Last modified on November 28th, 2024

chapter outline

 

Chebyshev Polynomials

Chebyshev polynomials, named after Russian mathematician Pafnuty Chebyshev, are a class of orthogonal polynomials that are widely used in numerical analysis, approximation theory and applied mathematics, with applications in error minimization, interpolation, solving differential equations, and signal processing. 

They are also used to solve the Chebyshev differential equation.

First Kind – Tn(x)

The Chebyshev polynomials of the first kind, denoted by Tn(x), are expressed as:

Tn(x) = cos(n arccos (x)), where x ∈ [-1, 1]

 Or equivalently, Tn(cos θ) = cos nθ

These polynomials are derived using trigonometric identities, making them closely related to cosine functions.

Second Kind – Un(x) 

The Chebyshev polynomials of the second kind, denoted by Un(x), are expressed as:

Un(x) = ${\dfrac{sin\left( \left( n+1\right) \arccos \left( x\right) \right) }{\sqrt{1-x^{2}}}}$, where x ∈ (-1, 1)

These polynomials extend the properties of the first kind and are widely applied in orthogonal expansions.

List

A few Chebyshev polynomials of the first and second kinds are listed below.

Here, 

  • All coefficients are integers
  • Tn(x) is an n-degree polynomial with the leading coefficient 2n – 1, for n ≥ 1
  • Un(x) is an n-degree polynomial with the leading coefficient 2n, for n ≥ 1

Problem: Finding T5(x) using the TRIGONOMETRIC FUNCTIONS

For n = 5, we have

T5(x) = cos (5θ), here, θ = arccos (x)

Now, using the cosine expansion formula,

cos 5θ

= cos(3θ + 2θ)

= (cos 3θ)(cos 2θ) – (sin 3θ)(sin 2θ) …..(i)

By the triple angle formula, 

  • cos 3θ = 4cos3 θ – 3cos θ …..(ii)
  • sin 3θ = 3sin θ – 4sin3 θ …..(iii)

By the double angle formula, 

  • cos 2θ = 2cos2 θ – 1 …..(iv)
  • sin 2θ = (sin θ)(cos θ) …..(v)

Now, substituting the values of (ii), (iii), (iv), and (v) in (i),

cos 5θ = (cos 3θ)(cos 2θ) – (sin 3θ)(sin 2θ)

= (4cos3 θ – 3cos θ)(2cos2 θ – 1) – (3sin θ – 4sin3 θ)(2sin θcos θ)

= 4cos3 θ(2cos2 θ – 1) – 3cos θ(2cos2 θ – 1) – 6sin2 θcos θ + 8sin4 θcos θ

= 8cos5 θ – 4cos3 θ – 6cos3 θ + 3cos θ – 6(1 – cos2 θ)cos θ + 8(1 – cos2 θ)2cos θ

= 8cos5 θ – 10cos3 θ + 3cos θ – 6cos θ + 6cos3 θ + 8(1 – 2cos2 θ + cos4 θ)cos θ

= 8cos5 θ – 10cos3 θ + 3cos θ – 6cos θ + 6cos3 θ + 8cos θ – 16cos3 θ + 8cos5 θ

= 16cos5 θ – 20cos3 θ + 5cos θ

Replacing cos θ by x, we get

T5(x) = 16x5 – 20x3 + 5x

Find T6(x) using the trigonometric definition of Chebyshev polynomials.

Solution:

For n = 6, we have
T6(x) = cos (6θ), here, θ = arccos (x)
Now, using the cosine expansion formula,
cos 6θ = 2cos2 3θ – 1 …..(i)
Since cos 3θ = 4cos3 θ – 3cos θ
From the equation (i), we get
cos 6θ = 2(4cos3 θ – 3cos θ)2 – 1
⇒ cos 6θ = 2(16cos6 θ – 24cos4 θ + 9cos2 θ) – 1
⇒ cos 6θ = 32cos6 θ – 48cos4 θ + 18cos2 θ – 1
Now, replacing cos θ by x, we get
T6(x) = 32x6 – 48x4 + 18x2 – 1

Properties

Chebyshev polynomials follow the given properties:

Orthogonality

Chebyshev polynomials are orthogonal over the closed interval [-1, 1] with respect to specific weight function w(x) = ${\dfrac{1}{\sqrt{1-x^{2}}}}$:

  • First Kind: ${\int ^{1}_{-1}\dfrac{T_{n}\left( x\right) T_{m}\left( x\right) }{\sqrt{1-x^{2}}}dx=\begin{cases}0, \  if \  n\neq m\\ \pi , \  if \  n=m=0\\ \dfrac{\pi }{2}, \  if \  n=m\neq 0\end{cases}}$
  • Second Kind: ${\int ^{1}_{-1}U_{n}\left( x\right) U_{m}\left( x\right) \sqrt{1-x^{2}}dx=\delta _{nm}}$, here, δnm is the Kronecker Delta

This orthogonal property is used in approximation theory as it ensures minimal overlap between terms, which leads to efficient function representations.

Show that T2(x) and T3(x) are orthogonal over the interval [-1, 1] with the weight function w(x) = ${\dfrac{1}{\sqrt{1-x^{2}}}}$.

Solution:

As we know, T2(x) = 2x2 – 1 and T3(x) = 4x3 – 3x
Now, T2(x)T3(x) = (2x2 – 1)(4x3 – 3x) = 8x5 – 6x3 – 4x3 + 3x = 8x5 – 10x3 + 3x
Here,
${\int ^{1}_{-1}\dfrac{T_{2}\left( x\right) T_{3}\left( x\right) }{\sqrt{1-x^{2}}}dx}$
= ${\int ^{1}_{-1}\dfrac{\left( 2x^{2}-1\right) \left( 4x^{3}-3x\right) }{\sqrt{1-x^{2}}}dx}$
= ${\int ^{1}_{-1}\dfrac{\left( 8x^{5}-10x^{3}+3x\right) }{\sqrt{1-x^{2}}}dx}$
= ${8\int ^{1}_{-1}\dfrac{x^{5}}{\sqrt{1-x^{2}}}dx-10\int ^{1}_{-1}\dfrac{x^{3}}{\sqrt{1-x^{2}}}dx+3\int ^{1}_{-1}\dfrac{x}{\sqrt{1-x^{2}}}dx}$
= 0 – 0 + 0 
= 0
Thus, ${\int ^{1}_{-1}\dfrac{T_{2}\left( x\right) T_{3}\left( x\right) }{\sqrt{1-x^{2}}}dx=0}$, which proves T2(x) and T3(x) are orthogonal over the interval [-1, 1]

Recurrence Relation

Higher-degree Chebyshev polynomials can be computed easily using recurrence relations:

  • First Kind: Tn + 1(x) = 2xTn(x)  – Tn – 1(x)
  • Second Kind: Un + 1(x) = 2xUn(x) – Un – 1(x)

Using the recurrence relation, compute the following polynomials when T0(x) = 1 and T1(x) = x:
a) T2(x)
b) T3(x) 
c)T4(x)

Solution:

The recurrence relation is: Tn + 1(x) = 2xTn(x) – Tn – 1(x)
Given, T0(x) = 1 and T1(x) = x
a) Here, for n = 1,
T1 + 1(x) = 2xT1(x) – T1 – 1(x)
⇒ T2(x) = 2xT1(x) – T0(x)
⇒ T2(x) = 2x ⋅ x – 1
⇒ T2(x) = 2x2 – 1
b) Here, for n = 2,
T2 + 1(x) = 2xT2(x) – T2 – 1(x)
⇒ T3(x) = 2xT2(x) – T1(x)
⇒ T3(x) = 2x(2x2 – 1) – x
⇒ T3(x) = 4x3 – 2x – x
⇒ T3(x) = 4x3 – 3x
c) Here, for n = 3,
T3 + 1(x) = 2xT3(x) – T3 – 1(x)
⇒ T4(x) = 2xT3(x) – T2(x)
⇒ T4(x) = 2x(4x3 – 3x) – (2x2 – 1)
⇒ T4(x) = 8x4 – 6x2 – 2x2 + 1
⇒ T4(x) = 8x4 – 8x2 + 1

Extrema and Zeros

  • The extrema of Tn(x) occur at ${x=\cos \left( \dfrac{k\pi }{n}\right)}$; here, k = 0, 1, 2, …, n
  • The zeros of Tn(x) are ${x_{k}=\cos \left( \dfrac{2k-1}{2n}\pi \right)}$; here, k = 1, 2, …, n

These zeros are often used as interpolation nodes in numerical methods.

Using the definition of Chebyshev polynomials of the first kind, find the zeros of T6(x), where Tn(x) = cos(n arccos (x))

Solution:

The zeros of the Chebyshev polynomial Tn(x) are given by:
${x_{k}=\cos \left( \dfrac{2k-1}{2n}\pi \right)}$; here, k = 1, 2, …, n
For T6(x), the zeros are
${x_{k}=\cos \left( \dfrac{2k-1}{12}\pi \right)}$; here, k = 1, 2, …, n …..(i)
Substituting k = 1, 2, …, 6 in the formula (i), we get
For k = 1, ${x_{1}=\cos \left( \dfrac{2\left( 1\right) -1}{12}\pi \right)}$ = ${\cos \left( \dfrac{\pi }{12}\right)}$
For k = 2, ${x_{2}=\cos \left( \dfrac{2\left( 2\right) -1}{12}\pi \right)}$ = ${\cos \left( \dfrac{\pi }{4}\right)}$
For k = 3, ${x_{3}=\cos \left( \dfrac{2\left( 3\right) -1}{12}\pi \right)}$ = ${\cos \left( \dfrac{5\pi }{12}\right)}$
For k = 4, ${x_{4}=\cos \left( \dfrac{2\left( 4\right) -1}{12}\pi \right)}$ = ${\cos \left( \dfrac{7\pi }{12}\right)}$
For k = 5, ${x_{5}=\cos \left( \dfrac{2\left( 5\right) -1}{12}\pi \right)}$ = ${\cos \left( \dfrac{3\pi }{4}\right)}$
For k = 6, ${x_{6}=\cos \left( \dfrac{2\left( 6\right) -1}{12}\pi \right)}$ = ${\cos \left( \dfrac{11\pi }{12}\right)}$
Thus, the zeros of T6(x) are
${x_{1}=\cos \left( \dfrac{\pi }{12}\right)}$, ${x_{2}=\cos \left( \dfrac{\pi }{4}\right)}$, ${x_{3}=\cos \left( \dfrac{5\pi }{12}\right)}$, ${x_{4}=\cos \left( \dfrac{7\pi }{12}\right)}$, ${x_{5}=\cos \left( \dfrac{3\pi }{4}\right)}$, and ${x_{6}=\cos \left( \dfrac{11\pi }{12}\right)}$

Graphical Representation

For small values of n, graphs of Chebyshev polynomials Tn(x) show oscillatory behavior, with their amplitudes bounded within the interval [-1, 1]. The number of oscillations increases as n increases.

Following is a list of a few polynomials and their graphs:

nTn(x) Nature of Graph
01Constant
1xLinear
22x2 – 1Parabolic
34x3 – 3xCubic

In the above graph,

  • T0(x) and T2(x) are symmetric about the y-axis (even functions)
  • T1(x) and T3(x) are symmetric about the origin (odd functions)
  • The number of oscillations increases with the degree n
  • Tn(x) oscillates n times between [-1, 1]
  • The extrema of each polynomial are bounded within [-1, 1], and their locations and zeros are symmetrically distributed

Generating Functions – First Kind

Generating functions provide a compact representation of Chebyshev polynomials, enabling efficient derivation of their properties.

The generating function for Tn(x) is ${\dfrac{1-xt}{1-2xt+t^{2}}=\sum ^{\infty }_{n=0}T_{n}\left( x\right) t^{n}}$

Generating Functions – Second Kind

Similarly, the generating function for Un(x) is ${\dfrac{1}{1-2xt+t^{2}}=\sum ^{\infty }_{n=0}U_{n}\left( x\right) t^{n}}$

These representations are used to derive recurrence relations or analyze series expansions.

Problem: APPROXIMATION Using CHEBYSHEV POLYNOMIAL

Approximate f(x) = ex on [-1, 1] using the first 3 Chebyshev polynomials

Solution:

As we know, the first 3 Chebyshev polynomials are
T0(x) = 1, T1(x) = x, and T2(x) = 2x2 – 1
The approximation formula is P(x) = c0T0(x) + c1T1(x) + c2T2(x)
Here, ${c_{k}=\dfrac{2-\delta _{k_{0}}}{\pi }\int ^{1}_{-1}\dfrac{f\left( x\right) T_{k}\left( x\right) }{\sqrt{1-x^{2}}}dx}$
Calculating coefficients numerically,
c0 ≈ 1.26607, c1 ≈ 0.21936, and c2 ≈ 0.09799
The approximated polynomial is P(x) ≈ 1.26607 + 0.21936x + 0.09799(2x2 – 1)
Now, simplifying, we get
P(x) ≈ 1.26607 + 0.21936x + 0.19598x2 – 0.09799
⇒ P(x) ≈ 1.16808 + 0.21936x + 0.19598x2 Thus, f(x) = ex is approximated to P(x) ≈ 1.16808 + 0.21936x + 0.19598x2

Last modified on November 28th, 2024