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SPECIAL FUNCTIONSUNIT – 12: LEGENDRE POLYNOMIAL AND FUNCTIONS
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh BishtAcademic Consultant
Department of Mathematics
Uttarakhand Open University, Haldwani
CONTENTS
1. LEGENDRE’S EQUATION
1.1. LEGENDRE’S POLYNOMIAL Pn(x)
1.2. LEGENDRE’S FUNCTION OF THE SECOND KIND i.e. Qn(x)
1.3 GENERAL SOLUTION OF LEGENDRE’S EQUATION
2. RODRIGUE’S FORMULA
3. A GENERATING FUNCTION OF LEGENDRE’S POLYNOMIAL
4. ORTHOGONALITY OF LEGENDRE POLYNOMIALS
5. RECURRENCE FORMULAE
6. NUMERICAL PROBLEMS
7. LEGENDRE’S POLYNOMIALS APPLICATIONS
8. NUMERICAL PROBLEMS
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
1. LEGENDRE’S EQUATION
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Now,
1 2
0 1 2
0
0
2
2
( ...) ...(2)
dyso that ( )
dx
d y (
dx
m
m r
r
r
m r l
r
r
r
y x a a x a x
y a x
a m r
and a m
− −
−
=
− −
=
= + + +
=
= −
= −
2
0
)( ) m r
r
r m r l x
− −
=
− −
Substituting these values in (1), we have
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
2 2
0 0 0
2
0
(1- ) ( )( ) 2 ( ) ( 1) 0
( )( ) ( 1) 2( ) ( )( ) 0
( )(
m r m r l m r
r r r
r r r
m r m r
r
r
x a m r m r l x x a m r x n n a x
m r m r l x n n m r m r m r l x a
m r
− − − − −
= = =
− − −
=
− − − − − + + =
− − − + + − − − − − − =
−
2
0
) ( 1) ( ) ( ) 0 ...(3)m r m r
r
r
m r l x n n m r m r l x a
− − −
=
− − + + − − − − − =
The equation (3) is an identity and therefore coefficients of various powers of x
must variable. Now equating to zero the coefficient of xm i.e. by substituting r = 0
in the second summation we get,
a0 {n(n + 1) – m (m + 1)} = 0
But a0 ≠ 0, as it is the coefficient of the very first term in the series
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Here, n(n + 1) – m (m + 1) = 0 …(4)
n2 + n – m2 – m = 0 ⇒ (n2 – m2) + (n – m) = 0
i.e., (n – m)(n + m + 1) = 0, This is the indicial equation.
⇒ which gives m = n or m = - n – 1 …(5)
Next equating to zero the coefficient of xm – 1 by putting r = 1, in the second
summation a1[n(n + 1) – (m – 1)m] = 0
⇒ a1(n2 + n – m2 + m) = 0 ⇒ a1[(n2 – m2) + n + m] = 0
⇒ a1((m + n) (m – n – l)] = 0
which gives a1 = 0 …(6)
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Since (m + n)(m – n – l) ≠ 0, by (5)
Again to find a relation in successive coefficients ar, etc., equating the
coefficient of xm – r - 2 to zero, we get
(m – r)(m – r – l) ar + [n(n + 1) – (m – r – 2)(m – r – l)] ar + 2 = 0 …(7)
Now, n(n + 1) – (m – r – 2)(m – r – l) = n2 + n – (m – r – l – l)(m – r – l)
= − 𝑚 − 𝑟 − 𝑙 2 − 𝑚 − 𝑟 − 𝑙 − 𝑛2 − 𝑛
= − 𝑚 − 𝑟 − 𝑙 + 𝑛 𝑚 − 𝑟 − ! − 𝑛 − (𝑚 − 𝑟 − 𝑙 + 𝑛)
= − 𝑚 − 𝑟 − 𝑙 + 𝑛 𝑚 − 𝑟 − 𝑙 − 𝑛 − 𝑙
= − 𝑚 − 𝑟 + 𝑛 − 𝑙 𝑚 − 𝑟 − 𝑛 − 2
On simplification (7) becomes
⇒ (m – r)(m – r – l) ar – (m – r + n – l)(m – r – n – 2) ar + 2 = 0
⇒ 𝑎𝑟+2 = 𝑚−𝑟 𝑚−𝑟−𝑙
𝑚−𝑟+𝑛−𝑙 𝑚−𝑟−𝑛−2 𝑎𝑟 …(8)
Now since a1 = a3 = a5 = a1 = … = 0
For the two values given by (5) there arises following two cases.
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Case I: When m = n
𝑎𝑟+2 = − 𝑛−𝑟 𝑛−𝑟−𝑙
2𝑛−𝑟−𝑙 𝑟+2 𝑎𝑟 [From(8)]
If r = 0 𝑎2 = −𝑛 𝑛−𝑙
2𝑛−𝑙 2𝑎0
If r = 2, 𝑎4 = − 𝑛−2 𝑛−3
2𝑛−3 ×4𝑎2 =
𝑛 𝑛−1 𝑛−2 𝑛−3
2𝑛−1 2𝑛−3 2.4𝑎0
and so on and a1 = a3 = a5 = … = 0
Hence the series (2) becomes
𝑦 = 𝑎0 𝑥𝑛 𝑛 𝑛−1
2𝑛−1 .2𝑥𝑛−2 +
𝑛 𝑛−1 𝑛−2 (𝑛−3)
2𝑛−1 2𝑛−3 2.4. 𝑥𝑛−2 − ⋯
Which is a solution of (1).
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Case II: When m = − (n + 1), we have
𝑎𝑟+2 = − 𝑛+𝑟+1 𝑛+𝑟+2
𝑟+2 2𝑛+𝑟+3 𝑎𝑟 [From(8)]
If r = 0, 𝑎2 = − 𝑛+1 𝑛+2
2 2𝑛+3 𝑎0;
If r = 2, 𝑎4 = − 𝑛+3 𝑛+4
4. 2𝑛+5 𝑎2 =
𝑛+1 𝑛+2 𝑛+3 𝑛+4
2.4 2𝑛+3 2𝑛+5 𝑎0 and so on.
Hence the series (2) in this case becomes
𝑦 = 𝑎0 𝑥−𝑛−1 + 𝑛+1 𝑛+2
2. 2𝑛+3 𝑥−𝑛−3 +
𝑛+1 𝑛+2 𝑛+3 𝑛+4
2.4 2𝑛+3 2𝑛+5 . 𝑥−𝑛−5 + + ⋯ …(9)
This gives another solution of (1) in a series of descending powers of x.
Note. If we want to integrate the Legendre’s equation in a series of ascending
powers of x, we any proceed by taking
4 1 1 2
0 1 2
0
...k k r
r
r
y a x a x a x a x
+ + +
=
= + + + =
But integration in descending powers of x is more important than that in ascending
powers of x.
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
1.1. LEGENDRE’S POLYNOMIAL Pn(x)
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
1.2. LEGENDRE’S FUNCTION OF THE SECOND KIND i.e. Qn(x)
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
1.3 GENERAL SOLUTION OF LEGENDRE’S EQUATION
Since Pn(x) and Qn(x) are two independent solution of Legendre’s equation,
Therefore the most general solution of Legendre’s equation is
y = APn(x) + Qn(x)
Where A and B are two arbitrary constants
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
2. RODRIGUE’S FORMULA
𝑃𝑛 𝑥 = 1
2𝑛 .𝑛 !
𝑑𝑛
𝑑𝑥 𝑛 𝑥2 − 1 𝜋
Proof. Let v = (x2 – 1)n …(1)
Then 𝑑𝑣
𝑑𝑥= 𝑛 𝑥2 − 1 𝑛−1 2𝑥
Multiplying both sides by (x2 – 1), we get
𝑥2 − 1 𝑑𝑣
𝑑𝑥= 2𝑛 𝑥2 − 1 𝑛𝑥.
⇒ 𝑥2 − 1 𝑑𝑣
𝑑𝑥= 2𝑛𝑣𝑥 [Using (1)] …(2)
Now differentiating (2), (n + 1) times Leibnitz’s theorem, we have
𝑥2 − 1 𝑑𝑛+2𝑣
𝑑𝑥 𝑛+2+ 𝑛+1 𝐶1 2𝑥
𝑑𝑛+1𝑣
𝑑𝑥 𝑛+1+ 𝑛+1 𝐶2 2
𝑑𝑛𝑣
𝑑𝑥 𝑛= 2𝑛 𝑥
𝑑𝑛+1𝑣
𝑑𝑥 𝑛+1+ 𝑛+1 𝐶1 𝑙
𝑑𝑛𝑣
𝑑𝑥 𝑛
⇒ 𝑥2 − 1 𝑑𝑛+2𝑣
𝑑𝑥 𝑛+2+ 2𝑥 𝐶1 − 𝑛
𝑛+1
𝑑𝑛+1𝑣
𝑑𝑥 𝑛+1+ 2 𝐶2 − 𝑛. 𝑛+1 𝑛+1
𝐶1 𝑑𝑛𝑣
𝑑𝑥 𝑛= 0
⇒ 𝑥2 − 1 𝑑𝑛+2𝑣
𝑑𝑥 𝑛+2+ 2𝑥
𝑑𝑛+1𝑣
𝑑𝑥 𝑛+1− 𝑛 𝑛 + 1
𝑑𝑛𝑣
𝑑𝑥 𝑛= 0 …(3)
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
If we put 𝑑𝑛𝑣
𝑑𝑥 𝑛= 𝑦, (3) becomes
𝑥2 − 1 𝑑2𝑦
𝑑𝑥 2+ 2𝑥
𝑑𝑦
𝑑𝑥− 𝑛 𝑛 + 1 𝑦 = 0
⇒ 1 − 𝑥2 𝑑2𝑦
𝑑𝑥 2− 2𝑥
𝑑𝑦
𝑑𝑥+ 𝑛 𝑛 + 1 𝑦 = 0
This shows that 𝑦 =𝑑𝑛𝑣
𝑑𝑥 𝑛 is a solution of Legendre’s equation.
∴ 𝐶𝑑𝑛𝑣
𝑑𝑥 𝑛= 𝑃𝑛 𝑥 …(4)
Where C is a constant.
But v = (x2 – 1)n = (x + 1)n (x – 1)n
so that 𝑑𝑛𝑣
𝑑𝑥 𝑛= 𝑥 + 1 𝑛 𝑑𝑛
𝑑𝑥 𝑛 𝑥 − 1 𝑛 + 𝐶1. 𝑛 𝑥 + 1 𝑛 𝑛−1 𝑑𝑛−1
𝑑𝑥 𝑛−1 𝑥 − 1 𝑛 + ⋯ +
𝑥 − 1 𝑛 𝑑𝑛
𝑑𝑥 𝑛 𝑥 + 1 𝑛 = 0
when x = 1, then 𝑑𝑛𝑣
𝑑𝑥 𝑛= 2𝑛 . 𝑛!
All the other terms disappear as (x – 1) is a factor in every term except first.
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Therefore when x = 1, (4) gives
C.2n.n! = Pn(1) = 1 [Pn(1) = 1]
𝐶 =1
2𝑛 .𝑛 ! …(5)
Substituting the value of C from (5) in (4), we have
𝑃𝑛 𝑥 =1
2𝑛 .𝑛 ! 𝑑𝑛𝑣
𝑑𝑥 𝑛
𝑃𝑛 𝑥 =1
2𝑛 .𝑛 !
𝑑𝑛
𝑑𝑥 𝑛 𝑥2 − 1 𝑛 ∵ 𝑣 = 𝑥2 − 1 𝑛
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
3. A GENERATING FUNCTION OF LEGENDRE’S POLYNOMIAL
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Thus coefficient of zn in the expansion of (1) is sum of (2), (3) and (4) etc.
=1.3.5.. 2𝑛−1
𝑛 ! 𝑥𝑛 −
𝑛 𝑛−1
2 2𝑛−1 . 𝑥𝑛−2 +
𝑛 𝑛+1 𝑛−2 𝑛−3
2.4 2𝑛−1 2𝑛−3 𝑥𝑛−4 − ⋯ = 𝑃𝑛(𝑥)
Thus coefficient of z, z2, z3 … etc. in (1) are P1(x), P2(x), P3(x) …
Hence
(1 – 2xz + z2)-1/2 = P0(x) + zP1(x) + z2P2(x) + z3P3(x) + … + zn Pn(x) + …
i.e.,
( )
1/ 22
0
1 2 ( ). Proved.n
n
n
n
xz z P x z=
−
=
− + =
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
5. RECURRENCE FORMULAE
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
6. NUMERICAL PROBLEMS
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
7. LEGENDRE’S POLYNOMIALS APPLICATIONS
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Similarly 𝑃3 𝑥 =1
2 5𝑥3 − 3𝑥
𝑃4 𝑥 =1
8 35𝑥4 − 30𝑥4 + 3
𝑃5 𝑥 =1
8 63𝑥5 − 70𝑥3 + 15𝑥
……………………………………………….
2
0
( 1) (2 2 )!( )
2 . !( )!( 2 )!
rnn r
n nr
n rP x x
r n r n r
−
=
− −=
− −
where 𝑁 =𝑛
2 if n is even
𝑁 =1
2 𝑛 − 1 if n is odd
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Note. We can evaluate Pn(x) by differentiating (x2 – 1)n, n times.
( )
2 2 2 2
0 0
2 2 2
0
2
0
!( 1) ( ) ( 1) ( 1)
!( )!
1 1 !( ) ( 1) ( 1)
2 . ! 2 . ! !( )!
( 1) (2 2 )!
!( )!( 2 )!
n r n r nn n n r r r n r
rnr r
n r nn n n r
n n n nr
rNn r
r
d nx C x x
dx r n r
d nP x x x
n dx n r n r
n rx
r n r n r
= =− −
= =
=−
=
−
=
− = − = −−
= − = −−
− −=
− −
Either x0 or x1 is in the last term.
∴ n – 2r = 0 or 𝑟 =𝑛
2 (n is even)
N – 2r = 1 or 𝑟 =1
2 𝑛 − 1 (n is odd)
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
8. NUMERICAL PROBLEMS
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Example 6. Prove that Pn(1) = 1.
Solution. We know that
(1 – 2xz + z2)-1/2 = 1 + zP1(x) + z2 P2(x) + z3 P3(x) + … + zn Pn(x) + …
Substituting 1 for x in the above equation, we get
(1 – 2 z + z2)-1/2 = 1 + zP1(1) + z2 P2(1) + z3 P3(1) + … + zn Pn(1) + …
( )
( )
1/ 2 12
0
1 2 3
(1 ) (1) 1 (1)
(1) 1 1 ... ...
n n
n n
n
n n
n
z z P z z P
z P z z z z z
− −
=
−
− = − =
= − = + + + + + +
Equating the coefficient of zn on both sides, we get
Pn(1) = 1
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Example 7. Prove that
0
1( ) .
2 2n
n
P xx
=
=−
Solution. We know that
1
2 2
0
(1 2 ) ( ) ...(1)n
n
n
xz z z P x−
=
− + =
Putting z = 1 in (1), we get
( )1
2
0
0
1 2 1 ( )
1( ) Proved.
2 2
n
n
n
n
x P x
P xx
−
=
=
− + =
=−
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Example 8. Prove that:
1
0
1 1( ) log
1 2 1
n
n
n
x xP x
n x
+
=
+ =
+ −
Solution. We know that
( )
1/ 22
0
( ) 1 2n
n
n
h P x xh h
−
=
= − +
Integrating both sides w.r.t. h from 0 to h, we get
( ) ( )
1
2 2 20 0 0
( ) ; 1 Here x is constant h is variable.1 1 2 1
h hn
n
n
h dh dhP x if x
n hx h h x x
+
=
= = + − + − + −
= log ℎ−𝑥 + ℎ2−2ℎ𝑥+1
1−𝑥
𝑑ℎ
ℎ2+𝑎2= log
ℎ+ ℎ2+𝑎2
𝑎
Putting h = x in the expression, we get
1 2
0
1 1 1( ) log log Proved.
1 1 2 1
n
n
n
x x xP x
n x x
+
=
− + = = + − −
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Example 9. Show that
Pn(-x) = (-1)n Pn (x) and Pn(-1) = (-1)n.
Solution. We know that
( )
12 2
0
1 2 ( ) ...(1)n
n
n
xz z z P x
−
=
− + =
Putting –x for x in both sides of (1), we get
( )
12 2
0
1 2 ( ) ...(2)n
n
n
xz z z P x
−
=
+ + = −
Again putting –z for z in (1), we obtain
( )
12 2
0
1 2 ( 1) ( ) ...(3)n n
n
n
xz z z P x
−
=
+ + = −
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
From (2) and (3), we have
0 0
( ) ( 1) ( )n n n
n n
n n
z P x z P x
= =
− = −
Comparing the coefficients of zn from both sides of (4), we obtain …(4)
Pn(-x) = (-1)n Pn(x)
Pn(-1) = (-1)n Pn(1) = (1)(-1)n …(5)
Putting x = 1 in (5), we get [Pn(1)n1]
(i) If n is even, then from (5)
Pn(-x) = Pn (x),
So, Pn(x), is even function of x.
(ii) If n is odd, then from (5)
Pn(-x) = -Pn(x), so Pn(x) is odd function.
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Example 12. Show that
( )
3/ 22
0
1(2 1) .
1 2
in
n
n
zn P z
xz z
=
−= +
− +
Solution. We know that
( )
12 2
0
1 2 ( ) ...(1)n
n
n
xz z z P x
=
− + =
Differentiating both sides of (1) with respect to z, we get
( )3/ 2
2 1
0
1
2 3/ 20
11 2 ( 2 2 ) . ( )
2
( ) ...(2)(1 2 )
n
n
n
n
n
n
xz z x z nz P x
x znz P x
xz z
−
−
=
−
=
− − + − + =
−=
− +
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Multiplying both sides of (2) by 2z, we get
21
2 3/ 20
2 22 ( ) ...(3)
(1 2 )
n
n
n
xz znz P x
xz z
−
=
−=
− +
2 2
20
2
2 3/ 20
1 2 2 2 (2 1) ( )
(1 2 )
1 (2 1) ' Proved.
(1 2 )
n
n
n
n
n
n
xz z xz zn z P x
xz z
zn z P
xz z
=
=
− + + − +
− +
− +
− +
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Example 13. Prove that
( )
12
0
1 1( + )
1 2
n
n n
n
zP P z
zz xz z
+
=
−− =
− +
Solution.
1 1
0 0 0
. . . ( + ) +n n n
n n n n
n n n
R H S P P z z P z P
+ +
= = =
= =
1
1
0 0
1 ...(1)n n
n n
n n
z P z Pz
+
+
= =
+
2 3
0 1 2 3
0
1 2 3
1 1 2 3
0
But ...
And ...
n
n
n
n
n
n
z P P zP z P z P
z P zP z P z P
=
+
+
=
= + + + +
= + + +
= -P0 + P0 + zP1 + z2P2 + z3P3 + … = -P0 + 𝑧𝑛𝑃𝑛
⇒ log1+sin
𝜃
2
sin𝜃
2
= 1 +1
2𝑃1 cos 𝜃 +
1
3𝑃2 cos 𝜃 +
1
4𝑃3 cos 𝜃 + …
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Example 17. Assuming that a polynomial f(x) of degree n can be written as
0
( ) ( ).x
m m
m
f x C P x=
=
show that 𝐶𝑚 =2𝑚+1
2 𝑓 𝑥 𝑃𝑚 𝑥 𝑑𝑥
1
1
Solution. We have,
0
( ) ( )m m
m
f x C P x
=
=
= C0P0(x) + C1P1(x) + C2P2(x) + C3P3(x) + C4P4(x) + … + CmPn(x)+…+
Multiplying both sides by Pm(x), we get
Pm(x)f(x)=C0P0(x) Pm(x) + C1P1(x) Pm(x) + C2P2(x) Pm(x) + … + Cm𝑃𝑚2 (x) + …
𝑓(𝑥)+1
−1 Pm(x) dx = [
+1
−1C0P0(x) Pm(x) + C1P1(x) Pm(x) + C2P2(x) Pm(x) + …
Cm𝑃𝑚2 (x) + …]
= 0 + 0 + ⋯ + 𝐶𝑚2
2𝑚+1+ ⋯ =
2𝐶𝑚
2𝑚+1
𝐶𝑚2𝑚+1
2 𝑓 𝑥 𝑃𝑚 𝑥 𝑑𝑥
+1
−1
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875
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THANKS
Dr. Kamlesh Bisht(Mathematics)
Mob. No.- 8279829875