

this we can show in another way as

$e^{x^{2}} \frac{\partial^{n}}{\partial t^{n}} e^{\left\{-(t-x)^{2}\right\}}=H_{n}(x)+H_{n+1}(x) t+H_{n+2}(x) . t^{2}+\ldots \ldots$

differentiating

$e^{ \left.-(t-x)^{2}\right\}$

with respect to t gives

$\frac{\partial}{\partial t} e^{\left\{-(t-x)^{2}\right\}}=-2(t-x) e^{\left\{-(t-x)^{2}\right\}}$

taking limit t tends to zero

$\lim _{t \rightarrow 0} \frac{\partial}{\partial t} e^{\left\{-(t-x)^{2}\right\}}=2 x e^{-x^{2}}$

now differentiating with respect to x

$\frac{\partial}{\partial x} e^{\left\{-(t-x)^{2}\right\}}=(-1)^{2}(t-x) e^{\left\{-(t-x)^{2}\right\}}$

taking limit t tends to zero

$\lim _{t \rightarrow 0} \frac{\partial}{\partial x} e^{\left\{-(t-x)^{2}\right\}}=-2 x e^{-x^{2}}$

from these two expressions we can write

$\left.\lim _{t \rightarrow 0} \frac{\partial}{\partial t} e^{\left\{-(t-x)^{2}\right\}}=(-1)^{1} \lim _{t \rightarrow 0} \frac{\partial}{\partial x} e^{\left\{-(t-x)^{2}\right.}\right\}$

in the same way we can write

$\left.\lim _{t \rightarrow 0} \frac{\partial^{2}}{\partial t^{2}} e^{\left\{-(t-x)^{2}\right\}}=(-1)^{2} \lim _{t \rightarrow 0} \frac{\partial^{2}}{\partial x^{2}} e^{\left\{-(t-x)^{2}\right.}\right\}$

$\lim _{t \rightarrow 0} \frac{\partial^{n}}{\partial t^{n}} e^{\left\{-(t-x)^{2}\right\}}=(-1)^{n} \lim _{t \rightarrow 0} \frac{\partial^{n}}{\partial x^{n}} e^{\left\{-(t-x)^{2}\right\}}=(-1)^{n} \frac{d^{n}}{d x^{n}} e^{-x^{2}}$

differentiating n times put t=0, we get

$\lim _{t \rightarrow 0} e^{x^{2}} \frac{\partial^{n}}{\partial t^{n}} e^{\left\{-(t-x)^{2}\right\}}=H_{n}(x)$

from these values we can write

$\begin{array}{l} (-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}} e^{-x^{2}}=H_{n}(x) \\ H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}} e^{-x^{2}} \\ n=0 \end{array}$

from these we can get the values

$\begin{array}{l} n=0\\ H_{0}(x)=(-1)^{0} e^{x^{2}} e^{-x^{2}}=1 \\ H_{0}(x)=1 \end{array}$

$\begin{array}{l} n=1\\ H_{1}(x)=(-1)^{1} e^{x^{2}} \frac{d}{d x} e^{-x^{2}}=-e^{x^{2}}(-2 x) e^{-x^{2}}=2 x \\ H_{1}(x)=2 x \\ n=2 \end{array}$

\begin{aligned} H_{2}(x) &=(-1)^{2} e^{x^{2}} \frac{d^{2}}{d x^{2}} e^{-x^{2}}=e^{x^{2}} \frac{d}{d x}\left(-2 x e^{-x^{2}}\right) \\ &=e^{x^{2}}\left[-2 e^{x^{2}}-2 x(-2 x) e^{-x^{2}}\right.\\ &=-2+4 x^{2} \\ & H_{2}(x)=4 x^{2}-2 \\ n=3 \end{aligned}

\begin{aligned} H_{3}(x) &=(-1)^{3} e^{x^{2}} \frac{d^{3}}{d x^{3}}\left(e^{-x^{2}}\right)=-e^{x^{2}} \frac{d^{2}}{d x^{2}}\left(-2 x e^{-x^{2}}\right) \\ &=-e^{x^{2}} \frac{d}{d x}\left(-2 e^{-x^{2}}+(-2 x)(-2 x) e^{-x^{2}}\right) \\ &=-e^{x^{2}} \frac{d}{d x}\left(-2+4 x^{2}\right) e^{-x^{2}}=-e^{x^{2}}\left[8 x e^{-x^{2}}+\left(4 x^{2}-2\right)(-2 x) e^{-x^{2}}\right] \end{aligned}

$\begin{array}{l} =-\left[8 x+\left(4 x^{2}-2\right)(-2 x)\right]=-8 x+8 x^{3}-4 x=8 x^{3}-12 x \\ H_{3}(x)=8 x^{3}-12 x \\ H_{4}(x)=16 x^{4}-48 x^{2}+12 \end{array}$

$\begin{array}{l} H_{5}(x)=32 x^{5}-160 x^{3}+120 x \\ H_{6}(x)=64 x^{6}-480 x^{4}+720 x^{2}-120 \\ H_{7}(x)=128 x^{7}-1344 x^{5}+3360 x^{3}-1680 x \end{array}$

## Example on Hermite Polynomial

1. Find the ordinary polynomial of

$2 H_{4}(x)+3 H_{3}(x)-H_{2}(x)+5 H_{1}(x)+6 H_{0}$

Solution: using the Hermite polynomial definition and the relations we have

$\begin{array}{l} 2 H_{4}(x)+3 H_{3}(x)-H_{2}(x)+5 H_{1}(x)+6 H_{0} \\ \quad=2\left[16 x^{4}-48 x^{2}+12\right]+3\left\{8 x^{3}-12 x\right\}-\left(4 x^{2}-2\right)+5(2 x)+6(1) \\ \quad=32 x^{4}-96 x^{2}+24+24 x^{3}-36 x-4 x^{2}+2+10 x+6\\ =32 x^{4}+24 x^{3}-100 x^{2}-26 x+32 \end{array}$

2. Find the Hermite polynomial of the ordinary polynomial

$64 x^{4}+8 x^{3}-32 x^{2}+40 x+10$

Solution: The given equation we can convert to Hermite as

\begin{aligned} 64 x^{4}+8 x^{3} &-32 x^{2}+40 x+10=\mathrm{AH}_{4}(x)+\mathrm{BH}_{3}(x)+\mathrm{CH}_{2}(x)+\mathrm{DH}_{1}(x)+\mathrm{EH}_{0}(x) \\ &=\mathrm{A}\left(16 x^{4}-48 x^{2}+12\right)+\mathrm{B}\left(8 x^{3}-12 x\right)+\mathrm{C}\left(4 x^{2}-2\right)+\mathrm{D}(2 x)+\mathrm{E}(1) \\ &=16 \mathrm{~A} x^{4}+8 \mathrm{~B} x^{3}(-48 \mathrm{~A}+4 \mathrm{C}) x^{2}+(-12 \mathrm{~B}+2 \mathrm{D}) x+12 \mathrm{~A}-2 \mathrm{C}+\mathrm{E} \end{aligned}

and from this equation equating the same powers coefficient

\begin{aligned} 16 \mathrm{~A}=64 & \Rightarrow \mathrm{A}=4 \\ 8 \mathrm{~B}=8 & \Rightarrow \mathrm{B}=1 \\ -48 \mathrm{~A}+4 \mathrm{C}=-32 & \Rightarrow 4 \mathrm{C}=-32+192 \Rightarrow \mathrm{C}=40 \\ -12 \mathrm{~B}+2 \mathrm{D}=40 & \Rightarrow-12+2 \mathrm{D}=40 \Rightarrow 2 \mathrm{D}=52 \Rightarrow \mathrm{D}=26 \\ 12 \mathrm{~A}-2 \mathrm{C}+\mathrm{E}=10 & \Rightarrow 12 \times 4-2(40)+\mathrm{E}=10 \Rightarrow \mathrm{E}=42 \end{aligned}

hence the Hermite polynomial will be

$4 \mathrm{H}_{4}(x)+\mathrm{H}_{3}(x)+40 \mathrm{H}_{2}(x)+26 \mathrm{H}_{1}(x)+42 \mathrm{H}_{0}(x)$

## Orthogonality of Hermite Polynomial | Orthogonal property of Hermite Polynomial

The important characteristic for Hermite polynomial is its orthogonality which states that

$\int_{-\infty}^{\infty} e^{-x^{2}} H_{m}(x) H_{n}(x) d x=\left\{\begin{array}{ll} 0, & m \neq n \\ 2^{n} n ! \sqrt{\pi}, & m=n \end{array}\right.$

To prove this orthogonality let us recall that

$e^{\left\{x^{2}-\left(t_{1}-x\right)^{2}\right\}}=\sum \frac{H_{n}(x)}{n !} t_{1}^{n}$

which is the generating function for the Hermite polynomial and we know

$e^{\left\{x^{2}-\left(t_{2}-x\right)^{2}\right\}}=\sum \frac{H_{m}(x)}{m !} t_{2}^{m}$

so multiplying these two equations we will get

\begin{aligned} e^{\left\{x^{2}-\left(t_{1}-x\right)^{2}\right\}} \cdot e^{\left\{x^{2}-\left(t_{2}-x\right)^{2}\right\}} &=\left[\sum_{n=0}^{\infty} \frac{H_{n}(x)}{n !} t_{1}^{n}\right]\left[\sum_{m=0}^{\infty} \frac{H_{m}(x)}{m !} t_{2}^{m}\right] \\ &=\sum_{n=0}^{\infty}\left[H_{n}(x)\left|H_{m}(x)\right|\right] \frac{t_{1}^{n} \cdot t_{2}^{m}}{n ! m !} \end{aligned}

multiplying and integrating within infinite limits

$\begin{array}{l} \left.\left.\sum_{n m}\left[\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x\right] \frac{t_{1}^{n} t_{2}^{m}}{n ! m !}=e^{-x^{2}} \int_{-\infty}^{\infty} e^{\left\{x^{2}-\left(t_{1}-x\right)^{2}\right.}\right\}_{.} e^{\left\{x^{2}-\left(t_{2}-x\right)^{2}\right.}\right\}_{d x} \\ =\int_{-\infty}^{\infty} e^{\left\{x^{2}-\left(t_{1}-x\right)^{2}\right\}-\left(t_{2}-x\right)^{2}} d x \\ =e^{\left(-\left(t_{1}^{2}+t_{2}^{2}\right)\right\}} \int_{-\infty}^{\infty} e^{\left\{-x^{2}+2 x\left(t_{1}+t_{2}\right)\right\}} d x \end{array}$

and since

$\int_{-\infty}^{\infty} e^{\left\{-a x^{2}+2 b x\right\}} d x=\sqrt{\frac{\pi}{2} e^{\frac{b^{2}}{a}}} \quad$

so

$\int_{-\infty}^{\infty} e^{\left\{-x^{2}+2 x\left(t_{1}+t_{2}\right)\right\}} d x=\sqrt{\pi} e^{\left(t_{1}+t_{2}\right)^{2}}$

using this value in above expression we have

\begin{aligned} e^{\left\{-\left(i+1+r_{2}\right)^{2}\right\}} \cdot \sqrt{\pi} e^{\left(t_{1}+t_{2}\right)^{2}} &=\sqrt{\pi} e^{-t^{2}-t_{2}^{2}+t_{1}^{2}+t_{2}^{2}+2 \uparrow r_{2}}=\sqrt{\pi} e^{2 l_{1} l_{2}} \\ &=\sqrt{\pi}\left[1+2 t_{1} t_{2}+\frac{\left(2 t_{1} t_{2}\right)^{2}}{2 !}+\frac{\left(2 t_{1} t_{2}\right)^{3}}{3 !}+\ldots \ldots .\right]=\sqrt{\pi} \sum \frac{\left(2 t_{1} t_{2}\right)^{n}}{n !} \\ &=\sqrt{\pi} \sum \frac{2^{n} t_{1}^{n} t_{2}^{n}}{n !}=\sqrt{\pi} \sum_{m=0 \atop n=0}^{\infty} 2^{n} t_{1}^{n} t_{2}^{m} \delta_{m, n} \quad\left[t_{2}^{n}=t_{2}^{m} \delta_{n, m}\right] \end{aligned}

which gives

$\sum_{n m}\left[\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x\right] \frac{t_{1}^{n} t_{2}^{m}}{n ! m !}=\sqrt{\pi} \sum_{n m} \frac{2^{n}}{n !} e^{n} t_{2}^{m} \delta_{n, m}$

now equate the coefficients on both the sides

$\begin{array}{ll} & \int_{-\infty}^{\infty} e^{-x^{2}} \frac{H_{n}(x) H_{m}(x)}{n ! m !} d x=\frac{\sqrt{\pi} 2^{n}}{n !} \delta_{n, m} \\ \Rightarrow & \int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} m \mid \delta_{n, m} \\ \Rightarrow & \int_{-\infty}^{\infty} e^{-x^{2}} H_{m}(x) H_{n}(x) d x=\left\{\begin{array}{ll} 0 & m \neq n\left[\delta_{n, m}=0, \text { if } m \neq n\right. \\ 2^{n} n ! \sqrt{\pi}, & m=n \end{array}\left[\begin{array}{l} =1, \text { if } m=n \end{array}\right]\right. \end{array}$

which shows the orthogonal property of Hermite polynomial.

The result of orthogonal property of Hermite polynomial can be shown in another way by considering the recurrence relation

## Example on orthogonality of Hermite Polynomial

1.Evaluate the integral

$\int_{-\infty}^{\infty} e^{-x^{2}} H_{2}(x) H_{3}(x) d x$

Solution: By using the property of orthogonality of hermite polynomial

$\begin{array}{l} \int_{-\infty}^{\infty} e^{-x^{2}} H_{m}(x) H_{n}(x)=0 \text { if } m \neq n \end{array}$

since the values here are m=3 and n=2 so

$\int_{-\infty}^{\infty} e^{-x^{2}} H_{2}(x) H_{3}(x)=0$

2. Evaluate the integral

$\int_{-\infty}^{\infty} e^{-x^{2}}\left[H_{2}(x)\right]^{2} d x$

Solution: Using the orthogonality property of Hermite polynomial we can write

$\begin{array}{l} \int_{-\infty}^{\infty} e^{-x^{2}}\left[H_{n}(x)\right]^{2} d x=2^{n}(n) ! \sqrt{\pi} \\ \int_{-\infty}^{\infty} e^{-x^{2}}\left[H_{2}(x)\right]^{2} d x=2^{2}(2 !) \sqrt{\pi}=8 \sqrt{\pi} \end{array}$

## Recurrence relations of Hermite polynomial

The value of Hermite polynomial can be easily find out by the recurrence relations

These relations can easily obtained with the help of definition and properties.

Proofs:1. We know the Hermite equation

y”-2xy’+2ny = 0

$y^{\prime \prime}-2 x y^{\prime}+2 n y=0$

and the relation

$e^{2 tx-t^{2}}=\sum_{n=0}^{\infty} \frac{H_{n}(x) t^{n}}{n !}$

by taking differentiation with respect to x partially we can write it as

$2 t e^{2t x-t^{2}}=\sum_{n=0}^{\infty} H_{n}^{‘}(x) \frac{r^{m}}{n !}$

from these two equations

$\quad 2 t \sum_{n=0}^{\infty} \frac{H_{n}(x) t^{n}}{n !}=\sum_{n=0}^{\infty} H_{n}^{‘}(x) \frac{t^{n}}{n !}$

$\quad 2 \sum_{n=0}^{\infty} \frac{H_{n}(x) t^{n+1}}{n !}=\sum_{n=0}^{\infty} H_{n}^{\prime}(x) \frac{t^{n}}{n !}$

now replace n by n-1

$2 \frac{H_{\mathrm{m}-1}(\mathrm{x}) t^{n}}{(n-1) !}=H_{n}^{‘}(x) \frac{t^{n}}{n !}$

$\quad \frac{2 n H_{n-1}(x) t^{n}}{n !}=H_{n}^{\prime}(x) \frac{t^{n}}{n !}$

by equating the coefficient of tn

$2 \frac{n !}{(n-1) !} H_{n-1}(x)=H^{\prime}{ }_{n}(x)$

$\quad 2 n H_{n-1}(x)=H_{n}^{\prime}(x)$

so the required result is

$\mathbf{2 n H_{n-1}(x)=H_{n}^{\prime}(x)}$

2. In the similar way differentiating partially with respect to t the equation

$e^{2 tx-t^{2}}=\sum_{n=0}^{\infty} \frac{H_{n}(x) t^{n}}{n !}$

we get

$2(x-t) e^{2 tx-t^{2}}=\sum_{n=0}^{\infty} H_{n}(x) \frac{n t^{n-1}}{(n-1) !}$

$2(x-t) e^{2tx-t^{2}}=\sum_{n=1}^{\infty} H_{n}(x) \frac{t^{n-1}}{(n-1) !}$

n=0 will be vanished so by putting this value of e

$2(x-t) \sum_{n=0}^{\infty} H_{n}(x) \frac{t^{n}}{n !}=\sum_{n=1}^{\infty} H_{n}(x) \frac{t^{n-1}}{(n-1) !}$

$\quad 2 x \sum_{n=0}^{\infty} H_{n}(x) \frac{t^{n}}{n !}-2 \sum_{n=0}^{\infty} H_{n}(x) \frac{t^{n+1}}{n !}=\sum_{n=1}^{\infty} H_{n}(x) \frac{r^{n-1}}{(n-1) !}$

now equating the coefficients of tn

$2 x \frac{H_{n}(x)}{n !}-2 \frac{H_{n-1}(x)}{(n-1) !}=\frac{H_{n+1}(x)}{n !} \quad$

thus

$\quad \mathbf{2 x H_{n}(x)=2 n H_{n-1}(x)+H_{n+1}(x)}$

3. To prove this result we will eliminate Hn-1 from

$2 x H_{n}(x)=2 n H_{n-1}(x)+H_{n+1}(x)$

and

$2 n H_{n-1}(x)=H_{n}^{\prime}(x)$

so we get

\begin{aligned} 2 x H_{n}(x)=2 n H_{n-1}(x)+H_{m+1}(x) &(x) \\ 2 x H_{n}(x)=H_{n}^{r}(x)+H_{n+1}(x) \\\ldots \ldots\end{aligned}

thus we can write the result

$\mathbf{H_{n}^{\prime}(x)=2 x H_{n}(x)-H_{n+1}(x)}$

4. To prove this result we differentiate

$H_{n}^{\prime}(x)=2 x H_{n}(x)-H_{n+1}(x)$

we get the relation

$H_{n}^{\prime \prime}(x)=2 x H_{n}^{‘}(x)+2 H_{n}(x)-H_{n+1}^{\prime}(x)$

substituting the value

$H_{n+1}^{‘}(x)=2(n+1) H_{n}(x)$

and replacing n by n+1

$H_{n}^{‘}(x)=2 \mathrm{x} H_{n}^{\prime}(x)+2 H_{n}(x)-2(n+1) H_{n}(x)$

$\quad H_{n}^{‘}(x)-2 x H_{n}^{\prime}(x)+2 n H_{n}(x)=0$

which gives

$\mathbf{H_{n}^{\prime \prime}(x)-2 x H_{n}^{1}(x)+2 n H_{n}(x)=0]}$

## Examples on Recurrence relations of Hermite polynomial

1.Show that

H2n(0) = (-1)n. 22n (1/2)n

$H_{2 n}(0)=(-1)^{n} \cdot 2^{2 n}\left(\frac{1}{2}\right)^{n}$

Solution:

To show the result we have

H2n(x) =

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

taking x=0 here we get

\begin{aligned} H_{2 n}(0) &=\frac{(-1)^{n}(2 n) !}{(n) !}=(-1)^{n} \frac{(2 n)(2 n-1)(2 n-2) \cdot \ldots}{n(n-1)(n-2) \ldots \ldots 1} \\ &=(-1)^{n} \frac{2(2 n-1) 2(2 n-3) 2(2 n-5) 2 \cdot \ldots 2.1}{n !} n ! \\ &=(-1)^{n} 2^{n} \cdot 2^{n} \frac{(2 n-1)}{2} \frac{(2 n-3)}{2} \frac{(2 n-5)}{2} \\ &=(-1)^{n} 2^{2 n}\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)\left(\frac{7}{2}\right) \ldots \ldots\left(\frac{2 n-3}{2}\right)\left(\frac{2 n-1}{2}\right) \\&=(-1)^{n} 2^{2 n}\left(\frac{1}{2}\right)^{m} \end{aligned}

2. Show that

H’2n+1(0) = (-1)n 22n+1 (3/2)2

$H^{\prime}{ }_{2 n+1}(0)=(-1)^{n} 2^{2 n+1}\left(\frac{3}{2}\right)^{2}$

Solution:

Since from the recurrence relation

H’n(x) = 2nHn-1(X)

$H_{n}^{\prime}(x)=2 n H_{n-1}(x)$

here replace n by 2n+1 so

H’2n-1(x) = 2(2n+1) H2n(x)

$H_{2 n+1}^{\prime}(x)=2(2 n+1) H_{2 n}(x)$

taking x=0

\begin{aligned} H_{2 n+1}^{\prime}(0) &=2(2 n+1) H_{2 n}(0) \\ &=2(2 n+1)(-1)^{n} 2^{2 n}\left(\frac{1}{2}\right)^{n} \\ &=(2 n+1)(-1)^{n} 2^{2 n+1}\left[\frac{(2 n-1)(2 n-3) \ldots \ldots 3.1}{2^{n}}\right]\\ &=(-1)^{n} 2^{2 n+1}\left[\frac{3}{2}\left(\frac{3}{2}+1\right) \ldots \ldots\left(\frac{3}{2}+n-1\right)\right] \\ &=(-1)^{n} \cdot 2^{2 n+1}\left(\frac{3}{2}\right)^{n} \end{aligned}

3. Find the value of

H2n+1(0)

$H_{2 n+1}(0)$

Solution

Since we know

$H_{2 n+1}(x)=\sum_{k=0}^{2 n+1 / 2} \frac{(-1)^{k}(2 n+1) !(2 x)^{2 n+1-2 k}}{k !(2 n+1-2 k)}$

use x=0 here

H2n-1(0) = 0

$\therefore H_{2 n+1}(0)=0$

4. Find the value of H’2n(0).

Solution :

we have the recurrence relation

H’n(x) = 2nHn-1(x)

$H^{\prime}{ }_{n}(x)=2 n H_{n-1}(x)$

here replace n by 2n

H’2n(x) = =2(2n)H2n-1(x)

$H^{\prime}_{2 n}(x)=2(2 n) H_{2 n-1}(x)$

put x=0

H’2n(0) = (4n)H2n-1(0) = 4n*0=0

$H^{\prime}_{2 n}(0)=(4 n) H_{2 n-1}(0)=4n*0=0$

5. Show the following result

$\frac{d^{m}}{d x^{m}}\left\{H_{n}(x)\right\}=\frac{2^{n}(n) !}{(n-m) !} H_{n-m} \quad m<n$

Solution :

Using the recurrence relation

H’n(x) = 2nHn-1 (x)

$H^{\prime}{ }_{n}(x)=2 n H_{n-1}(x)$

so

\begin{aligned} \quad \frac{d}{d x}\left\{H_{n}(x)\right\} &=2 m H_{n-1}(x) \\ \quad \frac{d^{2}}{d x^{2}}\left\{H_{n}(x)\right\} &=2 n \frac{d}{d x}\left[H_{n-1}(x)\right] \\ &=2 n H^{\prime} n-1 \atop(x) \\ &=2 n\left[2(n-1) H_{n-2}(x)\right] \\ &=2^{2} n(n-1) H_{n-2}(x) \end{aligned}

and

d3/dx3 {Hn(x)} = 23n(n-1)(n-2)Hn-3(x)

$\frac{d^{3}}{d x^{3}}\left\{H_{n}(x)\right\}=2^{3} n(n-1)(n-2) H_{n-3}(x)$

differentiating this m times

$\frac{d^{m}}{d^{m}}\left\{H_{n}(x)\right\}=2^{m} n(n-1) \ldots \ldots(n-m+1) H_{n-m}(x)\\=\frac{2^{\prime m}}{(n-m) !} H_{n-w}(x), m<n$

which gives

$\frac{d^{m}}{d x^{m}}\left{H_{n}(x)\right}=\frac{2^{n}(n) !}{(n-m) !} H_{n-m} \quad m<n$

6. Show that

Hn(-x) = (-1)n Hn(x)

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

Solution :

we can write

$\sum_{n=0}^{\infty} \frac{H_{n}(x) t^{n}}{n !}=e^{2 n-t^{2}}=e^{2 \pi} e^{-t^{2}}=\sum_{n=0}^{\infty} \frac{(2 x)^{n} t^{n}}{n !} \times \sum_{n=0}^{\infty} \frac{(-1) t^{2 n}}{n !}$

$=\sum_{n=0}^{\infty} \sum_{k=0}^{n / 2} \frac{(-1)^{k}(2 x)^{n-2 k}}{k(n-2 k) !}$

from the coefficient of tn we have

$H_{n}(x)=\sum_{k=0}^{n / 2} \frac{(-1)^{k} n !(2 x)^{n-2 k}}{k !(n-2 k) !}$

and for -x

\begin{aligned} H_{n}(-x) &=\sum_{k=0}^{\pi / 2} \frac{(-1)^{k} n !(-2 x)^{n-2 k}}{k(n-2 k) !} \\ &=\sum_{k=0}^{n / 2} \frac{(-1)^{k}(-1)^{n-2 k} n !(2 x)^{n-2 k}}{k(n-2 k) !} \\ &=(-1)^{n} \sum_{k=0}^{n / 2} \frac{(-1)^{k} n !(2 x)^{n-2 k}}{k(n-2 k) !}=(-1)^{n} H_{n}(x) \end{aligned}

7. Evaluate the integral and show

$\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{x}\left[2^{n-1} m \mid 8_{m, n-1}+2^{n}(n+1) \delta_{n * 1, m}\right] .$

Solution : For solving this integral use integration parts as

$\begin{array}{l} \int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\left[-\frac{1}{2} e^{-x^{2}} H_{n}(x) H_{m}(x) d x\right]_{-\infty}^{\infty} \\ \quad+\frac{1}{2} \int_{-\infty}^{\infty} e^{-x^{2}} \frac{d}{d x}\left\{H_{n}(x) H_{m}(x)\right\} d x \\ =0+\frac{1}{2} \int_{-\infty}^{\infty} e^{-x^{2}} \frac{d}{d x}\left\{H_{n}(x) H_{m}(x)\right\} d x \text { (Orthogonality property) } \end{array}$

Now differentiation under the Integral sign differentiate with

respect to x

$=\frac{1}{2} \int_{-\infty}^{\infty} e^{-x^{2}}\left\{H_{n}^{\prime}(x) H_{m}(x)+H_{n}(x) H_{m}^{\prime}(x)\right\} d x$

using

H’n(x) = 2nHn-1 (x)

$H_{n}^{\prime}(x)=2 n H_{n-1}(x)$

and

H’m(x) = 2mHm-1 (x)

$H_{m}^{\prime}(x)=2 m H_{m-1}(x)$

we have

$\begin{array}{l} =\frac{1}{2} \int_{-\infty}^{\infty} e^{-x^{2}}\left[2 n H_{n-1}(x) H_{m}(x)+2 m H_{n}(x) H_{m-1}(x)\right] d x \\ =n \int_{-\infty}^{\infty} e^{-x^{2}} H_{n-1}(x) H_{m}(x) d x+m \int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m-1}(x) d x \\ =n \sqrt{\pi} 2^{n-1}(n-1) ! \delta_{m, n-1}+m \sqrt{\pi} 2^{n} n ! \delta_{n, m-1} \end{array}$

and since

𝝳 n,m-1 = 𝝳n+1, m

$\delta_{n, m-1}=\delta_{n+1, m}$

so the value of integral will be

$=\sqrt{\pi}\left[2^{n-1} n ! \delta_{m, n-1}+2^{n}(n+1) ! \delta_{n+1, m}\right]$

## Conclusion:

The specific polynomial which frequently occurs in application is Hermite polynomial, so the basic definition, generating function , recurrence relations and examples related to Hermite Polynomial were discussed in brief here   , if you require further reading go through

https://en.wikipedia.org/wiki/Hermite_polynomials