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
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 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
by following this put the values of a0,a1,a2,a3,a4 and a5 we have
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
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)=Ay1(x)+By2(x) where y1(x) and y2(x) are the series terms as discussed above,
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
[n/2] is the greatest integer less than or equal to n/2 so it follows the value of Hn(x) as
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
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
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 tn
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 tn
thus
3. To prove this result we will eliminate Hn-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
H2n(0) = (-1)n. 22n (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 22n+1 (3/2)2
Solution:
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
3. Find the value of
H2n+1(0)
Solution
Since we know
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
Solution :
Using the recurrence relation
H’n(x) = 2nHn-1 (x)
so
and
d3/dx3 {Hn(x)} = 23n(n-1)(n-2)Hn-3(x)
differentiating this m times
which gives
6. Show that
Hn(-x) = (-1)n Hn(x)
Solution :
we can write
from the coefficient of tn 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) = 2nHn-1 (x)
and
H’m(x) = 2mHm-1 (x)
we have
and since
𝝳 n,m-1 = 𝝳n+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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