Hermite Polynomial: 9 Complete Quick Facts

  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

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

with the help of series solution of differential equation

101 1

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

image 136

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

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


and equating the coefficients of xm+k to zero,

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

we can write it as

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

if m=0

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

if m=1

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

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


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

image 140

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

image 141

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

image 142

so the solution will be

image 143

so the complete solution is

image 144

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,

image 145
image 146

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

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

image 150
image 149

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

image 151
image 152

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

Hn(x) = 2nxn + πn-2 (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)

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

Generating function of Hermite polynomial by Rodrigue Formula

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

image 153

since the relation of generating function

image 154

  Using the Maclaurin’s theorem, we have

image 155


by putting z=x-t and

for t=0,so z=x gives

this we can show in another way as


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           

  1. 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


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

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

Hermite polynomial
Hermite polynomial 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

image 174

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

image 175

from these two equations

image 176
image 177

now replace n by n-1

image 178
image 179

by equating the coefficient of tn

image 180
image 181

so the required result is

image 182

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

image 183

we get

image 184
image 185

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

image 186
image 187

now equating the coefficients of tn

image 188


image 189

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

image 190


image 191

so we get

image 192

thus we can write the result

image 193

4. To prove this result we differentiate

image 194

we get the relation

image 195

substituting the value

image 196

and replacing n by n+1

image 197

which gives

image 173

Examples on Recurrence relations of Hermite polynomial

1.Show that

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


To show the result we have

image 172

H2n(x) =

taking x=0 here we get

image 171

2. Show that

H’2n+1(0) = (-1)n 22n+1 (3/2)2


Since from the recurrence relation

H’n(x) = 2nHn-1(X)

here replace n by 2n+1 so

H’2n-1(x) = 2(2n+1) H2n(x)

taking x=0

image 170

3. Find the value of



Since we know

image 169

use x=0 here

H2n-1(0) = 0

4. Find the value of H’2n(0).

Solution :

we have the recurrence relation

H’n(x) = 2nHn-1(x)

here replace n by 2n

H’2n(x) = =2(2n)H2n-1(x)

put x=0

H’2n(0) = (4n)H2n-1(0) = 4n*0=0

5. Show the following result

image 168

Solution :

Using the recurrence relation

H’n(x) = 2nHn-1 (x)


image 167


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

differentiating this m times

image 166

which gives

image 165

6. Show that

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

Solution :

we can write

image 163
image 164

from the coefficient of tn we have

image 162

and for -x

image 161

7. Evaluate the integral and show

Solution : For solving this integral use integration parts as

image 160

Now differentiation under the Integral sign differentiate with

respect to x

image 159


H’n(x) = 2nHn-1 (x)


H’m(x) = 2mHm-1 (x)

we have

image 157

and since

???? n,m-1 = ????n+1, m

so the value of integral will be

image 156


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


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