Page 1
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
Page 2
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
Page 3
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
Page 4
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
Page 5
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
Page 6
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
Page 7
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
Page 8
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
Page 9
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
Page 10
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
Page 11
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
Page 12
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
Page 13
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
Page 14
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
Page 15
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
Page 16
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
Page 17
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
Page 18
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
Page 19
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
Page 20
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
Page 21
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
Page 22
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
Page 23
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
Page 24
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
Page 25
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
Page 26
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
Page 27
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
Page 28
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
Page 29
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
Page 30
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
Page 31
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
Page 32
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
Page 33
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
Page 34
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
Page 35
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
Page 36
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
Page 37
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
Page 38
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
Page 39
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
Page 40
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
Page 41
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
Page 42
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
Page 43
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
Page 44
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
Page 45
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
Page 46
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
Page 47
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
Page 48
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
Page 49
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
Page 50
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
Page 51
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
Page 52
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
Page 53
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
Page 54
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
Page 55
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
Page 56
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
Page 57
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
Page 58
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
Page 59
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
Page 60
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
Page 61
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
Page 62
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
Page 63
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
Page 64
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
Page 65
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
Page 66
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
Page 67
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
Page 68
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
Page 69
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
Page 70
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
Page 71
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
Page 72
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
Page 73
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
Page 74
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
Page 75
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
Page 76
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
Page 77
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
Page 78
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
Page 79
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
Page 80
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
Page 81
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
Page 82
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
Page 83
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
Page 84
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
Page 85
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
Page 86
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
Page 87
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
Page 88
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
Page 89
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
Page 90
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
Page 91
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
Page 92
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
Page 93
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
Page 94
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
Page 95
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
Page 96
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
Page 97
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
Page 98
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
Page 99
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
Page 100
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
Page 101
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
Page 102
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
Page 103
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
Page 104
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
Page 105
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
Page 106
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
Page 107
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
Page 108
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
Page 109
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
Page 110
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
Page 111
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
Page 112
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
Page 113
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
Page 114
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
Page 115
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
Page 116
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
Page 117
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
Page 118
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
Page 119
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
Page 120
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
Page 121
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
Page 122
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
Page 123
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
Page 124
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
Page 125
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
Page 126
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
Page 127
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
Page 128
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
Page 129
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
Page 130
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
Page 131
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
Page 132
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
Page 133
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
Page 134
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
Page 135
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
Page 136
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
Page 137
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
Page 138
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
Page 139
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
Page 140
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
Page 141
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
Page 142
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
Page 143
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
Page 144
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
Page 145
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
Page 146
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
Page 147
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
Page 148
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
Page 149
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
Page 150
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
Page 151
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
Page 152
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
Page 153
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
Page 154
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
Page 155
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
Page 156
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
Page 157
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
Page 158
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
Page 159
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
Page 160
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
Page 161
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
Page 162
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
Page 163
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
Page 164
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