orthogonality of Legendre polynomials

orthogonality of Legendre polynomials

We start from the

first order

differential equation

(

1

-

x

2

)

⁢

d

⁢

u

d

⁢

x

+

2

⁢

n

⁢

x

⁢

u

=

0

,

(1)

where one can separate the variables (

http://planetmath.org/SeparationOfVariables

) and then get the

general solution

u

=

C

⁢

(

1

-

x

2

)

n

.

(2)

Differentiating

n

+

1

times the equation (1) it takes the form

(

1

-

x

2

)

⁢

d

n

+

2

⁢

u

d

⁢

x

n

+

2

-

2

⁢

x

⁢

d

n

+

1

⁢

u

d

⁢

x

n

+

1

+

n

⁢

(

n

+

1

)

⁢

d

n

⁢

u

d

⁢

x

n

=

0

or

(

1

-

x

2

)

⁢

d

2

⁢

y

d

⁢

x

2

-

2

⁢

x

⁢

d

⁢

y

d

⁢

x

+

n

⁢

(

n

+

1

)

⁢

y

=

0

(3)

where

y

=

d

n

⁢

u

d

⁢

x

n

=

C

⁢

d

n

d

⁢

x

n

⁢

(

1

-

x

2

)

n

.

Especially, the particular solution

y

=

P

n

⁢

(

x

)

:=

1

2

n

⁢

n

!

⁢

d

n

d

⁢

x

n

⁢

(

1

-

x

2

)

n

,

(4)

which which is the

Legendre polynomial

of degree

n

, has been seen to satisfy the Legendre’s differential equation (3).

The equality (4) is

Rodrigues formula

(

http://planetmath.org/RodriguesFormula

).  We use it to find the

leading coefficient

of

P

n

⁢

(

x

)

and to show the orthogonality (

http://planetmath.org/OrthogonalPolynomials

) of the Legendre polynomials

P

0

⁢

(

x

)

,

P

1

⁢

(

x

)

,

P

2

⁢

(

x

)

,

…

0.1

The coefficient of

x

n

By the

binomial theorem

,

P

n

⁢

(

x

)

=

1

2

n

⁢

n

!

⁢

d

n

d

⁢

x

n

⁢

∑

j

=

0

n

(

n

j

)

⁢

x

2

⁢

(

n

-

j

)

⁢

(

-

1

)

j

=

1

2

n

⁢

n

!

⁢

∑

j

=

0

n

(

n

j

)

⁢

(

2

⁢

n

-

2

⁢

j

)

⁢

(

2

⁢

n

-

2

⁢

j

-

1

)

⁢

⋯

⁢

(

2

⁢

n

-

2

⁢

j

-

n

+

1

)

⁢

x

n

-

2

⁢

j

⁢

(

-

1

)

j

.

From the term with

j

=

0

we get as the coefficient of

x

n

the following:

1

2

n

⁢

n

!

⁢

(

n

0

)

⁢

(

2

⁢

n

)

⁢

(

2

⁢

n

-

1

)

⁢

(

2

⁢

n

-

2

)

⁢

⋯

⁢

(

2

⁢

n

-

n

+

1

)

⁢

(

-

1

)

0

=

1

2

n

⁢

n

!

⋅

(

2

⁢

n

)

!

(

2

⁢

n

-

n

)

!

=

(

2

⁢

n

)

!

2

n

⁢

(

n

!

)

2

(5)

0.2

Orthogonality

Let

f

m

⁢

(

x

)

:=

a

0

+

a

1

⁢

x

+

…

+

a

m

⁢

x

m

be any

polynomial

of degree

m

<

n

.  Integrating by parts (

http://planetmath.org/IntegrationByParts

)

m

times we obtain

∫

-

1

1

f

m

⁢

(

x

)

⁢

P

n

⁢

(

x

)

⁢

𝑑

x

=

1

2

n

⁢

n

!

⁢

∫

-

1

1

f

m

⁢

(

x

)

⁢

d

n

d

⁢

x

n

⁢

(

x

2

-

1

)

n

⁢

𝑑

x

=

1

2

n

⁢

n

!

⁢

/

-

1

1

⁡

f

m

⁢

(

x

)

⁢

d

n

-

1

d

⁢

x

n

-

1

⁢

(

x

2

-

1

)

n

-

1

2

n

⁢

n

!

⁢

∫

-

1

1

f

m

′

⁢

(

x

)

⁢

d

n

-

1

d

⁢

x

n

-

1

⁢

(

x

2

-

1

)

n

⁢

𝑑

x

⋯

⋯

=

(

-

1

)

m

⁢

a

m

⁢

m

!

2

n

⁢

n

!

⁢

∫

-

1

1

d

n

-

m

d

⁢

x

n

-

m

⁢

(

x

2

-

1

)

n

⁢

𝑑

x

=

(

-

1

)

m

⁢

a

m

⁢

m

!

2

n

⁢

n

!

⁢

/

-

1

1

⁡

f

m

⁢

(

x

)

⁢

d

n

-

m

-

1

d

⁢

x

n

-

m

-

1

⁢

(

x

2

-

1

)

n

=

0

,

since

x

=

±

1

are zeros of the

derivatives

d

n

-

k

d

⁢

x

n

-

k

⁢

(

x

2

-

1

)

n

.

If, on the other hand,

m

=

n

,  the calculation gives firstly

∫

-

1

1

f

n

⁢

(

x

)

⁢

P

n

⁢

(

x

)

⁢

𝑑

x

=

2

⁢

(

-

1

)

n

⁢

a

n

2

n

⁢

∫

0

1

(

x

2

-

1

)

n

⁢

𝑑

x

=

2

⁢

(

-

1

)

n

⁢

a

n

2

n

⋅

I

n

,

(6)

where the

integral

I

n

is gotten from

I

n

=

/

0

1

⁡

x

⁢

(

x

2

-

1

)

n

-

2

⁢

n

⁢

∫

0

1

x

2

⁢

(

x

2

-

1

)

n

-

1

⁢

𝑑

x

=

-

2

⁢

n

⁢

∫

0

1

[

(

x

2

-

1

)

n

+

(

x

2

-

1

)

n

-

1

]

⁢

𝑑

x

=

-

2

⁢

n

⁢

I

n

-

2

⁢

n

⁢

I

n

-

1

,

Thus we infer the recurrence relation

I

n

=

-

2

⁢

n

2

⁢

n

+

1

⁢

I

n

-

1

.

Using this and

I

0

=

1

one easily arrives at

I

n

=

(

-

1

)

n

⁢

2

⋅

4

⋅

6

⁢

⋯

⁢

(

2

⁢

n

)

3

⋅

5

⋅

7

⁢

⋯

⁢

(

2

⁢

n

+

1

)

=

(

-

1

)

n

⁢

[

2

⋅

4

⋅

6

⁢

⋯

⁢

(

2

⁢

n

)

]

2

(

2

⁢

n

+

1

)

!

=

(

-

1

)

n

⁢

2

2

⁢

n

⁢

(

n

!

)

2

(

2

⁢

n

+

1

)

!

.

(7)

If

f

n

⁢

(

x

)

also is a Legendre polynomial

P

n

⁢

(

x

)

, we can in (6) by (5) put

a

n

=

(

2

⁢

n

)

!

2

n

⁢

(

n

!

)

2

and taking into account (7), too, (6) reads

∫

-

1

1

[

P

n

⁢

(

x

)

]

2

⁢

𝑑

x

=

(

-

1

)

n

2

n

-

1

⋅

(

2

⁢

n

)

!

2

n

⁢

(

n

!

)

2

⋅

(

-

1

)

n

⁢

2

2

⁢

n

⁢

(

n

!

)

2

(

2

⁢

n

+

1

)

!

=

2

2

⁢

n

+

1

.

Our results imply the orthonormality (

http://planetmath.org/Orthonormal

) condition

∫

-

1

1

P

m

⁢

(

x

)

⁢

P

n

⁢

(

x

)

⁢

𝑑

x

=

2

2

⁢

n

+

1

⁢

δ

m

⁢

n

,

(8)

where

δ

m

⁢

n

is the

Kronecker delta

.

References

1

K. Kurki-Suonio:

Matemaattiset apuneuvot

.  Limes r.y., Helsinki (1966).

Title

orthogonality of Legendre polynomials

Canonical name

OrthogonalityOfLegendrePolynomials

Date of creation

2013-03-22 18:55:30

Last modified on

2013-03-22 18:55:30

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Derivation

Classification

msc 33C45

Related topic

OrthogonalPolynomials

Related topic

OrthogonalityOfChebyshevPolynomials

Related topic

SubstitutionNotation

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