The document discusses orthogonal polynomials, focusing on Legendre and Chebyshev polynomials. It introduces Hilbert spaces and self-adjoint operators, describing properties like Hermitian matrices having real eigenvalues. It defines the L2 space and shows how differential operators can act as self-adjoint. Legendre polynomials are defined using Rodrigue's formula and their generating function is explored. The first few Legendre polynomials are shown.

1. Orthogonal Polynomials
Indre Skripkauskaite
F10GP Project
supervised by
Dr. M. Dreher
March 31, 2016
1

2. 2 CONTENTS
Contents
1 Introduction 3
2 Hilbert Spaces and Self-Adjoint Operators 4
2.1 Hermitian Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 The Space L2[a, b] and Differential Operators . . . . . . . . . . . . . . . . . . . . 6
3 Legendre Polynomials 10
3.1 Generating Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.2 Recurrence Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.3 The Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.4 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4 Chebyshev’s Polynomials 20
4.1 Generating Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
4.2 The Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.3 Recurrence Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.4 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5 Conclusion 29

3. 3
1 Introduction
In mathematics, a set of polynomials is said to be orthogonal under some inner product if any
two of the polynomials from the given set are orthogonal, i.e. their scalar product equals zero.
There are quite a few families of Orthogonal polynomials, but in this project we will be focusing
only on Legendre and Chebyshev’s polynomials.
Legendre polynomials are widely used in physics and Chebyshev’s polynomials are applicable
in finance. However, by the use of our knowledge of various topics from Complex Analysis,
Functional Analysis, Linear Algebra as well as Calculus, we will be focusing only on the mathe-
matical, i.e. the theoretical part and understanding the behavior of such families of polynomials
in respective L2 Hilbert spaces, which is the aim of this project.

4. 4 2 HILBERT SPACES AND SELF-ADJOINT OPERATORS
2 Hilbert Spaces and Self-Adjoint Operators
In this section we will be discussing properties of Hilbert spaces and Self-Adjoint Operators,
where the understanding of Hermitian which we will be recalling constantly in later chapters.
We will also mention the famous Fourier Series, which plays an important role in orthogonality
of functions.
2.1 Hermitian Matrices
Definition 2.1.1. Let A be an n×n matrix and let AT be its transpose. A is said to be a
Hermitian matrix [2] if it is equal to its conjugate transpose. This means
AT = A.
and we write
AH
= A.
Definition 2.1.2. Let u, v ∈ Cn and k ∈ N. We define a scalar product in C as
u, v =
n
k=1
ukvk. (2.1)
Proposition 2.1.3. Let u, v ∈ Cn and let A = AH. Then A is said to be self-adjoint if we can
write Au, v = u, Av .
PROOF:
Assume A = AH. We claim that Au, v = u, Av , for all u, v ∈ Cn.
Then we have
Au, v =
n
k=1
(Au)kvk
=
n
k=1


n
j=1
akjuj

 vk
=
n
j=1
uj
n
k=1
akjvk
=
n
j=1
AHvj

5. 2.1 Hermitian Matrices 5
Therefore if A = AH, then
Au, v = u, Av . (2.2)
Proposition 2.1.4. Eigenvalues of a Hermitian matrix are real.
PROOF: Let A be a Hermitian matrix. Let λ be an eigenvalue of A and let u be an eigenvector
of A to the eigenvalue λ. Suppose λ ∈ C.
By (2.2) we have that
Au, u = u, Au ,
Au, u = λu, u = λ u, u = λ u 2
,
u, Au = u, λu = λ u, u = λ u 2
,
hence
λ u 2
= λ u 2
.
Note that u 2 > 0 since u = 0, therefore λ = λ ∈ R.
Theorem 2.1.5. Let A = AH, λ, µ be eigenvalues of A, λ = µ. Let u, v ∈ Cn be eigenvectors
of A to the eigenvalues λ, µ, hence
Au = λu
Av = µv,
then u, v = 0.
PROOF: Consider Au, v . From (2.2) we already know that

6. 6 2 HILBERT SPACES AND SELF-ADJOINT OPERATORS
Au, v = u, Av
λu, v = u, µv
λ u, v = µ u, v
= µ u, v
λ u, v = µ u, v
(λ − µ) u, v = 0
u, v = 0
since λ = µ.
Hence the eigenvectors of distinct eigenvalues of a hermitian matrix are orthogonal.
Theorem 2.1.6 (Spectral Theorem). [1] Let A be a Hermitian matrix and let V be a finite
dimensional inner product space. Then there exists an orthonormal basis of V consisting of
eigenvectors of A.
2.2 The Space L2
[a, b] and Differential Operators
Definition 2.2.1. Let a, b ∈ R and let f(x),g(x) be two functions in L2[a, b]. Then the scalar
product in L2[a, b] [2] is defined as
f, g =
b
a
f(x)g(x)dx. (2.3)
We can also set A to act as a differential operator.
Example 2.2.2. Let A be a differential operator A := d2
dx2 and gn(x) := sin(nx),
gn(x) ∈ L2[−π, π], n ∈ N, then we get Agn = −n2gn.
By the use of (2.3) and integration by parts

7. 2.2 The Space L2[a, b] and Differential Operators 7
Agn, gm =
π
−π
gn(x)gm(x)dx
= −
π
−π
gn(x)gm(x)dx + gn(x)gm(x)
π
−π
=
π
−π
gn(x)gm(x)dx − gn(x)gm(x)
π
−π
= gn, Agm ,
Agn, gm = gn, Agm
(2.4)
for n, m ∈ N.
Remark: If compare this result to the Proposition 2.1.3 in the previous section, we can see
that in this example A acts as a differential operator rather than being a Hermitian matrix and
gn, gm are being treated as eigenfunctions rather than being eigenvectors of A.
Continuing from 2.4 we have that
−n2
gn, gm = gn, −m2
gm ,
n2
gn, gm = m2
gn, gm ,
(n2
− m2
) gn, gm = 0,
gn, gm = 0
(2.5)
for n = m.
Definition 2.2.3. The set of functions {f1(x), f2(x), . . .} is orthogonal in L2[a, b] [2] if:
fn(x), fm(x) = 0 (2.6)
for n = m.
Proposition 2.2.4. The set {sin(nx), cos(mx)}, n ∈ N+, m ∈ N0} is an orthogonal set of
functions [2] in the Hilbert space L2[−π, π].
PROOF:
By use of (2.3) we can see that

8. 8 2 HILBERT SPACES AND SELF-ADJOINT OPERATORS
1, cos(nx) =
π
π
cos(nx)dx =
sin(nx)
n
π
−π
= 0,
1, sin(nx) =
π
π
sin(nx)dx = −
cos(nx)
n
π
−π
= 0.
(2.7)
By (2.5)
sin(nx), sin(mx) = 0 (2.8)
cos(nx), cos(mx) = 0 for n = m.
Note that cos(x) is an even function and sin(x) is an odd function.
Hence
cos(nx), sin(mx) =
π
−π
cos(nx) sin(mx)dx = 0
Theorem 2.2.5. Let n ∈ N and let f(t) be 2L-periodic function in Hilbert space L2[−L, L].
Then the Fourier Series expansion [4] of f(t) is represented as
f(t) = a0 +
∞
n=1
an cos
nπt
L
+
∞
n=1
bn sin
nπt
L
, (2.9)
where
an =
1
L
L
−L
f(t) cos
nπt
L
dt,
bn =
1
L
L
−L
f(t) sin
nπt
L
dt
a0 =
1
2L
L
−L
f(t)dt.
Remark: We can make a comparison of the above to the Theorem 2.1.5, i.e. we can say that
cos nπt
L , sin nπt
L are the eigen functions to the eigen values an, bn, a0 in the Hilbert space
L2[−L, L]. Although, we do not claim that we can span full L2[−L, L] space, since it requires
deeper analysis, so we will not be focusing on that in this project.

9. 2.2 The Space L2[a, b] and Differential Operators 9
Example 2.2.6 (Heaviside step function).
f(t) =



−1 for −π < t < 0
1 for 0 < t < π.
(2.10)
Our function f(t) is 2π periodic and it can be approximated by the use of (2.9) as
f(t) =
4c
π
sin(t) +
4c
3π
sin(3t) +
4c
5π
sin(5t) + · · · +
4c
(2n + 1)π
sin((2n + 1)t) + . . . (2.11)
n ∈ N.
We can now compare the original function with the approximated result where the blue line
indicates the Heaviside step function f(t) and the red line indicated the Fourier approximation
with n = 6.
Figure 1: Fourier series approximation of the Heaviside step function (2.10) with n = 6

10. 10 3 LEGENDRE POLYNOMIALS
3 Legendre Polynomials
3.1 Generating Function
Legendre Polynomials were introduced by Legendre in the theory of potential, where they are
related to the expansion of the reciprocal of the distance 1
R , where R is the distance between two
points r and r .
Figure 2:
R = |r − r | = (r2
+ r 2
− 2rr cos θ)
1
2 ,
where θ is the angle between r and r . If we let t = r
r , x = cos(θ) we have that
1
R
=
1
r
(1 − 2xt + t2
)−1
2 (3.1)
for −1 x 1. If we rewrite
(1 − 2xt + t2
)−1
2
as
(t − x − x2 − 1)−1
2 (t − x + x2 − 1)−1
2 (3.2)
and treat it as a function of t, we see that (3.2) has two singularities at
x ± x2 − 1,
for −1 x 1. Therefore, if |t| ≤ min |x ±
√
x2 − 1|, we have the following Taylor Series
expansion:
(1 − 2xt + t2
)−1
2 =
∞
n=0
Pn(x)tn
.
t ∈ C, is is called the generating function [3] for some function Pn(x) which will be discussed
now.

11. 3.1 Generating Function 11
Definition 3.1.1. The Legendre polynomials are defined by Rodrigue’s formula [5]
Pn(x) =
1
2nn!
dn
dxn
(x2
− 1)n
, n ∈ N, (3.3)
for arbitrary values of x.
Later we will see that these Pn’s are the same as in section (3.3).
To obtain the general expression for the nth Legendre polynomial we will use the binomial
expression
(x2
− 1)n
=
n
k=0
(−1kn!)
k!(n − k)!
x2n−2k
. (3.4)
Substituting (3.4) into (3.3) implies
Pn(x) =
[n/2]
k=0
(−1)k(2n − 2k)!
2nk!(n − k!)(n − 2k)!
xn−2k
. (3.5)
By the use of (3.5) we can see that the first eleven Legendre polynomials [8] are:
P0(x) = 1,
P1(x) = x,
P2(x) =
1
2
(3x2
− 1),
P3(x) =
1
2
(5x3
− 3x),
P4(x) =
1
8
(35x4
− 30x2
+ 3),
P5(x) =
1
8
(63x5
− 70x3
− 5),
P6(x) =
1
16
(231x6
− 315x4
+ 105x2
− 5),
P7(x) =
1
16
(429x7
− 693x5
+ 315x3
− 35x),
P8(x) =
1
128
(6435x8
− 12012x6
+ 6930x4
− 1260x2
+ 35),
P9(x) =
1
128
(12155x9
− 25740x7
+ 18018x5
− 4620x3
+ 315x),
P10(x) =
1
256
(46189x10
− 109395x8
+ 90090x6
− 30030x4
+ 3465x2
− 63).

12. 12 3 LEGENDRE POLYNOMIALS
Figure 3: Legendre polynomials of degree 0 through 10
We can also approach the generating function via Cauchy Integral Formula in the following way:
Theorem 3.1.2 (Cauchy Integral Formula [5]). Let f(z) be analytic in a simply connected
domain D. Let C be a closed contour going in the counter-clockwise direction inside D, and let
z be an interior point of D. Then
1
2πi C
f(ζ)
ζ − z
dζ = f(z).
n!
2πi C
f(ζ)
(ζ − z)n+1
dζ =
dn
dzn
f(z) (3.6)
Proposition 3.1.3. For n ∈ N, t ∈ C, |x| 1, x ∈ R, (1 − 2xt + t2) = 0, hence |t| < 1
3,
g(x, t) = (1 − 2xt + t2
)−1
2 (3.7)
is the generating function for Legendre polynomials.
PROOF: We wish to show that
∞
n=0
Pn(x)tn
= g(x, t). (3.8)

13. 3.1 Generating Function 13
Replace Pn(x) by the Rodrigues formulas and use the Cauchy integral theorem (3.6) we obtain
Pn(x) =
(−1)n
2nn!
dn
dxn
(1 − x2
)n
=
(−1)n
2n
1
2πi C
(1 − z2)n
(z − x)n+1
dz, (3.9)
where C is any curve enclosing x, going in a counter-clockwise direction. Now inserting (3.9)
into (3.8) we get
∞
n=0
Pn(x)tn
=
∞
n=0
(−1)n
2n
1
2πi
tn
C
(1 − z2)n
(z − x)n+1
dz. (3.10)
Now interchanging summation and integration in (3.10) we get
∞
n=0
Pn(x)tn
=
1
2πi C
dz
z − x
∞
n=0
(−1)n
2n
tn (1 − z2)
z − x
n
. (3.11)
The resulting series is a simple geometric series, which is readily summed. We then obtain the
following integral:
−
1
2πi C
2
t
dz
z2 − 2
t z − 1 − 2
t x
. (3.12)
Now we evaluate this integral by using the familiar residue integration technique. We can clearly
see that the denominator of (3.12) has two roots:
z1 =
1
t
+
1
t2
−
2
t
x + 1 =
1
t
+
1
t
1 − 2xt + t2
and
z2 =
1
t
−
1
t2
−
2
t
x + 1 =
1
t
−
1
t
1 − 2xt + t2.
We can rewrite z2 as
z2 =
1
t2 − 1
t2 x + 1
1
t + 1
t2 − 2
t x + 1
=
2
t x − 1
1
t + 1
t2 − 2
t x + 1
(3.13)
.
Now if we multiply both numerator and denominator of (3.13) by t we obtain the following:
z2 =
2x − t
1 +
√
1 − 2xt + t2
≈ x (3.14)

14. 14 3 LEGENDRE POLYNOMIALS
for t ≈ 0.
We now choose C to be a path surrounding the point x and z2 and by applying Residue Theorem
we see that
−
2
t
lim
z→z2
z − 1
t − 1
t2 − 2
t x + 1
z2 − 2
t z − 1 − 2
t x
=
1
√
1 − 2xt + t2
.
Consequently,
∞
n=0
Pn(x)tn
=
1
√
1 − 2xt + t2
. (3.15)
3.2 Recurrence Relation
Proposition 3.2.1. Legendre Polynomials satisfy the Recurrence Relation [5]
(n + 1)Pn+1(x) − (2n + 1)xPn(x) + nPn−1(x) = 0 (3.16)
for n ∈ N+.
PROOF:
We are going to show it by first differentiating the generating function (3.7) with respect to t:
∂g
∂t
= −
1
2
(1 + t2
− 2xt)−3
2 (2t − 2x) =
x − t
(1 + t2 − 2xt)
3
2
=
∞
n=0
Pn(x)tn−1
(3.17)
Multiply (3.17) by (1 + t2 − 2xt) to obtain
(1 + t2
− 2xt)
∞
n=0
nPn(x)tn−1
= (x − t)(1 + t2
− 2xt)−1
2 = (x − t)
∞
n=0
Pn(x)tn
,
(1 + t2
− 2xt)
∞
n=0
nPn(x)tn−1
= (x − t)
∞
n=0
Pn(x)tn
,
(1 + t2
− 2xt)
∞
n=0
nPn(x)tn−1
− (x − t)
∞
n=0
Pn(x)tn
= 0.
Setting the coefficient of tn equal to zero, we find that
(n + 1)Pn+1(x) − 2nxPn(x) + (n − 1)Pn−1(x) − xPn(x) = 0,

15. 3.3 The Differential Equation 15
or equivalently
(n + 1)Pn+1(x) − (2n + 1)xPn(x) + nPn−1(x) = 0, n ∈ N.
Legendre polynomials can be calculated step by step, starting from P0(x) = 1, P1(x) = x.
3.3 The Differential Equation
Proposition 3.3.1. Legendre Polynomials solve the differential equation [7]
((1 − x2
)Pn(x)) + n(n + 1)Pn(x) = 0, n ∈ N0. (3.18)
PROOF:
By the use of the generating function (3.15)
∂
∂x
:
∞
n=0
Pn(x)tn
= −
1
2
−2t
(1 − 2xt + t2)
3
2
=
t
(1 − 2xt + t2)
3
2
∂
∂t
:
∞
n=0
nPn(x)tn−1
= −
1
2
−2x + 2t
(1 − 2xt + t2)
3
2
=
x − t
(1 − 2xt + t2)
3
2
(3.19)
∂2
∂t2
:
∞
n=0
n(n − 1)Pn(x)tn−2
= −
1
(1 − 2xt + t2)
3
2
+ (x − t) −
3
2
−2x + 2t
(1 − 2xt + t2)
5
2
= −
1
(1 − 2xt + t2)
3
2
+
3(x − t)2
(1 − 2xt + t2)
5
2
(3.20)
(1 − x2
)
∂
∂x
:
∞
n=0
(1 − x2
)Pn(x)tn
=
t(1 − x2)
1 − 2xt + t2)
3
2
(1 − x2
)
∂
∂x
:
∞
n=0
(1 − x2
)Pn tn
=
−2xt
(1 − 2xt + t2)
3
2
+ t(1 − x2
) −
3
2
−2t
(1 − 2xt + t2)
5
2
= −
2xt
(1 − 2xt + t2)
3
2
+
3t2(1 − x2)
(1 − 2xt + t2)
5
2
(3.21)
Now we multiply (3.19) by t and obtain
∞
n=0
nPn(x)tn
=
t(x − t)
(1 − 2xt + t2)
3
2
(3.22)

16. 16 3 LEGENDRE POLYNOMIALS
If we multiply (3.20) by t2 we get
∞
n=0
n(n − 1)Pn(x)tn
= −
t2
(1 − 2xt + t2)
3
2
+
3t2(x − t)2
(1 − 2xt + t2)
5
2
(3.23)
Now if we multiply (3.22) by 2 an add it to (3.23) we obtain the following:
∞
n=0
2nPn(x)tn
+
∞
n=0
n(n − 1)Pn(x)tn
=
2t(x − t)
(1 − 2xt + t2)
3
2
−
t2
(1 − 2xt + t2)
3
2
+
3t2(x − t)2
(1 − 2xt + t2)
5
2
∞
n=0
(2n + n(n − 1))Pn(x)tn
=
2xt − 3t2
(1 − 2xt + t2)
3
2
+
3t2(x − t)2
(1 − 2xt + t2)
5
2
∞
n=0
n(n + 1)Pn(x)tn
=
2xt − 3t2
(1 − 2xt + t2)
3
2
+
3t2(x − t)2
(1 − 2xt + t2)
5
2
(3.24)
Now by adding (3.21) to (3.24) we see that
∞
n=0
(1 − x2
)Pn(x) tn
+
∞
n=0
n(n + 1)Pn(x)tn
= −
2xt
(1 − 2xt + t2)
3
2
+
3t2(1 − x2)
(1 − 2xt + t2)
5
2
+
2xt − 3t2
(1 − 2xt + t2)
3
2
+
3t2(x − t)2
(1 − 2xt + t2)
5
2
∞
n=0
(1 − x2
)Pn(x) tn
+ n(n + 1)Pn(x)tn
=
−2xt + 2xt − 3t2
(1 − 2xt + t2)
3
2
+
3t2((1 − x2) + (x − t)2)
(1 − 2xt + t2)
5
2
=
−3t2
(1 − 2xt + t)
3
2
+
3t2(1 − 2xt + t2)
(1 − 2xt + t2)
5
2
=
−3t2 + 3t2
(1 − 2xt + t2)
3
2
= 0.
Hence
∞
n=0
(1 − x2
)Pn(x) tn
+ n(n + 1)Pn(x)tn
= 0.
Now setting the coefficients of tn equal to zero, we finally obtain
(1 − x2
)Pn(x) + n(n + 1)Pn(x) = 0, n ∈ N,
(3.25)
Remark: Let us compare the above result to Theorem 2.1.5, which states the following:
Au = λu (3.26)

17. 3.4 Orthogonality 17
for all u ∈ Cn.
We shal prove the following proposition first:
Proposition 3.3.2. A = d
dx (1 − x2) d
dx is a self-adjoint operator in Hilbert space L2[−1, 1],
i.e. Au, v = u, Av , for all twice differentiable functions u(x), v(x), x ∈ R.
PROOF: Using differentiation by parts we see that
Au, v =
1
−1
(Au) · v
dx
√
1 − x2
=
1
−1
1 − x2
d
dx
( 1 − x2
d
dx
u(x)) · v(x)
1
√
1 − x2
dx
=
1
−1
1 − x2u vdx
= 1 − x2u v
1
−1
−
1
−1
1 − x2u v dx
= 0 −
1
−1
u · 1 − x2v dx
= u 1 − x2v
1
−1
+
1
−1
u · 1 − x2v dx
= 0 +
1
−1
u · (Av)
1
√
1 − x2
dx
= u, Av ,
hence A is indeed self-adjoint
If we set λ = −n(n + 1), u = Pn(x) and treat A as a differential operator in L2, i.e.
A = d
dx (1 − x2) d
dx in (3.26), then comparing to Theorem 2.1.5 we can write (3.25) as
d
dx
(1 − x2
)
d
dx
Pn(x) = −n(n + 1)Pn(x), (3.27)
Consequently,
((1 − x2
)Pn(x)) = −n(n + 1)Pn(x) (3.28)
which is the self-adjoint form of Legendre Differential Equation.
3.4 Orthogonality
Proposition 3.4.1. Legendre polynomials are orthogonal in the Hilbert space L2[(−1, 1), dt].

18. 18 3 LEGENDRE POLYNOMIALS
PROOF:
We wish to show
1
−1 Pn(x)Pm(x)dx = 0 for n = m. If we multiply both sides by Pm(x) and take
the scalar product as defined in (2.3) of both sides we get
n(n + 1)
1
−1
Pn(x)Pm(x) = −
1
−1
Pm(x) (1 − x2
)Pn(x) dx
= − Pm(x) (1 − x2
)Pn(x)
1
−1
+
1
−1
Pm(x) (1 − x2
)Pn(x) dx
= −
1
−1
Pm(x)(1 − x2
) Pn(x)dx
= m(m + 1)
1
−1
Pm(x)Pn(x)dx
=
1
−1
Pm(x)Pn(x)dx = 0 (3.29)
whenever n = m.
Proposition 3.4.2. If n = m, then the scalar product or two Legendre polynomials satisfy
PROOF:
1
−1
P2
n(x)dx =
2
2n + 1
.
We substitute (3.15) into L2([−1, 1]) scalar product by the use of (2.3):
∞
n=0
Pn(x)tn
,
∞
m=0
Pm(x)tm
=
1
√
1 − 2xt + t2
,
1
√
1 − 2xt + t2
∞
n=0
∞
m=0
Pn(x), Pm(x) tn+m
=
1
−1
1
1 − 2xt + t2
dx.
By use of (3.34) we drop the terms where n = m and consider only the cases where n = m

19. 3.4 Orthogonality 19
obtaining the following:
∞
n=0
Pn(x), Pn(x) t2n
=
1
−1
1
1 − 2xt + t2
dx
∞
n=0
t2n
1
−1
P2
n(x)dx =
1
−1
1
1 − 2xt + t2
dx
= −
1
2t
ln(1 − 2xt + t2
)
1
−1
= −
1
2t
ln(1 − 2t + t2
) − ln(1 + 2t + t2
) .
Note that ln(t)α = α ln(t)
∞
n=0
t2n
1
−1
P2
n(x)dx = −
1
2t
ln(1 − t)2
− ln(1 + t)2
= −
1
t
(ln(1 − t) − ln(1 + t)) .
Note that ln(a) − ln(b) = ln(a
b )
∞
n=0
t2n
1
−1
P2
n(x)dx = −
1
t
ln
1 − t
1 + t
=
1
t
ln
1 − t
1 + t
−1
=
1
t
ln
1 + t
1 − t
.
Using Taylor expansion we obtain
ln
1 + t
1 − t
= 2 +
2t3
3
+
2t5
5
+
2t7
7
+ · · · +
2t2n+1
2n + 1
+ · · · =
∞
n=1
2t2n+1
2n + 1
(3.30)
and so
1
t
ln
1 + t
1 − t
= 2 +
2t2
3
+
2t4
5
+
2t6
7
+ · · · +
2t2n
2n + 1
+ · · · =
∞
n=1
2t2n
2n + 1
. (3.31)
Finally
∞
n=0
2t2n
2n + 1
=
∞
n=0
t2n
1
−1
P2
n(x)dx,

20. 20 4 CHEBYSHEV’S POLYNOMIALS
hence
1
−1
P2
n(x)dx =
2
2n + 1
Remark:
We have shown that Legendre polynomials are orthogonal, hence, linearly independent. We can
refer back to Theorem 2.1.6. and make a comparison. We can say that in the Hilbert space
L2[(−1, 1), dt], Legendre polynomials act as eigen functions. However, we do not claim that Pns
span full L2[(−1, 1), dt] space, because proving it would require much deeper understanding so
we will not be focusing on that in this project.
4 Chebyshev’s Polynomials
Definition 4.0.1. Chebyshev’s polynomials are defined as
Tn(x) = cos(n arccos(x)), n ∈ N+, −1 ≤ x ≤ 1. (4.1)
The first eleven Chebyshev’s polynomials [9] are:
T0(x) = 1
T1(x) = x
T2(x) = 2x2
− 1
T3(x) = 4x3
− 3x
T4(x) = 8x4
− 8x2
+ 1
T5(x) = 16x5
− 20x3
+ 5x
T6(x) = 32x6
− 48x4
+ 18x2
− 1
T7(x) = 64x7
− 112x5
+ 56x3
− 7x
T8(x) = 128x8
− 256x6
+ 160x4
− 32x2
+ 1
T9(x) = 256x9
− 576x7
+ 432x5
− 120x3
+ 9x
T10(x) = 512x1
0 − 1280x8
+ 1120x6
− 100x4
+ 50x2
− 1

21. 4.1 Generating Function 21
Figure 4: Chebyshev’s polynomials of degree 0 through 10
4.1 Generating Function
Proposition 4.1.1. Let x = cos(θ) and Tn(x) = cos(nθ). Then the generating function [5] for
Chebyshev’s polynomials is
1 − xt
1 − 2xt + t2
=
∞
n=0
Tn(x)tn
for t ∈ C, |t| 1, n ∈ N0.
PROOF:
Let |t| < 1, then
∞
n=0
Tn(x)tn
=
∞
n=0
cos(nθ)tn
. (4.2)
By the use of Euler’s formulas:
einθ
= cos nθ + i sin nθ (4.3)
e−inθ
= cos nθ − i sin nθ (4.4)
If we add (4.3) to (4.4) we get
einθ
+ e−inθ
= 2 cos(nθ) (4.5)

22. 22 4 CHEBYSHEV’S POLYNOMIALS
Now dividing both sides of (4.5) by 2, we can see that
∞
n=0
Tn(x)tn
=
1
2
∞
n=0
(teiθ
)n
+ (te−iθ
)n
=
1
2
∞
n=0
(teiθ
)n
+
∞
n=0
(te−iθ
)n
Note that ∞
n=0(teiθ)n = 1
1−teiθ and ∞
n=0(teiθ)n = 1
1−te−iθ , so
∞
n=0
Tn(x)tn
=
1
2
1
1 − teiθ
+
1
1 − te−iθ
=
1
2
1 − te−iθ + 1 − teiθ
(1 − teiθ)(1 − te−iθ)
=
1
2
2 − 2t(eiθ + e−iθ)
1 − (eiθ + eiθ) + t2
=
1 − t cos(θ)
1 − 2t cos(θ) + t2
.
And since x = cos(θ)
∞
n=0
Tn(x)tn
=
1 − xt
1 − 2xt + t2
. (4.6)
4.2 The Differential Equation
Proposition 4.2.1. Chebyshev’s polynomials solve Chebyshev’s differential equation [6]
(1 − x2
)Tn(x) − xTn(x) + n2
Tn(x) = 0, −1 ≤ x ≤ 1 (4.7)
PROOF:

23. 4.2 The Differential Equation 23
We differentiate the generating function (4.6) with respect to x and t:
∂
∂x
:
∞
n=0
Tn(x)tn
=
2t(1 − tx)
(1 − 2xt + t2)2
−
t
1 − 2xt + t2
=
t(1 − t2)
(1 − 2xt + t2)2
(4.8)
∂2
∂x2
:
∞
n=0
Tn (x)tn
=
4t2(1 − t2)
(1 − 2xt + t2)3
(4.9)
∂
∂t
:
∞
n=0
nTn(x)tn−1
= −
x
1 − 2xt + t2
−
(2t − 2x)(1 − xt)
(t2 − 2xt + 1)2
=
xt2 − 2t + x
(1 − 2xt + t2)2
(4.10)
∂2
∂t2
:
∞
n=0
n(n − 1)Tn(x)tn−2
=
2xt − 2
(t2 − 2xt + 1)2
−
2(2t − 2x)(xt2 − 2t + x)
(1 − 2xt + t2)3
= −
2(xt3 − 3t2 + 3xt − 2x2 + 1)
(t2 − 2xt + 1)3
(4.11)
∞
n=0
(1 − x2
)Tn (x)tn
=
4t2(1 − x2)(1 − t2)
(1 − 2xt + t2)3
(4.12)
∞
n=0
xTn(x)tn
=
xt(1 − t2)
(1 − 2xt + t2)2
. (4.13)
Multiplying (4.10) by t we get
∞
n=0
nTn(x)tn
=
t(xt2 − 2t + x)
(1 − 2xt + t2)2
, (4.14)
then multiplying (4.11) by t2
∞
n=0
n(n − 1)Tn(x)tn
= −
2t2(xt3 − 3t2 + 3xt − 2x2 + 1)
(1 − 2xt + t2)3
(4.15)
Now adding (4.14) to (4.15) we obtain the following:
∞
n=0
n2
Tn(x)tn
=
t(xt2 − 2t + x)
(1 − 2xt + t2)2
−
2t2(xt3 − 3t2 + 3xt − 2x2 + 1)
(1 − 2xt + t2)3
(4.16)

24. 24 4 CHEBYSHEV’S POLYNOMIALS
Adding (4.12), subtracting (4.14) and then adding (4.16) we see that
∞
n=0
(1 − x2
)Tn (x) − xTn(x) + n2
Tn(x) tn
=
4t2(1 − x2)(1 − t2)
(1 − 2xt + t2)3
(4.17)
−
xt(1 − t2)
(1 − 2xt + t2)2
+
t(xt2 − 2t + x)
(1 − 2xt + t2)2
−
2t2(xt3 − 3t2 + 3xt − 2x2 + 1)
(1 − 2xt + t2)3
Now writing everything under common denominator we obtain the following:
∞
n=0
(1 − x2
)Tn (x) − xTn(x) + n2
Tn(x) tn
=
4t2(1 − t2 − x2 + x2t2)
(1 − 2xt + t2)3
−
xt(1 − t2)(1 − 2xt + t2)
(1 − t2 + x2t2)3
(4.18)
+
t(xt2 − 2t + x)(1 − 2xt + t2)
(1 − 2xt + t2)3
−
2t2(xt3 − 3t2 + 3xt + 2x2 + 1)
(1 − 2xt + t2)3
= (4t2 − 4t4 − 4t2x2 + 4x2t4 − xt + 2x2t2xt3 − xt3 − 2x2t4 + xt5 + xt3 − 3x2t4 + xt5 − 2t2 + 4xt3 −
2t4 + tx − 2x2t2 + t3x − 2xt5 − t4 − 6xt3 + 4x2t2 − 2t2)/(1 − 2xt + 2x2 + 1)3
0/(1 − 2xt + 2x2 + 1) = 0
Hence
∞
n=0
(1 − x2
)Tn (x) − xTn(x) + n2
Tn(x) tn
= 0
Now setting the coefficients of tn equal to zero we obtain
(1 − x2
)Tn (x) − xTn(x) + n2
Tn(x) = 0
Proposition 4.2.2. Let u(x), v(x) be continuous and twice differentiable functions, x ∈ R. Let
A be the differential operator
1 − x2
d
dx
1 − x2
d
dx
.
Then
Au, v = u, Av ,
in L2 [−1, 1], dx√
1−x2
i.e. A is self-adjoint.

25. 4.2 The Differential Equation 25
PROOF:
Au, v =
1
−1
(Au(x)) · v(x)
dx
√
1 − x2
=
1
−1
1 − x2
d
dx
( 1 − x2
d
dx
u(x) · v(x)
dx
√
1 − x2
=
1
−1
1 − x2u (x) · v(x)dx
= 1 − x2u (x)v(x)
1
−1
−
1
−1
1 − x2u (x)v (x)dx
= 0 −
1
−1
u (x) · 1 − x2v (x) dx
= 0 +
1
−1
u(x) · (Av(x))
dx
√
1 − x2
= u, Av
i.e. A is self-adjoint.
Remark:
If we divide (4.7) by
√
1 − x2 we obtain the following:
1 − x2Tn (x) −
x
√
1 − x2
Tn(x) +
n2
√
1 − x2
Tn(x) = 0 (4.19)
Proposition 4.2.3. The expression
1 − x2Tn(x) +
n2
√
1 − x2
Tn(x) = 0 (4.20)
is the self-adjoint form of the Chebyshev differential equation.
PROOF:
If we multiply (4.20) by
√
1 − x2, apply A as in Proposition 4.2.2 and let u(x) = Tn(x)
√
1−x2
and
λ(x) = −n2, we obtain

26. 26 4 CHEBYSHEV’S POLYNOMIALS
Au = λu,
d
dx
1 − x2
d
dx
Tn(x) = −
n2
√
1 − x2
Tn(x)
1 − x2Tn(x) = −
n2
√
1 − x2
Tn(x), (4.21)
hence A is a self-adjoint operator and (4.21) is a self-adjoint form of Chebyshev’s differential
equation.
Remark: We compare the above to Theorem 2.1.5 and see that the same idea repeats
once again, but instead of having an eigenvector u, we have an eigenfunction u(x) and an
eigenvalue λ = −n2 ∈ R.
4.3 Recurrence Relation
Proposition 4.3.1. Chebyshev’s Polynomials satisfy the following recurrence relation [7]:
Tn+1(x) = 2xTn(x) − Tn−1(x)
PROOF:
Let’s introduce the notation θ = arccos(x).
Then (4.1) becomes Tn(θ(x)) = Tn(θ) = cos(nθ), where 0 θ 2π.
We observe that replacing n by n + 1
Tn+1(θ) = cos((n + 1)θ) = cos(nθ) cos(nθ) − sin(nθ) sin(nθ), (4.22)
Tn−1(θ) = cos((n − 1)θ) = cos(nθ) cos(θ) + sin(nθ) sin(θ). (4.23)
Now adding (4.22) to (4.23) we obtain the following
Tn+1 + Tn−1 = 2 cos(nθ)cos(θ) + sin(nθ) sin(θ),
Tn+1(θ) = 2 cos(nθ) cos(θ) − Tn−1(θ),
Tn+1(x) = 2xTn(x) − Tn−1(x),

27. 4.4 Orthogonality 27
or equivalently
Tn+1(x) = 2xTn(x) − Tn−1(x), (4.24)
which is the recurrence relation for Chebyshev’s Polynomials.
Observation: Now we know that Tn’s are indeed polynomials.
4.4 Orthogonality
Proposition 4.4.1. Chebyshev’s Polynomials are orthogonal in L2 [−1, 1], 1√
1−x2
.
PROOF:
We shall prove this by using (4.20) for
d
dx
1 − x2Tn +
n2
√
1 − x2
Tn = 0 (4.25)
d
dx
1 − x2Tm +
m2
√
1 − x2
Tm = 0 (4.26)
Multiplying (4.25) by Tm and (4.26) by Tn and then subtracting the results we obtain
d
dx
1 − x2Tn Tm −
d
dx
1 − x2Tm Tn +
n2 − m2
√
1 − x2
TmTn = 0
d
dx
1 − x2(TnTm − TmTn) −
n2 − m2
√
1 − x2
TmTn = 0 (4.27)
Now if we integrate (4.27) over the interval [−1, 1] with respect to x we get
1
−1
TmTn
√
1 − x2
dx =
√
1 − x2
n2 − m2
TnTm − TmTn
1
−1
(4.28)
1
−1
TmTn
√
1 − x2
dx = 0
for n = m.
We can clearly see that the value of (4.28) is zero if m = n = 0. But what happens
when m = n = 0 and m = n = 0?

28. 28 4 CHEBYSHEV’S POLYNOMIALS
Let
θ = arccos(x)
dθ = −
dx
√
1 − x2
.
Tn(x) = cos(n arccos(x)) = cos(nθ) (4.29)
Tm(x) = cos(m arccos(x)) = cos(mθ) (4.30)
where θ ∈ [0, π], n, m ∈ N.
Now if we take a scalar product in L2[0, π] of (4.29) and (4.30) we obtain the following:
Tn(x), Tm(x) =
π
0
cos(nθ) cos(mθ)dθ (4.31)
=
π
0
cos(n + m)θ − cos(n − m)θ
2
dθ (4.32)
Now if n = m = 0 we can see that (4.32) equals to
1
2
π
0
(cos((2n)θ − cos(0)) dθ =
1
2
sin(2n)θ − x
2n
π
0
=
π
2
(4.33)
and if n = m = 0 we have
1
2
π
0
(cos((2n)θ − cos(0)) dθ =
1
2
sin(θ)
2n
− 2x
π
0
= π (4.34)
Now summarizing (4.28), (4.33) and (4.34) we finally get
1
−1
Tn(x)Tm(x)
√
1 − x2
=



0, for m = n,
π
2 , for m = n = 0,
π, for m = n = 0
Remark:
We can refer back to Theorem 2.1.6 once again and and to summarize the above we can say that
Tn’s are eigen functions, since they are orthogonal, i.e. linearly independent. However, similarly
as for Legendre polynomials, we do not claim that they span entire L2 [−1, 1], dx√
1−x2
, since
showing it would require much further analysis and we shall not be focusing on that in this
project.

29. 29
5 Conclusion
We can see that we have been following the same pattern through out the paper. Either we are
examining Hermite matrices, Legendre polynomials or Chebyshev’s polynomials, in each case we
have some scalar product and orthogonality in some Hilbert space. We have also noticed a strong
connection between Hermite matrices and Orthogonal polynomials in general, where Hermmite
matrices are self-adjoint and Orthogonal polynomials can be also expressed in their self adjoint
form. It is also important to stress that the Spectral Theorem plays an important role in the
analysis of Orthogonal polynomials.

30. 30 REFERENCES
References
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