legendre polynomials

Some Special Functions of Some Special Functions of Mathematical Physics

1 Manoj Kumar Pandey, Dept. of Mathematics

Special Functions

� Legendre Polynomials

� Bessel Functions and Gamma Function

2 Manoj Kumar Pandey, Dept. of Mathematics

Legendre Polynomials� The Legendre Equation is

wheren is a non negative integer.

We have found solution of this equation near the ordinary pointx = 0.

,0)1(2)1( 2 =++′−′′− ynnyxyx

Manoj Kumar Pandey, Dept. of Mathematics3

which is valid for -1<x< 1.

y(x) = a0

1− n(n +1)2!

x + n(n − 2)(n +1)(n + 3)3!

x4 +⋯

+ a1

x − (n −1)(n + 2)3!

x3 + (n −1)(n − 3)(n + 2)(n + 4)5!

x5 +⋯

� Howeverthe solutions most useful in the applications arethose bounded near x =1.


� Here we will solve the Legendre’s equation nearx =1, interms of hypergeometric functionF(a, b, c, x).

We change the independent variable fromx to t by

In terms oft the equation becomes

(1 ) (1 2 ) ( 1) 0 .....(*)t t y t y n n y′′ ′− + − + + =

1(1 )

2t x= −


Heret = 0 corresponds tox = 1 of the Legendre equation.(*) is a hypergeometric equation with

⇒ a = -n, b = n+1, c = 1

(1 ) (1 2 ) ( 1) 0 .....(*)t t y t y n n y′′ ′− + − + + =

1, 1 2, ( 1)c a b ab n n= + + = = − +

Therefore, the solution of (*) near t = 0 is

( ) ( , 1,1, )y t F n n t= − +

Since c=1 which implies the exponentsm1= 0 andm2 = 1- c = 0henceonly onesolutionis possibleat thismoment.


henceonly onesolutionis possibleat thismoment.

To find a second linearly independent solution we take

� y2 = vy1, where

−=

−=

∫=∫=′ −−−−

)1(

11

)1(

1

11

22

)1(/)12(

21

21

tyttty

ey

ey

vdttttdxP


� Sincey12 is a polynomial with constant term 1, the bracketed

expression on the right is an analytic function of the form

−− )1()1( 2

12

1 tyttty

⋯+++ 2211 tata

We can write

This yields

Hence the general solution of (*) near t=0 is

⋯+++=′ taat

v 21

1

⋯++= tatv 1log )(log 112 ⋯++= tatyy⇒


The solution will bebounded near t =0 if and only if c2 = 0.

.2211 ycycy +=

If we replacet by (½)(1-x), we obtain the solution of Legendre’s

equation nearx=1.

The solution of the Legendre’s equation which is bounded near


x = 1 are precisely constant multiplies of the polynomial

F(-n, n+1, 1, ½(1-x)).

Legendre Polynomials� Thenth Legendre polynomialdenoted byPn(x) is defined by

Pn(x) = F −n,n +1,1,

1− x

2

(−n)(n +1) 1− x (−n)(−n +1)(n +1)(n + 2) 1− x 2


=1+ (−n)(n +1)

(1!)2

1− x

2

+

(−n)(−n +1)(n +1)(n + 2)

(2!)2

1− x

2

+⋯

+ (−n)(−n +1)⋯[−n + (n −1)](n +1)(n + 2)⋯(2n)

(n!)2

1− x

2

n

Which gives

Pn(x) =1+ n(n +1)

(1!)22(x −1)+ n(n −1)(n +1)(n + 2)

(2!)222(x −1)2 +⋯

+ (2n)!

(n!)22n(x −1)n.


with Pn(1)=1.

This is not a suitable form and very inconvenient tool. So we

look for a simple form.

+(n!)22n

(x −1) .

Rodrigues FormulaThe nth Legendre polynomial is a polynomial of degree n

satisfying the Legendre’s equation with Pn(1) =1.

This is given by the Rodrigues formula

21( ) ( 1) , 0,1,2,....

2 !

nn

n n n

dP x x n

n dx= − =


Remark: Any polynomial of degree n satisfying Legendre equation

with Pn(1) = 1 is called Legendre polynomial .

( ) ( 1) , 0,1,2,....2 !n n n

P x x nn dx

= − =

2(1 ) 2 ( 1) 0x y xy n n y′′ ′− − + + =

� To show which is a polynomial of degreen, is a solution of the Legendre equation.

� Leibnitz’s rule for higher order derivative

( )2( ) 1n

n

n

dy x x

dx= −


� Leibnitz’s rule for higher order derivative

( ) ( )

( ) ( ) ( 1) (1) ( 1) (1) (1) ( 1) ( )

0

( 1)( )

2!n n n n n n

nn r r

r

n nuv u v nu v u v nu v uv

nu v

r

− − −

−

=

−= + + + + +

=

∑

⋯⋯

Legendre Polynomials

� Rodrigues formula provides a

relatively easy method for computing the successive Legendre

polynomials.

21( ) ( 1)

2 !

nn

n n n

dP x x

n dx= −


� Some of the first few Legendre Polynomials

( ) ( )

0 1

2 32 3

4 2 5 34 5

( ) 1, ( ) ,

1 1( ) (3 1), ( ) (5 3 ),

2 21 1

( ) 35 30 3 , ( ) 63 70 15 ,....8 8

P x P x x

P x x P x x x

P x x x P x x x x

= =

= − = −

= − + = − +

Properties of Legendre PolynomialsOrthogonalityOrthogonality

The most important property of the Legendre polynomials

is

0 1( ), ( ), , ( ),nP x P x P x⋯ ⋯


is

that is Legendre polynomials forms an orthogonal sequence offunctions in the interval [-1, 1].

∫−

=+

≠=

1

1 .12

2,0

)()(nmif

n

nmifdxxPxP nm

ProofProof:: Let f(x) be any function with at leastn continuousderivatives on the interval , and consider the integral

By Rodriguesformula

11 ≤≤− x1

1( ) ( )nI f x P x dx

−= ∫


By Rodriguesformula

after integration by parts, we get

1 2

1

1( ) ( 1)

2 !

nn

n n

dI f x x dx

n dx−= −∫

11 112 21 11

1

1 1( ) ( 1) ( ) ( 1)

2 ! 2 !

n nn n

n n n n

d dI f x x f x x dx

n dx n dx

− −

− −−−

′= − − −

∫0

Therefore, we get

continuing to integrate by parts, we obtain

11 2

11

( 1)( ) ( 1)

2 !

nn

n n

dI f x x dx

n dx

−

−−

− ′= −∫

1 ( ) 2( 1)( )( 1)

nn nI f x x dx

−= −∫


If f(x) = Pm(x) with m < n, thenf(n)(x) = 0 and consequentlyI = 0.

which proves the first part i.e. the case whenm ≠ n (????)

For the second part lets takef(x) = Pn(x).

1 ( ) 2

1

( 1)( )( 1)

2 !n n

nI f x x dx

n −

−= −∫

Since we get,!2

)!2()()(

n

nxP

n

nn =

)!2(2

)1()!(2

)!2(

1

1

1

222

n

dxxn

nI n

n

∫

∫ −=−


If we change the variable by , and recall the formula

We conclude that in this case , hence the proof.

.)1()!(2

)!2(2 1

0

222

dxxn

n nn ∫ −=

θsin=x

.)12()!2(

)!(2cos

222/

0

12

+=∫

+

nn

nd

nnπ

θθ

)12/(2 += nI

Every polynomial of degree k can be expressed as a linear

combination of first k+1 Legendre polynomials:

That is

0 1( ) , ( ) , , ( )kP x P x P x⋯


.)()(0∑

=

=k

nnn xPaxp

For example let

( )0 11 , ( ),P x x P x= =

p(x)=1+2x+3x2+5x3


2 22 2 0 2

3 33 3 1 3

1 1 2 1 2( ) (3 1) ( ) ( ) ( ) ,

2 3 3 3 31 3 2 3 2

( ) (5 3 ) ( ) ( ) ( )2 5 5 5 5

P x x x P x P x P x

P x x x x x P x P x P x

= − ⇒ = + = +

= − ⇒ = + = +

Therefore

0 1 0 2 1 3

0 1 2 3

3

1 2 3 2( ) ( ) 2 ( ) 3 ( ) ( ) 5 ( ) ( )

3 3 5 5

2 ( ) 5 ( ) 2 ( ) 2 ( )

( ).

p x P x P x P x P x P x P x

P x P x P x P x

a P x

= + + + + +

= + + +

=∑


Therefore, any polynomialp(x) of degreek can be written as

0

( ).n nn

a P x=

=∑

.)()(0∑

=

=k

nnn xPaxp

Legendre SeriesWhat about an arbitrary function?

An arbitrary functionf(x) can be expressed asLegendre series

Need to calculate coefficientsan for the above expression.0

( ) ( )n nn

f x a P x∞

=

=∑


n

Multiply Pm(x) on both sides and integrate from –1 to 1, we get

Using orthogonality of Legendre polynomials

1 1

1 10

( ) ( ) ( )m n m nn

f x P dx a P x P x dx∞

− −=

=∑∫ ∫

1

1

2( )

2 1m

m

af x P dx

m−=

+∫

Therefore, we have

.)(2

1 1

1dxPxfna nn ∫−

+=


Problem: Find the first three terms of the Legendre Series of

0 1 0( ) ( )

0 1.

if xi f x

x if x

− ≤ <= ≤ ≤

Let f(x) be a function defined in [-1, 1], now consider the problem

of approximating f(x) as closely as possible in the sense of least

square by polynomials p(x) of degree ≤ n.1 2

1[ ( ) ( )] .I f x p x dx

−= −∫

Least Square Approximation


The Problem is to minimize I:

It turns out that the minimizing polynomial is:

With

0 0 1 1( ) ( ) ( ) ( )n np x a P x a P x a P x= + + +…

.)(2

1 1

1dxPxfna nn ∫−

+=

is called the generating function of the Legendre polynomials.

20

1( )

1 2

nn

n

P x txt t

∞

=

=− +

∑

Generating Function and Recursion Formula


Recursion Formula :

1 1( 1) ( ) (2 1) ( ) ( ), 1,2,....n n nn P x n xP x nP x n+ −+ = + − =

Show that

Recursion Formula

2 1

2

( ) (1) 1, ( ) ( 1) ( 1) ( ) (0) 0,

1 3 (2 1)( ) (0) ( 1) .

2 !

nn n n

nn n

i P ii P iii P

niv P

n

+= − = − =⋅ − = − ⋅

⋯


2( ) (0) ( 1) .2 !n n

iv Pn

= − ⋅

� If p(x) is a polynomial of degree such that1

1≥n


Show thatp(x) = c Pn(x) for some constant c.

1

1( ) 0 for 0,1, , 1.kx p x dx k n

−= = −∫ ⋯

legendre polynomials

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