Dr. Kamlesh Bisht - Uttarakhand Open University · 2020. 5. 12. · M 0, as it is the coefficient...
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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
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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
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1. LEGENDRE’S EQUATION
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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
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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
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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)
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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.
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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).
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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.
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1.1. LEGENDRE’S POLYNOMIAL Pn(x)
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1.2. LEGENDRE’S FUNCTION OF THE SECOND KIND i.e. Qn(x)
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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
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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)
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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.
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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
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3. A GENERATING FUNCTION OF LEGENDRE’S POLYNOMIAL
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Mob. No.- 8279829875
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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=
−
=
− + =
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5. RECURRENCE FORMULAE
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6. NUMERICAL PROBLEMS
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7. LEGENDRE’S POLYNOMIALS APPLICATIONS
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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
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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)
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8. NUMERICAL PROBLEMS
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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
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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
−
=
=
− + =
=−
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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
+
=
− + = = + − −
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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
−
=
+ + = −
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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.
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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
−
−
=
−
=
− − + − + =
−=
− +
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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
=
=
− + + − +
− +
− +
− +
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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 + …
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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
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Mob. No.- 8279829875