CONTENTS
 INTRODUCTION
 HERMITE EQUATION AND ITS SOLUTIONS
 PROPERTY OF HERMITE SERIES
* GENERATING FUNCTION
* RODRIGUE FORMULA
* ORTHOGONALITY
* EVEN AND ODD FUNCTIONS
* RECURRENCE RELATION
 REFERENCES
2

A differential equation that arises from the theory of the linear harmonic
oscillator is closely related to the Hermite equation is
𝑑2 𝑤
𝑑𝑥2 + 2𝑛 + 1 − 𝑥2 𝑤 = 0 , … 𝐴 , where n is a constant.
We are interested in the solution of (A) .
To find the solution of the (A) ,we have to simplify (A), for simplify (A) we
introduce a new dependent variable y by means of
𝑤 = 𝑦 𝑒−𝑥2 2 … (B)
This transform (A) into
𝑑2 𝑦
𝑑𝑥2 − 2x
𝑑𝑦
𝑑𝑥
+ 2𝑛𝑦 = 0 …(C) , where n is
constant
which is called Hermite’s Equation .
INTRODUCTION
3

HERMITE EQUATION AND ITS SOLUTIONS
Hermite equation is
𝑑2 𝑦
𝑑𝑥2 − 2x
𝑑𝑦
𝑑𝑥
+ 2𝑛𝑦 = 0 , where n is constant …(1)
We now solve the second order homogeneous equation in a series,
Let the series solution of (1) be y = 𝑚=0
∞
𝐶ₘ 𝑥 𝑘+𝑚
, C₀≠0. …(2)
Differentiating (2) and then putting the value of y , dy/dx and d2y/dx2 in (1) , we get
𝑚=0
∞
𝐶ₘ(𝑘 + 𝑚)(𝑘 + 𝑚 − 1)𝑥 𝑘+𝑚−2 − 2x𝐶ₘ(𝑘 + 𝑚)𝑥 𝑘+𝑚−1 + 2n 𝑚=0
∞
𝐶ₘ 𝑥 𝑘+𝑚 = 0
𝑚=0
∞
𝐶 ₘ(𝑘 + 𝑚)(𝑘 + 𝑚 − 1)𝑥 𝑘+𝑚−2 −2[ 𝑚=0
∞
𝐶ₘ (𝑘 + 𝑚)𝑥 𝑘+𝑚− 𝑚=0
∞
𝐶ₘ 𝑛𝑥 𝑘+𝑚]=0
𝑚=0
∞
𝐶ₘ (𝑘 + 𝑚)(𝑘 + 𝑚 − 1)𝑥 𝑘+𝑚−2
− 2 𝑚=0
∞
𝐶ₘ (𝑘 + 𝑚 − 𝑛) 𝑥 𝑘+𝑚
= 0 …(3)
To get the indicial equation , we equate to zero the coefficients of the smallest power of
𝑥 𝑘−2
in (3) and obtain
𝐶0 𝑘 𝑘 − 1 = 0 or 𝑘 𝑘 − 1 = 0 𝑎𝑠 𝐶0≠0. …(4)
so the roots of indicial equation (4) are 𝑘 = 0 , 1.
4

The next smallest power of 𝑥 is 𝑘 − 1 . So equating to zero the coefficients of 𝑥 𝑘−1 in (3) , we get
𝐶1 𝑘 + 1 𝑘 = 0
When 𝑘 = 0 ( One of the roots of the indicials equation ) , which shows that 𝐶1 is indeterminant .
Hence 𝐶0 and 𝐶1 may be taken as arbitrary constants . Equating to zero the coefficient of 𝑥 𝑘+𝑚−2
, (3) gives
𝐶 𝑚 k + m k + m − 1 − 2𝐶 𝑚−2 𝑘 + 𝑚 − 2 − 𝑛 = 0
𝐶ₘ =
2 (𝑘+𝑚−𝑛−2)
(𝑘+𝑚)(𝑘+𝑚−1)
𝐶 𝑚−2 …(6)
Putting 𝑘 = 0 in (6) gives
𝐶ₘ =
2 (𝑚−𝑛−2)
𝑚(𝑚−1)
𝐶 𝑚−2 …(7)
5

put m=2,4,6,…,2m in (7), we have
𝐶2=
−2𝑛
2!
𝐶0
𝐶4=
−1 222 𝑛(𝑛−2)
4!
𝐶0
…
𝐶2𝑚 =
−1 𝑚2 𝑚 𝑛 𝑛−2 …(𝑛−2𝑚+2)
2𝑚!
Put m=3,5,7,…,2m+1 in (7) , we have
𝐶3=
−1 121(𝑛−1)
3!
𝐶1
𝐶5=
−1 222(𝑛−1)(𝑛−3)
5!
𝐶1
… 𝐶2𝑚+1=
−1 𝑚2 𝑚 𝑛−1 𝑛−3 …(𝑛−2𝑚+1)
2𝑚+1 !
𝐶1
Now put above values in (2) with k=0 , we get 𝑦 = 𝐶0 + 𝐶1 𝑥 + 𝐶2 𝑥2 + 𝐶3 𝑥3 + …
𝑦 = 𝐶0[1 −
2𝑛
2!
𝑥2 +
−1 222 𝑛(𝑛−2)
4!
𝑥4 + ⋯ +
−1 𝑚2 𝑚 𝑛 𝑛−2 … 𝑛−2𝑚+2
2𝑚!
𝑥2𝑚] +
𝐶1[𝑥 −
21 𝑛−1
3!
𝑥3 +
22 𝑛−1 𝑛−3
5!
𝑥5 + ⋯ +
−1 𝑚2 𝑚 𝑛−1 𝑛−3 … 𝑛−2𝑚+1
2𝑚+1 !
𝑥2𝑚+1]
𝑦 = 𝐶0 𝑢 + 𝐶1v
Replacing m by m+2 in (7) , 𝐶 𝑚+2 =
2(𝑚−𝑛)
(𝑚+1)(𝑚+2)
𝐶 𝑚 …(10)
6

Now , we shall obtain the series solution of y = 𝑚=0
∞
𝐶ₘ 𝑥 𝑘+𝑚 in descending power of x by
assuming n to be a non negative integer . Re-writing (2) for k=0
𝑦 = 𝐶 𝑛 𝑥 𝑛 + 𝐶 𝑛−2 𝑥 𝑛−2 + 𝐶 𝑛−4 𝑥 𝑛−4 + … …(11)
𝐶 𝑚=
−(𝑚+1)(𝑚+2)
2(𝑛−𝑚)
𝐶 𝑚+2 …(12)
Put 𝑚 = 𝑛 − 2 , 𝑛 − 4 , … 𝑖𝑛 12
𝐶 𝑛−2 = −
𝑛(𝑛−1)
2.2
𝐶 𝑛 , 𝐶 𝑛−4 =
𝑛(𝑛−1)(𝑛−2)(𝑛−3)
22.2.4
𝐶 𝑛 ,…
Put these in (11) , we get 𝑦 = 𝐶 𝑛 𝑥 𝑛 −
𝑛 𝑛−1
2.2
𝑥 𝑛−2 +
𝑛 𝑛−1 𝑛−2 𝑛−3
22.2.4
𝑥 𝑛−4 + ⋯ +
7

𝐻0 𝑥 = 1
𝐻1 𝑥 = 2𝑥
𝐻2 𝑥 = 4𝑥2 − 2
𝐻3 𝑥 = 8𝑥3
− 12𝑥
𝐻4 𝑥 = 16𝑥4
− 48𝑥2
+ 12
𝐻5 𝑥 = 32𝑥5 − 160𝑥3 + 120𝑥
.
.
.
𝐻 𝑛 𝑥 = 𝑟=0
[ 𝑛 2] (−1) 𝑟 𝑛!
𝑟! 𝑛−2𝑟 !
(2𝑥) 𝑛−2𝑟
By putting the value of n=0,1,2,3,… in 𝐻 𝑛 𝑥
8

Property of Hermite Series :-
Generating Function
Even and Odd functions
Rodrigue formula
Orthogonality
Recurrence Relation
9

GENERATING FUNCTION OF HERMITE SERIES
Show that 𝑒2𝑡𝑥−𝑡2
= 𝑛=0
∞ 𝑡 𝑛
𝑛!
𝐻 𝑛(𝑥)
Proof :
𝑒2𝑡𝑥−𝑡2
=𝑒2𝑡𝑥 𝑒−𝑡2
= 𝑟=0
∞ (−𝑡2) 𝑟
𝑟! 𝑠=0
∞ (2𝑡𝑥) 𝑠
𝑠!
= 𝑟=0
∞
𝑠=0
∞ (−1) 𝑟 𝑡 𝑠+2𝑟 (2𝑥) 𝑠
𝑟! 𝑠!
put 𝑠 + 2𝑟 = 𝑛
= 𝑛=0
∞
[ 𝑟=0
𝑛/2 (−1) 𝑟 (2𝑥) 𝑛−2𝑟
𝑟! (𝑛−2𝑟)!
]
𝑡 𝑛
𝑛!
= 𝑛=0
∞
𝐻 𝑛 𝑥
𝑡 𝑛
𝑛!
10

EVEN AND ODD FUNCTIONS
Whether a Hermite Polynomial is an even or odd function depends on its degree n based on
𝐻 𝑛 −𝑥 = −1 𝑛
𝐻 𝑛(𝑥) .
Show that 𝐻 𝑛 −𝑥 = −1 𝑛
𝐻 𝑛(𝑥)
Proof:
𝐻 𝑛 𝑥 = 𝑟=0
[ 𝑛 2] (−1) 𝑟 𝑛!
𝑟! 𝑛−2𝑟 !
(2𝑥) 𝑛−2𝑟
Replace x by –x
𝐻 𝑛 𝑥 = 𝑟=0
[ 𝑛 2] (−1) 𝑟 𝑛!
𝑟! 𝑛−2𝑟 !
(2𝑥) 𝑛−2𝑟
(−1) 𝑛−2𝑟
𝐻 𝑛 𝑥 = 𝑟=0
[ 𝑛 2] (−1) 𝑟
(−1) 𝑛 𝑛!
𝑟! 𝑛−2𝑟 !
(2𝑥) 𝑛−2𝑟
𝐻 𝑛 −𝑥 = −1 𝑛 𝐻 𝑛(𝑥)
13

RODRIGUE FORMULA
To Show that 𝐻 𝑛 𝑥 = (−1) 𝑛 𝑒(𝑥2) 𝑑 𝑛(𝑒−𝑥2
)
𝑑𝑥 𝑛 ….. n=0,1,2,3…, -∞<t<∞
PROOF :
The generating function of hermite polynomial is given by 𝑒2𝑡𝑥−𝑡2
= 𝑛=0
∞ 𝑡 𝑛
𝑛!
𝐻 𝑛(𝑋)
𝑒2𝑡𝑥−𝑡2
= 𝐻0 𝑥 +
𝑡 𝐻1(𝑥)
1!
+
𝑡2
2!
𝐻2 𝑥 + ⋯ +
𝑡 𝑛
𝑛!
𝐻 𝑛 𝑥 +
𝑡 𝑛+1
𝑛+1 !
𝐻 𝑛+1 𝑥 + ⋯
Differentiate both sides wrt to t at n times then put t=0
𝜕 𝑛(𝑒2𝑡𝑥−𝑡2
)
𝜕𝑡 𝑛 =
𝑛!
𝑛!
𝐻 𝑛(𝑥) at t=0
𝐻 𝑛(𝑥) =
𝜕 𝑛(𝑒2𝑡𝑥−𝑡2+𝑥2−𝑥2
)
𝜕𝑡 𝑛 = 𝑒(𝑥2)
[
𝜕 𝑛(𝑒2𝑡𝑥−𝑡2+𝑥2
)
𝜕𝑡 𝑛 ]
𝐻 𝑛(𝑥) = 𝑒 𝑥2
[
𝜕 𝑛 𝑒− 𝑥−𝑡 2
𝜕𝑡 𝑛 ]
Now put x-t = u
𝜕
𝜕𝑡
=
𝜕
𝜕𝑢
𝜕𝑢
𝜕𝑡
𝐻 𝑛(𝑥) = 𝑒 𝑥2
(−1) 𝑛
[
𝜕 𝑛
𝜕𝑡 𝑛 (𝑒−𝑥2
)]
𝐻 𝑛 𝑥 = (−1) 𝑛 𝑒 𝑥2 𝑑 𝑛(𝑒−𝑥2
)
𝑑𝑥 𝑛
11

ORTHOGONAL PROPERTIES OF THE HERMITE POLYNOMIALS
To show that 𝑥=−∞
∞
𝑒−𝑥2
𝐻 𝑛 𝑥 𝐻 𝑚 𝑥 𝑑𝑥 =
0 𝑖𝑓 𝑚 ≠ 𝑛
П 2 𝑛 𝑛! 𝑖𝑓 𝑚 = 𝑛
Proof :Using the generating functions , we have
𝑛=0
∞ 𝑡 𝑛
𝑛!
𝐻 𝑛(𝑥) = 𝑒2𝑡𝑥−𝑡2
and 𝑚=0
∞ 𝑠 𝑛
𝑚!
𝐻 𝑚(𝑥) = 𝑒2𝑠𝑥−𝑠2
Multiplying their corresponding sides ,gives 𝑛=0
∞
𝑚=0
∞ 𝐻 𝑛 𝑥 𝐻 𝑚(𝑥)
𝑛! 𝑚!
𝑡 𝑛
𝑠 𝑚
= 𝑒2𝑡𝑥 − 𝑡2+2𝑠𝑥 − 𝑠2
Multiplying both sides by 𝑒−𝑥2
and then integrating both sides w.r.t. ‘x’ from −∞ 𝑡𝑜 ∞ , we have
𝑛=0
∞
𝑚=0
∞
−∞
∞ 𝑒−𝑥2
𝐻 𝑛 𝑥 𝐻 𝑚(𝑥)
𝑛! 𝑚!
𝑑𝑥 𝑡 𝑛
𝑠 𝑚
= 𝑒−𝑥2+2𝑡𝑥 − 𝑡2+2𝑠𝑥 − 𝑠2
𝑛=0
∞
𝑚=0
∞
−∞
∞ 𝑒−𝑥2
𝐻 𝑛 𝑥 𝐻 𝑚(𝑥)
𝑛! 𝑚!
𝑑𝑥 𝑡 𝑛
𝑠 𝑚
= −∞
∞
𝑒−𝑥2+2𝑡𝑥 − 𝑡2+2𝑠𝑥 − 𝑠2
𝑑𝑥
𝑛=0
∞
𝑚=0
∞
−∞
∞ 𝑒−𝑥2
𝐻 𝑛 𝑥 𝐻 𝑚(𝑥)
𝑛! 𝑚!
𝑑𝑥 𝑡 𝑛
𝑠 𝑚
= −∞
∞
𝑒−𝑥2+2𝑥 𝑡+𝑠 −(𝑡+𝑠)2
× 𝑒(𝑡+𝑥)2−(𝑡2+ 𝑥2)
𝑑𝑥
=𝑒2𝑡𝑠
−∞
∞
𝑒−[𝑥− 𝑡+𝑠 ]2
𝑑𝑥 = 𝑒2𝑡𝑠
−∞
∞
𝑒−𝑦2
𝑑𝑦 = П𝑒2𝑡𝑠
putting x− 𝑡 + 𝑠 = 𝑦 so that dx=dy and −∞
∞
𝑒−𝑦2
𝑑𝑦 = П
𝑛=0
∞
𝑚=0
∞
−∞
∞ 𝑒−𝑥2
𝐻 𝑛 𝑥 𝐻 𝑚(𝑥)
𝑛! 𝑚!
𝑑𝑥 𝑡 𝑛 𝑠 𝑚 = П (1 + 2ts +
(2𝑡𝑠)2
2!
+ … +
(2𝑡𝑠) 𝑛
𝑛!
+ ⋯ )
Hence , 𝑥=−∞
∞
𝑒−𝑥2
𝐻 𝑛 𝑥 𝐻 𝑚 𝑥 𝑑𝑥 =
0 𝑖𝑓 𝑚 ≠ 𝑛
П 2 𝑛 𝑛! 𝑖𝑓 𝑚 = 𝑛
12

RECURRENCE RELATION
A Hermite Polynomial at one point can be expressed by neighbouring
Hermite Polynomials at the same point.
(a) 𝐻 𝑛+1 𝑥 = 2𝑥𝐻 𝑛 𝑥 − 2𝑛𝐻 𝑛−1 𝑥
(b) 𝐻 𝑛
′
𝑥 = 2𝑛𝐻 𝑛−1 𝑥
(a)To show that 𝐻 𝑛
′ 𝑥 = 2𝑛𝐻 𝑛−1 𝑥
Proof:
𝑒2𝑡𝑥−𝑡2
= 𝑛=0
∞ 𝑡 𝑛
𝑛!
𝐻 𝑛(𝑥)
Diff with respect to x
𝑒2𝑡𝑥−𝑡2
2t = 𝑛=0
∞ 𝑡 𝑛
𝑛!
𝐻 𝑛′(𝑥)
2 𝑛=0
∞ 𝑡 𝑛+1
𝑛!
𝐻 𝑛 𝑥 = 𝑛=0
∞ 𝑡 𝑛
𝑛!
𝐻 𝑛′(𝑥) and compare the coeffficent of 𝑡 𝑛
2𝑛𝐻 𝑛−1 𝑥 = 𝐻 𝑛′(𝑥)
14

(b) To show that 𝐻 𝑛+1 𝑥 = 2𝑥𝐻 𝑛 𝑥 − 2𝑛𝐻 𝑛−1 𝑥
Proof :
we know that 𝑒2𝑡𝑥−𝑡2
= 𝑛=0
∞ 𝑡 𝑛
𝑛!
𝐻 𝑛(𝑥)
Differentiating both sides w.r.t ‘t’ gives
(2x-2t)𝑒2𝑡𝑥−𝑡2
= 𝑛=0
∞ 𝑛𝑡 𝑛−1
𝑛!
𝐻 𝑛(𝑥)
(2𝑥 − 2𝑡) 𝑛=0
∞ 𝑡 𝑛
𝑛!
𝐻 𝑛 𝑋 = 𝑛=0
∞ 𝑛 𝑡 𝑛−1
𝑛!
𝐻 𝑛(𝑥)
2𝑥 𝑛=0
∞ 𝑡 𝑛
𝑛!
𝐻 𝑛(𝑥) − 2 𝑛=0
∞ 𝑡 𝑛+1
𝑛!
𝐻 𝑛(𝑥) = 𝑛=0
∞ 𝑛 𝑡 𝑛−1
𝑛!
𝐻 𝑛(𝑥)
∴ Compare the coefficient of 𝑡 𝑛
2𝑥
𝑛!
𝐻 𝑛 𝑥 −
2
𝑛−1 !
𝐻 𝑛−1 𝑥 =
𝑛+1
𝑛+1 !
𝐻 𝑛+1(𝑥)
2𝑥𝐻 𝑛 𝑥 − 2𝑛𝐻 𝑛−1 𝑥 = 𝐻 𝑛+1 𝑥
15

REFERENCES
(1) Differential Equations With Applications and Historical Notes , CRC Press
By George F . Simmons
(2) Ordinary and Partial Differential Equations , S chand Publication by Dr.
M.D. Raisinghania
(3) Series with Hermite Polynomials and Application By Khristo N.
Boyadzhiev and Ayhan Dil
16

THANK YOU …
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