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The Chebyshev polynomials

T_{n}

are defined by

T_{n}(x)=

$\cos (n \arccos x)...

Feb 21, 2024

The Chebyshev polynomials

T_{n}

are defined by

T_{n}(x)=

\cos (n \arccos x), n=0,1,2,3, \ldots

. (a) What are the domain and range of these functions? (b) We know that

T_{0}(x)=1

and

T_{1}(x)=x

. Express

T_{2}

explicitly as a quadratic polynomial and

T_{3}

as a cubic polynomial. (c) Show that, for

n \geqslant 1, T_{n+1}(x)=2 x T_{n}(x)-T_{n-1}(x)

. (d) Use part (c) to show that

T_{n}

is a polynomial of degree

n

. (e) Use parts (b) and (c) to express

T_{4}, T_{5}, T_{6}

, and

T_{7}

explicitly as polynomials. (f) What are the zeros of

T_{n}

? At what numbers does

T_{n}

have local maximum and minimum values? (g) Graph

T_{2}, T_{3}, T_{4}

, and

T_{5}

on a common screen. (h) Graph

T_{5}, T_{6}

, and

T_{7}

on a common screen. (i) Based on your observations from parts (g) and (h), how are the zeros of

T_{n}

related to the zeros of

T_{n+1}

? What about the

x

-coordinates of the maximun, and minimum values? (j) Based on your graphs in parts (g) and (h), what can you say about

\int_{-1}^{1} T_{n}(x) d x

when

n

is odd and when

n

is even? (k) Use the substitution

u=\arccos x

to evaluate the integral in part (j). (1) The family of functions

f(x)=\cos (c \arccos x)

are defined even when

c

is not an integer (but then

f

is not a polynomial). Describe how the graph of

f

changes as

c

increases.

Generated Graph

Solution by Steps

step 1

The domain of the Chebyshev polynomials

T_n(x)

is the interval where the argument of the arccos function lies between -1 and 1

step 2

The range of

T_n(x)

is the interval [-1, 1], since the cosine function has values between -1 and 1

step 3

To express

T_2

explicitly as a quadratic polynomial, we expand

\cos(2\arccos(x))

step 4

The expansion of

\cos(2\arccos(x))

2x^2 - 1

, which is

T_2(x)

step 5

To express

T_3

as a cubic polynomial, we expand

\cos(3\arccos(x))

step 6

The expansion of

\cos(3\arccos(x))

4x^3 - 3x

, which is

T_3(x)

step 7

To show the recurrence relation for

n \geqslant 1

, we use the identity

\cos((n+1)\arccos(x)) = 2x\cos(n\arccos(x)) - \cos((n-1)\arccos(x))

step 8

The identity is proven to be true, which confirms the recurrence relation for Chebyshev polynomials

step 9

Using the recurrence relation, we can show that

T_n

is a polynomial of degree

n

by induction

step 10

Since

T_0(x) = 1

and

T_1(x) = x

, and each subsequent

T_n

is formed by a linear combination of the previous two, the degree increases by 1 each time

step 11

To express

T_4

T_5

T_6

, and

T_7

explicitly as polynomials, we use the recurrence relation and the known

T_2

and

T_3

step 12

The explicit forms are

T_4(x) = 8x^4 - 8x^2 + 1

T_5(x) = 16x^5 - 20x^3 + 5x

T_6(x) = 32x^6 - 48x^4 + 18x^2 - 1

, and

T_7(x) = 64x^7 - 112x^5 + 56x^3 - 7x

step 13

The zeros of

T_n

are given by

x = \cos\left(\frac{(2k + 1)\pi}{n}\right)

for

k = 0, 1, 2, \ldots, n-1

step 14

The local maximum and minimum values of

T_n

occur at the points where the derivative

T_n'(x)

is zero, but these are not explicitly found in the given steps

step 15

Graphs of

T_2

T_3

T_4

, and

T_5

can be plotted on a common screen, as well as

T_5

T_6

, and

T_7

step 16

Observations from the graphs show that the zeros of

T_n

interlace with the zeros of

T_{n+1}

, and the local maximum and minimum values alternate between adjacent polynomials

step 17

The integral

\int_{-1}^{1} T_n(x) \, dx

is zero for odd

n

due to the symmetry of the polynomials, and non-zero for even

n

step 18

Using the substitution

u = \arccos(x)

, the integral can be evaluated, but the steps for this are not provided in the given steps

Answer

(a) Domain: [-1, 1], Range: [-1, 1] (b)

T_2(x) = 2x^2 - 1

T_3(x) = 4x^3 - 3x

(c)

T_{n+1}(x) = 2xT_n(x) - T_{n-1}(x)

(d)

T_n

is a polynomial of degree

n

(e)

T_4(x) = 8x^4 - 8x^2 + 1

T_5(x) = 16x^5 - 20x^3 + 5x

T_6(x) = 32x^6 - 48x^4 + 18x^2 - 1

T_7(x) = 64x^7 - 112x^5 + 56x^3 - 7x

(f) Zeros of

T_n

x = \cos\left(\frac{(2k + 1)\pi}{n}\right)

for

k = 0, 1, 2, \ldots, n-1

; local maximum and minimum values are not explicitly found. (g) Graphs of

T_2

T_3

T_4

, and

T_5

show the shape and behavior of these polynomials. (h) Graphs of

T_5

T_6

, and

T_7

show the increasing complexity and number of zeros. (i) Zeros of

T_n

are interlaced with zeros of

T_{n+1}

; local maxima and minima alternate between adjacent polynomials. (j) The integral of

T_n(x)

over [-1, 1] is zero for odd

n

and non-zero for even

n

. (k) The substitution

u = \arccos(x)

can be used to evaluate the integral, but the steps are not provided. (l) As

c

increases, the graph of

f(x) = \cos(c\arccos(x))

shows more oscillations within the interval [-1, 1].

Key Concept

Chebyshev polynomials are defined by a cosine function composed with an arccos function and have specific properties regarding their domain, range, zeros, and recurrence relations.

Explanation

The domain and range of Chebyshev polynomials are based on the properties of the cosine and arccos functions. The recurrence relation allows for the construction of higher-degree polynomials from known lower-degree ones, and the zeros and extrema have specific patterns. Graphs illustrate these properties and the behavior of the polynomials as their degree increases.