Legendre Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

Legendre’s Function

N. B. Vyas

Department of MathematicsAtmiya Institute of Technology and Science

Department of Mathematics

N. B. Vyas Legendre’s Function

The differential equation

(1− x2)y′′− 2xy′+ n(n+ 1)y = 0is called Legendre’s differential equation,n is real constant

(1− x2)y′′− 2xy′+ n(n+ 1)y = 0

is called Legendre’s differential equation,n is real constant

(1− x2)y′′− 2xy′+ n(n+ 1)y = 0is called Legendre’s differential equation,n is real constant

Legendre’s Polynomials:

⇒ P0(x) = 1

⇒ P1(x) = x

⇒ P2(x) =1

2(3x2 − 1)

⇒ P3(x) =1

2(5x3 − 3x)

⇒ P4(x) =1

8(35x3 − 30x2 + 3)

⇒ P5(x) =1

8(63x5 − 70x3 + 15x)

⇒ P0(x) = 1

⇒ P1(x) = x

⇒ P2(x) =1

2(3x2 − 1)

⇒ P3(x) =1

2(5x3 − 3x)

⇒ P4(x) =1

8(35x3 − 30x2 + 3)

⇒ P5(x) =1

8(63x5 − 70x3 + 15x)

⇒ P0(x) = 1

⇒ P1(x) = x

⇒ P2(x) =1

2(3x2 − 1)

⇒ P3(x) =1

2(5x3 − 3x)

⇒ P4(x) =1

8(35x3 − 30x2 + 3)

⇒ P5(x) =1

8(63x5 − 70x3 + 15x)

⇒ P0(x) = 1

⇒ P1(x) = x

⇒ P2(x) =1

2(3x2 − 1)

⇒ P3(x) =1

2(5x3 − 3x)

⇒ P4(x) =1

8(35x3 − 30x2 + 3)

⇒ P5(x) =1

8(63x5 − 70x3 + 15x)

⇒ P0(x) = 1

⇒ P1(x) = x

⇒ P2(x) =1

2(3x2 − 1)

⇒ P3(x) =1

2(5x3 − 3x)

⇒ P4(x) =1

8(35x3 − 30x2 + 3)

⇒ P5(x) =1

8(63x5 − 70x3 + 15x)

⇒ P0(x) = 1

⇒ P1(x) = x

⇒ P2(x) =1

2(3x2 − 1)

⇒ P3(x) =1

2(5x3 − 3x)

⇒ P4(x) =1

8(35x3 − 30x2 + 3)

⇒ P5(x) =1

8(63x5 − 70x3 + 15x)

Ex.1 Express f (x) in terms ofLegendre’s polynomials wheref (x) = x3 + 2x2 − x− 3.

Solution:

⇒ P0(x) = 1∴ 1 = P0(x)

⇒ P1(x) = x∴ x = P1(x)

⇒ P3(x) =1

2(5x3 − 3x)

∴ 2P3(x) = (5x3 − 3x)

∴ 2P3(x) + 3x = 5x3

∴ 2P3(x) + 3P1(x) = 5x3 {∵ x = P1(x)}

∴ x3 =2

5P3(x) +

5P1(x)

Solution:

⇒ P0(x) = 1∴ 1 = P0(x)

⇒ P1(x) = x∴ x = P1(x)

⇒ P3(x) =1

2(5x3 − 3x)

∴ 2P3(x) = (5x3 − 3x)

∴ 2P3(x) + 3x = 5x3

∴ 2P3(x) + 3P1(x) = 5x3 {∵ x = P1(x)}

∴ x3 =2

5P3(x) +

5P1(x)

Solution:

⇒ P0(x) = 1∴ 1 = P0(x)

⇒ P1(x) = x∴ x = P1(x)

⇒ P3(x) =1

2(5x3 − 3x)

∴ 2P3(x) = (5x3 − 3x)

∴ 2P3(x) + 3x = 5x3

∴ 2P3(x) + 3P1(x) = 5x3 {∵ x = P1(x)}

∴ x3 =2

5P3(x) +

5P1(x)

Solution:

⇒ P0(x) = 1∴ 1 = P0(x)

⇒ P1(x) = x∴ x = P1(x)

⇒ P3(x) =1

2(5x3 − 3x)

∴ 2P3(x) = (5x3 − 3x)

∴ 2P3(x) + 3x = 5x3

∴ 2P3(x) + 3P1(x) = 5x3 {∵ x = P1(x)}

∴ x3 =2

5P3(x) +

5P1(x)

Solution:

⇒ P0(x) = 1∴ 1 = P0(x)

⇒ P1(x) = x∴ x = P1(x)

⇒ P3(x) =1

2(5x3 − 3x)

∴ 2P3(x) = (5x3 − 3x)

∴ 2P3(x) + 3x = 5x3

∴ 2P3(x) + 3P1(x) = 5x3 {∵ x = P1(x)}

∴ x3 =2

5P3(x) +

5P1(x)

Solution:

⇒ P0(x) = 1∴ 1 = P0(x)

⇒ P1(x) = x∴ x = P1(x)

⇒ P3(x) =1

2(5x3 − 3x)

∴ 2P3(x) = (5x3 − 3x)

∴ 2P3(x) + 3x = 5x3

∴ 2P3(x) + 3P1(x) = 5x3 {∵ x = P1(x)}

∴ x3 =2

5P3(x) +

5P1(x)

Solution:

⇒ P0(x) = 1∴ 1 = P0(x)

⇒ P1(x) = x∴ x = P1(x)

⇒ P3(x) =1

2(5x3 − 3x)

∴ 2P3(x) = (5x3 − 3x)

∴ 2P3(x) + 3x = 5x3

∴ 2P3(x) + 3P1(x) = 5x3 {∵ x = P1(x)}

∴ x3 =2

5P3(x) +

5P1(x)

Solution:

⇒ P0(x) = 1∴ 1 = P0(x)

⇒ P1(x) = x∴ x = P1(x)

⇒ P3(x) =1

2(5x3 − 3x)

∴ 2P3(x) = (5x3 − 3x)

∴ 2P3(x) + 3x = 5x3

∴ 2P3(x) + 3P1(x) = 5x3 {∵ x = P1(x)}

∴ x3 =2

5P3(x) +

5P1(x)

⇒ P2(x) =1

2(3x2 − 1)

∴ 2P2(x) = (3x2 − 1)

∴ 2P2(x) + 1 = 3x2

∴ 2P2(x) + P0(x) = 3x2 {∵ 1 = P0(x)}

∴ x2 =2

3P2(x) +

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

5P3(x) +

5P1(x) +

3P2(x) +

3P0(x)− P1(x)− 3P0(x)

⇒ P2(x) =1

2(3x2 − 1)

∴ 2P2(x) = (3x2 − 1)

∴ 2P2(x) + 1 = 3x2

∴ 2P2(x) + P0(x) = 3x2 {∵ 1 = P0(x)}

∴ x2 =2

3P2(x) +

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

5P3(x) +

5P1(x) +

3P2(x) +

3P0(x)− P1(x)− 3P0(x)

⇒ P2(x) =1

2(3x2 − 1)

∴ 2P2(x) = (3x2 − 1)

∴ 2P2(x) + 1 = 3x2

∴ 2P2(x) + P0(x) = 3x2 {∵ 1 = P0(x)}

∴ x2 =2

3P2(x) +

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

5P3(x) +

5P1(x) +

3P2(x) +

3P0(x)− P1(x)− 3P0(x)

⇒ P2(x) =1

2(3x2 − 1)

∴ 2P2(x) = (3x2 − 1)

∴ 2P2(x) + 1 = 3x2

∴ 2P2(x) + P0(x) = 3x2 {∵ 1 = P0(x)}

∴ x2 =2

3P2(x) +

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

5P3(x) +

5P1(x) +

3P2(x) +

3P0(x)− P1(x)− 3P0(x)

⇒ P2(x) =1

2(3x2 − 1)

∴ 2P2(x) = (3x2 − 1)

∴ 2P2(x) + 1 = 3x2

∴ 2P2(x) + P0(x) = 3x2 {∵ 1 = P0(x)}

∴ x2 =2

3P2(x) +

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

5P3(x) +

5P1(x) +

3P2(x) +

3P0(x)− P1(x)− 3P0(x)

⇒ P2(x) =1

2(3x2 − 1)

∴ 2P2(x) = (3x2 − 1)

∴ 2P2(x) + 1 = 3x2

∴ 2P2(x) + P0(x) = 3x2 {∵ 1 = P0(x)}

∴ x2 =2

3P2(x) +

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

5P3(x) +

5P1(x) +

3P2(x) +

3P0(x)− P1(x)− 3P0(x)

⇒ P2(x) =1

2(3x2 − 1)

∴ 2P2(x) = (3x2 − 1)

∴ 2P2(x) + 1 = 3x2

∴ 2P2(x) + P0(x) = 3x2 {∵ 1 = P0(x)}

∴ x2 =2

3P2(x) +

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

5P3(x) +

5P1(x) +

3P2(x) +

3P0(x)− P1(x)− 3P0(x)

Ex.2 Express x3 − 5x2 + 6x + 1 interms of Legendre’s polynomial.

Ex.3 Express 4x3 − 2x2 − 3x + 8 interms of Legendre’s polynomial.

Generating Function for Pn(x)∞∑n=0

Pn(x)tn =

1√1− 2xt + t2

= (1− 2xt + t2)−12

The function (1− 2xt + t2)−12 is

called Generating function ofLegendre’s polynomial Pn(x)

Ex Show that

(i)Pn(1) = 1

(ii)Pn(−1) = (−1)n

(iii)Pn(−x) = (−1)nPn(x)

Ex Show that

(i)Pn(1) = 1

(ii)Pn(−1) = (−1)n

Ex Show that

(i)Pn(1) = 1

(ii)Pn(−1) = (−1)n

Ex Show that

(i)Pn(1) = 1

(ii)Pn(−1) = (−1)n

Solution:

(i) We have

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−12

Putting x = 1 in eq(1), we get

∞∑n=0

Pn(1)tn = (1− 2t + t2)−12 = (1− t)−1

∴∞∑n=0

Pn(1)tn =1

1− t= 1 + t + t2 + t3 + ...

∴∞∑n=0

Pn(1)tn =∞∑n=0

Comparing the coefficient of tn both the sides, we get

Pn(1) = 1

Solution:

(i) We have

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−12

Putting x = 1 in eq(1), we get∞∑n=0

Pn(1)tn = (1− 2t + t2)−12 = (1− t)−1

∴∞∑n=0

Pn(1)tn =1

1− t= 1 + t + t2 + t3 + ...

∴∞∑n=0

Pn(1)tn =∞∑n=0

Pn(1) = 1

Solution:

(i) We have

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−12

Pn(1)tn = (1− 2t + t2)−12 = (1− t)−1

∴∞∑n=0

Pn(1)tn =1

1− t= 1 + t + t2 + t3 + ...

∴∞∑n=0

Pn(1)tn =∞∑n=0

Pn(1) = 1

Solution:

(i) We have

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−12

Pn(1)tn = (1− 2t + t2)−12 = (1− t)−1

∴∞∑n=0

Pn(1)tn =1

1− t= 1 + t + t2 + t3 + ...

∴∞∑n=0

Pn(1)tn =

∞∑n=0

Pn(1) = 1

Solution:

(i) We have

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−12

Pn(1)tn = (1− 2t + t2)−12 = (1− t)−1

∴∞∑n=0

Pn(1)tn =1

1− t= 1 + t + t2 + t3 + ...

∴∞∑n=0

Pn(1)tn =

∞∑n=0

Pn(1) = 1

Solution:

(i) We have

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−12

Pn(1)tn = (1− 2t + t2)−12 = (1− t)−1

∴∞∑n=0

Pn(1)tn =1

1− t= 1 + t + t2 + t3 + ...

∴∞∑n=0

Pn(1)tn =

∞∑n=0

Pn(1) = 1

(ii) Putting x = −1 in eq(1), we get

∞∑n=0

Pn(−1)tn = (1 + 2t + t2)−12 = (1 + t)−1

∴∞∑n=0

Pn(−1)tn =1

1 + t= 1− t + t2 − t3 + ...

∴∞∑n=0

Pn(−1)tn =∞∑n=0

(−1)ntn

Comparing coefficients of tn, we get

Pn(−1) = (−1)n

(ii) Putting x = −1 in eq(1), we get∞∑n=0

Pn(−1)tn = (1 + 2t + t2)−12 = (1 + t)−1

∴∞∑n=0

Pn(−1)tn =1

1 + t= 1− t + t2 − t3 + ...

∴∞∑n=0

Pn(−1)tn =∞∑n=0

(−1)ntn

Pn(−1) = (−1)n

Pn(−1)tn = (1 + 2t + t2)−12 = (1 + t)−1

∴∞∑n=0

Pn(−1)tn =1

1 + t= 1− t + t2 − t3 + ...

∴∞∑n=0

Pn(−1)tn =∞∑n=0

(−1)ntn

Pn(−1) = (−1)n

Pn(−1)tn = (1 + 2t + t2)−12 = (1 + t)−1

∴∞∑n=0

Pn(−1)tn =1

1 + t= 1− t + t2 − t3 + ...

∴∞∑n=0

Pn(−1)tn =

∞∑n=0

(−1)ntn

Pn(−1) = (−1)n

Pn(−1)tn = (1 + 2t + t2)−12 = (1 + t)−1

∴∞∑n=0

Pn(−1)tn =1

1 + t= 1− t + t2 − t3 + ...

∴∞∑n=0

Pn(−1)tn =

∞∑n=0

(−1)ntn

Pn(−1) = (−1)n

Pn(−1)tn = (1 + 2t + t2)−12 = (1 + t)−1

∴∞∑n=0

Pn(−1)tn =1

1 + t= 1− t + t2 − t3 + ...

∴∞∑n=0

Pn(−1)tn =

∞∑n=0

(−1)ntn

Pn(−1) = (−1)n

(iii) Now replacing x by −x in eq(1), we get

∞∑n=0

Pn(−x)tn = (1 + 2xt + t2)−12 —(a)

Now, replacing t by −t in eq(1), we get∞∑n=0

Pn(x)(−t)n = (1 + 2xt + t2)−12 —(b)

from equation (a) and (b)∞∑n=0

Pn(−x)(t)n =

∞∑n=0

Pn(x)(−1)n(t)n

Comparing the coefficients of tn , both sides, we get

Pn(−x) = (−1)nPn(x)

(iii) Now replacing x by −x in eq(1), we get∞∑n=0

Pn(−x)tn = (1 + 2xt + t2)−12 —(a)

Pn(x)(−t)n = (1 + 2xt + t2)−12 —(b)

Pn(−x)(t)n =

∞∑n=0

Pn(x)(−1)n(t)n

Pn(−x) = (−1)nPn(x)

Pn(−x)tn = (1 + 2xt + t2)−12 —(a)

Now, replacing t by −t in eq(1), we get

∞∑n=0

Pn(x)(−t)n = (1 + 2xt + t2)−12 —(b)

Pn(−x)(t)n =

∞∑n=0

Pn(x)(−1)n(t)n

Pn(−x) = (−1)nPn(x)

Pn(−x)tn = (1 + 2xt + t2)−12 —(a)

Pn(x)(−t)n = (1 + 2xt + t2)−12 —(b)

Pn(−x)(t)n =

∞∑n=0

Pn(x)(−1)n(t)n

Pn(−x) = (−1)nPn(x)

Pn(−x)tn = (1 + 2xt + t2)−12 —(a)

Pn(x)(−t)n = (1 + 2xt + t2)−12 —(b)

from equation (a) and (b)

∞∑n=0

Pn(−x)(t)n =

∞∑n=0

Pn(x)(−1)n(t)n

Pn(−x) = (−1)nPn(x)

Pn(−x)tn = (1 + 2xt + t2)−12 —(a)

Pn(x)(−t)n = (1 + 2xt + t2)−12 —(b)

Pn(−x)(t)n =

∞∑n=0

Pn(x)(−1)n(t)n

Pn(−x) = (−1)nPn(x)

Pn(−x)tn = (1 + 2xt + t2)−12 —(a)

Pn(x)(−t)n = (1 + 2xt + t2)−12 —(b)

Pn(−x)(t)n =

∞∑n=0

Pn(x)(−1)n(t)n

Pn(−x) = (−1)nPn(x)

Pn(−x)tn = (1 + 2xt + t2)−12 —(a)

Pn(x)(−t)n = (1 + 2xt + t2)−12 —(b)

Pn(−x)(t)n =

∞∑n=0

Pn(x)(−1)n(t)n

Pn(−x) = (−1)nPn(x)

Rodrigue’s Formula

Pn(x) =1

dxn[(x2 − 1)n]

Proof:

Let y = (x2 − 1)n

Differentiating wit respect to x

∴ y1 = n(x2 − 1)n−1(2x)

∴ y1 =2nx(x2 − 1)n

(x2 − 1)=

x2 − 1

∴ (x2 − 1)y1 = 2nxy

Differentiating with respect to x,

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

Proof:

Let y = (x2 − 1)n

∴ y1 = n(x2 − 1)n−1(2x)

∴ y1 =2nx(x2 − 1)n

(x2 − 1)=

x2 − 1

∴ (x2 − 1)y1 = 2nxy

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

Proof:

Let y = (x2 − 1)n

∴ y1 = n(x2 − 1)n−1(2x)

∴ y1 =2nx(x2 − 1)n

(x2 − 1)=

x2 − 1

∴ (x2 − 1)y1 = 2nxy

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

Proof:

Let y = (x2 − 1)n

∴ y1 = n(x2 − 1)n−1(2x)

∴ y1 =2nx(x2 − 1)n

(x2 − 1)=

x2 − 1

∴ (x2 − 1)y1 = 2nxy

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

Proof:

Let y = (x2 − 1)n

∴ y1 = n(x2 − 1)n−1(2x)

∴ y1 =2nx(x2 − 1)n

(x2 − 1)=

x2 − 1

∴ (x2 − 1)y1 = 2nxy

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

Proof:

Let y = (x2 − 1)n

∴ y1 = n(x2 − 1)n−1(2x)

∴ y1 =2nx(x2 − 1)n

(x2 − 1)=

x2 − 1

∴ (x2 − 1)y1 = 2nxy

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

Proof:

Let y = (x2 − 1)n

∴ y1 = n(x2 − 1)n−1(2x)

∴ y1 =2nx(x2 − 1)n

(x2 − 1)=

x2 − 1

∴ (x2 − 1)y1 = 2nxy

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

dxn(UV ) = nC0UVn + nC1U1Vn−1 + ... + nCnUnV

→ dn

dxn((x2−1)y2) = nC0(x

2−1)yn+2+nC1(2x)yn+1+nC2(2)yn

→ dn

dxn(2xy1) = nC0(2x)yn+1 + nC1(2)yn

→ dn

dxn(2nxy1) = nC0(2nx)yn+1 + nC1(2n)yn

→ dn

dxn(2ny) = 2nyn

Also nC0 = 1, nC1 = n, nC2 =n(n− 1)

}→ dn

dxn((x2−1)y2) = nC0(x

2−1)yn+2+nC1(2x)yn+1+nC2(2)yn

→ dn

dxn(2ny) = 2nyn

Also nC0 = 1, nC1 = n, nC2 =n(n− 1)

}→ dn

dxn((x2−1)y2) = nC0(x

2−1)yn+2+nC1(2x)yn+1+nC2(2)yn

→ dn

dxn(2ny) = 2nyn

Also nC0 = 1, nC1 = n, nC2 =n(n− 1)

}→ dn

dxn((x2−1)y2) = nC0(x

2−1)yn+2+nC1(2x)yn+1+nC2(2)yn

→ dn

dxn(2ny) = 2nyn

Also nC0 = 1, nC1 = n, nC2 =n(n− 1)

}→ dn

dxn((x2−1)y2) = nC0(x

2−1)yn+2+nC1(2x)yn+1+nC2(2)yn

→ dn

dxn(2ny) = 2nyn

Also nC0 = 1, nC1 = n, nC2 =n(n− 1)

}→ dn

dxn((x2−1)y2) = nC0(x

2−1)yn+2+nC1(2x)yn+1+nC2(2)yn

→ dn

dxn(2ny) = 2nyn

Also nC0 = 1, nC1 = n, nC2 =n(n− 1)

∴ (x2 − 1)yn+2 + 2nxyn+1 + n(n− 1)yn+2xyn+1 + 2nyn = 2nxyn+1 + n(2n)yn + 2nyn

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

Equation (2) is a Legendre’s equation in variables v and x

⇒ Pn(x) is a solution of equation (2)

Also, v = f(x) is a solution of equation (2)

Pn = cv where c is constant

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

∴ (x2 − 1)yn+2 + 2xyn+1 + (n2 − n + 2n− 2n2 − 2n)yn = 0

∴ (x2 − 1)yn+2 + 2xyn+1 − n(n + 1)yn = 0

∴ (1− x2)yn+2 − 2xyn+1 + n(n + 1)yn = 0

Let v = yn =dny

∴ (1− x2)v2 − 2xv1 + n(n + 1)v = 0 ——(2)

∴ Pn(x) = cdny

dxn= c

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+dn((x + 1)n)

dxn(x− 1)n

∴ Pn(x) = cdny

dxn= c

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+dn((x + 1)n)

dxn(x− 1)n

∴ Pn(x) = cdny

dxn= c

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+dn((x + 1)n)

dxn(x− 1)n

∴ Pn(x) = cdny

dxn= c

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+dn((x + 1)n)

dxn(x− 1)n

∴ Pn(x) = cdny

dxn= c

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+dn((x + 1)n)

dxn(x− 1)n

∴ Pn(x) = cdny

dxn= c

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+dn((x + 1)n)

dxn(x− 1)n

∴ Pn(x) = cdny

dxn= c

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+dn((x + 1)n)

dxn(x− 1)n

Recurrence Relations for Pn(x) : −

(1) (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

Proof: We have∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

Differentiating equation (i) partially with respect to t, weget∞∑n=1

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

= (1− 2xt + t2)−1(1− 2xt + t2)−12 (x− t)

=(1− 2xt + t2)−

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

(1) (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

Proof: We have

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

= (1− 2xt + t2)−1(1− 2xt + t2)−12 (x− t)

=(1− 2xt + t2)−

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

∞∑n=0

Pn(x)tn

(1) (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

= (1− 2xt + t2)−1(1− 2xt + t2)−12 (x− t)

=(1− 2xt + t2)−

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

∞∑n=0

Pn(x)tn

(1) (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

Differentiating equation (i) partially with respect to t, weget

∞∑n=1

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

= (1− 2xt + t2)−1(1− 2xt + t2)−12 (x− t)

=(1− 2xt + t2)−

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

∞∑n=0

Pn(x)tn

(1) (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

= (1− 2xt + t2)−1(1− 2xt + t2)−12 (x− t)

=(1− 2xt + t2)−

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

∞∑n=0

Pn(x)tn

(1) (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

= (1− 2xt + t2)−1(1− 2xt + t2)−12 (x− t)

=(1− 2xt + t2)−

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

∞∑n=0

Pn(x)tn

(1) (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

= (1− 2xt + t2)−1(1− 2xt + t2)−12 (x− t)

=(1− 2xt + t2)−

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

∞∑n=0

Pn(x)tn

(1) (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

= (1− 2xt + t2)−1(1− 2xt + t2)−12 (x− t)

=(1− 2xt + t2)−

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

∞∑n=0

Pn(x)tn

(1) (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

= (1− 2xt + t2)−1(1− 2xt + t2)−12 (x− t)

=(1− 2xt + t2)−

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

∞∑n=0

Pn(x)tn

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

∞∑n=0

Pn(x)tn −∞∑n=0

Pn(x)tn+1

replacing n by n+1 in 1st term, n by n-1 in 3rd term inL.H.S.

replacing n by n-1 in 2nd term in R.H.S∞∑n=0

(n + 1)Pn+1(x)tn − 2x∞∑n=1

nPn(x)tn +∞∑n=2

1)Pn−1(x)tn = x

∞∑n=0

Pn−1(x)tn

comparing the coefficients of tn on both the sides

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

∞∑n=0

Pn(x)tn+1

(n + 1)Pn+1(x)tn − 2x∞∑n=1

nPn(x)tn +∞∑n=2

1)Pn−1(x)tn = x

∞∑n=0

Pn−1(x)tn

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

∞∑n=0

Pn(x)tn+1

replacing n by n-1 in 2nd term in R.H.S

∞∑n=0

(n + 1)Pn+1(x)tn − 2x∞∑n=1

nPn(x)tn +∞∑n=2

1)Pn−1(x)tn = x

∞∑n=0

Pn−1(x)tn

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

∞∑n=0

Pn(x)tn+1

(n + 1)Pn+1(x)tn − 2x

∞∑n=1

nPn(x)tn +

∞∑n=2

1)Pn−1(x)tn = x

∞∑n=0

Pn−1(x)tn

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

∞∑n=0

Pn(x)tn+1

(n + 1)Pn+1(x)tn − 2x

∞∑n=1

nPn(x)tn +

∞∑n=2

1)Pn−1(x)tn = x

∞∑n=0

Pn−1(x)tn

∴ (n + 1)Pn+1(x)− 2xnPn(x) + (n− 1)Pn−1(x) =xPn(x)− Pn−1(x)

∴ (n + 1)Pn+1(x) = (2n + 1)xPn(x)− (n− 1 + 1)Pn−1(x)

∴ (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (n + 1)Pn+1(x) = (2n + 1)xPn(x)− (n− 1 + 1)Pn−1(x)

∴ (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (n + 1)Pn+1(x) = (2n + 1)xPn(x)− (n− 1 + 1)Pn−1(x)

∴ (n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

(2) nPn(x) = xP ′n(x)− P ′n−1(x)

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

Differentiating equation (i) partially with respect to x, weget∞∑n=0

P ′n(x)tn = −1

2(1− 2xt + t2)−

32 (−2t)

∴∞∑n=0

P ′n(x)tn =t(1− 2xt + t2)−

(1− 2xt + t2)————-(ii)

⇒ Differentiating equation (i) partially with respect to t, weget∞∑n=1

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

∞∑n=1

nPn(x)tn−1 =(x− t)(1− 2xt + t2)−

(1− 2xt + t2)

(2) nPn(x) = xP ′n(x)− P ′n−1(x)Proof: We have

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

P ′n(x)tn = −1

2(1− 2xt + t2)−

32 (−2t)

∴∞∑n=0

P ′n(x)tn =t(1− 2xt + t2)−

(1− 2xt + t2)————-(ii)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

∞∑n=1

nPn(x)tn−1 =(x− t)(1− 2xt + t2)−

(1− 2xt + t2)

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

P ′n(x)tn = −1

2(1− 2xt + t2)−

32 (−2t)

∴∞∑n=0

P ′n(x)tn =t(1− 2xt + t2)−

(1− 2xt + t2)————-(ii)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

∞∑n=1

nPn(x)tn−1 =(x− t)(1− 2xt + t2)−

(1− 2xt + t2)

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

Differentiating equation (i) partially with respect to x, weget

∞∑n=0

P ′n(x)tn = −1

2(1− 2xt + t2)−

32 (−2t)

∴∞∑n=0

P ′n(x)tn =t(1− 2xt + t2)−

(1− 2xt + t2)————-(ii)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

∞∑n=1

nPn(x)tn−1 =(x− t)(1− 2xt + t2)−

(1− 2xt + t2)

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

P ′n(x)tn = −1

2(1− 2xt + t2)−

32 (−2t)

∴∞∑n=0

P ′n(x)tn =t(1− 2xt + t2)−

(1− 2xt + t2)————-(ii)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

∞∑n=1

nPn(x)tn−1 =(x− t)(1− 2xt + t2)−

(1− 2xt + t2)

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

P ′n(x)tn = −1

2(1− 2xt + t2)−

32 (−2t)

∴∞∑n=0

P ′n(x)tn =t(1− 2xt + t2)−

(1− 2xt + t2)————-(ii)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

∞∑n=1

nPn(x)tn−1 =(x− t)(1− 2xt + t2)−

(1− 2xt + t2)

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

P ′n(x)tn = −1

2(1− 2xt + t2)−

32 (−2t)

∴∞∑n=0

P ′n(x)tn =t(1− 2xt + t2)−

(1− 2xt + t2)————-(ii)

⇒ Differentiating equation (i) partially with respect to t, weget

∞∑n=1

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

∞∑n=1

nPn(x)tn−1 =(x− t)(1− 2xt + t2)−

(1− 2xt + t2)

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

P ′n(x)tn = −1

2(1− 2xt + t2)−

32 (−2t)

∴∞∑n=0

P ′n(x)tn =t(1− 2xt + t2)−

(1− 2xt + t2)————-(ii)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

∞∑n=1

nPn(x)tn−1 =(x− t)(1− 2xt + t2)−

(1− 2xt + t2)

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

2 ——–(i)

P ′n(x)tn = −1

2(1− 2xt + t2)−

32 (−2t)

∴∞∑n=0

P ′n(x)tn =t(1− 2xt + t2)−

(1− 2xt + t2)————-(ii)

nPn(x)tn−1 = −1

2(1− 2xt + t2)−

32 (−2x + 2t)

∞∑n=1

nPn(x)tn−1 =(x− t)(1− 2xt + t2)−

(1− 2xt + t2)N. B. Vyas Legendre’s Function

∞∑n=1

nPn(x)tn−1 =(x− t)

∞∑n=0

P ′n(x)tn {by eq. (ii)

∴ t∞∑n=1

∞∑n=0

P ′n(x)tn

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=0

P ′n(x)tn+1

Replacing n by n-1 in 2nd term in R.H.S.

∴∞∑n=1

nPn(x)tn = x∞∑n=0

P ′n(x)tn −∞∑n=1

P ′n−1(x)tn

comparing the coefficients of tn on both sides, we get

nPn(x) = xP ′n(x)− P ′n−1(x)

∞∑n=1

∞∑n=0

∞∑n=1

∞∑n=0

P ′n(x)tn

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=0

P ′n(x)tn+1

∴∞∑n=1

P ′n(x)tn −∞∑n=1

P ′n−1(x)tn

nPn(x) = xP ′n(x)− P ′n−1(x)

∞∑n=1

∞∑n=0

∞∑n=1

∞∑n=0

P ′n(x)tn

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=0

P ′n(x)tn+1

∴∞∑n=1

P ′n(x)tn −∞∑n=1

P ′n−1(x)tn

nPn(x) = xP ′n(x)− P ′n−1(x)

∞∑n=1

∞∑n=0

∞∑n=1

∞∑n=0

P ′n(x)tn

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=0

P ′n(x)tn+1

∴∞∑n=1

P ′n(x)tn −∞∑n=1

P ′n−1(x)tn

nPn(x) = xP ′n(x)− P ′n−1(x)

∞∑n=1

∞∑n=0

∞∑n=1

∞∑n=0

P ′n(x)tn

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=0

P ′n(x)tn+1

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=1

P ′n−1(x)tn

nPn(x) = xP ′n(x)− P ′n−1(x)

∞∑n=1

∞∑n=0

∞∑n=1

∞∑n=0

P ′n(x)tn

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=0

P ′n(x)tn+1

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=1

P ′n−1(x)tn

nPn(x) = xP ′n(x)− P ′n−1(x)

∞∑n=1

∞∑n=0

∞∑n=1

∞∑n=0

P ′n(x)tn

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=0

P ′n(x)tn+1

∴∞∑n=1

nPn(x)tn = x

∞∑n=0

P ′n(x)tn −∞∑n=1

P ′n−1(x)tn

nPn(x) = xP ′n(x)− P ′n−1(x)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

Proof: We have ( from relation (1) )

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

differentiating equation (a) partially with respect to x, Weget

∴ (2n+1)Pn(x)+(2n+1)xP ′n(x) = (n+1)P ′n+1(x)+nP ′n−1(x)—(b)

Also from relation (2)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

Substituting the value of (c) in equation (b), we get

∴ (2n + 1)Pn(x) + (2n + 1)[nPn(x) + P ′n−1(x)] =(n + 1)P ′n+1(x) + nP ′n−1(x)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

(3) (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)n

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

∴ (2n + 1)xPn(x) = (n + 1)Pn+1(x) + nPn−1(x) — (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

∴ xP ′n(x) = nPn(x) + P ′n−1(x)—– (c)

∴ (2n + 1)(n + 1)Pn(x) + (2n + 1)P ′n−1(x) =(n + 1)P ′n+1(x) + nP ′n−1(x)

∴ (2n + 1)(n + 1)Pn(x) = (n + 1)P ′n+1(x)− (n + 1)P ′n−1(x)

∴ dividing by (n + 1), we get

∴ (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)

∴ (2n + 1)(n + 1)Pn(x) = (n + 1)P ′n+1(x)− (n + 1)P ′n−1(x)

∴ (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)

∴ (2n + 1)(n + 1)Pn(x) = (n + 1)P ′n+1(x)− (n + 1)P ′n−1(x)

∴ (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)

∴ (2n + 1)(n + 1)Pn(x) = (n + 1)P ′n+1(x)− (n + 1)P ′n−1(x)

∴ (2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

Proof: We have (from relation (3) ),

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

Also we have (from relation (2) ),

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

Taking (a) - (b), we get

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

replacing n by n− 1, we get

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(4) P ′n(x) = xPn−1(x) + nPn−1(x)

(2n + 1)Pn(x) = P ′n+1(x)− P ′n−1(x) —–(a)

∴ nPn(x) = xP ′n(x)− P ′n−1(x) ——(b)

∴ (n + 1)Pn(x) = P ′n+1(x)− xP ′n(x)

∴ nPn−1(x) = P ′n(x)− xP ′n−1(x)

∴ P ′n(x) = xP ′n−1(x) + nPn−1(x)

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

Proof: We have (from relation (4) )

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

also we have (from relation (2) )

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

taking (a) - x X (b), we get

(1−x2)P ′n(x) = xP ′n−1(x)+nPn−1(x)−nxPn(x)−xP ′n−1(x)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(5) (1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

P ′n(x) = xP ′n−1(x) + nPn−1(x) ——- (a)

nPn(x) = xP ′n(x)− P ′n−1(x)

xP ′n(x) = nPn(x) + P ′n−1(x) ——– (b)

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

(n + 1)Pn+1(x) = (n + 1)xPn(x) + nxPn(x)− nPn−1(x)

(n + 1)Pn+1(x) = (n + 1)xPn(x) + n[xPn(x)− Pn−1(x)]

n(Pn−1(x)− xPn(x)) = (n + 1)(xPn(x)− Pn+1(x))

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(6) (1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

(1− x2)P ′n(x) = n[Pn−1(x)− xPn(x)] ——(a)

(n + 1)Pn+1(x) = (2n + 1)xPn(x)− nPn−1(x)

from equation (a),

(1− x2)P ′n(x) = (n + 1)[xPn(x)− Pn+1(x)]

Legendre Function

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CHAPTER 12 LEGENDRE FUNCTIONS