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The Chebyshev polynomials are defined by the recurrence formula of equa- tion (7.59) for |x| < 1. This means that x2 < 1, and therefore x² – 1 = iV1 – x², where i = V-1. (7.67) Consequently, 22 – 1 = x ±i/1 – a² = e±i¢¸ (7.68) where V1- 22 tan o(x) (7.69) APPLICATIONS 223 1 V1-x X FIGURE 7.1: Definition of the angle o. and (x + Va2 – 1)* + (x – Væ2 – 1)* = eikø -iko = 2 cos(kø). (7.70) This last result means that Tk(x), as given by equation (7.66), can be written in the following form: cos[kø(x)] Tr(x) (7.71) 2k-1 A representation of the angle o, defined by equation (7.69), is given in Figure 7.1. Consideration of this diagram allows us to immediately conclude that Co (7.72) cos o = x or = cos x. Therefore, equations (7.71) and (7.72) jointly imply that the kth Chebyshev polynomial can also be expressed as cos(k cos x) TR(x) = |x| < 1, k = 0, 1, 2, 3, .... (7.73) 2k-1

The Chebyshev polynomials are defined by the recurrence formula of equa- tion (7.59) for |x| < 1. This means that x2 < 1, and therefore x² – 1 = iV1 – x², where i = V-1. (7.67) Consequently, 22 – 1 = x ±i/1 – a² = e±i¢¸ (7.68) where V1- 22 tan o(x) (7.69) APPLICATIONS 223 1 V1-x X FIGURE 7.1: Definition of the angle o. and (x + Va2 – 1)* + (x – Væ2 – 1)* = eikø -iko = 2 cos(kø). (7.70) This last result means that Tk(x), as given by equation (7.66), can be written in the following form: cos[kø(x)] Tr(x) (7.71) 2k-1 A representation of the angle o, defined by equation (7.69), is given in Figure 7.1. Consideration of this diagram allows us to immediately conclude that Co (7.72) cos o = x or = cos x. Therefore, equations (7.71) and (7.72) jointly imply that the kth Chebyshev polynomial can also be expressed as cos(k cos x) TR(x) = |x| < 1, k = 0, 1, 2, 3, .... (7.73) 2k-1

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 56EQ
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ol stc ksa 6:21 PM C @ 1 94% Tg (2) = () [(x + Vx? – 1)* + (x – Va2 – 1)*]. k (7.66) A check of equation (7.66) for k 2,3, and 4 shows that it reproduces the rOsults nroviously civon hy oauation (7 60) Cancel Actual Size (445 KB) Choose
Transcribed Image Text:ol stc ksa 6:21 PM C @ 1 94% Tg (2) = () [(x + Vx? – 1)* + (x – Va2 – 1)*]. k (7.66) A check of equation (7.66) for k 2,3, and 4 shows that it reproduces the rOsults nroviously civon hy oauation (7 60) Cancel Actual Size (445 KB) Choose
The Chebyshev polynomials are defined by the recurrence formula of equa- tion (7.59) for |x| < 1. This means that x2 < 1, and therefore x² – 1 = iv1 – x², where i V-1. (7.67) Consequently, (*+ Vx? – 1 = x ±iv1– a² = e+i¢, (7.68) where VI - x2 tan ø(x) : (7.69) APPLICATIONS 223 V1-x Ф X FIGURE 7.1: Definition of the angle ø. and -ikø (x + Vx2 – 1)* + (x – Væ² – 1)* = etkø 2 cos(kø). (7.70) +e This last result means that Tr(x), as given by equation (7.66), can be written in the following form: cos[kø(x)] 2k-1 Tr(x) (7.71) A representation of the angle ¢, defined by equation (7.69), is given in Figure 7.1. Consideration of this diagram allows us to immediately conclude that cos o = x or o -1 = COS x. (7.72) Therefore, equations (7.71) and (7.72) jointly imply that the kth Chebyshev polynomial can also be expressed as TR (x): cos(k cos-1 2k-1 |æ| < 1, k = 0, 1, 2, 3, .... (7.73) It should be clear that although equations (7.66) and (7.73) are equivalent, the form given by equation (7.66) is to be used if an explicit expression is needed for any particular value of k. As an elementary application of the use of Chebyshev polynomials, we show how to expand the function f (x) = 2x4 – 3x² + x + 7 (7.74) in terms of these polynomials. First, we must invert the Chebyshev polynomi- als and express the various powers of x in terms of them. This is easily done,
Transcribed Image Text:The Chebyshev polynomials are defined by the recurrence formula of equa- tion (7.59) for |x| < 1. This means that x2 < 1, and therefore x² – 1 = iv1 – x², where i V-1. (7.67) Consequently, (*+ Vx? – 1 = x ±iv1– a² = e+i¢, (7.68) where VI - x2 tan ø(x) : (7.69) APPLICATIONS 223 V1-x Ф X FIGURE 7.1: Definition of the angle ø. and -ikø (x + Vx2 – 1)* + (x – Væ² – 1)* = etkø 2 cos(kø). (7.70) +e This last result means that Tr(x), as given by equation (7.66), can be written in the following form: cos[kø(x)] 2k-1 Tr(x) (7.71) A representation of the angle ¢, defined by equation (7.69), is given in Figure 7.1. Consideration of this diagram allows us to immediately conclude that cos o = x or o -1 = COS x. (7.72) Therefore, equations (7.71) and (7.72) jointly imply that the kth Chebyshev polynomial can also be expressed as TR (x): cos(k cos-1 2k-1 |æ| < 1, k = 0, 1, 2, 3, .... (7.73) It should be clear that although equations (7.66) and (7.73) are equivalent, the form given by equation (7.66) is to be used if an explicit expression is needed for any particular value of k. As an elementary application of the use of Chebyshev polynomials, we show how to expand the function f (x) = 2x4 – 3x² + x + 7 (7.74) in terms of these polynomials. First, we must invert the Chebyshev polynomi- als and express the various powers of x in terms of them. This is easily done,
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