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: rmathpal@uou.ac.in

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