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The Chebyshev polynomials

T_{n}

are defined by

T_{n}(x)=

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

Feb 13, 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

T_n(x)

is the set of all

x

for which

\arccos(x)

is defined

step 2

Since

\arccos(x)

is defined for

-1 \leq x \leq 1

, the domain of

T_n(x)

-1 \leq x \leq 1

step 3

The range of

T_n(x)

is the set of all possible values of

\cos(n \arccos(x))

step 4

Since the cosine function has a range of

-1 \leq \cos(\theta) \leq 1

for any angle

\theta

, the range of

T_n(x)

is also

-1 \leq T_n(x) \leq 1

step 5

To express

T_2

explicitly as a quadratic polynomial, we use the expansion of

\cos(2\arccos(x))

step 6

From the asksia-ll calculator,

T_2(x) = 2x^2 - 1

step 7

To express

T_3

as a cubic polynomial, we use the expansion of

\cos(3\arccos(x))

step 8

From the asksia-ll calculator,

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

step 9

To show that

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

, we use the cosine addition formula

step 10

The asksia-ll calculator confirms that

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

is true

step 11

Using part (c), we can recursively find

T_4, T_5, T_6,

and

T_7

by applying the relation

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

with

T_0(x) = 1

and

T_1(x) = x

step 12

The zeros of

T_n

are the values of

x

for which

\cos(n\arccos(x)) = 0

step 13

From the asksia-ll calculator, the zeros of

T_n

are given by

x = \cos\left(\frac{2\pi c_1 - \pi/2}{n}\right)

where

c_1

is an integer

step 14

The local maxima and minima of

T_n

occur at the values of

x

where the derivative of

T_n

is zero. However, the asksia-ll calculator did not find any local maxima or minima

step 15

To graph

T_2, T_3, T_4,

and

T_5

, we plot the respective polynomials on the same axes

step 16

Similarly, to graph

T_5, T_6,

and

T_7

, we plot these polynomials on the same axes

step 17

Observing the graphs, we can infer the relationship between the zeros of

T_n

and

T_{n+1}

, as well as the

x

-coordinates of the maximum and minimum values

step 18

The integral

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

can be evaluated using the substitution

u = \arccos(x)

step 19

For odd

n

, the integral of

T_n(x)

over

[-1, 1]

is zero due to the symmetry of the function. For even

n

, the integral is non-zero

Answer

(a) Domain:

-1 \leq x \leq 1

, Range:

-1 \leq T_n(x) \leq 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)

is true. (d)

T_n

is a polynomial of degree

n

. (e)

T_4, T_5, T_6,

and

T_7

can be found using the recursive relation. (f) Zeros of

T_n

x = \cos\left(\frac{2\pi c_1 - \pi/2}{n}\right)

, no local maxima or minima found. (g) Graphs of

T_2, T_3, T_4,

and

T_5

show the polynomials on a common screen. (h) Graphs of

T_5, T_6,

and

T_7

show the polynomials on a common screen. (i) Zeros of

T_n

and

T_{n+1}

are interlaced; the

x

-coordinates of the maximum and minimum values follow a similar pattern. (j)

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

is zero for odd

n

and non-zero for even

n

. (k) The substitution

u = \arccos(x)

can be used to evaluate the integral.

Key Concept

Chebyshev polynomials and their properties, including domain, range, explicit forms, zeros, and behavior under integration.

Explanation

The Chebyshev polynomials

T_n(x)

are defined by

\cos(n\arccos(x))

and have specific properties such as a fixed domain and range, recursive relationships that define their polynomial form, and a predictable pattern of zeros and extrema. Their integrals over

[-1, 1]

are zero for odd

n

due to symmetry.