1 Introduction 2 Legendre Equation€¦ · their main properties and some applications are...

Dr. Akul Rana, Department of Mathematics Narajole Raj College

1 Introduction

In this chapter, two important differential equations viz. Legendre differentialequation and Bessel’s differential equation are studied. Properties of Legendrepolynomial and Bessel’s function are discussed. These functions has applicationin mathematical physics and engineering. Here, the definition of these functions,their main properties and some applications are presented.

2 Legendre Equation

Legendre polynomial is used in many different mathematical and physical prob-lems. These polynomials may be constructed as a consequence of demandinga complete, orthogonal set of functions over the interval [−1, 1]. In quantummechanics they represent angular momentum eigenfunctions.

2.1 Legendre Differential Equation

The equation

(1− z2)d2w

dz2− 2z dw

dz+ n(n+ 1)w = 0 (1)

where n is a real parameter, is known as Legendre equation of order n. Thisequation has regular singularities at z = ±1. The physically interesting rangefor |z| is generally |z| ≤ 1 (usually z is the cosine of some angle). The equation(1) can be written as

d

dz

[(1− z2)dw

dz

]+ n(n+ 1)w = 0 (2)

To check the singularity at infinity, we transform the equation by substitutingz = 1/t. The transformed equation is

(t2 − 1)d2w

dt2+ 2t

dw

dt+

1

t2n(n+ 1)w = 0 (3)

which shows that the point ∞ is a regular singularity.Further, this equation is a particular type of hypergeometric equation. This

is done by substituting z1 = (1 − z)/2 which transforms z = 1 to z1 = 0 andz = −1 to z1 = 1. Also, 2dz1 = −dz

Under this transformation the Legendre equation becomes(1− z2

)(1 + z

2

)d2w

dz21+[1− (−n+ n+ 1 + 1)z1

] dwdz1

+ n(n+ 1)w = 0

or z1(1− z1)d2w

dz21+[1− (−n+ n+ 1 + 1)z1

] dwdz1

− (−n)(n+ 1)w = 0

PAPER-DSE4T, UNIT -I, Topic-LEGENDRE and BESSEL’s Eq.


Here a = −n, b = n + 1, c = 1 and z = z1. Therefore the solution of thisequation is

w = F (−n, n+ 1, 1; z1) = F (−n, n+ 1, 1; (1− z)/2)

which shows that when n is integer it reduces to a polynomial of degree n andthis is known as Legendre function of first kind.

2.2 Solution of Legendre Equation at z = 0

It is obvious that z = 0 is the ordinary point of the Legendre equation.Let

w =∞∑k=0

ckzk

be the solution of (1). Using this substitution (1) becomes

(1− z2)∞∑k=0

ckk(k − 1)zk−2 − 2z∞∑k=0

ckkzk−1 + n(n+ 1)

∞∑k=0

ckzk = 0

The coefficient of zk leads to the following equation

ck+2(k + 1)(k + 2)− ck(k − 1)k − 2ckk + n(n+ 1)ck = 0

and this equation can be written as

ck+2 =n(n+ 1)− k(k + 1)

(k + 1)(k + 2)ck = −

(n− k)(n+ k + 1)(k + 1)(k + 2)

ck

Thus

c2 = −n(n+ 1)

2!c0, c3 = −

(n− 1)(n+ 2)2.3

c1

c4 = −(n− 1)(n+ 3)

3.4c2, c5 = −

(n− 3)(n+ 4)4.5

c3

and so on.Thus the solution of (1) at z = 0 is

w = c0

[1− n(n+ 1)

2!z2 +

n(n+ 1)(n− 2)(n+ 3)4!

z4 − · · ·]

+c1

[z − (n− 1)(n+ 2)

3!z3 +

(n− 3)(n− 1)(n+ 2)(n+ 4)5!

− · · ·]

From this solution, it may be noted that whether n be even or odd one ofthe series terminates. When n is even, the first one terminates while the secondgives an infinite series and otherwise. When n is odd, first one is infinite andthe second expression is a polynomial. The series which terminates is calledLegendre polynomial of degree n and is denoted by Pn(z). The interval ofconvergence of the infinite series is |z| < 1. Moreover, |z| ≥ 1 gives divergentseries. Hence the general solution of Legendre equation is

w = APn(z) +BQn(z)

where Qn(z) is an infinite series and satisfies (1).



2.2.1 General expression of Pn(z)

Irrespective of the value of n(odd or even) the polynomial Pn(z) of degree n canbe written as

Pn(z) = cn

[zn +

cn−2cn

zn−2 +cn−2cn

cn−4cn−2

zn−4 + · · ·]

where

ck+2 = −(n− k)(n+ k + 1)(k + 1)(k + 2)

ck k = n− 2, n− 4, . . .

Thus

Pn(z) = cn

[zn − n(n− 1)

2(2n− 1)zn−2 +

n(n− 1)(n− 2)(n− 3)2.4(2n− 1)(2n− 3)

zn−4 + · · · p terms]

where p = n/2 + 1, when n is even and p = (n− 1)/2 + 1, when n is odd.

Since the constant cn is arbitrary, we can choose cn =a.3.5 . . . (2n− 1)

n!and

hence the Legendre polynomial becomes

Pn(z) =1.3.5. . . . (2n− 1)

n!

[zn − n(n− 1)

2(2n− 1)zn−2 +

n(n− 1)(n− 2)(n− 3)2.4(2n− 1)(2n− 3)

zn−4 + · · ·+ p terms]

=(2n)!

2n(n!)2

[zn − n(n− 1)

2(2n− 1)zn−2 +

n(n− 1)(n− 2)(n− 3)2.4(2n− 1)(2n− 3)

zn−4 + · · ·+ p terms]

The last term is constant when n is even or contain z when n is odd.

Some lower order Legendre polynomials are

P0(z) = 1

P1(z) = z

P2(z) =1

2(3z2 − 1)

P3(z) =1

2(5z3 − 3z)

P4(z) =1

8(35z4 − 30z2 + 3)

P5(z) =1

8(63z5 − 70z3 + 15z)

P6(z) =1

16(231z6 − 315z4 + 105z2 − 5)

P6(z) =1

16(429z7 − 693z4 + 315z3 − 35z)

P7(z) =1

128(6435z8 − 1201z6 + 6930z4 − 1260z2 + 35)



Conversely the terms 1, z, z2 etc. can be expressed using Legendre polyno-mial. Thus every algebraic function is expressible in terms of Legendre polyno-mial

1 = P0(z)

z = P1(z)

z2 =1

3[2P2(z) + 1]

z3 =1

5[2P3(z) + 3P1(z)]

z4 =1

35[8P4(z) + 20P2(z) + 7]

z5 =1

63[8P5(z) + 28P3(z) + 27P1(z)] and so on.

Example 2.1 Show that

4z3 − 5z2 − 3z = 85P3(z)−

10

3P2(z)−

3

5P1(z)−

5

3P0(z)

Solution. 4z3 − 5z2 − 3z = 45 [2P3(z) + 3P1(z)]−53 [2P2(z) + 1]− 3P1(z)

= 85P3(z)−103 P2(z)−

35P1(z)−

53P0(z)

2.3 Rodrigues Formula

The Legendre polynomial is

Pn(z) =(2n)!

2n(n!)2

[zn − n(n− 1)

2(2n− 1)zn−2 +

n(n− 1)(n− 2)(n− 3)2.4(2n− 1)(2n− 3)

zn−4 + · · ·+ p terms]

The last term being a constant or multiple of z according as m is even or odd.When n is even number of term is n2 + 1 and when n is odd number of term isn−12 + 1. So the number of term is denoted by p.

Pn(z) =1.3.5. . . . (2n− 1)

n!

d

dz

[ zn+1n+ 1

− n(n− 1)2(2n− 1)(n− 1)

zn−1

+n(n− 1)(n− 2)(n− 3)

2.4(2n− 1)(2n− 3)(n− 3)zn−3 + · · ·+ p terms

]

=1.3.5. . . . (2n− 1)

(n+ 1)!

d

dz

[zn+1 − (n+ 1)n

2(2n− 1)zn−1

+(n+ 1)n(n− 1)(n− 2)2.4(2n− 1)(2n− 3)

zn−3 + · · ·+ p terms]



=1.3.5. . . . (2n− 1)

(n+ 2)!

d2

dz2[zn+2 − (n+ 2)(n+ 1)

2(2n− 1)zn

+(n+ 2)(n+ 1)n(n− 1)2.4(2n− 1)(2n− 3)

zn−2 + · · ·+ p terms]

=1.3.5. . . . (2n− 1)

(2n)!

dn

dzn[z2n − 2n(2n− 1)

2(2n− 1)z2n−2

+2n(2n− 1)(2n− 2)(2n− 3)

2.4(2n− 1)(2n− 3)z2n−4 + · · ·+ p terms

]The last term being a multiple of zn or zn+1 according as n is even or odd.Therefore

Pn(z) =1.3.5. . . . (2n− 1)

(2n)!

dn

dzn[z2n − 2n(2n− 1)

2(2n− 1)z2n−2

+2n(2n− 1)(2n− 2)(2n− 3)

2.4(2n− 1)(2n− 3)z2n−4 + · · ·+ (n+ 1) terms

]The terms that are added within the third bracket to make (n+1) terms are

at most of degree zn−2 when n is even and zn−1 when n is odd. So that theseterms when differentiated n times with respect to z become zero. Therefore,

Pn(z) =1.3.5. . . . (2n− 1)

(2n)!

dn

dzn[(z2 − 1)n

]=

1

2.4.6. . . . (2n)

dn

dzn[(z2 − 1)n

]=

1

2nn!

dn

dzn[(z2 − 1)n

](4)

The equation (4) is called Rodrigues formula for Legendre polynomial.The Legendre polynomial has many interesting properties, some of them can

be deduced from Rodrigues formula.Recall that Pn(z) is a solution of Legendre equation

(1− z2)w′′ − 2zw′ + n(n+ 1)w = 0

and since the equation is unchanged on changing n to −n− 1, therefore

Pn(z) = P−n−1(z)

The Legendre polynomial is even or odd function according as n is even orodd, i.e.,

Pn(−z) =(−1)n

2nn!

dn

dzn[(z2 − 1)n

]= (−1)nPn(z) (5)



The lower order polynomial can be deduced from Rodrigues formula bysubstitutingn = 0, 1, 2, . . ..

The value of the Legendre polynomial at z = 1 is unity, i.e.,

Pn(1) = 1

This proposition is justified in the following.

Pn(1) =1

2nn!

dn

dzn[(z + 1)n(z − 1)n

]z=1

Differentiating[(z + 1)n(z − 1)n

], n times using Leibnitz theorem

Pn(1) =1

2nn!

[n!(z + 1)n + n(n− 1)!(z + 1)n−1(z − 1) + · · ·+ n!(z − 1)n

]z=1

=1

2nn![n!2n + 0] = 1

Example 2.2 Use Rodrigues formula to show that

P ′n+1(z)− P ′n− 1(z) = (2n+ 1)Pn(z),

where prime represents the derivative with respect to z

Solution.P ′n+1(z)− P ′n− 1(z)

=d

dz

[1

2n+1(n+ 1)!

dn+1

dzn+1[(z2 − 1)n+1

]]+

d

dz

[1

2n−1(n− 1)!dn−1

dzn−1[(z2 − 1)n−1

]]=

1

2n+1(n+ 1)!

dn+2

dzn+2[(z2 − 1)n+1

]+

1

2n−1(n− 1)!dn

dzn[(z2 − 1)n−1

]=

1

2nn!

dn

dzn

[1

2(n+ 1)

d2

dz2[(z2 − 1)n+1

]− 2n(z2 − 1)n−1

]=

1

2nn!

dn

dzn

[d

dz

[z(z2 − 1)n

]− 2n(z2 − 1)n−1

]=

1

2nn!

dn

dzn[(z2 − 1)n + 2n(z2 − 1)n−1 − 2n(z2 − 1)n−1

]=

1

2nn!

dn

dzn[(z2 − 1)n + 2n(z2 − 1)(z2 − 1)n−1

]=

1

2nn!

dn

dzn[(2n+ 1)(z2 − 1)n

]=

(2n+ 1)

2nn!

dn

dzn[(z2 − 1)n

]= (2n+ 1)Pn(z)

The Legendre polynomial of order n contains n zeros and these zeros liebetween −1 and 1. This proposition is proved in the following theorem.



Theorem 1 The zeros of Legendre polynomial are all real, distinct and lie be-tween −1 and 1.

Proof: Let us consider the Rodrigues formula

Pn(z) =1

2nn!

dn

dzn[(z2 − 1)n

]Rewriting this equation becomes

Pn(z) =1

2nn!

dn

dzn[(z + 1)n(z − 1)n]

and applying Leibnitz theorem n times we obtain

Pn(z) =1

2nn!

[n!(z + 1)n + n(n− 1)!(z − 1)(z + 1)n−1 + · · ·+ n!(z − 1)n

]If z be real and |z| > 1 all the terms in the above are positive, i.e., Pn(z) isalways positive. Hence there cannot be any zero for all z with |z| > 1. Hencethe zeros lie between −1 and 1.

Again, the equation (z2 − 1)n = 0 has 2n roots.Among these roots n roots are positive (1) and n roots are negative (-1).

Therefore by Rolle’s theorem, the equationd

dz

[(z2 − 1)n

]= 0 must have

at least one root between −1 and 1 also this is the same as

(z + 1)n−1n(z − 1)n + n(z + 1)n(z − 1)n−1 = 0or (z + 1)n−1(z − 1)n−1.2z = 0

Hence the equationd

dz

[(z2 − 1)n

]= 0 has (2n− 1) roots, n− 1 roots equal

to 1 and n− 1 roots equal to −1 and one root, say α0 (= 0) between −1 and 1.

Again, from the above result it follows that the equationd2

dz2[(z2 − 1)n

]=

0 must have at least one root, say α1 (say) between −1 and α0 and at least

one root α2 (say) between α0 and 1. Henced2

dz2[(z2 − 1)n

]= 0 has two roots

between −1 and 1.But,

d2

dz2[(z2 − 1)n

]= (z − 1)n−2(z + 1)n+2f2(z) = 0

where f2(z) is a polynomial of degree 2 of z. Thus the equation has n− 2 rootsequal to 1, n− 2 roots equal to −1 and two roots α1 and α2 between −1 and 1.

Similarly,d3

dz3[(z2 − 1)n

]= 0 has three roots α′1, α

′2, α

′3 between −1 and

α1; α1 and α2; α2 and 1.

Proceeding this way,dn

dzn[(z2 − 1)n

]= 0 has n roots all real, distinct

and lie between −1 and 1. Hence the theorem follws.



2.4 Orthogonality

Let z be real and Pm(z) and Pn(z) are the Legendre polynomial of degree mand n respectively. Since the Legendre polynomial is the solution of Legendreequation, therefore

d

dz

[(1− z2)dPm

dz

]+m(m+ 1)Pm = 0 (6)

and

d

dz

[(1− z2)dPn

dz

]+ n(n+ 1)Pn = 0 (7)

Multiplying (6) by Pn and (7) by Pm and subtracting we have

d

dz

[(1− z2)(P ′mPn − P ′nPm)

]+ [m(m+ 1)− n(n+ 1)]PmPn = 0

Integrating between the limits −1 and 1 with respect to z

[(1− z2)(P ′mPn − P ′nPm)

]1−1 +

1∫−1

[m(m+ 1)− n(n+ 1)]PmPndz = 0

or [m(m+ 1)− n(n+ 1)]1∫

−1

PmPndz = 0

or

1∫−1

PmPndz = 0 since m ̸= n

It can be shown that

1∫−1

Pn(z)2dz =

2

2n+ 1

This fact is justified by considering the Rodrigues formula in the following.

1∫−1

Pn(z)2dz = A

1∫−1

u(n)n u(n)n dz

where A =1

(2nn!)2and un = (z

2 − 1)n, u(n)n denotes the nth derivative of



un

=[u(n−1)n u

(n)n

]1−1

−A1∫

−1

u(n−1)n u(n+1)n dz

= 0 + (−1)A1∫

−1

u(n−1)n u(n+1)n dz

= (−1)nA1∫

−1

unu(2n)n dz

= (−1)n(2n)!A1∫

−1

undz since u(2n)n = (2n)!

Now,1∫

−1

(z2 − 1)ndz =1∫

−1

(z + 1)n(z − 1)ndz

Putting 1− z = 2y ⇒ −dz = 2dy

Therefore the integral = (−1)n0∫

1

22n+1yn(1− y)n(−dz)

= (−1)n22n+1B(n+ 1, n+ 1) = (−1)n22n+1 (n!)2

(2n+ 1)!

Hence1∫

−1

P 2n(z)dz = A(−1)n(2n)!(−1)n22n+1(n!)2

(2n+ 1)!

Note: The orthogonality of Legendre polynomial in terms of trigonometric formis

Pm(cos θ)Pn(cos θ) sin θdθ =

{0, if m ̸= n

2

2n+ 1, if m = n

.

2.5 Expansion of Functions: Legendre Series

If f(z) be a piecewise continuous function then it can be expanded in Legendreseries. Let us assume that the series

∞∑n=0

anPn(z) = f(z) (8)



in the sense of convergence in the interval [−1, 1]. This means that the functionf(z) and f ′(z) be at least peace wise continuous in this interval. To find an,multiplying (8) by Pm(z), integrating term by term and then using orthogonalproperty, we obtain

2

2m+ 1am =

1∫−1

f(z)Pm(z)dz

Changing the variable of integration z by t and the index m by n, we have

an =2n+ 1

2

1∫−1

f(t)Pn(t)dt (9)

The constant an are called Legendre constants.

Using (9), (8) becomes

f(z) =∞∑

n=0

2n+ 1

2

1∫−1

(f(t)Pn(t)dt)Pn(z) (10)

This expansion in a series of Legendre polynomial is usually referred to as aLegendre series.

Legendre polynomial is also satisfy the following properties.

Property 1 Prove that

1∫−1

Pn(z)dz =

{0, if n ̸= 02, if n = 0

.

Proof: Since Pn(z) is a solution of Legendre equation, then

d

dz

[(z2 − 1)dPn

dz

]+ n(n+ 1)Pn = 0



Integrating between the limits −1 and 1, we get

1∫−1

d

dz

[(z2 − 1)dPn

dz

]dz +

1∫−1

n(n+ 1)Pn(z)dz = 0

or

[(z2 − 1)dPn

dz

]1−1

+ n(n+ 1)

1∫−1

Pn(z)dz = 0

or 0 + n(n+ 1)

1∫−1

Pn(z)dz = 0

or

1∫−1

Pn(z)dz = 0, n ̸= 0

When n = 0,

1∫−1

Pn(z)dz =

1∫−1

P0(z)dz =

1∫−1

dz = 2

Property 2 For integers m and n

1∫−1

zmPn(z)dz =

0, if m < n2n+1(n!)2

(2n+ 1)!, if m = n

.

Proof: Let A =1

2nn!and un = (z

2 − 1)n



Then

1∫−1

zmPn(z)dz = A

1∫−1

zmu(nn)dz

= A[u(n−1)n z

m]1−1

−A1∫

−1

mzm−1u(n−1)n dz

= (−1)A1∫

−1

mzm−1u(n−1)n dz

= (−1)2Am1∫

−1

(m− 1)zm−2u(n−2)n dz

= (−1)mAm(m− 1)(m− 2) . . . 2.11∫

−1

u(n−m)n dz (11)

= (−1)mAm![u(n−m−1)n

]1−1

= (−1)mAm!0 = 0, since n > m,n−m− 1 ≥ 0, so u(n−m−1)n contains at least one z2 − 1.

From equation (11), for n = m we have

1∫−1

zmPn(z)dz = (−1)nAn!1∫

−1

un(z)dz

= (−1)n 12nn!

n!(−1)n (n!)2

(2n)!

22n+1

2n+ 1=

2n+1(n!)2

(2n+ 1)!

2.6 Generating Function

The generating function for Legendre polynomial is

g(z, t) = (1− 2zt+ t2)− 12 , | z |≤ 1, | t |< 1

That is, the coefficient of tn is the Legendre polynomial of order n is (1 −2zt+ t2)−

12 . This generating function generates the Legendre polynomial of all

orders.For | z |≤ 1 and | t |< 1, the expansion of (1− 2zt+ t2)− 12 is shown below.Since | z |≤ 1 and | t |< 1, (1−2zt+ t2)− 12 can be expanded in power series.

(1− 2zt+ t2)− 12 = [1− t(2z − t)]− 12

= 1 +1

2t(2z − t) + 1.3

2.4t2(2z − t)2 + · · ·+ 1.3. . . . (2n− 1)

2.4 . . . 2ntn(2z − t)n + · · ·



Now the coefficient of tn in the term

1.3. . . . (2n− 1)2.4 . . . 2n

tn(2z − t)n is 1.3. . . . (2n− 1)2.4 . . . 2n

(2z)n

=1.3. . . . (2n− 1)

2.4 . . . 2n2nzn

=1.3. . . . (2n− 1)

n!zn

Again the coefficient of tn in the term

1.3. . . . (2n− 3)2.4 . . . (2n− 2)

tn−1(2z − t)n−1 is 1.3. . . . (2n− 3)2.4 . . . (2n− 2)

{−(n− 1)(2z)n−2}

= − 1.3. . . . (2n− 3)2n−11.2 . . . (n− 1)

2n− 1n

n

2n− 1{(n− 1)(2)n−2zn−2}

= −1.3. . . . (2n− 2)n!

n(n− 1)2(2n− 1)

zn−2

and so on.Hence the coefficient of tn in the expression of (1− 2zt+ t2)− 12 is

1.3. . . . (2n− 2)n!

[zn − n(n− 1)

2(2n− 1)zn−2 + · · ·

]i.e., the coefficient of tn in the expression of (1− 2zt+ t2)− 12 is Pn(z).

Thus (1− 2zt+ t2)− 12 =∞∑

n=0

tnPn(z) (12)

2.7 Special values

The generating function gives more information about Legendre polynomials.For z = 1 the equation (12) becomes

(1− 2t+ t2)− 12 =∞∑

n=0

tnPn(1)

or1

1− t=

∞∑n=0

tnPn(1)

or∞∑

n=0

tn =∞∑

n=0

tnPn(1)

Hence Pn(1) = 1 for all values of n.For z = 0 the equation (12) becomes

(1 + t2)−12 =

∞∑n=0

Pn(z)tn



Also, by binomial expansion

(1 + t2)−12 = 1− 1

2t2 +

3

8t4 + · · ·+ (−1)n 1.3. . . . .(2n− 1)

2nn!t2n + · · ·

Hence P2n(0) =coefficient of t2n in the expansion of (1 + t2)−

12

= (−1)n 1.3. . . . .(2n− 1)2nn!

=(−1)n(2n)!22n(n!)2

P2n+1(0) =coefficient of t2n+1 in the expansion of (1 + t2)−

12 is 0.

The generating function remains unchanged, if we replace z by −z and t by−t.

That is,

g(t, z) = g(−t,−z) = [1− 2(−t)(−z) + (−z)2]− 12 =∞∑

n=0

Pn(−z)(−t)n

i.e.,∞∑

n=0

Pn(z)(t)n =

∞∑n=0

(−1)nPn(−z)(t)n

Hence

Pn(−z) = (−1)nPn(z) (13)

That is, the Legendre polynomials are odd or even with respect to z = 0 ac-cording to whether the index is odd or even. This is called parity or reflectionproperty of Legendre polynomial.

From the relation (13), it is easy to observe that

P2n+1 = 0

The Legendre polynomial can be expressed in terms of cosine function. To findthe expression for Pn(cos θ), recall the generating function (12).

Substituting z = cos θ =eiθ + e−iθ

2to

(1− 2zt+ t2)− 12 =∞∑

n=0

Pn(z)(t)n

we get

∞∑n=0

Pn(cos θ)(t)n =

[1− t(eiθ + e−iθ) + t2

]− 12 = (1− teiθ)− 12 (1− te−iθ)− 12Using binomial expansion, we have

(1− teiθ)− 12 = 1 + 12teiθ +

1.3

2.4t2e2iθ + · · ·+ 1.3. . . . .(2n− 1)

2.4. . . . (2n)tneinθ + · · ·



(1− te−iθ)− 12 = 1 + 12te−iθ +

1.3

2.4t2e−2iθ + · · ·+ 1.3. . . . .(2n− 1)

2.4. . . . (2n)tne−inθ + · · ·

Multiplying the above two expression and equating the coefficient of tn we obtain

Pn(cos θ) =1.3. . . . .(2n− 1)2.4. . . . (2n)

2 cosnθ +1

2

1.3. . . . .(2n− 3)2.4. . . . (2n− 2)

2 cos(n− 2)θ+ (14)

Example 2.3 If (R,ϕ, z) and (r, θ, ϕ) are cylindrical polar and spherical polarcoordinates of the same point, show that

Pn(cos θ) = (−1)nrn+1

n!

∂n

∂zn

(1

r

)

Solution. We have r2 = x2+y2+z2 = R2+z2. So1

r= (R2+z2)−

12 = f(R, z).

Then f(R, z + h) =[R2 + (z + h)2

]− 12 = [R2 + z2 + 2zh+ h2]− 12= [r2 + 2hr cos θ + h2]−

12 =

1

r

[1 +

2h cos θ

r+

h2

r2

]− 12=

1

r

∞∑n=0

(−1)n(h

r

)nPn(cos θ) (15)

Again, by Taylor’s series, for several variables

f(R, z + h) =

∞∑n=0

hn

n!

∂n

∂znf(R, z) =

∞∑n=0

hn

n!

∂n

∂zn

(1

r

)(16)

From the equation (15) and (16) we have

1

r

∞∑n=0

(−1)n(h

r

)nPn(cos θ) =

∞∑n=0

hn

n!

∂n

∂zn

(1

r

)Equating the coefficient of hn on both sides of the above equation, we get

(−1)n

rn+1Pn(cos θ) =

1

n!

∂n

∂zn

(1

r

)That is,

Pn(cos θ) =(−1)n

n!rn+1

∂n

∂zn

(1

r

)

2.8 Bounds of Pn(cos θ)

The generating function may be used to find the bounds of Pn(cos θ).We have for z = cos θ[

1− t(eiθ + e−iθ) + t2]− 12 = (1− teiθ)− 12 (1− te−iθ)− 12



=

(1 +

1

2teiθ +

1.3

2.4t2e2iθ + · · ·

)×(1 +

1

2te−iθ +

1.3

2.4t2e−2iθ + · · ·

)(17)

It may be noted that all the coefficient of above expression are positive. Sothe each term of right hand side of (17) is positive. Each term of the (17) is ofthe form

am(eimθ + e−imθ)

2= am cosmθ

where am is positive. Then (17) reduces to

Pn(cos θ) =n∑

m=0 or 1am cosmθ (18)

The right hand side of (18) is clearly maximum when θ = 0 and mθ = 1.That is, the maximum value of Pn(cos θ) is Pn(1). But for z = cos θ = 1,Pn(1) = 1.

Therefore |Pn(cos θ)| ≤ Pn(1) = 1 i.e., −1 ≤ Pn(cos θ) ≤ 1From the equation (14) and(18) we may conclude that the Legendre polyno-

mial is a linear combination of cosmθ. This means that the Legendre polyno-mials form a complete set for any functions that may be expanded by a Fouriercosine series over the interval (0, π).

2.9 Recurrence Relations

The generating function for Legendre polynomials provides a better way todeduce the recurrence relations and some special properties. Differentiating therelation

(1− 2zt+ t2)− 12 =∞∑

n=0

tnPn(z) (19)

with respect to t we obtain

−12(1− 2zt+ t2)− 32 (−2z + 2t) =

∞∑n=0

ntn−1Pn(z)

or (z − t)(1− 2zt+ t2)− 12 = (1− 2zt+ t2)∞∑

n=0

ntn−1Pn(z)

or (z − t)∞∑

n=0

tnPn(z) = (1− 2zt+ t2)∞∑

n=0

ntn−1Pn(z)

Equating the coefficient of tn from both sides, we get

zPn(z)− Pn−1(z) = (n+ 1)Pn+1(z) + nPn−1(z)



On simplification,

(2n+ 1)zPn(z) = (n+ 1)(n+ 1)Pn+1(z) + nPn−1(z) (20)

Again differentiating the relation (19) with respect to t we obtain

−12(1− 2zt+ t2)− 32 (−2z + 2t) =

∞∑n=0

ntn−1Pn(z) (21)

Differentiating the relation (19) with respect to z we obtain

−12(1− 2zt+ t2)− 32 (−2t) =

∞∑n=0

tnP ′n(z) (22)

Dividing (21) by (22)

t− z−t

=

∞∑n=0

ntn−1Pn(z)

∞∑n=0

tnP ′n(z)

That is,∞∑

n=0

tn+1P ′n(z)− z∞∑

n=0

tnP ′n(z) +∞∑

n=0

ntnPn(z)

Equating coefficient of tn to zero,

P ′n−1(z)− zP ′n(z) + nPn(z) = 0

On simplification we get

nPn(z) = zP′n(z)− P ′n−1(z) (23)

Another recurrence relation can be obtained by differentiating the relation (19)with respect to z. Differentiating the relation (19) with respect to z we obtain

−12(1− 2zt+ t2)− 32 (−2t) =

∞∑n=0

tnP ′n(z)

Differentiating the relation (19) with respect to z we obtain

t

∞∑n=0

tnPn(z) = (1− 2zt+ t2)∞∑

n=0

tnP ′n(z)

Equating the coefficient of tn+1 in both sides we obtain

Pn(z) = zP′n+1(z)− 2zP ′n(z) + P ′n−1(z) (24)



From (23),zP ′n(z) = nPn(z) + P

′n−1(z)

Substituting this value in (24), we get

Pn(z) = zP′n+1(z)− 2[nPn(z) + P ′n−1(z)] + P ′n−1(z)

or (2n+ 1)Pn(z) = P′n+1(z)− P ′n−1(z) (25)

Combination of (23) and (24) yields

Pn(z) = P′n+1(z)− 2zP ′n(z) + [zP ′n(z)− nPn(z)]

i.e., (n+ 1)Pn(z) = P′n+1(z)− zP ′n(z) (26)

Replacing n by n− 1 we get

nPn−1(z) = P′n(z)− zP ′n−1(z) (27)

Multiplying (26) by z and adding with (27) we get

nPn−1(z) + z(n+ 1)Pn(z) = zP′n+1(z)− z2P ′n(z) + P ′n(z)− zP ′n−1(z)

= (1− z2)P ′n(z)− z(P ′n+1(z)− P ′n−1(z))= (1− z2)P ′n(z) + z(2n+ 1)Pn(z) [Using (25)]

This relation can be written as

(1− z2)P ′n(z) = n[(Pn−1(z)− zPn(z)] (28)

Differentiating (28) we have

(1− z2)P ′′n (z)− 2zP ′n(z) = nP ′n−1(z)− nPn(z)− nzP ′n(z)= n[zP ′n(z)− nPn(z)]− nPn(z)− nzP ′n(z) [Using (23)]

Thus

(1− z2)P ′′n (z)− 2zP ′n(z) + n(n+ 1)Pn(z) = 0 (29)

The equation (20), (23), (25), (26) and (28) are the recurrence relations forLegendre polynomials.

The equations (23), (25), (26) and (28) are all first order differential equa-tions but with two different indices. the indices of (29) are equal but of secondorder differential equation and this equation is Legendre equation.

It may be noted that the recurrence relations are deduced from the gen-erating function of the Legendre polynomial without considering the Legendredifferential equation and finally we obtain the Legendre differential equation(29).



The equation (29) can be written as

d

dz

[(1− z2)dPn(z)

dz

]+ n(n+ 1)Pn(z) = 0

Substituting z = cos θ,dz

dθ= − sin θ

d

dθ

[(1− z2)dPn(z)

dθ

dθ

dz

]+ n(n+ 1)Pn(z) = 0

1

sin θ

d

dθ

[sin θ

dPn(cos θ)

dθ

]+ n(n+ 1)Pn(cos θ) = 0 (30)

2.10 Integral Representation

Schlafli’s IntegralThe Rodrigues formula is

Pn(z) =1

2nn!

dn

dzn{(z2 − 1)n}

From the theory of analytic function, it is known that if f(z) be analytic thanits nth derivative can be represented in the integral form

dn

dzn{f(z)} = n!

2πi

∫C

f(t)

(t− z)n+1dt

where C is a closed contour within and on which f(z) is analytic.The function (z2 − 1)n is analytic for all values of z. Therefore

dn

dzn{(z2 − 1)n} = n!

2πi

∫C0

(t2 − 1)n

(t− z)n+1dt

where C0 is a closed contour in t-plane enclosing t = z.Therefore

Pn(z) =1

2n.2πi

∫C0

(t2 − 1)n

(t− z)n+1dt (31)

This is Schilafli’s integral for Legendre’s polynomial when n is an integer.For non-integral values of n, the solution of Legendre equation can also be

represented in terms of integration.Assume

w =1

2n.2πi

∫C

(t2 − 1)n

(t− z)n+1dt (32)



-

-

�

-

qr rt=-1 t=1

Figure 1: The contour C2

where the contour C is to be so chosen that this satisfies the Legendre differentialequation

d

dz

{(1− z2)dw

dz

}+ n(n+ 1)w = 0 (33)

the left hand side of (33) by substituting (32) is

1

2πi

∫C

(t2 − 1)n

2n

[d

dz

{(n+ 1)(1− z2)

(t− z)n+2

}+

n(n+ 1)

(t− z)n+1

]dt

=(n+ 1)

2n.2πi

∫C

(t2 − 1)n

(t− z)n+3[(n+ 2)(1− z2)− 2z(1− z) + n(t− z2)

]dt

=(n+ 1)

2n.2πi

∫C

(t2 − 1)n

(t− z)n+3[2t(n+ 1)(1− z)− (n+ 2)(t2 − 1)

]dt

=(n+ 1)

2n.2πi

∫C

d

dt

[(t2 − 1)n+1

(t− z)n+2

]dt

=(n+ 1)

2n.2πi

[(t2 − 1)n+1

(t− z)n+2

]C

(34)

If n be an integer then (34) vanishes when C is any closed curve surroundingthe point t = z and thus (32) satisfies the differential equation (32) for such acontour.

If n = λ (say) where λ is not an integer then the contour C is to be suchthat

(t2 − 1)λ+1

(t− z)λ+2

returns to its original value after describing the contour C. The singularities ofthe above expression are at t = ±1 and at t = z.

If we consider a circle or any closed curve C1 surrounding the singularitiest = z and t = 1 in the clockwise sense the expression

(t2 − 1)λ+1

(t− z)λ+2



is multiplied by

e−2πi(λ+1)

e−2πi(λ+2)= 1

when we go round the contour once.

Apart from C1, there is another contour C2 which is shown in Figure 10.1.C2 consists of two loops one surrounding t = −1 in clockwise sense and theother surrounding t = 1 surrounding t = 1 in clockwise sense. On moving roundthis contour the expression is multiplied by e−2πi(λ+1)e2πi(λ+1) = 1 so that thecontour C2 also serves our purpose. The function with this contour integralform is Legendre formula of second kind.

2.11 Laplace Integral

Recall the Schlafli’s integral representation (31) for Legendre polynomials

Pn(z) =1

2n.2πi

∫C0

(t2 − 1)n

(t− z)n+1dt

where C is the closed contour enclosing the point t = 1 and t = z but nott = −1.

Let us take for the contour C, the circle with the point z as centre and ofradius |z2− 1| 12 . In order that this circle may enclosed the point t = 1, we musthave |z − 1| < |z2 − 1| 12 i.e., |z − 1| < |z + 1|. Also, in order that this circlemay not enclose the point t = −1, i.e., t = −1 be outside the circle, we musthave |z2 − 1| 12 < |z + 1|.

In order that this circle may enclosed of point z = 1 and may not t = −1the point z not lie to the right of the imaginary axis. Therefore

| arg z| < π2

If we take

t = z + (z2 − 1) 12 eiϕ

be any point on the above circle with z as centre and radius√z2 − 1. Then t

describes C once in the positive sense ϕ changes from −π to π. Now,

t2 − 1 = z2 + (z2 − 1)e2iϕ + 2z(z2 − 1) 12 eiϕ − 1

= (z2 − 1) 12 e2iϕ[2z + (z2 − 1) 12 (eiϕ + e−iϕ)

]= 2(z2 − 1) 12 e2iϕ

[z + (z2 − 1) 12 cosϕ

]dt = (z2 − 1) 12 eiϕidϕ



Therefore

Pn(z) =1

2n.2πi

π∫−π

2n(z2 − 1)n2 einϕ{z + (z2 − 1) 12 cosϕ

}n(z2 − 1)n+1e(n+1)iϕ

(z2 − 1) 12 eiϕidϕ

=1

π

π∫0

[z + (z2 − 1) 12 cosϕ

]ndϕ

=1

2π

π∫−π

[z + (z2 − 1) 12 cosϕ

]ndϕ (35)

This is known as Laplace first integral.

3 Bessel Differential Equation

Bessel functions appear in a large class of physical problem. Bessel functionsand closely related functions form a rich area of applied mathematics with manyrepresentations, many interesting and useful properties and many interrelations.

The Bessel differential equation of order λ is

z2d2w

dz2+ z

dw

dz+ (z2 − λ2)w = 0 (36)

It is obvious that z = 0 is a regular singular point. To investigate the point

at ∞ replacing the independent variable z by 1t, the equation becomes

t4d2w

dt2+ t3

dw

dt+ (1− λ2t2)w = 0 (37)

This shows that the equation (40) has a regular singular point at t = 0 andhence the equation (36) has a singular point at z = ∞.

To obtain a series solution in the neighbourhood of z = 0 by Frobeniusmethod, let us put

w =∞∑k=0

ckzk in the equation (36).

Then (36) becomes

∞∑k=0

ck(ρ+k)(ρ+k−1)zρ+k+∞∑k=0

ck(ρ+k)zρ+k+

∞∑k=0

ckzρ+k+2−λ2

∞∑k=0

ckzρ+k = 0

The indicial equation is obtained by equating the coefficient of zρ to zero as

ρ(ρ− 1) + ρ− λ2 = 0



That is, ρ = ±λNow the coefficient of zρ+k is

ck[(ρ+ k)2 − λ2

]+ ck−2 = 0

i.e., ck = −ck−2

(ρ+ λ+ k)(ρ− λ+ k)Coefficient of zρ+1 is c1

[(ρ+ 1)2 − λ2

]= 0

But [(ρ+ 1)2 − λ2] ̸= 0 for ρ = ±λ. Thus c1 = 0.This leads to the relation c3 = c5 = c7 = · · · = 0, i.e., the coefficient of all

odd indices are zero and

c2k = −c2k−2

(ρ+ λ+ 2k)(ρ− λ+ 2k), k = 1, 2, . . .

Case I: ρ = λ. Without loss of generality we assume that λ ≥ 0.Therefore,

c2k = −c2k−2

2k(2λ+ 2k)= − 1

22c2k−2

k(λ+ k)

Hence c2 = −1

221!

1

(λ+ 1)c0

c4 = (−1)21

242!

1

(λ+ 1)(λ+ 2)c0

c6 = (−1)31

263!

1

(λ+ 1)(λ+ 2)(λ+ 3)c0 and so on.

Hence w = zλ[c0 + c2z2 + c4z

4 + c6z6 + · · · ]

= c0zλ

[1− z

2

22(λ+ 1)+

z4

242!(λ+ 1)(λ+ 2)− z

6

26.3!(λ+ 1)(λ+ 2)(λ+ 3)+ · · ·

]= c0z

λ

[1− (z/2)

2

(λ+ 1)+

(z/2)4

2!(λ+ 1)(λ+ 2)− (z/2)

6

3!(λ+ 1)(λ+ 2)(λ+ 3)+ · · ·

]With a special choice c0 =

1

2λΓ(λ+ 1), w is given by

w = (z/2)λ1

Γ(λ+ 1)

[1− (z/2)

2

(λ+ 1)+

(z/2)4

2!(λ+ 1)(λ+ 2)− (z/2)

6

3!(λ+ 1)(λ+ 2)(λ+ 3)+ · · ·

]=

∞∑k=0

(−1)k(z/2)λ+2k

Γ(k + 1)Γ(λ+ k + 1)

This is called Bessel function (of first kind) of order λ and is denoted by Jλ(z).Thus

Jλ(z) =∞∑k=0

(−1)k(z/2)λ+2k

Γ(k + 1)Γ(λ+ k + 1)(38)



This is a convergent series for all values of λ ≥ 0 and for all values of z.Case II: λ− (−λ) = 2λ is not an integer.

In this case the other independent solution is obtained by putting −λ for λ.Hence

J−λ(z) =∞∑k=0

(−1)k(z/2)2k−λ

Γ(k + 1)Γ(−λ+ k + 1)(39)

This series is also convergent for all values of z.It can be verified that both Jλ(z) and J−λ(z) are solution of (36). If 2λ is

not an integer then Jλ(z) and J−λ(z) are two independent solution of (36) andthe general solution of Bessel equation is

w = AJλ(z) +BJ−λ(z).

Case III: λ = 0In this case Jλ(z) and J−λ(z) are not independent. One solution is J0(z),

where

J0(z) =∞∑k=0

(−1)k(z/2)2k

Γ(k + 1)Γ(k + 1)=

∞∑k=0

(−1)k(z/2)2k

(k!)2.

To find the second solution let

w =

∞∑k=0

ckzρ+k

Proceeding in the similar way as in Case I, we have c2k+1 = 0, k = 0, 1, 2, . . .and

c2k = −c2k−2

(ρ+ λ+ 2k)(ρ− λ+ 2k)= − c2k−2

(ρ+ 2k)2, (as λ = 0) k = 1, 2, . . .

Therefore

w = zρ[c0 + c2z2 + c4z

4 + c6z6 + · · · ]

= c0zρ

[1 +

c2c0

z2 +c4c2

c2c0

z4 + · · ·]

= c0zρ

[1− z

2

(ρ+ 2)2+

z4

{(ρ+ 2)(ρ+ 4)}2− z

4

{(ρ+ 2)(ρ+ 4)(ρ+ 6)}2+ · · ·

]

= c0zρ

∞∑k=0

(−1)kz2k

{(ρ+ 2)(ρ+ 4) . . . (ρ+ 2k)}2

= c0zρ

∞∑k=0

(−1)kz2kF (ρ)



where

F (ρ) =1

{(ρ+ 2)(ρ+ 4) . . . (ρ+ 2k)}2

Now, logF (ρ) = −2{log(ρ+ 2) + log(ρ+ 4) + · · ·+ log(ρ+ 2k)}

Differentiating with respect to ρ we get

1

F (ρ)F ′(ρ) = −

[1

ρ+ 2+

1

ρ+ 4+ · · ·+ 1

ρ+ 2k

]That is,

F ′(ρ) = −2F (ρ)k∑

s=1

1

ρ+ 2s

This gives

F ′(0) = −F (0)k∑

s=1

1

s= − 1

(2.4. . . . 2k)2

k∑s=1

1

s

= − 122k(k!)2

k∑s=1

1

s

Hence

∂w

∂ρ= c0z

ρ log z∞∑k=0

(−1)kz2kF (ρ) + c0zρ log z∞∑k=0

(−1)kz2kF ′(ρ)

Therefore the second solution denoted by Y0(z) is given by

Y0(z) =

(∂w

∂ρ

)ρ=0

= c0 log z∞∑k=0

(−1)kz2kF (ρ) + c0∞∑k=0

(−1)kz2kF ′(ρ)

= log z

(c0

∞∑k=0

(−1)kz2k

22k(k!)2

)− c0

∞∑k=0

(−1)kz2k 122k(k!)2

k∑s=1

1

s

As c0 = 1 because λ = 0. Therefore

Y0(z) = J0(z) log z −∞∑k=1

(−1)k(z/2)2k

(k!)2

k∑s=1

1

s(40)

The function Y0(z) is called Neumann’s function of the second kind of orderzero. The function

Y0(z) =2

π[Y0(z)− (log 2− γ)J0(z)]



where γ is Euler’s constant, being obtained by adding to Y0(z) a function whichis a constant multiple of J0(z), will also satisfy the Bessel equation. It is calledWeber’s Bessel function of the second kind of order zero.

Therefore

Y0(z) =2

π{log(z/2) + γ}J0(z)−

2

π

∞∑k=1

(−1)k(z/2)2k

(k!)2

k∑s=1

1

s

The general solution of Bessel equation when λ = 0 is

w = AJ0(z) +BY0(z) or w = AJ0(z) +BY0(z)

Case IV: 2λ is an integer.In this case the general solution of the Bessel equation can be written as

w = AJλ(z) +BYλ(z)

where

Yλ(z) =1

π{log(z/2) + γ}Jλ(z)−

1

π

λ−1∑k=0

(λ− k − 1)!k!

(2

z

)λ−2k

− 1π

∞∑k=0

(−1)k(z/2)λ+2k

k!(λ+ k)!

{λ+k∑s=1

1

s+

k∑s=1

1

s

}The function Yλ(z) is called Weber’s Bessel function of the second kind of orderλ.

The Neumann Bessel function of second kind of order λ is

Yλ(z) = J0(z) log z −1

π

λ−1∑k=0

(λ− k − 1)!k!

(2

z

)λ−2k

− 1π

∞∑k=0

(−1)k(z/2)λ+2k

k!(λ+ k)!

{λ+k∑s=1

1

s+

k∑s=1

1

s

}The equation (39) is also valid for integer values of λ. Let λ = n where n is apositive integer. Then

J−n(z) =∞∑k=0

(−1)k(z/2)2k−n

k!(k − n)!

Since n is an integer,1

(k − n)!=

1

Γ(k − n+ 1)→ 0 for k = 0, 1, . . . , n − 1.

Hence the series may be considered to start with k = n. Replacing k by k + n,we obtain

J−n(z) =

∞∑k=0

(−1)k+n(z/2)2k+n

k!(k + n)!



Theorem 2 The Bessel functions Jn(z) and J−n(z) are not independent, butthey are related by

Jn(z) = (−1)nJ−n(z) for integral n.

Proof.

J−n(z) =

∞∑k=0

(−1)k+n(z/2)2k+n

k!(k + n)!= (−1)n

∞∑k=0

(−1)k(z/2)2k+n

k!(k + n)!= (−1)nJn(z)

3.0.1 Generating Function

The generating function has the advantage of focusing on the functions them-selves rather than on the differential equations they satisfy. The generatingfunction of Bessel differential equation is g(z, t) = e

z2 (t−

1t ). Thus when λ is a

positive integer Jλ(z) is equal to the coefficient of tλ in the expansion of e

z2 (t−

1t ),

i.e.,

ez2 (t−

1t ) =

∞∑−∞

tλJλ(z)

Proof. ez2 (t−

1t ) = e

zt2 e−

z2t

=

(1 +

zt

2+

1

2!

z2t2

22+ · · ·+ 1

λ!

zλtλ

2λ+

1

(λ+ 1)!

z(λ+1)t(λ+1)

2(λ+1)+ · · ·

)

×(1− z

2t+

1

2!

z2

22t2+ · · ·+ (−1)

λ

λ!

zλ

2λtλ+

(−1)λ+1

(λ+ 1)!

z(λ+1)

2(λ+1)t(λ+1)+ · · ·

)Coefficient of tλ in the product is

zλ

2λ.λ!− z

λ+2

2λ+2.(λ+ 1)!+

zλ+4

2!2λ+4.(λ+ 2)!− · · ·

=zλ

2λ.λ!

[1− 1

λ+ 1

(z2

)2+

1

2!(λ+ 1)(λ+ 2)

(z2

)4− · · ·

]= Jλ(z)

Coefficient of t−λ in the product is

(−1)λzλ

2λ.λ!− (−1)

λ+1zλ+2

2λ+2.(λ+ 1)!+

(−1)λ+2zλ+4

2!2λ+4.(λ+ 2)!+ · · ·

=(−1)λzλ

2λ.λ!

[1− 1

λ+ 1

(z2

)2+

1

2!(λ+ 1)(λ+ 2)

(z2

)4− · · ·

]= (−1)λJλ(z) = J−λ(z)

The following result can easily be deduced using the concept of generating func-tion rather than using the differential equation.




cos(z sin θ) = J0(z) + 2∞∑

n=1

J2n(z) cos 2nθ (41)

sin(z sin θ) = 2∞∑

n=0

J2n+1(z) sin(2n+ 1)θ (42)

What will be the cases when θ = π2 ?

Solution. From generating function

ez2 (t−

1t ) =

∞∑n=−∞

tnJn(z)

= (J0 + J1t+ J2t2 + · · · ) + {(−1)1J1t−1 + (−1)2J2t−2 + (−1)3J3t−3 + · · · }

= J0 + J1(t− t−1) + J2(t2 + t−2) + J3(t3 + t−3) + · · ·

Let t = cos θ + i sin θ.Then tn = cosnθ + i sinnθ and t−n = cosnθ − i sinnθ.Therefore tn + t−n = 2 cosnθ and tn − t−n = 2i sinnθ.Substituting these values to the above relation we get

eiz sin θ = J0 + 2iJ1 sin θ + 2J2 cos 2θ + 2iJ3 sin 3θ + 2J4 sin 4θ + · · ·

Equating the real and imaginary parts from both sides we have

cos(z sin θ) = J0(z) + 2

∞∑n=1

J2n(z) cos 2nθ

and

sin(z sin θ) = 2

∞∑n=0

J2n+1(z) sin(2n+ 1)θ

Second part: when θ = π2 then above relations become

cos z = J0(z) + 2∞∑

n=1

(−1)nJ2n(z)

and

sin z = 2∞∑

n=0

(−1)nJ2n+1(z)

This example suggest that sin z and cos z can be expressed in terms of Besselfunctions.



3.1 Recurrence Relation of Jλ(z)

Different recurrence relations are available for Bessel functions which are mostimportant to many applications. These are presented below.

Differentiating partially the relation ez2 (t−

1t ) =

∞∑n=−∞

tλJλ(z) with respect to z, we obtain

1

2

(t− 1

t

)e

z2 (t−

1t ) =

∞∑n=−∞

tλJ ′λ(z)

or

(t− 1

t

) ∞∑n=−∞

tλJλ(z) = 2

∞∑n=−∞

tλJ ′λ(z)

Equating the coefficient of tλ, we get

2Jλ(z) = Jλ−1(z)− Jλ+1(z) (43)

Again differentiating partially the relation

ez2 (t−

1t ) =

∞∑n=−∞

tλJλ(z)

with respect to t we obtain

z

2

(1 +

1

t2

)e

z2 (t−

1t ) =

∞∑n=−∞

λtλ−1Jλ(z)

This gives,

z

(1 +

1

t2

) ∞∑n=−∞

tλJλ(z) = 2

∞∑n=−∞

λtλ−1Jλ(z)

Equating the coefficient of tλ−1 from both sides we obtain

2λ

zJλ(z) = Jλ−1(z) + Jλ+1(z) (44)

Adding (43) and (44) we get

2J ′λ(z) +2λ

zJλ(z) = 2Jλ−1(z)

This relation can be written as

zJ ′λ(z) = zJλ−1(z)− λJλ(z) (45)

Subtracting (43) from (44) we get

2λ

zJλ(z)− 2J ′λ(z) = 2Jλ+1(z)



That is,

zJ ′λ(z) = λJλ(z)− zJλ+1(z) (46)

Adding (45) and (46) we obtain another recurrence relation as

Jλ−1(z)− Jλ+1(z) = 2J ′λ(z) (47)

Example 3.2 Prove that

d

dz

[zλJλ(z)

]= zλJλ−1(z)

d

dz

[z−λJλ(z)

]= −z−λJλ+1(z)

Solution. Multiplying both sides of (45) by zλ−1 we obtain

zλJ ′λ(z) = zλJλ−1(z)− λzλ−1Jλ(z)

or zλJ ′λ(z) + λzλ−1Jλ(z) = z

λJλ−1(z)

ord

dz

[zλJλ(z)

]= zλJλ−1(z)

Multiplying both sides of (46) by z−λ−1 we obtain

z−λJ ′λ(z) = λz−λ−1Jλ(z)− z−λJλ+1(z)

or z−λJ ′λ(z)− λz−λ−1Jλ(z) = −z−λJλ+1(z)

ord

dz

[z−λJλ(z)

]= −z−λJλ+1(z)

Example 3.3 Prove that(i) J ′0(z) = −J1(z)(ii) J2(z) = J

′′0 (z)− 1zJ

′0(z)

Solution. The recurrence relation (46) is

zJ ′λ(z) = λJλ(z)− zJλ+1(z)

Putting λ = 0 we havezJ ′0(z) = −zJ1(z)

i.e., J ′0(z) = −J1(z)Differentiating w. r. t. z, we get

J ′1(z) = −J ′′0 (z) (48)

Putting λ = 0 in (46) we have

zJ ′1(z) = J1(z)− zJ2(z)or z[−J ′′0 (z)] = −J ′0(z)− zJ2(z) [Using (48)]

or J2(z) = J′′0 (z)−

1

zJ ′0(z)



3.2 Bessel Differential Equation from Recurrence Rela-tions

Suppose we consider a set of functions wλ(z) which satisfies the recurrencerelation (45), but wλ(z) not necessarily given by the series (38). The equation(45) can be written as

zwλ(z) = zwλ−1(z)− λwλ(z) (49)

Differentiating w. r. t. z we have

zw′′λ(z) + (λ+ 1)w′λ(z)− zw′λ−1(z)− wλ−1(z) = 0 (50)

Multiplying this equation by z and then subtracting (48) multiplied by λ, weget

z2w′′λ(z) + zw′λ(z)− λ2wλ(z) + (λ− 1)zwλ−1(z)− z2w′λ−1(z) = 0 (51)

Using the equation (46) and replacing λ by λ− 1 we have

zw′λ−1 = (λ− 1)wλ−1 − zwλ

Substituting the value of zw′λ−1(z) to the last term of (51) we get

z2w′′λ(z) + zw′λ(z)− λ2wλ(z) + (λ− 1)zwλ−1(z)− z{(λ− 1)wλ−1 − zwλ} = 0

On simplification this equation becomes

z2w′′λ(z) + zw′λ(z) + (z

2 − λ2)wλ(z) = 0

This is well known Bessel equation. Hence any functions, wλ(z) that satisfy therecurrence relations (45) and (46), satisfy Bessel differential equation, that isthe unknown function wλ(z) is Bessel function.

3.3 Representation of Bessel Functions as Continued Frac-tion

The Bessel functions can be expressed as continued fraction. The equation (44)can be written as

Jλ−1 + Jλ+1 =2λ

zJλ

That isJλ−1Jλ

=2λ

z− Jλ+1

Jλ=

2λ

z− 1

JλJλ+1

=2λ

z− 1

2(λ+1)z −

12(λ+2)

z −1

2(λ+3)z

−···

(52)



Multiplying (52) by z we get

zJλ−1Jλ

= 2λ− z2

2(λ+ 1)− z22(λ+2)− z2

2(λ+3)−···

(53)

In particular when λ = 0, 12 we have

zJ−1J0

= − z2

2− z24− z26−···

zJ− 12J 1

2

= 1− z2

3− z25− z27−···

(54)


J 12(z) =

√2

πzsin z

J− 12 (z) =

√2

πzcos z

z cot z = 1− z2

3− z25− z27−···

{J 12(z)}2 + {J− 12 (z)}

2 =2

πz

Solution. We have

J 12(z) =

∞∑0

(−1)k(z/2) 12+2k

Γ(k + 1)Γ( 12 + k + 1)

=

√2

z

∞∑0

(−1)k(z/2)1+2k

Γ(k + 1)Γ( 32 + k)

=

√2

z

1

Γ(1/2)

[ z212

−( z2 )

3

1!3212

+( z2 )

5

2! 523212

− · · ·]

=

√2

πz

[z − z

3

3!+

z5

5!− · · ·

]=

√2

zsin z (55)

Similarly,

J− 12 (z) =

√2

zcos z (56)

Dividing (58) by (55) we obtain

J− 12 (z)

J 12 (z)

= cot z (57)



Combination of the results (59) and (54) give

z cot z = 1− z2

3− z25− z27−···

Squaring and adding (55) and (58) we get

{J 12(z)}2 + {J− 12 (z)}

2 =2

πz

3.4 Integral representation

Recall the equations (41) and (42) as

cos(z sin θ) = J0(z) + 2

∞∑n=1

J2n(z) cos 2nθ

sin(z sin θ) = 2∞∑

n=0

J2n+1(z) sin(2n+ 1)θ

By employing the orthogonality properties of cosine and sine functions

π∫0

cosnθ cosmθdθ =π

2δmn

π∫0

sinnθ sinmθdθ =π

2δmn

in which m and n are non negative integers. Now,

Now,1

π

π∫0

cos(z sin θ) cosnθdθ

=1

π

π∫0

[J0(z) + 2

∞∑m=1

J2m(z) cos 2mθ

]cosnθdθ

=1

π

π∫0

J0(z) cosnθdθ +2

π

∞∑m=1

J2m(z)

π∫0

cos 2mθ cosnθdθ

=1

πJ0(z)

[sinnθ

n

]π0

+2

π

∞∑m=1

J2m(z)π

2δ2mn

=

{Jn(z), when n is even0, when n is odd

Since δ2mn = 1 when m =n

2, otherwise δ2mn = 0



Similarly,

1

π

π∫0

sin(z sin θ) sinnθdθ =

{0, when n is evenJn(z), when n is odd

Addition of these two relatives gives

Jn(z) =1

π

π∫0

[cos(z sin θ) cosnθ + sin(z sin θ) sinnθ] dθ

=1

π

π∫0

cos(nθ − z sin θ)dθ n = 0, 1, 2, . . .

As a special case,

J0(z) =1

π

π∫0

cos(z sin θ)dθ (58)

Note that cos(z sin θ) repeats itself in all four quadrants (θ1 = θ, π−θ, π+θ,−θ).The equation (58) can be written as

J0(z) =1

2π

2π∫0

cos(z sin θ)dθ (59)

On the other hand, sin(z sin θ) reverses its sign in the third and fourth quadrantsso that

1

2π

2π∫0

cos(z sin θ)dθ = 0 (60)

Adding (59) and i times (60), we obtain

J0(z) =1

2π

2π∫0

eiz sin θdθ =1

2π

2π∫0

eiz cos θdθ (61)

The Bessel function can be expressed in terms of contour integration as follows.

Theorem 3 For a suitable contour C enclosing t = 0, the Bessel function oforder λ is given by

Jλ =1

2πi

∫C

ez2 (t−

1t )tλ+1dt



Proof: The Laurent series for an analytic function f(t) is

f(t) =

∞∑n=−∞

antn where an =

1

2πi

∫C

f(t)

tn+1dt (62)

Let f(t) = ez2 (t−

1t ). Again

ez2 (t−

1t ) =

∞∑n=−∞

tλJλ(z) (63)

For this f(t), (62) becomes

ez2 (t−

1t ) =

∞∑n=−∞

aλtnwhere aλ =

1

2πi

∫C

f(t)

tλ+1dt (64)

Comparing (63) and (64) we have

Jλ(z) = aλ =1

2πi

∫C

ez2 (t−

1t )

tλ+1dt

The contour C is shown in FigureThe contour C consists of1. −∞ to −1 along the real axis,2. the unit circle in the anticlockwise sense3. the real axis from −1 to −∞ in the opposite direction

When λ is not an integerIn this case, t−(λ+1) has branch point at t = 0 and ±∞ and a cut is necessary

along the negative real axis from 0 to −∞ and thus we get

Jλ(z) = aλ =1

2πi

∫C1

ez2 (t−

1t )

tλ+1dt

When λ is not an integerThe contour C becomes a unit circle. Since the two integrals along the

negative real axis cancel each other and finally

Jλ(z) = aλ =1

2πi

∫C0

ez2 (t−

1t )

tλ+1dt

where C0 is a unit circle containing t = 0.

Example 3.5 For an integral value of n, show that

Jn(z) =1

2π

∫ 2π0

ei(z sin θ−nθ)dθ



Solution. We have

Jn(z) =1

2πi

∫C

ez2 (t−

1t )

tn+1dt

where C is the contour represents a unit circle with centre at the origin.Let t = cos θ + i sin θ = eiθ

Then t− 1t = 2i sin θ and θ varies from 0 to 2π.Therefore,

Jn(z) =1

2πi

∫ 2π0

ez2 (2i sin θ)

(eiθ)n+1.ieiθdθ =

1

2π

1

2π

∫ 2π0

ei(z sin θ−nθ)dθ

EXERCISES

(i) Express the following expressions in terms of Legendre polynomials

(a) 3x− 2x2 + x3

(b) x4 − 8x2 − 1

(c) 1 + x+ x2

(d) sinx

(ii) Expand the following function in terms of Legendre polynomials

f(x) =

{0, if − 1 ≤ x < 01, if 0 < x ≤ 1 .

(iii) Show that

1+3P1(z)+5P2(z)+7P3(z)+ · · ·+(2n+1)Pn(z) =d

dz[Pn+1(z) + Pn(z)]

(iv) Prove the following recurrence relations

(a) P ′n+1(z)− zP ′n(z) = (n+ 1)Pn(z)

(b) (n+ 1)Pn+1(z)− (2n+ 1)zPn(z) + nPn−1(z) = 0

(c) zP ′n(z)− P ′n−1(z) = nPn(z)

(d) P ′n+1(z)− zP ′n−1(z) = (2n+ 1)Pn(z)

(e) (1− z2)P ′n(z) = nPn−1(z)− nzPn(z)

(v) Show that (1− 2zt+ t2)− 12 is the generating function of Legendre polyno-mials.



(vi) Show that Pn(z) is the coefficient of tn in the expansion of (1−2zt+t2)− 12 .

(vii) Show that the Legendre differential equation can be written in the form

1

sin θ

d

dθ

[sin θ

dPn(θ)

dθ

]+ n(n+ 1)Pn(cos θ) = 0

(viii) Show that

π∫0

Pn(cos θ)Pn(cos θ) sin θdθ =

{0, if m ̸= n

2

2n+ 1, if m = n

(ix) Show that

1 + t

t√1− 2tz + t2

− 1t=

∞∑n=0

[Pn(z) + Pn+1(z)] tn

(x) Prove that

1− t2

{1− 2tz + t2} 32=

∞∑n=0

(2n+ 1)Pn(z)tn

(xi) Show that

1 +

∞∑n=0

1

n+ 1Pn(cos θ) = log

[1 + sin θ/2

sin θ/2

].

(xii) Show that

Pn(z) =dn

dzn

[(1− z)n

{1− 1− z

2

}n](xiii) Show that

1∫0

P2n(z)P2n+1(z)dz =

1∫0

P2n(z)P2n−1(z)dz

(xiv) Prove that

(a)

1∫−1

[Pn(z)]2dz =

2

2n+ 1

(b)

1∫−1

[P ′n(z)]2dz = n(n+ 1)



(c)

1∫−1

zPn(z)P′n(z)dz =

2n

2n− 1

(d)

1∫−1

zPn(z)Pn−1(z)dz =2n

4n2 − 1

(xv) Show that

(a)

1∫−1

z2P 2n(z)dz =1

8

[1

2n− 1+

6

2n+ 1+

1

2n+ 3

]

(b)

1∫−1

(z2 − 1)Pn+1(z)P ′n(z)dz =2n(n+ 1)

(2n+ 1)(2n+ 3)

(xvi) Prove that

Pn(cos θ) =∞∑k=0

(−1)k(nCk)2 cos2n−2 θ sin2k θ

(xvii) Show thatπ∫

0

P2n(cos θ)dθ =

[Γ(n+ 12 )

Γ(n+ 1)

]2

(xviii) Prove that

1∫−1

z(1−z2)P ′n(z)P ′m(z)dz =

0, unless m = n± 12n(n2−1)4n2−1 δm,n−1, if m < n

2n(n+2)(n+1)(2n+1)(2n+3) δm,n+1, if m > n

(xix) Operating in spherical polar co-ordinates, show that

∂

∂z

[Pn(cos θ)

rn+1

]= −(n+ 1)Pn+1(cos θ)

rn+2

(xx) Prove that

P ′n(1) =1

2n(n+ 1)

(xxi) Prove that, for n ≥ 1,

1∫−1

log(1− z)Pn(z)dz = −2

n(n+ 1)



(xxii) Prove that

1∫−1

Pn(z)dz = −1

(2n+ 1)[Pn−1(z)− Pn+1(z)]

(xxiii) Prove that(a) Pn(0) = 0 for odd n and

(b) Pn(0) =(−1)

n2 n!

2n{(n/2)!}2 for even n.

(xxiv) Prove that

(a)

z∫0

znJn−1(z)dz = znJn(z)

(b)

z∫0

zn+1Jn(z)dz = zn+1Jn+1(z)

(c)

∞∫0

Jn(z)

zdz =

1

n

(d)

z∫0

J0(z)dz = 1

(xxv) Prove that (a) 4

∫Jn+1(z)dz =

∫Jn−1(z)dz − 2Jn(z)

(b) 4d2

dz2[Jn(z)] = Jn−2(z)− 2Jn(z) + Jn+2(z)

(c)

∞∫0

e−zJ0(z)dz =1√2

(xxvi) Show that by the use of recurrence formulae that (a) J ′′0 =J2 − J0

2

(b)J2(z)

J1(z)=

1

z− J

′′0 (z)

J ′0(z)

(c) J ′′′0 (z) + 3J′0(z) + J3(z) = 0


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