MATHEMATICAL PHYSICS
UNIT – 2
BESSEL’S EQUATION
DR. RAJESH MATHPAL
ACADEMIC CONSULTANT
SCHOOL OF SCIENCES
UTTARAKHAND OPEN UNIVERSITY
TEENPANI, HALDWANI
UTTRAKHAND
MOB:9758417736,7983713112
Email: [email protected]
STRUCTURE OF UNIT
• 2.1 INTRODUCTION
• 2.2 BESSEL’S EQUATION
• 2.4 BESSEL’S FUNCTIONS, Jn (x)
• 2.5 Bessel’s function of the second kind of order n
• 2.6 RECURRENCE FORMULAE
• 2.7 ORTHOGONALITY OF BESSEL FUNCTION
• 2.8 A GENERATING FUNCTION FOR Jn (x)
• 2.9 SOME EXAMPLES
2.1 INTRODUCTION
We find the Bessel’s equation while solving Laplace equation in polar coordinates by the needed of separation of variables. This equation has a number of applications in engineering.
Bessel’s function are involved in
• The Oscillatory motion of a hanging chain
• Euler’s theory of a circular membrane
• The studies of planetary motion
• The propagation of waves
• The Elasticity
• The fluid motion
• The potential theory
• Cylindrical and spherical waves
• Theory of plane waves
• Bessel’s function are also known as cylindrical and spherical function.
2.2 BESSEL’S EQUATION
The differential equation
𝑥2 𝑑2𝑦
𝑑𝑥2 + 𝑥𝑑𝑦
𝑑𝑥+ 𝑥2 − 𝑥𝑛 𝑦 = 0
is called the Bessel’s differential equation, and particular solutions of this equation are
called Bessel’s fraction of order n.
2.3 SOLUTION OF BESSEL’S EQUATION
𝑥2 𝑑2𝑦′
𝑑𝑥2 + 𝑥𝑑𝑦
𝑑𝑥+ 𝑥2 − 𝑥𝑛 𝑦 = 0. …(1)
Let σ𝑟=0∞ 𝑎𝑟𝑥
𝑚+𝑟 𝑜𝑟 𝑦 = 𝑎0𝑥𝑚 + 𝑎1𝑥
𝑚+1 + 𝑎2𝑥𝑚+2 + ⋯ …(2)
So that𝑑𝑦
𝑑𝑥= σ𝑟=0
∞ 𝑎𝑟 𝑚 + 𝑟 𝑥𝑚+𝑟−1
and 𝑑2𝑦
𝑑𝑥2 = σ𝑟=0∞ 𝑎𝑟 𝑚 + 𝑟 (𝑚 + 𝑟 −1)𝑥𝑚+𝑟−2
Substituting these values in (1), we get
𝑥2
𝑟=0
∞
𝑎𝑟 𝑚+𝑟 (𝑚+𝑟 −1)𝑥𝑚+𝑟−2 +𝑥
𝑟=0
∞
𝑎𝑟 𝑚 + 𝑟 𝑥𝑚+𝑟−1 + (
)
𝑥2
− 𝑛2
𝑟=0
∞
𝑎𝑟𝑥𝑚+𝑟 = 0
2 2
0 0 0 0
2 2
0 0
2 2 2
0 0
( )( 1) ( ) 0
[( )( 1) ( ) ] 0
[( ) ] 0.
m r m r m r m r
r r r r
r r r r
m r m r
r r
r r
m r m r
r r
r r
a m r m r x a m r x a x n a x
a m r m r m r n x a x
a m r n x a x
+ + + + +
= = = =
+ + +
= =
+ + +
= =
+ + − + + + − =
+ + − + + − + =
+ − + =
Equating the coefficient of lowest degree term of xm in the identity (3) to zero,
by putting r = 0 in the first summation we get the indicial equation.
a0[m+0)2 – n2] = 0. (r = 0)
⇒ m2 = n2 i.e. m = n, m = - n a0 ≠ 0
Equating the coefficient of the next lowest degree term xm+1 in the identity (3), we put r = 1 in the first summation
a1 [m + 1)2 – n2] = 0 i.e. a1 = 0, since m + 1)2 – n2 ≠ 0
Equating the coefficient of xm + r + 2 in (3) to zero, to find relation in successive coefficients, we get
ar +2[(m+r+2)2 – n2] +ar = 0
⇒ 𝑎𝑟+2 = −1
𝑚+𝑟+2 2−𝑛2 . ar
Therefore, a3 = a5 = a1 = …. = 0, since a1 = 0
If r = 0, 𝑎2 = −1
𝑚+2 2−𝑛2 . a0
If r = 2, 𝑎4 = −1
𝑚+4 2−𝑛2 a2 =1
𝑚+2 2−𝑛2 [(𝑚+4)2−𝑛2 a0 and so on.
On substituting the values of the coefficients a1, a2, a3, a4 …….. in (2), we have
y = a0xm = −
𝑎0
𝑚+2 2−𝑛2 𝑥𝑚+2 +𝑎0
𝑚+2 2−𝑛2 [ 𝑚+4)2−𝑛2 𝑥𝑚+4 + ⋯
y = a0xm = 1 −
1
𝑚+2 2−𝑛2 𝑥2 +1
𝑚+2 2−𝑛2 [ 𝑚+4)2−𝑛2 𝑥4 − ⋯
For m = n
y = a0xn 1 −
1
4 𝑛+1𝑥2 +
1
42.2! 𝑛+1 𝑛+2𝑥4 − ⋯
where a0 is an arbitrary constant.
For m = − n
y = a0x-n 1 −
1
4 −𝑛+1𝑥2 +
1
42.2! −𝑛+1 −𝑛+2𝑥4 − ⋯
2.4 BESSEL’S FUNCTIONS, Jn (x)
The Bessel’s equation is 𝑥2 𝑑2𝑦
𝑑𝑥 2+ 𝑥
𝑑𝑦
𝑑𝑥+ (𝑥2 − 𝑥𝑛)𝑦 = 0. …(1)
Solution of (1) is
y = a0x-n 1 −𝑥2
2.2(𝑛+1)+
𝑥4
2.4.22(𝑛+1)(𝑛+2)− ⋯ + (−1)𝑟 𝑥2𝑟
(2𝑟𝑟!).2𝑟(𝑛+1)(𝑛+2)…(𝑛+𝑟)+ ⋯
2
0 20
( 1)2 . !( 1)( 2)...( )
rn r
rr
xa x
r n n n r
=
= −+ + +
where a0 is an arbitrary constant.
If a0 = 1
2𝑛 (𝑛+1)
The above solution is called Bessel’s function denoted by Jn (x).
Thus 𝐽𝑛(𝑥) = 1
2𝑛 (𝑛+1)σ(−1)𝑟 𝑥𝑛+2𝑟
22𝑟 .𝑟!(𝑛+1)(𝑛+2)…(𝑛+𝑟) 𝑛 + 1 = 𝑛!
⇒ 𝐽𝑛(𝑥) = x
2
n
1
(𝑛+1)−
1
1! (𝑛+2)
x
2
2
+1
2! (𝑛+3)
x
2
4
−1
3! (𝑛+4)
x
2
6
+ ⋯
⇒ 𝐽𝑛(𝑥) =𝑥𝑛
2𝑛 𝑛+1 1 −
𝑥2
2.(2𝑛+2)+
𝑥4
2.4.(2𝑛+2)(2𝑛+4)+ ⋯ …(2)
2 2
0 0
( 1) ( 1) ( ) ( )
2 2! ( 1) ! ( )!
n r n rr r
n n
r r
x xJ x J x
r n r r n r
+ +
= =
− − = =
+ + +
If n = 0, J0 (x) = σ(−1)𝑟
(𝑟!)2 𝑥
2
2𝑟
⇒ J0 (x) = 1 −𝑥2
22+
𝑥4
22 .42−
𝑥6
22 .42 .62+ ⋯
If n = 1, J1 (x) = 𝑥
2−
𝑥3
22.4+
𝑥5
22.42.6− ⋯
We draw the length of these two functions. Both the functions are oscillatory with a varying period and a decreasing amplitude.
Replacing n by – n in (2), we get J-n (x) = σ𝑟=0∞ −1 𝑟
𝑟! −𝑛+𝑟+1
𝑥
2
−𝑛+2𝑟
Case I. If n is not integer or zero, then complete solution of (1) is
Case II. If n = 0, then y1 = y2 and complete solution of (1) is the Bessel’s function of order zero.
Case III. If n is positive integer, then y2 is not solution of (1). And y1 fails to give a solution for negative values of n. Let us find out the general solution when n is an integer.
2.5 Bessel’s function of the second kind of
order n
𝑥2 𝑑2𝑦
𝑑𝑥2 + 𝑥𝑑𝑦
𝑑𝑥+ 𝑥2 − 𝑛2 𝑦 = 0 …(1)
Let y = u(x) Jn (x) be the second of the Bessel’s equation when n integer.
𝑑𝑦
𝑑𝑥= u’ Jn + u J’n
𝑑2𝑦
𝑑𝑥2 = u’’ Jn + 2u’ J’n + u J’’n
Substituting these values of y, y’, yn in (1), we get
x2 (u’’ Jn + 2u’ J’n + u J’’n) + x(u’ Jn + u.J’n) + (x2 – n2) u Jn = 0
⇒ u [x2 Jnn + x J’n + (x2 – n2) Jn] + x2 u’’ Jn + 2x2 u’ Jn + x u’ Jn = 0 …(2)
⇒ x2 J’’n + x J’n + (x2 – n2) Jn = 0 [Since Jn is a solution of (1)]
(2) becomes x2 u’’ Jn + 2x2 u’ J’n + xu’ Jn = 0 …(3)
Dividing (3) by x2 u’ Jn, we have
𝑢𝑛
𝑢′ + 2𝐽𝑛′
𝐽𝑛+
1
𝑥= 0
(4) Can also be written as …(4)
𝑑
𝑑𝑥[log𝑢′ + 2
𝑑
𝑑𝑥[log 𝐽𝑛] +
𝑑
𝑑𝑥(log 𝑥) = 0
⇒𝑑
𝑑𝑥[log𝑢′ + 2 log 𝐽𝑛 + log 𝑥] = 0
⇒𝑑
𝑑𝑥log(𝑢′. 𝐽𝑛
2 𝑥 = 0 …(5)
Integrating (5), we get
log 𝑢′. 𝐽𝑛2 . 𝑥 = log 𝐶1
⇒ 𝑢′. 𝐽𝑛2. 𝑥 =𝐶1 ⇒ 𝑢′ =
𝐶1
𝐽𝑛2.𝑥
…(6)
On integrating (6), we obtain
𝑢 = 𝐶1
𝐽𝑛2 .𝑥
𝑑𝑥 +𝐶2
Putting the value of 𝑢 in the assumed solution y = u (x). 𝐽𝑛2(𝑥), we get
2.6 RECURRENCE FORMULAE
These formulae are very useful in solving the questions. So, they are to be
committed to memory.
1. x 𝐽𝑛′ = 𝑛𝐽𝑛 − 𝑥 𝐽𝑛+1
2. x 𝐽𝑛′ = −𝑛𝐽𝑛 + 𝑥 𝐽𝑛−1
3. 2 𝐽𝑛′ = 𝐽𝑛−1 − 𝐽𝑛+1
4. 2𝑛 𝐽𝑛 = 𝑥 𝐽𝑛−1 + 𝐽𝑛+1
5. 𝑑
𝑑𝑥𝑥−𝑛𝐽𝑛 = −𝑥−𝑛 𝐽𝑛+1
6. 𝑑
𝑑𝑥𝑥−𝑛𝐽𝑛 = 𝑥𝑛 𝐽𝑛−1
Formula I. x 𝐽𝑛′ = 𝑛𝐽𝑛 − 𝑥𝐽𝑛+1
Proof. We know that
𝐽𝑛 = σ𝑟=0∞ −1 𝑟
𝑟! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟
Differentiating with respect to x, we get
𝐽𝑛′ = σ
−1 𝑟 𝑛+2𝑟
𝑟! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟−1 1
2
⇒ 𝑥𝐽𝑛′ = 𝑛σ
−1 𝑟
𝑟! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟+ 𝑥 σ
−1 𝑟.2𝑟
2.𝑟! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟−1
= 𝑥𝐽𝑛 + 𝑥 σ𝑟=0∞ −1 𝑟
𝑟−1 ! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟−1
= 𝑛𝐽𝑛 + 𝑥 σ𝑠=0∞ −1 𝑠+1
𝑠! 𝑛+𝑠+2
𝑥
2
𝑛+2𝑠−1[Putting r – 1 = s]
= 𝑛𝐽𝑛 − 𝑥 σ𝑠=0∞ −1 𝑠
𝑠! 𝑛+1 +𝑠+1
𝑥
2
𝑛+1 +2𝑠
𝑥𝐽𝑛′ = 𝑛𝐽𝑛 − 𝑥𝐽𝑛+1 Proved.
Formula II. 𝑥𝐽𝑛′ = −𝑛𝐽𝑛 + 𝑥𝐽𝑛−1
Proof. We know that 𝐽𝑛 = σ𝑟=0∞ −1 𝑟
𝑟! 𝑛+𝑟+2
𝑥
2
𝑛+2𝑟
Differentiating w.r.t. ‘x’, we get 𝐽𝑛′ = σ𝑟=0
∞ −1 𝑟 𝑛+2𝑟
𝑟! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟−1 1
2
𝐽𝑛′ = σ𝑟=0
∞ −1 𝑟 𝑛+2𝑟
𝑟! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟= σ𝑟=0
∞ −1 𝑟 2𝑛+2𝑟 −𝑛
𝑟! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟
= σ𝑟=0∞ −1 𝑟 2𝑛+2𝑟
𝑟! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟− 𝑛 σ𝑟=0
∞ −1 𝑟
𝑟! 𝑛+𝑟+1
𝑥
2
𝑛+2𝑟
= σ𝑟=0∞ −1 𝑟2
𝑟! 𝑛+𝑟
𝑥
2
𝑛+2𝑟− 𝑛𝐽𝑛
= 𝑥
𝑟=0
∞−1 𝑟
𝑟! 𝑛 − 1 + 𝑟 + 1
𝑥
2
𝑛−1+2𝑟
− 𝑛𝐽𝑛
⇒ 𝒙𝑱𝒏′ = 𝒙𝑱𝒏−𝟏 − 𝒏𝑱𝒏
Formula III. 𝟐𝑱𝒏′ = 𝑱𝒏−𝟏 − 𝑱𝒏+𝟏
Proof.
We know that
𝑥𝐽𝑛′ = 𝐽𝑛 − 𝑥𝐽𝑛+1 …(1) (Recurrence formula I)
𝑥𝐽𝑛′ = −𝑛𝐽𝑛 + 𝑥𝐽𝑛−1 …(2) (Recurrence formula II)
Adding (1) and (2), we get
2𝑥𝐽𝑛′ = −𝑥𝐽𝑛+1 + 𝑥𝐽𝑛−1 ⇒ 2𝐽𝑛
′ = 𝐽𝑛−1 − 𝐽𝑛+1
Formula IV. 2𝑛𝐽𝑛 = 𝑥 (𝐽𝑛−1 + 𝐽𝑛+1)
Proof.
We know that
𝑥𝐽𝑛′ = 𝑛𝐽𝑛 − 𝑥𝐽𝑛+1 …(1) (Recurrence formula I)
𝑥𝐽𝑛′ = −𝑛𝐽𝑛 + 𝑥𝐽𝑛−1 …(2) (Recurrence formula II)
subtracting (2) from (1), we get
0 = 2 𝑛 𝐽𝑛 −𝑥𝐽𝑛+1 −𝑥𝐽𝑛−1
⇒ 2 𝑛 𝐽𝑛 = 𝑥 (𝐽𝑛−1 +𝐽𝑛+1) …(3)
Formula V.𝒅
𝒅𝒙(x-n. Jn) = -x-n Jn+1
Proof. We know that 𝑥𝐽𝑛′ = 𝑛𝐽𝑛 − 𝑥𝐽𝑛+1
(Recurrence formula I)
Multiplying by x-n-1, we obtain x-n 𝐽𝑛′ = nx-n-1 Jn – x-n Jn+1
i.e., x-n 𝐽𝑛′ = nx-n-1 Jn =– x-n Jn+1
⇒𝑑
𝑑𝑥(x-n Jn) = - x-n Jn + 1
Formula VI.𝒅
𝒅𝒙(xn Jn) = xn Jn-1
Proof.
We know that x-n 𝐽𝑛′ = -nJn + x Jn-1 (Recurrence formula II)
Multiplying by xn+1, we have
xn 𝐽𝑛′ = -nxn-1 Jn + xn Jn-1 i.e., xn 𝐽𝑛
′ +nxn-1 Jn = xn Jn-1
⇒𝒅
𝒅𝒙(xn Jn) = xn Jn-1
2.7 ORTHOGONALITY OF BESSEL
FUNCTION
Proof. We know that
𝑥2 𝑑2𝑦
𝑑𝑥2 + 𝑥𝑑𝑦
𝑑𝑥+ 𝛼2𝑥2 − 𝑛2 𝑦 = 0 …(1)
⇒ 𝑥2 𝑑2𝑧
𝑑𝑥2 + 𝑥𝑑𝑧
𝑑𝑑𝑥+ 𝛽2𝑥2 − 𝑛2 𝑧 = 0 …(2)
Solution of (1) and (2) are y = Jn (𝛼 𝑥), z = Jn (𝛽 𝑥) respectively.
Multiplying (1) by 𝑧
𝑥and (2) by –
𝑦
𝑥and adding, we get
𝑥 𝑧𝑑2𝑦
𝑑𝑥2 − 𝑦𝑑2𝑧
𝑑𝑥2 + 𝑧𝑑𝑦
𝑑𝑥− 𝑦
𝑑𝑧
𝑑𝑥+ 𝛼2 −𝛽2 𝑥𝑦𝑧 = 0.
⇒𝑑
𝑑𝑥𝑥 𝑧
𝑑𝑦
𝑑𝑥− 𝑦
𝑑𝑧
𝑑𝑥+ 𝛼2 −𝛽2 𝑥𝑦𝑧 = 0 …(3)
Integrating (3) w.r.t. ‘x’ between the limits 0 and 1, we get
𝑑
𝑑𝑥𝑥 𝑧
𝑑𝑦
𝑑𝑥− 𝑦
𝑑𝑧
𝑑𝑥 0
1+ 𝛼2 −𝛽2 0
1𝑥 𝑦 𝑧 𝑑𝑥 = 0
⇒ 𝛽2 −𝛼2 01𝑥 𝑦 𝑧 𝑑𝑥 = 𝑥 𝑧
𝑑𝑦
𝑑𝑥− 𝑦
𝑑𝑧
𝑑𝑥 0
1= 𝑧
𝑑𝑦
𝑑𝑥− 𝑦
𝑑𝑧
𝑑𝑥 𝑥=1…(4)
Putting the values of y = Jn (𝛼 𝑥), 𝑑𝑦
𝑑𝑥= 𝛼 𝐽𝑛
′ 𝛼𝑥 , 𝑧 = 𝐽𝑛 𝛽𝑥 ,𝑑𝑧
𝑑𝑥= 𝛽, 𝐽𝑛
′ 𝛽𝑥 in (4), we get
𝛽2 − 𝛼2 01𝑥𝐽𝑛 𝛼𝑥 . 𝐽𝑛 𝛽𝑥 𝑑𝑥 = 𝛼𝐽𝑛
′ 𝛼𝑥 𝐽𝑛 𝛽𝑥 = 𝛽𝐽𝑛′ 𝛽𝑥 𝐽𝑛 𝛼𝑥 𝑥=1
= 𝛼𝐽𝑛′ 𝛼 𝐽𝑛 𝛽 − 𝛽𝐽𝑛
′ 𝛽 𝐽𝑛 𝛼 …(5)
Since 𝛼, 𝛽 are the roots of Jn (x) = 0, so Jn 𝛼 = Jn 𝛽 = 0
Putting the values of Jn (𝛼) = Jn (𝛽) = 0 in (5), we get
(𝛼2−𝛽2) 𝑥1
0 Jn (𝛼𝑥). Jn (𝛽𝑥) dx = 0
⇒ 𝑥1
0 Jn (𝛼𝑥). Jn (𝛽𝑥) dx = 0 Proved.
We also know that Jn (𝛼) = 0. Let 𝛽 be a neighboring value of 𝛼, which tends to 𝛼.
Then
1 '
2 2
0
0 ( ). ( )lim ( ). ( ) lim n n
n n
J JxJ x J x dx
→ →
+=
−
As the limit is of the form 0
0, we apply L’ Hopital’s rule
1 ' '2
2 '
0
0 ( ). ( ) 1( ) lim ( )
2 2
n nn n
J JxJ x dx J
→
+ = = =
Proved.
2.8 A GENERATING FUNCTION FOR
Jn (x)
Prove that Jn (x) is the coefficient of zn in the expansion of 𝑒𝑥
2𝑧−
1
𝑧
Proof. We know that et = 1 + t + 𝑡2
2!+
𝑡3
3!+ ⋯
𝑒𝑥𝑧
2 = 1 +𝑥𝑧
2+
1
2!
𝑥
2𝑧
2−
1
3!
𝑥
2𝑧
3+ ⋯ …(1)
𝑒𝑥
2𝑧 = 1 −𝑥
2𝑧+
1
2!
𝑥
2𝑧
2−
1
3!
𝑥
2𝑧
3+ ⋯ …(2)
On multiplying (1) and (2), we get
𝑒𝑥
2𝑧1
𝑧 = 1 +𝑥𝑧
2+
1
2!
𝑥𝑧
2
2+
1
3!
𝑥𝑧
2
3+ ⋯ × 1 −
𝑥
2𝑧+
1
2!
𝑥
2𝑧
2−
1
3!
𝑥
2𝑧
3+ ⋯ …(3)
The coefficient of zn in the product of (3), we get
=1
𝑛 ! 𝑥
2 𝑛
−1
(𝑛+1)! 𝑥
2 𝑛+2
+1
2!(𝑛+2)! 𝑥
2 𝑛+4
− ⋯ = Jn (x)
Similarly, coefficient of z-n in the product of (3) = J-n(x)
∴ 𝑒𝑥
2 𝑧
1
𝑧 = J0 + z J1 + z2 J2 + z3 J3 + … + z-1 J-1 + z-2 J-2 + z-3 J-3 + …
1
2 ( )
xz
nz
n
n
e z J x
=−
=
For this reason 𝑒𝑥
2 𝑧
1
𝑧 is known as the generating function of Bessel’s functions.
Proved.
2.9 SOME EXAMPLES
Example 1. Show that Bessel’s Function Jn(x) is an even function when n is even and is
odd function when n is odd.
Solution. We know that
2
0
( 1)( ) ...(1)
2! 1
n rr
n
r
xJ x
r n r
+
=
− =
+ +
Replacing x by – x in (1), we get
2
0
( 1)( ) ...(2)
2! 1
n rr
n
r
xJ x
r n r
+
=
− − − =
+ +
Case I. If n is even, then n + 2r is even ⇒ −𝑥
2 𝑛+2𝑟
= 𝑥
2 𝑛+2𝑟
Thus (2), becomes
2
0
( 1)( )
2! 1
n rr
n
r
xJ x
r n r
+
=
− − =
+ +
= Jn (x) 𝐹𝑜𝑟 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑓(−𝑥) = 𝑓(𝑥)
Hence, Jn (x) is even function
Case II. If n is odd, then n + 2r is odd ⇒ −𝑥
2 𝑛+2𝑟
= − 𝑥
2 𝑛+2𝑟
Thus (2). Becomes
2
0
( 1)( )
2! 1
n rr
n
r
xJ x
r n r
+
=
− − = −
+ +
= − Jn (x) 𝐹𝑜𝑟 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑓(−𝑥) = −𝑓(𝑥)
Proved.
Hence, Jn (x) os odd function.
Example 2. Prove that:
x 0
( ) 1lim ;( 1).
2 1
n
n n
J xn
x n→= −
+
Solution. From the equation (2) of Article 29.3 on page 798, we know that
Jn (x) = 𝑥𝑛
2𝑛 𝑛+1 1 −
𝑥2
2.(2𝑛+2)+
𝑥4
2.4(2𝑛+2)(2𝑛+4)− ⋯
On taking limit on both sides when x → 0, we get
2 4
x 0 x 0
( ) 1lim lim 1 ...
2.(2 2) 2.4.(2 2)(2 4)2 1
n
n n
J x x x
x n n nn→ →
= − + −
+ + ++
= 1
2𝑛 𝑛+1
Example 3. Find the value of J-1 (x) + J1 (x).
Solution. By using Recurrence relation IV for Jn (x) is
2n Jn = x (Jn – 1 + Jn + 1)
Jn – 1 (x) + Jn + 1 (x) = 2𝑛
𝑥Jn (x)
Put n = 0
J-1(X) + J1(x) = 0
Example 4.Prove that
Formula V.𝒅
𝒅𝒙(x-n. Jn) = -x-n Jn+1
Proof. We know that 𝑥𝐽𝑛′ = 𝑛𝐽𝑛 − 𝑥𝐽𝑛+1 (Recurrence formula I)
Multiplying by x-n-1, we obtain x-n 𝐽𝑛′ = nx-n-1 Jn – x-n Jn+1
i.e., x-n 𝐽𝑛′ = nx-n-1 Jn =– x-n Jn+1
⇒𝑑
𝑑𝑥(x-n Jn) = - x-n Jn + 1
THANKS