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

d^{2}y/dx^{2} – 2x dy/dx + 2xy = 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

d^{2}y/dx^{2} – 2x dy/dx + 2ny = 0

with the help of series solution of differential equation

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

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 x^{m-2} take k=0 in the first equation as the second gives negative value, so the coefficient x^{m-2} is

a_{0}m (m-1)=0 ⇒ m=0,m=1

as a_{0} ≠ 0

now in the same way equating the coefficient of x^{m-1} from the second summation

and equating the coefficients of x^{m+k} to zero,

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

we can write it as

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

if m=0

a_{k+2} = 2(k-n)/(k+2)(k+1) a_{k}

if m=1

a_{k+2} = 2(k+1-n)/(k+3)(k+2) a_{k}

for these two cases now we discuss the cases for k

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

If, $k=0 a_{2} =-2 n/2 a_{0}=-n a_{0}$

$k=1, a_{3}=2(1-n)/6 a_{1} =-2(n-1)/3 ! a_{1}$

If $k=2, a_{4} =2(2-n)/12 a_{2} =2 (2-n)/12 (-n a_{0}) =2^{2 }n(n-2)/4 ! a_{0}$

so far m=0 we have two conditions when a_{1}=0, then a_{3}=a_{5}=a_{7}=….=a_{2r+1}=0 and when a_{1} is not zero then

by following this put the values of a_{0},a_{1},a_{2},a_{3},a_{4} and a_{5} we have

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

a_{k+2} = 2(k+1-n)/(k+3)(k+2)a_{k}

so the solution will be

so the complete solution is

where A and B are the arbitrary constants

**Hermite Polynomial**

The Hermite’s equation solution is of the form y(x)=Ay_{1}(x)+By_{2}(x) where y_{1}(x) and y_{2}(x) are the series terms as discussed above,

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

1,x,1-2x^{2}, x-2/3 x^{3}, 1-4x^{2}+4/3x^{4}, x-4/3x^{3}+ 4/15x^{5}

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 2^{n}x^{n} denoted by H_{n}(x) is known as **Hermite polynomial**

**Generating function of Hermite polynomial**

Hermite polynomial usually defined with the help of relation using generating function

[n/2] is the greatest integer less than or equal to n/2 so it follows the value of **H _{n}(x)** as

this shows that **H _{n}(x)** is a polynomial of degree n in x and

H_{n}(x) = 2^{n}x^{n} + π_{n-2} (x)

where *π*_{n-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

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

some of the starting Hermite polynomials are

H_{0}(x) = 1

H_{1}(x) = 2x

H_{2}(x) = 4x^{2} – 2

H_{3}(x) = 8x^{3}-12

H_{4}(x) = 16x^{4} – 48x^{2}+12

H_{5}(x) = 32x^{2} – 160x^{3}+120x

## Generating function of Hermite polynomial by Rodrigue Formula

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

since the relation of generating function

Using the Maclaurin’s theorem, we have

or

by putting z=x-t and

for t=0,so z=x gives

this we can show in another way as

differentiating

with respect to t gives

taking limit t tends to zero

now differentiating with respect to x

taking limit t tends to zero

from these two expressions we can write

in the same way we can write

differentiating n times put t=0, we get

from these values we can write

from these we can get the values

**Example on Hermite Polynomial**

- Find the ordinary polynomial of

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

2. Find the Hermite polynomial of the ordinary polynomial

Solution: The given equation we can convert to Hermite as

and from this equation equating the same powers coefficient

hence the Hermite polynomial will be

**Orthogonality of Hermite Polynomial | Orthogonal property of Hermite Polynomial**

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

To prove this orthogonality let us recall that

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

so multiplying these two equations we will get

multiplying and integrating within infinite limits

and since

so

using this value in above expression we have

which gives

now equate the coefficients on both the sides

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

Solution: By using the property of orthogonality of hermite polynomial

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

2. Evaluate the integral

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

**Recurrence relations of Hermite polynomial**

**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

and the relation

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

from these two equations

now replace n by n-1

by equating the coefficient of t^{n}

so the required result is

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

we get

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

now equating the coefficients of t^{n}

thus

3. To prove this result we will eliminate H_{n-1} from

and

so we get

thus we can write the result

4. To prove this result we differentiate

we get the relation

substituting the value

and replacing n by n+1

which gives

## Examples on Recurrence relations of Hermite polynomial

1.Show that

H_{2n}(0) = (-1)^{n}. 2^{2n} (1/2)^{n}

**Solution:**

To show the result we have

H2n(x) =

taking x=0 here we get

2. Show that

H’_{2n+1}(0) = (-1)^{n} 2^{2n+1} (3/2)^{2}

**Solution:**

Since from the recurrence relation

H’_{n}(x) = 2nH_{n-1}(X)

here replace n by 2n+1 so

H’_{2n-1}(x) = 2(2n+1) H_{2n}(x)

taking x=0

3. Find the value of

H_{2n+1}(0)

**Solution**

Since we know

use x=0 here

H_{2n-1}(0) = 0

4. Find the value of H’_{2n}(0).

**Solution** :

we have the recurrence relation

H’_{n}(x) = 2nH_{n-1}(x)

here replace n by 2n

H’_{2n}(x) = =2(2n)H_{2n-1}(x)

put x=0

H’_{2n}(0) = (4n)H_{2n-1}(0) = 4n*0=0

5. Show the following result

**Solution** :

Using the recurrence relation

H’_{n}(x) = 2nH_{n-1} (x)

so

and

d^{3}/dx^{3 }{H_{n}(x)} = 2^{3}n(n-1)(n-2)H_{n-3}(x)

differentiating this m times

which gives

6. Show that

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

**Solution** :

we can write

from the coefficient of t^{n} we have

and for -x

7. Evaluate the integral and show

**Solution** : For solving this integral use integration parts as

Now differentiation under the Integral sign differentiate with

respect to x

using

H’_{n}(x) = 2_{n}H_{n-1} (x)

and

H’_{m}(x) = 2mH_{m-1} (x)

we have

and since

𝝳 _{n,m-1} = 𝝳_{n+1, m }

so the value of integral will be

**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

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