Orthogonal Polynomial

1. Orthogonal Polynomial
Dr. Varun Kumar
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 1 / 9

2. Outlines
1 Introduction
2 Inner product and orthogonal polynomial
3 Orthogonal properties of Legendre’s polynomial
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 2 / 9

3. Introduction:
⇒ An orthogonal polynomial sequence is a family of polynomials such
that any two different polynomials in the sequence are orthogonal to
each other under some inner product.
⇒ The most widely used orthogonal polynomials are the classical
orthogonal polynomials, consisting of the
Hermite polynomials,
Laguerre polynomials
Jacobi polynomials
Chebyshev polynomials
Legendre polynomials
⇒ Developed in the late 19th century from a study of continued
fractions by P. L. Chebyshev and was pursued by A. A. Markov
and T. J. Stieltjes.
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 3 / 9

4. Inner product and orthogonal polynomial
Property of Inner product
1 (a, b) = (b, a)
2 (a, b + c) = (a, b) + (a, c)
3 (a, a) > 0 ∀ a 6= 0
4 (~
a, ~
b) = |~
a|.|~
b| cos θ
Orthogonal polynomial
(p, q) =
Z 1
−1
p(x)q(x)dx (1)
Example:
(A) Let p(x) = 1 and q(x) = 1
(1, 1) =
Z 1
−1
1 × 1dx = 2 → Non-orthogonal (2)
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 4 / 9

5. Continued–
(B) Let p(x) = 1 and q(x) = x
(1, x) =
Z 1
−1
1 × xdx = 0 → Orthogonal (3)
(C) Let p(x) = x and q(x) = x
(x, x) =
Z 1
−1
x2
dx =
2
3
→ Non-Orthogonal (4)
(D) Let p(x) = x and q(x) = x2
(x, x) =
Z 1
−1
x3
dx = 0 → Orthogonal (5)
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 5 / 9

6. Orthogonal properties of Legendre’s polynomial
Prove that Z 1
−1
pm(x)pn(x)dx = 0, m 6= n
=
2
2n + 1
m = n
(6)
Proof:
Case 1: when m 6= n By Legendre’s DE
(1 − x2)d2y
dx2 − 2x dy
dx + n(n + 1)y = 0
or
d
dx
(1 − x2
)
dy
dx
+ n(n + 1)y = 0 (7)
Let pm(x) and pn(x) are solution of (7)
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 6 / 9

7. Continued–
Here, y = pm(x) or pn(x)
d
dx
(1 − x2
)
dpn
dx
+ n(n + 1)pn = 0 (8)
and
d
dx
(1 − x2
)
dpm
dx
+ m(m + 1)pm = 0 (9)
Multiply (8) by pm(x) and (9) by pn(x) and subtracting them
pm
d
dx
(1−x2
)
dpn
dx
−pn
d
dx
(1−x2
)
dpm
dx
+ n(n+1)−m(m+1)
pmpn = 0
(10)
Integrating (10) between -1 to 1
Z 1
−1
pm
d
dx
(1 − x2
)
dpn
dx
dx −
Z 1
−1
pn
d
dx
(1 − x2
)
dpm
dx
dx
+ n(n + 1) − m(m + 1)
Z 1
−1
pm(x)pn(x)dx = 0
(11)
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 7 / 9

8. Applying integration by parts in integral (11)
pm
(1 − x2
)
dpn
dx

11. 1
−1
−
Z 1
−1
dpm
dx
(1 − x2
)
dpn
dx
dx −
h
pn
(1 − x2
)
dpm
dx

14. 1
−1
−
Z 1
−1
dpn
dx
(1 − x2
)
dpx
dx
dx
i
+ n(n + 1) − m(m + 1)
Z 1
−1
pm(x)pn(x)dx = 0
(12)
⇒ n(n + 1) − m(m + 1)
R 1
−1
pm(x)pn(x)dx = 0
⇒ Since, m 6= n Z 1
−1
pm(x)pn(x)dx = 0 m 6= n (13)
Case 2: As per Legendre’s relation
(1 − 2xh + h2
)−1/2
=
∞
X
n=0
hn
pn(x) = p0(x) + hp1(x) + ... (14)
Squaring on both side
(1 − 2xh + h2
)−1
=
∞
X
n=0
h2n
p2
n(x) +
∞
X
m,n=0,m6=n
2hm+n
pm(x)pn(x) (15)
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 8 / 9

15. Integrating (15) between -1 to 1
Z 1
−1
1
(1 − 2xh + h2)
dx =
∞
X
n=0
h2n
Z 1
−1
p2
n(x)dx +
∞
X
m,n=0,m6=n
2hm+n
Z 1
−1
pm(x)pn(x)dx (16)
or ∞
X
n=0
h2n
Z 1
−1
p2
n(x)dx = − log(1 − 2xh + h2
)

18. 1
−1
=
1
h
h
log(1 + h) − log(1 − h)
i
(17)
Here, log(1 + h) = h − h2
2 + h3
3 − .... and log(1 − h) = −h − h2
2 − h3
3 − .....
From (17)
∞
X
n=0
h2n
Z 1
−1
p2
n(x)dx =
2
h
h
h +
h3
3
+ ....
i
=
∞
X
n=0
h2n
2n + 1
(18)
Hence, Z 1
−1
p2
n(x)dx =
1
2n + 1
(19)
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-5 9 / 9