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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Explain the determine blue
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb06891c1-7310-49aa-a044-9ddf722456d6%2Fd87596e8-8ace-41f1-981b-876ccb56a8f8%2Fw04h7m_processed.png&w=3840&q=75)
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,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb06891c1-7310-49aa-a044-9ddf722456d6%2Fd87596e8-8ace-41f1-981b-876ccb56a8f8%2Flkrug3_processed.jpeg&w=3840&q=75)
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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