Computational Physics - Department of Physics

120 5 Numerical Integration

ForL 1 (x)we have the general expression

L 1 (x) =a+bx,

and using the orthogonality relation
∫ 1
− 1

L 0 (x)L 1 (x)dx= 0 ,

we obtaina= 0 and with the conditionL 1 ( 1 ) = 1 , we obtainb= 1 , yielding

L 1 (x) =x.

We can proceed in a similar fashion in order to determine the coefficients ofL 2

L 2 (x) =a+bx+cx^2 ,

using the orthogonality relations
∫ 1
− 1

L 0 (x)L 2 (x)dx= 0 ,

and ∫ 1

− 1

L 1 (x)L 2 (x)dx= 0 ,

and the conditionL 2 ( 1 ) = 1 we would get

L 2 (x) =^1

2

(

3 x^2 − 1

)

. (5.12)

We note that we have three equations to determine the three coefficientsa,bandc.

Alternatively, we could have employed the recursion relation of Eq. (5.11), resulting in

2 L 2 (x) = 3 xL 1 (x)−L 0 ,

which leads to Eq. (5.12).
The orthogonality relation above is important in our discussion on how to obtain the
weights and mesh points. Suppose we have an arbitrary polynomialQN− 1 of orderN− 1 and a
Legendre polynomialLN(x)of orderN. We could representQN− 1 by the Legendre polynomials
through

QN− 1 (x) =

N− 1

∑

k= 0

αkLk(x), (5.13)

whereαk’s are constants.
Using the orthogonality relation of Eq. (5.10) we see that
∫ 1
− 1

LN(x)QN− 1 (x)dx=

N− 1

∑

k= 0

∫ 1

− 1

LN(x)αkLk(x)dx= 0. (5.14)

We will use this result in our construction of mesh points andweights in the next subsection.
In summary, the first few Legendre polynomials are

L 0 (x) = 1 ,

L 1 (x) =x,

L 2 (x) = ( 3 x^2 − 1 )/ 2 ,