Assignment V
11 Construct orthogonal polynomial of degree 0, 1, 2 on the interval (0, 1) with the weight function
w(x) = − ln x
Ans: p0 (x) = 1 and p1 (x) = x − a1 where
R ∞ −2t
R1
te dt
< xp0 , p0 >
1 Γ(2)
1
0 − ln(x) x dx
a1 =
= R0∞ −t
=
= R1
= ,
< p0 , p0 >
4
Γ(2)
4
te
dt
− ln(x) dx
0
0
where we have used x = e−t and Γ(p) =
R∞
0
e−t tp−1 dt and Γ(p + 1) = pΓ(p) and Γ(1) = 1.
Hence p1 (x) = x − 1/4. Now p2 (x) = (x − a2 )p1 (x) − b2 p0 (x), where
R1
− ln(x) x(x − 1/4)2 dx
< xp1 , p1 >
Γ(2)/16 − Γ(2)/18 + Γ(2)/64
13/576
13
a2 =
= R0 1
=
=
= ,
< p1 , p1 >
Γ(2)/9 − Γ(2)/8 + Γ(2)/16
7/144
28
− ln(x)(x − 1/4)2 dx
0
< xp1 , p0 >
b2 =
=
< p0 , p0 >
R1
0
− ln(x) x(x − 1/4) dx
Γ(2)/9 − Γ(2)/16
7/144
=
=
= 7/144,
R1
Γ(2)
1
0 − ln(x) dx
Hence p2 (x) = (x − 13/28)(x − 1/4) − 7/144 = x2 − 5x/7 + 17/252
14 The Lagurre polynomials are orthogonal in the interval (0, ∞) with respect to the weight function
w(x) = e−x . Find the first four Lagurre polynomials such that the coefficient of the leading term
of φk (x) is (−1)k /k!.
Ans: p0 (x) = 1 and p1 (x) = x − a1 where
R ∞ −x
xe dx
Γ(2)
< xp0 , p0 >
= R0 ∞ −x
=
= 1,
a1 =
< p0 , p0 >
Γ(1)
e
dx
0
Hence p1 (x) = x − 1. Now p2 (x) = (x − a2 )p1 (x) − b2 p0 (x), where
R∞
x(x − 1)2 e−x dx
Γ(4) − 2Γ(3) + Γ(2)
< xp1 , p1 >
=
a2 =
= 0 R∞
=3
2 dx
< p1 , p1 >
Γ(3)
− 2Γ(2) + Γ(1)
(x
−
1)
0
R∞
x(x − 1)e−x dx
Γ(3) − Γ(2)
< xp1 , p0 >
b2 =
= 0 R ∞ −x
=
=1
< p0 , p0 >
Γ(1)
dx
0 e
Hence p2 (x) = (x − 3)(x − 1) − 1 = x2 − 4x + 2. Similarly, we find p3 (x) = x3 − 9x2 + 18x − 6.
Now the Lagurre polynomials Lk (x) have leading coefficient (−1)k /k!. Hence,
L0 (x) = 1, L1 (x) = −x + 1, L2 (x) = (x2 − 4x + 2)/2, L3 (x) = (−x3 + 9x2 − 18x + 6)/6
15 The Hermite polynomials are orthogonal in the interval (−∞, ∞) with respect to the weight
2
function w(x) = e−x . Find the first four Hermite polynomials such that the coefficient of the
leading term of φk (x) is 2k .
Ans: p0 (x) = 1 and p1 (x) = x − a1 where
R∞
2
xe−x dx
< xp0 , p0 >
= R−∞
= 0.
a1 =
∞ −x2
< p0 , p0 >
dx
−∞ e
Hence p1 (x) = x. Now p2 (x) = (x − a2 )p1 (x) − b2 p0 (x), where
R ∞ 3 −x2
x e
dx
< xp1 , p1 >
a2 =
= R−∞
= 0.
∞
2
2 −x dx
< p1 , p1 >
−∞ x e
R ∞ 2 −x2
R ∞ 3/2−1 −t
dx
t
e dt
Γ(3/2)
< xp1 , p0 >
1
−∞ x e
b2 =
= R ∞ −x2
=
= R0∞ 1/2−1 −t =
< p0 , p0 >
Γ(1/2)
2
e dt
dx
0 t
−∞ e
Hence p2 (x) = x2 − 1/2. Similarly, we find p3 (x) = x3 − 3x/2.
Now the Hermite polynomials Hk (x) have leading coefficient 2k . Hence,
H0 (x) = 1, H1 (x) = 2x, H2 (x) = 4x2 − 2, H3 (x) = 8x3 − 12x.