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

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

The differential equation

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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

The differential equation

(1− x2)y′′− 2xy′+ n(n+ 1)y = 0

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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

The differential equation

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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

3

5P1(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

3

5P1(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

3

5P1(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

3

5P1(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

3

5P1(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

3

5P1(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

3

5P1(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

3

5P1(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

1

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

=2

5P3(x) +

3

5P1(x) +

4

3P2(x) +

2

3P0(x)− P1(x)− 3P0(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

1

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

=2

5P3(x) +

3

5P1(x) +

4

3P2(x) +

2

3P0(x)− P1(x)− 3P0(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

1

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

=2

5P3(x) +

3

5P1(x) +

4

3P2(x) +

2

3P0(x)− P1(x)− 3P0(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

1

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

=2

5P3(x) +

3

5P1(x) +

4

3P2(x) +

2

3P0(x)− P1(x)− 3P0(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

1

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

=2

5P3(x) +

3

5P1(x) +

4

3P2(x) +

2

3P0(x)− P1(x)− 3P0(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

1

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

=2

5P3(x) +

3

5P1(x) +

4

3P2(x) +

2

3P0(x)− P1(x)− 3P0(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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) +

1

3P0(x)

Now, f(x) = x3 + 2x2 − x− 3

f(x) = x3 + 2x2 − x− 3

=2

5P3(x) +

3

5P1(x) +

4

3P2(x) +

2

3P0(x)− P1(x)− 3P0(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

Generating Function for Pn(x)∞∑n=0

Pn(x)tn =

1√1− 2xt + t2

= (1− 2xt + t2)−12

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

The function (1− 2xt + t2)−12 is

called Generating function ofLegendre’s polynomial Pn(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

Ex Show that

(i)Pn(1) = 1

(ii)Pn(−1) = (−1)n

(iii)Pn(−x) = (−1)nPn(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

Ex Show that

(i)Pn(1) = 1

(ii)Pn(−1) = (−1)n

(iii)Pn(−x) = (−1)nPn(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

Ex Show that

(i)Pn(1) = 1

(ii)Pn(−1) = (−1)n

(iii)Pn(−x) = (−1)nPn(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

Ex Show that

(i)Pn(1) = 1

(ii)Pn(−1) = (−1)n

(iii)Pn(−x) = (−1)nPn(x)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

tn

Comparing the coefficient of tn both the sides, we get

Pn(1) = 1

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

tn

Comparing the coefficient of tn both the sides, we get

Pn(1) = 1

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

tn

Comparing the coefficient of tn both the sides, we get

Pn(1) = 1

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

tn

Comparing the coefficient of tn both the sides, we get

Pn(1) = 1

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

tn

Comparing the coefficient of tn both the sides, we get

Pn(1) = 1

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

tn

Comparing the coefficient of tn both the sides, we get

Pn(1) = 1

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

Rodrigue’s Formula

Pn(x) =1

2nn!

dn

dxn[(x2 − 1)n]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)=

2nxy

x2 − 1

∴ (x2 − 1)y1 = 2nxy

Differentiating with respect to x,

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)=

2nxy

x2 − 1

∴ (x2 − 1)y1 = 2nxy

Differentiating with respect to x,

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)=

2nxy

x2 − 1

∴ (x2 − 1)y1 = 2nxy

Differentiating with respect to x,

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)=

2nxy

x2 − 1

∴ (x2 − 1)y1 = 2nxy

Differentiating with respect to x,

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)=

2nxy

x2 − 1

∴ (x2 − 1)y1 = 2nxy

Differentiating with respect to x,

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)=

2nxy

x2 − 1

∴ (x2 − 1)y1 = 2nxy

Differentiating with respect to x,

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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)=

2nxy

x2 − 1

∴ (x2 − 1)y1 = 2nxy

Differentiating with respect to x,

(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

{dn

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)

2!

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

{dn

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)

2!

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

{dn

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)

2!

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

{dn

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)

2!

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

{dn

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)

2!

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

{dn

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)

2!

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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

dxn

∴ (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

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ Pn(x) = cdny

dxn= c

dn

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dn

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+...

+dn((x + 1)n)

dxn(x− 1)n

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ Pn(x) = cdny

dxn= c

dn

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dn

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+...

+dn((x + 1)n)

dxn(x− 1)n

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ Pn(x) = cdny

dxn= c

dn

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dn

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+...

+dn((x + 1)n)

dxn(x− 1)n

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ Pn(x) = cdny

dxn= c

dn

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dn

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+...

+dn((x + 1)n)

dxn(x− 1)n

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ Pn(x) = cdny

dxn= c

dn

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dn

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+...

+dn((x + 1)n)

dxn(x− 1)n

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ Pn(x) = cdny

dxn= c

dn

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dn

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+...

+dn((x + 1)n)

dxn(x− 1)n

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ Pn(x) = cdny

dxn= c

dn

dxn(x2 − 1)n ——(3)

Now y = (x2 − 1)n

= (x + 1)n(x− 1)n

∴dny

dxn= (x + 1)n

dn

dxn((x− 1)n)

+n(x + 1)n−1dn−1

dxn−1((x− 1)n)

+...

+dn((x + 1)n)

dxn(x− 1)n

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

Recurrence Relations for Pn(x) : −

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

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)−

1

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)−

12

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

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)−

1

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)−

12

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

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)−

1

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)−

12

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

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)−

1

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)−

12

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

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)−

1

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)−

12

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

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)−

1

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)−

12

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

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)−

1

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)−

12

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

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)−

1

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)−

12

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

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)−

1

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)−

12

(1− 2xt + t2)(x− t)

from (i)

(1− 2xt + t2)

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑n=0

Pn(x)tn

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

x

∞∑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

(n−

1)Pn−1(x)tn = x

∞∑n=0

Pn(x)tn −∞∑n=1

Pn−1(x)tn

comparing the coefficients of tn on both the sides

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

x

∞∑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

(n−

1)Pn−1(x)tn = x

∞∑n=0

Pn(x)tn −∞∑n=1

Pn−1(x)tn

comparing the coefficients of tn on both the sides

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

x

∞∑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

(n−

1)Pn−1(x)tn = x

∞∑n=0

Pn(x)tn −∞∑n=1

Pn−1(x)tn

comparing the coefficients of tn on both the sides

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

x

∞∑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

(n−

1)Pn−1(x)tn = x

∞∑n=0

Pn(x)tn −∞∑n=1

Pn−1(x)tn

comparing the coefficients of tn on both the sides

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴∞∑n=1

nPn(x)tn−1 − 2x

∞∑n=1

nPn(x)tn +

∞∑n=1

nPn(x)tn+1 =

x

∞∑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

(n−

1)Pn−1(x)tn = x

∞∑n=0

Pn(x)tn −∞∑n=1

Pn−1(x)tn

comparing the coefficients of tn on both the sides

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

(2) nPn(x) = xP ′n(x)− P ′n−1(x)

Proof: We have∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

1

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)−

12

(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)−

12

(1− 2xt + t2)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

1

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)−

12

(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)−

12

(1− 2xt + t2)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

1

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)−

12

(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)−

12

(1− 2xt + t2)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

1

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)−

12

(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)−

12

(1− 2xt + t2)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

1

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)−

12

(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)−

12

(1− 2xt + t2)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

1

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)−

12

(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)−

12

(1− 2xt + t2)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

1

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)−

12

(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)−

12

(1− 2xt + t2)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

1

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)−

12

(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)−

12

(1− 2xt + t2)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

∞∑n=0

Pn(x)tn = (1− 2xt + t2)−

1

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)−

12

(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)−

12

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

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∞∑n=1

nPn(x)tn−1 =(x− t)

t

∞∑n=0

P ′n(x)tn {by eq. (ii)

∴ t∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∞∑n=1

nPn(x)tn−1 =(x− t)

t

∞∑n=0

P ′n(x)tn {by eq. (ii)

∴ t

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∞∑n=1

nPn(x)tn−1 =(x− t)

t

∞∑n=0

P ′n(x)tn {by eq. (ii)

∴ t

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∞∑n=1

nPn(x)tn−1 =(x− t)

t

∞∑n=0

P ′n(x)tn {by eq. (ii)

∴ t

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∞∑n=1

nPn(x)tn−1 =(x− t)

t

∞∑n=0

P ′n(x)tn {by eq. (ii)

∴ t

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∞∑n=1

nPn(x)tn−1 =(x− t)

t

∞∑n=0

P ′n(x)tn {by eq. (ii)

∴ t

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∞∑n=1

nPn(x)tn−1 =(x− t)

t

∞∑n=0

P ′n(x)tn {by eq. (ii)

∴ t

∞∑n=1

nPn(x)tn−1 = (x− t)

∞∑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. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

∴ (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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function

Legendre’s PolynomialsExamples of Legendre’s Polynomials

Generating Function for Pn(x)Rodrigue’s Formula

Recurrence Relations for Pn(x)

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

Proof: We have (from relation (5) )

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

also we have (from relation (1) )

(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)]

N. B. Vyas Legendre’s Function