Legendre Function

Legendre’s Polynomials
Examples 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 Mathematics
Atmiya Institute of Technology and Science

Department of Mathematics

N. B. Vyas Legendre’s Function


The diﬀerential equation



2
(1 − x )y − 2xy + n(n + 1)y = 0



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



Legendre’s Polynomials:
⇒ P0 (x) = 1



⇒ P0 (x) = 1
⇒ P1 (x) = x



⇒ P0 (x) = 1
⇒ P1 (x) = x
1
⇒ P2 (x) = (3x2 − 1)
2



⇒ P0 (x) = 1
⇒ P1 (x) = x
1
⇒ P2 (x) = (3x2 − 1)
2
1
⇒ P3 (x) = (5x3 − 3x)
2



⇒ P0 (x) = 1
⇒ P1 (x) = x
1
⇒ P2 (x) = (3x2 − 1)
2
1
⇒ P3 (x) = (5x3 − 3x)
2
1
⇒ P4 (x) = (35x3 − 30x2 + 3)
8



⇒ P0 (x) = 1
⇒ P1 (x) = x
1
⇒ P2 (x) = (3x2 − 1)
2
1
⇒ P3 (x) = (5x3 − 3x)
2
1
⇒ P4 (x) = (35x3 − 30x2 + 3)
8
1
⇒ P5 (x) = (63x5 − 70x3 + 15x)
8



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



Solution:



Solution:
⇒ P0 (x) = 1
∴ 1 = P0 (x)



Solution:
⇒ P0 (x) = 1
∴ 1 = P0 (x)
⇒ P1 (x) = x
∴ x = P1 (x)



Solution:
⇒ P0 (x) = 1
∴ 1 = P0 (x)
⇒ P1 (x) = x
∴ x = P1 (x)
1
⇒ P3 (x) = (5x3 − 3x)
2



Solution:
⇒ P0 (x) = 1
∴ 1 = P0 (x)
⇒ P1 (x) = x
∴ x = P1 (x)
1
⇒ P3 (x) = (5x3 − 3x)
2
∴ 2P3 (x) = (5x3 − 3x)



Solution:
⇒ P0 (x) = 1
∴ 1 = P0 (x)
⇒ P1 (x) = x
∴ x = P1 (x)
1
⇒ P3 (x) = (5x3 − 3x)
2
∴ 2P3 (x) = (5x3 − 3x)
∴ 2P3 (x) + 3x = 5x3



Solution:
⇒ P0 (x) = 1
∴ 1 = P0 (x)
⇒ P1 (x) = x
∴ x = P1 (x)
1
⇒ P3 (x) = (5x3 − 3x)
2
∴ 2P3 (x) = (5x3 − 3x)
∴ 2P3 (x) + 3x = 5x3
∴ 2P3 (x) + 3P1 (x) = 5x3 { x = P1 (x)}



Solution:
⇒ P0 (x) = 1
∴ 1 = P0 (x)
⇒ P1 (x) = x
∴ x = P1 (x)
1
⇒ P3 (x) = (5x3 − 3x)
2
∴ 2P3 (x) = (5x3 − 3x)
∴ 2P3 (x) + 3x = 5x3
∴ 2P3 (x) + 3P1 (x) = 5x3 { x = P1 (x)}
2 3
∴ x3 = P3 (x) + P1 (x)
5 5



1
⇒ P2 (x) = (3x2 − 1)
2



1
⇒ P2 (x) = (3x2 − 1)
2
∴ 2P2 (x) = (3x2 − 1)



1
⇒ P2 (x) = (3x2 − 1)
2
∴ 2P2 (x) = (3x2 − 1)
∴ 2P2 (x) + 1 = 3x2



1
⇒ P2 (x) = (3x2 − 1)
2
∴ 2P2 (x) = (3x2 − 1)
∴ 2P2 (x) + 1 = 3x2
∴ 2P2 (x) + P0 (x) = 3x2 { 1 = P0 (x)}



1
⇒ P2 (x) = (3x2 − 1)
2
∴ 2P2 (x) = (3x2 − 1)
∴ 2P2 (x) + 1 = 3x2
∴ 2P2 (x) + P0 (x) = 3x2 { 1 = P0 (x)}
2 1
∴ x2 = P2 (x) + P0 (x)
3 3



1
⇒ P2 (x) = (3x2 − 1)
2
∴ 2P2 (x) = (3x2 − 1)
∴ 2P2 (x) + 1 = 3x2
∴ 2P2 (x) + P0 (x) = 3x2 { 1 = P0 (x)}
2 1
∴ x2 = P2 (x) + P0 (x)
3 3
Now, f (x) = x3 + 2x2 − x − 3



1
⇒ P2 (x) = (3x2 − 1)
2
∴ 2P2 (x) = (3x2 − 1)
∴ 2P2 (x) + 1 = 3x2
∴ 2P2 (x) + P0 (x) = 3x2 { 1 = P0 (x)}
2 1
∴ x2 = P2 (x) + P0 (x)
3 3
Now, f (x) = x3 + 2x2 − x − 3
f (x) = x3 + 2x2 − x − 3
2 3 4 2
= P3 (x) + P1 (x) + P2 (x) + P0 (x) − P1 (x) − 3P0 (x)
5 5 3 3



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



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



Generating Function for Pn(x)
∞
1
Pn(x)tn = √
n=0
1 − 2xt + t2
1
= (1 − 2xt + t2)− 2



1
The function (1 − 2xt + t2)− 2 is
called Generating function of
Legendre’s polynomial Pn(x)



Ex Show that



Ex Show that
(i)Pn(1) = 1



Ex Show that
(i)Pn(1) = 1
(ii)Pn(−1) = (−1)n



Ex Show that
(i)Pn(1) = 1
(ii)Pn(−1) = (−1)n
(iii)Pn(−x) = (−1)nPn(x)



Solution:
∞
1
(i) We have Pn (x)tn = (1 − 2xt + t2 )− 2
n=0
Putting x = 1 in eq(1), we get



Solution:
∞
1
n=0
∞
1
Pn (1)tn = (1 − 2t + t2 )− 2 = (1 − t)−1
n=0



Solution:
∞
1
n=0
∞
1
Pn (1)tn = (1 − 2t + t2 )− 2 = (1 − t)−1
n=0
∞
1
∴ Pn (1)tn = = 1 + t + t2 + t3 + ...
1−t
n=0



Solution:
∞
1
n=0
∞
1
Pn (1)tn = (1 − 2t + t2 )− 2 = (1 − t)−1
n=0
∞
1
∴ Pn (1)tn = = 1 + t + t2 + t3 + ...
1−t
n=0
∞ ∞
n
∴ Pn (1)t = tn
n=0 n=0



Solution:
∞
1
n=0
∞
1
Pn (1)tn = (1 − 2t + t2 )− 2 = (1 − t)−1
n=0
∞
1
∴ Pn (1)tn = = 1 + t + t2 + t3 + ...
1−t
n=0
∞ ∞
n
∴ Pn (1)t = tn
n=0 n=0
Comparing the coeﬃcient of tn both the sides, we get



Solution:
∞
1
n=0
∞
1
Pn (1)tn = (1 − 2t + t2 )− 2 = (1 − t)−1
n=0
∞
1
∴ Pn (1)tn = = 1 + t + t2 + t3 + ...
1−t
n=0
∞ ∞
n
∴ Pn (1)t = tn
n=0 n=0
Comparing the coeﬃcient of tn both the sides, we get
Pn (1) = 1



(ii) Putting x = −1 in eq(1), we get



∞
1
Pn (−1)tn = (1 + 2t + t2 )− 2 = (1 + t)−1
n=0



∞
1
Pn (−1)tn = (1 + 2t + t2 )− 2 = (1 + t)−1
n=0
∞
1
∴ Pn (−1)tn = = 1 − t + t2 − t3 + ...
1+t
n=0



∞
1
Pn (−1)tn = (1 + 2t + t2 )− 2 = (1 + t)−1
n=0
∞
1
∴ Pn (−1)tn = = 1 − t + t2 − t3 + ...
1+t
n=0
∞ ∞
∴ Pn (−1)tn = (−1)n tn
n=0 n=0



∞
1
Pn (−1)tn = (1 + 2t + t2 )− 2 = (1 + t)−1
n=0
∞
1
∴ Pn (−1)tn = = 1 − t + t2 − t3 + ...
1+t
n=0
∞ ∞
∴ Pn (−1)tn = (−1)n tn
n=0 n=0
Comparing coeﬃcients of tn , we get



∞
1
Pn (−1)tn = (1 + 2t + t2 )− 2 = (1 + t)−1
n=0
∞
1
∴ Pn (−1)tn = = 1 − t + t2 − t3 + ...
1+t
n=0
∞ ∞
∴ Pn (−1)tn = (−1)n tn
n=0 n=0
Comparing coeﬃcients of tn , we get
Pn (−1) = (−1)n



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



∞
1
Pn (−x)tn = (1 + 2xt + t2 )− 2 —(a)
n=0



∞
1
Pn (−x)tn = (1 + 2xt + t2 )− 2 —(a)
n=0
Now, replacing t by −t in eq(1), we get



∞
1
Pn (−x)tn = (1 + 2xt + t2 )− 2 —(a)
n=0
∞
1
Pn (x)(−t)n = (1 + 2xt + t2 )− 2 —(b)
n=0



∞
1
Pn (−x)tn = (1 + 2xt + t2 )− 2 —(a)
n=0
∞
1
Pn (x)(−t)n = (1 + 2xt + t2 )− 2 —(b)
n=0
from equation (a) and (b)



∞
1
Pn (−x)tn = (1 + 2xt + t2 )− 2 —(a)
n=0
∞
1
Pn (x)(−t)n = (1 + 2xt + t2 )− 2 —(b)
n=0
∞ ∞
n
Pn (−x)(t) = Pn (x)(−1)n (t)n
n=0 n=0



∞
1
Pn (−x)tn = (1 + 2xt + t2 )− 2 —(a)
n=0
∞
1
Pn (x)(−t)n = (1 + 2xt + t2 )− 2 —(b)
n=0
∞ ∞
n
Pn (−x)(t) = Pn (x)(−1)n (t)n
n=0 n=0
Comparing the coeﬃcients of tn , both sides, we get



∞
1
Pn (−x)tn = (1 + 2xt + t2 )− 2 —(a)
n=0
∞
1
Pn (x)(−t)n = (1 + 2xt + t2 )− 2 —(b)
n=0
∞ ∞
n
Pn (−x)(t) = Pn (x)(−1)n (t)n
n=0 n=0
Comparing the coeﬃcients of tn , both sides, we get
Pn (−x) = (−1)n Pn (x)




1 dn
Pn(x) = n [(x2 − 1)n]
2 n! dxn



Proof:
Let y = (x2 − 1)n



Proof:
Let y = (x2 − 1)n
Diﬀerentiating wit respect to x



Proof:
Let y = (x2 − 1)n
∴ y1 = n(x2 − 1)n−1 (2x)



Proof:
Let y = (x2 − 1)n
∴ y1 = n(x2 − 1)n−1 (2x)
2nx(x2 − 1)n 2nxy
∴ y1 = = 2
(x2 − 1) x −1



Proof:
Let y = (x2 − 1)n
∴ y1 = n(x2 − 1)n−1 (2x)
2nx(x2 − 1)n 2nxy
∴ y1 = = 2
(x2 − 1) x −1
∴ (x2 − 1)y1 = 2nxy



Proof:
Let y = (x2 − 1)n
∴ y1 = n(x2 − 1)n−1 (2x)
2nx(x2 − 1)n 2nxy
∴ y1 = = 2
(x2 − 1) x −1
∴ (x2 − 1)y1 = 2nxy
Diﬀerentiating with respect to x,



Proof:
Let y = (x2 − 1)n
∴ y1 = n(x2 − 1)n−1 (2x)
2nx(x2 − 1)n 2nxy
∴ y1 = = 2
(x2 − 1) x −1
∴ (x2 − 1)y1 = 2nxy
Diﬀerentiating with respect to x,
(x2 − 1)y2 +2xy1 = 2nxy1 + 2ny



dn
(U V ) = nC0 U Vn + nC1 U1 Vn−1 + ... + nCn Un V
dxn



dn
dxn
dn
→ ((x2 −1)y2 ) = nC0 (x2 −1)yn+2 +nC1 (2x)yn+1 +nC2 (2)yn
dxn



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



dn
dxn
dn
dxn
dn
→ (2xy1 ) = nC0 (2x)yn+1 + nC1 (2)yn
dxn
dn
→ (2nxy1 ) = nC0 (2nx)yn+1 + nC1 (2n)yn
dxn



dn
dxn
dn
dxn
dn
→ (2xy1 ) = nC0 (2x)yn+1 + nC1 (2)yn
dxn
dn
→ (2nxy1 ) = nC0 (2nx)yn+1 + nC1 (2n)yn
dxn
dn
→ (2ny) = 2nyn
dxn



dn
dxn
dn
dxn
dn
→ (2xy1 ) = nC0 (2x)yn+1 + nC1 (2)yn
dxn
dn
→ (2nxy1 ) = nC0 (2nx)yn+1 + nC1 (2n)yn
dxn
dn
→ (2ny) = 2nyn
dxn
n(n − 1)
Also nC0 = 1, nC1 = n, nC2 =
2!



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



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



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



∴ (x2 − 1)yn+2 + 2nxyn+1 + n(n − 1)yn
∴ (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



∴ (x2 − 1)yn+2 + 2nxyn+1 + n(n − 1)yn
∴ (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
dn y
Let v = yn = n
dx



∴ (x2 − 1)yn+2 + 2nxyn+1 + n(n − 1)yn
∴ (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
dn y
Let v = yn = n
dx
∴ (1 − x2 )v2 − 2xv1 + n(n + 1)v = 0 ——(2)



∴ (x2 − 1)yn+2 + 2nxyn+1 + n(n − 1)yn
∴ (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
dn y
Let v = yn = n
dx
∴ (1 − x2 )v2 − 2xv1 + n(n + 1)v = 0 ——(2)
Equation (2) is a Legendre’s equation in variables v and x



∴ (x2 − 1)yn+2 + 2nxyn+1 + n(n − 1)yn
∴ (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
dn y
Let v = yn = n
dx
∴ (1 − x2 )v2 − 2xv1 + n(n + 1)v = 0 ——(2)
⇒ Pn (x) is a solution of equation (2)



∴ (x2 − 1)yn+2 + 2nxyn+1 + n(n − 1)yn
∴ (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
dn y
Let v = yn = n
dx
∴ (1 − x2 )v2 − 2xv1 + n(n + 1)v = 0 ——(2)
Also, v = f (x) is a solution of equation (2)



∴ (x2 − 1)yn+2 + 2nxyn+1 + n(n − 1)yn
∴ (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
dn y
Let v = yn = n
dx
∴ (1 − x2 )v2 − 2xv1 + n(n + 1)v = 0 ——(2)
Also, v = f (x) is a solution of equation (2)
Pn = cv where c is constant



dn y dn
∴ Pn (x) = c = c n (x2 − 1)n ——(3)
dxn dx



dn y dn
∴ Pn (x) = c = c n (x2 − 1)n ——(3)
dxn dx
Now y = (x2 − 1)n



dn y dn
∴ Pn (x) = c = c n (x2 − 1)n ——(3)
dxn dx
Now y = (x2 − 1)n
= (x + 1)n (x − 1)n



dn y dn
∴ Pn (x) = c = c n (x2 − 1)n ——(3)
dxn dx
Now y = (x2 − 1)n
= (x + 1)n (x − 1)n
dn y dn
∴ = (x + 1)n n ((x − 1)n )
dxn dx



dn y dn
∴ Pn (x) = c = c n (x2 − 1)n ——(3)
dxn dx
Now y = (x2 − 1)n
= (x + 1)n (x − 1)n
dn y dn
∴ = (x + 1)n n ((x − 1)n )
dxn dx
dn−1
+n(x + 1)n−1 n−1 ((x − 1)n )
dx



dn y dn
∴ Pn (x) = c = c n (x2 − 1)n ——(3)
dxn dx
Now y = (x2 − 1)n
= (x + 1)n (x − 1)n
dn y dn
∴ = (x + 1)n n ((x − 1)n )
dxn dx
dn−1
+n(x + 1)n−1 n−1 ((x − 1)n )
dx
+...



dn y dn
∴ Pn (x) = c = c n (x2 − 1)n ——(3)
dxn dx
Now y = (x2 − 1)n
= (x + 1)n (x − 1)n
dn y dn
∴ = (x + 1)n n ((x − 1)n )
dxn dx
dn−1
+n(x + 1)n−1 n−1 ((x − 1)n )
dx
+...
dn ((x + 1)n )
+ (x − 1)n
dxn



Recurrence Relations for Pn(x) : −



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



(1) (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
Proof: We have



(1) (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
Proof: We have
∞ 1
−
n 2 ) 2 ——–(i)
Pn (x)t = (1 − 2xt + t
n=0



(1) (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
Proof: We have
∞ 1
−
n 2 ) 2 ——–(i)
Pn (x)t = (1 − 2xt + t
n=0
Diﬀerentiating equation (i) partially with respect to t, we
get



(1) (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
Proof: We have
∞ 1
−
n 2 ) 2 ——–(i)
Pn (x)t = (1 − 2xt + t
n=0
get
∞
1 3
nPn (x)tn−1 = − (1 − 2xt + t2 )− 2 (−2x + 2t)
2
n=1



(1) (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
Proof: We have
∞ 1
−
n 2 ) 2 ——–(i)
Pn (x)t = (1 − 2xt + t
n=0
get
∞
1 3
nPn (x)tn−1 = − (1 − 2xt + t2 )− 2 (−2x + 2t)
2
n=1
1
= (1 − 2xt + t2 )−1 (1 − 2xt + t2 )− 2 (x − t)



(1) (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
Proof: We have
∞ 1
−
n 2 ) 2 ——–(i)
Pn (x)t = (1 − 2xt + t
n=0
get
∞
1 3
nPn (x)tn−1 = − (1 − 2xt + t2 )− 2 (−2x + 2t)
2
n=1
1
= (1 − 2xt + t2 )−1 (1 − 2xt + t2 )− 2 (x − t)
1
(1 − 2xt + t2 )− 2
= (x − t)
(1 − 2xt + t2 )



(1) (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
Proof: We have
∞ 1
−
n 2 ) 2 ——–(i)
Pn (x)t = (1 − 2xt + t
n=0
get
∞
1 3
nPn (x)tn−1 = − (1 − 2xt + t2 )− 2 (−2x + 2t)
2
n=1
1
= (1 − 2xt + t2 )−1 (1 − 2xt + t2 )− 2 (x − t)
1
(1 − 2xt + t2 )− 2
= (x − t)
(1 − 2xt + t2 )
from (i)



(1) (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
Proof: We have
∞ 1
−
n 2 ) 2 ——–(i)
Pn (x)t = (1 − 2xt + t
n=0
get
∞
1 3
nPn (x)tn−1 = − (1 − 2xt + t2 )− 2 (−2x + 2t)
2
n=1
1
= (1 − 2xt + t2 )−1 (1 − 2xt + t2 )− 2 (x − t)
1
(1 − 2xt + t2 )− 2
= (x − t)
(1 − 2xt + t2 )
from (i)
∞ ∞
n−1
(1 − 2xt + t2 ) nPn (x)t = (x − t) Pn (x)tn
n=1 n=0


∞ ∞ ∞
∴ nPn (x)tn−1 − 2x nPn (x)tn + nPn (x)tn+1 =
n=1 n=1 n=1
∞ ∞
x Pn (x)tn − Pn (x)tn+1
n=0 n=0



∞ ∞ ∞
n=1 n=1 n=1
∞ ∞
n=0 n=0
replacing n by n+1 in 1st term, n by n-1 in 3rd term in
L.H.S.



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



∞ ∞ ∞
n=1 n=1 n=1
∞ ∞
n=0 n=0
L.H.S.
∞ ∞ ∞
(n + 1)Pn+1 (x)tn − 2x nPn (x)tn + (n −
n=0 n=1 n=2
∞ ∞
1)Pn−1 (x)tn = x Pn (x)tn − Pn−1 (x)tn
n=0 n=1



∞ ∞ ∞
n=1 n=1 n=1
∞ ∞
n=0 n=0
L.H.S.
∞ ∞ ∞
(n + 1)Pn+1 (x)tn − 2x nPn (x)tn + (n −
n=0 n=1 n=2
∞ ∞
1)Pn−1 (x)tn = x Pn (x)tn − Pn−1 (x)tn
n=0 n=1
comparing the coeﬃcients of tn on both the sides



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



(2) nPn (x) = xPn (x) − Pn−1 (x)



(2) nPn (x) = xPn (x) − Pn−1 (x)
Proof: We have



(2) nPn (x) = xPn (x) − Pn−1 (x)
Proof: We have
∞ 1
−
n
Pn (x)t = (1 − 2xt + t2 ) 2 ——–(i)
n=0



(2) nPn (x) = xPn (x) − Pn−1 (x)
Proof: We have
∞ 1
−
n
Pn (x)t = (1 − 2xt + t2 ) 2 ——–(i)
n=0
Diﬀerentiating equation (i) partially with respect to x, we
get



(2) nPn (x) = xPn (x) − Pn−1 (x)
Proof: We have
∞ 1
−
n
Pn (x)t = (1 − 2xt + t2 ) 2 ——–(i)
n=0
get
∞
1 3
Pn (x)tn = − (1 − 2xt + t2 )− 2 (−2t)
2
n=0



(2) nPn (x) = xPn (x) − Pn−1 (x)
Proof: We have
∞ 1
−
n
Pn (x)t = (1 − 2xt + t2 ) 2 ——–(i)
n=0
get
∞
1 3
Pn (x)tn = − (1 − 2xt + t2 )− 2 (−2t)
2
n=0
∞ 1
n t(1 − 2xt + t2 )− 2
∴ Pn (x)t = ————-(ii)
(1 − 2xt + t2 )
n=0



(2) nPn (x) = xPn (x) − Pn−1 (x)
Proof: We have
∞ 1
−
n
Pn (x)t = (1 − 2xt + t2 ) 2 ——–(i)
n=0
get
∞
1 3
Pn (x)tn = − (1 − 2xt + t2 )− 2 (−2t)
2
n=0
∞ 1
t(1 − 2xt + t2 )− 2
n
∴ Pn (x)t = ————-(ii)
(1 − 2xt + t2 )
n=0
⇒ Diﬀerentiating equation (i) partially with respect to t, we
get



(2) nPn (x) = xPn (x) − Pn−1 (x)
Proof: We have
∞ 1
−
n
Pn (x)t = (1 − 2xt + t2 ) 2 ——–(i)
n=0
get
∞
1 3
Pn (x)tn = − (1 − 2xt + t2 )− 2 (−2t)
2
n=0
∞ 1
t(1 − 2xt + t2 )− 2
n
∴ Pn (x)t = ————-(ii)
(1 − 2xt + t2 )
n=0
get
∞
1 3
nPn (x)tn−1 = − (1 − 2xt + t2 )− 2 (−2x + 2t)
2
n=1



(2) nPn (x) = xPn (x) − Pn−1 (x)
Proof: We have
∞ 1
−
n
Pn (x)t = (1 − 2xt + t2 ) 2 ——–(i)
n=0
get
∞
1 3
Pn (x)tn = − (1 − 2xt + t2 )− 2 (−2t)
2
n=0
∞ 1
t(1 − 2xt + t2 )− 2
n
∴ Pn (x)t = ————-(ii)
(1 − 2xt + t2 )
n=0
get
∞
1 3
nPn (x)tn−1 = − (1 − 2xt + t2 )− 2 (−2x + 2t)
2
n=1
∞ 1
n−1 (x − t)(1 −Legendre’s )− 2
N. B. Vyas
2xt + t2 Function


∞ ∞
n−1 (x − t)
nPn (x)t = Pn (x)tn {by eq. (ii)
t
n=1 n=0



∞ ∞
n−1 (x − t)
t
n=1 n=0
∞ ∞
∴ t nPn (x)tn−1 = (x − t) Pn (x)tn
n=1 n=0



∞ ∞
n−1 (x − t)
t
n=1 n=0
∞ ∞
n=1 n=0
∞ ∞ ∞
∴ nPn (x)tn = x Pn (x)tn − Pn (x)tn+1
n=1 n=0 n=0



∞ ∞
n−1 (x − t)
t
n=1 n=0
∞ ∞
n=1 n=0
∞ ∞ ∞
n=1 n=0 n=0
Replacing n by n-1 in 2nd term in R.H.S.



∞ ∞
n−1 (x − t)
t
n=1 n=0
∞ ∞
n=1 n=0
∞ ∞ ∞
n=1 n=0 n=0
∞ ∞ ∞
n n
∴ nPn (x)t = x Pn (x)t − Pn−1 (x)tn
n=1 n=0 n=1



∞ ∞
n−1 (x − t)
t
n=1 n=0
∞ ∞
n=1 n=0
∞ ∞ ∞
n=1 n=0 n=0
∞ ∞ ∞
n n
n=1 n=0 n=1
comparing the coeﬃcients of tn on both sides, we get



∞ ∞
n−1 (x − t)
t
n=1 n=0
∞ ∞
n=1 n=0
∞ ∞ ∞
n=1 n=0 n=0
∞ ∞ ∞
n n
n=1 n=0 n=1
comparing the coeﬃcients of tn on both sides, we get
nPn (x) = xPn (x) − Pn−1 (x)



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
Proof: We have ( from relation (1) )



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
∴ (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x) — (a)



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
∴ (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x) — (a)
diﬀerentiating equation (a) partially with respect to x, We
get



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
∴ (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x) — (a)
get
∴ (2n + 1)Pn (x) + (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x)
—(b)



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
∴ (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x) — (a)
get
—(b)
Also from relation (2)



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
∴ (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x) — (a)
get
—(b)
nPn (x) = xPn (x) − Pn−1 (x)



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
∴ (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x) — (a)
get
—(b)
nPn (x) = xPn (x) − Pn−1 (x)
∴ xPn (x) = nPn (x) + Pn−1 (x)—– (c)



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
∴ (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x) — (a)
get
—(b)
nPn (x) = xPn (x) − Pn−1 (x)
∴ xPn (x) = nPn (x) + Pn−1 (x)—– (c)
Substituting the value of (c) in equation (b), we get



(3) (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)n
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
∴ (2n + 1)xPn (x) = (n + 1)Pn+1 (x) + nPn−1 (x) — (a)
get
—(b)
nPn (x) = xPn (x) − Pn−1 (x)
∴ xPn (x) = nPn (x) + Pn−1 (x)—– (c)
Substituting the value of (c) in equation (b), we get
∴ (2n + 1)Pn (x) + (2n + 1)[nPn (x) + Pn−1 (x)] =
(n + 1)Pn+1 (x) + nPn−1 (x)


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



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



∴ (2n + 1)(n + 1)Pn (x) + (2n + 1)Pn−1 (x) =
(n + 1)Pn+1 (x) + nPn−1 (x)
∴ (2n + 1)(n + 1)Pn (x) = (n + 1)Pn+1 (x) − (n + 1)Pn−1 (x)
∴ dividing by (n + 1), we get



∴ (2n + 1)(n + 1)Pn (x) + (2n + 1)Pn−1 (x) =
(n + 1)Pn+1 (x) + nPn−1 (x)
∴ (2n + 1)(n + 1)Pn (x) = (n + 1)Pn+1 (x) − (n + 1)Pn−1 (x)
∴ dividing by (n + 1), we get
∴ (2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x)



(4) Pn (x) = xPn−1 (x) + nPn−1 (x)



(4) Pn (x) = xPn−1 (x) + nPn−1 (x)
Proof: We have (from relation (3) ),



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



(4) Pn (x) = xPn−1 (x) + nPn−1 (x)
(2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x) —–(a)
Also we have (from relation (2) ),



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



(4) Pn (x) = xPn−1 (x) + nPn−1 (x)
(2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x) —–(a)
∴ nPn (x) = xPn (x) − Pn−1 (x) ——(b)
Taking (a) - (b), we get



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



(4) Pn (x) = xPn−1 (x) + nPn−1 (x)
(2n + 1)Pn (x) = Pn+1 (x) − Pn−1 (x) —–(a)
∴ nPn (x) = xPn (x) − Pn−1 (x) ——(b)
∴ (n + 1)Pn (x) = Pn+1 (x) − xPn (x)
replacing n by n − 1, we get



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



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



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



(5) (1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)]
Proof: We have (from relation (4) )



(5) (1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)]
Pn (x) = xPn−1 (x) + nPn−1 (x) ——- (a)



(5) (1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)]
Pn (x) = xPn−1 (x) + nPn−1 (x) ——- (a)
also we have (from relation (2) )



(5) (1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)]
Pn (x) = xPn−1 (x) + nPn−1 (x) ——- (a)
nPn (x) = xPn (x) − Pn−1 (x)



(5) (1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)]
Pn (x) = xPn−1 (x) + nPn−1 (x) ——- (a)
nPn (x) = xPn (x) − Pn−1 (x)
xPn (x) = nPn (x) + Pn−1 (x) ——– (b)



(5) (1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)]
Pn (x) = xPn−1 (x) + nPn−1 (x) ——- (a)
nPn (x) = xPn (x) − Pn−1 (x)
xPn (x) = nPn (x) + Pn−1 (x) ——– (b)
taking (a) - x X (b), we get



(5) (1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)]
Pn (x) = xPn−1 (x) + nPn−1 (x) ——- (a)
nPn (x) = xPn (x) − Pn−1 (x)
xPn (x) = nPn (x) + Pn−1 (x) ——– (b)
(1 − x2 )Pn (x) = xPn−1 (x) + nPn−1 (x) − nxPn (x) − xPn−1 (x)



(5) (1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)]
Pn (x) = xPn−1 (x) + nPn−1 (x) ——- (a)
nPn (x) = xPn (x) − Pn−1 (x)
xPn (x) = nPn (x) + Pn−1 (x) ——– (b)
(1 − x2 )Pn (x) = xPn−1 (x) + nPn−1 (x) − nxPn (x) − xPn−1 (x)
(1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)]



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



(6) (1 − x2 )Pn (x) = (n + 1)[xPn (x) − Pn+1 (x)]
(1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)] ——(a)



(6) (1 − x2 )Pn (x) = (n + 1)[xPn (x) − Pn+1 (x)]
(1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)] ——(a)
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)



(6) (1 − x2 )Pn (x) = (n + 1)[xPn (x) − Pn+1 (x)]
(1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)] ——(a)
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
(n + 1)Pn+1 (x) = (n + 1)xPn (x) + nxPn (x) − nPn−1 (x)



(6) (1 − x2 )Pn (x) = (n + 1)[xPn (x) − Pn+1 (x)]
(1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)] ——(a)
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
(n + 1)Pn+1 (x) = (n + 1)xPn (x) + n[xPn (x) − Pn−1 (x)]



(6) (1 − x2 )Pn (x) = (n + 1)[xPn (x) − Pn+1 (x)]
(1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)] ——(a)
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
n(Pn−1 (x) − xPn (x)) = (n + 1)(xPn (x) − Pn+1 (x))



(6) (1 − x2 )Pn (x) = (n + 1)[xPn (x) − Pn+1 (x)]
(1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)] ——(a)
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
from equation (a),



(6) (1 − x2 )Pn (x) = (n + 1)[xPn (x) − Pn+1 (x)]
(1 − x2 )Pn (x) = n[Pn−1 (x) − xPn (x)] ——(a)
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)
from equation (a),
(1 − x2 )Pn (x) = (n + 1)[xPn (x) − Pn+1 (x)]


Legendre Function

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Legendre Function